Talk:Momentum/Archive 1

Momentum is a vector
I added a couple vector notation sigs on the variables on this page - promptly deleted by WAS 4.something. I *don't* need a source to know that momentum is a vector - you can look it up yourself WAS. In any case, this page needs to address the vector issues of momentum, and I would like to start by adding the vector symbol (e.g. $$ \vec V $$ ) to the equations on this page. *talk* to me WAS. Fresheneesz 01:25, 22 November 2005 (UTC)

Your statement "A little knowlege is a dangerous thing" is utterly ridiculous and abhoradly against what wikipedia is about. That quote sounds like it might come out of the mouthes of old aristocrats and rulers wanting to keep people down. Have you ever read 1984? That world comes from thinking knowlege is dangerous. *Stuipidity* is the real danger, and you're displaying some of that in your blind reverts. Fresheneesz 01:29, 22 November 2005 (UTC)

"A little knowledge" and "Knowledge" are two different things, as are having knowledge and thinking you do. The point of the saying is that more harm can be caused by incomplete knowledge than by total ignorance, although the originator of the quotation would probably agree that complete knowledge is preferable to those alternatives.

Gamma
What is &#947;?

It may be gamma, or something else, all I know is that it is Greek.

If that symbol is indeed Greek, it is gamma.. except that it doesn't exactly look like gamma : ). It should really have a loop at the bottom.

It is gamma, and it's the tex representation of gamma, it doesn't have to have a loop at the bottom. It is the lorentz factor in special relativity Phyrexicaid

relativistic quantum momentum
Is there a relativistic 'four dimensional' analogon to the operator associated with momentum in nonrelativistic quantum mechanics?

SI
Are there any particular units for momentum? Like, is it usually kg m /s, or g m/s, etc.? 65.96.72.221 18:50, 30 July 2005 (UTC)

No, no really. You can use (m kg)/(s) or (N)(s). These would be "kilogram meters per second" and "newtons of (times) seconds" respectively. You could also use (J)(s)/(m), or "joule second per meter." On that same note, you could probably use "watt second squared per meter." But usually people use the first one. g m/s would be an odd mix of CGS (in general, outdated) and MKS (in use).

Any SI unit can be expressed as a combination of other unit (on on my biggest pet peeves with the system. mass could correctly be expressed as (N)(s)(s)/(m) [Newton seconds squared per metre]. why anyone would do this is beyond a guess. most other derivations of units arise from specific examples, (such as an impused being expressed as a force integrated with a time).

Merge Impulse physics article here
The article about Impulse (physics) and the section about Impulse in the Momentum article are almost identical (the Impulse article just contains the term definitions which are present at the top of the Momentum article).

I suggest:

- Merging the Impulse (physics) article into Momentum#Impulse, adding any new information thats present on the Impulse article to the existing Impulse section on the Momentum article - Redirecting Impulse (physics) article into Impulse (disambiguation) - Linking the physics definition of Impulse on the disambiguation page into Momentum#Impulse

- Elvarg, August 20 15:57 PSt

Good idea. --Jeepien 06:14:16, 2005-08-22 (UTC)

So let it be written...so let it be done!

great idea it would help a ton

Impulse and Momentum, while similar topics and based around the same idea must be kept seperate in the mind of the reader. I agree that within the article the ideas should be linked, but I feel it would be confusing to someone unfamiliar with the topic to merge them. Wheatleya 22:11, 8 November 2005 (UTC)


 * Well, there's already a lot of information about impulse in this article. Should it be taken out, and replaced with a link to the impulse article?  You don't really need to know about impulse to understand momentum.  Pfalstad 01:34, 9 November 2005 (UTC)

I would help a lot for physics students like me.


