# I Jetpacking above a Black Hole

1. May 9, 2017

### greswd

Let's say that I'm hovering using a jetpack at a fixed distance above the event horizon of a black hole. And that I'm perfectly well shielded from any radiation and whatnot.

Its a supermassive black hole with a super large Schwarzschild radius, so tidal forces are minimal.

Approximately how much thrust does the jetpack need to provide in order for me to hover?

2. May 10, 2017

### Orodruin

Staff Emeritus
Depends on how close you are to the event horizon. You can find out by computing the proper acceleration of a worldline with fixed spatial coordinates in the Schwarzschild metric.

3. May 10, 2017

### greswd

Is there a simple, approximate formula for the thrust needed? In terms of the mass of the black hole and the height above the event horizon.

4. May 10, 2017

### Ibix

There's an exact solution in terms of the Schwarzschild r coordinate. But that's not really distance from the horizon, which is a fairly arbitrary concept. It depends what coordinate system you use (similar to how distance depends on your reference frame in special relativity), and there's no particular significance to the choice. You can't even bounce a radar signal off the horizon in order to get some kind of frame-independent measure.

5. May 10, 2017

### greswd

let's say everything's at rest with respect to each other.

just curious as to whether one could hover above an event horizon.

6. May 10, 2017

### Staff: Mentor

Yes

7. May 10, 2017

### Ibix

As Dale says, yes, at least in theory. The thrust requirements are fairly splatter-inducing. The relevant formula for acceleration needed to hover is $$a=\frac 1{\sqrt{1-R_S/r}}\frac{GM}{r^2}$$where $r$ is the Schwarzschild radial coordinate and $R_S=2GM/c^2$ is the Schwarzschild radius. Careful about $r$! Lower $r$ means closer to the horizon, but it isn't a straightforward measure of distance as the $r$ in Newtonian gravity is.

That said, at large $r$ the square root tends to one and the result is just Newton. This is where you can ignore the little bit in the middle where non-Euclidean effects are significant. Note also that the acceleration needed to hover goes to infinity at the Schwarzschild radius. You're stuck at that point.

8. May 10, 2017

### greswd

In this simplified scenario, are they equivalent?

9. May 10, 2017

### Staff: Mentor

No. It isn't a matter of simplification. They are defined differently.

10. May 10, 2017

### greswd

I mean, in the case of Ibix's formula, can we take r to be the sum of the height at which one hovers above the event horizon and the Schwarzschild radius?

11. May 10, 2017

### Staff: Mentor

No. In that equation r is defined such that the surface area of the sphere at that r is $A=4\pi r^2$

12. May 10, 2017

### Ibix

No. If you build a (massless) sphere around the black hole and its total area is A, then an observer standing on the sphere would use $r=\sqrt {A/4\pi}$ to calculate the gravitational acceleration they feel. If spacetime were flat then $r$ would be the radius, yes. But spacetime isn't flat and there isn't a unique way to define the distance to the event horizon.

13. May 12, 2017

### pervect

Staff Emeritus
It depends on how you define distance, really. You can't use radar methods, for obvious reasons.

The formula for the proper acceleration required to hover at a Schwarzschild r coordinate r is

$$a = \frac{GM}{r^2\sqrt{1-\frac{r_s}{r}}} = \frac {c^2 r_s}{2 r^2 \sqrt{1-\frac{r_s}{r}}}$$

The later expression comes from utilizing the realtionship $r_s = 2GM/c^2$ to express GM in terms of $r_s$.

If we measure distances by rulers held by static observers (rulers held by observers maintaining a constant Schwarzschild r-coordinate), the distance from $r = r_s + \epsilon$ to the horizon by this definition would be:

$$d = \int_{r_s}^{r_s+\epsilon} \frac{dr}{\sqrt{ 1-r_s/r} }$$

If we let $r_s = 1$, so that my computer algebra package doesn't throw fits, I get

$$a \approx \frac{c^2}{2 \sqrt{\epsilon}}$$
$$d \approx 2 \sqrt{\epsilon}$$

which means that the product a*d $\approx c^2$. Thus we conclude that $a \approx c^2/d$.

