- #1
center o bass
- 545
- 2
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?
\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?
