0
$\begingroup$

It is my understanding that just as special relativity contracts length with velocity general relativity contracts length with gravity. Would this mean the radius of a BH is smaller than it would appear without the gravitational effects?

$\endgroup$
6
  • 2
    $\begingroup$ What do you mean by "without the gravitational effects"? $\endgroup$ Commented May 21 at 20:33
  • $\begingroup$ I believe what the OP means is that the black hole might appear to have a smaller radius as observed by somebody nearby vs. somebody farther away. $\endgroup$ Commented May 21 at 23:18
  • $\begingroup$ So what I mean is that under SR if you go close to the speed of light you would appear, from the view point of someone that is stationary, shorter in the direction of travel. I am asking if similarly a black hole looks "shorter" (smaller radius) because it is deep in the gravitational field from the view point of someone way outside the field. Also I would like to know that if this is the case how big is the black hole actually if you adjust out the radius length contraction? $\endgroup$
    – Joe
    Commented May 21 at 23:53
  • $\begingroup$ I believe Joe is asking if the radius is different to the Newtonian expectation for a body of the same mass. $\endgroup$
    – KDP
    Commented May 22 at 10:14
  • $\begingroup$ See Sec. 42-3 here. $\endgroup$
    – J.G.
    Commented May 22 at 10:27

2 Answers 2

3
$\begingroup$

In this answer, I am ignoring optical effects and illusions of how a body would appear to look from a distance and focussing on actual measurements of length.

Starting with the Schwarzschild metric, and using the (-,+,+,+) signature convention, we can set $dt=0$ and $d\Omega = 0$ and obtain $$ds = \frac{d r}{\sqrt{1-2M/r}}$$ using units such that c=1 ang G=1. This is the gravitational length contraction factor and is measured vertically. $dr$ can be thought of as the proper length of a infinitesimal ruler measured locally and $dS $ as the coordinate length of the ruler measured by an observer at 'infinity'. To calculate the vertical ruler length between two radial coordinates we simply have to calculate the definite integral of the above equation using suitable radial coordinates for the limits. When we do this we find the ruler distance between two radial coordinates that are one unit apart is greater, the nearer we get to the event horizon, because the rulers are length contracting and it requires more of the rulers to span the distance between the radial coordinates.

Just for info, the proper ruler distance is approximately the average of the radar distance measured from the upper coordinate and the radar distance measured from the lower coordinate.

Surprisingly, we can use the event horizon as the lower limit, but if we try to try to use a lower altitude than 2m as the lower limit we start getting complex results, representing the fact we cannot have stationary observers or stationary rulers below the event horizon, making it difficult to calculate the proper ruler distance from somewhere inside the event horizon, to somewhere outside the event horizon.

I came up with a way we can get an idea of what the proper ruler length measured from the centre is.

We start with the interior Schwarzschild solution which is the metric for a fluid body that has a physical surface. (A small low density object could float on this surface.) For r>2m*, (before the black hole has formed) it is still possible to have stationary rulers and observers all the way to the centre, so a physical proper ruler distance can be defined.

Setting $dt=0$ and $d\Omega = 0$, we obtain $$dS = \frac{dr}{\sqrt{1 - 2M r^2/R^3}},$$

where R is the radius of the fluid body. When integrated with respect to r, we get the distance from the centre (r=0) to the surface (r=R) as:

$$ S(r) = \int_0^{R}{\frac{dr}{\sqrt{1 - 2M r^2/R^3}}} = \sqrt{\frac{R^3}{2M}} \sin^{-1}\left(\sqrt{\frac{2Mr^2}{R^3}}\right)$$

Setting r=R and M=1, we can define the distance S as a function of R as:

$$ S(R) = \sqrt{\frac{R^3}{2}} \sin^{-1}\left(\sqrt{\frac{2}{R}}\right)$$

and the plot of the function gives:

enter image description here

where the orange curve represents the imaginary part of the distance. If we set the coordinate R arbitrarily close to 2M, then the proper radius becomes arbitrarily close to $\pi M$. The Newtonian expectation for the circumference of an object with radius $R = \pi M$ is $2 \pi R = 2M \pi^2$, but the Schwarzschild circumference is $4M \pi $ which is smaller. In this way, we can consider the coordinate radius of the object to be length contracted. For $R<2M$ the result has a mon zero imaginary part. We cannot get a sensible answer for R<2M, because the interior metric is a static solution and for R<2M the radius is not static. More technically the result is only valid for R<18M/8 because the force at the centre becomes infinite at that radius and the object has to collapse, but nevertheless this analysis gives an idea of how the proper and coordinate radii differ. A more exact analysis could probably be done on a Tolman dust model where internal pressure does not complicate the issue.

In summary, in Schwarzschild metric, restricting ourselves to bodies with radii greater than 2M, proper vertical coordinate lengths are shorter than their proper lengths, similar to how in SR, proper lengths parallel to the motion are shorter than their proper lengths.

$\endgroup$
2
  • $\begingroup$ The first part of this answer based on the vacuum solution properly describes the gravitational length contraction. The second part based on the interior solution is more specific and complex, but still may be helpful to some. Good answer +1 $\endgroup$
    – safesphere
    Commented May 25 at 5:19
  • $\begingroup$ we cannot have stationary observers or stationary rulers below the event horizon” - More precisely, we cannot have things stationary in $r$, which is the time coordinate inside the horizon. Well, we cannot have things stationary in time outside the horizon either, so there is no physical difference. We can have things stationary in space outside. On the inside, the $t$ coordinate measures space and we sure can have things static in $t$ there. So again, no physical difference. Therefore the popular quoted statement is misleading, as it neglects the changing nature of the coordinates. $\endgroup$
    – safesphere
    Commented May 25 at 5:32
1
$\begingroup$

It is rather inaccurate to say that gravity contracts length in the same way as special relativity. Gravity is the curvature of spacetime as a whole which is a different concept from special relativistic length contraction, which occurs on flat spacetime. The time dimension is also involved. Due to spacetime curvature, the apparent "disk", or shadow, of a Schwarzschild black hole actually spans a radius of $3\sqrt{3}r_s/2 \approx 2.6r_s$ where $r_s$ is the Schwarzschild radius. This is demonstrated in this post and this animation. This phenomenon is known as gravitational lensing.

Even for highly dense objects such as neutron stars, gravitational lensing enables more than one hemisphere to be seen at the same time. So spacetime curvature causes it to appear larger rather than smaller.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.