Talk:Centrifugal force/Archive 3

New article
--Cleon Teunissen 09:42, 18 Jan 2005 (UTC)

I have replaced the article on centrifugal force with a new article. The first section is written with people with very little physics background in mind. As the article goes on, the level increases. Most of the article is aimed at explaining why centrifugal force is so confusing. The dynamics of rotation is counter-intuitive. It would already be confusing with only newtonian dynamics around, but inevitably people start invoking relativistic dynamics, multiplying the    confusion. Relativistic dynamics views centrifugal force the same as newtonian dynamics does, but not discussing relativistic physics would allow people room to assume that relativistic physics has not been taken into account. --Cleon Teunissen 09:42, 18 Jan 2005 (UTC)


 * Smyth has done some essential cleanup. Thanks! --Cleon Teunissen 11:56, 18 Jan 2005 (UTC)


 * Duk, thanks for correcting the spelling. Actually, I spelled physically on purpose with 'ph', referring to Physics. But *pedia articles are not the place to use non-standard spelling. Cleon Teunissen 21:16, 22 Jan 2005 (UTC)

"Centrifugal force" is not a force
 One often hears of ``centrifugal force.'' This is the apparent force that throws you to the outside of a turn during cornering. If there is anything loose in the car, it will immediately slide to the right in a left hand turn, and vice versa. Perhaps you have experienced what happened to me once. I had omitted to remove an empty Pepsi can hidden under the passenger seat. During a particularly aggressive run (something for which I am not unknown), this can came loose, fluttered around the cockpit for a while, and eventually flew out the passenger window in the middle of a hard left hand corner.

I shall attempt to convince you, in this month's article, that centrifugal force is a fiction, and a consequence of the fact first noticed just over three hundred years ago by Newton that objects tend to continue moving in a straight line unless acted on by an external force.

When you turn the steering wheel, you are trying to get the front tires to push a little sideways on the ground, which then pushes back, by Newton's third law. When the ground pushes back, it causes a little sideways acceleration. This sideways acceleration is a change in the sideways velocity. The acceleration is proportional to the sideways force, and inversely proportional to the mass of the car, by Newton's second law. The sideways acceleration thus causes the car to veer a little sideways, which is what you wanted when you turned the wheel. If you keep the steering and throttle at constant positions, you will continue to go mostly forwards and a little sideways until you end up where you started. In other words, you will go in a circle. When driving through a sweeper, you are going part way around a circle. If you take skid pad lessons (highly recommended), you will go around in circles all day.

If you turn the steering wheel a little more, you will go in a tighter circle, and the sideways force needed to keep you going is greater. If you go around the same circle but faster, the necessary force is greater. If you try to go around too fast, the adhesive limit of the tires will be exceeded, they will slide, and you will not stick to the circular path-you will not ``make it.''

From the discussion above, we can see that in order to turn right, for example, a force, pointing to the right, must act on the car that veers it away from the straight line it naturally tries to follow. If the force stays constant, the car will go in a circle. From the point of view of the car, the force always points to the right. From a point of view outside the car, at rest with respect to the ground, however, the force points toward the center of the circle. From this point of view, although the force is constant in magnitude, it changes direction, going around and around as the car turns, always pointing at the geometrical center of the circle. This force is called centripetal, from the Greek for ``center seeking.'' The point of view on the ground is privileged, since objects at rest from this point of view feel no net forces. Physicists call this special point of view an inertial frame of reference. The forces measured in an inertial frame are, in a sense, more correct than those measured by a physicist riding in the car. Forces measured inside the car are biased by the centripetal force.

Inside the car, all objects, such as the driver, feel the natural inertial tendency to continue moving in a straight line. The driver receives a centripetal force from the car through the seat and the belts. If you don't have good restraints, you may find yourself pushing with your knee against the door and tugging on the controls in order to get the centripetal force you need to go in a circle with the car. It took me a long time to overcome the habit of tugging on the car in order to stay put in it. I used to come home with bruises on my left knee from pushing hard against the door during an autocross. I found that a tight five- point harness helped me to overcome this unnecessary habit. With it, I no longer think about body position while driving-I can concentrate on trying to be smooth and fast. As a result, I use the wheel and the gearshift lever for steering and shifting rather than for helping me stay put in the car!

The `forces' that the driver and other objects inside the car feel are actually centripetal. The term centrifugal, or ``center fleeing,'' refers to the inertial tendency to resist the centripetal force and to continue going straight. If the centripetal force is constant in magnitude, the centrifugal tendency will be constant. There is no such thing as centrifugal force (although it is a convenient fiction for the purpose of some calculations).

Let's figure out exactly how much sideways acceleration is needed to keep a car going at speed in a circle of radius. We can then convert this into force using Newton's second law, and then figure out how fast we can go in a circle before exceeding the adhesive limit-in other words, we can derive maximum cornering speed. For the following discussion, it will be helpful for you to draw little back-of-the-envelope pictures (I'm leaving them out, giving our editor a rest from transcribing my graphics into the newsletter).

Consider a very short interval of time, far less than a second. Call it ( stands for ``delta, a Greek letter mathematicians use as shorthand for ``tiny increment). In time, let us say we go forward a distance and sideways a distance. The forward component of the velocity of the car is approximately. At the beginning of the time interval, the car has no sideways velocity. At the end, it has sideways velocity. In the time, the car has thus had a change in sideways velocity of. Acceleration is, precisely, the change in velocity over a certain time, divided by the time; just as velocity is the change in position over a certain time, divided by the time. Thus, the sideways acceleration is How is  related to, the radius of the circle? If we go forward by a fraction of the radius of the circle, we must go sideways by exactly the same fraction of  to stay on the circle. This means that. The fraction is, however, nothing but. By this reasoning, we get the relation We can substitute this expression for  into the expression for, and remembering that , we get the final result  This equation simply says quantitatively what we wrote before: that the acceleration (and the force) needed to keep to a circular line increases with the velocity and increases as the radius gets smaller.

What was not appreciated before we went through this derivation is that the necessary acceleration increases as the square of the velocity. This means that the centripetal force your tires must give you for you to make it through a sweeper is very sensitive to your speed. If you go just a little bit too fast, you might as well go much too fast-your're not going to make it. The following table shows the maximum speed that can be achieved in turns of various radii for various sideways accelerations. This table shows the value of the expression which is the solution of  for, the velocity. The conversion factor 15/22 converts from feet per second to miles per hour, and 32.1 converts  from gees to feet per second squared. We covered these conversion factors in part 3 of this series.

For autocrossing, the columns for 50 and 100 feet and the row for 1.00 are most germane. The table tells us that to achieve 1.00 sideways acceleration in a corner of 50 foot radius (this kind of corner is all too common in autocross), a driver must not go faster than 27.32 miles per hour. To go 30 mph, 1.25 is required, which is probably not within the capability of an autocross tire at this speed. There is not much subjective difference between 27 and 30 mph, but the objective difference is usually between making a controlled run and spinning badly.

The absolute fastest way to go through a corner is to be just over the limit near the exit, in a controlled slide. To do this, however, you must be pointed in just such a way that when the car breaks loose and slides to the exit of the corner it will be pointed straight down the optimal racing line at the exit when it ``hooks up'' again. You can smoothly add throttle during this maneuver and be really moving out of the corner. But you must do it smoothly. It takes a long time to learn this, and probably a lifetime to perfect it, but it feels absolutely triumphal when done right. I have not figured out how to drive through a sweeper, except for the exit, at anything greater than the limiting velocity because sweepers are just too long to slide around. If anyone (Ayrton Senna, perhaps?) knows how, please tell me!

The chain of reasoning we have just gone through was first discovered by Newton and Leibniz, working independently. It is, in fact, a derivation in differential calculus, the mathematics of very small quantities. Newton keeps popping up. He was perhaps the greatest of all physicists, having discovered the laws of motion, the law of gravity, and calculus, among other things such as the fact that white light is made up of multiple colors mixed together.

It is an excellent diagnostic exercise to drive a car around a circle marked with cones or chalk and gently to increase the speed until the car slides. If the front breaks away first, your car has natural understeer, and if the rear slides first, it has natural oversteer. You can use this information for chassis tuning. Of course, this is only to be done in safe circumstances, on a rented skid pad or your own private parking lot. The police will gleefully give you a ticket if they catch you doing this in the wrong places.

Rewrite of the article, 30 jan
I have rewritten the article. The first section is written for people with no phyisics background at all. The first section aims to stay as close to everyday experience of motion as possible. As the article goes on, the level increases. The final sections assume that the reader has a lot of physics knowledge. Cleon Teunissen 07:04, 30 Jan 2005 (UTC)

How circular motion is to be understood
202.156.2.82 wrote:
 * The centrifugal force may be understood as arising from the tendency of the stone to move in a straight line, or a path tangential to its position on the circle.

I have removed this remark because it leads to an inconsistency. If there would be a centrifugal force then the centripetal force would be prevented from maintaining the circular motion. The circular motion is maintained, so there is only one force at play: centripetal force.

Because of inertia a force is required to make an object deviate from moving in a straight line. Manifestation of inertia opposes the centripetal force, but it does not prevent the centripetal acceleration.

Without inertia, objects would be instantly accelerated to lightspeed. Because of inertia acceleration is proportional to the exerted force. The bottom line: inertia is not a force. Inertia involves opposition to change, but inertia never prevents the change. --Cleon Teunissen | Talk 14:49, 22 Apr 2005 (UTC)

Centrifugal Force *is* a force
I'm dismayed to see the amount of damage that has been done to physics literacy by the "Fictitious Force" falacy.

Centrifugal force is just as "real" a force as the one a wall exerts on you if you bang your head against it. And that's what should be done to the so-called educational "experts" who, at some point, decided that there were such things as fictitious, fake, or phantom forces.

How all this started, as best I can determine, is this: an effort was made in middle school curriculum development to "correct" the understanding of the forces that give rise to circular motion. "They" (you know, Them) decided it was misleading to describe that pail on a rope as pulling on the rope, when in fact it was the rope that was pulling on the pail. And that's fine as far as it goes. It is a change in point of view. It describes circular motion as caused by the acceleration of an object, changing its direction of motion from linear to circular, due to the application of a centripetal force. It's true, it's a good way to describe it, and it probably does help in gaining an understanding of circular motion; or would have, except for what happened next.

Somehow, while the new concept was making its way into the hands of middle-school educators, somebody overreacted. Rather than switching the emphasis from the centrifugal to the centripetal force, the thought pendulum swung too far, and people began teaching that there was no centrifugal force. That's rubbish.

Newton's Third Law tells us that unmatched forces do not exist in the universe. In any force interaction between two objects, the forces exist in pairs. Each force of each pair is the same magnitude as, and opposite in direction to, its partner. Furthermore, it does not matter whether any motion occurs as a result of the interaction, or not. The 3r d law holds perfectly whether you are pushing on a brick wall or whether you are pushing on a shopping cart (accelerating it). The force back on your hand by the object is the same as the force of your hand on the object. The cart accelerates not because there is something wrong with the 3rd law, but because the sum of all the forces acting on it (your hand is but one) is non-zero.

If centrifugal force did not exist, it would be the only known example of 3rd law failure. Thankfully, that's not the case. Centrifugal force and Centripetal force are a 3rd-law force pair, just like any other. It's certainly true that centripetal force is the force that maintains circular motion. And it's a wise way to approach teaching the matter. But it's a simple physical fact that when the rope pulls on the pail, the pail does pull back. It has no choice. Forces exist in pairs.

And that's the way centrifugal force should be taught. It is the Newton's 3rd law reaction force to the centripetal force. That doesn't make it false, fake, fictitious, or phantasmic. It's real. All reaction forces are real. If they weren't, rockets would not take off, and you wouldn't hurt your head if you banged it on the wall.

It does not help anyone's understanding of physics to imply that there are phantom or magic forces around us. And the damage done by teaching that centrifugal force does not "really" exist when it can obviously be directly experienced by anyone, fosters a distrust of empirical evidence.

Beyond that, it's just wrong.

The wiki article needs to be fixed, but I hesitate to take a crack at it myself, because it's not a simple edit.

Anyone have suggestions? --68.239.184.113 17:08, 31 July 2005 (UTC)

I will try to answer this.

The thing to recognize, I think, is that inertia is real. Inertia is always there, it is omnipresent. And paradoxically, because it is never absent, inertia tends to be overlooked.

We have that a force is required to change the velocity of an object. Newtons literal formulation of his third law looks very much like statics, but in presenting his dynamics newton used his third law in a form that today is known as conservation of momentum. If we have object floating in space, and object A is exerting a force on object B, then both will accelerate, in opposite directions. The center of mass will not start moving.

So it is a bit awkward to use Newtons third law as a law of statics.

We do have a law of statics that looks the same: if a force is being exerted, and that does not result in acceleration, then the thing being pushed against is pushing back.

Now, inertia.

Can we say that inertia is pushing back? Well no, for if inertia would be pushing back then there would be no acceleration, and inertia does not prevent acceleration. So it would lead to an inconsistency if we would decree: inertia is pushing back.

Inertia seems a bit like friction, but it is also very different. When there is friction and you can exert only so much force, then the velocity reaches a "ceiling". A car's engine can exert a lot of forward force, but at several hundred kilometers per hour there is so much drag that the car cannot go faster.

But with inertia it is a different story. There is inertia, but there is no "ceiling" to velocity (at least not until you go at a relativistic speed)

Because of the existence of inertia, we have the property that the amount of acceleration is proportional to the force. Twice as much force means accelerating twice as hard.

In all, friction relates to velocity, but inertia is a level higher, inertia relates to acceleration.

It is important to recognize the tremendous importance of inertia. But remarkably it is not listed among the fundamental forces of nature: gravitation, electromagnetism, weak nuclear force, strond nuclear force. But then, it would lead to an inconsistency to declare by decree that inertia is 'a force'.

The four fundamental forces have in common that they are interactions between pairs of object. It is a force when momentum is exchanged between two objects, for example in a gravitational slingshot

Inertia, important as it is, is not an interaction between two objects; you cannot use inertia by itself for a slingshot.

In the general theory of relativity, the description of inertia and the description of gravitation are unified at a most profound level. In that sense, general ralativity has acknowledged how fundamentally important inertia is. In general relativity inertia is described as an interation between matter and space itself. You cannot call it friction, but space does oppose change of velocity. (Acceleration is opposed by inertia, but not prevented. Even the slightest force will get the largest object moving, it just takes more time with a weaker force.)

So it is best to recognize centrifugal force for what it is: inertia. When an object is being moved in a circle, it is constantly being accelerated towards the center. The acceleration towards the center is opposed, but not prevented, by inertia.  --Cleon Teunissen | Talk 18:51, 31 July 2005 (UTC)

I'm sorry, but there's still a good deal of misconception here. Let's address this in Newtonian terms, with non-quantum, macroscopic objects like pails and ropes.

For background, let us agree that inertia is not a force. Inertia is, for all intents and purposes the same thing as mass. To determine the inertia of an object, it is sufficient to know its mass. It depends upon nothing else. So let us drop the word inertia and speak of mass.

Force is the product of a mass and an acceleration. This is Newton's second law. $$(F=ma)$$ Although he originally framed it as a change in momentum, we can, through a simple mathematical theorem show the two to be equivalent. The SI unit of force, the newton, is defined as the force necessary to accelerate a kilogram at a rate of 1 m/s^2. So a newton is 1 m·kg·s^-2. Force is therefore dimensionally different from mass, and cannot be compared to it (i.e, added subtracted or added). We will avoid using them interchangeably, or otherwise confusing them.

Mass (inertia) does not "push". Force pushes; mass resists pushing. It resists in the sense that a larger mass under a given force does not accelerate as much.

Newton's third law says that whatever force I exert on an object, the object will always exert an equal and opposite force on me. Note well the use of the word "always". There are no exceptions. In particular, there is no exception for static vs. kinetic situations.

Consider a 1 kg mass glued to the table. If I apply a 1 N force to it, it will push back with a 1 N force. Newton is vindicated, at least in the static case.

Now consider a 1 kg mass in space. If I apply a 1 N force on it, it will still push back with a 1 N force. It will also begin to accelerate away from me at a predictable 1 m/s^2 rate (and I from it, at a somewhat slower rate), so I will have to use some ingenuity to continue to apply that constant 1 N force. As a thought experiment, I might use a tiny, relatively massless, calibrated, 1 N rocket engine, carefully adjusted to act on the center of mass. The rocket applies a 1 N force to the mass; the mass applies a 1 N force to the rocket. Newton is vindicated again. At the end of the day, if the rocket hasn't run out of thought-fuel, the forces are both unchanged, but of course the mass is now quite far away and moving at a high speed.

Note well, that the reason the first mass did not accelerate, in fact did not move, was not because the 1 N forces matched. All reaction pairs match, all the time. The reason the mass did not move was because the 1 N force I was applying to it was not the only force applied to it. There was an additional force applied by the glue and table. The mass did not move because the total forces on the mass summed to zero.

Similarly, the reason that the mass in space did accelerate was not because the reaction force did not match the applied force. Of course they must be equal and opposite, or the 3rd law fails. The reason it moved was because the 1 N force I applied to it was the only force applied to it. Absent some other force, such as friction, this motive force was unbalanced, and the mass accelerated in the direction of the force applied to it.

Just as in the static case, the reaction force is irrelevant in determining the motion of the object, since it was not exerted on the object, but rather by the object.

So, some of your statements are incorrect:

You said inertia seems sowhat like friction, but we've already dispensed with that idea. Inertia is mass, friction is a force. The two are not comparable. You said, "if a force is being exerted, and that does not result in acceleration, then the thing being pushed against is pushing back." In fact, it means no such thing. The object must always be pushing back equally, whether it accelerates or not. What the lack of acceleration truly means is that there must be one or more other forces on the object, such that vector sum of all of these is zero. The push back is irrelevant, since it is not a force applied to the object. The push back on me will, of course, determine how I move, when summed with all the other forces acting on me, but that's relevant to my motion, not the object's. This is why free-body diagrams only show the forces on a single object at a time, and not the forces they exert.

Your statement,"if inertia would be pushing back then there would be no acceleration," is incorrect for the same reason. Inertia does push back, precisely equal, precicely opposite, to the applied force. This does not mean that inertia is a force, or even proportional to the force. It is neither. The force with which the object pushes back does not depend on its mass. Be it a gram or a megagram, the reaction force will be equal and opposite to the applied force. What will differ is that the gram will accelerate much faster away from you than the megagram will. Thus it seems that the megagram is pushing back "harder". It is not. They both push back according to the 3rd law, just as hard as you push on them. The difference is that the gram, when pushed on, rapidly leaves. Once it does so, it is no longer possible for you to exert a force on it, unless you follow it, and continue to push on it with the same force, which means you have to accelerate to keep up with it. But if you succeed, the forces remain equal and opposite.

The same is true of an object in circular motion, with one important difference. Since you are applying the force perpendicular to the object's motion, it does not move toward or away from you, but curves around you at a constant radius. The object does not leave the area, so you can continue to pull on it. You exert a constant centripetal force on it, which results in a constant centripetal acceleration. The force, by the 2nd law, is equal to the product of the mass of the object and the centripetal acceleration. Since there is no other force on the object, the vector sum is nonzero, and the object must be under constant acceleration, which it is. Its speed does not change because there is no component of this acceleration parallel to its velocity.

And just as in every other case, there is a reaction force, applied not to the object, but by the object. It is exerted on the rope (or in the case of a circular orbit, by gravity, on the planet). It does not affect the motion of the object because it is not exerted on the object; it is exerted on whatever is pulling on the object. By the 3rd law, this force is equal in magnitude, and opposite in direction to the centripetal force. It is called the centrifugal force.

Centripetal (center seeking) acceleration is inwardly directed, and caused strictly by the inwardly directed centripetal force. There is no disagreement on that. Since it is the only force on the object, it accounts for the only acceleration of the object. The centrifugal force does not affect it, not because it's fictitious, but because it's exerted on a different object!

The perfectly real and non-fictitious centrifugal (center fleeing) force is outwardly directed, and equal in magnitude to the centripetal force. It does not affect the motion of the object, but might well affect the motion of the rope, unless something pulls on the other end of it with equal force.

In the case of orbital motion, it is the very real centrifugal force that causes the planet to wobble as the moon revolves around it, such that the pair, viewed as an isolated syatem,revolve around their common center of mass.

And there is nothing in the least bit fictitious about any of it.

--Jeepien 02:31:43, 2005-08-01 (UTC)

P.S. There are also two problems with your final paragraph above. You say that centrifugal force is actually inertia, which we had agreed was impossible, as mass and force are dimensionally distinct quantities. You also say that the centripetal acceleration is "opposed but not prevented" by inertia. This is similarly confusing because it encourages the reader to assume that there is some outward-directed something "opposing" the acceleration toward the center. In fact, there is but a single force acting on the object, and it alone (and unopposed) is fully responsible for the acceleration of the object. This is, of course, the centripetal force.

By saying that it is opposed in some way, you are dancing on the edge of the pit of the very fallacy you are seeking to debunk. There is nothing opposing the centripetal force on the object. If there were, its acceleration would be less than that predicted by $$F=ma$$. It isn't, because there is no force pulling outward on the object, not even a "phantom" one. And certainly not a "centrifugal" one.

Remember, the centrifugal force does not pull outward on the object (and therefore cannnot oppose the centripetal force), for the simple reason that the centrifugal force does not act on the object in the first place. It is the force exerted by the object upon whatever is constraining it to a circular path.

As long as we keep this straight, we can allow the centrifugal force to be real (after all, the pail does pull outward on the rope, and we can feel this force and measure it with a scale). But it does not pull outward on the pail. The only force upon the pail is inward, as confirmed by its strictly inward acceleration in accordance with the second law.

This entire muddle, which is all too common in textbooks, well-meaning popularized science explanations, and even on some college .edu web sites, boils down to a confusion between Newton's second and third laws. In his second law, all the forces on a single object are considered, and determine the change in momentum of the object. In his third law, pairs of forces between pairs of objects are considered, and two forces are never exerted on the same object, so they can not be said to cancel, or balance or otherwise oppose each other.