This is the same answer I got by more intuitive method, in which we approximate the black hole metric by the Rindler metric, and the black hole horizon by the Rindler horizon, and assume that the difference in Rindler coordinates serves as a reasonable definition of distance.

If you don't like the intuitive way, by all means use the more formal way, though it's much messier. The biggest issue is evaluating the messy integral for distance, (and taking the limit to remove the infinities), that's the most likely spot for an error.

14. May 13, 2017

### timmdeeg

If of interest, the radial proper distance is given here in equation (9.4.21).

15. May 13, 2017

### pervect

Staff Emeritus
Thanks, that is helpful. Using this textbook formula, we can drop the assumption that $r_s=1$. With a fair amount of computer algebra, I get:

$$a \approx \frac{c^2}{2\,r_s\, \sqrt{\frac{\epsilon}{r_s}}} \quad d \approx 2 r_s \sqrt{\frac{\epsilon}{r_s}}$$

Here $r_s$ is the Schwarzschild radius of the black hole, and $r_s + \epsilon$ is the Schwarzschild coordinate of the static (hoovering) observer. We see that the product of the proper acceleration a, and the proper distance in the static frame, d, is approximately equal to c^2, independent of the value of $r_s$.

It's still messy, but most of the equations have a textbook reference now. I tracked down the reference to an expression for the proper acceleration in Wald's "General Realtivity", it's on pg 152. Wald's expression is in geometric units, so G and c are assumed to be equal to 1.

Google books can poentitally find this, a search for "wald general realtivity acceleration of static observer" worked for me.

16. May 15, 2017

### pervect

Staff Emeritus
I thought of a way to mathematically illustrate the close tie between the Scwarzschild metric and the Rindler metric near the event horizoin based on the above.

Consider the transformation

$$r = r_s + \frac{R^2}{4r_s}$$

or, equivalently

$$R = 2 \sqrt{r_s \left(r-r_s \right) }$$

Then we can write

$$dr = \frac{R}{2 \, r_s}\,dR$$

In geometric units, with c=1, the line element for the Schwarzschild metric in the r-t plane is:

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

Using the above, we can write the line element in terms of dt and dR as:

$$ds^2 = -\left(1-\frac{r_s}{r} \right) dt^2 + \frac{R^2}{4\,r_s^2 \left(1-\frac{r_s}{r} \right) } \, dR^2$$

Substituting the expression for r in terms of R and simplifying, we get

$$ds^2 = -\frac{R^2}{R^2 + 4\,r_s^2}\,dt^2 + \left(1 + \frac{R^2}{4\,r_s^2} \right) dR^2$$

When $R << r_s$ this is approximately equal to

$$ds^2 = -\frac{1}{4\,r_s^2} \,R^2 \, dt^2 + dR^2$$

which is just one form of the Rindler metric $-a^2 R^2 dt^2 + dR^2$ with $a = 1 / 2\,r_s$. Note that a is just a parameter of the Rindler metric, it's not equal to the proper acceleration.

17. May 16, 2017

### timmdeeg

I wonder if "the close tie" means that the Equivalence Principle holds if we consider two observer in a uniformly accelerating rocket. Could they (one at the bottom, the other at the top) distinguish experimentally whether they are in a rocket or are shell observers close to a black hole?
The acceleration at the bottom is larger than that at the top. Having these values together with the proper distance between them, could they calculate pairs of r-coordinates and values for M which fit to their observation (acceleration, proper distance) or is there only one possibility for r and M?

18. May 18, 2017

### Staff: Mentor

As an aside, that could read "close to any spherically symmetric mass". The Schwarzschild solution describes the vacuum region outside of things that aren't black holes as well as outside the event horizon of black holes.

19. May 18, 2017

### timmdeeg

My question is related to the Schwarzschild solutuion.

20. May 18, 2017

### Staff: Mentor

Right - so whether there's a black hole involved or not is irrelevant. We have our experimenters in a windowless compartment, and your question is whether they can tell whether they're in an accelerating spaceship or sitting at a constant Schwarzschild $r$ coordinate near some gravitating mass (perhaps even parked on the surface of the earth). And the answer is that it depends on whether the compartment is large enough for tidal effects to be detectable - no matter how sensitive their instruments, we can always make the compartment small enough that they cannot tell the difference.