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Probably Fairly Simple Special Relativity Calculation

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to show that the velocity of an ultra-relativistic particle can be approximated by the following expressions:

    [tex]

    v \approx c \left[1-\frac{1}{2}\left(\frac{mc^2}{E}\right) ^2 \right]

    [/tex]

    and


    [tex]

    \frac{1}{v} \approx \frac{1}{c} \left[1+\frac{1}{2}\left(\frac{mc^2}{E}\right) ^2 \right]

    [/tex]

    ...but I'm struggling. I know it can't be that hard, but I just can't quite get there. It seems I can get close, but I miss the factor of 1/2 out the front of the mass/energy term. I'm definitely doing something wrong, so any help would be much appreciated. Here's how I went about trying to obtain the first expression:

    3. The attempt at a solution

    So we have our usual relativistic expression:

    [tex]
    E^2 = p^2 c^2 + m^2 c^4
    [/tex]

    Then, I divided through by m^2 c^4 to get:

    [tex]
    \frac{E^2}{(mc^2)^2}-1 = \frac{p^2 c^2}{(mc^2)^2}
    [/tex]

    Now, using a relation that I pulled from somewhere - [tex] (pc)^2 = E^2 \frac{v^2}{c^2}[/tex]

    I simply sub that in for (pc)^2, getting

    [tex]
    \frac{E^2}{(mc^2)^2}-1 = \frac{E^2}{(mc^2)^2}. \frac{v^2}{c^2}
    [/tex]

    and dividing through by [tex]\frac{E^2}{(mc^2)^2}[/tex], get

    [tex]
    \frac{v^2}{c^2} = 1 - \frac{(mc^2)^2}{E^2}
    [/tex]

    so

    [tex]
    \frac{v}{c} = 1 - \frac{(mc^2)^2}{E^2}
    [/tex]

    i.e

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

    Hmmmm - the problem is that a) this is an exact equality not an approximation, b) I'm missing the damn factor of 1/2 out the front of the mass-energy fraction, and c) nowhere have I used the fact that the particle is highly relativistic. These facts are almost certainly all related, but I can't see how and I've been staring at it for some time now.

    Any help or suggestions, or more ideas for more rigorous derivations would be exceedingly welcome!
     
  2. jcsd
  3. Sep 14, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I didn't look over the rest of your derivation, but this step:
    ...doesn't follow. (You forgot to take the square root of the RHS!)

    Hint: Use a Taylor series expansion to approximate the square root of 1 - x, where x << 1.
     
  4. Sep 14, 2009 #3
    "(You forgot to take the square root of the RHS!)"

    Der! Must be tired - or just plain dumb...

    Thanks a lot Doc Al - I got it out easily after that little hint. I appreciate it a lot!
     
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