It is easy to tell whether someone is a victim of this confusion. Propose the following to them: "Since by the 3rd law, all forces occur in strictly balanced pairs, all acceleration is impossible." If they are tripped up by that, or begin to deny that the 3rd law applies universally, they are victims. If they can easily point out the fallacy, they have their concepts straight.

--Jeepien

The siginificance of Newton's Third Law
Jeepien Wrote:
 * It is easy to tell whether someone is a victim of this confusion. Propose the following to them: "Since by the 3rd law, all forces occur in strictly balanced pairs, all acceleration is impossible."   If they are tripped up by that, or begin to deny that the 3rd law applies universally, they are victims.  If they can easily point out the fallacy, they have their concepts straight. --Jeepien 02:31:43, 2005-08-01 (UTC)

Hi Jeepien, I think you and I do not really disagree. I may use rather different metaphores but I doubt there is a real difference.

It seems to me that Newton's third law is the key element here. Newtons third law is intimately connected to the principle of relativity of inertial motion. In the Principia the third law's formulation is very reminiscent of statics, but in its application in mechanics it is about interactons between pairs of objects.

The law of equilibrium of forces in the context of statics is of course quite unrelated to newtons third law of motion. If an object is glued to a table then the glue transmits (if stressed) the counterforce of the table. The table can be bolted to the floor, etc etc

When an observer and an object are both free-floating in space and the observer pushes an object, then according to newtonian mechanics you will both accelerate,(away from each other) but the common center of mass of the two of you will remain in inertial motion. Strictly speaking none of these assertions of Newtonian mechanics is provable, but it all works so well it would be absurd to demand more corroborating evidence. Does the object push back? Srictly speaking we don't know, what we see is that Newtons third law holds good.

Any violation of Newtons third law is at the same time a violation of the principle of relativity of inertial motion. The principle of relativity of inertial motion makes asserting Newtons third law inevitable.

Newtons's third law does not trip me up. In fact, I use Newton's third law to provide a distinction between force and non-force. Force is what happens in physical interactions between pairs of objects. Inertia involves an interaction, but there is only a single object, so important as it is, inertia is not to be categorized as a force

I will have another good look at what you wrote, to see if I should stop using particular metaphores. --Cleon Teunissen | Talk 10:22, 8 August 2005 (UTC)

Inertia and Mass
clearly, the expressions 'ínertia' and 'mass' are interchangable. The amount of mass is measured by measuring the ratio of amount of force and the resulting acceleration (m=F/a).

Is there anything we can say about the origin of inertia? Not all particles have mass. Photons have no mass, and they are instantly accelerated to lightspeed. Particles with inertia on the other hand, have an interesting property: the acceleration is proportional to the applied force.

That is somewhere between the extremes: if an object would have infinite inertia, then every finite amount of force would not move it. If an object has zero inertia, then it instantly jumps to lightspeed. But inertia is somewhere in between that: acceleration is proportional to the applied force.

There is an interesting analogy with the phenomenon of inductance. A current circuit with a coil with self-induction in it will not particularly resist current strength; it will conduct a wide range of current strengths. If the current circuit is super-conducting it will conduct current without the necessity to keep applying a voltage. Suppose some voltage is applied, a voltage that tends to increase the current strength. The very first change in current strength leads to a change of the magnetic field of the coil. The change of the magnetic field induces an electric field, and this induced electric field opposes the applied voltage. Inductance opposes change of current strength, but does not prevent change of current strenght, for the mechanism can only come into action when there is some change of current strength. In operation it is a self-tuning mechanism.

When there is a coil with self-induction in a current circuit, then the current strength is by good approximation proportional to the applied voltage. By contrast, without the coil with self-induction, applying a voltage on a superconducting conductor causes an immediate jump in current strenght.

It is not clear how far the analogy between inertia and inductance extends, but I find it an interesting analogy. --Cleon Teunissen | Talk 10:59, 8 August 2005 (UTC)

The force that is exerted on the other object

 * Centripetal (center seeking) acceleration is inwardly directed, and caused strictly by the inwardly directed centripetal force. There is no disagreement on that. Since it is the only force on the object, it accounts for the only acceleration of the object.  The centrifugal force does not affect it, not because it's fictitious, but because it's exerted on a different object --Jeepien 02:31:43, 2005-08-01 (UTC)

This is a quite astute remark. I see now how my words can seem to teeter on the brink of the fallacy I seek to debunk.

When I am swinging around a weight on the end of a rope, then the tension in the rope is just regular tension. So the weight is exerting a centrifugal force on ME, and if I am to remain stationary I must brace myself. The weight, unattached to anything but the rope, is being accelerated.

The frequent fallacy is that when an arrow for a centrifugal force is added to a diagram representing forces it is all too often added to the center of mass of the swinging weight. It is my habit to never add any arrow for a "centrifugal force" in any diagram, because its not a force, and the diagram is supposed to represent forces. This habit of mine "protected" me from the fallacy, but I had never spotted the nature of the mistake in sticking an arrow for a centrifugal force on the wrong object --Cleon Teunissen | Talk 12:08, 8 August 2005 (UTC)


 * Thank you.


 * Yes, the tension in the rope is just regular tension. And we know that tension in a rope or string is considered to "act" in either direction, depending on where we focus our attention.


 * Rather than an object in circular motion, consider exerting a force on a massive object at rest in space. We use our calibrated 1 N thought rocket hooked to a strong massless string to tow the object.  The force on the object will cause it to accelerate in inverse proportion to its mass.  The only force  on the object will be that applied by the string, so the acceleration vector will be toward the rocket.  The acceleration will continue as long as the force is applied.


 * All the while, the object will exert a third-law reaction force of 1 N back on the string. This is what we would call the centrifugal force, if the motion were circular.  It is this force that keeps the string under tension in both scenarios.


 * We know that a string that sustains a force on only one of its ends will not show any tension. It can't.  It will simply move limply in the direction of the force.  What is sometimes less clear is that all Newtonian force pairs behave like the tension in a string.  They either exist in both directions equally, or they don't exist.  A somewhat Zen-like rephrasing of the third law might be, "A force can be exerted only to the extent that it is resisted."


 * I illustrate this to my Physics students by having the strongest guy in the class (yes, I do get football players in Honors Physics) push on the wall as hard as hard as he can with his flat palm. If I like, I can measure the peak force with a gauge, but the apparent effort is enough to quantify it.  I ask him to remember what it felt like.  After he's demonstrated his strength, I tell him that, through my advanced mastery of the forces of the universe, I can prevent him him from pushing on my hand with even a small fraction of that force, and challenge him to do so.  I make a show of holding up my palm and bracing myself to resist.


 * When he pushes, of course, I don't resist, but move my hand back, gently yielding to all but a small amount of force. It is naturally impossible for him to push harder than I am willing to push back.  "Come on," I say, "that was weak. Can't you push harder than that?"  But, of course, there's nothing he can do to exert a larger force, and the class gets the point: Forces must exist in matched pairs.  You cannot touch without being touched, you cannot push (or pull) without being pushed (or pulled) in return. --Jeepien 15:05:30, 2005-08-08 (UTC)

So, recapitulating: A stone is being swung around on the end of a rope: the stone is exerting a force in centrifugal direction on the central pivot point.

Additional remarks: In the case of for example a hammer thrower in olympic hammer throwing. The only way for the hammer thrower to swing around the hammer is to make the hammer and his own body both circle around their common center of mass. Then there are two centripetal forces, pointing in opposite directions. (The two stars of a double star system both rotate around their common center of mass. That looks like a no-centrifugal-force scenario to me.)


 * Not at all, unless you're repealing the third law. It is a case of two centrifugal forces.  And this is not unique to hammer throwers and binary stars.  Every real-world case of two-body circular motion will have an element of this sort of motion, which will be less pronounced when the masses of the objects are very different.  Only a perfectly fixed center point could produce a perfectly circular path and such points do not exist.  Each of the forces between two stars, F1,2 and F2,1 is both a centripetal and a centrifugal force, depending on your point of view.  What matters is whether you consider the force to be exerted inward on a star (centripetal) or exerted outward by a star (centrifugal).  Of course each force is truly both, since each is exerted by one star upon its companion.--Jeepien 17:55:48, 2005-08-09 (UTC)

The assertion: 'the centrifugal force does not exist' refers to the fallacy of conceptually putting the point of action of centrifugal force in the wrong place. For example: what causes the weights of a centrifugal governor to swing out? Answer: an increase in angular velocity causes the weights to swing out. The weights of the centrifugal governor do exert a centrifugal force on the vertical axis. However, this centrifugal force is not involved in the weights swinging out.


 * Quite so, and the remedy, in my view, is not to repeat the false statement (that it does not exist), but rather to emphasize the true statement (that people tend to get confused about where it is exerted). In the case of the centrifugal governor, the centrifugal force is not exerted on the weights, and so can't be the cause of their increased radius.  It is exerted by the weights upon the central mechanism. The centrifugal force is what closes the steam valve. The weights swing out for the usual reason: their inertial tendency to move in straight lines, unless constrained by an increased centripetal force.  --Jeepien 17:55:48, 2005-08-09 (UTC)

We have that rotational dynamics is counterintuitive. So I feel that it is important to try an be helpful to people with little physics background. Sometimes the formally correct is at odds with the "common sense". I think then an effort must be made to reconcile the two; to explain to people what the tempting fallacy is. --Cleon Teunissen | Talk 17:33, 8 August 2005 (UTC)

Depending on your point of view
(The two stars of a double star system both rotate around their common center of mass. That looks like a no-centrifugal-force scenario to me.) --Cleon Teunissen | Talk 17:33, 8 August 2005 (UTC)


 * Each of the forces between two stars, F1,2 and F2,1 is both a centripetal and a centrifugal force, depending on your point of view. --Jeepien 17:55:48, 2005-08-09 (UTC)


 * Let me verify I understand you correctly:As seen from the point of view of star (1) the force that is exerted upon it is a centripetal force, and the force it exerts on star (2) is in a direction that is the centrifugal direction for star (1).Likewise, as seen from the point of view of star (2) the force that is exerted upon it is a centripetal force, etc.


 * The diagram representing forces being exerted in the double star system contains two arrows. --Cleon Teunissen | Talk 05:41, 10 August 2005 (UTC)


 * Yes. Each star is pulled upon by a centripetal force arising from its companion, and reacts, according to the third law, with a centrifugal force pulling back upon the companion star.  There are only two forces, but their names change depending upon which star you observe from.  When considering third-law force pairs, the question of which force is "really" the action force and which is the reaction force is rarely clear-cut, and often varies with the observer.


 * An outside observer might look at the force diagram and consider each of the arrows to be centripetal (=) until it reached the center of mass (X), and centrifugal (-) thereafter:

========>         X  <========


 * The same would be true if the stars were of different masses, shifting the center of mass:

===->    X  <---=============


 * This mass difference biases our thinking as to which force should "truly" be called centripetal. This is essentially the situation of our hammer-throw competitor, or me, with my rope and bucket.


 * And, of course, if you consider the center of mass itself, the vector sum of all forces toward and away from it is equal to zero, so the binary system as a whole would move with uniform (inertial) motion. --Jeepien 15:58:41, 2005-08-10 (UTC)


 * Well the verification was superfluous, the matter is clear to me, to my full satisfaction. Newton was sharp-eyed indeed in formulating his third law. It is, contrary to what I thought, possible to formulate one law (the third law) that covers both statics and dynamics, the key is to always recognize the reciprocal pairs. --Cleon Teunissen | Talk 17:21, 10 August 2005 (UTC)

The origin of the expression fictitious force
In the context of a rotating coordinate system the equation of motion contains a term for a centrifugal force (and a term for a coriolis force). In the context of the rotating coordinate system these fictitious forces perform the job that in the context of a non-rotating coordinate system is performed by F=ma.

Sometimes it is formulated as follows: "Centrifugal force is present only in a rotating coordinate system." That sounds very odd to me.

The confusion here is that a habit has grown to use the same label 'centrifugal force' for two quite different situations. Some people blindly assume that since in both situations the same name is used, it must actually be the same.

I try to spot that kind of confusion, and avoid it. When thinking about rotational dynamics, for example the dynamics of a gravitational slingshot I do all the thinking from/in the perspective of the non-rotating frame of reference. --Cleon Teunissen | Talk 17:28, 10 August 2005 (UTC)

Two Definitions
There are two completely different definitions of centrifugal force, that were mixed and matched in the entry. The author sounded confused. I kept some of their examples, and emphasised the distinction between the two definitions, and which one is more commonly used. I also removed some confusing extraneous material from the latter half of the entry. The entry could use some maths, perhaps deriving the centrifugal force from the chain rule and a time dependent coordinate change.


 * Can I define the terms "fictitious centrifugal force" and "reactive centrifugal force" and use them in this article, or would that count as original research? --Ihope127 17:49, 25 September 2005 (UTC)


 * Those terms seem very uncommon, if there is a better way to refer to a definition, try to find it. I'd suggest breaking the article into sections, one being titled "Fictitious force" and one being titled "Reaction force", or something like that.  &mdash; siro  &chi;  o  23:47, 26 September 2005 (UTC)

What does "plate spinning" use?
Can somebody help a Wikipedia-editor-wannabe (me) who barely got through high school physics? while checking up on some WikiLinks in other articles, I linked the article on Plate spinning to Centrifugal force. After reading through these comments, now I'm not so sure that was a good idea. Can one of you "physics" eggheads look at that article, figure out what's right, and either let me know or fix it yourself? Thanks. Joe 02:23, 3 October 2005 (UTC)

random_factor 10:23, 7 October 2005 (GMT) Centrifugal force is an illusion of angular momentum. The friction of an item against the spinning object that it is in or on causes it to accererate in a straight line, because forces act in straight lines. this causes the item to move along a chord of the circular path followed by any given point of the spinning object.this means that the friction force is now acting in a different direction, so the item is accelerated in that direction. This causes it to move along another chord, this one slightly further from the centre of spin and at a slight angle to the first one (because of how the item has moved). This process is repeated so that the item follows a spiral path outwards from the centre of the spinning object, until it reaches the edge. When it reaches the edge, it is able to do what it physics have been trying to make it do all along - fly off in a directon that is tangential to the circular direction of spin. If you are the item in the spinning object (say, one of those fairground rides where you get pressed against the wall and then the floor drops out and you stay stuck against the wall) then you feel as if there is a force flinging you outwards because you are being accelerated at a tangent to the circular path of the spinning fairground ride. Because the force is a straight line and the wall is not, an arrow used to show this force would intersect the wall. This means that there is a force pressing you against the wall, but that force is not some mystical power pushing outwards from the centre of the circle, but instead it is your own momentum pressing you sideways into the wall. You feel pushed back, not pushed sideways, because your inertia cancels out the feel of sideways motion. The simplest way to show that centrifugal force cannot exist is to note that there is no way for a force to push outwards from the centre without there being a solid object for it to push with (except magnetism, and humans aren't magnetic). The centrifugal force that everyone believes in is a fictitious kind of un-gravity, some force that is somehow created by spinning objects. If it did work like that, the people standing next to the fairground ride would be pushed back as well.


 * That's a great answer. But I'm not sure if it answered my question.  Simply put, is the link to Centrifugal force appropriately used in the article on Plate spinning?  Yes?  No?  If not, can somebody fix it or tell me what is more appropriate?  Joe 13:53, 12 October 2005 (UTC)


 * I changed it to "Plate spinning relies on the gyroscopic effect, like it also applies for a spinning top. This makes balancing easier." I think that is more relevant than centrifugal force.--Patrick 16:06, 12 October 2005 (UTC)

More on reaction to centrifugal force
The following unsigned text is from 220.233.107.29:


 * For every action there is an equal and opposite reaction. I am not the first wise person to say that. Who said it before me ? Ah, Newton did. Um, centrifigual force, reaction force to centripedal force.


 * Inertia is a force. F=MA. M is inertial, so therefor F is to !


 * The problem here is that there are too many sophomores who are repeating the answers that their 2nd year tutors expect based on the musings (not publifications) of a professor in astrophysics.


 * In the context of motion on earth, centrifigual force is quite real.

I have previously edited this article anonymously from IP addreses 85.81.19.235 and 192.38.79.130). I have mostly reverted the article edit by 220.233.107.29 which attempted to assert that centrifugal force is only the reaction force to the centripetal force, and that the fictitious force meaning is wrong. Nonwithstanding its factual merits (I think it reflects a grave misunderstanding of the physics), the style was decidely non-encyclopedic, containing internal commentary and gems such as "For this reason, teachers of science have gone bizzare and started to *emphasize* ..."

In general I have become less and less convinced that the "the real reaction force that complements the centripetal force" meaning is actually legitimate. I have been able to locate no authoritative sources that show this meaning actually being used in physics, whereas I think that every university-level textbook in mechanics will describe the fictitious force meaning. I'm considering to be bold and edit to present the reaction-force meaning as a common mistake rather than a legitimate meaning, unless somebody can point to respectiable source of the reaction-force usage. Henning Makholm 23:16, 22 December 2005 (UTC)


 * Be bold, give it a go, people will complain if they don't like it! William M. Connolley 15:52, 23 December 2005 (UTC).


 * So done. On further thought it does make sense to identify the reaction force with the centrifugal force in statics, so I have added an explanation of this in lieu of the "two definitions" concept. Henning Makholm 11:38, 26 December 2005 (UTC)


 * Good work. I have just one question. You write:
 * There is tension in the rope, pulling inwards on the ball (the centripetal force) and simultaneously pulling outwards on the pivot (the reaction force). The tension is real, so these two forces still exist if we move to a corotating frame. However, in the rotating frame there is also a centrifugal force that pulls outwards on the ball.
 * So we have three forces, the odd number arising because the centrifugal force is fictitious and has no third-law counterpart. So far so good. But then you write:
 * When solving statics problems in the rotating frame, [...] one often considers a force "the same" before and after it has been conveyed by a structural element, so according to this view the reaction force on the pivot is the centrifugal force.
 * This can't mean what it says. We have the centripetal reaction force, and the centrifugal force. They have identical directions and magnitudes, and because their objects are connected by a structural element, we consider them both as being applied to the pivot. But if we combine the two, the combined force on the pivot should be exactly double the centrifugal force, not equal to it. – Smyth\talk 14:51, 26 December 2005 (UTC)


 * We don't combine the two, we identify them. We could also measure the force thrice forces, because the outer half of the rope pulls the inner half of the rope with the same force, and so on.


 * How can I explain? The way I think myself is that force means a flow of momentum, by Newton's 2nd law. In a typical statics problem, gravity makes vertical momentum seep into the parts that make up a building, and we try to calculate how this momentum flows through beams and columns and ends up in the foundation; then we can check whether more momentum needs to flow through each element than it can carry (i.e. if the force on the element is larger than its strength). In the rotation problem, the centrifugal force magically makes momentum appear in the ball, and this momentum flows through the rope and into the pivot. We can then say that it is the same flow of momentum whether we look at it as it arises in the ball or as it arrives at the pivot. This is similar to the Thames carrying 65 cubic metres of water per second past London Bridge and 65 cubic metres of water per second past Greenwich. We do not add up the two numbers to find the water that arrives at the North Sea; the two measurements are of the same flow of water, and in a similar sense one can view the centrifugal force and the centripetal-force reaction as the same flow of momentum, i.e., the same force.


 * Of course, viewing force as momentum flux, while true, is not a suitable mode of explanation for an elementary encyclopedia article. I'm still trying to think of a way to explain this that does not create confusion more than it enlightens. Henning Makholm 00:55, 27 December 2005 (UTC)


 * I think you've done an excellent job – the article is definitely better now than it's ever been before. – Smyth\talk 21:20, 27 December 2005 (UTC)


 * Thanks :-) Henning Makholm 00:10, 28 December 2005 (UTC)

improve first paragraph?
Somewhere in the text it is written: "the reaction force on the pivot is the centrifugal force. This identification often leads to confusion about the "fictitious" nature of the centrifugal force, because the pull on the pivot is a perfectly real force". IMO that is the best description/explanation; but this is not apparent in the intro to the article, which even erroneously suggests that a centrifugal force isn't a contact force. My suggestion: rework the intro to include the clear and correct description from the body of the article, which explains that the centrifugal force is a reaction force to a ("real", in the sense of causal) contact force.

BTW, I'm a bit surprised by the apparently erroneous claim that this would be true only in the rotational frame. In my vocabulary, in a washing machine the centripetal force of the wall on the clothes is met by a centrifugal reaction force of the contents on the wall; this is true in the inertial frame. Harald88 18:20, 2 January 2006 (UTC)

The problem I think is that different people use the expression 'centrifugal force' in fundamentally different ways. I will present two examples of two totally different usages of the expression 'centrifugal force'.

Example 1: Take a pilot training centrifuge, and put a stack of three weighing scales in the pod of the pilot training centrifuge. Let the weighing scales be 1 kilogram of weight each. Start the pilot training centrifuge, and spin up until the pod is pulling 1 G of acceleration. The weighing scale at the bottom of the stack will read 2 kilograms of weight, for it is supporting two weighing scales, the middle weighing scale will read 1 kilogram of weight. The pattern of weight readings for the stack shows that the stack is transmitting accelerating force upwards. At every level of the stack the force and the reaction force are equal in strength. Of course, because the the pilot training centrifuge is attached to the Earth it is so much heavier that any motion due to the reaction force is negligable. (Then again, a badly loaded household centrifuge will wobble violently.)

Example 2: If you are inside a rotating space-station, co-rotating with the station, and you trow an object, then the trajectory of the object with respect to the rotating space-station will be curvilinear. (Of course, the actual motion of the object will be along a straigh line as inertially moving objects move along a straight line.) The curvilinear trajectory can be described by inferring a centrifugal acceleration factor.

