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Cyclotron formula and Transverse Mass

  1. Oct 24, 2003 #1

    pmb

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    The cyclotron forumla is derived here
    See -www.geocities.com/physics_world/cyclotron.htm

    That's a lot of work for something so simple. Recall that the relativistic force can be written as
    equation. Recall that

    F = M_t A_t + A_L M_L

    where

    M_t = transverse mass
    A_t = transverse acceleration

    M_L = longitudinal mass
    A_L = longitidinal acceleration

    It can be shown that transverse mass = (relativistic) mass = m. For the cyclotron motion described in above link A_L = 0 and a = A_t = v^2/r. Therefore

    F = dp/dt = M_t A_t = m A_t = ma

    where a = transverse acceleration = v^2/r and p = mv = gamma*m_o*v. Therefore it follows that

    F = ma = e[E + (v/c)xB]

    So, in this particular instance, F = ma is 100% correct if m = relativistic mass and a is transverse acceleration. Plugging vxB = vB (B is now a magnitude rather than a vector)

    ma = m(v^2/r) = vB

    Canceling a factor of v and moving the r to the otherside gives

    mv = vBr

    p = m_o*v/sqrt[1-(v/c)^2] = mv

    p = vBr

    which is the relativistically correct relation for cyclotron motion.

    Someone claims that this is not simpler than the normal derivation derived above [Then again the same people didn't know that F = q[E + vxB] either]. Does anyone know of a simpled/easier way to derive the cyclotron formula?

    Pete
     
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