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Homework Help: 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.
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