Calculating Black Hole Gravity: Velocity > Light Speed?

In summary, an observer in Newtonian gravity would slowly fall into a black hole once his acceleration exceeded the speed of light. An observer in gtr would not be able to maintain constant r, but would instead slowly fall in radially.
  • #1
the one
13
0
Hi
I have a simple question
in GR we can claculate the gravitational acceleration of the black hole with the Equation :
[tex]
\[
a = \frac{{GM}}{{r^2 \sqrt {1 - \frac{{r_s }}{r}} }}
\]
[/tex]
now , the acceleration of a mass which falls into the black hole at the event horizon (when[tex]
\[
r = r_s
\]
[/tex])
goes to infinity . So the velocity of this mass must be greater than the speed of light.
How ?
 
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  • #2
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
Acceleration vector of a hovering observer in a simple BH model

the one said:
in GR we can claculate the gravitational acceleration of the black hole with the Equation :
[tex]
\[
a = \frac{{GM}}{{r^2 \sqrt {1 - \frac{{r_s }}{r}} }}
\]
[/tex]

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
[tex]
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),
[/tex]
[tex]
-\infty < t < \infty, \; 2m < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi
[/tex]
(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
[tex]
\nabla_{\vec{e}_1} \vec{e}_1 = \frac{m/r^2}{\sqrt{1-2m/r}} \, \vec{e}_2
[/tex]
where [itex]\vec{e}_1 = 1/\sqrt{1-2m/r} \, \partial_t[/itex] 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 [itex]\vec{e}_2 = \sqrt{1-2m/r} \, \partial_r[/itex] 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 [itex]m/r^2[/itex]. (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 [itex]A = 4 \pi \, r^2[/itex], 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!

the one said:
now , the acceleration of a mass which falls into the black hole at the event horizon (when[tex]
\[
r = r_s
\]
[/tex])
goes to infinity . So the velocity of this mass must be greater than the speed of light.
How ?

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
[tex]
\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
[/tex]
and determine the function [itex]\alpha(r)[/itex] by demanding that the acceleration be [itex]\nabla_{\vec{f}_1} \vec{f}_1 = m/r^2 \, \vec{f}_2[/itex]. Then finding the integral curve of [itex]\vec{f}_1[/itex] 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:

1. How do you calculate the gravity of a black hole?

The gravity of a black hole can be calculated using the formula F = (G * M) / r^2, where F is the gravitational force, G is the gravitational constant, M is the mass of the black hole, and r is the distance from the center of the black hole.

2. Can the gravity of a black hole be greater than the speed of light?

According to Einstein's theory of relativity, nothing can travel faster than the speed of light. Therefore, the gravity of a black hole cannot be greater than the speed of light.

3. How does the velocity of a black hole affect its gravity?

The velocity of a black hole does not directly affect its gravity. However, as an object approaches the event horizon (the point of no return), its velocity will increase due to the immense gravitational pull of the black hole.

4. Is the gravity of a black hole the same throughout its entire structure?

No, the gravity of a black hole is not the same throughout its entire structure. As you get closer to the center of the black hole, the gravitational pull becomes stronger due to the concentration of mass at the singularity.

5. How does the gravity of a black hole compare to other objects in space?

The gravity of a black hole is significantly stronger than any other object in space. This is due to the massive amount of mass concentrated in a relatively small area, which creates a strong gravitational pull.

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