# How much extra distance to an event horizon?

How much extra distance would I have to travel through space to get from Earth to a stellar mass event horizon? (compared to the same point in space without a black hole)

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Wouldn't the point in space be the same in both cases, regardless of the existence of the black hole? – Kyle Kanos Nov 9 '13 at 20:00
I don't think so, there is curvature outside th EH. The smaller the BH the greater the curvature at the EH? – Jitter Nov 9 '13 at 20:11

I suspect you're not asking the question you're really interested in, because the answer to your question is really boring. If you jump into a black hole you'll see the event horizon retreating before you, and you'll never cross it. The distance you've travelled is an ambiguous quantity since of course in your frame you're stationary and have travelled no distance at all. The time you take to cross the horizon then hit the singularity is finite, and for stellar mass black holes very short.

A far more interesting question is if you hover outside the horizon and let down a tape measure how long would it have to be to reach the horizon i.e. what do you get by integrating $dr$ in a radial direction towards the event horizon? The Schwarzschild metric is:

$$ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 d\Omega^2$$

Lets suppose we hover at a distance $r_1$ from the singularity and let down a tape measure to measure the distance to some point at a radial distance of $r_2$. Because $dt$ and $d\Omega$ are constant the equation for the line element simplifies to:

$$ds = \frac{dr}{\left(1-\frac{r_s}{r}\right)^{1/2}}$$

To get the length of the tape we just need to integrate this expression from $r_1$ to $r_2$:

\begin{align} s &= \int_{r_2}^{r_1} \frac{dr}{\left(1-\frac{r_s}{r}\right)^{1/2}} \\ &= \int_{r_2}^{r_1} \frac{r^{1/2}dr}{\left(r-r_s\right)^{1/2}} \end{align}

To integrate this we cheat and look up the answer in a GR book, the result being:

$$s = \left[ z \sqrt{z ^2 - r_s} + r_s \ln \left( z + \sqrt{z ^2 - r_s} \right) \right]_{z_2}^{z_1}$$

where we've used the substitution $r = z^2$ to make the integral manageable.

To make this concrete lets take a black hole with the mass of the Sun, so $r_s$ = 2954m, and we'll start from 5km out i.e. $r_1 = 5000$. Let's graph the length of the tape as a function of $r_2$:

The magenta line is the Newtonian result, i.e. if space was flat, and the blue line is what we actually measure. The tape measure distance from $r = 5km to the event horizon is about 4,780m compared to the Newtonian calculation of 2046m. So the affect of the curvature is to make the distance measured radially greater than$r_1 - r_2\$. However I must emphasise that this is not what you'd measure if I threw you into the black hole. This is the distance measured by an observer hovering far from the event horizon.

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That was the question I was thinking of. You missed my edits. How far out from the EH does the curvature extend, to earth. – Jitter Nov 9 '13 at 20:08
@Jitter: no, you're asking about travelling to the event horizon and that's completely different from hovering outside it. The former is a rainfall frame and the latter is a shell frame. They measure completely different distances. – John Rennie Nov 9 '13 at 20:12
I did not read/think very clearly did I. My question still stands. Is there anything I should know about measuring by traveling from Earth over measuring from orbit? – Jitter Nov 9 '13 at 20:21
@Jitter: have a look at my answer to physics.stackexchange.com/questions/83127/… and see if that is relevant to your question. If you so might want to edit your question or maybe ask a new one. – John Rennie Nov 9 '13 at 20:24
Ok, I check the link and see if my answer will be more or less than a light year;-) – Jitter Nov 9 '13 at 20:34