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Proper acceleration of cyclotron particle.

  1. Jun 1, 2012 #1
    Hey! I've been working on a problem here for which I have found the four acceleration of a particle going in a circle with radius R to be

    [tex]\underline{A} = -R\omega^2 \gamma^2 (0,\cos(\omega \gamma \tau), \sin(\omega \gamma \tau), 0)[/tex]

    and in going from this to find the proper acceleration I reasoned as follows: The four acceleration is valid in all inertial frames and in the instantaneous rest frame of the particle it self it will have a spatial four acceleration which is the proper acceleration of the particle, i.e

    [tex]\underline{A} = (0, \vec a_0)[/tex]
    in the instantaneous particle. In this frame [itex]\gamma[/itex]=1 so i reasoned that if i just sett [itex]\gamma[/itex]=1 in the above expression I get the proper acceleration. I thus obtain

    [tex]\vec a_0 = -\omega ^2 R(\cos(\omega \tau), \sin(\omega \tau),0)[/tex]
    which means that
    [tex] a_0 = \omega^2 R.[/tex]
    but laboratory frame of the cyclotron
    [tex]v = \omega R \ \ \ \ a = \frac{v^2}{R}[/tex]
    so that
    [tex] a_0 = a[/tex]
    So this is clearly wrong.
    However another line of reasoning would be to say that in the instantaneous rest frame of the particle
    [tex]\underline{A} = (0, \vec a_0)[/tex]
    so the invariant norm is equal to

    [tex]\vec a_0^2 = A_\mu A^\mu = (R\omega^2 \gamma^2)^2[/tex]
    such that
    [tex] a_0 = R\omega^2 \gamma^2 = \gamma^2 a[/tex]
    Which is correct. So my question is, what is wrong about the first line of reasoning?
     
  2. jcsd
  3. Jun 1, 2012 #2
    I think I spotted my error. The four acceleration is not valid in all inertial frames. One must transform it to another reference frame in order to use it there.
     
    Last edited: Jun 1, 2012
  4. Jun 1, 2012 #3
    Less than 5 minutes later you spotted your error? Ok.
     
  5. Jun 1, 2012 #4
    Yes, i'm sorry. It is amazing how your subconsciousness works on the problem when you are trying to formulate it.
     
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