# Faster than light

1. Nov 22, 2007

### the one

Hi
I have a simple question
in GR we can claculate the gravitational acceleration of the black hole with the Equation :
$$$a = \frac{{GM}}{{r^2 \sqrt {1 - \frac{{r_s }}{r}} }}$$$
now , the acceleration of a mass which falls into the black hole at the event horizon (when$$$r = r_s$$$)
goes to infinity . So the velocity of this mass must be greater than the speed of light.
How ???????????????????????

2. Nov 22, 2007

### Staff: Mentor

Acceleration is the derivative of velocity. There is a limit on speed, not acceleration. An arbitrarily large acceleration does not imply an arbitrarily large speed.

3. Nov 22, 2007

### Chris Hillman

Acceleration vector of a hovering observer in a simple BH model

This is the path curvature of the world line of an observer who uses his rocket engine to hover over a non-rotating massive object (not neccessarily a black hole!) whose exterior gravitational field is modeled by the Schwarzschild vacuum solution
$$ds^2 = -\sqrt{1-2m/r} \, dt^2 + \frac{dr^2}{\sqrt{1-2m/r}} + \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),$$
$$-\infty < t < \infty, \; 2m < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
(Good practice demands that you should always write out the line element with the intended range of the coordinates like this.) Or more precisely, the magnitude of the acceleration vector
$$\nabla_{\vec{e}_1} \vec{e}_1 = \frac{m/r^2}{\sqrt{1-2m/r}} \, \vec{e}_2$$
where $\vec{e}_1 = 1/\sqrt{1-2m/r} \, \partial_t$ is the tangent vector (a timelike unit vector) to the world line of our observer (which appears as a vertical coordinate line in our chart, but is nonetheless a curved path in the spacetime geometry), and where $\vec{e}_2 = \sqrt{1-2m/r} \, \partial_r$ is a radially and outward pointing spacelike unit vector. Notice that our observer needs to accelerate radially outward, so he must fire his engine radially inward, with a thrust slightly greater than the Newtonian expression $m/r^2$. (To compare the two, note that the Schwarzschild radial coordinate is defined so that the area of the static sphere at the observer's location is $A = 4 \pi \, r^2$, which also makes sense in Newtonian gravitation, except that in gtr, the "radial distance" is not given by the Schwarzschild radial coordinate!)

BTW, correct me if I am wrong in assuming that you are an advanced undergraduate physics major who is currently studying gtr!

Rather, the thrust required to maintain the position of an observer hovering over a hole diverges as his (static) position approaches the horizon. Nothing mysterious there! All this says is that close to the horizon, no real rocket ship can possibly maintain constant r but must start to fall in radially.

I just said that in gtr, slightly greater radially inward thrust is required for a hovering observer to maintain his position than in Newtonian gravity. This suggests asking what happens if we consider an observer who fires his rocket engine with the Newtonian value. Does he slowly fall into the hole? To find out, consider the new frame field
$$\vec{f}_1 = \cosh(\alpha) \vec{e}_1 - \sinh(\alpha) \vec{e}_2, \; \vec{f}_2 = \cosh(\alpha) \vec{e}_2 - \sinh(\alpha) \vec{e}_1, \; \vec{f}_3 = \vec{e}_3, \; \vec{f}_4 = \vec{e}_4$$
and determine the function $\alpha(r)$ by demanding that the acceleration be $\nabla_{\vec{f}_1} \vec{f}_1 = m/r^2 \, \vec{f}_2$. Then finding the integral curve of $\vec{f}_1$ we see that such an observer does indeed slowly fall in. Look through the posts listed in the post in my current sig (22 Nov 2007) for posts in which I discuss static, slowfall, Lemaitre, and Frolov observers in several charts, including charts valid in both exterior and future interior regions.

Last edited: Nov 22, 2007