Talk:Momentum/Archive 1

message: 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)
thread_title: Momentum is a vector

message: 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)
thread_title: SI

message: 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)
replies: text: 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)

text: What would help?  Pfalstad 14:25, 22 December 2005 (UTC)
thread_title:  Merge Impulse physics article here 

message: It would probably be a good idea to provide a link to the article Four-momentum somewhere in the Momentum#Momentum in relativistic mechanics 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)
replies: text: If you believe  it -- do it!!--Light current 00:59, 29 October 2005 (UTC)

text: 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)
thread_title:  Link to "Four-momentum" article and possible error 

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

text: 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#Lack_of_an_absolute_reference_frame.  Pfalstad 04:23, 30 October 2005 (UTC)
replies: text: 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)
thread_title:  Momentum and frames 

message: 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)
thread_title:  Curved space 

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

text: 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)
thread_title:  Origin of momentum 

message: 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)
replies: text: 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)
thread_title:  Removed paragraph 

message: 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)
thread_title:  "How does momentum arise from invariance under translation?" 

message: 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)
replies: text: 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)
thread_title:  photon momentum 

message: "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)
thread_title:  Intro 

message: Use a ref frame in which the initial velocity is zero! Why make it complicated?--Light current 13:42, 11 April 2006 (UTC)
replies: text: Make it even less complicated and take out the wrong information.  Anyway, I fixed it.  Pfalstad 01:29, 12 April 2006 (UTC)

text: 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)

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

text: 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)
replies: text:  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)

text: 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)

text: I'll fix that error on the inertia page. Fresheneesz 02:39, 17 April 2006 (UTC)
replies: text:  Of course, : ) . I think I can argue my point well enough tho. Fresheneesz 09:07, 20 April 2006 (UTC)
thread_title:  Intuitive description