Some people insist that the expression 'centrifugal force' should be used exclusively to express what is portrayed in example 2. It is my understanding that when people insist that centrifugal force is present exclusively 'in a rotating frame' they are referring to the context of example 2, not to the reaction force of example 1. --Cleonis | Talk 19:21, 2 January 2006 (UTC)


 * Cleon, thanks for the clarification; but the second interpretation is a contradiction in terms with "centrifugal", for the curvilinear motion does not generally correspond to a centrifugal force... Hmm, I think that Newton wrote about it; then it may help to cite him on this! Harald88 19:31, 2 January 2006 (UTC)


 * The transformed equation for motion with respect to a rotating coordinate system has a socalled 'centrifugal term' and a 'coriolis term'. It is my understanding that some people follow this line of thought: there is a centrifugal term present only in the equation of motion for a rotating coordinate system, hence there is only a centrifugal force 'in the rotating frame'.
 * The centrifugal term in the transformed equation for motion with respect to a rotating coordinate system does point in centrifugal direction.


 * Unfortunately, Newton did not express himself clearly when making statements about centrifugal force. Citing Newton is probably not helpful.
 * A somewhat baffling aspect of the pilot training centrifuge scenario is that there is a reaction force, but inertia itself is not a force. If you would categorize inertia as a force you would end up with self-contradiction. If you make a technical drawing of the pilot training centrifuge scenario, and you put in arrows to represent forces, then you do not put in an arrow for inertia. --Cleonis | Talk 20:37, 2 January 2006 (UTC)


 * I will look out for Newton's comments. In the training centrifuge scenario, the centriptal force is the causal force; according to you this is not due to inertia? According to me that force is due to the change of velocity of the chair in an inertial frame. Where is the self-contradiction? Harald88 23:08, 2 January 2006 (UTC)


 * As you state, in the pilot training centrifuge, the first in the causal chain is the centripetal force. The floor of the centrifuge pod exerts a centripetal force on any object resting on that floor, let's say a ball rests on the floor. The centripetal force causes acceleration towards the center (but there is also sideways velocity, so the distance to the center does not decrease.) Next in the causal chain is that inertia opposes the change of velocity, so the ball exerts a reaction force on the floor. I avoid equating inertia with the reaction force that the ball exerts on the floor. I prefer to define inertia as the property that a force is required to cause acceleration. Following that definition, inertia is not a force. --Cleonis | Talk 01:25, 3 January 2006 (UTC)

Harald, my contention is that the term "centrifugal force" is only used by serious physicists to refer to the fictitious force that appears in the rotating system. The reaction to the centripetal force is in itself perfectly respectable, but it is a different beast and not the thing physicists do actually call centrifugal force. If you disagree, feel free to cite respectable sources. Henning Makholm 00:30, 3 January 2006 (UTC)


 * Henning, your claim about "serious physicists"is tendentious (and I know it to be wrong). What are your respectable sources? Harald88 00:36, 3 January 2006 (UTC)


 * From what I have access to right at the moment: ScienceWorld, as cited in the article, and the Feynman Lectures on Physics, volume 1, section 12-5. The relevant paragraph from the latter reads, Another example of pseudo force is what is often called "centrifugal force". An observer in a rotating coordinate system, e.g., in a rotating box, will find mysterious forces, not accounted for by any known origin of force, throwing things outward toward the walls. These forces are due merely to the fact that the observer does not have Newton's coordinate system, which is the simplest coordinat system. Also the article centrifugalkraft in Den Store Danske Encyklopædi (which, however, is in Danish, and so may not be individually convincing). Henning Makholm 00:58, 3 January 2006 (UTC)


 * The centrifugal force that Feynman describes there corresponds quite well to my vocabulary: it appears as a mysterious force (a pseudo-causal force) in the rotating frame; and it's a perfectly physical reaction force in the inertial frame. Harald88 13:03, 3 January 2006 (UTC)


 * No it does not: Feynman clearly describes a force that appears to act on the object iself. That is the force that people in general call centrifugal force. The reaction force also appears in the rotating frame, but there is nothing mysterious about that in any frame. Henning Makholm 13:44, 3 January 2006 (UTC)


 * Hi Henning, I gather that you assert that the meaning of the expression 'centrifugal force' that you have in mind entails that it is most definitely not a force. I rather like to compare it to the expression 'retrograde motion of a planet'. The planet Mars does not actually change from prograde to retrograde motion, it only seems so as seen from the Earth. The true motion of Mars is its eccentric orbit around the Sun, the retrograde motion is apparent motion.
 * I think the name 'centrifugal force' is unfortunate, since what you have in mind is not a force at all. --Cleonis | Talk 02:24, 3 January 2006 (UTC)
 * Addendum. I should define here what I mean by force. I define force as a physcial interaction between two objects; if the objects can move freely, momentum is exchanged. Examples: electrostatic attraction (or repulsion) between two charged particles, and gravitational slingshot. --Cleonis | Talk 03:12, 3 January 2006 (UTC)


 * You are quite welcome to think that the centrifugal force should not be called a force, but in an encyclopedia it is our taske to describe things with the terminology that people out there actually use. Remember that truth per se is not our objective; an encyclopedic compilation of ideas that are already current independently of Wikipedia itself is. (Not because there is anything wrong with truth, but because Wikipedia is not an appropriate medium for conducting a discussion about what the truth is). Henning Makholm 13:44, 3 January 2006 (UTC)

Centrifugal force: 2 citations [] --24.202.163.194 06:23, 4 January 2006 (UTC) --Aïki 05:12, 10 January 2006 (UTC)

The concept of coordinate acceleration
In physics, it is customary to refer to acceleration with respect to the local inertial frame of reference as true acceleration. On the other hand, when motion/acceleration with respect to an accelerating coordinate system is considered, it is recognized that there is some apparent motion.

Example: the retrograde motion of Mars. The keplerian orbit of Mars around the Sun is its true trajectory. The motion of Mars as seen from the Earth is a linear combination of its true trajectory and the curvilinear motion of the Earth. In other words: a linear combination of Mars' true trajectory and a coordinate acceleration.

Formulated as generally as possible: if the motion/acceleration with respect to the local inertial frame is considered, then all the motion/acceleration can be accounted for as a consequence of action of one or more of the four fundamental forces of Nature: gravitational interaction, electromagnetic interaction, strong nuclear interaction, weak nuclear interaction. If motion with respect to a non-inertial coordinate system is considered, the motion is accounted for in terms of one or more forces, plus the coordinate acceleration that is involved. The coordinate acceleration is an artifact of having chosen a non-inertial coordinate system.

The only way to formulate laws of motion at all is to consider the motion with respect to the local inertial frame of reference. All laws of motion refer to the local inertial frame of reference. For example, the correction term for centrifugal coordinate acceleration: Let $$\omega_i$$ be the rotation rate with respect to the local inertial frame of reference.
 * $$ F_c = - m \omega_i \times  (\omega_i \times   r )$$

The correction term must use $$\omega_i$$, and $$\omega_i$$ refers to the local inertial frame of reference.

So of course it is possible to transform the equation of motion to a rotating coordinates system, but it should be recognized that the transformed equation of motion is still referring to the inertial frame of reference. Any law of motion that actually works is in one way or another referring to the inertial frame.

Using a rotating coordinate system instead of an inertial coordinate system introduces a coordinate acceleration, which must duly be taken into account. I regret that the expression 'centrifugal force' is used for the concept of coordinate acceleration. I think that using the expression 'centrifugal force' in that way is an open invitation to confusion. --Cleonis | Talk 11:28, 3 January 2006 (UTC)


 * I might even join you in that regret, but it is not our job here to change the terminology that people already use. We can explain the confusion as best we can, but we should not claim that words mean something else than the thing they are actually being used for, whether or not the actual use is desirable or not. Henning Makholm 13:47, 3 January 2006 (UTC)

That makes me unsuitable for editing wikipedia, for I have a severe dislike of misleading terminology.


 * You can edit all you want, as long as you do not try to use Wikipedia as a tool to change established terminology. That is not what an encyclopedia is for. Henning Makholm 14:55, 3 January 2006 (UTC)


 * I can be of help: against all odds I found Newton's definition of "centrifugal force". And guess what? His definiton is the same as that of Huygens, Jeepeen here above and myself. That should help to avoid changing established terminology! See below. Harald88 19:19, 3 January 2006 (UTC)

Currently, the article contains the statement: "Whenever a body is stationary in a rotating frame, there must be some force that cancels out the centrifugal force, or it would be seen to accelerate away from the center." That is a very misleading statement. Coordinate acceleration is an artifact of employing a non-inertial coordinate system. The suggestion that coordinate acceleration can either cancel a physical force, or be canceled by a physical force does not apply. Coordinate acceleration is frame-dependent, an exerted physical force is frame-independent. --Cleonis | Talk 14:18, 3 January 2006 (UTC)


 * In the rotating frame, the movement of the body is described by a modified second law:
 * m·a = fictitious forces + real forces
 * When the body is stationary in the rotating frame, the left-hand-side of this equation is zero; thus so must the right-hand side be. Two vectors that add to zero are said to cancel each other; this is a mathematical concept and is used irrespective of the physical interpretation (or lack of same) of the two vectors. We know the fictitious force, and so we deduce that a real force that is opposite to it somewhere in the system, and we call that force, whatever its origin is, the centripetal force. We could also have deduced the centripetal force by working in an inertial frame, but my point in this paragraph was to work within the rotating frame (because I'm building up to doing statics there, and in statics you don't want to be moving between frames) and show how the existence of the real centripetal force can be shown there, and how it happens to be minus the "fictitious force" term. Henning Makholm 14:55, 3 January 2006 (UTC)

The concept of 'being in a rotating frame'
In physics, only things that can be measured are considered to be things with physical meaning. For example, we can measure the physical effects of electromagnetism, therefore physicist deem it justified to postulate the existence of an electromagnetic field. Conversely, if some postulated field has no observable effects, then a physicist should not bother to postulate it, much like physicists do not bother anymore to postulate a luminiferous ether.

Clearly, there is no such thing as measuring whether you are 'in an inertial frame of reference' or 'in a rotating frame of reference'. There are no observable differences, any difference between the two is merely psychological, not physics

Let a tethered ball be circling a stationary pivot. The measurable quantities are the centripetal force, exerted by the pivot on the ball, and the reaction force, exerted by the ball on the pivot. The centripetal force and the reaction force are present both 'in the inertial frame' and 'in the rotating frame'. Any postulated centrifugal force that is supposed to be present only "in the rotating frame" is inherently unmeasurable.

Of course, the equation of motion for a rotating coordinate system is a good tool in the physicist's mathematical toolbox. But it should be recognized that their is no ground for ascribing physical meaning to the concept of 'being in a rotating frame of reference'.

Excellent time-saving calculational tools without physical meaning are not unusual in physics. For example, calculating the current in electronic oscillator components. To obtain a solution, second order differential equations must be solved, and a powerful mathematical tool to do that is to extend the problem to the number space of complex numbers. The complex number solutions show the way to real number solutions. That is an example of a powerful mathematical tool, without a physical counterpart to the actual mathematical operations. --Cleonis | Talk 15:25, 3 January 2006 (UTC)

Long time established terminology -- according to Newton
I found back some definitions by others. For example, The Columbia Electronic Encyclopedia   :

centripetal force and centrifugal force, action-reaction force pair associated with circular motion.

But also Newton, Principia : And by such propositions, Mr. Huygens, in his excellent book De Horologio Oscillatorio, has compared the force of gravity with the centrifugal forces of revolving bodies. The preceding Proposition may be likewise demonstrated after this manner. In any circle suppose a polygon to be inscribed of any number of sides. And if a body, moved with a given velocity along the sides of the polygon, is reflected from the circle at the several angular points, the force, with which at every reflection it strikes the circle, will be as its velocity: and therefore the sum of the forces, in a given time, will be as that velocity and the number of reflections conjunctly; that is (if the species of the polygon be given), as the length described in that given time, and increased and diminished in the ratio of the same length to the radius of the circle; that is, as the square of that length applied to the radius; and therefore the polygon, by having its sides diminished in infinitum, coincides with the circle, as the square of the arc described in a given time applied to the radius. This is the centrifugal force, with which the body impels the circle; and to which the contrary force, wherewith the circle continually repels the body towards the centre, is equal. (emphasis mine).

Assuming that the translation is not erroneous, I think this settles the issue about improving the article: just as centripetal, it has nothing to do with "frames". I suggest to stick to this straightforward and traditional definition in the introduction as well as in most of the article, and to put alternative "modern" interpretations in a separate paragraph (it would still be interesting to find out who kooked them up). Harald88 19:37, 3 January 2006 (UTC)

PS I already adapted the intro. Harald88 19:56, 3 January 2006 (UTC)

multiple meanings

 * Clearly there are multiple meanings of the expression 'centrifugal force' in circulation. Wikipedia neutral point of view policy calls for paying due attention to each of the meanings that is 'out there' in wide circulation. --Cleonis | Talk 20:22, 3 January 2006 (UTC)


 * Sure. I also left both meanings in the intro, just as it was before. Any idea where the alternative meaning originated? Harald88 21:03, 3 January 2006 (UTC)

In the Newton's example above, the two forces doesn't act on the same object: the centrifugal acts on the circle, and the centripetal acts on the object.

In a rotational reference frame, the two forces act on the same object.

This make a big difference and ... a lot of confusion if we don't see that difference! --24.202.163.194 04:40, 5 January 2006 (UTC)

Usage as "occasional"?
It is not beyond the bounds of possibility that the meaning of the term among scientists has become clarified or even modified in the last 300 years. I agree that both of the meanings described in the article are reasonable, but it is a fact that even the most obvious search shows that current scientific usage of the term is overwhelmingly limited to rotational reference frames. For the article to describe this usage as "occasional" is completely misleading. – Smyth\<sup style="color:gray;">talk 13:36, 7 January 2006 (UTC)


 * I agree, and there is no reason for such a word at all. Correct immediately. Harald88 15:22, 7 January 2006 (UTC)


 * Do you mean that the section 'Rotating frame of reference' of the article should be enlarged, or that the article suggest that it is use occasionally, or say it is so, or is it another meaning? If you could clarified a little bit more what you mean exactly, it would help. Thank you.   --Aïki 15:37, 7 January 2006 (UTC)

I mean that the article should be reverted to the version of 14:21, 3 January 2006. Scientists do not use the phrase "centrifugal force" to mean "any force oriented away from the center"; they use it to refer to one specific force, operating on one specific object, in a rotating reference frame. Just because such precision is not found in simplified explanations for non-scientists, or other encyclopedias, or even Principia Mathematica, does not mean that the other meanings have equal status. – Smyth\<sup style="color:gray;">talk 16:35, 7 January 2006 (UTC)

Alright, perhaps not completely reverted, because the statement "sloppy labeling can obscure which forces are acting upon which objects in a system" is an important one which escaped me for some time. But the statements about which forces are correctly referred to as "centrifugal" should be restored to what that version said. – Smyth\<sup style="color:gray;">talk 17:08, 7 January 2006 (UTC)


 * Thanks for putting our attention on the inaccurcy of that latest addition, I will correct that (but according to me, that whole part is superflous). But note that the version that you refer to was full of errors and started with a derived (and rather obscure) meaning of the term. In physics, we do not commonly regard rotating frames as valid physical reference frames. Harald88 17:54, 7 January 2006 (UTC)

"Derived and rather obscure"? That is the definition used by 9 out of the first 10 results of the above-referenced search. Even the 2 sites aimed at children correctly identify the centrifugal force as being suffered by the rotating object, not the fixed point. The one exception is from a meterology course, where of course all calculation is done from within a rotating reference frame, and a little oversimplification is forgivable.

Consider this an answer to your request above for sources about "serious physicists". This is the only meaning those physicists use, and the article should identify all other meanings as being simplifications or informal usage. – Smyth\<sup style="color:gray;">talk 18:31, 7 January 2006 (UTC)


 * Well, I have to disappoint you: I am a physical engineer, and I always used it in the Newtonian way. Jeepien above is a high school physics teacher and he obviously teaches the same. Thus your claims are outrageous. Harald88 03:33, 8 January 2006 (UTC)

The «meaning» and the «using»

 * I agree that the centrifugal force is mostly used in rotating reference frames.

But I think there is a distinction to be made between the «meaning» and the «using» of a word or expression. In problems to be solved using a rotating frame, the meaning of the centrifugal force is not usually express explicity by scientists, because for them, the meaning is evident. It is like the word «inertia», they don't give a definition of it each time they use it! The context is sufficient for them to know in what sense the word must be used. But here, we are in an encyclopedia open to every one. Therefore, we have to explain a little more, at least, what means the terms we use.

Regarding equal status or not, the official politics of wikipedia (NPOV) is to give a fair share to each one i.e. more space for the more important, and lesser space to the lesser important. --Aïki 18:51, 7 January 2006 (UTC)

In most of the pages I found, it is explicitly stated that "centrifugal force" only exists in an accelerating frame. In the others, the phrase is used in such a way as to imply the same thing.

There is no confusion here: it's clear which meaning scientists consider to be the "more important" one, and it's not the one currently asserted by the article. It was the one asserted by the version of several days ago. – Smyth\<sup style="color:gray;">talk 19:10, 7 January 2006 (UTC)


 * You show which opinion is apparently most talkative on internet. But the version of a few days ago made many categorical general claims that are factually erroneous in classical physics, and rotating frames are little used. In any case, there is a long paragraph about the derived meaning of centrifugal force in a rotating frame, so where is your problem? Harald88 03:33, 8 January 2006 (UTC)


 * I show which opinion is overwhelmingly expressed on .edu sites on the Internet. What "categorical general claims" do you mean? – Smyth\<sup style="color:gray;">talk 11:10, 8 January 2006 (UTC)

In my point of view, of the four texts that you have submited, #6 is the best one, and even the best one that I have seen up to now among all those vulgarisation texts.

Again, I agree with you that the centrifugal force is mostly used in rotating reference frames, the major raison being that from the standpoint of an inertial reference frame, it does not exist at all, meaning that we don't need it to explain the situation, while we absolutely need it in a rotating frame. But, this said, its utilisation in a rotating frame, and not in a inertial frame, does not change at all its meaning! Moreover, there is other utilisations of the concept of centrifugal force, one of them being in the Newton's Book 1.

Thus what is clearly more important is not the meaning, for the simple reason that there is only one, but its utilisation, which is with the rotating frame, (agreed again!), at least, as much as we can see from the outside of the physicists world. --Aïki 03:52, 8 January 2006 (UTC)

Proposition of agreement

 * Aiki, are you saying that the meaning of "centrifugal force" is "any force directed away from the center of rotation"? In that case, I think you are confusing meaning with derivation. Adrenaline is so called because it is produced by the adrenal glands. But this does not mean that every substance produced by the adrenal glands can properly be referred to as "adrenaline", even if scientists in the past may have done so. The defining characteristic of adrenaline is its molecular structure, not its place of origin, just as the defining characteristic of centrifugal force is its relationship to accelerating reference frames, not merely its orientation. At least, that is what the above search shows.


 * So let us reorganize the article again, distinguishing the two outward-directed forces as the "fictitious" and the "reactive" centrifugal forces. We will make no statement about which meaning is more "popular" or "important"; we will simply state the facts of usage. Do you agree? – Smyth\<sup style="color:gray;">talk 11:10, 8 January 2006 (UTC)


 * I agree, except for the logical sequence: it is evident which is the original meaning that is still in use, and which the secondary meaning, from improvising Newtonian physics for rotational frames. Putting that on its head is illogical and therefore confusing. Harald88 11:48, 8 January 2006 (UTC)

I agree to say that the words "fictitious" and "reactive" are the ones in usage. But if someone proves us the contrary, or give us serious informations to the contrary, we will have to reconsidered the situation. --Aïki 04:54, 10 January 2006 (UTC)

Newton's meaning
But I don't agree on the meaning of it.

Harald88, at the very beginning of this discussion, gave us the example brought by Newton himself: 'This is the centrifugal force, with which the body impels the circle'. (I suggest that you read too what it is write by Newton above this quotation.)

In this example gave by Newton, there is no rotational frame of reference at all, first of all, and secondly, his centrifugal force does not act on the object at all, but on the circle !

This make a big difference with the usual acceptation of the meaning. --Aïki 01:44, 21 January 2006 (UTC)

Centrifugal force without a rotating frame
A piece of rock coming from outer the solar system is now entering in it and is attracted by the sun. Depending on its initial direction, it will crash on the sun (or in!), or will turn around it and go away in the outer space. In the second case, his path will be a curved one called hyperbola.

In his coming approach, the object will go nearer and nearer of the sun, till a certain point where it will begin to go further and further away of the sun. At the approaching stage, the force acting on it is directed toward the sun; at the go away stage, the force acting on it is no more directed toward the sun but outward. Taking the sun as the center of the curve, we can then called the force toward the sun the centripetal force, because it is directed toward a center, and the force outward the sun the centrifugal force, because it is directed outward the center.

In this example, there is no need of a rotating frame to have a centrifugal force. The only thing that it is needed, it's to know the direction of the force. That's all!

What is the reason to call a force, any force, by the adjective centrifugal, it is not because that it is viewed from a rotational reference frame, or an inertial frame, or else, but because it is directed outward the center. Nothing else count! Note too, that in a rotating reference frame, the centripetal force is also in the same frame and it is not called centrifugal just because it is in that frame! --Aïki 19:41, 8 January 2006 (UTC)


 * You are obviously confused. In Newtonian physics, the force acting on the object is the gravitational force of the sun, and it's all the time directed towards the sun. And yes, it's a centripetal force. Harald88 20:59, 8 January 2006 (UTC)

Citation: 'And if the parabolic section of the cone (by changing the inclination of the cutting plane to the cone) degenerates into an hyperbola, the body will move in the perimeter of this hyperbola, having its centripetal force changed into a centrifugal force.' Newton. --Aïki 04:31, 9 January 2006 (UTC)

The four references by Smyth
In most of the pages I found, it is explicitly stated that "centrifugal force" only exists in an accelerating frame [4] [5] [6] [7]. Smyth. See:

I just review these four references and I did not found where it is explicitly stated that the "centrifugal force" only exists in an accelerating frame. Would you be kind enough to show us where exactly it is explicitly stated? (or someone else) --Aïki 02:22, 21 January 2006 (UTC)

Newton's BOOK I. OF THE MOTION OF BODIES. SECTION II. Of the invention of Centripetal Forces.
1- First, thanks to our collaborator Harald88 for having bring to our attention this remarquable demonstration made by Newton.

