starthaus said:
a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'}(\frac{dr'}{dt'})=\frac{d}{dt}(\frac{dr'}{dt'})\frac{dt}{dt'}=<br />
\gamma_g\frac{d}{dt}(\gamma_g^2\frac{dr}{dt})=\gamma_g^3\frac{d^2r}{dt^2}+2(\gamma_g\frac{dr}{dt})^2\frac{d\gamma_g}{dr}
OK, thanks for that. It agrees with the expansion given by Altabeh although I am not sure if Altabeh just expanded on your differentiation. Anyway, I will expand on your result in a different way to obtain a result in terms of radial velocity.
a' = a\:\gamma_g^3+2\left(\gamma_g\frac{dr}{dt}\right)^2\frac{d\gamma_g}{dr} = a\:\gamma_g^3+2 \left(\gamma_g\frac{dr}{dt}\right)^2\left( \frac{-M }{c^2r^2}\right)\gamma_g^3 = a\:\gamma_g^3\: - \:\frac{2M}{r^2} \left(\frac{dr}{dt}\right)^2 \gamma_g^5 = a\:\gamma_g^3 -\frac{2M}{r^2} \left(\frac{dr'}{dt'}\right)^2 \gamma_g
Now we should be in a position to correct my previous derivation of the general expression for proper radial acceleration.
The following is a rehash of my first attempt in #318 (using units of G=c=1):
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For easy visualisation, consider the case of a small rocket descending into a gravitational well, with its thrust (Fo) directed in the opposite direction to the local gravitational acceleration (a'_g) so that the descent acceleration of the rocket at r, as measured by a stationary observer at r, is d^2r'/dt'^2 using his local clocks and rulers. (All primed variables indicate measurements made by a stationary local observer at r in the Schwarzschild coordinates). I can now say:
F_o = \left( \frac{d^2r'}{dt' ^2} - a'_g \right) m_o
where Fo is the proper force directed upwards and m_o is the rest mass of the rocket .
Dividing both sides by m_o gives the proper acceleration a_o of the rocket as:
a_o = \frac{d^2r'}{dt' ^2} - a'_g
where positive acceleration is defined here as upwards and negative acceleration means downwards.
A combination of http://www.mathpages.com/rr/s6-07/6-07.htm" gives the velocity dependent Schwarzschild coordinate gravitational acceleration a_g as:
a_g = - \frac{M}{r^2}(1-2M/r)\left(1-\frac{3(dr/dt)^2}{(1-2M/r)^2}\right) = - \frac{M}{r^2}(1-2M/r)(1-3(dr'/dt')^2)
Assuming the following relationships between the local vertical measurements (dr’, dt’, a’) made by a stationary observer at r and the coordinate measurements (dr, dt, a) made by a Schwarzschild observer at infinity using the gravitational boost factor \gamma_g = 1/\sqrt{(1-2M/r)}:
dr' = dr \; \gamma_g
dt' = dt \; \gamma_g^{-1}
dr'/dt' = dr/dt \; \gamma_g^{2}
a' = a\: \gamma_g^3 -\frac{2M}{r^2} \left(\frac{dr'}{dt'}\right)^2 \gamma_g
it is fairly easy to obtain:
a'_g = - \frac{M}{r^2}\left(\frac{1-(dr'/dt')^2}{\sqrt{1-2M/r}}\right)
where dr’/dt’ is the descent velocity of the rocket as it passes radial coordinate r, as measured by a stationary observer located at r.
The more general proper acceleration can now be expressed in the form:
a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{d^2r'}{dt' ^2} + \frac{M}{r^2}\left(\frac{1-(dr'/dt')^2}{\sqrt{1-2M/r}}\right)
When the rocket is holding position at r by using its thrust to hover at r, dr’/dt’ = 0 and d^2r'/dt' ^2 =0 and its proper acceleration is:
a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{M}{r^2}\frac{1}{\sqrt{1-2M/r}}
which is the expected result.
When the rocket turns off its engines and falls freely in a vacuum, the proper acceleration of the rocket is:
a_o = \frac{d^2{r}'}{d{t}' ^2} - a'_g = 0
for all R>2m and r>2m.
That, I hope is now a correct result, but I can not find any references to check it against for the case where the particle is neither free falling nor stationary.
, not 4-vectors. I will use the metric, Christoffel symbols, etc. given https://www.physicsforums.com/showpost.php?p=2712746&postcount=38". The worldline of a particle moving only in the radial direction in these coordinates is given by: