Space-like trajectory in Schwarzschild spacetime

[crime9894]

Homework Statement

For Schwarzschild metric, consider a mass M following space-like trajectory.
The relativistic energy of the mass at infinity is E.
Show that there is smallest value of r, the radial coordinate, reachable by the mass, and find its value.

Relevant Equations

As shown below

I'm not sure how to approach this question.

So I start off with the fact the path taken is space-like,
$$ds^2>0$$
Input the Schwarzschild metric,
$$−(1−\frac{2GM}{r})dt^2+(1−\frac{2GM}{r})^{−1}dr^2>0$$
Where I assume the mass doesn't move in angular direction.
How should I continue?

 

[PeroK]

Homework Statement:: For Schwarzschild metric, consider a mass M following space-like trajectory.
The relativistic energy of the mass at infinity is E.
Show that there is smallest value of r, the radial coordinate, reachable by the mass, and find its value.
Relevant Equations:: As shown below

I'm not sure how to approach this question.

So I start off with the fact the path taken is space-like,
$$ds^2>0$$
Input the Schwarzschild metric,
$$−(1−\frac{2GM}{r})dt^2+(1−\frac{2GM}{r})^{−1}dr^2>0$$
Where I assume the mass doesn't move in angular direction.
How should I continue?

Are you assuming any space-like path or space-like geodesics?

 

[vela]

Did you mean to use $M$ in the metric? Isn't $M$ the mass of the moving body?

 

[ergospherical]

I assume that the limiting case of these trajectories is a null trajectory, i.e. a trajectory followed by a light ray. As per usual, it would help to cast the problem in terms of an effective potential. Assuming the trajectory is confined to the equatorial plane $\theta \equiv \pi/2$ (which is always possible to achieve by means of a rotation of the coordinate system), write\begin{align*}
0 = g_{\mu \nu} u^{\mu} u^{\nu} &= g_{tt} \left( \dfrac{dt}{d\tau} \right)^2 + g_{rr} \left( \dfrac{dr}{d\tau} \right)^2 + g_{\phi \phi} \left( \dfrac{d\phi}{d\tau} \right)^2 \\
&= -f(r) \left( \dfrac{dt}{d\tau} \right)^2 + f(r)^{-1} \left( \dfrac{dr}{d\tau} \right)^2 + r^2 \left( \dfrac{d\phi}{d\tau} \right)^2 \ \ \ (\dagger)
\end{align*}with $f(r) \equiv 1-2m^*/r$. Recall that this metric implies two Killing vectors $\xi = \partial / \partial t$ and $m = \partial / \partial \phi$, and therefore two conserved quantities \begin{align*}
E &\equiv -u \cdot \xi = -g_{\mu \nu} u^{\mu} \xi^{\nu} = -g_{tt} u^t \xi^t = f(r) \dfrac{dt}{d\tau} \ \ \ (\sim \mathrm{energy})\\
L &\equiv u \cdot m = g_{\mu \nu} u^{\mu} m^{\nu} = g_{\phi \phi} u^{\phi} m^{\phi} = r^2 \dfrac{d\phi}{d\tau} \ \ \ (\sim \mathrm{angular \ momentum})
\end{align*}which allows you to substitute $\dfrac{dt}{d\tau} = \dfrac{E}{f(r)}$ and $\dfrac{d\phi}{d\tau} = \dfrac{L}{r^2}$ in $(\dagger)$, \begin{align*}
\boxed{\dfrac{1}{2} \dot{r}^2 + \dfrac{L^2 f(r)}{2r^2} = \dfrac{1}{2} E^2 }
\end{align*}with $\dfrac{L^2 f(r)}{2r^2} \equiv V(r)$ an effective potential. Can you figure out how to use this equation to investigate the positions $r_0$ of closest approach?