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Massive particles in SR

  1. Sep 10, 2006 #1
    I don't have any SR texts to hand, and was wondering if someone could prove the following conjecture:

    all particles with positive rest-mass must move at speed less than c.

    I understand that there are a multitude of physical reasons that support the above conjecture. But is there a good (mathematical) reason why the boundary condition at [itex]\sigma = 0[/itex] (not [itex]\tau,[/itex] as that becomes a useless parameter in this case) the particle cannot be travelling at c already?

    Thanks in advance.
     
    Last edited: Sep 10, 2006
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  3. Sep 10, 2006 #2

    HallsofIvy

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    Surely that's a misprint- you meant "if someone could prove that all particles with positive rest-mass must move at speed less than c".

    An object with positive rest-mass, moving at speed c would have infinite gravitational mass, disrupting the entire universe!
     
  4. Sep 10, 2006 #3
    Thank you for the correction; I have subsequently corrected my original post.

    I am aware of the various physical arguments for the above conjecture. Howeever, I would prefer an answer relying on the mathematical machinery of SR only...
     
  5. Sep 10, 2006 #4

    robphy

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    There are probably many demonstrations of this... which may or may not be satisfactory to you.

    One approach is this:
    a particle with positive rest-mass has a future-timelike 4-velocity, which means that its tangent vector always points into the interior of its future light cone... it can't travel at the speed of light.

    Another approach:
    in terms of the rapidity parameter, that is, the Minkowski arc-length on the future unit hyperbola,
    the asymptotic point corresponding to its light-cone is infinitely far away from any point on that hyperbola, corresponding to the tip of a future-timelike 4-vector.
     
  6. Sep 10, 2006 #5

    HallsofIvy

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    The "mathematical machinery" of SR? Like the Lorenz transformation formulas? That's exactly what says that if a body with non-zero rest mass were to go at light speed, its relativistic mass would be infinite. No such infinite mass has ever been observed!
     
  7. Sep 10, 2006 #6
    Well, the "infinite gravitational mass" argument doesn't really answer your question because you said: is there a good (mathematical) reason why the ... the particle cannot be travelling at c already?

    To answer your question, there is nothing in SR's postulates that prevents "faster than light" particles. In fact, they have been written about quite a bit (see http://en.wikipedia.org/wiki/Tachyons for an introduction). However if you allow "physical arguements", I believe they would destroy an "invariant" concept of causality.

    Some consider the notion of time order so important in physics, that they feel this DOES count as violating SR ( http://scienceworld.wolfram.com/physics/Tachyon.html ). However, many believe the fundemental equations of physics make no distinction of time order (all current fundemental laws already have this feature). So I would argue that strictly SR doesn't require the non-existance of tachyons (its postulates don't preclude it), but to preserve an "invariant" causality they would require you to adopt the notion of a "preferred" frame. Some physicists even go so far as to argue that since the physics would still be the same in each frame, it is just a "reinterpretation" of events, not a break of causality (but that sounds to me more like redefining the word).
     
    Last edited: Sep 10, 2006
  8. Sep 11, 2006 #7

    jtbell

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    Using the definitions of relativistic momentum and energy

    [tex]p = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}[/tex]

    [tex]E = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}[/tex]

    one can derive the following expression for the speed:

    [tex]\frac{v}{c} = \frac{pc}{E}[/tex]

    Going a step further, using [itex]E^2 = (pc)^2 + (m_0 c^2)^2[/itex], gives us

    [tex]\frac{v}{c} = \frac{\sqrt{E^2 - (m_0 c^2)^2}}{E} = \sqrt{1 - \left(\frac{m_0 c^2}{E}\right)^2}[/tex]

    Therefore, if [itex]m_0 > 0[/itex], then [itex]v/c < 1[/itex], that is, [itex]v < c[/itex].
     
    Last edited: Sep 11, 2006
  9. Sep 11, 2006 #8
    Thanks for all your responses. I have some further questions.

    robphy:
    I liked your first approach (and not entirely sure if I understood your second approach). However it led to the following question: why must all massive particles move on timelike paths?

    jtbell:
    Haven't your definitions for p and E assumed that the particle was massive in the first place (thus ensuring that v < c)? I ask this because those definitions of p and E don't work for relativistic particles.

    ---

    I understand that particles on timelike/spacelike/null paths remain on timelike/spacelike/null paths. I suppose, then, my new question is: is there a mathematical reason (in the context of SR kinematics only) why all massive particles are travelling along timelike paths in the first place?
     
  10. Sep 11, 2006 #9

    jtbell

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    I think in the second sentence you meant to say "massless" rather than "relativistic."

    Nevertheless, your original question was specifically about massive particles!

     
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