# Communication near a black hole

One ship is in a stable orbit outside of the event horizon of a very large black hole so that there is significant time dilation.

Another ship is in its own orbit around the same black hole, but at a much greater distance, so that time dilation is minimal.

If someone on the ship much farther away turns a very bright light on and off again every 10 seconds, what will the observer on the ship near the black hole see?

• I don't understand why this got downvoted. This seems like a perfectly reasonable question to me. – Ben Crowell Mar 23 at 18:57
• Then why not up-vote it to cancel the downvote? – Rohit Pandey Mar 23 at 19:36
• Just to clarify, since someone identified an ambiguity below, what I meant was that someone briefly turns the light on and off again every 10 seconds. However, it does not matter whether it is that or that the light gets turned on for ten seconds and then off for ten seconds, repeatedly. The point is to figure out what happens to the timing. – maude_mouse Mar 24 at 19:07

This was a nice question. I enjoyed working it out, and it turned out to be more nontrivial than I might have thought initially. It's totally weird that there were two unexplained drive-by downvotes on the question -- I hope you won't be driven away from the site by that.

This is essentially the same question as if you ask what the Doppler shift is for a distant star, as seen by an observer in a circular orbit. People usually think of the Doppler effect in a context like this as changing the color of light, but it is also the factor by which the 10 second period of the flashes will vary.

For answers to a similar question, see What will the universe look like for anyone falling into a black hole? The Doppler shift is dependent on the direction from which the light is arriving. The amount of the effect depends on the angular position of the observer in the close orbit relative to the observer in the distant orbit.

In my answer to the other question, what I simulated in the video was the view of an observer infalling radially from rest at infinity. In that case, we get both redshifts and blueshifts, depending on the angle. This is simpler to see for sufficiently large orbital radii, because for those radii the result is well approximated by a Newtonian Doppler shift.

For a circular orbit, I don't know of any simple way to prove, without a calculation, that both red and blue shifts occur, even at small radii. However, a calculation shows this to be true. An outline of the calculation is as follows. All units are such that $$G=1$$, $$c=1$$, the Schwarzschild radius is also 1, and vectors are expressed in Schwarzschild coordinates $$(t,r,\theta,\phi)$$. The orbiting observer is in the plane $$\theta=\pi/2$$. The $$\pm$$ signs refer to the extreme cases of the orbiting observer detecting a ray of light from the forward direction and the backward direction. The velocity vector of the orbiting observer is

$$u'=\left(1-\frac{3}{2r}\right)^{-1/2}(1,0,0,2^{-1/2}r^{-3/2})$$.

Let the velocity vector of the ray at detection, with an arbitrary choice of affine parameter, be

$$v'=(1,0,0,\pm (1-1/r)^{1/2}r^{-1})$$.

The velocity vector of the distant observer emitting the ray is

$$u=(1,0,0,0)$$.

We would also like to extrapolate backward in time to find $$v$$, the velocity vector of the ray upon emission by the distant observer. The complete vector probably can't be found in closed form, but because there is a conserved energy, we can get the only component we need in closed form as

$$v=(1-1/r,\ldots)$$.

The Doppler shift is

$$\frac{\omega'}{\omega} = \frac{u_av'^a}{u_bv^b}=\left(1-\frac{3}{2r}\right)^{-1/2}\left[1\mp\left[2r(1-1/r)\right]^{-1/2}\right]$$.

The graph of this result as a function of $$r$$ looks like this:

So it is true, all the way down to the radius of the innermost circular orbit (not just the innermost stable circular orbit), that the observer will sometimes see the flashes at intervals shorter than 10 s, and sometimes at longer intervals. It will vary as they go through each orbit.

For anyone interested in seeing slightly more detail, I have this written up as a homework problem, with the solution in the back of the book, in ch. 7 of my general relativity book.

As a numerical example, suppose that the observer is in the innermost stable circular orbit (ISCO), with $$r=3$$. Then the Doppler shift factor varies between 0.71 and 2.12, which means that the 10 second period of the signals is perceived as varying from 4.7 s to 14.1 s.

Unstable circular orbits are possible for any $$r>1.5$$. As an extreme example, $$r=1.5001$$ gives a period varying from 0.041 s to 820 s.

Because of time dilation the near black hole spacecraft will perceive the intervals as shorter than ten seconds.

If the spaceship far away are situated at a negligible depth in the gravitational potential and has a negligible velocity in relation to the black hole and shines the light for ten seconds and then shuts it off for ten seconds and the spaceship near the black hole was "hoovering" with no velocity in relation to the black hole the spaceship near the black hole would experience the light to be shining for an amount of time equal to:

$$\text{time of light shining}= \text{ten seconds} \times{\sqrt{1-\frac{2GM}{rc^2}}}$$

and then be swithched off for the same amount of time. The light will also be blueshifted (wavelentgh shortened) by the same factor. This effect is known as "gravitational time dilation" The people on the spaceship near the black hole will think that the light has shorter frequency than the people on the spaceship far away. The factor above goes by names as "gravitational time dilation" or "gravitational redshift/blueshift".

If the spaceship near the black hole is in circular orbit the time of light shining should instead on average be:

$$\text{time of light shining}= \text{ten seconds}\times{\sqrt{1-\frac{2GM}{rc^2}}\sqrt{1-\frac{v^2}{c^2(1-\frac{2GM}{rc^2})}}}$$

Using $$v=\sqrt{\frac{GM}{r}}$$ this could be written as:

$$\text{time of light shining}= \text{ten seconds}\times{\sqrt{1-\frac{3GM}{rc^2)}}}$$

Besides from this effect of time dilation you also have doppler shift due to the fact that the near black hole spacecraft sometimes moves at least somewhat towards the distant aircraft and sometimes somewhat away from it.