2- In the two examples that are present in this text, the first one at the beginning, and the second, the Newton-Huygens one, in the SCHOLIUM of PROPOSITION IV. THEOREM IV., the body never goes along a circular path. It always goes in a straight line, and at a constant speed along this line i.e. in an uniform rectilinear motion, except at the points of deviation, of course. More than that, in the first one, there is not even a circle!

3- In the first one, the body receive impulses at equal amount of times, thus at same lenght of straight line and same areas of triangle. Each triangle have their summit at the same point S, the focal point. Each impulse comes from the outside of the triangles and is directed inward in the direction of the focal point S. This impulse is called here: the centripetal force, and is apply on the body.

4- In the Newton-Huygens one, it is the circle which, at first, receive an impulse when the body hurts this circle, coming from a straight line i.e. any side of the polygone. The force generated by this impulse is directed outward the circle and polygone, and it is called here: the centrifugal force, and it is not at all fictitious or fallacious or imaginary or other seemingly expression.

It's differ from the centripetal force of the first example, by its direction and its point of application: in the first example, the force is directed inward and applied on the body; in the Newton-Huygens one, the force is outward and applied on the circle. The name centripetal or centrifugal, depends only on the direction: if outward, it is centrifugal; if inward, it is centripetal. It does not depend at all on which body (we can see the circle as a wall i.e. a body) a force, whatever its nature could be, is applied. In both examples, each force is as real as the other one. ( to be continued) --24.202.163.194 20:20, 6 January 2006 (UTC)

New terminology -- according to whom?
Smyth writes here above: "the defining characteristic of centrifugal force is its relationship to accelerating reference frames". What is the source, and who started this alternative definition? &ndash Harald88 11:59, 8 January 2006 (UTC)


 * I have given a very large number of sources that show this to be a current definition, though I can't answer your historical question. – Smyth\<sup style="color:gray;">talk 17:46, 11 January 2006 (UTC)


 * Currently the article has no reference list. Perhaps you'd like to contribute by putting your selection WP:CITE of good sources there. Harald88 19:10, 11 January 2006 (UTC)

Note that apparently also in some modern texts the meaning of the word "force" differs from the standard meaning of "force"; apparentely some texts mean with "force" active force. &ndash Harald88 11:59, 8 January 2006 (UTC)


 * What do you mean by an "active force"? – Smyth\<sup style="color:gray;">talk 17:46, 11 January 2006 (UTC)


 * A force that deviates an object from inertial motion. Harald88 19:10, 11 January 2006 (UTC)

Some common misconceptions
--- ''I moved this Discussion item to here: no point should be discussed between editors in the article space. IMO such points should be discussed here until the discussion is over and consensus is reached about the opinions according to the two different meanings. Below is the current state of discussion, I number the points for ease.'' ---

Some commonly encountered misconceptions about centrifugal force include:

1 No matter how objects move, there is never any centrifugal force in an intertial frame of reference. No, that's overstating the case. Although an object's inertia should not be mistaken for a force acting upon itself, that doesn't stop it from acting upon other objects. If there is any force acting inward upon an object, then it exerts an equal and opposite force that may be acting outward at the point of contact -- upon a separate object.

2 Centrifugal forces always occur when objects are viewed in a rotating frame of reference, irrespective of whether the object follows the frame's rotation. No -- Inertia may be mistaken for a centrifugal force in a rotating frame, but that is not the only application of the term "centrifugal". In an inertial reference frame, an object forced inward will push back.

3 It is always wrong to speak about centrifugal force. No, This misconception is a very common over-reaction against the misunderstanding of inertia. The key here is not to eliminate "centrifugal" from our vocabulary but to detach it from what it is not and use it for what it fits. Since centrifugal force describes any force whose vector rotates with an object of interest to act continuously outwardly, it accurately describes forces like the passenger acting upon a car door during a turn.

4 Centrifugal force is not fictitious but real. That depends on what apparent force one labels "centrifugal". Used loosely to describe inertia within an rotating reference frame, the "force" is artificial (a product of an accelerating coordinate system). On the other hand (literally), if one spins around with a brick held at arm's length, while no real force pulls the brick outward, the brick does exert a very real centrifugal force upon one's hand.

5 Centrifugal force is just another word for inertia. In one application, but not all. Aside from the misunderstanding of inertia, there is also opposition to centripetal force -- but acting upon another object. Centrifugal is merely a descriptor indicating direction (like the "normal" force exerted by an inclined plane). "Centrifugal force" becomes a useful term when used to describe the force with which an object pushes back when forced into circular motion.

6 An outward reaction force against an object exerting centripetal force is a real centrifugal force. Correct.

Upto here was the discussion (without my comments). Harald88 08:08, 11 January 2006 (UTC)

comments
My comments: 1. Looks OK to me, after deleting the biased "No, that is overstating the case"; but superfluous now. 2. Messed up: probably "always" should be "only"? And superfluous now. 3. Perhaps OK but superfluous now. 4. Looks OK but superfluous now. 5. Perhaps useful; but I think there is disagreement about "Centrifugal is merely a descriptor indicating direction". 6. That is not a misconception, as now explained right from the start. Thus superflous. Harald88 08:55, 11 January 2006 (UTC)

[moved Aiki's comments from above text block below point 1 to this comment space]

Text 1
In the last sentence of no.1, does the 'it' refer to the 'force' or to the 'object'? --Aïki 19:25, 12 January 2006 (UTC)


 * The object: a force can't exert a force... Harald88 21:47, 14 January 2006 (UTC)
 * Then, does no.1 can be applied to the case of the attraction between two objects, like the Sun and a planet, where one object exert a force on the other, while this last one, in turn, exerts an equal and opposite force on the first one? --Aïki 04:23, 15 January 2006 (UTC)


 * No.1 is a sweeping, erroneously general statement. It would be correct if only pointing to such limited cases where all forces are pointing inwards. Do you think that there is a misconception about that, which is worth mentioning? Then please propose a reformulation. Harald88 10:22, 15 January 2006 (UTC)

No.1 is right and wrong

 * If a satellite turning around the Earth is hurted by an object coming from the inside space of the orbit i.e. between the Earth and the satellite, it will get an impulse sending it outward on a new orbit. In other terms, as seen from the standpoint of an inertial reference frame, a centifugal force will act on it, a brief one, but centrifugal. So, in this case, the no.1 statement is wrong.

On the other hand, if we take only the case of a satellite in orbit, without any outside interference, and we look at this situation from the standpoint of an inertial reference frame, there is never in that case and all the cases alike, a centrifugal force. So, in those cases, we can say that statement no.1 is right.

We have too, the case pointed out by Newton, of an object turning around the Sun in a hyperbolic path where we have alternatively a cenripetal force and a centrifugal force. So, in this case, the statement no.1 is wrong.

For more details on my point of view, see too: the 'Pivot-ball example' ; the section 'Reaction force and opposite force' ; 'Centrifugal force: 2 citations' ; 'Newton's BOOK I. OF THE MOTION OF BODIES. SECTION II. Of the invention of Centripetal Forces.' --Aïki 17:57, 15 January 2006 (UTC)

The common usage of physics shorthand
I copy and paste from above:
 * I show which opinion is overwhelmingly expressed on .edu sites on the Internet. Smyth 11:10, 8 January 2006 (UTC)

I would like to point out that it is very common in Physics to use physics shorthand. <BR> For example: Tidal force. Every physicist knows that tidal force does not exist as an independent entity. Tidal force is a gradient in a gravitational field. It would be tedious to keep writing, 'the comet Shoemaker-Levy was ripped apart by the gradient in Jupiter's gravitational field'. A search will find numerous places where the expression 'tidal forces' is used, and in all thos places it is intended as physics shorthand.

The expression 'centrifugal force' in the context of 'exists exclusively in a rotating frame of reference' is physics shorthand. It does not refer to something that actually exists.

'Centrifugal force' in the context of 'exists exclusively in a rotating frame of reference' is not ranked as a physics operative factor.

The primary assumption of physics is that all physics taking place can be accounted for in terms of one or more of the fundamental interactions of of nature: Gravitational interaction, electromagnetic interaction, weak nuclear interaction, strong nuclear interaction.

No physicist does even begin to think that the list of fundamental interactions of nature ought to be amended, that 'centrifugal force' ought to be ranked among the Fundamental Forces of Nature, bringing the number up to five.

It is obvious that the expression 'centrifugal force' in the context of 'exists exclusively in a rotating frame of reference' is physics shorthand. It is a tool, a device that is applied in calculations. Problems arise when people start taking the shorthand literally. --Cleonis | Talk 15:37, 12 January 2006 (UTC)


 * Do we have to understand from what you write, that 'centripetal force' is a shorthand for 'gravitational interaction' ? --Aïki 18:58, 12 January 2006 (UTC)


 * I reserve the expression 'gravitational interaction' for the interaction between a pair of objects. As they exert a force on each other, they each change the other's velocity with respect to their common center of mass.
 * I take 'centrifugal force' as physics shorthand for either of the following two things (depending on the context):
 * 1) If the trajectory of an object is circular, then the reaction force to the exerted centripetal force acts in centrifugal direction (So in the case of a ball circling a pivot the pivot exerts a centripetal force on the ball, and subsequently the ball exerts a reaction force on the pivot)
 * 2) If motion of objects is mapped to a rotating coordinate system, then coordinate acceleration must be taken into account. The acceleration with respect to the rotating coordinate system is a vector addition of the true acceleration and the coordinate acceleration. Within the context of a calculation, the coordinate acceleration shares with gravity the property that the force required to "counteract" the coordinate acceleration is proportional to the mass (within a calculation coordinate acceleration is like a gravitational field). --Cleonis | Talk 20:03, 12 January 2006 (UTC)


 * Do we have to understand from what you write, that 'centripetal force' is a shorthand for 'gravitational interaction' ? --Aïki 18:58, 12 January 2006 (UTC)
 * I think my previous answer came from slightly misreading your question, so I will try again.
 * When the expression 'centripetal force' used it always refers to the function it performs. The question is to identify:"Which force is providing the centripetal force in this situation?" In the case of a ball circling a pivot the wire exerts the centripetal force on the ball, in the case of planetary motion the centripetal force is provided by gravity. --Cleonis | Talk 20:26, 12 January 2006 (UTC)

Don't use Wikipedia talk pages as references
To Smyth: Being a contributor who does not know all the rules of wikipedia yet, I have therefore to ask you these following questions: 1- Is it a rule of wikipedia? If it is, where is the reference? 2- Why take off the italics: is it also a rule to not put italics in the article? If it is so, please give us the reference. 3- Why don't you make your comments in the talk-page: is it not a rule too? (+ the usual reference). Thank you. --Aïki 20:05, 13 January 2006 (UTC)


 * On a wiki-technical note: You can start a new entry on the Talk page by clicking on the plussign that is among the tabs at the top of the page (between the tab 'Edit this page' and the tab 'History'. What you did works, but in the History of the Talk page your entry looks like a reply that is appended to the section "The common usage of physics shorthand". --Cleonis | Talk 22:37, 13 January 2006 (UTC)
 * You are right. It woud have been much better. Thank you for your advise. --Aïki 23:50, 13 January 2006 (UTC)

Reaction force and opposite force
An opposite force is not a reaction force. The expression «reaction force» appears in the action-reaction law where we have two objects and two forces, with each force applied on the OTHER object, NOT on the same one.

In the examples of an object moving in a curved path used in this talk page and the article, when the situations are seen from a rotating reference frame, and not from an inertial one, we have at least two forces, sometimes three, applied on the object rotating around the center.

In the case of the pivot-ball example, seen from a rotating reference frame, if the rope is oblique, we must split the tension of the rope in two components (even in an inertial reference frame), one vertical which is OPPOSED to the gravitational force (mg), and one horizontal that is OPPOSED to the centrifugal force, for a total, now, of FOUR forces applied on the SAME object. The two vertical forces are not in reaction with each other, they are 'opposite'; same thing for the two horizontal forces.

In physics, a force does not react to another force when both act on the SAME object: they just add to produced a net force, that is it, one and only one, force. There is no reaction force at all in this matter. Reaction force is used in a totally different situation! --Aïki 17:23, 8 January 2006 (UTC)

For a force to be opposite to another force, we need that these two forces be exerted on the same object. For a force to be a reaction to another force, we need that one force be exerted on a object, and the other force be exerted on another object. --Aïki 14:57, 16 January 2006 (UTC)

Rotation of two balls and a spring

 * Maybe the talking is helped by having a picture to it.<BR>
 * The two balls are connected by a spring, and they are circling their common center of mass. Here are two ways of looking at it:
 * Each ball exerts a force on the other ball, and in reaction the other ball exerts a force in the opposite direction.
 * The center point of the spring exerts a centripetal force on each of the balls. Each ball exerts a reaction force on the center point of the spring.


 * That is the pattern of force that is at play. Alternatively, you can map the motion of the circling balls in a rotating coordinate system, but that will not add force. Remapping to another coordinate system is just that: you go from one way of representing it to another, remappping is something that happens inside the brain. --Cleonis | Talk 18:37, 15 January 2006 (UTC)


 * Excellent initiative! Where did you found that? Is it a kind of 'computer design' experiment? --Aïki 23:25, 15 January 2006 (UTC)


 * I made that animation myself. I used Excel spreadsheets to perform the calculatons, and then I made the frames of the animation one by one. This animation consists of 30 frames, and there are 15 different lengths of the spring in it. I am quite pleased with the result, but let's stick to the subject: there is no need to wreck one's brains over what things look like as seen from a rotating point of view. That is unnecessary complexity. To understand the physics taking place, you identify the inertial frame of reference. Using a rotating frame for the purpose of calculation is another matter. Using a rotating coordinate system is a tool in the physicist's mathematical toolbox. --Cleonis | Talk 01:27, 16 January 2006 (UTC)

All roads lead to the equivalence class of inertial coordinate systems
I copy and paste from above.
 * From the standpoint of a rotational reference frame, the description of the situation is different.<BR> But for the two standpoints, the description must be done using the system of forces of classical physics with its three laws of motion, which includes the use of components and resultant for the forces. --Aïki 18:05, 16 January 2006 (UTC)

It is inherently impossible to directly apply the system of classical physics on motion that is mapped to a rotating coordinate system. For example, for motion that is mapped to a rotating coordinate system the principle of conservation of momentum does not hold. And so on and so on. In order to apply the system of classical physics to motion that is mapped to a rotating coordinate system, adjustments must be put in place. With the adjustments in place, the adjusted versions of the laws of motion refer once again to the inertial coordinate system, and are mathematically equivalent to the laws of motion for an inertial coordinate system, only more elaborate.

Try as you might, there is no getting away from the supremacy of the inertial coordinate system. No matter what you do, putting the necessary adjustments in place will loop you back to the inertial coordinate system. --Cleonis | Talk 10:29, 17 January 2006 (UTC)


 * Maybe a little more precision on my part would help to clarify what I said. The system that I was talking about, is the system of forces which uses component and resultant forces, based on the three laws of motion, all which we find in the classical system: nothing else.


 * On your part, if you can give us one or two examples of what you call 'adjustements', it could surely be helpful too. --Aïki 14:20, 17 January 2006 (UTC)


 * The diagrams that I just found here, may be helpful too. (Note: I forgot to mentionned the equilibrant force in my text. Sorry!) --Aïki 14:40, 17 January 2006 (UTC)


 * Another question: Is your 'equivalence class of inertial coordinate systems' the same thing as the 'equivalent class of inertial reference frames', or are you speaking of the different coordinate systems that we may use in an inertial reference frame, such as the cartesian coordinates, the polar coordinates, the cylindrical polar coordinates and the spherical polar coordinates? --Aïki 15:01, 17 January 2006 (UTC)
 * In newtonian dynamics the equivalence class is all coordinate systems that are related to each other in the form of Galilean transformation. Physics shorthand for the entire set of all members of this equivalence class is 'the inertial coordinate system'.
 * As an example of adjustments the transformed equation of motion as mappped in a rotating coordinate system:
 * $$ F_{fict} = - 2 m \omega_{inertial} \times v_{rot} - m \omega_{inertial} \times (\omega_{inertial} \times r )$$
 * Where $$\omega_{inertial}$$ is the rotation rate of the rotating coordinate system with respect to the inertial coordinate system. All terms in that force law contain the rotation rate with respect to the inertial coordinate systems.
 * As you state, the laws of motion apply in the case of motion. Principles of statics do not necessarily apply for motion, usually they don't. When a ball is suspended from a mounting point, hanging straight down, then that is statics; it is a closed loop. The mounting point is a pole that is part of the solid earth. None of the parts of that loop is in motion relative to another part of that loop.
 * The ball circling the pivot is dynamics. No loop.
 * In the case of a circling ball with the wire oblique, then the vertical components of the forces are statics, and the horizontal components are dynamics. --Cleonis | Talk 16:22, 17 January 2006 (UTC)


 * Do we use the system of forces, with its component, resultant and equilibrant forces, both in statics and dynamics, or in only one of them, or in neither one of them? --Aïki 18:24, 17 January 2006 (UTC)
 * Statics is where there is no relative motion of the involved parts. it is statics when all the forces are balanced. Statics is entirely frame-independent. In dynamics, with unbalanced forces, you find that you can formulate laws of motion only for motion with respect to the inertial coordinate system.
 * The reason it is possible to formulate laws of motion is that for motion relative to the inertial coordinate system inertia is uniform. You can formulate laws when behavior is repeatable. --Cleonis | Talk 19:40, 17 January 2006 (UTC)

Reformulation of #7 last sentence
The wording of the last sentence of #7 seeming to be the start point of all the discussion above, I will reformulate here in another way.

But for the two standpoints, the description must be done using the system of forces with its component, resultant and equilibrant forces.

1- Is it that system which is used in statics and dynamics, or is it another one?

2- Is it that system wich is used in an inertial reference frame and a rotational reference frame, or is it another one? --Aïki 02:27, 18 January 2006 (UTC)

Statics and dynamics; opposite force and reaction force
Statics is when there is no relative motion between the parts. Suppose you pushing a railway car, but the brake has been applied. Then you are not going to move the railway car; if you push at it with 500 newtons of force, the thing is just going to push back with a 500 newtons of force. If you push with 1000 newtons of force the railway car will just exert an opposite force of 1000 newtons on you.

Now try the same without the brake applied. Push at the railway car with 500 newtons of force. Then the railway car will still push back with 500 newton (reaction force), but the railway car without the brake applied will yield to the applied force. If you increase your push to 1000 newtons of force, then the reaction force will match that increase, and also the railway car will yield more.

So how come I'm talking about an unbalanced force? The reaction force always matches what I'm throwing at it!

Of course, the reason for calling the accelerating force unbalanced in dynamics, is the fact that the object that is at the recieving end of the unbalanced force is yielding to it. --Cleonis | Talk 20:39, 17 January 2006 (UTC)


 * In the example above, there is only two forces which acts on the train: the push, if I may call it like that, and the friction of the wheels on the track, the two being opposite.


 * If the push is greater than the friction, we will have a net force in the direction of the push, and the train will move in that direction. This net force is also called an unbalanced force.


 * The reaction force do not act on the train but on the object which is pushing, and it is coming from the train.


 * If it is a second train which is pushing, this pushing force will come from the friction of its wheels on the track, and it will be oppose by the friction force of the first one. Therefore, on the second train, we will have two forces acting on it: the reaction force of the first train, and the pushing force of the second. If there is no net force resulting, nothing will move, (except perhaps the spinning of the wheels of the second train).


 * The reaction force and the pushing force of the second train, its friction force, are opposite forces acting on the second train. The reaction force is not applied on the first train, thus not opposite to the friction force of the first train, but opposite to the friction force of the second train.


 * If this reaction force is less than the friction force of the second train, then the second train will move, and, by way of consequence, the first will move too.

Do we agree with that? --Aïki 03:28, 18 January 2006 (UTC)
 * I think you have introduced unnecessary complexity by weaving friction into the picture. To focus on the core of the matter, a situation must be discussed in which friction can be switched between overpoweringly large and negligably small.
 * Think of the television cameras in television studios. Those huge things weigh hundreds of pounds, an during operation they rest on an air cussion of a milimeter of air. The studio floor is very smooth, and compressed air is fed to the base of the television camera. The camera-operator needs to push all the mass of the camera around, but friction is negligable.
 * Statics is when no air is fed to the camera: then both the operator and the camera will firmly stick to the floor. Dynamics is when the operator has grip, and the camera is lifted from the floor by the escaping air underneath. --Cleonis | Talk 11:28, 19 January 2006 (UTC)

Reaction force matching
1- I did not knew about the television cameras, but I know about the 'Hovercraft' and other things suspended on cushion air. The case of negligible friction is also a familiar one. So, no problem there.

2- The friction force has been introduce by you in your example, but whitout stating it directly.

3- Your last sentence does not match with your two last paragraphs of your initial text. In one, you speaks of dynamics with a railway car not lift by air, and in the other, dynamics with a object lifted by a cushion air. Perhaps you should clarify this matter.

4- Note: The 'yield' that you are speaking of in your 'without the brake applied' case, can also happen with the breaks on.

5- In the second paragraph of your initial text, we have: 'Then the railway car will still push back with 500 newton (reaction force)'. This reaction force acts on the object which exert the pushing, not on the railway car. This object can not go forward, not advance, till the net force on it is positive in the forward direction, i.e. till the pushing force become superior to the reaction force. That is why it has to push harder. At a certain degree of pushing force, this one will overcome the reaction force, and the object will move ahead.

