OK, if I understand correctly, because of the t' contained in the cos and sin functions in the last terms, the expressions for (6) should be:
\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - R\gamma^2\omega^2\cos(\gamma\omega t')\right) (Eq6x)
\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right) (Eq6y)
Is that correct? As I said before, I am no math expert. Should the same be done for the other terms in the expression?
If they are correct, then when t' = 0, \sin(\gamma\omega t') = 0 and \cos(\gamma\omega t') = 1 and the equations simplify to:
a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}-\frac{dy'}{dt'}\gamma^{2}\omega- R\gamma^2\omega^2\right) (Eq7x)
a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(\gamma\omega \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\right) (Eq7y)
Now if the particle is stationary in the rotating frame and located on the y'=0 axis, which is the radial line from the centre of rotation to the particle, the result is:
a_x = \frac{d^2x}{dt^2}= -R\omega^2\right)
a_y = \frac{d^2y}{dt^2}= 0
which is a good result, because that is what we expect in the non-rotating lab frame. This is the coordinate acceleration.
Equations (7x) and (7y) can be rearranged to give:
a'_x = \frac{d^2x'}{dt'^2}=\gamma^{2}\left(\frac{d^2x}{dt^2}+\frac{dy'}{dt'}\omega+ R\omega^2\right)
a'_y = \frac{d^2y'}{dt'^2}=\gamma\frac{d^2y}{dt^2}-\omega \frac{dx'}{dt'}
If the particle is released so that it flies off in a straight line from the point of view of the lab frame at time t’=0, its initial velocity in the rotating frame was zero so that upon release dx\prime/dt\prime \approx 0 and dy\prime/dt\prime \approx 0 then:
a'_x = \frac{d^2x'}{dt'^2} =\gamma^{2}\left(\frac{d^2x}{dt^2}+\frac{dy'}{dt'}\omega+ R\omega^2\right) <br />
=\gamma^{2}\left(0+\frac{dy'}{dt'}\omega+ R\omega^2\right)<br />
\approx \gamma^{2}\left(R\omega^2\right)
a'_y = \frac{d^2y'}{dt'^2}=\gamma\frac{d^2y}{dt^2}-\omega \frac{dx'}{dt'}<br />
=0-\omega \frac{dx'}{dt'} \approx 0
So in the lab frame the centripetal acceleration of the particle when it is traveling in a circle is R\omega^2 and when it is released the acceleration of the particle according to the observer in the rotating frame is initially \gamma^{2} R\omega^2. This magnitude of this measurement coincides with with the magnitude of the four acceleration given by Pervect and Dalespam (Thanks for the imput by the way
), which is the proper acceleration as measured by an accelerometer.
Note:The x and x' axes always pass through the center of rotation so a_x and a_x' represent the inward pointing centripetal acceleration at time t'=0, which these calculations are based on. a_y is the angular acceleration directed tangentially.
Conclusion
Coordinate centripetal acceleration:
a_x = R\omega^2 (Proven by Starthaus)
Proper centripetal acceleration:
As measured by an accelerometer in the rotating frame:
a\prime_x = \gamma^2 R\omega^2 (Proven by Dalespam and Pervect)
As measured by initial “fall rate acceleration” in the rotating frame:
a\prime_x \approx \gamma^2 R\omega^2 (Proven by Starthaus)
The above is basically in agreement with everything I said in #1 but the methods of measurement have been more carefully defined now.
This has significance because it tells us how a particle behaves in a gravitational field, by applying the equivalence principle.
