Talk:Momentum/Archive 2

Why cite poor books
I do not understand why at the end of the articles books like serways Physics for Scientists and Engineers are cited and books like Kittel or Kleppner Mechanics are not if the last ones are better. Kids I think don't know what momentom is, and this is adult language. My perspective is getting fun with the writing! --69.248.244.83 (talk) 23:12, 10 December 2007 (UTC)Mary Michelle

Earlier talk archived at
Talk:Momentum/archive1

Notation
The notation in the momentum page that used to describe collisions is different than that that on the page for elastic collision. The former uses an f subscript, the latter uses the prime (') superscript. I like the latter, it is more compact.

Impulse - step function?

 * On this page it says: impulse is a step function change in momentum.
 * By definition, an impulse happens instantaneously,(but it may transfer the same amount of momentum as a slower interaction does)
 * An (ideal)impulse is "an infinite force applied for zero time but with finite momentum change"
 * Impulse is plain and simple an instantaneous change in momentum (a step function change)

--Light current 02:45, 17 April 2006 (UTC)


 * Huh? Can you provide a reference for this?  The impulse page says nothing about a step function.  In my textbook (Sears, Zemansky, and Young), impulse is defined as force x time, and it says nothing about instantaneous or ideal impulses.  Another book defines an impulse function as a synonym for step function, but that has nothing to do with momentum.  Pfalstad 14:33, 17 April 2006 (UTC)


 * No. I didnt put it in! but Ive always thought of an impulse as a step change in momentum. Hence my statements above. Thats what this article page says. Are you saying that's wrong? If so, please feel free to correct the page and my misunderstanding of the definition of impulse will be corrected by your doing so ;-)--Light current 15:44, 17 April 2006 (UTC)


 * I'm saying that's wrong. Pfalstad 19:25, 17 April 2006 (UTC)


 * Looks like the word has different meanings in signal theory/math and mechanics! How interesting! I cant believe that mech engs dont use the dirac delta function in working out collisions taking v short times. But we'll never know as we dont seem to meet many mech engs here. I suppose we'll have to rely on physicists. Oh well...--Light current 15:51, 17 April 2006 (UTC)


 * When working out collisions, esp. when treating them as instantaneous events, I don't think you deal with force at all. At least not in freshman physics.  You deal with conservation of momentum and kinetic energy.  No need to introduce impulse or delta functions.  Pfalstad 19:25, 17 April 2006 (UTC)


 * You're the boss on this one!--Light current 19:47, 17 April 2006 (UTC)


 * What about a hammer striking a nail (when its done properly!)? The time of contact is very small (cos the hammer bounces off) yet most of the hammer's momentum is transferred to the nail. Do we not call this an impulse? -- A delta function?. Have mechanical engineers not heard of this yet? Shame!--Light current 20:48, 17 April 2006 (UTC)

Well, looking up stuff about impulse very quickly, I can't find anything about instantaneous change. Defining it that way makes no sense to me. In the hammer example, yes thats an impulse, but not because of the "short" amount of time, its simply because of the change in momentum - the momentum tranfer. I'll change the part about impulse. User:fresheneesz 68.6.112.70 09:46, 18 April 2006 (UTC)

I fixed it, I reccommend we remove the disputed facts tag - but I'll leave that up to you guys. User:fresheneesz 68.6.112.70 09:52, 18 April 2006 (UTC)


 * Ok Its not disputed now so Ill remove tag--Light current 13:11, 18 April 2006 (UTC)

Funny thing, today, my ECE prof mentioned an "impulse" that is infact a step functionish sort of thing. However, it is distinctly different from the "change in momentum" impulse. Look at delta function - they give a maybe incorrect reference to a unit impulse function. I think impulse can mean instantaneous change, but not in the context of momentum. Fresheneesz 05:04, 19 April 2006 (UTC)

Ahh, actually look up impulse response, and that gives more insight (but not much more). The disambiguation page for impulse gives it - I wonder if the page impulse should be turned into a full disambig page. Fresheneesz 05:09, 19 April 2006 (UTC)


 * "Impulse" in the classical physics sense is different than "an impulse" in the electronics sense. In the physics sense, you don't say "an impulse is...".  You say "the impulse of this force is equal to..."  Or you measure impulse.  You don't measure an impulse.  I think there are a bunch of problems with this section; it's confusing the electronics meaning with the physics meaning.  Unfortunately impulse is used so little in physics that my sources don't talk much about it.  (I'll let LC indent this comment appropriately.)  Pfalstad 05:30, 19 April 2006 (UTC)