• How did you arrive at the second expression? This is assuming that the velocity "v" is calculated by a far away observer. This isn't really a definition of $v$. – Ben Crowell Mar 23 at 23:14
• If you want to calculate for instance the motions of all major celestial bodies in our solar system over time you do not want to bother with the fact that people with clocks on the different planets will disagree on the rate of time. Is it something wrong with the expression? It is meant to be the time dilation due to velocity in a gravitational field where the velocity of light according to a distant observer is less than c. – Agerhell Mar 23 at 23:44
• Is it something wrong with the expression? I don't know if your expression is right or wrong, but you haven't said where it came from. You say it's some kind of averege, but you haven't shown how you averaged. And your verbal descriptions of $v$ are not enough to define what you mean by $v$. – Ben Crowell Mar 24 at 0:08
• The denominator can be written as $\sqrt{1-\frac{3GM}{rc^2}}$ using $v=\sqrt{GM/R}$ as you have a circular orbit. Velocity v as measured from an observer at rest with respect to the black hole and located infinitely far away so you do not have to bother with the gravitational potential of the observer. If you switch the lamp on and of enough times sometimes the near black hole spaceship will have a net velocity towards the incoming light and sometimes away from it, changing the expression I wrote. That the time average of this effect is zero remains to be proven. – Agerhell Mar 24 at 1:14
• The expression you're using for a circular orbit is a Newtonian one, and velocities in GR are not in general things that can be expressed except locally. – Ben Crowell Mar 24 at 20:25

So we have Alice and Bob, Bob is in an orbit close to the black hole (just outside of the event horizon) and Alice is orbiting the black hole at a large radial distance.

Irrespective of the orbit, Bob has to accelerate away from the black hole in order to prevent himself from falling beyond the event horizon (from Alice's perspective (the orbit provides the acceleration)). The net acceleration of Bob from Alice's perspective is zero (because he is forcing a stable orbit with fixed radial distance from Alice's perspective) and he is travelling outward radially at constant velocity. This means that any light which has a negative radial component to it's momentum (travelling towards the event horizon) will be blue shifted from Bob's perspective because he is travelling towards it.

In addition, gravitational time dilation will cause a clock in Bob's frame of reference to run slower relative to Alice's clock. This means that the period between the switching from Alice will appear to occur faster in Bob's frame of reference and that light will be further blue shifted due to the time dilation.

So the light emitted from Alice is blue shifted in Bob's frame of reference and the period of the switching is smaller in Bob's frame relative to Alice.

• Irrespective of the orbit, Bob has to accelerate away from the black hole in order to prevent himself from falling beyond the event horizon (from Alice's perspective (the orbit provides the acceleration)). The question states that they're both in stable orbits. This is possible without any thrust (assuming they're outside the radius of the innermost stable circular orbit). – Ben Crowell Mar 23 at 18:51

As per the correct comments, my answer is based on the fact that you want the orbit to be near the innermost stable circular orbit. If you want the orbits be different (radially infalling observer), the effect could be either red or blueshift.

It is a misconception to think that time dilation exist at a single point in space. Time dilation is defined so, that you can interpret it between two points in space.

In you case these two points in space are the two spaceships. One spaceship is near the black hole's event horizon, and the other one is far away from the gravitational zone of the black hole.

It is another misconception to think that mass causes gravity (and gravitational time dilation). In reality it is stress-energy.

In your case there is a significant difference of stress-energy between the two spacehips's location. The spaceship closer to the event horizon will experience greater stress-energy from the black hole, and the spaceship far away will experience significantly less.

Now in your case, the clock on the spaceship near the event horizon will run slower then the clock on the spaceship far away, if they compare them.

So you are saying that the light on the spaceship far away (where the clock relatively runs faster) turns on and off every 10 seconds. It is not clear whether the light will be on for 10 sec and then off for 10 sec or it will just shine for a moment every 10 sec. But I will take it as the light being on for 10 sec and then off for 10 sec.

Now the clock on the spaceship far away, where the lightsource is, runs relatively faster, so 10 sec there will mean maybe only 1 sec on the spaceship near the horizon.

Photons emitted from the lightsource will be emitted for 10 sec measured locally far away. But measured near the horizon, the photons will be emitted only for 1 sec. And then no photons for 1 sec. And then again photons emitted for 1 sec.

So the observer near the horizon will see as if the photons were emitted only for very short periods. And then the periods when no photons are emitted are very short too. So they will see a fast flashing light.

Now the person at the lightsource, far away from the black hole, will see the light shine for 10 sec, and then no light for 10 sec. Se they will see a slower flashing light.

• The beginning of this answer talks about two misconceptions, but the OP hasn't actually said anything that would make this relevant. You seem to be claiming that the Doppler shift is always a blueshift, but have you done any calculations to back up this claim? I don't think it's true for all radii, although it's possible that it is true for small radii, near the innermost stable circular orbit. In the case of a radially infalling observer, the shift can be either a redshift or a blueshift, depending on the geometry. – Ben Crowell Mar 23 at 19:17
• Thank you I edited. – Árpád Szendrei Mar 23 at 22:26
• A calculation is required if you want to support your claim that the Doppler shift is always a blueshift for an observer near the ISCO -- and the OP didn't say they wanted the observer to be near the ISCO. – Ben Crowell Mar 23 at 23:15