 * What would help? Pfalstad 14:25, 22 December 2005 (UTC)

I don't think impulse should be merged in here. They should be separate. They are different concepts; why merge them? Momentum is a fundamental concept; impulse, I haven't used since freshman physics. Impulse is the change in momentum, but so what? Should we merge "velocity" and "acceleration" into one article? Also, the information on impulse in this article should be removed, since it is identical to the information in the Impulse article. Pfalstad 14:25, 22 December 2005 (UTC)

Agreed, but both articles should mention the link between impulse and momentum. Wheatleya 16:21, 28 February 2006 (UTC)

I do not think they should be merged, if for no other reason then it might be confusing for a reader to look up Impulse and end up on the Momentum page from a redirect. --Falcorian (talk) 00:13, 11 March 2006 (UTC)

Strongly oppose. They are different concepts entirely. Let's not get carried away with merge-o-philia. By that same rationale, momentum is closely tied to the concept of kinetic energy; shall we merge that also? and kinetic energy and potential energy are conceptually linked, so let's merge them also. heck, let's merge all of classical mechanics into one, big, unnavigable page. i know i'm wandering a bit here, but i'm trying to make my point as clearly as possible. if you don't understand physics, *don't* modify a physics page. period. OPPOSE VIGOROUSLY. lesotho


 * Strongly oppose. These are different quantities, easier to distinguish with different articles.  Of course, the shenanicans proposed for redirect are also silly, totally unnecessary if we keep the articles separate as they should be.  Gene Nygaard 03:48, 7 April 2006 (UTC)

Vector Notation
-- Please: the impulse or momentum is directed, as the velocity is. The formulas presented on this page are incorrect in that point, we need to correct these, as the impulse P results from a scalar multiplication of the vector V with the mass M. --213.196.203.206 2005-08-22


 * The formulas in this article are fine. The vectors are all shown in bold and the scalars in italics.  This follows NIST style, and is the way modern physics texts treat vectors and scalars typographically.  The old form of:


 * $$ \vec F = m \vec a $$


 * is not seen much any more.  Actually in NIST style the vectors should be bold italics but this Wikipedia lex knockoff thingamajig does not seem able to produce bold italics, so most authors appear to settle for bold alone.  --Jeepien 04:11:56, 2005-08-24 (UTC)

Link to "Four-momentum" article and possible error
It would probably be a good idea to provide a link to the article Four-momentum somewhere in the Momentum section, as it explains the concept more thoroughly.

Also, I believe there is an error in the expression given for the quantity in four momentum that remains constant, given as $$ 	\mathbf{p}\cdot\mathbf{p} - E^2 $$ where it should be $$ 	\mathbf{p}\cdot\mathbf{p} - \left ( \frac{E^2}{c^2} \right ) $$ This is the square of the Minkowski norm of the momentum four vector (basically; the signs are backwards), which is conserved. This agrees with the Four-momentum article; the idea of the Minkowski norm could also be explained in this section of the article, although there is a link to the Minkowski space article.

If someone backs me up on this, I'll make the changes. Thats it for my first post.

AtomBum 21:26, 25 September 2005 (UTC)


 * If you believe it -- do it!!--Light current 00:59, 29 October 2005 (UTC)


 * That looks right. Relativity people often set c=1, so that's probably why the c^2 is not there.  Pfalstad 03:17, 30 October 2005 (UTC)

Momentum and frames
If an object has a certain momentum in one reference frame, it also has a certain kinetic energy in that frame. However, the momentum in another frame may be different and so may its kinetic energy. So what is its real momentum and kinetic energy????--Light current 01:44, 28 October 2005 (UTC)


 * Sorry, it's all relative. Pfalstad 03:13, 30 October 2005 (UTC)

If I have energy in one frame, are you saying I may have none in another? If so, where has my energy gone? I thought it was conserved!--Light current 03:47, 30 October 2005 (UTC)


 * Yep. Your energy didn't go anywhere.  If you pick a frame, total energy is conserved in that frame (if the system is isolated).  There is no "real" energy or "real" velocity; see Special_relativity.  Pfalstad 04:23, 30 October 2005 (UTC)


