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For practical purposes, I'm considering a non-rotating black hole or neutron star, because I find Kerr black holes a little confusing.

The ratio between orbital velocity and escape velocity in Newtonian physics is $$ \frac{v_\text{escape}}{v_\text{orbit}} = \sqrt{2}$$

I've read several times that the escape velocity at the event horizon is exactly the speed of light $c$, mentioned here, among other places.

At the photon sphere at 1.5 times the Schwarzschild radius, the orbital speed is $c$, so a photon can orbit the black hole. But this is weird, because it seems to be the case that near a black hole orbital speed is greater than the escape velocity.

Can this be explained in layman terms? Is there an error in my train of thought?

If somebody wants to do the math, that's cool, but I'm not sure I'd understand it.

Assuming the premise of the first question isn't flawed, does this happen around a dense neutron star as well? Dense neutron stars may also have photon-spheres. Wiki says it's possible, Andrew says it's not. Whether neutron stars can or can't make photons fly in circles isn't strictly relevant to this question, but they are dense enough to not be far off and perhaps get relativistic effects and, perhaps, that funky relation where orbital velocity approaches and even surpasses escape velocity (which makes absolutely no sense).

Is there a neat and tidy mathematical relation between orbital speed and escape velocity for these ultra-high gravity situations?

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    $\begingroup$ BTW I think it is just possible for a neutron star to have a photon sphere, though that's probably best posted as a separate question. $\endgroup$ – John Rennie Sep 19 '15 at 13:58
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For a static black hole described by the Schwarzschild metric the escape velocity is:

$$ v_e = c\left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}} \tag{1} $$

and the orbital velocity is:

$$ v_o = c\sqrt{\frac{r_s}{2(r - r_s)}} \tag{2} $$

where $r_s$ is the Schwarzschild radius:

$$ r_s = \frac{2GM}{c^2} $$

If we graph these we get:

escape velocity etc

Note that the $x$ axis is in units of $r/r_s$ i.e. $2$ means $2r_s$, $3$ means $3r_s$ and so on. The ratio $v_e/v_o$ is plotted on the left hand axis and the speeds $v_e$ and $v_o$ on the right hand axis.

At large distances the ratio $v_e/v_o$ approaches $\sqrt{2}$, which is the Newtonian value, but as we approach the black hole the ratio falls. It becomes less than one at about $r/r_s = 4.8$, and at the last possible orbit where the orbital velocity is $c$ it has fallen to about $0.27$.

This probably seems odd at first glance, but it's due the well known fact that any infalling object slows as it approaches the event horizon and in fact its velocity tends to zero at the horizon. The object would take an infinite time to reach, let along cross, the event horizon. See the question How can anything ever fall into a black hole as seen from an outside observer? for more on this.

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    $\begingroup$ (1) converges to 0 when approaching the event horizon, while (2) converges to 1 while appoaching the photon sphere. So it seems to me that (1) gives the escape velocity as it is observed by an coordinate observer at infinity, while (2) is the orbital velocity in terms of a local shell observer. The local escape velocity in radial direction should converge to 1 when r=2, otherwise it wouldn't be the event horizon! For (2): An observer at infinity would see the object orbiting at newtonian orbital velocity since the extra term and and time dilation cancel out, at r=3 we see less than c (Shapiro) $\endgroup$ – Yukterez Jun 12 '16 at 3:02
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The local escape velocity is

$$v_{esc} = \sqrt{\text{2 G M}/\text{r}}$$

At infinty you observe that velocity slower by a factor of

$$1-\text{r}_s/\text{r}$$

so at infinity you observe

$$\text{v}_{esc} = \sqrt{\text{2 G M}/\text{r}} \cdot (1-\text{r}_s/\text{r})$$

because of gravitational length contraction radial to the mass and gravitational time dilation in all directions.

The local orbital velocity is

$$v_{orb} = \sqrt{\text{G M}/\text{r}}/\sqrt{1-\text{r}_s/\text{r}}$$

which is at infinity observed to be slower by a factor of

$$\sqrt{1-\text{r}_s/\text{r}}$$

At infinity you observe simply

$$\text{v}_{orb} = \sqrt{\text{G M}/\text{r}}$$

So the observer at infity should observe an orbiting particle orbit with it's newtonian velocity, while locally this velocity is higher.


enter image description here

The Plots show the velocities in terms of v/c and the Schwarzschild r coordinate in units of GM/c². Left is the system of a local observer, and right like an observer at infinity would see the Shapiro-delayed velocities.

