Escape velocity near a black hole: Newtonian or Special Relativity?

[Trollfaz]

For a spherical body mass M and radius r, the escape velocity at the surface is $\sqrt{\frac{2GM}{r}}$. At the surface, an object mass m has GPE $-\frac{GMm}{r}$ and 0 GPE at infinity. So $\frac{GMm}{r}$ KE must be converted to GPE which means the minimum take off speed must be $\sqrt{\frac{2GM}{r}}$ for any m. This is the case of Newtonian view of gravity.

Now Newtonian mechanics of KE don't work well at speeds near c when Special Relativity (SR) dominates. SR says that a body with rest mass m travelling at speed v has total mass energy $\gamma mc^2$.
$$\gamma=\frac{c}{\sqrt{c^2-v^2}}$$
So KE is $(\gamma-1)mc^2$. This gives
$$ (\gamma-1)c^2=\frac{GM}{r}$$
at r from the black hole or intense source of gravity.
$$\gamma=\frac{GMc^2}{r}+1$$
And $v=c\sqrt{1-\frac{1}{\gamma^2}}$
So is the classical interpretation of gravity or the relativistic interpretation of gravity a better model. If the relativistic model dominates then Schwarzchild radius will not be $\frac{\sqrt{2GM}}{c}$.

 

[PeterDonis]

So is the classical interpretation of gravity or the relativistic interpretation of gravity a better model.

General relativity is the best model for gravity that we have. Special relativity, which is what you mention in your post, is not. A mismash of Newtonian gravity and SR, which is what the equations you give are, is certainly not.

If the relativistic model dominates then Schwarzchild radius will not be $\frac{\sqrt{2GM}}{c}$.

The Schwarzschild radius of a non-rotating black hole in GR is $2 G M / c^2$. Your equations, as noted above, are not the correct GR equations. That is why they are giving you nonsensical answers.

 

[Ibix]

So is the classical interpretation of gravity or the relativistic interpretation of gravity a better model. If the relativistic model dominates then Schwarzchild radius will not be $\frac{\sqrt{2GM}}{c}$.

Or you've done several things wrong. The correct maths is in chapter 7 of Carroll's lecture notes, leading to equations 7.47 and 7.48 which, for the case of a radial ($L=0$) stops-at-infinity ($E=1$) timelike ($\epsilon=1$, $\lambda=\tau$) trajectory, simplify (Edit: with minor correction pointed out by JimWhoKnew below) to$$\frac{dr}{d\tau}=\sqrt{\frac{2GM}{r}}$$Note that this is in terms of the proper time of a free falling observer. Using the time of an observer on the surface at constant $r$ this becomes$$\frac{dr}{dt}=\sqrt{\frac{2GM}{\left(1-\frac{2GM}{c^2r}\right)r}}$$which does indeed diverge as $r$ approaches the Schwarzschild radius.

More generally, if you get a result that contradicts over a century of study of a topic, particularly at the beginning of one of the most well-studied parts of it, it is overwhelmingly more likely that you've messed up than that the Schwarzschild radius isn't what it is.

 

[Vanadium 50]

Calculating incorrectly does not make relativity wrong.

 

[JimWhoKnew]

The correct maths is in chapter 7 of Carroll's lecture notes, leading to equations 7.47 and 7.48 which, for the case of a radial ($L=0$) stops-at-infinity ($E=1$) timelike ($\epsilon=1$, $\lambda=\tau$) trajectory, simplify to$$\frac{dr}{d\tau}=\sqrt{\frac{GM}{r}}$$Note that this is in terms of the proper time of a free falling observer.

Seems like a little typo. The result should be $$\frac{dr}{d\tau}=\sqrt{\frac{2M}{r}}$$(with $~G=c=1~$). It is valid in all charts that share the same definition of the $r$ coordinate (Schwarzschild, E-F, G-P, K-S).

Using the time of an observer on the surface at constant $r$ this becomes$$\frac{dr}{dt}=\sqrt{\frac{GM}{\left(1-\frac{2GM}{c^2r}\right)r}}$$which does indeed diverge as $r$ approaches the Schwarzschild radius.

I think that the more meaningful quantity is the "tetrad" escape velocity $~\frac{d\hat r}{d\hat t}~$ that is measured by an observer at rest at the constant initiial $r$ , using his clocks and rods. The result (if I've calculated correctly) is $$\frac{d\hat r}{d\hat t}=\sqrt{\frac{2M}{r}}$$which approaches $c$ near the horizon, as expected (and also happens to coincide with the Newtonian result).

Another interesting quantity is the escape velocity $~\frac{dr}{dt}~$ in the global coordinates, which "paradoxically" tends to zero near the horizon, as expected from the "frozen star" effect.

 

[Ibix]

Seems like a little typo.

Mental arithmetic fail, I think. Will correct.

 

[JimWhoKnew]

So KE is $(\gamma-1)mc^2$. This gives
$$ (\gamma-1)c^2=\frac{GM}{r}$$
at r from the black hole or intense source of gravity.

Actually, your argument, although incorrect, is not that far off. In this particular case, the energy of the escaping object is indeed conserved (not an obvious feature in GR) and satisfies$$E_{kin}+E_{pot}=0$$as in OP. Moreover, $~E_{kin}~$ has the same form as deduced from SR in OP. Only $~E_{pot}~$ is modified. This energy conservation condition becomes:$$(\gamma-1)mc^2+\left(\sqrt{1-\frac{2GM}{c^2r}}-1\right)~\gamma~mc^2=0$$which gives$$\beta=\sqrt\frac{2GM}{c^2r}$$so the horizon is at $~r=2GM~$ as advertised.

Edit: corrected terms containing Schwarzschild radius to include $c^2$ in denominator.

