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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?

Community
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Hypnosifl
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2 Answers2

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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 $f^2$.

John Rennie
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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.

stuffu
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