On the railway car, there is also two forces: the pushing force and the resisting force, this last one NOT being the reaction force but the friction force existing between the wheels and the track. When the pushing force will be greater than the resisting force, the railway car will move forward. This will happen at the same moment the pushing object will move forward, because the pushing force on the railway car is also the pushing force on the object, and the resisting force on the railway car is also the reaction force on the object. Thus, when the set of forces on the object becomes unbalanced, the set of forces on the railway car becomes AUTOMATICALLY unbalanced too. That is why the two objects seem to move as if they were a unique object, and why this kind of example often produce confusion when we don't see it with the two set of forces.

6- When the unbalanced force appears, the railway car is yielding (and the pushing object too!). An unbalanced force means that a force is greater than another one. In the railway car example, it happens when the pushing force is greater than the resisting force, this last one being, for the object pushing, the reaction force, and for the railway car, the friction force. Remember that there is no reaction force acting on the railway car.

In the set of forces where the reaction force appears i.e. in the pushing object, this reaction force is finally overcome by the pushing force. This means that the reaction force no more balance the pushing force, i.e. the reaction force no more match the increased force. If it was still matching the other force, there would not be any yielding at all.

7- Therefore it is not correct to say that 'The reaction force always matches what I'm throwing at it!'. Sometimes it does, some times it does not, depending on the situation, as we can see from the example of the raiway car. Aïki 15:51, 19 January 2006

Another presentation
Another way to present the situation is with a pushing object which could be another railway car, no.2, coming at a speed v with the engine stopped i.e. no force applied to make the wheels turning. We neglect the friction between the wheels and the track and the friction with the air. In this situation, there is no force acting on no.2. It is then a case of motion in an uniform velocity (Newton's first law).

Then no.2 hurts no.1. At this moment, no.2 exerts an action force on no.1 of x newtons, and no.1 exerts a reaction force of the same x newtons on no.2 (Newton's second law). In this case we can say that the reaction force has always the same magnitude than the action force, and I think that was what Cleonis wanted to say.

For no.2, at the moment of impact, there is now a force acting on it (the reaction force) and not two as in the first part above. This force is now acting against the direction of the motion, so it will change the velocity v of no.2, i.e. will diminish it. Being of the magnitude of F =ma (the action force), where a is the difference between the velocity of no.2 before the impact and the new velocity of no.2 after the impact, and m the mass of no.2, this force will then stop no.2, whatever if no.1 is moving or not after the impact. --Aïki 19:26, 19 January 2006 (UTC)

A man in a swivel chair, holding a rope
I copy and paste from above:
 * In this case we can say that the reaction force has always the same magnitude than the action force, and I think that was what Cleonis wanted to say. --Aïki 19:26, 19 January 2006 (UTC)

I think you have introduced unnecessary complexity by weaving collisions (hard impacts) into the picture.

I only use examples in which the physics takes place slowly and smoothly, so it is easy to keep track of what is going on. I like the example of television studio cameras. They are very bulky, and when on their air cushion, friction is negligable.

Let a man be strapped in a swivel chair, and let a 200 kg camera be circling around the man, with the man holding a rope in his hands, maintaining the circular motion of the camera. The man in the swivel chair exerts a force on the camera, and the camera exerts a reaction force on the man in the swivel chair. For each radius of circular motion, and each angular velocity, you can calculate how much force the man in the swivel chair must exert, and the reaction force exerted on the man is in opposite direction and equal in magnitude.

if the camera is not on an air cushion, but stuck to the ground, and the man in the chair is pulling the rope, then the man is exerting a force, and an equal and opposite force is exerted on him. (But nothing happens, because nothing will budge now.)

In the above two situations there is no need to come up with a 'rotating frame of reference' with the property that there is a particular kind of 'centrifugal force' that exists 'only in that rotating frame of reference'. That is all unnecessary complexity. --Cleonis | Talk 00:44, 20 January 2006 (UTC)


 * Why should someone need to come up with a 'rotating frame of reference' in those two situations? --Aïki 01:46, 20 January 2006 (UTC)


 * I don't know. I don't see such a need.<BR>
 * If I remember correctly, but it is many kilobytes ago, you have insisted that when a ball is circling a pivot, then as seen from a rotating frame of reference there is a centrifugal force 'that exists only in the rotating frame of reference'. If that wasn't you, then probably you agree with me. --Cleonis | Talk 08:47, 20 January 2006 (UTC)


 * 1- If it becomes possible for you to remember it correctly, please come back here and give it correctly. In this way, we will be able to discuss on what I say, if there is something to discuss. Thank you.


 * 2- For the example of the 'swivel chair', the situation can be seen from an inertial frame of reference as well as from a rotational frame of reference.
 * If someone, for a reason or another, choose the rotational one instead of the inertial one, to analyse and explain the situation, he will then have to include in the system of forces acting on the ball, a force which bear the name of 'the centrifugal force'. It is as simple as that, and I don't see any problem at all there.


 * But if a reader or another contributor see a problem in what it is said above, he or she can expose this problem here, and we will see what we can do for him or her to clear the things out. --Aïki 23:07, 20 January 2006 (UTC)


 * I have only read part of the above, but about your question: A rotating frame is a handy reference frame to describe things in, when measurements are done with that frame. Then one maps Newton's laws to that frame. That is often done on the earth for example: that rotating frame is mapped from the Newtonian Earth Centered Inertial frame. Harald88 12:31, 21 January 2006 (UTC)

Usage of physics shorthand should be stated explicitly
I copy and paste from above:
 * If someone, for a reason or another, choose the rotational one instead of the inertial one, to analyse and explain the situation, he will then have to include in the system of forces acting on the ball, a force which bear the name of 'the centrifugal force'. --Aïki 23:07, 20 January 2006 (UTC)

There is a self-contradiction in the intention of choosing a "rotational frame" for analysis. For this someone, in choosing to use a rotational frame, first ascertains how fast it rotates with respect to the inertial frame.

All analysis that yields workable results refers (either directly or indirectly) to the inertial frame. Analysis that does not refer at all to the inertial frame doesn't get anywhere.

You may be using the word 'force' as one or other physics shorthand.<BR> If you do, please state so explicitly.<BR> What you write is consistent with usage (in this particular context) of of the word 'force' for 'coordinate acceleration'. See this Sandbox article of mine to see what I mean with 'coordinate acceleration'.

Unless you explicitly state you are using physics shorthand, it is unclear to the reader what you have in mind. In physics, there are only the four fundamental forces of nature (gravitation, electromagnetism, weak nuclear, strong nuclear). Any force that acts in any particular situation can be traced back to either of those four. Anything else is not a force.

By the way, there is the lovely Java applet by Robert Bryll, of a fountain inside a wheelshaped space station. Once a squirt of water has been ejected from the nozzle, the squirt of water is moving inertially. It helped me, and still helps me, to visualize motion as mapped in a rotating coordinate system. --Cleonis | Talk 10:07, 21 January 2006 (UTC)


 * In addition: Cleonis's own use here of the word "force" happens to be physics shorthand as well: In Newtonian physics, a reaction force is a force too. Harald88 12:24, 21 January 2006 (UTC)


 * The point is to trace back which of the four fundamental forces is the mediator. In the animation with the two balls connected by a spring, they each exert a reaction force on the other. The spring acts as the mediator of that force. The elacticity of the spring can be traced back to electromagnetism.
 * When I use the expression 'reaction force' in newtonian context, then I am refering to what is depicted in the animation: object A exerting a force on object B (and vice versa).
 * Communicating about physics with words is next to impossible, most words have different meanings for different people. Diagrams and animations can help. --Cleonis | Talk 13:41, 21 January 2006 (UTC)

Force vector diagram

 * To help distinguish the four fundamental forces of nature (gravitation, electromagnetism, weak nuclear, strong nuclear) from the forces used in what I have called a system of forces, I intend to rename this last expression a force vector diagram, which one is used in making diagrams of the forces acting on an object, using components, resultant, equilibrant and net force or unbalanced force. Cleonis, do you think that this renaming will help to distinguish between the two meanings? --Aïki 02:24, 22 January 2006 (UTC)


 * That seems quite crisp and useful terminology to me. The force vector diagram represents functions that are being performed. Example, a high tech train that is magneticly levitated. In magnetic levitation the magnetic repulsion provides the required normal force. The expression 'normal force' describes the function to be performed, and in the case of Maglev magnetic levitation does the job.


 * In the case of the pivot and ball, and the wire at 45&deg;, that force vector diagram is two independent cases. The force and component of force acting in vertical direction are adequately modeled with a one-dimensional case: a ball hanging straight down. The force and component of force in horizontal direction are adequately modeled with a two-dimensional case: a ball circling a pivot in the absence of gravitation. So effectively you have two separate force vector diagrams: a one-dimensional diagram, and a two-dimensional diagram. --Cleonis | Talk 07:51, 22 January 2006 (UTC)
 * In your 'two-dimensional case', the tension of the rope, in a force vector diagram, will be divided in two component forces acting on the ball: one vertical component and one horizontal component (the centripetal force). Having no gravitation force at all, the ball will move up till the rope will be horizontal. In this last situation, as seen from a force vector diagram point of view, the tension will have the vertical component force equal to zero, and the horizontal component force will then be equal to the tension, which means that the centripetal force will then be equal to the tension, and only in this situation, but not in the 'oblique situation'. --Aïki 16:12, 24 January 2006 (UTC)

Force vector diagram represents functions
To Cleonis: In wich book(s) did you find that the force vector diagram represents functions that are being performed? I looked at some books but did not found it yet (at least in the index), neither in the article normal force that you have mentionned. Thank you. --Aïki 00:55, 25 January 2006 (UTC)

Centripetal force in the pivot-ball example

 * 1- In the pivot-ball example, the centripetal force on the ball is not along the oblique rope, but it is situated horizontally in the plane of rotation of the ball, along the radius between the ball and the center of rotation which lies in the pivot: how do you trace that back 'to either of those four'?


 * 2- On another hand, if the theory of the big-bang correspond to the reality, we can say that everything which exist now can be traced back to the big-bang. This is to say that I don't see the usefulness of this 'tracing back' thing of yours. --Aïki 15:25, 21 January 2006 (UTC)


 * You must be joking. The pivot exerts a centripetal force on the ball, and the tension of the wire mediates that force. The wire resists elongation because of molecular bonds in the material. Those molecular bonds can be traced back to electromagnetism. --Cleonis | Talk 15:37, 21 January 2006 (UTC)


 * The centre toward which the centripetal force is directed is not situated in the top of the pivot where the rope is attached, but inside of the pivot, and in the plane of rotation of the ball.

The centripetal force acting on the ball, is not directed along the rope but along the radius of the circular path followed by the ball.

This centripetal force is a component of the tension in the rope, and thus we can say it comes from the rope, but it is not equal in magnitude to this tension, nor in the same direction, it is only a part of it, a component.

The other part i.e. the other component of the tension force, is situated vertically, and is opposed to the gravitational force, which is usually represented by 'mg', and it is equal in magnitude to 'mg' when the ball stay at he same height over the ground. --Aïki 16:36, 21 January 2006 (UTC)

Theories organize our perceptions
I copy and paste from above:
 * Any theory is a view of the mind. --Aïki 18:05, 16 January 2006 (UTC)

Your remark is a philosophical one, and an interesting one (well, I think it is interesting, and worth discussing.)

There have been philosophers, even philosophers of science, who believed that the laws of nature as they are used by scientists are deduced from observations.

It has been shown that the scientific process is more a process of inferring laws. In breaking new ground in physics, theoretical physicists make educated guesses at laws of nature. If such a conjectured law of nature holds good time and again, then the perception of that property of nature shifts from seeing it as a conjecture to more and more seeing it as a law.

All theories unavoidably contain elements that cannot be directly observed. For example, we cannot directly percieve the presence of a magnetic field. Charged particles behave differently when what we call a 'magnetic field' is present, that is the observable. The reason we take it for granted that such an entity as a 'magnetic field' actually exists is that it provides an efficient way to organize the body of knowledge about magnetism.

Any theory contains unobservable entities, that serve to organize the perception of what is taking place. As a rule of thumb, it is general practice to not multiply entities beyond necessity. --Cleonis | Talk 16:10, 21 January 2006 (UTC)


 * Good comment!
 * To see that a theory is just a view of the mind, we don't have to go very far. Just seeing the historical part of the article 'Inertia' and its talk-page, and all those theories which came up along the centuries, is sufficient to convince any one. --Aïki 16:52, 21 January 2006 (UTC)


 * In newtonian mechanics, the importance of inertia tends to be overlooked, because of its omnipresence. Inertia is an example of a physical entity that cannot be directly observed, but we take its existence as a certainty, because the concept is so efficient in organizing our perceptions.
 * So a list of the fundamental operative factors of physics looks like this: Inertia, Gravitation, electromagnetism, weak nuclear, strong nuclear.
 * In the general theory of relativity, the way the perceptions are organized is different, so the entities are different then. In general relativity, the description of gravitation and the description of inertia are unified in the properties and behavior of a single entity.
 * I like general relativity better: one less entity.--Cleonis | Talk 17:43, 21 January 2006 (UTC)

Archiving
I recently archived a lot of this page, without bias or reason. It was reverted. I'm sorry, but I fail to understand what the problem with archiving conversations older than 7 days is. Specific discussions can still be referenced, or be revived, while older discussions, which may have outlived their usefulness, are allowed to rest.

Would someone else like to try? -- Ec5618 02:29, 22 January 2006 (UTC)


 * It appeared that the archiving had been done carelessly. Very old discussion (1 year old, completely outdated) remained on the talk page, interspersed with less than one day old text. I agree that archiving conversatios older than 7 days is badly necessary for this Talk page. --Cleonis | Talk 07:25, 22 January 2006 (UTC)


 * Carelessly or not, any discussion still relevant could have easily been recovered. This reversion was pointless. All I did was start at the top, and work my way down, moving any discussion that had comments less than 7 days old to the archive.
 * Technically, all content could be moved to an archive page (it's not unheard of). Actually relevant discussions can then be revived (begun anew) by interested parties. It cleans up the page, while preserving the timeline. -- Ec5618 14:06, 22 January 2006 (UTC)
 * There are two main methods for archiving a talk page, detailed below. Regardless of which method you choose, you should leave current, ongoing discussions on the existing talk page. How to archive a talk page --Aïki 14:26, 22 January 2006 (UTC)

I have tried the second method of archiving a talk page (with the section « # 2 "Centrifugal force" is not a force»), found in How to archive a talk page. If you don't like it, just revert it. --Aïki 14:22, 22 January 2006 (UTC)


 * Hai Aiki, I think I favor the first method of archiving a talk page. How_to_archive_a_talk_page I propose that the existing Talk:centrifugal_force/Archive_2 page is used, and that all old discussions (older than 7 days) are moved there with a 'cut 'n paste' operation. The previous move unfortunately left some old discussion, making it unclear what had happened --Cleonis | Talk 10:13, 23 January 2006 (UTC)
 * I prefer the second method, but if the majority choose the first one, I am not opposed. For the 7 days, it seems to me very short. Not every contributor comes each day to make edits: we have to let them some time to contribute. I suggest one month. And keep the ongoing discussions in the active talk page. --Aïki 13:05, 24 January 2006 (UTC)
 * Month? Most Talk pages aren't archived more than once a month, so that wouldn't be a problem unless every discussion was archived the second it was a week old. Anyone interested can always reopen an archived discussion.
 * And I too prefer the first method of Archiving. -- Ec5618 13:24, 24 January 2006 (UTC)

Reasons why it is so
The definition of the centrifugal force which appears in the first sentence of the article:

In Newtonian physics it is the inertial reaction force to the centripetal force that induces curved motion.

is still very problematic for the following reasons:

1. The structure of the sentence is problematic because it's not clear whether the words "that induces curved motion" refer to the centrifugal force or the centripetal force.


 * Thanks, that may need improvement Harald88 19:40, 26 January 2006 (UTC)
 * I agree. It must be improved. --Aïki 15:54, 29 January 2006 (UTC)

2. I'm a physicist and I don't know what you mean by attributing to the force the description "inertial". Maybe wikilink to inertial force instead of to reaction.


 * I'm a physicist too; without inertia no reaction force can happen. Do you think that it may help to leave that obvious fact out? And thanks for the suggestion about linking, I'll look into that. Harald88 19:40, 26 January 2006 (UTC)
 * Reaction force goes with action force, not with inertia. Secondly, the concept inertia and its definition, right now in wikipedia, is a very litigious one as you can see in the talk page of the article Inertia. --Aïki 15:54, 29 January 2006 (UTC)
 * I see that the link to inertial force starts with the non-Newtonian definition; thus, linking to that would cause confusion. Also, you state that you don't know what is meant there with "inertial'; and the Newtonian definitions we have, don't explicitly use that term. Thus I will, at least for now, scrap it. Harald88 23:58, 26 January 2006 (UTC)

3. I think it is not a good Wiki style to include an external reference in the main part of the definition.


 * I agree, it was a left-over from a discussion on this Talk page. Harald88 19:40, 26 January 2006 (UTC)
 * I don't agree. It is much better than an internal one. It is in the essence of wikipedia to have external references, and it is a good wiki style. --Aïki 15:54, 29 January 2006 (UTC)


 * See --Aïki 02:24, 31 January 2006 (UTC)

4. Current use of the expression "centrifugal force" in science and in textbooks is according to the second definition (fictitious force). IMO the first definition (cited above), which seems to be known almost exclusively to historians of science, should be moved to a separate section.


 * I learned the first definition from casual use (centrifuge etc.) as well as by extension from the definition of centrifugal acceleration. About the second definition: I díd not learn it at all, likely because it's incompatible with Newtonian mechanics, which I did learn. And I'm not very old, I assure you!
 * Note that it's according to Wikipedia rules to put all notable word meanings in the lead, and to make a disambiguation when the meanings are conflicting. Especially in such cases as this one where in literature heated discussions can be found because people use different definitons without knowing it, such a disambiguation is essential and avoids edit wars as well as confusion or disagreement by readers. Harald88 19:40, 26 January 2006 (UTC)
 * I agree that the current use of the expression "centrifugal force" in science and in textbooks is according to the definition of a 'fictitious force' (which concept of 'fictitious' is the subject of a litigious debate in the talk page of the article Centrifugal force). --Aïki 15:54, 29 January 2006 (UTC)

Please express your opinion. Yevgeny Kats 18:38, 26 January 2006 (UTC)

No match with Newton
The reference provided with the first definition does not contain at all the expression inertial reaction force, nor that this inertial reaction force induces curved motion.

But in 'PROPOSITION II. THEOREM II.' we find this: ''Every body that moves in any curve line ... is urged by a centripetal force directed to that point.''

This being from Newton hinself, and the definition in the article being said to be from Newton, the conclusion is that the definition does not match with its corresponding reference, and it is not from Newton. Therefore it should be removed to conform with the NPOV official politics of wikipedia, unless we find a correct reference supporting this definition. --Aïki 23:15, 26 January 2006 (UTC)


 * The definition of centripetal force is correct. But centripetal force is not centrifugal force (is it necessary to explain that?!). In the same proposition we also find a description of centrifugal force, and that's what this article is about, and it's cited together with the corresponding encyclopdia definition higher up on this page. In short here again:


 * Newton: "This is the centrifugal force, with which the body impels the circle; and to which the contrary force, wherewith the circle continually repels the body towards the centre, is equal."


 * Columbia Electronic Encyclopedia: Centripetal force and centrifugal force: action-reaction force pair associated with circular motion.


 * PS at Yevgeny's suggestion I already removed the superfluous precision that the centripetal force induces curved motion...
 * Harald88 23:49, 26 January 2006 (UTC)


 * 1- Good thing that you remove it!
 * 2- Action and reaction are two forces acting on two different objects while centripetal and centrifugal forces (as understanded in the sense that they are utilised usually and generally in most books and here in the talk page, which is a specialised utilisation) are acting on the same object. Thus they cannot be an action-reaction force pair. They are opposite forces. This is very different.
 * In the Newton's quotation that you cite just above, the body does not move in a circle but on many straight lines along the sides of a polygon.
 * When Newton writes that the body impels the circle and that the circle continually repels the body, the action force acts on the circle, and the reaction force acts on the body, and if he calls the reaction force a centrifugal force it is only because it is directed toward the center. (It is the basic meaning of it, and not the specialised one.) And it is for a body moving in a straight line and colliding with the circle, not for a body circulating in a circle and not colliding with whatsoever else. In this last case the two forces does not act on the body because there is only one force acting on the body in this situation, and it is the centripetal force.


 * Those are two different cases. --Aïki 00:43, 27 January 2006 (UTC)


 * In classical (Newtonian) mechanics, centripetal forces and accelerations point toward the centre, and centrifugal forces and accelerations point away from the centre. That's indeed all there is to it -- just as basic as the meanings of "left" and "right".
 * Action-reaction forces are the forces of two objects working on each other, by definition and in all textbooks that I know. If that's not well explainend in the article, then that must be corrected.
 * Harald88 19:44, 27 January 2006 (UTC)


 * I agree. It seems to me that what you says in the first paragraph, is the same thing that what it is said in this sentence: 'More simply, and more generally, centrifugal force is any force which, instead of being oriented toward the center of a curved path, is oriented in the opposite direction.' Do you agree?


 * I think it is too the same thing that was being said in
 * 'The name centripetal or centrifugal, depends only on the direction: if outward, it is centrifugal; if inward, it is centripetal.'
 * --Aïki 20:17, 27 January 2006 (UTC)


 * Yes that looks fine to me -- the only problem I have with all this: where is the problem?! ... Harald88 20:39, 27 January 2006 (UTC)


 * One of the problem has been stated in the first paragraph of this discussion, but it is now partially solved.


 * Another problem is that a centrifugal force and a centripetal force, which act on the same object in a rotational frame of reference, and are not both present in an inertial frame of reference, cannot therefore be qualified of a pair of action-reaction force, which doesn't act on the same object. This lead to the problem of the actual definition in the article, which definition speaks of a reaction force, and to the definition of the 'Columbia Electronic Encyclopedia' which also use the action-reaction concept.


