1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Slow speed?

  1. Aug 1, 2007 #1
    1. The problem statement, all variables and given/known data
    On page 41 of Ryder's QFT, just below eqn (2.84), it says: [tex]\gamma = E/m[/tex] I was unable to verify this, unless it is meant to be true only for small speeds.

    2. Relevant equations
    [tex]E = \pm(m^2c^4 + p^2c^2)^{1/2}[/tex] (2.24) page 29, but as suggested n the book, we let c = 1, so
    [tex]E = \pm(m^2 + p^2)^{1/2}[/tex]

    3. The attempt at a solution
    Well, I'm not sure this is legal, but I replaced p with mv as in the classical case, and then v with [tex]\beta[/tex].
    [tex]E = \pm(m^2 + m^2v^2)^{1/2} = \pm m(1 + \beta^2)^{1/2} \approx \pm m\gamma[/tex]
    But that approximation is only good when v is small.
  2. jcsd
  3. Aug 1, 2007 #2
    In relativistic physics momentum is not given by classical formula [itex] p = mv [/itex]. Try using

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

  4. Aug 1, 2007 #3
    Got it. Thanks Eugene. Now I have:
    [tex]E = \pm(m^2 + \frac{m^2v^2}{1 - v^2})^{1/2} = \pm m(\frac{1 - v^2 + v^2}{1 - v^2})^{1/2} = \pm m\gamma[/tex]
    Now my only problem is the matter of the [tex]\pm[/tex].
  5. Aug 1, 2007 #4
    You should always take the positive sign. Energy is positive, by definition.

  6. Aug 1, 2007 #5
    Thanks Eugene.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?