 * In classical mechanics the kinetic energy of a system depends on the inertial frame of reference. It is lowest with respect to the center of mass, i.e., in a frame of reference in which the center of mass is stationary. In another frame of reference the additional kinetic energy is that corresponding to the total mass and the speed of the center of mass.--Patrick 10:50, 30 October 2005 (UTC)

Yes I suppose thats right. My KE wrt another planetary body could be tremendous, but in my frame here on earth, I have very little energy. I certainly feel like that at the moment (tired)! --Light current 12:53, 30 October 2005 (UTC)

The problem with relativity is that there is no absolute reference (except the speed of light, which isn't really a reference at all, considering it's fixed values). The different results that are obtained from preforming the same experiment in different frames of reference are both correct and accurate, within the margin of error of your equipment.

Curved space
It is interesting to note from the page that momentum is not defined in curved spacetime. Therfore, since all space is curved to some extent (by the presence of mass), momentum appears to be a figment of the imagination. Comments?--Light current 18:38, 22 April 2006 (UTC)

Origin of momentum
''"Momentum arises from the condition that an experiment must give the same results regardless of the position or relative velocity of the observer. More formally, it is the requirement of invariance under translation. Classical momentum is the result of this invariance in three dimensions. The definition of momentum was changed when Einstein formulated special relativity, so that its magnitude would remain invariant under relativistic transformations." ''

What is this talking about? Momentum is just a definition, isn't it? How does momentum arise from invariance under translation? Total momentum is not invariant under classical or relativistic transformations. Momentum is not conserved if forces are present. The definition of momentum was changed in relativity to preserve conservation of momentum when interacting particles are moving near or at the speed of light. This is the real deficiency with classical momentum. The last sentence might be talking about 4-vectors, but that is a separate thing. They are handy, but hardly necessary. Pfalstad 03:38, 30 October 2005 (UTC)


 * Yes I think this para is wrong and needs rewording.--Light current 03:45, 20 January 2006 (UTC)

Removed paragraph
I think there was something important in this para that is probably worth keeping altohugh I cant quite put my finger on what it is. Something to do with the fact that momentum is conserved is a natural consequence of the principle of relativity?--Light current 16:14, 20 January 2006 (UTC)

Ahh I ve just seen the post below. This confirms my suspicion. So perhaps we should reinstate this para(rewritten of course)--Light current 16:15, 20 January 2006 (UTC)


 * Well I was going to reword it and move it to the "Conservation of Momentum" section, but I noticed that section already had a simpler explanation of why momentum is conserved, so figured we didn't need an additional explanation. Pfalstad 04:47, 21 January 2006 (UTC)

"How does momentum arise from invariance under translation?"
In the Lagrangian formulation of classical mechanics the momentum "conjugate to a dynamical variable" is defined as the partial derivative of the Lagrangian ( L(x,v) ) w.r.t. v, the velocity.

If the Lagrangian is translationally invariant (i.e. not a function of x) then the Euler-Lagrange equations tell us that momentum is conserved.


 * Oh yes, that rings a bell. Thanks.  So conservation of momentum arises from a Lagrangian that is invariant under translation, not momentum.  I think what I said about relativistic momentum above is still correct.  Pfalstad 17:15, 20 December 2005 (UTC)

Origin of p as symbol for momentum
How did the letter p become the variable for momentum? Is it like a Greek symbol or relation to physics, or...? Mattderojas 22:00, 18 February 2006 (UTC)


 * Something to do with Isaac Newtons original work would be my best guess (remember that his inital equations were written in calculus)There may have been a slight change when it was changed to an algebraic sense. Or it may be that the letter (m) was already taken (possibly?)


