We have images of stars orbiting black holes or black holes destroying near stars, but why do we see the stars moving normally? I mean, if time dilation does exist, shouldn't we see that stars slow down and speed up? Why do we see stars orbiting at a normal rate?

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    $\begingroup$ Seems like all you're doing is looking at an animation an assuming things should be going slower than they look like they're moving. The only way to say the effect isn't being observed is with a quantitative analysis which I'm sure would reveal that they are slowed down exactly as much as predicted. You're just imagining more slowdown than there actually is. $\endgroup$ Commented Jul 26, 2015 at 22:30
  • $\begingroup$ @BrandonEnright i though that being so close to a super-massive black hole it would show a stronger time dilatation effect , also the time scale are just a few years. But as your said with out and analysis we wouldn't exactly know $\endgroup$ Commented Jul 26, 2015 at 22:50
  • $\begingroup$ I think the easiest thing to see would be the precession of the perihelion of their orbits. $\endgroup$ Commented Jul 26, 2015 at 23:12
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    $\begingroup$ @BrandonEnright: be aware that there is expected to be a large amount of non-luminous matter around these objects, so perehelion precession will also be governed by this superious matter. $\endgroup$ Commented Jul 27, 2015 at 1:41

3 Answers 3


Those objects are orbiting closely to SgrA${}^{*}$, certainly, but they are not orbiting closely enough to exhibit significant time dilation effects. In particular, consider the Schwarzschild spacetime. The inner most stable circular orbit around the central obect is at $r = 6M$, three Schwarzschild radii away. This makes the time dilation factor:

$$\sqrt{1-\frac{2M}{r}}= \sqrt{1-1/3} = \sqrt{2/3} = .82$$

So, even the farthest in stable orbit is only running 18% slower than a distant clock. You can cheat at this a bit by giving the central black hole spin, which will draw in the innermost orbit, but generically, you don't see huge time dilation effects for orbiting bodies.

Wikipedia gives the orbit of the closest of those stars, S2, as being 17 light hours. We can now compare this distance to the schwarzschild radius of the black hole to guess how much time dilation we should see.

$$\begin{align} r_{s} &= \frac{2GM}{c^{2}} \\ &= \frac{2\times\bigl(6.11*10^{-11}\; {\rm N \cdot m^{2}/kg^{2}}\bigr)\times\bigl({10^6}\times(2\times 10^{30}\;{\rm kg})\bigr)}{(3*10^8\;{\rm m/s})^{2}} = 2.7 \times 10^{9}\; {\rm m} \\ r_\text{S2} &= 17 \;\text{light-hours} \times (3\times 10^{8} \;{\rm m/s})(60 {\rm \;s/min})(60 \;{\rm min/h}) = 1.8\times10^{13}\; {\rm m} \end{align}$$

So, S2 is roughly ten thousand Schwarzschild radii away from SgrA${}^{*}$, and no significant time dilation is expected. Now, you might ask "why is this evidence that there is a black hole there, then?" The reason why is that this is still a HUGE amount of mass in an area roughly the size of the solar system. General relativity predicts that there is no possible stable configuration of matter of this density that is NOT a black hole.

  • $\begingroup$ fantastic answer with all the hardcore maths. thanks that explain me alot $\endgroup$ Commented Jul 27, 2015 at 0:36
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    $\begingroup$ @eli.rodriguez It's important to understand that Jerry's first paragraph- i.e. the ISCO notion - is rather general and holds for any Schwarzschild metric, it's not an observation confined to the special case of ${\rm SgR^*}$. (Sorry Jerry - I'm making an assumption that the OP will slip up the first time on meeting this concept in the way that I did - no slight meant on your writing style). There's a good, clear, simple writeup of this notion in Box 33.5, Section 33.8 in Misner Thorne and Wheeler "Gravitation". As Jerry says, other metrics can vary from this a bit, but its a .... $\endgroup$ Commented Jul 27, 2015 at 2:53
  • $\begingroup$ ..good, back of the envelope estimate. The ISCO is the radius at which the eigenvalues of the linearized equations of motion of a test particle cross into the right half plane, thus meaning that perturbations to an orbit grow with time rather than being damped. $\endgroup$ Commented Jul 27, 2015 at 2:53

