# Proper acceleration of cyclotron particle.

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1. Jun 1, 2012

### center o bass

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

$$\underline{A} = -R\omega^2 \gamma^2 (0,\cos(\omega \gamma \tau), \sin(\omega \gamma \tau), 0)$$

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

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

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

$$\vec a_0^2 = A_\mu A^\mu = (R\omega^2 \gamma^2)^2$$
such that
$$a_0 = R\omega^2 \gamma^2 = \gamma^2 a$$
Which is correct. So my question is, what is wrong about the first line of reasoning?

2. Jun 1, 2012

### center o bass

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
3. Jun 1, 2012

### dimension10

Less than 5 minutes later you spotted your error? Ok.

4. Jun 1, 2012

### center o bass

Yes, i'm sorry. It is amazing how your subconsciousness works on the problem when you are trying to formulate it.