 

[PeterDonis]

In this particular case, the energy of the escaping object is indeed conserved

More precisely, the energy at infinity is conserved, because it is a constant of the motion associated with the timelike Killing vector field of the spacetime. In a spacetime without such a KVF, there is no such conserved quantity.

(not an obvious feature in GR)

It's not a feature of GR in general, it's only a feature of particular spacetimes, the ones that have a timelike KVF. For those spacetimes, it is indeed obvious, by Noether's Theorem.

 

[JimWhoKnew]

It's not a feature of GR in general, it's only a feature of particular spacetimes, the ones that have a timelike KVF. For those spacetimes, it is indeed obvious, by Noether's Theorem.

A timelike KVF guarantees energy conservation, but by itself is not sufficient to make it obvious. For that, you also need the ability to identify the KVF.

 

[PeterDonis]

For that, you also need the ability to identify the KVF.

Why do you think that isn't obvious in the case of Schwarzschild spacetime?

 

[JimWhoKnew]

Why do you think that isn't obvious in the case of Schwarzschild spacetime?

I meant it in this sense: Suppose, for example, that I apply a complicated symmetry-wrecking coordinate transformation to the Schwarzschild solution, and hand you only the line element in these new coordinates (for simplicity, let's assume that I start from a static solution for a spherically symmetric star, to avoid the horizon and singularity). Are you sure you'll be able to identify a timelike KVF? (either by expressing it, or its integral curves, as functions of the new coordinates, or by finding a further coordinate transformation that will carry it back to a form where the KVF is manifest).
So if you know that the given line element is Schwarzschild in disguise, it is obvious that it has a timelike KVF. Otherwise, it may not be obvious, although it is still a Schwarzschild spacetime with its 3 linearly independent KVFs.

 

[PeterDonis]

Suppose, for example, that I apply a complicated symmetry-wrecking coordinate transformation to the Schwarzschild solution

A coordinate transformation can't change any invariants, and the KVF is an invariant.

if you know that the given line element is Schwarzschild

Which we do in this discussion, since the OP specified a spherical body surrounded by vacuum.

 

[JimWhoKnew]

I think we are going in circles now.
In post #7 I wrote "(not an obvious feature in GR)". Post #7 was a reply to the OP, who seems to be acquainted with SR but not GR. In Newtonian Mechanics and in SR, energy conservation always apply (because of the time-translation symmetry). Not so in the general case in GR, as you explained in more details and precision in post #8.

When I wrote in post #11 "if you know that the given line element is Schwarzschild", I was referring to the line element in the "bizarre" coordinates that I've described in the preceding paragraph.

A coordinate transformation can't change any invariants, and the KVF is an invariant.

In post #11 I wrote "although it is still a Schwarzschild spacetime with its 3 linearly independent KVFs". Pretty much the same, isn't it?
By "symmetry wrecking" I meant that no symmetry remains manifest, or easily recognizable, in the line element when presented in the new "bizarre" coordinates.

 

[Nugatory]

A coordinate transformation can't change any invariants, and the KVF is an invariant.

Yes, but obviousness is not an invariant and I think that was @JimWhoKnew's point.

 

[PeterDonis]

I think we are going in circles now.

My point is that the OP said the spacetime geometry was Schwarzschild. And if you know the spacetime geometry is Schwarzschild, you know it has the timelike KVF that goes with Schwarzschild spacetime geometry. In other words, when you specify a physical condition on a scenario, like "spherically symmetric mass surrounded by vacuum", which is what the OP specified, your specification is invariant, and you know whatever follows from that invariant specification. And that, in this case, is enough to make it obvious, in this case, that the spacetime the OP specified has a timelike KVF.

 

[PeterDonis]

obviousness is not an invariant

It is if the specification of the scenario told you the necessary invariants to make it obvious. See my post #15 just now.

Yes, if all someone tells you is "the line element looks like this in the coordinates I just made up", then it might not be obvious what spacetime geometry that line element describes. But the OP in this thread didn't do that. The OP told us directly what the spacetime geometry was. That makes discussion of weird coordinate charts irrelevant to this thread.

 

[JimWhoKnew]

Yes, but obviousness is not an invariant and I think that was @JimWhoKnew's point.

A concise and elegant summary

 

[pervect]

This is a somewhat old thread, but I was thinking about it and I came to the conclusion that General Relativity doesn't have quite the same concept of "escape velocity" that Newtonian mechanics does. Specifically, in Newtonian mechanics, the direction of the velocity doesn't matter - as long as you don't collide with something , if you have escape velocity your orbit will escape to infinity, if you don't, you won't.

The best example of this is the photon sphere. At 1.5 times the Schwarzschild radius or closer, a photon moving in a direction with no radial component won't escape the black hole, it will instead have an (unstable) orbit at the photon sphere, or spiral into the black hole inside the photon sphere. So the direction of the velocity matters, it's not simply a question of having enough energy to escape.

Note that if a photon can't escape, nothing else can, as any material object will be slower.

I'm not aware of any specific reference on the topic pointing out this difference, though the behavior of the photon sphere is well known. See https://en.wikipedia.org/wiki/Photon_sphere. It serves as a counter-example to the idea that the direction of the velocity doesn't matter in GR. In Newtonian mechanics, it doesn't matter, but in GR, it does.

On the general topic of orbits around black holes, I rather like https://www.fourmilab.ch/gravitation/orbits/, which borrows a lot of the math from MTW's gravitation and has some graphics as well. However, don't expect to learn the theory from it - it's a quick reference, not a thourough explanation. A book would be a better choice to learn the theory, there are a few such as "Exploring Black Holes" that are at an undergraduate level. This particular book happens to be available online for a free download from Taylor's website. There are other choices of course. It's been a while since I've looked at it, I don't think it uses quite the same approach as the fourmilab website / MTW does.