 * I haven't got a clue what you are talking about, I know of no such thing! Such claims can be found in which publication, and on which page? Without proper drawings such arguments are hard to understand... Harald88 08:36, 28 January 2006 (UTC)
 * I have already give two references in this same talk page in the last edit at the end of.

It would be a good idea too, to read the article Fictitious force and its talk page, which are talking much of the centrifugal force. --Aïki 14:49, 29 January 2006 (UTC)

Problem with sentence

 * Another problem is, now, with this sentence: More simply, and more generally, centrifugal force is any force which, instead of being oriented toward the center of a curved path, is oriented in the opposite direction., with wich you now agree, and which sentence has already been put in the article and was rejected by 100% of the others contributors, and has then be removed from the article. This lead to the problem of correcting what is in the article, which cannot be done whitout a certain consensus, thus all the discussions we have here on the talk page.
 * --Aïki 00:11, 28 January 2006 (UTC)


 * As I recall it, it was I who removed that from the lead because it was disputed and unsourced, and it doesn't fit well there anyway. That didn't mean at all that 100% of the editors thought that it was really wrong! If nowadays "left" isn't in general the opposite of "right" anymore, things are really getting messy... Harald88 08:36, 28 January 2006 (UTC)
 * First, I did not said 'that 100% of the editors thought that it was really wrong', I just said that it was rejected by 100% of them.
 * Secondly, does concepts such as left and right needed to be sourced? (In fact, I gave sources, but it was indirect ones, because no physicist says that 'the definition is ... so and so', for the simple raison that the meaning is obvious for them, beginning with Newton.)
 * Third, now that we both agree on that sentence, do you agree to put it back in the article, in one manner or the other? --Aïki 15:13, 29 January 2006 (UTC)

Definiton in the lead
Now it's clearer I think; but following the above discussion, the definiton in the lead looks too restrictive to me. There is no doubt that the term is mostly understood in the two ways that are indicated; but IMO it has in reality only one meaning (as Aïki suggested), but with two specific applications. Do we really have no source that states that centrifugal force in English means what it already means in latin?! Anyway, I'll adjust the intro in order that it isn't pertinently against the most obvious general meaning. Harald88 09:10, 28 January 2006 (UTC)

I now also included the Columbia Encyc. in the ref., as it describes both usus in a nutshell. Harald88 09:50, 28 January 2006 (UTC)


 * This reference being too an encyclopedia, they should, them too, have to give references: do you know which references they gave for their article? If so, did you go to verify them? --Aïki 15:24, 29 January 2006 (UTC)

which textbooks use pseudo forces?
According to some editors, the term centrifugal force is used in science to indicate a pseudo force. However, none of the textbooks I know of, uses any pseudo force or pseudo acceleration, instead they all stick to Newtonian mechanics. Who knows textbooks that teach pseudo-Newtonian mechanics? Harald88 00:16, 27 January 2006 (UTC)

Potential energy in the case of rotating mass
I have replaced the section 'potential energy of the centrifugal force'. What is referred to as 'centrifugal force' is not a force, and this shows in the fact that there isn't an associated potential energy.

The potential energy that is associated with rotating motion of a body of mass is associated with the centripetal force.

In physics education, it is unhelpful to present a concept of 'potential energy of the centrifugal force', because that imagery does not correspond to anything physical. --Cleonis | Talk 15:00, 29 January 2006 (UTC)


 * Dear Cleonis, I reverted your replacement. If you don't know how to use the concept of centrifugal force to your advantage, or if you think that it is not a useful concept for whatever other reason, you are welcome not to contribute to this article. (And if you really have something to say about the potential energy of the centripetal force, please do it in that article, not here.) Yevgeny Kats 17:50, 29 January 2006 (UTC)


 * Cleonis, my more particular reasons for reverting your replacement are as follows:
 * 1. The section you erased explained why the heavier substances go farther away from the axis of rotation. Your alternative treatment did not explain even this simple fact.
 * 2. The section you erased described other uses of the potential energy of the centrifugal force. You simply erased them.
 * 3. Your content is out-of-topic here. The appropriate article would be either centripetal force or centrifuge.
 * 4. Your description is flawed. I will discuss exactly why, if you suggest it to an appropriate article on its talk page.
 * Yevgeny Kats 18:28, 29 January 2006 (UTC)


 * I have created an new article, discussion about these matters can be held at the Talk page of that article. The article is about the shape of a rotating fluid: the shape of a Mercury mirror. The physics explanation that I have provided in that article is correct. --Cleonis | Talk 21:03, 29 January 2006 (UTC)


 * I have to retract some of the things I wrote earlier. Contrary to what I remarked before the process of centrifugation can quite meaningfully be interpreted as setting up a potential associated with pulling G's in the centrifuge. So I retreat, and only reject the concept of centrifugal force (and any associated potential) in for example celestial mechanics --Cleonis | Talk 22:37, 29 January 2006 (UTC)


 * I think your retreat was hastily, see below. Harald88 21:45, 4 February 2006 (UTC)

Pivot-ball example
Usually, when the pivot-ball example is utilized, the rope is not perpendicular to the pivot but oblique. Therefore the tension in the rope can be splitted into two component forces, one that is vertical and opposed to mg, and one that is horizontal, this last one being the centripetal force.

In a rotational reference frame, the horizontal component is in opposition with the centrifugal force. In an inertial reference frame, it is not, simply because there is no centrifugal force in that frame. --24.202.163.194 05:26, 5 January 2006 (UTC) --->new name: --Aïki 04:09, 7 January 2006 (UTC)

Component forces
In the inertial reference frame, as well as in the rotational reference frame, we must split the tension in the rope into two component forces to explain the behavior of the ball.

One component is horizontal and directed toward the pivot where the center of the plane of rotation is situated, and which component constitute the centripetal force acting on the ball.

The second component is vertical and directed upward, which component constitute the opposite force to the gravitational force (mg).

These two vertical forces explain why the ball doesn't move upward or downward i.e. stay at the same height.

Remark: As the two component forces replace the tension in the rope in the new scheme of forces acting on the ball, so we can say that these two forces are not real, at least as real as the rope, or the pivot and the ball, and thus are fictitious! The centripetal force, being the horizontal component, can therefore be called a fictitious force. --Aïki 05:27, 17 January 2006 (UTC)

View of the mind
To split the tension in the rope into two component forces, is like to replace this force, the tension in the rope, by two other different forces. To replace, means that the tension in the rope is no more there, and that in its place we have two forces. If the tension in the rope would still be there, we would have then three forces, which would not be a replacement but an addition.

As these two forces are not materialised by any body, on the contrary of the tension which is materialised by the rope, so we could attribute to these two forces the name 'fictitious'. The same thing can be said of a resultant force replacing two forces seen as components.

In using a system of forces, with its net force, and resultant, component, and equilibrant forces, to analyse and explain the behavior of an object, it is not necessary that each of these forces must represent a material thing.

To use a system of forces in the way exposed above, is simply to adopt a certain view of the mind on the situation under study. --Aïki 00:59, 21 January 2006 (UTC)


 * Aiki, for sure your last edit is in error:

In Newton's physics, the ball moving without rope is its natural motion. Connecting the rope makes first of all the rope pull on the ball -> thus that is the action force.
 * Harald88 23:38, 29 January 2006 (UTC)


 * also, I now understand that you confused the Newtonian description with the pseudo description. I changed it, combining with your comments. Maybe it's clear now. Harald88 23:55, 29 January 2006 (UTC)

Where is the center?
To have a centripetal or a centrifugal force, we must first have a center! But a center of what? It could be any curved path, even a partial one, and not necessarily a circle, or any polygon, even a partial one. If we take the case of a body rotating in a circle, the center will be the center of the plane of rotation of this body.

If we take the pivot-ball example, and if the ball is rotating, say, at 5 feet over the ground, while the top of the pivot is, say, at 10 feet above the ground, the plane of rotation will be at 5 feet over the ground and its center will be at 5 feet over the ground, and situated inside the pivot at 5 feet over the ground. Nowhere else, meaning that this center will not be at the top of the pivot. This is what we can called the 'oblique case'.

The top of the pivot will be the center, only when the plane of rotation will be at 10 feet over the ground i.e. when the rope and the ball will rotate horizontally, with the plane of rotation parallel to the ground. This is what we can called the 'horizontal case'.

Does everybody see it the same way?

--Aïki 00:19, 30 January 2006 (UTC)

Application of action-reaction
In the pivot-ball example, there is 3 places where we can apply the Newton's third law: 1- between the ball and the rope; 2- betweeen the rope and the top of the pivot (where the rope is attached); 3- between the base of the pivot and the ground.

We cannot apply it between the ball and the top of the pivot because they are not in direct contact with each other.

Who agrees and who does not? --Aïki 03:41, 30 January 2006 (UTC)

Cleaning required
The article is very messy:

1. While the article starts by presenting two different definitions, later the discussion is mixed and in some places it's difficult to understand which definition it describes. I would suggest to make a clear and elegant distinction between the discussions of the two meanings.

2. Probably due to multiple edits, there are many problems in the text: paragraphs which are not connected one to the other, repetitions, reference to unexisting drawing, etc.

Could anyone (who understands both definitions) make some order? (I can do it, if you agree. I don't promise to keep every vague sentence though :))

Yevgeny Kats 02:36, 3 February 2006 (UTC)


 * I have tried. I'm still unclear about the section on statics, though. – Smyth\<sup style="color:gray;">talk 12:11, 4 February 2006 (UTC)


 * Someone messed up a little with the cleaning: the essential explanation that rotating frames are used both in Newtonian approaches and fictitious approaches has been partly deleted, so that the intro of the article again wrongly suggests that one observes fictituous forces when using a rotating frame, while that's only the case if one ignores the rotation.

Harald88 13:26, 4 February 2006 (UTC)


 * I think it is unpractical to use a distinction between 'newtonian approaches' and 'fictitious approaches'.
 * The distinction is whether newtonian dynamics is applied in the context of a rotating coordinate system, or in the context of an inertial coordinate system.
 * If the laws of newtonian dynamics are applied in the context of a rotating coordinate system, then coordinate acceleration must be taken into account. The coordinate acceleration is taken into account by adding the centrifugal term and the coriolis term to the equation of motion. This is commonly referred to as: "Adding centrifugal force and coriolis force".
 * I want to emphasize that in using the transformed equation of motion (transformed to hold good in a rotating coordinate system) is still straightforward newtonian dynamics. --Cleonis | Talk 13:39, 4 February 2006 (UTC)


 * To the contrary: it's necessary as that confusion is the primary cause of all disagreements and disputes, as witnessed from the discussions on this Talk page as well as the edits of the article. What you here "want to emphasize" is now obscured in the article, while it was clearly pointed out. Harald88 14:37, 4 February 2006 (UTC)


 * The article's explanation is exactly what Cleonis just said. What do you suggest is added? – Smyth\<sup style="color:gray;">talk 15:09, 4 February 2006 (UTC)


 * I will reinsert some lost disambiguations; even the classical standard mechanics approach to moving frames has been omitted! Harald88 17:30, 4 February 2006 (UTC)


 * What? You've just reinserted a copy of that section as it was before I revised it. We now have two virtually identical sections, one marked "reactive" and one marked "fictitious". Explain please. – Smyth\<sup style="color:gray;">talk 22:07, 4 February 2006 (UTC)


 * You removed the classical physics instead of the fictive physics which can't exist without it, so that what remained was the misleading pretention that one needs fictituous forces in order to deal with rotating frames. Not so, and in fact one can't calculate those fictive forces without Newtonian physics in the first place. What can be done here, is to delete the duplicate derivation section - which is what I had done, but next it was turned on its head. As a result, it may be that for clarity, both must be there. Please tell me if it's clear enough when we take away the second acceleration derivation, which doesn't really belong there and which is superfluous.

That chapter may become simply as below. Harald88 23:25, 4 February 2006 (UTC)

Fictitious centrifugal force
An alternative way of dealing with a rotating frame of reference (which is not an inertial reference frame), is to make Newton's laws of motion artificially valid in it by adding fictitious forces.

The centrifugal acceleration as maesured in that frame is then not atributed to rotation, but instead attributed to the fictitious centrifugal force which is given by:




 * $$F_{centrifugal} \,$$
 * $$= m a_{centrifugal} \,$$
 * $$=m \omega^2 r_\perp \,$$
 * }
 * $$=m \omega^2 r_\perp \,$$
 * }
 * }

where $$m\,$$ is the mass of the object.

The centrifugal force is a sufficient correction to Newton's second law only if the body is stationary in the rotating frame. For bodies that move with respect to the rotating frame it must be supplemented with the Coriolis force. For example, a body that is stationary relative to the non-rotating frame, will be rotating when viewed from the rotating frame. The centripetal force of $$-m \omega^2 r_\perp$$ required to account for this rotation is the sum of the centrifugal force ($$m \omega^2 r_\perp$$) and the Coriolis force ($$-2m \omega \times v = -2m \omega^2 r_\perp$$). (This example also shows the difference from the reactive centrifugal force, which is zero in this case.)

Alternatively, from my Talk page:

The expression for the centripetal acceleration is derived in the article "Centripetal force". Why do you need to rederive it in section "Reactive centrifugal force" of the article "Centrifugal force"? Yevgeny Kats 23:28, 4 February 2006 (UTC) No, I did not derive it, it was there all the time, similarly underived. Thus I added no derivation, but for the classical physics part I used the standard notation of my textbook. Harald88 23:35, 4 February 2006 (UTC) The "reactive centrifugal force" is simple. State the formula for the centripetal force (derived in centripetal force) and reverse the sign according to Newton's third law. No other formulas are needed in that section. Yevgeny Kats 23:41, 4 February 2006 (UTC)


 * Indeed, we can also rewrite it like that. I'll do an attempt right now. Harald88 00:45, 5 February 2006 (UTC)

That's much better, thanks. – Smyth\<sup style="color:gray;">talk 11:42, 5 February 2006 (UTC)

sign convention problem
With my rewrite of the classical mechanics part I used the definition of Alsonso&Finn, according to which the centripetal acceleration = omega x omega x r. But someone else defines that furtheron in the article as -omega x omega x r. Comments please! Harald88 21:43, 4 February 2006 (UTC)

It's not a matter of convention. It's just that the expressions for the centripetal and the centrifugal accelerations have opposite signs. Yevgeny Kats 22:20, 4 February 2006 (UTC)


 * The problem is that according to my book and thus my text, the definition is as above; while it's half the time opposite according to another author in the next chapter. I think my notation is consistent, and the other's not; but also I can make mistakes of course. To be checked! Harald88 23:08, 4 February 2006 (UTC)

[moved from my Talk page, Harald88 02:21, 5 February 2006 (UTC) :] All our equations are vectorial (they involve cross products etc). Since $$r_\perp$$ goes from the axis of rotation to the object, an expression which is proportional to $$r_\perp$$ points out of the axis of rotation if the sign in front of it is plus, and towards the axis of rotation if the sign is minus. There are no issues of convention here (unless your book defines $$r_\perp$$ differently, which is not a usual thing to do). Yevgeny Kats 02:04, 5 February 2006 (UTC)


 * Alonso&Finn (see refs) has the sign as explained above: centripetal = +
 * Thus as I thought: there is difference in conventions. Harald88 02:21, 5 February 2006 (UTC)


 * Maybe their equation is not vectorial and they explain the meaning of positive/negative separately in the text. In the article the equation is vectorial, so the sign is unambiguous. Yevgeny Kats 02:33, 5 February 2006 (UTC)


 * Not so: their equations are vectorial throughout. Centripetal acceleration omega x omega x r is positive, using the cork-screw rule for defining omega. Now the equations of the article are according to their sign convention until half-way... Which book do you use? And what are the signs of omega and r according to that book? Harald88 21:36, 5 February 2006 (UTC)


 * Of course $$\omega \times (\omega \times r)$$ for the centripetal acceleration comes with a plus sign, but $$\omega \times (\omega \times r) = - \omega^2 r_\perp$$. The minus arises in the transition from one expression to the other. Yevgeny Kats 21:51, 5 February 2006 (UTC)


 * OK, I see: for consistency of sign, my book uses r*e_N instead. And yours is indeed inverse. Then only one sign error remains, in one of the two F_centrifugal. I'll deal with that now. Harald88 22:12, 5 February 2006 (UTC)

I added a derivation of the centrifugal "force" that kinda explains some of the sign confusion. Everything turns out nice and peachy if you use $$\vec A \times \vec B = -\vec B \times \vec A$$ to turn $$-\vec \omega \times \left ( \vec \omega \times \vec r \right )$$ into $$\vec \omega \times \left ( \vec r \times \vec \omega \right )$$. And if $$\vec r$$ is pointing away from you to the front, and $$\vec \omega$$ is pointing straight up, $$\vec r \times \vec \omega$$ will point to the right, which when you finish the cross products will make the whole thing point outward, as things should be. Ssiruuk25 02:44, 18 February 2006 (UTC)

claim of "usefulness" of centrifugal force's "potential energy"
The article has a paragraph in which the kinetic energy is presented as potential energy, thanks to the inversion of cause and effect. See also the comments above by Cleonis. It suggests that this "is useful, for example, in calculating the form of the water surface in a rotating bucket", etc.

However, it's just a matter of approach: no fictious force is needed at all, and I doubt that the calculations are simplified either (it's easily shown that the hydrostatic pressure results in delta_potential energy = delta_kinetic energy). Should we perhaps present the same calculation without using a fictituous force? Harald88 21:56, 4 February 2006 (UTC)

Calculation without a fictitious force would not be relevant to this article. (By the way, an attempt to do such a calculation is presented in Mercury mirror and it doesn't make much sense to me.) Yevgeny Kats 22:25, 4 February 2006 (UTC)


 * Ok, I don't mind; as long as no false impression is given that it's better to use such (or, alternatively, such should be demonstrated or cited). Thanks, Harald88 23:04, 4 February 2006 (UTC)


 * Example of a problem that is much easier to solve using a centrifugal potential: A heavy object moves frictionlessly along a smoothly curving track on a flat horizontal carousel that rotates with constant speed. At t=0 the object is released from rest with respect to the carousel, at point A in the track. It gets pushed by the track and starts to move along it in a complicated manner. Some unknown time later the object passes point B in the track for the first time. What is its speed then? If you insist on doing things in the inertial frame you have to solve a equation of motion which can get arbitrarily complex according to the shape of the track. In a co-rotating system, however, you can just apply conservation of energy: The contact forces between the track and the object are always normal to the motion (only in the corotating system!) and so do no work. Thus the kinetic energy of the object at point B is given directly by the difference in centrifugal potential between A and B. Henning Makholm 23:46, 18 April 2006 (UTC)


 * You reveal a misconception about how classical mechanics deals with such problems - evidently the article isn't clear enough yet, we should clarify more. But before I (or Cleon?) explain how this is done without fictituous forces, please clarify how the centrifugal potential is calculated in the above example. Harald88 06:47, 21 April 2006 (UTC)


 * Henning Makholm provided a great example. Maybe we could even include it in the article. Clarification to Harald88: the centrifugal potential energy is calculated as $$E_p = -\frac{1}{2} m \omega^2 r_\perp^2$$, where $$r_\perp$$ is the distance from the axis of rotation (exactly as explained in the article). A small correction to Henning Makholm: the gravitational potential energy difference between A and B should also be included (I'm sure he meant it). Yevgeny Kats 12:06, 21 April 2006 (UTC)


 * I explicitly specified that the carousel and track be horizontal such that gravity would not muddle the issue. As gravity can easily be handled in the rotating system, it would be unfair to have it add extra complexity to the non-rotating calculation. :-) Henning Makholm 20:29, 21 April 2006 (UTC)


 * OK it took me a little time to reflect on this; meanwhile, Cleon indicates below the classical mechanics approach. Now my opinion about this example is mixed:
 * 1. If I'm not mistaken, the (optimal) calculations are quite similar.
 * 2. The centrifugal force approach will probably be intuitively simpler for most people.
 * Thus, this example may indeed be a valuable addition to the article. Harald88 14:11, 23 April 2006 (UTC)

Calculation strategy: a conservative force-field
The calculation strategy that Henning Makholm proposes uses the theorem that if a force-field is conservative, then the amount of work that is done on an object is path-independent. By considering energy conversion, rather than velocity with respect to this or that coordinate system, the entire problem of choosing a coordinate system is avoided. The variable here the distance to the center of rotation; energy as a function of distance to the center of rotation. The distance to the center of rotation is a frame-independent variable, so the strategy proposed by Henning Makholm niftly avoids commitment to this or that coordinate system.

The key to the force-field strategy is the cause-effect inversion. As we all know, the actual physics is that an object that is forced along a trajectory that curves (any shape of curve) around some point will exert a reactive force in centrifugal direction. For the sake of efficient calculation, this cause-effect chain is inverted. It would be addition of unnecessary complexity to present the cause-effect inversion strategy in terms of "using a rotating coordinate system", because the very strength of the cause-effect inversion strategy is its frame-independence. --Cleonis | Talk 12:04, 23 April 2006 (UTC) Addendum: I credit Harald88 for pointing out that the cause-effect inversion explains both why the calculational strategy works so well, and why it goes against physics intuition. --Cleonis | Talk 12:18, 23 April 2006 (UTC)


 * That makes no sense. The potential is only valid when we are working in the co-rotating system. In any other frame of reference, (a) the potential is meaningless; it has nothing to do with the kinetic energy measured in that frame, and (b) even if one found a meaningful potential, it would be useful in itself, because the contact forces between track and object do work in the non-corotating frame. This work would have to be calculated and added to the potential difference. Henning Makholm 20:39, 23 April 2006 (UTC)

Personally, I don't use the metaphors such as 'in a co-rotating frame' and 'in an inertial frame' because it doesn't fit in with the common practice of physics thinking. There is no such thing as measuring whether you are 'in a co-rotating frame' or whether you are 'in an inertial frame'; the idea of either being 'in a rotating frame' or 'in an inertial frame' is merely psychological, it's not physics.