 * I think all mention of impulse should be removed, and instead just have a link to impulse in the "see also" section. It's easier than cleaning it up.  Impulse is just not important enough to warrant such a long discussion in this article.  My physics text goes on and on about momentum but barely mentions impulse, and I can't remember using it since freshman physics.  Pfalstad 05:34, 19 April 2006 (UTC)

I actually never used impulse, not once. Its really a useless concept as far as I'm concerned. I'm going to merge the info on this page with the page on impulse - but I'll leave it up to someone else to actually delete it and put it in the see also section. Fresheneesz 06:58, 19 April 2006 (UTC)

Impulse is an extremely important concept in the correct application of the Impulse-Momentum equation, it is the Impulse that causes the change in or transfer of momentum. From the discussion above there seems to be confusion about the definition of what Impulse actually is. Seeing how this article deals primarily with linear momentum the appropriate definition of Impulse is the integral of force with respect to time, or in the case where the magnitude and direction of the force vector remains constant over the time period used in analysis than the Impulse reduces simply to force multiplied by the change in time. The key thing to realize here is that the time period of analysis does not have to be instantaneous, in fact it can have any magnitude. In most cases the impulse defined by the Dirac Delta function has no applicability because although there are many cases in mechanics where the time interval used is closely approximated by zero (i.e impacts) the magnitude of the force causing the impulse is rarely infinite. As for those who state that they never used impulse in solving there simple problems in entry level courses they are confusing themselves and making simplifications without realizing them. There are two cases where the impulse of of a force can be neglected, 1. the force is "non-impulsive", 2. a system is defined in such a way that the impulsive force can be viewed as an internal force. In the first case a force can be considered "non-impulsive" if they are very small in comparison to other impulsive forces, and/or the time period over which the force acts is very small. A force is also "non-impulsive" if it acts in a direction orthogonal to the momentum being considered. The second instance in which Impulse is neglected is a consequence of Newton's Third Law which essentially states in this case that the internal impulsive force between the masses of the system are acting in equal and opposite directions and therefore "cancel" each other out causing no change in the momentum of the system itself. The second case is commonly referred to as the Law of conservation of momentum. Someone needs to revise this article including the appropriate definition and use of impulse in an Impulse-Momentum analysis because there are many many physical problems that cannot be solved or can only partially be solved correctly without taking into account the impulse imposed on the system from external forces. —Preceding unsigned comment added by 64.126.190.120 (talk) 02:44, 5 January 2009 (UTC)

removed one of the sections on impulse
I removed the larger section on impulse beause it almost fully duplicates the article impulse. Fresheneesz 08:22, 19 April 2006 (UTC)

Lede
Restore balance and focus of para to the more conventionally accepted one. --Light current 15:12, 20 April 2006 (UTC)

Formatting of equations
Whats with the fancy formatting? Is this in th MoS?--Light current 15:14, 20 April 2006 (UTC)
 * I doubt it. I originally did it to utilize more page space, and separate the variable definitions from the rest of the text. Now its just for the ladder reason, I think it works well. You don't like it? Fresheneesz 02:53, 21 April 2006 (UTC)

Canonical momentum X mechanical momentum
I want to put forth only a thing: is there really a distinction about canonical momentum as opposed (though related to) mechanical momentum, as we could see in Sakurai,J.J - Modern Quantum Mechanics page 130 (between formulae 2.6.22 and 2.6.23) 1994?

Momentum operator in QM
The article states:
 * For a single particle with no electric charge and no spin, the momentum operator can be written in the position basis as


 * $$\mathbf{p}={\hbar\over i}\nabla=-i\hbar\nabla$$


 * where $$\nabla$$ is the gradient operator. This is a commonly encountered form of the momentum operator, though not the most general one.

Why the qualification about not having spin or electric charge? Surely the formula is valid more generally? If not then let's say what the more general formula is (perhaps we wish to mention conjugate momenta?). --Michael C. Price talk 13:59, 6 October 2006 (UTC)

As far as I know there is no reason to assume no spin and no electrical charge. Because these two operators commute with momentum, they are completely independent (i.e. one can know precisely the momentum of a particle and its spin) of the momentum value.

