Suppose I have a pulsar at some radius $r_p$ from a black hole and an observer at a radius $r_o$ where $r_o \gg r_p$.

Assume that both the pulsar and observer are stationary. The pulsar 'pulses' at $\tau_1$ and emits a photon which travels along a null geodesic to an observer. At time $\tau_2$, a period $\Delta \tau_p$ later, it pulses again, emitting a second photon which travels along the same geodesic.

We can integrate the photon path from $r_p \rightarrow r_o$ and so determine the total coordinate flight time $t$.

Photons 1 and 2 then arrive at the observer at time $\tau_1 +t$ and $\tau_2 +t$

My question is: What period does the observer say that the pulsar has? The pulsar is in a region of stronger gravity, so its clock should run slower compared to the observer's clock, but it seems to me that since the photons both travel along the same geodesic, the observer will say that the period is also $\Delta \tau_p$?


1 Answer 1


You are overcomplicating this.

For convenience let's have both observers zero their clocks when the first photon is emitted, then the second photon is emitted at time $\tau_p$ for the pulsar observer and time $\tau_p/\sqrt{1-2M/r}$ for the distant observer, so the distant observer observes the period to be $\tau_p/\sqrt{1-2M/r}$. (I'm assuming the distant observer is far enough from the black hole for their time dilation to be negligible.)

Both photons take the same coordinate time to make the trip (though the two observers will disagree about the value of this travel time) so the travel time has no bearing on the observed time between the two pulses. That is, if the travel time measured by the distant observer is $T$ then the distant observer will receive the two photons at time $T$ and $\tau_p/\sqrt{1-2M/r}+T$.


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