If I'm interpreting your post correctly, you may be misunderstanding time dilation. Time dilation will not cause the stars to seem to move more slowly. The apparent velocity of a star in your frame of reference is the apparent velocity, and relativity will not change it. What time dilation would change is the apparent rate at which a clock moving with the star ticks. So if there was a clock following one of those stars around and we had a very powerful telescope with which to see the time on this clock, we would be able to see it ticking more slowly than our clocks.

I'm not entirely certain, but relativity may cause the apparent velocity of a star in an orbit to change very close to a black hole compared to what we expect from Newtonian mechanics, but this is not just time dilation. And my educated guess is that this would be almost unnoticeable except on very extreme scales (very close to the black hole). For a radius at which a stable orbit is possible, Newtonian mechanics should give a very good approximation so that nothing seems strange.

  • $\begingroup$ i read that if you see a person fall in a black hole you will see him slowly stop and freeze that person, I know that is a very extreme case but in this case i would hope that the black hole in the middle of our galaxy , a super-massive one would be massive enough to show some time dilatation effect to starts orbiting around it $\endgroup$ Commented Jul 26, 2015 at 22:59
  • $\begingroup$ But also we have images of black holes tearing apart stars and gas clouds and i don't see any time dilatation effect $\endgroup$ Commented Jul 26, 2015 at 23:01
  • $\begingroup$ @eli.rodriguez You're talking about effects of presence of event horizon - sth else than time dilation. $\endgroup$
    – Mithoron
    Commented Jul 26, 2015 at 23:28
  • $\begingroup$ @Mithoron Well, the effect he describes does involve time dilation, but as I mentioned, if these are orbiting stably they're definitely far enough that time dilation won't be significant enough to see in a .gif like this, as Jerry showed in his answer. $\endgroup$
    – JohnnyMo1
    Commented Jul 27, 2015 at 0:11
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    $\begingroup$ @eli.rodriguez: also note what I said in my answer when talking about accretion disks. Largely, they are still in orbits until they do the final plunge, but the final plunge is very fast, so the bulk of the luminosity of the disk will come from the plasma orbiting at a radius greater than $6M$ for a Schwarzschild hole. $\endgroup$ Commented Jul 27, 2015 at 0:25

We do "see it" but the effect is smaller than you imagine. As already pointed out, the effects of time dilation are too small to have any visible apparent influence on the orbits in that animation. This is because even the closest orbits are still $>1000$ Schwarzschild radii from the black hole. Nevertheless, the effects have been detected in two ways.

  1. The stars do carry their own observable clocks, in the form of the radiation they emit. For the star S2, which comes within 1400 Schwarzschild radii of the black hole, this causes a relativistic redshift of its spectral lines, part of which is caused by gravitational time dilation (Gravity Collaboration 2018). The time dilation component of this redshift, evidenced by the "clocks" on the stars amounts to an equivalent redshift of 200 km/s in its spectral lines (about 2 parts in 3000).

  2. In slightly more general terms, the time dilation caused by the non-Euclidean spacetime around the black hole has also been found in the non-Newtonian shape of the orbit of this star. The orbit is not a perfect ellipse and the line joining the black hole to the point of closest approach precesses by 12 arcminutes every orbit, as predicted by General Relativity (Gravity Collaboration 2020).

A final point, not made by existing answers is that the orbital velocity measured by a distant observer is essentially the same as the Newtonian prediction. i.e. For an object in a circular orbit at radial coordinate $r$ around a black hole of mass $M$, a distant observer would estimate the orbital speed to be $(GM/r)^{1/2}$ (see here for a proof), which is the same as the Newtonian expectation.


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