I have added a few lines to the QM Momentum section in an attempt to improve the generality of the topic by giving an example of the operator in a momentum basis, but you are correct that additional generality could be gained by removing those unnecessary restrictions. I will make this change and cite this discussion. Digemedi (talk) 22:04, 27 February 2010 (UTC)

I've added some generality but I would *really* like to add a link to the section in the quantum states article that talks about bases in the quantum sense (http://en.wikipedia.org/wiki/Quantum_state#Basis_states_of_one-particle_systems) but I can't figure out how to add that without it being an external link. Could someone please add this link to the first iteration of "basis" in the QM section?

Finally I should mention that my motivation, other than experience, for removing that spin/charge restriction comes from the lack of a commutation relation between momentum and these operators. This is a loose argument as that page is poorly cited but I can't think of any citations to improve it at the moment. Digemedi (talk) 22:04, 27 February 2010 (UTC)

Photon momentum
In the article it states that $$\vec p = m\vec v + q\vec A$$, accuratly represents a particle's momentum, but this still doesn't cover photons because photons have no net charge! Am I missing something here?


 * Is it really true that $$\vec p = m\vec v + q\vec A$$ accuratly represents a particle's momentum? I thought this was a replacement done in Schroedinger's equation to incorporate the (electric) potential energy and Lorentz force (from motion in a magnetic field).
 * Since photons have no charge, the $$q \vec A$$ term is zero for a photon. This is the same as saying that photons don't interact with the electromagnetic field. HEL 03:13, 5 November 2006 (UTC)


 * Not to mention the fact that the $$m\vec v$$ term is zero as well since the photon has no mass. No, the $$\vec p = m\vec v + q\vec A$$ holds only for a charged massive particle. PAR 03:35, 5 November 2006 (UTC)


 * The photon's momentum follows from the properties of the electromagnetic field, for which it is the quantum. See e.g. Jackson for a discussion of field momentum. Bkalafut 05:19, 28 June 2007 (UTC)

If the capture of a photon results in the capture of the photon's momentum, why then wouldn't the backwards reflection of the photon result in the creation of an equal additional amount on the part of the reflective object? Is that the theory of the solar sail?WFPM (talk) 18:35, 9 June 2010 (UTC)

Possibly missing information from the page
The article at present describes the effects of momentum. But at present it doesn't explain what it actually is. (For example knowing the effects of smoke inhalation is different from knowing the chemical composition of CO2 and the biology behind breathing.) Would it be more accurate (and hence saying in the article) that this 'law' is actually a 'theory that past observations of CoM apply to all things'? This is an idea I found in a popular science book. (Actually, no one really knows what momentum is, we just understand its effects.) Is it an accurate thing to say, or was something lost in the translation? ANTIcarrot 14:16, 13 December 2006 (UTC)

I think that the description on this page is misleading.

Note: Newton's First Law of Motion is often called the Law of Inertia. It reads, "An object will not accelerate unless acted on by an unbalanced force". Another way to phrase it would be, "An object will stay at constant velocity unless acted upon by an unbalanced force".

The description of momentum on this page sounds a lot like Newton's First Law of Motion. "the tendency of an object to continue to move in its direction of travel, unless acted on by a net external force" To me, this sounds more like a description of an object's inertia: "how difficult it is to accelerate the object", or "the object's natural tendency to resist changes in its motion".

I teach that an object's momentum is a measure of how difficult it is to stop the object. This depends on two factors multiplied togther the object's inertia and the object's speed.

If an object has more inertia (natural tendency to resist changes in its motion), then it will be more difficult to stop.

If an object is moving faster, then it will be more difficult to stop.

I hope this is clear and can be used to improve the page.

