<p>The answer is yes. The time dilation between the observer and the neutron star is a function of both the neutron star speed in its orbit around the black hole and its position in the gravitational potential. The two combined result in a &quot;redshift&quot; in the sense that <em>everything</em> will appear to slow down, including light emitted from the neutron star system and the orbital period of the planet going around it.</p>
<p>Note though that the neutron star will itself appear to move with the same speed as predicted by Newtonian gravity for its orbital radius. This may sound paradoxical but I think the root of this is that the &quot;orbital radius&quot;, while accurately giving the circumference of a circular orbit in Schwarzschild geometry is not the distance to the centre.</p>
<p>The effects will be stronger at periastron (neutron star closest to the black hole), both because it is deeper in the gravitational potential and because it is moving faster.</p>
<p>These effects have now been confirmed observationally using stars that orbit close to the supermassive black hole at the Galactic centre (<a href="https://arxiv.org/abs/1807.09409" rel="nofollow noreferrer">Gravity collaboration: 2018</a>).</p>

CC BY-SA 4.0
2023-01-12T19:38:24.813Z
@ProfRob This isn't *quite* a Schwarzschild vis-viva equation, but Jim Branson gives a nice derivation for the effective potential: https://hepweb.ucsd.edu/ph110b/110b_notes/node79.html He also gives an (elliptic) integral for photon trajectories.
PM 2Ring
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2022-12-01T12:37:30.357Z
thank you for that answer! So, for the observer, the Neutron Star and its Trabant will slow down at Periastron instead of getting faster, as orbiting objects would normally do. Would it look streched while doing so? I mean, it passes a rapidly changing gravitational environment, so I would guess so
Confused Merlin
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2022-12-01T16:58:33.657Z
"Note though that the neutron star will itself appear to move with the same speed as predicted by Newtonian gravity for its orbital radius." I don't think that holds for eccentric orbits.
TimRias
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2022-12-01T17:09:35.627Z
@TimRias I'm sure you are right. What is the equivalent of the vis-viva equation?
ProfRob
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2022-12-01T11:12:52.533Z
2022-12-01T12:59:41.810Z
2022-12-01T12:59:41.810Z
ProfRob
<p>This is a mostly hypothetical question and thought experiment, and I am aware of some aspects plausibility of it are open for debate. Also, this is Version 2 of the question, as the first question was a bit unclear and quite long.</p>
<p><strong>Set Up</strong></p>
<p>Take a Black Hole. It is orbited by a Neutron Star, which itself is orbited by a Planet (close enough to stay with it, but not so close that tidal forces would damage it).</p>
<p>The Neutron Star's orbit is eccentric, so apoapsis and periapsis are sufficiently different to allow different relativistic effects at both. But periapsis is not so close, that tidal forces would affect the Neutron Star and allow it to keep its planet.</p>
<p>Add an observer, who can observe all of that from a point at the ecliptic, at a distance, where the motion of the Neutron Star's Planet still can be detected with optical means.</p>
<p><strong>Question</strong></p>
<p>Will the Observer notice differences in the Neutron Star's Planet orbital speed?</p>
<p>I would guess &quot;yes, but it might be too minor to notice without proper equipment&quot;, but then I get lost in a follow up question, which clogged my initial  question too much. Like &quot;what is with the orbital velocity, if the Neutron Star seems to slow down?&quot;, but that is another question (maybe).</p>

CC BY-SA 4.0
2022-12-01T10:08:16.557Z
TLDR: Wall of text. Be aware that we can see (classical) keplerian motion of doppler-shifted gas around supermassive of neighbouring galaxies. General relativistic effects only come into play at ~3 Schwarzschild-radii from the BH.
AtmosphericPrisonEscape
CC BY-SA 4.0
2022-12-01T10:36:56.733Z
@ProfRob, like this? I agree that the observers position can be simplified, and most of my thoughts on that topic do not help clarifying what I actually want to know. The eccentric orbit, at the other hand, should stay, because the actual differences in the motion of the Planet is what I am care about now.
Confused Merlin
<black-hole><orbit><general-relativity>
Observable Speed of objects orbiting a black hole
Confused Merlin
# Observable Speed of objects orbiting a black hole

(Asked by: username_0 on 2022-12-01)

This is a mostly hypothetical question and thought experiment, and I am aware of some aspects plausibility of it are open for debate. Also, this is Version 2 of the question, as the first question was a bit unclear and quite long.

**Set Up**

Take a Black Hole. It is orbited by a Neutron Star, which itself is orbited by a Planet (close enough to stay with it, but not so close that tidal forces would damage it).

The Neutron Star's orbit is eccentric, so apoapsis and periapsis are sufficiently different to allow different relativistic effects at both. But periapsis is not so close, that tidal forces would affect the Neutron Star and allow it to keep its planet.

Add an observer, who can observe all of that from a point at the ecliptic, at a distance, where the motion of the Neutron Star's Planet still can be detected with optical means.

**Question**

Will the Observer notice differences in the Neutron Star's Planet orbital speed?

I would guess "yes, but it might be too minor to notice without proper equipment", but then I get lost in a follow up question, which clogged my initial question too much. Like "what is with the orbital velocity, if the Neutron Star seems to slow down?", but that is another question (maybe).

**Tags:** black-hole, orbit, general-relativity
> **Comments:**
> **username_1**: TLDR: Wall of text. Be aware that we can see (classical) keplerian motion of doppler-shifted gas around supermassive of neighbouring galaxies. General relativistic effects only come into play at ~3 Schwarzschild-radii from the BH.  
> **username_0**: @ProfRob, like this? I agree that the observers position can be simplified, and most of my thoughts on that topic do not help clarifying what I actually want to know. The eccentric orbit, at the other hand, should stay, because the actual differences in the motion of the Planet is what I am care about now.  

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## 1 answers

### Answer by username_2 on 2022-12-01. Score: 7

The answer is yes. The time dilation between the observer and the neutron star is a function of both the neutron star speed in its orbit around the black hole and its position in the gravitational potential. The two combined result in a "redshift" in the sense that *everything* will appear to slow down, including light emitted from the neutron star system and the orbital period of the planet going around it.

Note though that the neutron star will itself appear to move with the same speed as predicted by Newtonian gravity for its orbital radius. This may sound paradoxical but I think the root of this is that the "orbital radius", while accurately giving the circumference of a circular orbit in Schwarzschild geometry is not the distance to the centre.

The effects will be stronger at periastron (neutron star closest to the black hole), both because it is deeper in the gravitational potential and because it is moving faster.

These effects have now been confirmed observationally using stars that orbit close to the supermassive black hole at the Galactic centre ([Gravity collaboration: 2018](https://arxiv.org/abs/1807.09409)).

> **Comments:**
> **username_0**: thank you for that answer! So, for the observer, the Neutron Star and its Trabant will slow down at Periastron instead of getting faster, as orbiting objects would normally do. Would it look streched while doing so? I mean, it passes a rapidly changing gravitational environment, so I would guess so  
> **username_4**: "Note though that the neutron star will itself appear to move with the same speed as predicted by Newtonian gravity for its orbital radius." I don't think that holds for eccentric orbits.  
> **username_2**: @TimRias I'm sure you are right. What is the equivalent of the vis-viva equation?  
> **username_3**: @ProfRob This isn't *quite* a Schwarzschild vis-viva equation, but Jim Branson gives a nice derivation for the effective potential: https://hepweb.ucsd.edu/ph110b/110b_notes/node79.html He also gives an (elliptic) integral for photon trajectories.  

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