I follow the physics tradition of teaching physics in such a way that independence of how things are mapped in this or that coordinate system is emphasized. Example: when two objects collide perfectly non-elastically, then the amount of kinetic energy that is converted to heat is independent of the choice of coordinate system.

I argue that that the metaphors 'in a rotating frame' and 'in an inertial frame' carry no weight in arguing the validity of a calculational strategy. I do think it is valid to employ the cause-effect inversion strategy in calculating the carrousel problem that Henning Makholm described. Here is how I argue that it is a valid calculation strategy. (This was taught to me by Jeepien)

In teaching first year physics, Jeepien asks the heaviest, strongest guy in the class to come forward, and exert a strong force on Jeepien (pushing Jeepien backwards). First Jeepien does not brace himself at all. That makes it impossible for the strong guy to exert a strong force on Jeepien. Then Jeepien braces himself, and suddenly the strong guy can exert a strong force on him. Thus Jeepien demonstrates that the amount of work that can be done on a body is proportional to how much force is required to bring that body in motion.

That is where inertia comes in. The larger the inertia of an object, the larger the amount of work that can be done on it. In the carrousel example described by Henning Makholm: the object is constrained to follow a track that is fixed onto the carrousel. At any point in time, the amount of force that the carrousel can exert on the object is equal to the amount of reactive force that the object exerts on the carrousel. That is the physical explanation (a frame-independent explanation) of why the centrifugal force-field potential method is guaranteed to produce the correct outcome. --Cleonis | Talk 09:34, 24 April 2006 (UTC)


 * Thanks for the explanation, I must say that my insight has increased with this example.
 * If we try to say what Cleon explained in simpler terms, would it be correct to state in general that contact forces don't change the relative kinetic energy between the bodies?
 * Taking that rule, we can immediately understand that the gain in relative kinetic energy when moving from point A to point B must be equal to the difference in kinetic energy of points A and B, as determined in the inertial frame - right? Harald88 20:49, 24 April 2006 (UTC)

Reprise: the original problem
Henning Makholm wrote: "* Example of a problem that is much easier to solve using a centrifugal potential: A heavy object moves frictionlessly along a smoothly curving track on a flat horizontal carousel that rotates with constant speed. At t=0 the object is released from rest with respect to the carousel, at point A in the track. It gets pushed by the track and starts to move along it in a complicated manner. Some unknown time later the object passes point B in the track for the first time. What is its speed then? [...] Henning Makholm 23:46, 18 April 2006 (UTC)"

Henning Makholm asks for the speed. That would be relatively easy to answer if the problem is stated as follows: distance B to the center of rotation is the outer edge of the carrousel, so after leaving the track (at point B) the object continues to move inertially. Assuming that the initial velocity with respect to inertial space was zero or negligable, then the total kinetic energy of the object follows from "work being done by the centrifugal force-field". What you cannot calculate then is the direction of the velocity, because you only know the gain of energy and that has no directional information.

In the version of the problem given by Henning Makholm the initial condition is that at distance A to the center of rotation the object is stationary with respect to the carrousel. The angular velocity of the carrousel is constant. I think that gives insufficient information to calculate the velocity of the object with respect to the carroussel at point B. The problem, I think, is lack of directional information. --Cleonis | Talk 15:13, 23 April 2006 (UTC)


 * I carefully asked for speed rather than velocity, and specified that the point in time of interest is the first time that point B is passed. Thus, the direction is the direction from A to B, not vice versa. The problem is only underdetermined if (1) the track is a closed loop, and (2) point A has minimal distance to the center of rotation among all points on the track. In that case the calculation in the rotation system still easily gives the two possible answers (which are both possible, depending on how the initial unstable equilibrium at A happens to break). Henning Makholm 20:35, 23 April 2006 (UTC)


 * As I already thought and as mentioned above, the optimal calculations are indeed similar. I can now explain better the non-fictitious approach. In a nutshell: We can calculate the kinetic energy gain relative to point B for the simplest case of only the centrifugal coordinate acceleration from r1 to r2 (thus, along r); and as anywhere on the circle path the kinetic energy is the same, that result is valid for any location of point B on r2, and independent of the path. Still, for many people the fictitious force approach may indeed be more intuitive.
 * And now, as this subject comes back several times per year, let's add this comparitive example! Harald88 17:11, 29 April 2006 (UTC)

strange remark about weather forecasting
Someone added: "weather forecasting is made considerably simpler by always remaining within the rotating frame of the Earth, and taking the fictitious Coriolis effect as given."

However, the Coriolis effect is not fictitious and also not a result of the assumption that the earth doesn't rotate; it seems some editors still don't understand that in classical mechanics one simply maps a rotating frame to a physical inertial reference frame, without invoking anything fictitious.

Thus I see no other solution then to completely delete that spurious remark. Harald88 09:46, 5 March 2006 (UTC)


 * The Coriolis force is fictitious. Why don't you just change that word instead of deleting the whole sentence? – Smyth\<sup style="color:gray;">talk 11:24, 5 March 2006 (UTC)


 * Where I live, no "Coriolis force" is mentioned in weather forecasting, at least not on the news. I don't know what jargon is used by the weather specialists; and if they do speak of Coriolis force, then that would be appropriate for mention in an article "Coriolis force". In any case, the whole argument was mistaken, and I replaced it with a contrary remark. Harald88 17:41, 5 March 2006 (UTC)


 * I disagree with Harald's deletion and his remark, and I completely agree with Smyth. Somehow, in his education Harald hasn't been exposed to the concepts of the fictitious forces and so he is always trying to present them here as useless and/or anti-educational. However, my experience shows that the fictitious forces are taught at university level in many places (e.g., Harvard University) and they make life easier (e.g., the example with the potential energy given in the article). I suggest that Harald should take into consideration also the opinions and the background of fellow editors, similarly to how I accepted the existence of that weird alternative definition of the centrifugal force as a reaction force (I will be extremely happy to remove it, but it doesn't seem like we have a concensus about that here). I think the sentence about the usefulness of the fictitious forces in weather forecasting should be returned. Yevgeny Kats 19:12, 5 March 2006 (UTC)


 * You are completely right that no fictitious forces were taught in my physics education (and I'm surprised by your claim that at Harvard they do use it).
 * Thanks to that I immediately notice it when a fake argument is provided according to which such would be asked for. As for your claim that they "make life easier" - I fully disagree: your energy example is misleading if it is meant to suggest that such would be easier by assuming fictituous forces; by chance I did roughly the same calculation by myself a few months ago, without using anything fictituous. I appreciate other opinions, but not misinformation that even isn't properly sourced. Harald88 11:53, 6 March 2006 (UTC)


 * I also was formally taught it as well as part of my physics degree at Southampton University in the UK. I find your attitude quite frankly bizarre on this issue, and your actions and comments do not seem to be entirely reaching NPOV. If you don't wish to use non inertial frames for calculating things: don't. But the wikipedia's job is to accurately describe these kinds of mathematics for those sorts of situations where it is convenient to employ this (perfectly accurate and valid) approach.WolfKeeper 00:33, 13 November 2006 (UTC)
 * Centrifugal forces are useful for calculating the shape of The Space Elevator. In that example the thickness of the cable is determined by the forces on the cable and the rotating earth forms a natural rotating reference frame. I believe that they are also used for calculations on rotor systems on helicopters; and they are convenient for formally describing the shape of weather systems. We do actually live in a rotating reference frame. Of course many other approaches are equally valid; you can always derive anything from first principles, but I'm not clear that you would normally want to.WolfKeeper 00:33, 13 November 2006 (UTC)
 * See the helicopter reference in this article: indeed it uses centrifugal force --the real one
 * I'm willing to check out a comparision between the two calculation methods in the case of space elevators. Do they give that somewhere? Harald88 22:56, 13 November 2006 (UTC)
 * See section 3.1 where they calculate the optimal tapering: in the natural rotating reference frame. I've never seen a derivation in an inertial frame; although I think I can see how to do it, but I 'leave that as an exercise to the reader'. :-) WolfKeeper 01:05, 14 November 2006 (UTC)
 * That's really interesting: centrifugal force is indeed mentioned in the text. Equation 1 uses F=m(g-a), in which F= true net force and both g and a are relative to the Earth Centered Inertial frame. The "centrifugal force" mentioned used there is thus Newton's centrifugal force as in standard Newtonian mechanics. I don't see that paper use a rotating reference frame with pseudoforces, instead everything is mapped to the inertial frame in which the elevator is rotating around the origin. Harald88 01:04, 15 November 2006 (UTC)
 * Correction and precision: The author mostly uses the inertial frame description in the text but at that point his word choice makes it sound as a fictitious force balance corresponding to the equations F1=mg, F2=-ma, F=F1+F2. Nevertheless he uses the (real) Newtonian force equation F=m(g-a) which directly computes the weight from the acceleration in the inertial frame. Harald88 10:06, 18 November 2006 (UTC)

Some time ago I uploaded this animation to illustrate the concept of coordinate acceleration. The motions of the planets of the solar system can be mapped in a coordinate system that is co-moving with the center of mass of the Sun, or the motions of the planets can be mapped in a coordinate system that is co-moving with the Earth. As seen from te Earth there is an apparant retrogradation of Mars at certain times. Of course, nobody will claim that Mars is "pushed back" back by a force. The apparent retrogradation is an artifact of mapping the motion in a coordinate system that is co-moving with the Earth. A coordinate system that is co-moving with the Earth is a coordinate system that is accelerating with respect to the inertial coordinate system, so you get the artifact of coordinate acceleration. The animation on the right illustrates two things: frictionless motion over the surface of a dish with a parabolic shape, and coordinate transformation. On the left the motion as mapped in an inertial coordinate system, the right the motion as mapped in a co-rotating coordinate system. Coordinate transformation and physics taking place must be discussed separately: coordinate transformation is not physics, and physics is the science of tracing back what the physical cause of a (pattern of) motion is. In the animation, the pattern of motion is that the ice-puck, sliding frictionless, is in two oscillations: an oscillation in the distance to the overall center of rotation, and an oscillation in the angular velocity. The physical cause of the motion of the puck is the centripetal force that is exerted on the puck. It would be absurd to claim that coordinate transformation affects the physics taking place. --Cleonis | Talk 12:36, 6 March 2006 (UTC)

Fuller example

 * I thought a fuller example might be useful for this article. If anyone agrees, please feel free to use it in the article.

Consider a ball on the back seat of a car. As the car rounds a corner the ball begins to roll accross the seat to the other side of the car. To an observer outside the car with a birdseye view, who is in an inertial frame of reference, it is apparent that the ball is continuing in a straight line at a constant speed and thus no force * is being exerted on it. To an observer inside the car, in a rotating reference frame, it appears that the ball is accelerating towards the other side of the car due to 'centrifugal force'. Since it is not an actual force, this usage of the term 'centrifugal force' is regarded as a fictitious force.

When the ball reaches the wall of the car and the car continues around the corner, the wall exerts a centripetal force to accelerate the ball towards the inside of the curve. The ball exerts an equal and opposite force on the wall. This reactive force is sometimes described as the 'reactive centrifugal force' and is an actual force.

* We are considering an idealized model of the scenario and therefore can ignore irrelevant forces such as friction, gravity and drag.
 * 220.237.34.36 15:47, 15 March 2006 (UTC)

Recently added misconceptions
Jmd01007 replaced the existing discussion of misconceptions by a new discussion. I reverted this, because the new discussion is incorrect. For example, it says:


 * Question: A standing passenger is riding on the bus as the bus takes a left hand turn.  What happens to the passenger?
 * Typical Student Misconception: The passenger is flung to the right of the bus because of a centrifugal force.

This is not a misconception at all. In the reference frame rotating with the bus, it is the correct description. Yevgeny Kats 16:14, 9 August 2006 (UTC)


 * I agree that Jmd01007 should not delete text like that. However, about his/her additions: the example you highlight is a misconception. The centrifugal force in that example is fictituous (=not true). "Because" is erroneous: Fictitious forces can't be true causes! Newton painstakenly explained that the passenger is flung to the right of the bus because of centripetal acceleration of the bus.
 * If Jmd01007 proposes to include text about this sensitive subject, I propose that he/she adds it to this page first for discussion. Harald88 21:34, 9 August 2006 (UTC)


 * I agree with Harald. The principle of causality comes first.
 * In medieval times some scholars argued that the temporary phases of retrograde motion of for example the planet Mars had to be caused by a retrograde pushing force. Of course, there is no "retrograde pushing force" being exerted on Mars; the retrograde motion of Mars is apparent.
 * A statement like "the standing passengers in the bus are being flung outwards by a centrifugal force" is an example of pre-newtonian thinking. It is pre-newtonian in the sense that it disregards the principle of causality.
 * The purpose of physics is to identify the physical causes of the physics taking place. --Cleonis | Talk 11:19, 10 August 2006 (UTC)


 * Addendum: I just discovered that the common usage of the expression 'principle of causality' is to refer to a notion that 'the cause must precede the effect'.
 * My point should be clear though. A causative force complies with Newton's third law: a force is an interaction between two or more objects in space, each object causing a change in the other object's momentum (with conservation of the momentum of the common center of mass). The pre-newtonian concept of a centrifugal force as a causative thing is inconsistent with the newtonian approach. --Cleonis | Talk 11:52, 10 August 2006 (UTC)
 * Cleon, thanks for the precision. IMO the common use also applies, although in an indirect way: the cause of this phenomenon is the turning of the wheels of the bus. According to accepted physics, the primary effect of that is a change of direction of the bus (=centripetal acceleration); and not the magical creation of an obscure centrifugal force that makes the whole universe (except the bus) accelerate due to the turning of the wheels. Harald88 20:58, 10 August 2006 (UTC)

The rotating frame misconception
The expression "in a rotating frame" is physics shorthand for: "if you map the motion onto a coordinate system that is rotating with respect to inertial space."

From a physics point of view, there is no such thing as a distinction between 'motion in an inertial frame of reference' or 'motion in a rotating frame of reference'. Such a distinction is a difference in how a particular observer happens to be mapping the motion, which translates to a difference in how a particular observer happens to organize his perceptions. In terms of physics taking place, objects are in motion in inertial space.

Transition
For cutting edge science, Newtonian dynamics has been superseded by relativistic dynamics, but newtonian dynamics is not and will not be subsituted as the theory of choice for first introduction to physics. There is no need for the centrifugal force article to discuss relativistic physics. At the same time, it is necessary of course, to teach newtonian dynamics in such a way that a later transition to relativistic dynamics is prepared for as much as possible.

While there are many differences between newtonian dynamics and relativistic dynamics, one of the things they have in common is Newton's third law. Newton's third law is in equal measure an axiom of relativistic physics. This can be shown as follows. Light carries momentum, and the energy has inertial mass. Relativistic physics asserts that if an emitter emits equal quantities of light in opposite directions, the momentum of the emitter (center of mass of the system) is conserved: That is: relativistic physics extends the validity of the principle of conservation of momentum to emission of photons. That illustrates that relativistic physics is a deeper theory than newtonian dynamics

Focussing on compliance with Newton's third law is double action: it clarifies the newtonian approach, and it prepares the ground for a later transition to relativistic physics.

It is very important to address the misconception that "in a rotating frame there is a centrifugal force". That misconception is a serious impediment to physics understanding. --Cleonis | Talk 08:03, 12 August 2006 (UTC)

Black holes, ants, and roller coasters
I remember reading a Discover article about how centrifugal force acts in odd ways around black holes, how it can even reverse itself. Link to a copy of the relevant article if anyone is interested. I for myself can't figure out how to put it in there, though. --Zemylat 20:44, 13 August 2006 (UTC)

Terminology- fictitious vs pseudo
Doesn't fictitious imply that it's completely non existent, whereas 'pseudo' means fake but real?

I'm therefore proposing to edit fictitious to pseudo throughout the article. Anyone disagree?WolfKeeper 19:39, 14 August 2006 (UTC)


 * The article uses the term 'fictitious force' because it is the term that is usually used in this context.Yevgeny Kats 20:22, 14 August 2006 (UTC)


 * Really? WolfKeeper 21:36, 14 August 2006 (UTC)


 * Results 1 - 10 of about 101,000 for coriolis pseudo. (0.22 seconds)
 * Results 1 - 10 of about 25,700 for coriolis fictitious. (0.21 seconds)


 * It would seem that google disagrees with you.WolfKeeper 21:36, 14 August 2006 (UTC)


 * OK, it seems that my personal impression wasn't representative. I have no objection. Let's see what others think. Yevgeny Kats 22:18, 14 August 2006 (UTC)

A more appropriate search, omitting the thousands of Wikipedia copies: – Smyth\<sup style="color:gray;">talk 07:15, 15 August 2006 (UTC)
 * 811 for "centrifugal force" "fictitious force" -wikipedia
 * 1,160 for "centrifugal force" "pseudo force" -wikipedia
 * 14,400 for "centrifugal force" "inertial force" -wikipedia

Hmm. I'm not totally happy about the quoting, you're being too specific to a particular phrase I think. Try:

[http://www.google.co.uk/search?num=20&hl=en&safe=off&q=centrifugal+force+pseudo+-wikipedia&btnG=Search&meta= Results 1 - 20 of about 122,000 for centrifugal force pseudo -wikipedia. (0.11 seconds)]

[http://www.google.co.uk/search?num=20&hl=en&safe=off&q=centrifugal+force+fictitious+-wikipedia&btnG=Search&meta= Results 1 - 20 of about 57,800 for centrifugal force fictitious -wikipedia. (0.07 seconds)]

[http://www.google.co.uk/search?num=20&hl=en&safe=off&q=centrifugal+force+%22inertial+forces%22+-wikipedia&btnG=Search&meta= Results 1 - 20 of about 30,200 for centrifugal force "inertial forces" -wikipedia. (0.11 seconds) ] WolfKeeper 02:51, 27 August 2006 (UTC)

Come to think of it, do forces exist anyway?
I mean, energy seems to be much more fundamental. So what does it mean for a force to be real anyway? :-) WolfKeeper 20:25, 20 August 2006 (UTC)

And is a 'centrifugal acceleration' any more real if the object is inertial? From the point of view of a rotating reference frame it's accelerating... :-) WolfKeeper 20:25, 20 August 2006 (UTC)


 * A force is mediated by a field and that field has a quantumphysical counterpart. The mediator of electromagnetic interaction is the electromagnetic field. The quantisized description of the electromagnetic field involves photons. It is not clear whether it will be possible to formulate a quantum theory of gravitation, but if it is possible, the quantisized description will involve gravitons.


 * By contrast, it would be meaningless to consider the possibility of formulating a quantum theory of "centrifugal force", that would involve "centrifugalions".


 * I think I agree that it's probably possible to formulate theories of physics in such a way that everything is described in terms of fields and energy. Four fields suffice to account for the known phenomena: the field that mediates the the weak nuclear interaction, the field that mediates the strong nuclear interaction, the field that mediates electromagnetic interaction, and the field that mediates gravitational interaction. --Cleonis | Talk 21:26, 20 August 2006 (UTC)

Generalise this article to both centrifugal and coriolis forces?
I think we should generalise this article to cover both. The centrifugal and coriolis accelerations and forces pop out of the derivation anyway, so I claim this should be moved to 'Centrifugal and Coriolis forces'. There's also a redirect for Coriolis force that currently goes to Coriolis effect that I think would be better to come here.WolfKeeper 14:14, 25 August 2006 (UTC)


 * Sigh. Just go ahead and duplicate all the work you care, seing how you have already systematically eradicated the crossreferences to fictitious force, which is where the general concept is and should be discussed. I have given up my hope of this article ever becoming more than misinformation. Henning Makholm 16:16, 27 August 2006 (UTC)


 * Vote: Do no move. An encyclopedia is not a textbook. "Centrifugal and Coriolis forces" isn't one single subject, they are two different effects eventhough the cause (rotation) is the same. Harald88 17:10, 27 August 2006 (UTC)


 * Don't move: you are apparently suggesting a merge of three large pages (centrifugal, Coriolis and fictitious force) which already have a sensible division of material between them. – Smyth\<sup style="color:gray;">talk 21:12, 27 August 2006 (UTC)

Problem

 * Any reasonable high-school physics textbook will clear several misconceptions in the article
 * First of all, Centrifugal force is not a force. It's a pseudoforce. It is never used to denote any "reaction to centripetal force". In fact, a complete formalism of Classical Mechanics does not even require the concept of centrifugal force (the article mentions this, but rewording is needed). It is not a "reaction to centripetal force". Centripetal force is a force that is generated by another source (one of four in the universe as we know it today) and the radial part of it is called the centripetal force=m*w*w*r.The "reaction" caused by this force acts on the agent of the force ie the motor or whatever's causing the rotation, not the guy who's in the rotating frame. The ref that the first para cites to show that the term is also used to denote reactions is false. The ref is non-notable, and any mainstream textbook of Classical Mechanics (Goldstein, landau-Lifshitz Volume 1 etc) will refute it. I also encountered this statement in a popular book on Mechanics by Isaac Asimov and was surprised because it is wrong.
 * The derivation in teh article is not general, 'cause it does not mention the Euler pseudoforce in the article. It only mentions the coriolis and the centrifugal (See Landau Lifshitz vol 1 p128). This needs to be corrected
 * The notation is poor and cumbersome. Better to use dots to denote derivatives wrt time and explain everything more concisely (again, per Landau Vol 1).

That's all for now. Please discuss so that we can make this a better article together.


 * The article clearly explains the above, except for the points in which your're wrong and which are cited with full reference and clarification. That Newton is non-notable for classical mechanics is an amazing and obviously self-refuting claim.
 * BTW, I'm gladly surprised that Isaac Asimov is well informed about this. Do you have a detailed reference?
 * As this article is about centrifugal force and not about Coriolis or Euler force, I am opposed to introducing more than is strictly necessary for the Coriolis force; a precision of constant rotation speed is already present in order to free the reader from unnecessary stuff. The Euler force is derived in the general article about Fictitious_forces, and I'm not against the idea to delete part of the derivation here and instead refer to that article for the intermediate steps.
 * This article has been through such discussions before, see the archives. Of course it's always possible to refine some phrasings, but sometimes such improvements were deleted by others... Perhaps you want to help with clarifying that centrifugal pseudoforces don't significantly reduce the calculation effort in the mentioned examples, contrary to the suggestion. Harald88 22:11, 3 September 2006 (UTC)


 * Perhaps I did not explain myself clearly enough. I was wrong in saying "non-notable source", being a bit careless on my part there, thanks for correcting me.