--Unksnogan 21:35, 3 January 2007 (UTC)
 * Correct me if i am wrong, but inertia cannot be multiplied as it is not a quantifiable value. Perhaps you mean the objects mass? 59.167.18.132 14:27, 28 August 2007 (UTC)

Generalized and canonical momenta vs Newtonian momenta
This article doesn't seem at all to cover generalized momenta--the partial derivative of the Lagrangian with respect to velocity--and thus completely misses the momentum of a charged particle interacting with an electromagnetic field, or angular momentum. Is there a reason for this? Bkalafut 05:16, 28 June 2007 (UTC)

Link to self-published paper?
I removed a link to a self-published paper on "Unified Momentum Theory", since it has all the earmarks of a crank/crackpot theory. (Its own webpage with the word "God" in the title, author's yahoo email address in the paper, etc.) —Preceding unsigned comment added by Langphysics (talk • contribs) 09:57, 16 October 2007 (UTC)

What's a Hartman?
I've removed the paragraph that claimed the unit of momentum was a Hartman. It was posted by some guy at my high school, naming it after one of our physics teachers.

Bold face Killa 20:45, 26 October 2007 (UTC)

Hey but there does need to units for momentum like everything else. Come on cant the scientists be lazy by having a simplified version of kg*m/s. Who wants to wright that out each time they solve for momentum. There are lots of derived units that just make it easier in our lives. So a Hartman is a time saver. --User: (talk) 23:35, 26 October 2007 (UTC)

It just makes sence to have a unit for momentum anyways. --User: (talk) 23:42, 26 October 2007 (UTC)

A unit is what you make it. I'm gonna use it. You know how the units of inch, foot, yard, etc. originated? The length of the king's body parts. Just because some group of scientists says it isn't an official unit doesn't mean it can't be used...it just shouldn't be used anywhere where official SI units are required. People make up units all the time. Hell, in math class, if you've ever used something like 'Let y = (2x^3)', you're inventing a unit...though only temporarily. 209.160.28.45 00:19, 27 October 2007 (UTC)


 * The Hartmann is the unit of wikipedia banning.WolfKeeper 00:48, 27 October 2007 (UTC)

So it's cool for kings to randomly decide what a measurement is, but experiments to create one democratically must be quashed? I mean, hell, even the SI system was created by a king. Of course, I do realize you have a job to keep things truthful. What if it is phrased as 'There is a movement to create the Hartman as the unit of momentum'?

71.61.37.175 01:13, 27 October 2007 (UTC)
 * That's not a notable point of view quoted by a reliable source.WolfKeeper 01:21, 27 October 2007 (UTC)

How can there be sources if it is used everyday by physics teachers and not on websites yet? It is used commonly but it is one of those things that is hard to source, like how does one cite something that is used by all? That what the Hartman is like. --User: (talk) 01:08, 28 October 2007 (UTC)


 * "Sucks to be you then".WolfKeeper 05:13, 28 October 2007 (UTC)

all of its all lies —Preceding unsigned comment added by 74.249.94.240 (talk) 22:41, 6 December 2007 (UTC)

Use of p for momentum
Removed from article: "The origin of the use of p for momentum is unclear. It has been suggested that, since m had already been used for mass, meter, and -milli, the p may be derived from the Latin petere ("to go") or from "progress" (a term used by Leibniz)."

I would love to see more cultural lore in these articles, and I was the one who expanded the history section and answered the template on Newton. I have tried my level best to find the origin of the use of p and I cannot do it! For me not to be able to find it is saying something. It appears in the late 19th century without a trace before then. Maybe it means P for the point at which the force applies or p for the power of momentum. The earlier writers generally did not assign a variable to momentum, they used descriptive words, such as momentum, and none of those seem to fit p. Well, let's keep looking, but in case we are wrong, let's not mislead the public, so I moved the statement to here. There must be a first use of p that caught on or else it would not be a standard variable. Whether it was considered important enough to note is another matter, so we would at worst case scenario be searching every physics book in the world since 1689 to find out the earliest who used p! Whew.Dave (talk) 14:39, 23 February 2008 (UTC)

Definition of Momentum
This article is about momentum in physics. “In classical mechanics, momentum (pl. momenta; SI unit kg•m/s, or, equivalently, N•s) is the product of the mass and velocity of an object (p = mv). For more accurate measures of momentum, see the section "modern definitions of momentum" on this page.”

