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This answer explains that the time dilation for an observer in a circular orbit around a Schwarzschild black hole, relative to a distant observer at rest relative to the black hole, is given by the factor $\sqrt{1 - \frac{3GM}{rc^2}}$. So if the orbiting observer experiences one full orbit to take a time of $\tau$ according to their own clock, and they receive a signal from a distant source once an orbit, presumably that means that time $t$ between sending successive signals as experienced by the distant observer is related to $\tau$ by $t = \frac{\tau}{\sqrt{1 - \frac{3GM}{rc^2}}}$. Since the period of a wave is the inverse of its frequency, I'd also presume (correct me if I'm wrong of course) that this means that if the source is sending light with a frequency of $f_{source}$, this will be related to the average frequency of the received signals $\overline{f_{rec}}$ measured by the orbiting observer over the course of one full orbit by this formula:

$f_{source} = \overline{f_{rec}} \sqrt{1 - \frac{3GM}{rc^2}}$

And if that's right, then since energy is proportional to frequency, if the source is emitting photons with radiant power $P_{source}$ and all the photons are received by the orbiting observer, would the radiant power received by the orbiting observer $\overline{P_{rec}}$ (again averaged over one full orbit) just be given by this formula?

$P_{source} = \overline{P_{rec}} \sqrt{1 - \frac{3GM}{rc^2}}$

Or, do you have to separately account for the fact that the individual photons are blueshifted by that factor and that the rate that new photons arrive is increased by the same factor? If so, would the formula be this?

$P_{source} = \overline{P_{rec}} (1 - \frac{3GM}{rc^2})$

Or something else entirely?

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  • $\begingroup$ I think your assumptions are incorrect since you have forgotten the transverse doppler effect, which does not cancel out in the average and is extremely important for close-in orbits. $\endgroup$ – Rob Jeffries Apr 2 '16 at 8:01
  • $\begingroup$ @Rob Jeffries - Do you mean I've forgotten it in calculating the shifted frequency, or just the power? If the frequency, I was thinking that the relation for successive wave peaks should be the same as the relations for successive discrete signals, like "happy birthday" messages the orbiting observer gets once every orbit...in this case of regularly-spaced discrete signals, if the orbiting observer receives N signals over a clock time T that's an integer multiple of the time they measure for one orbit, the distant observer experiences a period of $T/\sqrt{1 - 3GM/rc^2}$ to send N signals, no? $\endgroup$ – Hypnosifl Apr 2 '16 at 13:31
  • $\begingroup$ @Rob Jeffries - Is it also incorrect in the example I gave in the comment? Say the orbiting observer is at 2 Schwarzschild radii, so that $\sqrt{1 - 3GM/rc^2}$ works out to 1/2. And say the black hole's mass is such that one orbit takes 1 year of proper time. If the orbiting observer gets a discrete pulse once a year on their own clock for 10 years, isn't it the case that the distant observer measured a time of 20 years to send the same 10 pulses? And more generally if they received N pulses (or wave peaks) at regular intervals for 10 years, the distant observer sent those N over 20 years? $\endgroup$ – Hypnosifl Apr 2 '16 at 13:48
  • $\begingroup$ Well, the innermost stable orbit is 3 Schwarzschild radii, so you can't have your example. Your argument about the times is correct, but the two observers would disagree entirely on how much energy was in the pulses. I may have to backtrack on the time dilation factor, it may well be correct. In fact I think I have to have a sit down and think. $\endgroup$ – Rob Jeffries Apr 2 '16 at 13:59
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    $\begingroup$ Yes I see and agree. I think the instantaneous formula is just $f_{source} =f_{rec} \sqrt{1 - 3GM/rc^2} (1 + \vec{v}\cdot \vec{r}/rc)$, which I suppose averages to what you have. $\endgroup$ – Rob Jeffries Apr 2 '16 at 17:27
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While you should note Rob's comment I'd guess the real question is whether the power is multiplied by the time dilation or by the time dilation squared. The answer is that it's the latter.

You had obvious guessed this was the case, and your guess is correct. If the time dilation factor is $f$, where in this case:

$$ f = \frac{1}{\sqrt{1-\frac{3M}{r}}} $$

then the power as measured by the orbiting observer is the original power multiplied by $f^2$.

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  • $\begingroup$ I think this is mixing up time dilation and doppler effect and probably arriving at the right result on average. The observed frequency is I think equal to the time dilation factor multiplied by $(1 + \vec{v}\cdot \vec{r}/cr)$, where $\vec{v}$ is the instantaneous velocity of the observer and $\vec{r}$ is the displacement vector to the source. I guess this averages to the time dilation factor? $\endgroup$ – Rob Jeffries Apr 2 '16 at 17:15
  • $\begingroup$ @RobJeffries: if I understand the OP correctly the question is more general than just an orbiting observer. For example with a static observer is the power increased by $1/\sqrt{1-2M/r}$ or $1/(1-2M/r)$ (it's the latter)? This I think is the key point. I take your point that for an orbiting observer the power (from any non-axial point) will oscillate as the observer orbits, but I don't think that's the OP's main concern. $\endgroup$ – John Rennie Apr 2 '16 at 18:05
  • $\begingroup$ Yes, I agree with that example. But is the statement general? The time dilation factor for a uniformly moving observer is $\gamma$, but the power isn't increased by $\gamma^2$, it is in fact increased by the square of the doppler shift factor - which averages to just $\gamma$ for an observer going around in a circle with constant speed. So do you just multiply by $f^2$ here ? $\endgroup$ – Rob Jeffries Apr 2 '16 at 18:32
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Simple case:

A continuous laser beam is shot at the orbiting observer from a point that is at constant distance from the observer. The observer will receive a beam with constant frequency: source frequency times the time dilation between the source and the observer.

If the sender sends 1 MWh of his energy and that takes one hour of his time, then the receiver will receive more energy in shorter time, so clearly the power is proportional to time dilation squared in this case.

Difficult case:

The laser beam is shot from any other position than in the simple case. I'm guessing that the power is larger in this case, and I'm leaving the calculation for other posters.

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  • $\begingroup$ The frequency seen by the orbiting observer will be doppler shifted by an amount which depends on the phase of the orbit and its orientation with respect to the observer. It would not be constant. $\endgroup$ – Rob Jeffries Apr 2 '16 at 17:10

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