Locally the escape velocity equals the orbital velocity at r=4GM/c², while at infinity their equality is observed at r=2·(2+√2)=6.8284GM/c².


At the photon sphere a local observer would observe the orbiting particle with c, while an observer at infinity would measure it slower by a factor of √(1-2/3), so with 0.577c.

An escaping particle near the event horizon would need a local radial velocity of c, and seem to have zero velocity for the observer at infinity.

The different factors for the radial and the transversal components is due the graviational length contraction which is only in radial direction (there is more radius inside the circle than the circumference divided by 2π).


To sum it up:

Locally, the radial escape velocity for Einstein is the same as for Newton. At infinity it is observed slower than that.

Also locally, the angular orbital velocity for Einstein is higher than for Newton, but at infinity it is observed to be the same as it would be under Newton.

The escape velocity can be lower than the orbital velocity, therefore orbits near the photon sphere are unstable in a sense that if your velocity is not only transversal but splits into a transversal and a radial component you will escape to infinity. With Newton you would just get an elliptical orbit with everything else staying the same.


Radial:

enter image description here

Transversal:

enter image description here

Comparison with Newton:

enter image description here

The index is in german, but I'm sure you'll find the local $v$ and the external $\text{v}$ (the latter with Shapiro delay).

Further reading: Equation of Motion and Geodesics, page 4, eq(9)

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In general relativity the energy of a test-body in a spherically symmetric gravitational field can be written as:

$$E=mc^2\left(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})^2(\hat{r}\cdot\hat{v})^2+(1-\frac{2GM}{rc^2})|\hat{r}\times\hat{v}|^2\right)}}}\right).\tag1$$

or equivalently:

$E=mc^2\left(\frac{1-\frac{2GM}{rc^2}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2)+|\hat{r}\times\hat{v}|^2\right)}}}\right)\tag2$

The condition for escape velocity can be stated as that the velocity must be such that the energy of the test-body must be the same as for an object at rest at infinity, $E=mc^2$. This is the same as saying that the expression within the parenthesis must be equal to unity. For a pure radial motion (2) reduces to:

$E=mc^2\left(\frac{1-\frac{2GM}{rc^2}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}}}\right)\tag3$

For this expression to be the same as $E=mc^2$ at infinity we must have:

$1-\frac{2GM}{rc^2}=\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2(1-2GM/(rc^2))}}\tag4$

which gives the escape velocity if you initially move in a pure radial direction as:

$v=\sqrt{\frac{2GM}{r}}(1-\frac{2GM}{rc^2})\tag5$

which have already been pointed out in other answers. If you instead want to know the escape velocity if you start off moving in a pure non-radial direction, such as for a circular orbit, and want to know how much faster you need to go in the direction that you are already moving you can see that (2) in this case reduces to:

$$E=mc^2\left(\frac{1-\frac{2GM}{rc^2}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}}\right)\tag6$$

In order for (6) to reduce to $E=mc^2$ at infinity we must have:

$$1-\frac{2GM}{rc^2}=\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}}\tag7$$

Which gives the escape velocity if you are setting of in a pure non-radial direction as:

$$v=\sqrt{\frac{2GM}{r}(1-\frac{2GM}{rc^2})}\tag8$$

Depending on in what direction you are initially moving the expression for the escape velocity will vary. In this answer I assume a velocity as measured by a distant observer / in coordinate time.

The orbital velocity for an object in circular orbit in coordinate time is the same in GR when you have spherical symmetry as classically:

$v=\sqrt{GM/r}\tag9$


At the photon sphere at 1.5 times the Schwarzschild radius, the orbital speed of circular motion in coordinate time (as seen by a distant observer) is:

$v=c\sqrt{1-\frac{2GM}{rc^2}}$

This can be said to be equal to the speed of light because in coordinate time in a spherically symmetric gravitational field you have:

$v_{light}=c(1-2GM/(rc^2))$

in the radial direction and:

$v_{light}=c\sqrt{1-\frac{2GM}{rc^2}}$

in the pure non-radial direction.

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