I think these are ‘Operational definition’ to measure, and in case of physics we should better go for ‘Conceptual definition’ to describe a physical quantity, viz., ―→ "quantity of motion” as in ―→ “Newton's "Mathematical Principles of Natural History" when it first came out in 1686 showed a similar casting around for words to use for the mathematical momentum. His Definition II[6] defines quantitas motus, "quantity of motion," as "arising from the velocity and quantity of matter conjointly", which identifies it as momentum.[7] Thus when in Law II he refers to mutatio motus, "change of motion," being proportional to the force impressed, he is generally taken to mean momentum and not motion.[8]

It remained only to assign a standard term to the quantity of motion. The first use of "momentum" in its proper mathematical sense is not clear but by the time of Jenning's Miscellanea in 1721, four years before the final edition of Newton's Principia Mathematica, momentum M or "quantity of motion" was being defined for students as "a rectangle", the product of Q and V where Q is "quantity of material" and V is "velocity", s/t.[9]”

Tridib phukan (talk) 20:05, 25 July 2008 (UTC)Tridib_phukanTridib phukan (talk) 20:05, 25 July 2008 (UTC)

Incorrect diagram for relativistic momentum
The diagram http://en.wikipedia.org/wiki/File:Relativistic_Dynamics.svg contains an angle $$\gamma$$ which should be $$\gamma^{-1}$$ becase $$\gamma\rightarrow\infty$$ as $$v\rightarrow c$$. —Preceding unsigned comment added by 91.47.165.182 (talk) 18:12, 1 January 2009 (UTC)


 * As far as I can see, this diagram does not contain an angle $$\gamma$$. It contains angles $$\theta$$ and $$\phi$$. Perhaps you are referring to some other diagram? DVdm (talk) 22:14, 1 January 2009 (UTC)

why p??
why p for momentum???

can you explain that... —Preceding unsigned comment added by 210.213.96.67 (talk) 09:50, 13 January 2009 (UTC)


 * It shouldn't be p it should be the Greek letter rho which looks like a p and is depicted in the equations showing integration to get momentum.

HobbsO (talk) 12:44, 25 February 2011 (UTC)


 * No, it's not a Greek rho (&rho;); it's definitely a Latin letter "p". — Q uantling (talk &#124; contribs) 17:55, 28 February 2011 (UTC)

canonical momentum
The definition of momentum with the electromagnetic vector potential is not specific to quantum mechanics.Klinfran (talk) 22:59, 19 January 2009 (UTC)

Units
I have seen a recent number of IP edits attempting to create a viral shift from the accepted SI units of momentum to a unit named after an obscure high school teacher. I can assure you that (at present) there is no movement within the scientific community to rename the unit to a "Grover", "Hartman", nor after any other individual.

To those IP editors who are making these particular changes: As we are trying to write an encyclopaedia, please refrain from inserting unreferenced and non-factual information with the intent to glorify any particular secondary school system or instructor.

Thank you. — Archon Magnus (Talk 14:14, 2 February 2009 (UTC)

Kinetic energy and Momentum
I am confused with definitions of the kinetic energy and momentum. In inelastic collisions, there is some loss of kinetic energy, but not momentum, why ?--Tohyf (talk) 13:49, 30 March 2009 (UTC)


 * So long as there are no external impulses, total linear momentum of the system is conserved. Energy is conserved as well, but it can take on many forms. In anything but a perfectly elastic collision, some kinetic energy is converted to deformation or heat in the colliding objects. I don't believe linear momentum can be converted into heat or deformation. -AndrewDressel (talk) 22:33, 10 April 2009 (UTC)


 * Net linear momentum could be "lost"/converted to sound waves or light waves, if those waves are not isotropic. — Q uantling (talk &#124; contribs) 18:00, 28 February 2011 (UTC)


 * I agree with Andrew Dressel, but not with Quantling. Heat, sound and light are all examples of energy - they can be measured in Joules.  In a collision, some kinetic energy is usually lost because it is converted to another form of energy such as heat, sound and light.  Momentum is not a form of energy - it isn't even measured in units of energy such as Joules.  In a collision some kinetic energy will usually be lost but momentum will not change.  Andrew is correct - so long as there are no external forces acting, linear momentum is conserved.  Dolphin  ( t ) 21:58, 28 February 2011 (UTC)


 * I agree with Andrew Dressel; momentum cannot be converted to heat or deformation. But, I disagree with Dolphin; momentum can be converted to sound or light; what is a solar sail?  — Q uantling (talk &#124; contribs) 18:50, 2 March 2011 (UTC)


