pervect said:
I've come up with a somewhat simpler approach for presenting the solution for the EF geodesic equations.
Let r, v be the Eddington-Finklesein (EF) coordinates, which are presumed to be functions of proper time \lambda. Then let:
\dot{r} = \frac{dr}{d\lambda} \hspace{3cm} \dot{v} = \frac{dv}{d\lambda}
In EF coordinates, ##\dot{v} = {dv}/{d\lambda}## can be interpreted as the redshift from a light source originating at infinity and propagating radially, this is directly observable as the amount of doppler shift of the light source of a known spectral line of the source at infinity.
The Ingoing Eddington Finklestein metric (gemoetrized) is
http://en.wikipedia.org/w/index.php?title=Eddington–Finkelstein_coordinates&oldid=516198830
-(1-2m/r) dv^2 + 2\,dv\,dr
The Christoffel symbols are:
\Gamma^{v}{}_{vv} = \frac{m}{r^2}
\Gamma^{r}{}_{vv} = \frac{m(1-2m/r)}{r^2}
\Gamma^{r}{}_{vr} = \Gamma^{r}{}_{rv} = -\frac{m}{r^2}So we can write the geodesic equations as
<br />
\ddot{v} + \Gamma^{v}{}_{vv} \dot{v}^2 = 0<br />
<br />
\ddot{r} + \Gamma^{r}{}_{vv} \dot{v}^2 + 2 \,\Gamma^{r}{}_{vr} \dot{v} \dot{r} = 0<br />
It will be convenient to solve for ##\dot{r}## and ##\dot{v}## as a function of r, rather than the proper time ##\lambda##To this end, we use the chain rule:
\ddot{r} = \frac{d^2 r}{d \lambda^2} = \frac{d \dot{r}}{d r} \frac{d r}{d \lambda} = \frac{d \dot{r}}{d r} \dot{r}
\ddot{v} = \frac{d^2 v}{d \lambda^2} = \frac{d \dot{v}}{d \lambda} = \frac{d \dot{v}}{dr}\frac{dr}{d\lambda} = \frac{d \dot{v}}{dr} \dot{r}
Then we can write the geodesic equations as:
<br />
\ddot{v} + \Gamma^{v}{}_{vv} \dot{v}^2 = \frac{d \dot{v}}{dr} \dot{r} + \frac{m}{r^2} \dot{v}^2 = 0<br />
<br />
\ddot{r} + \Gamma^{r}{}_{vv} \dot{v}^2 + 2 \,\Gamma^{r}{}_{vr} \dot{v} \dot{r} = \frac{d \dot{r}}{dr} \dot{r} + (1-2m/r)\frac{m}{r^2} \dot{v}^2 - 2 \frac{m}{r^2} \dot{r}\dot{v} = 0<br />
Solving this, one way of expressing the general solution is
<br />
\dot{v} = C\frac{1 \pm \sqrt{2m/r} \sqrt{1-\frac{r-2m}{r_{max} - 2m}}}{1-2m/r}<br />
<br />
\dot{r} = -\frac{m \dot{v}^2}{r^2 \frac{d \dot{v}}{dr}}<br />
where C and ##r_{max}## are constants, and the solution has the property that ##\dot{r}=0## at ##r=r_{max}##.
Infalling geodesics use the minus sign - this can be confirmed by solving for ##\dot{r}## and taking the limit as r->2m and confirming that only this choice makes ##\dot{r}## negative, which is required for an infalling geodesic.
<br />
\dot{v} = C\frac{1 - \sqrt{2m/r} \sqrt{1-\frac{r-2m}{r_{max} - 2m}}}{1-2m/r}<br />
If we take ##r_{max}## = infinity, and let ##\dot{v}## be unity at infinity, representing a fall from rest at infinity (sometimes called a Lemaitre observer) then we get the expected solution:
<br />
\dot{v} = \frac{1-\sqrt{2m/r}}{1-2m/r} = \frac{1-\sqrt{2m/r}}{(1-\sqrt{2m/r})(1+\sqrt{2m/r}) } = \frac{1}{1+\sqrt{2m/r}}<br />
If we take a finite ##r_{max}## and set ##r = r_{max}##, we find that
<br />
\dot{v}|_{r=r_{max}} = \frac{C}{1-2m/r}<br />
so, it appears that the correct value for C is ##\sqrt{1-2m/r_{max}}## as the doppler shift at ##r=r_max## should be the same as the Schwarzschild gravitational time dilation at the same r.
As a partial check on this: the killing vector ##\xi^a## should be ##\frac{\partial}{\partial v}##. This makes ##\xi_a=(-1+2m/r) \partial v + \partial r## Since ##u^a \xi_a## should be a constant along the curve, ##u^a## being the four-velocity, (-1+2m/r) \dot{v} + dot{r} should be a constant along the curve. Also, said constant should be equal to the "energy at infinity".
The value of this conserved constant evaluated in the limit as r->2m and again as r->r_max is just C. So we do expect C<1 for a fall from at rest from a finite r_max, and C=1 for a fall from at rest from infinity.
This yields the final expression for the radial doppler shift to infinity as a function of r for a body starting at rest at ##r=r_{max}##:
<br />
\dot{v} =\frac{ \sqrt{1-2m/r_{max}}}{1-2m/r } \left( {1 - \left( \sqrt{2m/r} \right) \left( \sqrt{1-\frac{r-2m}{r_{max} - 2m}} \right) } \right) <br />