 * M is for mass, n is for number, o is zero, and p is momentum?--Light current 00:31, 21 March 2006 (UTC)

photon momentum
This article has a one liner on photon's momentum. And it doesn't explain it at all. It would be very useful to put up more explanation about this. Fresheneesz 00:04, 8 April 2006 (UTC)


 * Yes. But do we know anyone who knows anything about photon momentum?? I believe Uncle Albert is no longer with us.--Light current 00:09, 8 April 2006 (UTC)

Conceptual definition of momentum
I tried to put in a conceptual definition into the intro, but it was reverted. Granted my conceptual definition wasn't very good and was very vauge. However, I think putting a conceptual definition is very important, since momentum isn't simply "m times v". Momentum is a concept that is conserved in certain cases, and thats why we use it. If anyone can come up with a good conceptual definition, it would be much appreciated. Fresheneesz 04:03, 10 April 2006 (UTC)


 * By Newtons laws, momentum is the time integral of applied force (either the force to accelerate the body OR the force to bring it to rest)


 * p= int(F.dt) equalt to area under the Force/time graph--Light current 04:09, 10 April 2006 (UTC)


 * Or in very simple terms its:


 * the amount of force multiplied by the time for which its applied.


 * I dont think you can get conceptually any simpler than that! Any good?--Light current 13:21, 10 April 2006 (UTC)


 * I know the math, and thats what that is. That definition isn't in conceptual terms. For example, "force" is defined as the rate of change of momentum - but conceptually its usually thought of as the cause of accelleration. This is the sort of thing I'm looking for. Is there any underlying thing that momentum causes, or is the cause of, or the general way momentum is thought of. Its hard for me to think of something that momentum corresponds to, because it isn't energy, and it doesn't affect how hard it is to accellerate. Lets brainstorm. ? Fresheneesz 19:27, 10 April 2006 (UTC)


 * Force is the rate of change of momentum. I dont see how you can get any more fundamental! Momentum is the result of a force acting on a mass. However you say it- it doesnt get any more intuitive!--Light current 20:59, 10 April 2006 (UTC)


 * Perhaps not, but acceleration is also the result of a force, and so is pressure, and velocity, and kinetic energy. Momentum must be differentiated from those other concepts. Otherwise I would have put that conceptuality up.
 * But one more thing, momentum is conserved but then again, so is energy. Conservation of momentum can be used sort of like energy that has direction. I dunno what conceptuality can be easily put into a sentence. Fresheneesz 05:27, 15 April 2006 (UTC)


 * I have a feeling that you are struggling with the ideas of energy and momentum and the differences between them. Am I correct in this assumption? Momentum is (force times time) whereas energy is 1/2 mv^2.


 * Its a bit like charge Q on a capacitor being equivalent to momentum and the energy stored 1/2 CV^2 is like the kinetic energy. Now we both know that the charge and the energy stored in a capacitor are not the same thing but thay are related. You have conservation of charge and conservation of energy here also. Is this any help?--Light current 05:42, 15 April 2006 (UTC)


 * Heh, no, sorry I was misleading. I have the concept of momentum and energy down perfectly - so perfectly that I don't have the words to describe my concept of it. My concern is how to best describe momentum to those that don't know as much about it. In the past couple hours I've come to an epiphany about momentum though. My epiphany was that reactionless moters are actually against the current laws of physics, I was hoping they weren't : ( . Oh well.
 * Anyways, as far as I can tell, the main use of momentum is determining the velocities after a collision. In such a case, both energy an momentum are used to find the velocities. Momentum embodies the average direction of movement of an object. Any given closed system of objects will have an average drift velocity that is unchangable. This I think is what I was looking for in terms of conceptualiazation - that the "average velocity" of the system of objects is p/m_total and momentum is someting like the "total velocity". But that doesn't quite work. I think I'm getting closer, what do you think? Fresheneesz 07:12, 15 April 2006 (UTC)


 * When you say:Any given closed system of objects will have an average drift velocity that is unchangable. this can be formalised as: The total momentum in any closed system is constant.
 * Also, Every force or action in the universe must have an equal an opposite reaction.


