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I May I use this argument?

  1. Mar 10, 2016 #1
    In a relativistic treatment of mechanics one can say, that momentum and energy are correlatively conserved.
    The argument I would use, is that the length of the four-momentum is lorentz-invariant, and therefore, if E is conserved in any frame of reference, so the momentum.
    But I don't know, if this argument is fully valid. The length of the four-momentum ( I mean pμpμ) is creating a relation between the Energy and the magnitude of the momentum, not the vectorial momentum itself. But I want a vectorial conservation... Where is the catch?
     
    Last edited: Mar 10, 2016
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  3. Mar 10, 2016 #2

    Orodruin

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    The four-momentum transforms as a vector in Minkowski space. This is a stronger requirement than just having a frame independent magnitude, just as having a frame invariant magnitude is not enough to ensure that a quantity transforms as a vector in three-space.
     
  4. Mar 10, 2016 #3
    So in the end, if I get it right, I am only interested in [itex]\vec p=conserved \Rightarrow E=conserved[/itex]. And the proof of [itex]E=conserved \Rightarrow \vec p=conserved[/itex] is not necessary, since I assume that [itex]\vec p=conserved[/itex] is true under any circumstance.
     
  5. Mar 10, 2016 #4

    Dale

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    I would go the other way. The relativistic conservation of four momentum leads to the conservation of energy, momentum, and mass. The conservation of the four momentum is justified by the translation symmetry of the Lagrangian density.
     
  6. Mar 11, 2016 #5
    Ah ok. But just to be on the safe side, with "relativistic conservation of four momentum" you do NOT mean the lorentz-invariance, but the four-vectorial conservation in the case of interactions, right? I am just a little confused about the terminology.
     
  7. Mar 11, 2016 #6

    Dale

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    Yes. There is a distinction between conservation vs invariance. Conserved means that the value doesn't change over time. Invariant means that different reference frames agree on the value.
     
  8. Mar 16, 2016 #7

    vanhees71

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    As in Newtonian physics in special relativity there are 10 conservation laws from the space-time symmetries (Noether's theorem). For any closed system energy, momentum, angular momentum and the speed of the center of energy (in non-relativistic physics the center of mass) is conserved.
     
  9. Mar 16, 2016 #8
    I was also wondering if it is possible, to show
     
  10. Mar 16, 2016 #9

    Orodruin

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    This is not possible. There are situations when energy is conserved but momentum is not and vice versa. Energy and momentum are related to different symmetries.
     
  11. Mar 17, 2016 #10

    stevendaryl

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    There is an argument that if a collision of two particles conserves kinetic energy in every frame, then it also conserves a vector quantity, which we can identify with momentum. Let [itex]T(\vec{u})[/itex] be the kinetic energy associated with a mass [itex]m[/itex] traveling at velocity [itex]\vec{u}[/itex]. Let [itex]\vec{F}(\vec{u}, \vec{v})[/itex] the relativistic velocity-addition function (that is, if a particle has velocity [itex]\vec{u}[/itex] in one frame, then it will have velocity [itex]\vec{F}(\vec{u}, \vec{v})[/itex] in a second frame that is moving at velocity [itex]\vec{v}[/itex] relative to the first). Then if there is a collision of two or more particles (all of mass m for simplicity) such that:

    [itex]\sum T(u)[/itex] is conserved in every frame (where the sum is over all particles)

    then it's also the case that

    [itex]\sum \vec{p}[/itex] is conserved

    where [itex]\vec{p}[/itex] is defined via components:

    [itex]p^j = \sum_k \frac{\partial T}{\partial u^k} \frac{\partial F^k}{\partial v^j}|_{\vec{v}=0}[/itex]

    This formula works both nonrelativistically and relativistically, if you use the appropriate forms for [itex]T[/itex] and [itex]F[/itex]
     
  12. Mar 18, 2016 #11

    vanhees71

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    This is somewhat problematic. It's (maybe, I'm not sure whether I understand your idea correctly) right, if you consider the usual scattering process as from an asymptotic free initial to an asymptotic free final state (here of course in the sense of classical mechanics).

    Of course, the general case is covered by Noether's theorems. If the system is translation invariant (maybe only in specific directions) then momentum is conserved (or the components in the specific directions). If the system is time-translation invariant, energy is conserved. You can even say that energy and momentum are defined as the generators of these symmetries.
     
  13. Mar 18, 2016 #12
    Interesting. Do you have an example of that, for a closed system?
     
  14. Mar 18, 2016 #13

    Orodruin

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    Not if you require translational invariance, which I would assume is one of your prerequisites for calling the system "closed". Translational invariance (in space) is equivalent to conservation of momentum.
     
  15. Mar 19, 2016 #14

    vanhees71

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    A closed system by definition is one whose action (or more generally the 1st variation of the action) is Poincare (or in the non-relativistic case Galileo) inviariant und thus all 10 conservation laws of the space-time symmetry hold (maybe even more if there are additional dynamical symmetries as for the non-relativistic harmonic oscillator or the Kepler problem).
     
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