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Why is 4-momentum usually given as a covector?

  1. Oct 7, 2006 #1
    I understand the association between vectors and covectors on a Riemannian manifold, but it appears that 4-momentum is given naturally as a covector, instead of vector.

    4-position is clearly a vector (for that is the most natural representation of it). Similarly, 4-velocity is also vector, and given that 4-momentum is a scalar multiple of 4-velocity, one would expect 4-momentum to also be a vector. But it appears in various references as a covector by default. Is there any fundamental reason for this?
     
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  3. Oct 7, 2006 #2

    robphy

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    One motivation is that, mechanically speaking, momentum is conjugate to position.
     
  4. Oct 7, 2006 #3
    A covector maps vector to a scalar. Momentum is naturally dual to velocity since it usually makes sense to perform the matrix multiplication between momentum (as row vector) and velocity (as column vector) and gets a scalar with units of energy. So momentum can be viewed as covector.
     
  5. Oct 7, 2006 #4

    Hurkyl

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    This is interesting, because I've only come across 4-momentum as a vector, not a covector. (note: I'm a mathematician, not a physicist)

    Isn't it a rather important fact about momentum that it determines the velocity? momentum as a vector does that naturally, but momentum as a covector requires you to invoke the metric. From that POV, it seems most natural to treat momentum as a vector.


    Isn't that just a constant mc²?
     
  6. Oct 7, 2006 #5

    pervect

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    Momentum as a scalar conserved quantity is naturally a covector. It's a map from a velocity to a scalar quantity, a scalar quantity which has the important property that its conserved.

    So when we talk about, say, the conserved quantity associated with orbital motion in the Schwarzschild metric, we naturaly get E_0 (which is a conserved energy thus not directly relevant to the question) due to the fact that the metric coefficients are not function of time, and phi_0 (which is a conserved momentum) due to the fact that the metric coefficients are not function of phi.

    But when we focus on conserved scalar quantities, the natural model is a covector.
     
  7. Oct 7, 2006 #6
    I am not a physicst either :tongue2: Anyway, I don't understand what you mean by mc². All what I was saying is that if you have a velocity column vector (x y z)^T, you can view momentum as the row vector m(x y z). And then matrix multiply them will give you a scalar (a 1 by 1 matrix)
    with unit of energy kgm^2/s^2. So in that sense momentum is dual to velocity.

    Of course people usually just use dot product between velocity vector and momentum vector to get energy. In which case you can also view momentum as vector.
     
  8. Oct 7, 2006 #7

    Hurkyl

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    I'm talking about the 4-vector formalism. I.E. position as <t, x, y, z>^T. In that, the length of the velocity vector is always c, meaning that the product of velocity and momentum will always be mc².
     
  9. Oct 7, 2006 #8
    Only in flat spacetime in Lorentzian coordinates. The quantity R = (ct, x, y, z) is the prototype for a Lorentz 4-vector. Otherwise the prototype of a 4-vector is dR = (cdt, dx, dy, dz)
    You're starting with a false assumption here. 4-momentum is a 4-vector, not a 1-form (aka covector). You will note that the 4-momentum of a tardyon (i.e. a particle for which v < c) is the product of a scalar (i.e. proper mass) and a 4-vector (i.e. 4-velocity). You can't change a 4-vector to a 1-form by multiplying by a scalar.

    Best wishes

    Pete
     
  10. Oct 7, 2006 #9

    robphy

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  11. Oct 7, 2006 #10
  12. Oct 7, 2006 #11

    robphy

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    Sorry... Am I missing something? ...which paper? where?
     
  13. Oct 7, 2006 #12
    I'm the one should be sorry. I posted that paper in another thread. Oops! :biggrin:

    The paper I'm refering to is the one I wrote on the concept of mass in relativity. I placed it on my website here

    http://www.geocities.com/physics_world/mass_paper.pdf

    I tried to be all inclusive while trying to keep the length on a leash.

    Best wishes

    Pete
     
  14. Oct 7, 2006 #13

    pervect

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    I gather Pete has his fingers in his ears as far as I'm concerned, but I have a few comments about this paper anyway.

    1) The references I've read define the relativistic mass (when they define it at all, not all references use the term) as E/c, as I commented in

    https://www.physicsforums.com/showpost.php?p=1096850&postcount=28

    This isn't necessarily a major point, until Pete starts "correcting" people who use (as nearly as I can determine) standard usage.

    Note that this point affects a number of frequently used WWW documents which will should probably be corrected if Pete can be shown to be correct - the sci.physics.faq, and the Wikipedia.

    2) Besides this issue, there are problems with eq 19. The first equation by that label is more or less correct, the second is seriously wrong, even using Pete's defintions.

    i.e. if we have a volume dV moving to the right (in the x direction) with velocity v, the x component of the momentum is

    momentum-x =gamma ( rho_0 + pressure-x_0/c^2) * velocity, where rho_0 is the density in the rest frame and pressure-x_0 is the x component of the pressure in the rest frame.

    The second equation incorrectly includes all components of the pressure.

    This elementary error is apparently driven by Pete's desire to explain gravity in terms of "mass", so he apparently wants the answer to be rho+3P. But the correct expression is that only the x component of the pressure contributes to the momentum.

    I'm not quite sure what to do about the situation since
    1) Pete is quoting his own personal paper, which hasn't passed peer review.

    2) The paper, in my opinion, has errors - it's close to being right, but is wrong on some fairly important issues. I haven't been able to get anyone else to offer a second opinion on these issues.

    3) And just to make life more interesting, Pete's stated that he's not going to read my posts.

    https://www.physicsforums.com/showpost.php?p=1097890&postcount=39

    I think I need some input from the admins and the moderators on this particular issue. I don't want to get into a lot of bickering, but I think there may be some PF guideline issues here.
     
  15. Oct 8, 2006 #14

    robphy

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    IMHO, the discussion of relativistic mass is moving away (on a tangent :tongue2: ) of the original question concerning the 4-momentum as a covector. The point of my quoting the usenet posts was to support my first comment that the momentum can be thought of as conjugate to the position. Expressions like this
    [tex] p_a = \displaystyle\frac{\partial L}{\partial \dot x^a} [/tex], [tex] p_a = \displaystyle\frac{\partial S}{\partial x^a} [/tex], [tex] p_a = -\displaystyle\frac{\partial H}{\partial x^a}= -F_a [/tex],

    [tex] e^{i p_a x^a/\hbar}[/tex] (or more suggestively [tex] e^{i k_a x^a}[/tex])

    suggest this conjugate relationship. In this interpretation, the units of momentum would be something like "action per unit length".
     
    Last edited: Oct 8, 2006
  16. Oct 8, 2006 #15
    As I stated above, that tangent on rel-mass was a mistake on my part. However there is a part in my paper which addresses the differences in the 4-momentum and the corresponding 1-form. It is my assertion that the time-component of the 4-momentum should be "mc" where "m" is relativitic mass and the time component of the corresponding 1-form should be defined as en energy. This allows a strick definition regardless of what system of coordinates one is using, i.e. the definition is coordinate independant.

    Best wishes

    Pete
     
  17. Oct 8, 2006 #16

    selfAdjoint

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