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)
[Physics] How much extra distance to an event horizon
black-holescurvatureevent-horizongeneral-relativity
Related Solutions
Well, if I'm being pedantic, your question is self-contradicting. The mathematical definition of a black hole is determined by tracing back the points that escape to infinity at late times, and then taking their boundary--thus, the event horizon is PRECISELY those points that cannot escape to infinity. But that's a boring answer that doesn't tell you anything about physics.
A less boring answer states that there is a very strong theorem that says that the area of a black hole is necessarily nondecreasing (so long as you have ordinary matter and no quantum mechanics), and as a corollary, you can prove that the null generators* of the hole either are destined to sit on the black hole's horizon forever, or to fall into the hole. They cannot escape. And since these light rays can't escape, then neither can particulate matter nor rays farther into the hole.
*The null generators are the light beams that are the boundary between the outgoing light rays that can freely leave the hole outside the horizon, and the ones forced to fall in that like inside the horizon. They simply float in space, neither able to go out or go in.
I suspect that what has confused you is the difference between remaining a fixed distance from the black hole and falling freely into it. Let me attempt an analogy to illustrate what I mean.
Suppose you are carrying a large and heavy backpack. You can feel the gravitational force of the backpack weighing you down. However this only happens because you're staying a fixed distance from the centre of the Earth i.e. you're standing stationary on the Earth's surface. If you and the backpack were to leap from a cliff then (ignoring air resistance) you would feel no gravity as you plummeted downwards and the backpack wouldn't weigh anything.
If we now switch our attention to the black hole, if you attempt to stay a fixed distance from the black hole (presumably by firing the rocket motors on your spaceship) you'd feel the weight of the backpack, and the weight would get bigger and bigger as you approach the event horizon. In fact the weight is given by:
$$ F=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} \tag{1} $$
where $m$ is the mass of the backpack, $M$ is the mass of the black hole, $r_s$ is the event horizn radius and $r$ is your distance from the centre of the black hole. As you approach the event horizon, i.e. as $r \rightarrow r_s$, equation (1) tells us that the force goes to infinity. That's why once you reach the event horizon it is impossible to resist falling inwards.
But you only feel this force because you're trying to resist the gravity of the black hole. If you just fling yourself off your spaceship towards the black hole then you will feel no weight at all. You would fall through the event horizon without noticing anything special. In fact you would see an apparent event horizon retreating before you and you would never actually cross anything that looks like a horizon to you.
But there is another phenomenon that can cause you problems, and this is related to the phenomenon of spaghettification that you mention. At any moment some parts of you will be nearer the centre of the black hole than others. For example if you're falling feet first your feet will be nearer the centre than your head. That means your feet will be accelerating slightly faster than your head, and the end result is that you get slightly stretched. This is called a tidal force, and it happens with all sources of gravity, not just black holes. Even on the Earth the gravitational force on your feet it slightly higher than on your head, though the difference is so small that you'd never notice it.
The thing about a black hole is that because its gravity is so strong the tidal forces can get very strong indeed. In fact they can get so strog that they'd pull you out into a long thin strip like a piece of spaghetti - hence the term spaghettification.
But the tidal forces only become infinite right at the centre of the black hole. They are not infinite at the event horizon, and in fact for large enough black holes the tidal forces at the event horizon can be negligably small. The equation for the variation of gravitational acceleration with distance is:
$$ \frac{\Delta a}{\Delta r} = \frac{c^6}{(2GM)^2} \tag{2} $$
If we take a black hole with the mass of the Sun and use equation (2) to calculate the tidal force we get $\Delta a/\Delta r \approx 10^{9}g$/m. So if you're two metres tall the difference between the acceleration of your head and feet would be $2 \times 10^9g$, where $g$ is the gravitational acceleration at the Earth's surface. This would spaghettify you very effectively. However at the event horizon of a supermassive black hole with the mass of a million Suns the difference between your head and feet would be only 0.001$g$ and you'd struggle to feel it.
Best Answer
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 pass the distance $r=r_s$ 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.