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Extreme case of relativitic momentum

  1. Jan 8, 2009 #1

    KFC

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    The relativitic momentum is given by

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

    or

    [tex]
    p = \frac{\sqrt{E^2 - m_0^2c^4}}{c}
    [/tex]
    where [tex]m_0[/tex] is the rest mass.

    I have two questions about the momentum

    1) Are the momentum given by these two expression the same?

    2) In extreme case (when v->c), the first expression will give infinite large momentum? How would that happen? If we use the second expression to find the momentum in extreme case, what will energy E there look like? Anyway, I just wonder how to find the momentum in extremem case.

    Thanks
     
  2. jcsd
  3. Jan 8, 2009 #2
    Start by expressing the momentum of a particle as

    p=p'(1+V/u')/(1-VV/cc)^1/2 (1)
    In the limi case u'=c
    p=p'sqr[(1+V/c)/1-V/c]1/2 (2)
    equation (2) performing the transformation of the momentum of a photon.
     
  4. Jan 8, 2009 #3

    KFC

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    Thanks for your reply. But I have no idea how do you get this. What does u', p' and V represent?

     
  5. Jan 8, 2009 #4

    jtbell

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    Staff: Mentor

    Small correction: in the equation above, [itex]m[/itex] should be [itex]m_0[/itex].

    Yes. To show this, you need to use the following equation for energy:

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

    As [itex]v \rightarrow c[/itex], [itex]\sqrt{1 - v^2 / c^2} \rightarrow 0[/itex]. But this is in the denominator, so the whole expression [itex]\rightarrow \infty[/itex].

    The expression I gave for E also has [itex]\sqrt{1 - v^2 / c^2}[/itex] in its denominator, so [itex]E \rightarrow \infty[/itex] also.
     
  6. Jan 8, 2009 #5

    KFC

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    Thanks. But I think the momentum shouldn't be infinite large even when v->c. For example, a photon is moving at the speed of light, it still have a finite wavelength so the finite momentum, right? My question is how to find the finite momentum for extreme case.
     
  7. Jan 8, 2009 #6

    jtbell

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    Staff: Mentor

    Your first formula for p, and the formula I gave for E, are not usable for photons, which have [itex]m_0 = 0[/itex] and [itex]v = c[/itex], because they both give 0/0 which is undefined mathematically.

    Your second formula for p, on the other hand, is usable for photons, and gives p = E/c.
     
  8. Jan 8, 2009 #7

    clem

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    Science Advisor

    The general equation v=pc/E, may clarify things for you.
     
  9. Jan 8, 2009 #8

    Vanadium 50

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    The equation I find most useful is (E)2 - (pc)2 = (mc2)2. This is valid for all particles at all velocities.
     
  10. Jan 8, 2009 #9
    o.k. Consider that the same tardyon (a particle the speed of which never exceeds c) is detected from the inertial reference frames I and I'. I' moves with constant speed V in the positive direction of the x,x' axes/ Let p and E be momentum and the energy of the tardyon when detected from I and I' respectively. The tardyon moves with speed u relative to I and with speed u' relative to I'. The momentum p and the energy of the tardyon are related in I by
    p=Eu/cc (1)
    and I' by
    p'=E'u'/cc (2)
    p and p' are related by the transformation equation [1]
    p=[p'+VE'/cc]/sqr(1-VV/cc)]= p'[1+V/u']]/sqr(1-VV/cc) (3)
    Replacing the tardyon considered so far by a photon (u'=c) (3) becomes
    p=p'sqr[(1+V/c)/(1-V/c) (4)
    equation (4) performing the transformation of the momentum of photons.
    [1] Robert Resnick, Introduction to Special Relativity (John Wiley and Sons New York, 1968) p.144 (Probably there are newer editions)
     
  11. Jan 10, 2009 #10
    For question 1 my answer is: indeed these two equations are of the same family. Since T=mc^2-m_0c^2=E-m_c^2,
    you can refrase the formula for kinetic energy without the huge potential energy.
    Perhaps that will help?
     
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