 * Total momentum in a system is the vector sum of individual momenta. THe average velocity (if that means anything) will be the total momentum divided by the total mass. So the momentum is the total mass times the average velocity. If you consider a unit mass, then of course momentum is directly prop to speed of this mass. Does this help at all?

--Light current 14:58, 15 April 2006 (UTC)


 * So what would you suggest we put in the intro? Well we could say that the velocity of the center of mass of a system is the momentum of a system divided by its total mass (thats what i meant by average velocity). Could we also say that momentum is "the tendency for an object to continue to move in its direction of travel"? That I think works quite well, if a bit vaugue. What do you think? Again, i'm just trying to find a good non-mathematical way of describing it in the intro. Fresheneesz 22:50, 15 April 2006 (UTC)


 * Well its up to you Fresh' .I dont really know what youre trying to achieve here. Why not put something in and wait for comments?--Light current 22:53, 15 April 2006 (UTC)


 * Sounds like a plan, I think I'll do that. Sorry if i've been extremely vauge. Fresheneesz 00:55, 17 April 2006 (UTC)

Intro
"For a constant force acting on a constant mass, it (momentum) is numerically equal to the product of the force and the time the force is acting on the mass." The change in momentum is equal to force times time, not momentum itself. I don't think that's particularly helpful in the first paragraph to mention that though. Pfalstad 01:35, 11 April 2006 (UTC)

Intuitive description
Use a ref frame in which the initial velocity is zero! Why make it complicated?--Light current 13:42, 11 April 2006 (UTC)


 * Make it even less complicated and take out the wrong information. Anyway, I fixed it.  Pfalstad 01:29, 12 April 2006 (UTC)


 * That's not a very intuitive description. Far more intuitive to say that momentum is proportional to velocity, or mass times velocity.  I don't see how defining it as the integral of force wrt time is going to help people understand the concept.  Pfalstad 01:30, 12 April 2006 (UTC)

If you read the previous posts, youll see that another editor was trying to find a more fundamental intuitve concept of momentum. I was just trying to help him! If you felt like it, you could remove it - I dont mind but someone else may.--Light current 02:23, 12 April 2006 (UTC)


 * Ok! I'll remove it (again).  Pfalstad 02:42, 12 April 2006 (UTC)

Actually, thinking about it, I think this force times time thing is more intuitive to me than mass times velocity as it uses the input quantity of force (ie the cause) rather than the output quantity (velocity). Its also the same as impulse (sort of) is it not?--Light current 04:50, 12 April 2006 (UTC)


 * Well, the mass x velocity thing has to go first. That is the definition of momentum, and mass and velocity are more fundamental concepts than force.  For me, it seems very intuitive and simple to say that an object's momentum is proportional to its mass and velocity.  If it's moving fast and is heavy, it has a lot of momentum.  Pretty simple.  Doesn't matter how it got that way.  Multiplying a force by time...  Not as simple.  Especially since you can't calculate the force in collisions and such, or if you're dealing with potential energy rather than looking at forces.  But if you think it helps, you can put back what I removed.  Pfalstad 05:06, 12 April 2006 (UTC)

I agree on the definition. I think Ill leave F times t out. People can always look here if they are really interested! Funny tho'- if momentum is conserved, so must Ft be in a system. But this is just impulse so therse nothing new here.--Light current 05:13, 12 April 2006 (UTC)
 * What do you mean F times t must be conserved? The force on what? If you're talking about the total force of a closed system (including direction), it is always 0, not just conserved. Fresheneesz 22:52, 15 April 2006 (UTC)

THe force on any object the mass collides with! (ie its an impulse equal to the momentum of the mass)--Light current 01:01, 17 April 2006 (UTC)

Fresh' If you look at the Inertia page (and its talk) you may get more ideas--Light current 01:33, 17 April 2006 (UTC)


 * Ok I looked, and its loooong. But I did find one conceptual definition "the tendency for an object to resist change in velocity. However, I find that definition erroneous as it is just as easy (takes the same power) to accellerate a mass M with a momentum Mv as it does to accellerate the same mass M with a different momentum Mv2. Fresheneesz 02:21, 17 April 2006 (UTC)

Thats right. The only thing that resists the accn is the mass. Velocity, momentum, and KE are all relative to the reference frame. Mass, force and accn are absolute and exist as the same in any ref frame. (I think!)--Light current 02:34, 17 April 2006 (UTC)


 * I'll fix that error on the inertia page. Fresheneesz 02:39, 17 April 2006 (UTC)

Please be very careful there- its a very controversial and touchy subject!! Expect reversions and harsh criticism!--Light current 02:48, 17 April 2006 (UTC)


 * Of course, : ) . I think I can argue my point well enough tho. Fresheneesz 09:07, 20 April 2006 (UTC)

More on conceptual definition
I removed the description of momentum as "a tendency for an object to continue to move in its direction of motion". That's what inertia is. That does not describe momentum, at least not in a strict sense. To get an idea of the common usage, I looked up momentum in a dictionary and couldn't find anything that matched that. Pfalstad 14:49, 17 April 2006 (UTC)


 * Now, now Paul. I thought we had agreed that inertia was just a word?--Light current 16:05, 17 April 2006 (UTC)

Yes, but momentum isn't "just a word". :) Pfalstad 19:24, 17 April 2006 (UTC)


 * This is not about words, inertia is equivalent to mass, not momentum. The difficulty in accellerating an object (changing its velocity) is ONLY due to its mass, not its momentum. Given a momentum vector {A*xhat,0*yhat,0*zhat}, the greater the magnitude of that momentum vector (the greater A is) one has, the more likely that the x component of the direction of motion will overpower other directions. Thus.. ima revert back. Fresheneesz 08:29, 19 April 2006 (UTC)


 * Yes I agree momentum is something else as well!--Light current 16:32, 19 April 2006 (UTC)

I don't see the difference between "tendency to resist change in velocity" and "tendency to keep moving in the same direction". Can you provide a reference for your definition? It's misleading and wrong. If you can find a dictionary or physics text that defines it that way, fine. Otherwise, it should be removed. Pfalstad 17:57, 19 April 2006 (UTC)


 * Oh sorry, I didn't realize you were already discussing it here. Well, the difference between the "velocity" quote and the "direction" quote, is that if an object has more mass, it obviously takes more force to change its velocity - but not so for momentum. Momentum really has a direct connection to direction ONLY. Consider the conservation of momentum. One of the main consequences of that principle is that for any given system - the direction of motion will not - cannot change.
 * I'm sorry if that didn't help, but what better conceptual definition can you think of Pfalstad? I'm open to change as long as it doesn't also describe inertia or energy. Fresheneesz 09:25, 20 April 2006 (UTC)
 * Btw, its not a definition - its a conceptuality. Something to help people gain intuition with momentum. Fresheneesz 09:26, 20 April 2006 (UTC)

Your explanation didn't make any sense to me, but I found a similar lead sentence in another encyclopedia, so I think I'll just leave it.. Especially since "mass times velocity" is not a great definition either (it's only classical). Pfalstad 13:23, 20 April 2006 (UTC)


 * Hmm, put another way: CASE 1) if an object with mass M is moving in x direction 5 units per second, by putting a force on it in the -x direction, you can slow it down until it reverses direction. That takes a certain amount of energy = 1/2 M 25 J/kg. CASE 2) If the starting velocity is 40 units/second instead of 5, then it would take much more energy (1/2 M 1600 J/kg) to reverse its direction.
 * This sort of embodies what I mean about momentum and direction. In both cases, it takes the same power to accellerate the object (change its speed or velocity), but it takes much more energy to reverse its direction in case 2 than case 1.
 * Perhaps that helped? Anyway, i'm glad you agree with keeping it - but i'm surprised you found a similar explanation somewhere. Fresheneesz 03:01, 21 April 2006 (UTC)