 * I am not in the least confused! The kinetic energy of a body in motion increases to the square of the velocity, but is directly proportional to the mass. For equal mass, twice the velocity is four times the kinetic energy. ( eg. A car crash at 60 mph is four times as severe at a crash at 30 mph.) For equal velocity twice the mass is just twice the kinetic energy. The animated 'gif' is simply erroneous! and very misleading! It needs to be corrected! otherwise Joules are being stolen! (we obviously assume no heat was generated in the inelastic collision, as this is an idealised presentation.) 5 x 5 = 25. and 2.5 x 2.5 = 6.25, 6.25 x 2 = 12.5. 5 x 1/2^2 = 3.5355339059... or 3.536. (for sly drules;)and that velocity squared is 12.503296, good enough for folk. Kinetic energy is conserved! we would all be in trouble if that was NOT conserved! — Preceding unsigned comment added by Alastair Carnegie (talk • contribs) 04:35, 7 July 2011 (UTC)


 * I am an experimentalist, put a magnet on your trolley, and have it pick up an equal weight of steel shot/iron filings, so that it is twice as heavy. Or if you like have two trolleys in the middle that pass two stationary trolleys/wooden blocks/whatever? each side, with magnetic coupling between the stationary objects and moving trolleys. The advantage of magnets, is that very little heat is generated in the collision. The snowball analogy is spurious, as the impact generates significant heat! it may not feel 'hot' but energy is still lost to heat. (Alastair Carnegie (talk) 05:11, 7 July 2011 (UTC))


 * Fortunately or not, depending on your point of view, your experiments have no bearing on this article. The coin of the realm here is a reliable source, and this is what Georgia State University says on their Hyperphysics web site: "Macroscopic collisions are generally inelastic and do not conserve kinetic energy, though of course the total energy is conserved." That pretty much settles the issue, unless some other equally reliable source can be found that contradicts it. -AndrewDressel (talk) 06:25, 7 July 2011 (UTC)

Contradiction
Page inconstantly gives both $$\mathbf{p}$$ and $$\mathbf{P}$$, depending on the section. Craig Pemberton (talk) 06:04, 10 April 2009 (UTC)

...momentum is not defined in non-asymptoticaly minkowski space...
In the section Momentum in relativistic mechanics there is a note in the end that the momentum is not defined at all in spacetime that is not asymptoticaly Minkowsky. Can somebody support this claim by some reference or explain it? One can still multiply rest mass by four-velocity. Why sould not this be taken as the definition of momentum in such a case? —Preceding unsigned comment added by Grepe (talk • contribs) 21:53, 18 June 2009 (UTC)

"History of the Concept"
The section seems to start with a sentence that looks like half of it was eaten: "mōmentum was not merely the motion, which was mōtus, but was the power residing in a moving object, captured by today's mathematical definitions." That doesn't look like a full thought to me. What's up with it, and where did the rest go? I can't find anything in the recent history. -- ✶♏&#8237;ݣ 20:23, 15 July 2009 (UTC)
 * Agreed. There are quite a few issues with the opening of the "history" section, including this sentence: "The Romans, handicapped by the limitations inherent in the Roman numeral system which lacks a symbol for zero, took these ideas no further."  WHAT?!?  That's a complete non sequitur, and it's also misleading.  Roman numerals are cumbersome for calculation because they don't have use positional notation, so they had no need for a symbol for "zero" in the modern sense in which we use the digit.  They were perfectly capable of expressing the idea of the quantity "zero" in writing (using terms like "nihil" or "nulla").
 * But that's all irrelevant to the history of momentum. Or, rather, it's just as relevant as the fact that the Romans lacked modern algebraic methods, graphs, calculus, etc.  But the formalized mathematics is only part of the "concept" of momentum.  While formalized mathematics of momentum was not carried particularly far in ancient times, there is plenty of discussion of motion concepts.  The pre-Socratics had interesting concepts that could be related to momentum.  Aristotle developed complex theories of motion that included concepts that sometimes behaved like modern momentum.  Even if they weren't using equations, they investigated complicated philosophical issues having to deal with motion.
 * While this may not be the place in the article to go into a long description of ancient theories of motion, the fact is that there were plenty of motion words in Latin and Greek (and other languages) other than "motus," and their meanings and uses in describing the world were debated. The first paragraph of this section might better be called "etymology," followed by "history of the *mathematical* concept" since it only investigates one ancient word and then summarily dismisses entire ancient treatises on physics which could not possibly consider a concept of momentum... because they didn't have a symbol for zero. 65.96.161.79 (talk) 17:24, 14 October 2009 (UTC)