 * However, bear in mind that the language in which Newton wrote his research is the language of the Jacobean period, which is highly confusing to modern eyes and ears and easily subject to false interpretations. In addition, the proposition that Newton may have interpreted centrifugal "force" as a reaction would either be


 * A matter of semantics, or


 * A product of his relatively immature understanding of the field that he started (for all his genius and originality, Newton's knowledge of Classical Mechanics is not as refined as ours is today. Physics is about truth, not about the words of "Big People vs little people", I mean, what are we, Social Scientists???).

The fact is that the "reactive forces" to "Centripetal Force" do not act on an observer in the frame of reference, but on the agent of the force (that follows from Newton's Third law itself). If the force felt by the observer in the rootating frame is a reaction, then where is the action coming from? Nowhere, that's where, which is why it's not a real force. The centripetal force is being caused by something else (a motor, gravity,nuclear interaction, whatever) and the reaction acts on THAT.


 * Amendment: I see now that this is pretty much what the para says, so this part of the discussion is unnecessary,sorry abt the confusion. I'm a physicist so I don't sleep much & got confused myself... Densagueo 23:26, 3 September 2006 (UTC)


 * Second Amendment:Dang! I REALLY need to get some sleep. I see now that the historical qualification is also there in the last sentence, so the point below is not relevant either.Densagueo 23:30, 3 September 2006 (UTC)

Amendment: This part of the discussion (below) is still relevant because the historical qualification is not mentioned in the article. Densagueo 23:26, 3 September 2006 (UTC) I'd put modern texts over Newton's any day. Plus, Asimov's undestanding of Physics is messed up (he even gets confused betwn escape velocity and terminal velocity in his books. Please! But that's beside the point). I'm just saying that no current normative text on the subject (Goldstein, Landau-I, Greenwood, Fetter & Waleecka, anyone) ever makes the claim that centrifugal force is a "reaction to centripetal force" (If I am wrong on this then please cite book and pg number). As I see it now, at best, the interpretation is historical only (if that's really what Newton meant in the ref, which I doubt) and should be stated with that kind of qualification.Densagueo 23:01, 3 September 2006 (UTC)
 * <Previous statement also deleted>. I think it's ok now 'cause the general treatment is properly linked to in the article.Densagueo 23:36, 3 September 2006 (UTC)


 * You're welcome. Note that the reaction force is not just historical: it has educational value to make people understand that the reality of their sensation of a force in rotational motion is not denied by Newton nor by modern physics - which is an important, misguided argument by some people who defend the use of pseudo-forces. Instead of denying what they witnessed themselves, it's very helpful to explain that they are not mistaken about their sensation, but that according to physical theory they sense a reaction force and not a primary force (see for example above on this page (point 7) the protest that "Centrifugal Force *is* a force"). In fact, the confusion and edit wars of this article only faded after this essential point was made clear. As a result, this article may now be better than most other articles in explaining the issue. Harald88 06:54, 4 September 2006 (UTC)


 * This article is horribly misleading and full of misconceptions, starting with the utter crap about "reactive centrifugal force". Of course the centripetal force has a reaction, but nobody who understands these things call it a centrifugal force, and the article lies to the readers by telling them that it is. To the extent it suggests that the corrcect, fictitious meaning is wrong, it lies even more. Henning Makholm 07:14, 4 September 2006 (UTC)


 * I agree with Henning Makholm. I think the rare/misconceptual/historical use of the term as a reactive force deserves maybe a short comment in the article, but not more than that. Yevgeny Kats 11:50, 4 September 2006 (UTC)


 * As Harald88 points out, the non-physicist reader must be informed that newtonian physics does not deny that a passenger in a car that is following a curve will feel that he is pushing against the car's door. On the contrary, newtonian demands it (action-reactioi principle).
 * Concentrating on the statement "centrifugal force is fictitious" would be very misleading to the non-physicist reader. The non-physicist reader may think: "Hell, I know what I feel, if the door of the car would not be secure my weight against it would make the door pop open. Nothing fictitious about that." You've got to give the non-physicist reader a handle on how to rethink his folk-physics.
 * Many expressions have in physics lingo a meaning that is different from the everyday meaning. In everyday lingo, the expression centrifugal force is used to refer the sensation of feeling one's weight pushing against the door (or the seatbelt or whatever). You've got to give the non-physicist reader an opportunity to notice that the meaning in physics lingo differs from the everyday meaning.--Cleonis | Talk 17:44, 4 September 2006 (UTC)
 * It is a fact that physicists does use the term "fictitious" with a technical meaning, and that this technical use of the word "fictitious" does NOT mean that the centrifugal force or the sensation it describes is somehow wrong or the product of a delusion on the part of the car passenger. The article does a grave disservice to laymen and students alike by refusing to discuss or even use the term "fictitious". This refusal means that the reader of the article does NOT get any chance to notice that "fictitious" has a different meaning in physics lingo that what he may suppose. Henning Makholm 11:22, 5 September 2006 (UTC)
 * That's an interesting claim. IMO the technical use of fictitious is perfectly in line with the normal meaning (dictionary: "accepted or assumed for the sake of convention"). Indeed there is nothing wrong with the kind of centrifugal forces such as the force that is exerted by the periphery of a very fast rotating disc on the bulk of the disc. This inertial reaction to the centripetal acceleration of the bulk is a very real centrifugal force - it even can pull the disc apart.
 * BTW since a centripetal force can't rip a disc apart, I'm surprised to hear that this very obvious, Newtonian meaning of the term centrifugal force is rarely used - perhaps due to lack of understanding?
 * OTOH, a decription according to which acceleration isn't the cause, but instead there is an "induced centrifugal force" coming from outside and that somehow only works on the atoms of the rotating disc while it rotates, even depending on the distance to the centre, is entirely a fiction.
 * Probably we should stress this more in the article. Harald88 18:47, 5 September 2006 (UTC)
 * The expression 'fictitious force is used. The current version of the article has been in place for some time now, and the expression 'fictitious force' is mentioned, in its proper context, in the opening section.
 * When did you last refreshed your cache? The term "fictitious force" has been removed from the opening section recently. I put it back there, got reverted. Henning Makholm 16:25, 6 September 2006 (UTC)
 * The rest of the article is a patchwork of edits by contributors with opposing views, but at least the opening section is a bit helpful to the reader. I'd love to see a coherent article, but given the anyone-can-edit-wikipedia policy, I see little to no hope for things getting better. --Cleonis | Talk 17:14, 5 September 2006 (UTC)
 * I propose one refinement, in line with my remark just here above: eventhough the centrifugal force equals the centripetal force, it may be clearer (and arguably, more accurate) to state in the intro that the (real) centrifugal force is the reaction to the centripetal acceleration. Harald88 19:02, 5 September 2006 (UTC)
 * It is wrong and misunderstood that the centrifugal force is the reaction to the centripetal one, and it does not become any righter by your continuing to claim so. Henning Makholm 16:23, 6 September 2006 (UTC)
 * Please look into the mirror: you may suggest as much as you like that Newton didn't understand mechanics, but it won't make a difference. Harald88 19:54, 6 September 2006 (UTC)

Real centrifugal forces
As I remarked above, I found it hard to believe that the real centrifugal force as discussed by Newton, would not be used anymore in modern times. And my memory turns out to be correct: it is (of course!) used for such cases as fast rotating spindles and rotor blades. I'll add a few modern links of uses that correspond to that classical use to the definition in the lead. Interestingly (but again I'm not surprised), the Brittannica Concise Encylopdia uses the Newtonian definition when discussing centrifuges. Harald88 20:42, 7 September 2006 (UTC)


 * Additional remarks:
 * Throughout the centuries the expression centrifugal force has been used in both of the two meanings that are listed in the opening section. It is unfortunate that many textbooks do not warn the reader about the multiple meanings.


 * In the case of rotorblades, such as those on a helicopter, often the expression 'centrifugal force' is used to refer to the 'reaction force in centrifugal direction' that is exerted by the rotorblade on the mounting point. The blades exert a force in centrifugal direction, because the blades are firmly attached to their mounting points. At worst a text about rotorblades is so ambiguous that it is not possible to figure out which meaning of the expression 'centrifugal force' the writer had in mind.


 * Then there are contexts where rigid attachment is not at play, for example the case of a turbopump. The operating principle of an axial centrifugal pump is that the water entering the chamber is not in any way attached to the axis. So in the case of axial centrifugal pump 'reactive force in centrifugal direction' is definitely not involved in the motion of the water away from the axis of the chamber.


 * The defining criterium is rigidity. When an object is rigidly connected to an axis of rotation then a 'reactive force in centrifugal direction' is present, and a writer may have opted to use the expression 'centrifugal force' for that. When no rigid connection is present, then the expression 'centrifugal force' can only be taken as referring to the concept of fictitious centrifugal force. --Cleonis | Talk 17:01, 10 September 2006 (UTC)

Reactive Centrifugal Force
What is being described is the Normal Force,or the perpendicular force. Car pushes on person, person pushes on car but in an equal but opposite direction. How is this a force other than the Normal force? Normal Force Meg1064 04:20, 21 February 2006 (UTC)
 * What about when it's something pulling on a central thing, like a tether ball "pulling" on the pole to which the rope is attached? Then it's a tension force, not a normal/contact force.  Maybe Reactive Centrifugal force is a good way to encompass/denote forces involved in reacting to centripedal force. ChrisLawson 13:37, 6 July 2006 (UTC)

The term "reactive centrifugal force" is unheard-of in the mainstream physics literature. Introducing it into the wikipedia is either unsubstantiated opinion, or (at best) original research, and therefore not permissable. It is also quite unnecessary, since conventional ideas and terminology suffice to explain both of the common meanings of "centrifugal force". I have revised the article to remove references to "reactive centrifugal force". Jsd 16:16, 25 October 2006 (UTC)


 * I agree that it's not a conventional term, and we better find some formulation of the article that would avoid it but will still make a clear distinction between the two different meanings. (And actually I'd say that the historical/rare/misconceptual "reactive" meaning should be removed altogether or only briefly mentioned as an aside, but it seems that other editors won't let me do it). Yevgeny Kats 19:07, 25 October 2006 (UTC)

Elsewhere Yevgeny Kats says: first justify -all- of your suggestions in Talk; some of them are patently wrong. Well, according to http://en.wikipedia.org/wiki/Wikipedia:How_to_revert_a_page_to_an_earlier_version it would be better if you explain which suggestions are patently wrong, rather than performing a knee-jerk reversion, especially since you agree that the reverted version contains impermissible innovations. Also, please don't tell me what other editors won't permit, especially when you disagree with them. Why don't we just do the right thing, and let other editors learn the right answer at their own pace?

I doubt anything I wrote was patently wrong; if you see an opportunity for improvement, please be specific. Jsd 19:33, 25 October 2006 (UTC)


 * Hi Jsd, I apologize for talking this way, it's just that I was surprised that after User:Crunchy Numbers reverted you and said "please read the talk page before drastic rewrites of this article" you did these drastic rewrites again with only a brief comment in the Talk page, which addressed only one of the things you changed, and you didn't wait for anyone's response. As you can see from the history of the article and the talk page, many people spent their time to get a formulation that would optimally address the various issues, so a major rewrite should be considered carefully, since all the issues will rise again. I'm a bit busy at the moment to discuss all the details, but I'll be back later. Yevgeny Kats 20:32, 25 October 2006 (UTC)


 * Also, please start adding edit summaries. Major edits without summaries are likely to be reverted.-Crunchy Numbers 20:36, 25 October 2006 (UTC)


 * These are the things that I find wrong in Jsd's edit (and therefore reverted ):


 * 1. The force is called fictitious because it is really a force per unit mass, i.e. an acceleration.
 * This is not the reason that the force is called fictitious. The actual reason is that unlike the regular forces this force does not arise from an interaction with another body, and actually it is missing altogether in a non-rotating frame of reference. Furthermore, what sense does it make to call the force "acceleration"? Suppose in some situation there is another (real) force that cancels the centrifugal force, so there is no acceleration. Does it mean that there is no centrifugal force? Or does it mean that the other force is also an acceleration and the two accelerations cancel each other??


 * 2. It is profoundly analogous to a gravitational field. Like a gravitational field, it gives rise to a pseudo force...
 * No, on the contrary: the gravitational force, caused by the gravitational field, is a -real- force, due to an interaction with the Earth. (Maybe Jsd meant that in the framework of general relativity the gravitational force is also related to coordinates, like the centrifugal force. But I don't think any reader would be able to understand that from the sentence, so the sentence as it stands, in the framework of the simple mechanics, is wrong. Maybe it would be nice to expand about it in a separate section in the article Fictitious force.)


 * 3. the modern practice is to use Free body diagrams that emphasize the forces exerted on a body by its surroundings, rather than the other way around. This practice is only a matter of convention
 * In the context of that paragraph, this would mean that the modern practice is to include the centripetal force in free body diagrams, which isn't usually the case.


 * Another comment to Jsd:


 * In most cases people talk about the centrifugal force without introducing the notion of a "centrifugal field". The latter is not necessary, so let's not mention it in the most basic definition. (I don't mind though if the term "centrifugal field" is mentioned somewhere later in the article, even though I don't find it particularly important.)


 * Yevgeny Kats 23:48, 25 October 2006 (UTC)

Response from jsd : I apologize for not heeding the sensibilities of the contributors to the talk page. I did read the talk page. I found it repetitious, with wrong ideas promiscuously mixed with correct ideas, and no clear indication of what was a consensus and what was not ... except insofar as the wikipedia entry itself represented a consensus. And that wasn't much help, since the entry appears to revolve around OR that is both impermissible (from a rules point of view) and unhelpful (from a reader's point of view).

Now, concerning the points enumerated above:
 * Point 3 is just wrong. The centripetal force must be included in the free body diagram.  Otherwise the particle would not follow the required curved path.  (We're talking about conventional, mainstream FBDs here.)
 * Point 2 is also wrong, for roughly the reason surmised in parentheses. The modern-physics view, i.e. the post-1900 view, is that the centrifugal field is no less real than the gravitational field ... or no more real, depending on how you look at it.  According to Einstein's principle of equivalence, and in practice, they are locally indistinguishable. To say that one is real while the other is not is kinda comical.  I recognize that many high-school texts are stuck in a pre-1900 viewpoint, but that doesn't mean we should perpetuate or propagate that viewpoint.
 * Point 1 is hard to follow. It may be that different folks have different reasons for calling pseudo-forces pseudo-forces.  I don't much care.  If you want to improve this passage, why don't you improve this passage, rather than doing a knee-jerk reversion?

68.0.165.239 00:18, 26 October 2006 (UTC)


 * Reply to Jsd (68.0.165.239):
 * Point 3: Usually one includes in the diagram the various forces that contribute to the centripetal force, but not the centripetal force itself.
 * Point 2: Do you suggest that in order to understand what the centrifugal force is, the reader needs to understand general relativity first? It doesn't make any sense. (Furthermore, you might be interested to know, or possibly you know already, that gravity can be perfectly described in terms of gravitons rather than in terms of curved spacetime, and then it's not different from the other forces.)
 * Point 1: I don't need to improve anything because the correct thing was already written in the article: "This is the fictitious centrifugal force, so called because it is not an actual force exerted by some other object."
 * Yevgeny Kats 00:37, 26 October 2006 (UTC)

From jsd:


 * Point 2: Do you think I suggested teaching GR first, and centrifugal force second?  I am quite aware that that doesn't make sense.  Why did you bring it up?  And why did you bring up gravitons?  Am I supposed to be impressed that you know about gravitons?
 * Point 3: How are we supposed to distinguish between "including the centripetal force" and "including the various forces that contribute to the centripetal force"?  Never mind, don't answer that.  Just tell me why any of this has any bearing on my suggestion that students who are tempted to put the centrifugal force in their FBD should put the centriptetal force in instead?

You seem to have taken "ownership" of an article that is a mélange of OR and error. If you want me to help fix it, let me know.


 * Hi Jsd, if I owned the article it would look quite different. It's just that I'm happy about the stable state that it is now, because at times it was much-much worse. If you really have time to work on the article, including discussing various issues with other editors who come from a different background than yours, I would be happy if you improved this article. (Please take into consideration my comments from above, and please wait a bit so that other people have an opportunity to express their opinions. Also, it would be nice if you explained what exactly the errors in the article are.)
 * As for Point 2, understanding that gravitational force is a pseudoforce (unlike the electric force, for instance) requires some understanding of general relativity. This is why I said what I said. (And I mentioned gravitons in order to say that if you describe gravity with gravitons instead of curved spacetime, gravity is not a pseudo force any more.)
 * Yevgeny Kats 02:08, 26 October 2006 (UTC)

From jsd:

You say understanding centrifugal force requires "some" understanding of GR. Well, how much? The word "some" covers a lot of territory. As for me, I don't expect readers to understand GR before they understand centrifugal fields. However, I would hope that people who wish to act as authorities on the subject would understand enough of modern (i.e. post-1900) physics to avoid saying things like the gravitational field is real while the centrifugal field is not. The 1800s have been over for a while now.
 * But I keep saying that :) Yevgeny Kats 07:46, 26 October 2006 (UTC)


 * If the analogy between the centrifugal field and the gravitational field is such a bad idea, why does the existing article speak of a body that "does not gravitate toward the outside" under some circumstances. Doesn't that imply that the body might "gravitate toward the outside" under other circumstances?  Do you think that word was used metaphorically by somebody who had no idea what he was talking about, or was it perhaps put there by somebody who actually knew something?
 * I guess it was used metaphorically. Yevgeny Kats 07:46, 26 October 2006 (UTC)


 * The article speaks of analyzing the centrifugal force in the car's frame "while ignoring its rotation". That's rather odd, since this type of centrifugal force was defined as existing only in rotating frames.
 * I suppose it meant to say that in that frame the car isn't rotating (which is correct). Worth fixing to make the formulation more clear. Yevgeny Kats 07:46, 26 October 2006 (UTC)


 * The article says "the passenger's body pushes against the outer edge of the car". Is that a fact?  Sure, sometimes it might happen that way, but it very well might not.  In fact, the centrifugal field acts on the passenger whether or not the passenger is restrained by the "outer edge of the car" or by something else or not restrained at all.
 * It's not always the case, but it's a very common example. This paragraph talks about the "reactive" meaning. The reactive force is absent if the passenger is freely flying in the car.


 * The article says "the car's path curves to meet the passenger". If we charitably assume this passage makes any sense at all, we must assume there has been a surreptitious switch from discussing force to discussing position, in particular a switch from a restrained passenger to (evidently) an unrestrained passenger.
 * I don't understand what you're saying here, but I agree that this sentence could be formulated better. Yevgeny Kats 07:46, 26 October 2006 (UTC)


 * The article says "it looks like an external force". Have we defined what is an "external" force?  How does that differ from other kinds of force?  Is it the opposite of an "internal" force, whatever that might be?
 * I suggest replacing "an external force" by "a force". Yevgeny Kats 07:46, 26 October 2006 (UTC)


 * If one type of centrifugal force is to be called a "reactive" force, please give an example of a force that is not a reaction to some other force. Please explain how to reconcile this example with the third law.
 * All the fictitious forces are not a reaction to some other force. This is why they're called fictitious. Newton's laws are not valid in non-inertial reference frames. Yevgeny Kats 07:46, 26 October 2006 (UTC)

I understand that "at times" the article was "much-much worse". I believe that. But that's not exactly high praise, is it? I've seen lots of ghastly things that could have been much-much worse.

Jsd 05:55, 26 October 2006 (UTC)


 * The reactive centrifugal force is well sourced in the intro already. Thus the notice in the article that claims that it is unsourced is simply a misrepresentation. I remove that now. Harald88 20:17, 9 November 2006 (UTC)

Well sourced??? I searched all the sources cited in the introduction, and failed to find any mention of "reactive centrifugal force" or anything similar. If you think this is anything other than a neologism, please cite the source. Please be specific.

Secondly, given the huuuge number of authoritative sources that pass up the opportunity to use this term, please consider the possibility that it is, at best, controversial.


 * The term is centrifugal force. There is consensus among editors here that this article should disambiguate the different uses of the term, so that some ediotrs decided to indicate the fictitious or pseudo centrifugal force here with "fictitious centrifugal force" or "pseudo centrifugal force" or "pseudo force"; similarly, some editors chose to indicate the inertial centrifugal reaction force with "reactive centrifugal force". There is no neologism.
 * One way to make it clearer is to simply put those disambiguating words between brackets.
 * And please note also that this artice refers to no authorative textbook that uses the fictitious centrifugal force; the use of such pseudoforces is very controversial. Harald88 02:52, 11 November 2006 (UTC)
 * I agree that the way it is put now is very good as disambiguation but it easily can be misunderstood as neologism. I'll try to rephrase the intro a little in order to avoid that misunderstanding. Harald88 09:25, 11 November 2006 (UTC)


 * Note: I had not seen how much the intro had been altered, so much that it had expanded and became a bit incoherent. I now tried to make it coherent and compact again, and added (again) the essential explanation that the introduction of the fictitious force implies removing accounting for the frame's acceleration. Harald88 10:49, 11 November 2006 (UTC)

Inline citations
I now added more inline citations, in view of the need for that as demonstrated above. This is also good practice according to WP:CITE. A next step could be to transform them into footnotes which look better as they automatically appear in superscript. Harald88 00:13, 10 November 2006 (UTC)

For the citations I prefer similar to Scientific_citation_guidelines Harald88 10:56, 11 November 2006 (UTC)