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I understand that closer orbits around Schwarzschild black holes require faster speeds, culminating in a required speed of $c$ at the photon sphere ($r = 3M$ in Schwarzschild coordinates). I'd love to be able to work out the orbital speed at the ISCO ($6M$) and marginally bound orbit ($4M$) and others. Is there a formula for this orbital speed as a function of radius? (I don't mind that these orbits are unstable.)

I gather that there are some challenges of interpretation here for subluminal velocities as to which reference frame the speed is being measured with respect to. I suppose I'm most interested in measurement with respect to a stationary observer at that same radius and to a stationary observer a distance infinity.

I'd also love to have a formula for the amount of kinetic energy required to be in that orbit in terms of the rest mass of the object. I gather that these are all going to be relativistic speeds and thus that this can be a substantial fraction of the rest mass -- approaching an infinite multiple as radius approaches $3M$.

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  • $\begingroup$ Question, is something stopping you from performing the calculations yourself? As in have you done orbital mechanics analysis for systems that are not black holes for example. Just asking so you might Include more information on where you are coming from here. $\endgroup$
    – Triatticus
    Oct 18, 2023 at 18:41
  • $\begingroup$ Unfortunately I know far too little about GR and orbital mechanics to derive the formulae myself. $\endgroup$
    – Toby Ord
    Oct 18, 2023 at 20:15

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There are a couple of expressions you could consider for circular orbits in the Schwarzschild metric (at $r \geq 3r_s/2$) - see here.

A "shell" observer, at the same radial coordinate as the orbiting body, but at a fixed $\theta$ and $\phi$, will observe its speed as it passes to be given by $$ \frac{v^2}{c^2} = \frac{r_s}{2r-2r_s}\ ,$$ where $r_s = 2GM/c^2$. Thus $$(\gamma -1)mc^2 =\left[ \left(\frac{2r-2r_s}{2r-3r_s}\right)^{1/2} -1 \right]mc^2\ . $$ This expression tends to infinity as $r \rightarrow 3r_s/2$. This would be the kinetic energy with which something would need to be launched tangentially from a stationary platform at that radius to be put into a circular orbit.

A distant observer would infer an orbital speed given by $$ \frac{v^2}{c^2} = \frac{r_s}{2r}$$ and thus $$(\gamma -1)mc^2 = \left[ \left(\frac{2r}{2r -r_s}\right)^{1/2} - 1\right]mc^2\ . $$

Note that these two expressions do not agree since kinetic energy is not the same in different frames of reference.

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  • $\begingroup$ Thanks @ProfRob. This is a very helpful explanation of both the velocities required as measured wrt a hovering observer and one at infinity and how they relate (and to the respective kinetic energies). $\endgroup$
    – Toby Ord
    Oct 22, 2023 at 20:21
  • $\begingroup$ It is especially nice that the orbital speed needed at the marginally bound orbit (4M) wrt infinity is c/2 and the orbital speed needed at ISCO (6M) wrt a hovering observer at 6M is also c/2. $\endgroup$
    – Toby Ord
    Oct 22, 2023 at 20:23
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I believe for a circular orbit the orbital velocity is always going to be $$v=\sqrt{GM/r}\tag{1}$$ in "coordinate time", as per a distant observer at rest. This is the same as in Newtonian gravitation. At the photon sphere you will no longer be able to reach this speed because you would have to travel at the speed of light which in a graviational field in coordinate time, but not in proper time, is less than c.

Regarding the energy of the orbiting body if I remember things correctly you could, assuming $E=mc^2$ at rest at infinity, write:

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

($\hat{r}=\bar{r}/r,\hat{v}=\bar{v}/v$)

Notice that this is basically both the kinetic and the potential energy baked into the same expression. For a circular orbit $\hat{r}\cdot\hat{v}=0$ and $|\hat{r}\times\hat{v}|=1$. and the expression becomes simpler. Notice that this is basically both the kinetic and the potential energy baked into the same expression. For a circular orbit we can insert (1) into (2) and get: $$E_{circularorbit}=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{3GM}{rc^2}}}\right).\tag{3}$$

From (2) and (3) we see that we need infinite energy to sustain a circular orbit at the photon radius. You can take the derivatve of (3) with respect to $r$ and find that the expression have a minimum at the "innermost stable circular orbit". At closer radial distances than that circular orbits closer to the central mass require more energy to uphold.

I like (2) because you can by "inspection" see that the energy for a "hoovering" object close to the Schwarzschild radius goes to zero but that even a miniscule motion will push the expression towards infinity, thus suggesting the known property of objects slowing down and "freezing" at the Schwarzshild radius, when using coordinate time.

Now in general relativity we have a phenomenon causing time, frequency and energy to be perceived different within a gravitational field. Energy and frequency will be perceived to be a factor: $$1\over{\sqrt{1-{2GM\over{rc^2}}}}\tag{4}$$ higher for a stationary observer within the gravitational field than at infinity. Time will slow down with the same factor. It so happens that the slowing down of time for a local observer, measuring "proper time", exactly balances the observed slowing down of light that a distant observer notices so the locally measured velocity of light will always be c.

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  • $\begingroup$ Wouldn't that formula for orbital velocity reach 1 (presumably the speed of light) at r = 1M, rather than at the photon sphere (r = 3M)? Or am I missing something? $\endgroup$
    – Toby Ord
    Oct 19, 2023 at 11:09
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    $\begingroup$ @TobyOrd According to a distant observer measuring "coordinate time" light and time appear to slow down deep in a gravitational field and not reach "c" at the photon radius. According to a "hoovering" observer at the same radial distance as the orbiting object measuring "proper time" speed will reach c at the photon radius. $\endgroup$
    – Agerhell
    Oct 21, 2023 at 7:01
  • $\begingroup$ Thanks @Agerhell, I didn't know that. So they would see a photon orbiting at a speed they don't measure as c? That is surprising. In that case, I think I'm really interested in the measure of speed of orbit (and kinetic energy) relative to a hovering (stationary Schwarzschild r coordinate) observer at the same radial distance. A formula on which speed should approach c (and kinetic energy approach infinity) at 3M. $\endgroup$
    – Toby Ord
    Oct 22, 2023 at 8:06
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    $\begingroup$ @TobyOrd Do note that using the expression for energy above the energy does become infinite at $r=3GM/c^2$. Formally I do not think you can actually separate potential and kinetic energy for an object in motion in GR, write "total energy = potential energy + kinetic energy". A hoovering observer at radius r will have time slowed down by a factor of $1/\sqrt{1-2GM/(rc*2)}$ compared to an observer at infinity. This will cause a hoovering observer at r to measure the orbital velocity of a circular orbit to be c at the photon radius distance. $\endgroup$
    – Agerhell
    Oct 22, 2023 at 8:58
  • $\begingroup$ Thanks again @Agerhell. That is indeed interesting. I hadn't taken that time dilation into account and it resolve the puzzle. Thanks for solving the part of the question with respect to observers at infinity. $\endgroup$
    – Toby Ord
    Oct 22, 2023 at 20:18

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