1
$\begingroup$

In my last question I learned that movement inside event horizon is actually well defined when represented in Kruskal coordinates.

However, I still don't understand how it is working out. Can you help me to solve a primer that I have devised?

Let's imagine there's an event horizon around a singularity, of radius R. There's a small mass m at rest relative to this event horizon, initially located at 10R from the singularity. It will begin falling towards the event horizon.

We can roughly calculate how much time has passed on m when it approaches 5R using classical mechanics. Using general relativity we can calculate how much time has passed on m when it approaches 2R.

Can you please guide me how much time will pass on m when it crosses R (i.e. event horizon, as measured from outside)? ¾R? ½R? ¼R?

Is there maybe an online calculator of sorts, where one can input some values and get the result?

$\endgroup$
1
  • 1
    $\begingroup$ Schwarzschild coords are also defined inside the EH, but they can be a little confusing because the radial space coordinate is timelike (and the time coordinate is spacelike). But sure, you can't use them for stuff on both sides of the EH because they have a coordinate singularity at the EH. Also see Gullstrand-Painlevé aka raindrop coords. $\endgroup$
    – PM 2Ring
    Commented Dec 10, 2020 at 1:53

1 Answer 1

1
$\begingroup$

There are simple closed-form solutions for a particle that falls into a Schwarzschild black hole, starting from rest at a large distance. See https://en.wikipedia.org/wiki/Schwarzschild_geodesics#Orbits_of_test_particles , below the text "When $E = m c^2$ and $h = 0$, we can solve for $t$ and $τ$ explicitly." The proper time $\tau$ is what you want here. The Schwarzschild time coordinate $t$ is not a particularly meaningful thing to refer to inside the horizon (it isn't timelike).

For the proper time, the result (omitting a constant of integration) is $\tau=-(2/3)(R/c)(r/R)^{3/2}$, where $R$ is the Schwarzschild radius. So for example, the time it takes to go from $5R$ to $R/4$ on such a trajectory (already in motion at $5R$) is $7.4R/c$.

Here's a simulation I made of what you would see optically: https://www.youtube.com/watch?v=z-H-PipYCKc&feature=youtu.be . This is for a black hole with 10 times the mass of the sun. The total time until you hit the singularity is a few milliseconds.

$\endgroup$

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.