General equation for coordinate acceleration
Moving on to the general case for coordinate acceleration as promised, this is the derivation based on the one I started in post #48.
Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that \theta = \pi/2 and d\theta = 0
ds^2=\alpha dt^2-dr^2/\alpha-r^2d\phi^2, (\alpha=1-2M/r)
Divide both sides by \alpha dt^2 and rearrange so that the constant (1) is on the LHS:
L = 1 =\frac{1}{\alpha}\frac{ds^2}{dt^2} +\frac{1}{\alpha^2}\frac{dr^2}{dt^2}+\frac{r^2}{\alpha}\frac{d\phi^2}{dt^2}
The metric is independent of \phi and t, so there is a constant associated with coordinate angular velocity (H_c) which is obtained by finding the partial derivative of L with respect to d\phi/dt
\frac{\delta{L}}{\delta(d\phi/dt)} = \frac{r^2}{\alpha} \frac{d\phi}{dt} = H_c
The metric is independent of s and t, so there is a constant associated with time dilation (K_c) which is obtained by finding the partial derivative of L with respect to ds/dt
\frac{\delta{L}}{\delta(ds/dt)} = \frac{1}{\alpha} \frac{ds}{dt} = K_c
Substitute these constants into the equation for L
1 =\alpha K_c^2 +\frac{1}{\alpha^2}\frac{dr^2}{dt^2}+\frac{\alpha}{r^2}H_c^2}
and solve for (dr/dt)^2:
\frac{dr^2}{dt^2} = (1-2M/r)^2 -(1-2M/r)^3 (K_c^2 + H_c^2/r^2)
Differentiate the above with respect to r and divide by 2 to obtain the general coordinate radial acceleration of a freefalling particle in the metric:
\frac{d^2r}{dt^2}= \frac{\alpha^2 H_c^2}{r^4}(r-5M) +\frac{M}{r^2}(2\alpha-3K_c^2\alpha^2)
Re-inserting the full forms of H_c and K_c back in gives:
\frac{d^2r}{dt^2}= <br />
\frac{d\phi^2}{dt^2}<br />
(r-5M) +\frac{M}{r^2}(2\alpha-3<br />
\frac{ds^2}{dt^2} <br />
)
Now the (ds/dt)^2 term is a little inconvenient, but we can find an alternative form by solving the Schwarzschild metric to directly obtain:
(ds/dt)^2 = (\alpha - dr^2/(\alpha dt^2) - r^2d\phi^2/dt^2)
and substituting this form into the equation above it to obtain:
\frac{d^2r}{dt^2}= <br />
\frac{d\phi^2}{dt^2}<br />
(r-5M) +\frac{M}{r^2}(2\alpha-3<br />
(\alpha - dr^2/(\alpha dt^2) - r^2d\phi^2/dt^2)<br />
)
which after a bit of algebra simplifies to:
\Rightarrow \frac{d^2r}{dt^2}= \alpha\left(<br />
r \frac{d\phi^2}{dt^2}-\frac{M}{r^2}+<br />
\frac{3M}{\alpha^2 r^2} \frac{dr^2}{dt^2}\right)
This is the fully general form of acceleration in Schwarzschild coordinates in terms of coordinate time and gives Espen something to compare his results against in his latest document.
Some quick checks:
For the special case of circular motion, the radial acceleration is zero and dr/dt=0 and so:
0 = \alpha\left(r \frac{d\phi^2}{dt^2}-\frac{M}{r^2}\right)
\Rightarrow \frac{d\phi}{dt} = \sqrt{\frac{M}{r^3}}
Passed the first test.
For the special case of a particle at apogee, dr/dt=0 (momentarily) and d\phi/dt=0 and:
\frac{d^2r}{dt^2}= -\frac{M}{r^2}\alpha
Passed the second test.
For the special case of a particle in purely radial free fall d\phi/dt=0 and:
\frac{d^2r}{dt^2}= \alpha\left(-\frac{M}{r^2}+<br />
\frac{3M}{\alpha^2 r^2} \frac{dr^2}{dt^2}\right)
\Rightarrow \frac{d^2r}{dt^2}= -\frac{M}{r^2}\left(\alpha -<br />
\frac{3}{\alpha } \frac{dr^2}{dt^2}\right)
Passed the third test.
See equations (4) and (5) of mathpages
http://www.mathpages.com/rr/s6-07/6-07.htm for an alternative proof of this last equation.