Question. Can anyone help?
In the "Beam Propulsion" section of the article on Interstellar Travel there is this statement "In theory a lightsail driven by a laser or other beam from Earth can be used to decelerate a spacecraft approaching a distant star or planet, by detaching part of the sail and using it to focus the beam on the forward-facing surface of the rest of the sail" The argument apparently is that after the beam is reflected it has reverse momentum which when striking the forward-facing surface is transferred to vehicle deaccelerating it. It seems counter intuitive that you can shine a beam of light at a body and it will move towards the source based on some reflective actions inside the body. Can anybody show the fallacy based on the law of conservation of momentum? Specificallly, where does all that momentum end up?Paulkint (talk) 20:34, 13 August 2009 (UTC)


 * This is answered at talk:interstellar travel. I even drew a diagram for you. Momentum is exchanged between the two sails (big sail accelerates, small sail decellerates). --Christopher Thomas (talk) 17:31, 14 August 2009 (UTC)

Concerning "mathematically infinite rigidity
"A totally elastic collision exists only in theory, occurring between bodies with mathematically infinite rigidity". If the definition of "elastic" means deformation with no dissipated energy then I don't see why "mathematically infinite rigidity" is needed. You just need an ideal material which does not dissipate energy. Like a mass with an ideal spring attached. —Preceding unsigned comment added by Baxtrom (talk • contribs) 12:11, 28 October 2009 (UTC)

Concerning Elastic collisions
It currently says -

"Thus the more massive body does not change its velocity, and the less massive body travels in the opposite direction with twice the original speed of the more massive body."

Isn't this comment only true if the smaller mass had no velocity at collision? Shouldn't it be "the less massive body travels in the opposite direction with twice the original speed of the more massive body plus its own original speed.

Also - shouldn't the "u"s and "v"s in this section be in bold since we are talking about vectors? —Preceding unsigned comment added by Timmo999 (talk • contribs) 04:09, 16 November 2009 (UTC)

Uppercase P
The article states that the usual symbol for momentum is uppercase P, to distinguish it from pressure. Is this correct? Every textbook I've ever seen uses a lowercase p for momentum, and an uppercase P for pressure.--70.171.19.116 (talk) 01:55, 11 February 2010 (UTC)
 * I've removed it. If uppercase bold P is reserved for anything, it's for the total momentum of a system of particles each with some momentum pi. Strad (talk) 18:50, 11 February 2010 (UTC)

Deleted transformsas from infobox
When my typo had been fixed, so that that line actually displayed, it seems to have had the side-effect of causing a forced linefeed in the formulae in the derivations ("Expressed in other quantities") field. This, might be a bug in template:infobox physical quantity, and if so it should be fixed there. But, frankly, deleting the transformsas entry was easier, and it's my subjective assessment that it wasn't a big loss... not as big a problem, anyway, as the unfortunate newline.—PaulTanenbaum (talk) 18:02, 1 March 2010 (UTC)


 * Wholeheartedly agreed. Good idea, these infoboxes! DVdm (talk) 18:09, 1 March 2010 (UTC)

First paragraph needs attention from an expert in the subject
The following sentence appears in the first paragraph of the article.

"Angular momentum is a pseudovector quantity because it gains an additional sign flip under an improper rotation."

Is there any way somebody can reword this? As an ordinary reader with a bachelor's degree, this makes no sense to me whatsoever, and there are no context clues/links to explain what on Earth this means. Kansan (talk) 08:59, 14 February 2011 (UTC)

Few bits of clean up

 * Make definition slightly more obvious in lead paragraph (why bracketed like this?)
 * In the EM section, indicate the terms kinetic/canonical/potential momentum, add a link to kinetic momentum article,
 * the differentials are both italix and upright - they should be one or the other in the same article, according WP:MOSMATH to and this hilarious thread on "upright vs italic d for diffs" I encountered. (Although I prefer upright), for this article they will be made italic since its easier to replace \mathrm{d} with d than the other way round using the find-replace tool on Microsoft word (or any other word processor). It turns out italic d is more common anyway so best to stick to resources.

-- F = q(E + v × B) 15:08, 14 December 2011 (UTC)