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Does Conservation of 3-momentum and energy still work with RoS?

  1. Nov 2, 2015 #1
    If the 3-momentum γmu and energy γmc2 are conserved in one frame, they will be conserved in any other frame.

    But I was thinking about RoS, since events may not happen simultaneously in another frame, will 3-momentum and energy always be conserved with respect to time?
     
    Last edited: Nov 2, 2015
  2. jcsd
  3. Nov 2, 2015 #2

    mfb

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    In inertial frames, yes. The transformation between the frames stays the same, so it does not matter at which time you do the transformation.
     
  4. Nov 5, 2015 #3
    great my thread has been derailed.

    let's just focus on linear momentum please.
     
  5. Nov 5, 2015 #4
    I don't see where. We are talking about your question the whole time.
    Again, in the square structure, it seems that the center of mass moves up and down without external forces.
     
  6. Nov 5, 2015 #5

    Nugatory

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    You may have missed mfb's answer to your original question, back before the digression into this more complicated thought experiment: yes.
     
  7. Nov 6, 2015 #6
    I do want discussions to be about linear momentum only though.
     
  8. Nov 6, 2015 #7

    Nugatory

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    f you have a particular thought experiment involving linear momentum only, you might want to post it up for comments.

    But it should probably go in a new thread, as the hijack of this one is pretty much a fait accompli :smile:
     
  9. Nov 6, 2015 #8

    PeterDonis

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  10. Nov 6, 2015 #9

    Nugatory

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    Oooohhhh.... magic! I like. :smile:
    I'm sorry, my post crossed your thread cleanup.
     
  11. Nov 6, 2015 #10

    PeterDonis

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    I agree this would be a good idea, and now that the thread is cleaned up, it can be done here. :smile:
     
  12. Nov 6, 2015 #11
    Thanks Pete!

    @mfb Consider this:

    Two identical positive charges are moving towards each other head-on. As viewed from one frame, both charges come to rest simultaneously at the distance of closest approach. In this frame, the total momentum is always zero.

    At their closest approach, draw a line connecting both charges. This line is not parallel to the x-axis, it is at a certain angle.

    The direction of the x-axis is defined by the velocity vector of another inertial frame. Now, transform to that frame.

    At the point of closest approach, both charges were separated by a certain x-distance. When there is an x-separation of simultaneous events in one frame, the events will no longer be simultaneous in another frame.

    Since the line was not parallel, we can consider the momentum in the transverse direction. There will be some point in time for both charges when their transverse momentum drops to zero. However, due to the relativity of simultaneity, one charge will reach zero transverse momentum before the other.

    So the total transverse momentum in the 2nd frame will never be zero.

    Also, when we consider that the total transverse momentum in the first frame was always zero, we will find that in the 2nd frame the total transverse momentum won't be constant.
     
  13. Nov 6, 2015 #12

    PeterDonis

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    greswd, your analysis leaves out the momentum stored in the EM field. You have to include that for momentum conservation to work.
     
  14. Nov 6, 2015 #13
    ahh, I see.

    Is it the EM stress-energy tensor?
     
  15. Nov 6, 2015 #14

    PeterDonis

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    You can use that, yes. Or, if you pick an inertial frame, you can use the Poynting vector, which is usually referred to as "energy flux", but for EM fields, energy flux is the same as momentum density (possibly divided by ##c## if you aren't using units where ##c = 1##). Then you can just use Maxwell's Equations and the Lorentz force law to relate the motion of the particles to the values of the fields.
     
  16. Nov 6, 2015 #15

    mfb

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    The issues with the existing analyses have been highlighted multiple times.
    Each individual component of the system conserves angular momentum, the global conservation (as seen by a given reference frame) is a trivial consequence in special relativity. Picking arbitrary subsets of a system and saying "the angular momentum of this component changes" is pointless. The same applies to linear momentum.
    I'm out of the discussion.
     
  17. Nov 6, 2015 #16
    These methods were developed before SR right?

    They seem to be the perfect solution to RoS and conservation laws though. How intriguing. :wideeyed:
     
  18. Nov 6, 2015 #17

    PeterDonis

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    Yes.

    It's because Maxwell's Equations, unbeknownst to physicists when they were first discovered, are Lorentz invariant. In other words, they already have the right transformation properties for SR, not Newtonian physics. Einstein recognized this; it was part of what led him to postulate an invariant speed of light.

    (Basically, Einstein imagined himself riding along with a light beam, so that he was at rest relative to it. He knew that light was an EM wave, so if it were possible to be at rest relative to a light beam, the wave would be a purely spatial standing wave that did not change with time. But there is no solution of Maxwell's Equations with that property. So Einstein realized that it must be impossible to ever be at rest relative to a light beam; and in fact, if you carry the argument further and look at the details of Maxwell's Equations, you see that if those equations are correct, a light beam must be moving at the same speed, ##c##, relative to every inertial observer.)
     
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