Take a rocket travelling in a circle at less than $1.5R_s$ away from a black hole. At that distance, no orbit is possible and the rocket must produce continuous force to stay outside the black hole. Travelling in a circle can reduce this force, but not eliminate it. In addition, if the rocket is travelling a near the speed of light, time dilation will cause the proper acceleration from the perspective of the rocket to be MUCH larger than from the perspective of a distant observer.

At what speed will the required proper acceleration be minimised?


1 Answer 1


blademan9999 asked: "At what speed will the required proper acceleration be minimised."

In that example with a circular orbit in the $\theta$ direction, the coordinate acceleration is

$$\rm \frac{d^2 x^{\mu}}{d \tau^2}=\{ \ddot{t}, \ \ddot{r}, \ \ddot{\theta}, \ \ddot{\phi}\}=\{0,\ 0, \ 0, \ 0\}$$

and the 4-velocity

$$\rm \frac{d x^{\mu}}{d \tau}=u^{\mu}=\{ \dot{t}, \ \dot{r}, \ \dot{\theta}, \ \dot{\phi}\}=\{ \gamma \surd g^{tt}, \ 0, \ v \gamma /r, \ 0\}$$

so the force required for that is in natural units

$$\rm F= \surd \ |\sum_{\mu, \nu} g_{\mu \nu} \ a^{\mu} a^{\nu}| = \surd \ |\frac{(r-2) \ v^2-1}{\sqrt{(r-2) \ r^3 \left(v^2-1\right)^2}}|$$

where $\rm v$ is the local velocity relative to a stationary observer, and the proper acceleration

$$ \rm a^{\mu}= \frac{d^2 x^{\mu}}{d \tau^2}+\sum_{\alpha, \beta} \ \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha}u^{\beta}$$

so at $\rm r=4$ we get $\rm F=0$ with $\rm v= \sqrt{1/(r-2)}=\sqrt{1/2}$ as expected. At $\rm r=1.5 \ r_s=3$ the radial force $\rm F=1/\sqrt{27}$ (in units of $\rm c^2 m/M/G$) is independend of the angular velocity, and at $\rm r<3$ the required force to keep a constant $\rm r$ is smallest when $\rm v=0$.

That is due to the $-3 \dot{\theta}^2$ term in the geodesic equation

$$\rm \ddot{r}=-3 \dot{\theta}^2+ r \dot{\theta}^2 -1/r^2$$

which differs from the Newtonian expression and makes the angular velocity contribute negatively below the critical radius $\rm r=3$, and cancel itself at that exact radius.

This is assuming the Schwarzschild metric, for rotating black holes it would be a little bit more complicated. The solution might be shortened into a more elegant form, but since the coordinate acceleration is $0$ on a circular path the calculation is straightforward.


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