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Revision as of 12:30, 6 February 2006

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. ${\displaystyle {\vec {V}}}$ ) 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[REDACTED] 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)

Gamma

What is γ?

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 g m/s would be an odd mix of CGS (in general, outdated) and MKS (in use).

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)

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:

${\displaystyle {\vec {F}}=m{\vec {a}}}$

is not seen much any more. Actually in NIST style the vectors should be bold italics but this Misplaced Pages 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#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 ${\displaystyle \mathbf {p} \cdot \mathbf {p} -E^{2}}$ where it should be ${\displaystyle \mathbf {p} \cdot \mathbf {p} -\left({\frac {E^{2}}{c^{2}}}\right)}$ 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#Lack_of_an_absolute_reference_frame. 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)

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)