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Relativistic cyclotron frequency

  1. May 16, 2008 #1
    Since the acceleration is transverse to the velocity, should we consider the transverse mass in the formula [itex]mv^2/r[/itex] ie

    [itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?
  2. jcsd
  3. May 17, 2008 #2
    How can you express the centripetal force in terms of momentum (with other terms)?
    How is the relativistic momentum?
  4. May 17, 2008 #3
    Yes. But just to be clear, using your symbols, the transverse mass = gamma*m.

  5. May 17, 2008 #4


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    It's clearer in terms of momentum pv/r=qvB-->p=qBr.
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