# Black Holes and escape velocity

1. Jul 23, 2014

### willoughby

It is my understanding that at the most fundamental level, a black hole is simply an object with a gravitational field so strong that there exists a sphere that lies outside the body of mass of that object from which the escape velocity is equal to the speed of light. In other words, a body of mass smaller than its Schwarzchild radius. I believe that my problem is that I am thinking about this from a perspective based on concepts of classical physics. Here is the reason for my confusion :

The escape velocity from the surface of the Earth is about 11.2 km/s. However, this does not mean that an object must reach 11.2 km/s to escape the Earth's gravitational influence. As I'm sure those of you reading this know, it is simply the instantaneous velocity an object would need to escape the Earth under no further power (via free fall) after that initial instantaneous velocity was achieved from the surface. It does not prevent an object from building a ladder and climbing out the the Earth's gravitational influence at 1 m/s, or under some kind of propulsion system either at any velocity less than that escape velocity. My question is why then is it stated - time and time again - that nothing can escape a black hole? Other than the obvious things like violent tidal forces that would rip an engine (or a human) apart, what fundamental concept(s) of general relativity prevent someone from flying out of a black hole under power?

2. Jul 23, 2014

### Staff: Mentor

Because your assumption is wrong. It is true that if the classical escape velocity calculation gives you the speed of light at the Schwarzchild radius, but that is neither the most fundamental definition of a black hole nor an accurate description of what is really happening.

It's hard to come up with a good non-mathematical description of what is happening, but you could try this: There are no paths through space that start inside the event horizon and end outside of it, so there is no way of connecting a point inside the horizon to a point outside the horizon with a ladder.

3. Jul 23, 2014

### Mr-R

I think you can imagine what Nugatory said as space-time becoming so distorted that there wont be a path you can take to get outside the event horizon. Is that al right for a loose visualization, Nugatory?

4. Jul 23, 2014

### Staff: Mentor

As long as you keep that word "loose" front and center, yes.

5. Jul 23, 2014

### WannabeNewton

6. Jul 23, 2014

### Staff: Mentor

I think the issue can be answered fairly succinctly: moving slowly out of a gravity field requires applying a force to overcome the local gravitational acceleration. So what is the local gravitational (proper) acceleration at the event horizon?

7. Jul 23, 2014

### phinds

To add to what has already been pointed out, you have another flaw here (at least some of the time). Large black holes have small tidal forces at the EH. A supermassive BH has trivial tidal forces at the EH but small BH can have huge tidal forces at the EH.

8. Jul 23, 2014

### Staff: Mentor

But this isn't true, at least not if "paths through space" means what it intuitively seems to mean, namely "spacelike paths". A better way to say it is: there are no *timelike* paths through *spacetime* that start inside the horizon and end outside it.

9. Jul 23, 2014

### Staff: Mentor

That's correct, just didn't feel to me like the right level of explanation for that particular OP.

10. Jul 23, 2014

### willoughby

Let me clarify that I am not attempting to make a case that you could exit from within a black hole in a machine under power. I understand that it is not possible; I wanted an explanation as to why. Yours makes sense, and I understood this much already.

11. Jul 23, 2014

### willoughby

According to Newtonian physics, can't super-massive black holes have relatively mild proper accelerations at the event horizon? I understand that saying "according to Newtonian physics" pretty much is meaningless within the realm of black holes, but this is why I am asking. I understand that I am wrong. That much doesn't need to be pointed out. I understand that I am wrong because I take a non-applicable concept such as Newtonian gravity into a realm involving black holes. I get it. I just wanted to know what was going on. Thanks for the replies.

12. Jul 23, 2014

### phinds

Small tidal forces at the EH of a large BH does not change the fact that light still cannot escape from the EH, it's just that the gradient is small all the way out at the EH.

13. Jul 23, 2014

### Staff: Mentor

According to general relativity, the proper acceleration experienced by someone falling freely through the event horizon is zero, while the proper acceleration experienced by someone stationary near the horizon is enormous; this is true regardless of the size of the black hole. Newtonian physics, on the other hand, has no concept of "proper acceleration".

I suspect that you are thinking of tidal forces, which do indeed behave as you describe.

14. Jul 23, 2014

### willoughby

Nugatory - I really appreciate your response. I willingly admit that I am beyond undereducated in this area, but I do understand your explanations. Thank you for taking the time. I've never taken a physics course officially other than basics in high school; I'm just thoroughly interested in it, and folks like you keep me interested. Thanks again.

I wasn't actually talking about tidal forces. I was just simply wrong in my assumption that the accelerations could be small at the event horizon of larger black holes. I do understand how tidal forces work and that a 'small' or 'large' tidal force doesn't really have anything to do with the strength of a gravitational field.

15. Jul 24, 2014

### willoughby

Here is an example of what I am talking about :

Just so I don't get outrageous with the masses here, I looked up the largest known black holes. The largest I could find was one that is 23 billion times the mass of the sun. Please - anyone correct my mistakes. The Schwartzchild radius of a BH this massive is about 6.8 * 10^13 m. The acceleration due to gravity using only Newtonian equations at this distance is only about 658 m/s².

I understand that my continuing posts are potentially just an annoyance at this point, but can someone explain specifically how and why GR can logically or mathematically reconcile this? Or correct this?

Perhaps is don't grasp entirely the concept of proper acceleration, but it seems that it would be an acceleration relative to a free fall or some other inertial frame, and in this case, 658 m/s² just doesn't seem all that insurmountable. I know there is a flaw in my logic and even in my approach (because I'm using Newtonian concepts). I'm gonna have another beer...

16. Jul 24, 2014

### Ich

I don't see any, except "Schwarzschild".

What you are describing is known as "surface gravity". It is a useful concept in relativity, too, needed for example to calculate Hawking radiation. To get the real, proper acceleration at the EH, you have to divide it by the "redshift factor" $\sqrt{1-\frac{2GM}{c²r}}$.
I think this previous thread could be an interesting read for you.

17. Jul 24, 2014

### Staff: Mentor

Good beer is extremely important. I personally am partial to dark Mexican brews and suggest that you avoid Budweiser and other high-volume American brands.

And you are right that there's only so far that Newtonian concepts will carry you, but here's a bit more:

When you're thinking about escaping from the gravity of this super-massive black hole, do not confuse how steep the sides of a well are with how deep it is.

There's a formula (no deep physical significance, just a happy coincidence that is kinda fun to prove) showing that the energy needed to completely escape from a classical gravitating body, starting at a radius $R$, is equal to the energy needed to move an object a distance $R$ against a force equal to the gravity at that distance from the center. Now, 658 m/sec^2 may not sound like much on a cosmic scale (although it's about 70g, comparable to the impact forces in a serious automobile accident)... but $W=Fd$ will tell you that applying enough force to accelerate an object at that rate across a distance of $6.8\times 10^{13}$ meters is going to require an astoundingly stupendous amount of energy.

This calculation is of course completely bogus because it's applying classical physics in a domain where relativistic effects cannot be ignored... But it should help you see that even a classicist should expect to see enormous huge energies here.

18. Jul 24, 2014

### willoughby

This was perfect. I've been reading up on my own as well, but this comment turned on the light for this problem for me. Thanks!