Light that is emitted from objects under the effect of time dilation (from out point of view) is redshifted that much is clear. However I am interested if another effect occurs that I never seem to be able to be discussed or dismissed anywhere.

Say we have an object emitting light (or reflecting or whatever) normally we can calculate the amount of photons/ms by looking at the total amount of energy that is being emitted.

Would this mean, that if the same object is under the effect of, for example, gravitational time dilation the detected amount would decrease? The object should still emit x photons/ms but that ms might be twice or longer than that for us.

Therefore shouldn't the "brightness" (in addition to the redshift) change and therefore also the calculated energy output of the object at the point of the observer even if we compensate for the redshift?


You're right. Suppose to Observer A at a source, the source is seen to emit N photons per second, each photon having energy $E = hf$, resulting in a net energy output of $Nhf$. To a distant Observer B the source is seen to emit $kN$ photons per second $(0<k<1)$ because of time dilation. Also, to Observer B the photon frequency $f$ is reduced to $kf$ due to time dilation so the energy per photon is reduced to $kE$. So, to Observer B the energy output of the source is reduced to $kNhkf= k^2Nhf $.

  • $\begingroup$ For completeness, can you show the same for the classical wave? $\endgroup$ – safesphere Dec 30 '18 at 17:04
  • $\begingroup$ That seems like a different question. $\endgroup$ – S. McGrew Dec 30 '18 at 17:45
  • $\begingroup$ Not really. Think of it this way. Does you result apply to the quantum case only or is it more general? I'd be happy to upvote, if you include the classical formula :) $\endgroup$ – safesphere Dec 30 '18 at 17:54
  • $\begingroup$ Not worried about upvotes :) However, the k in my formulas is actually the "gamma" in special relativity. E transforms as E' = gamma x E, and energy density is proportional to the square of E, so energy density I of the E field transforms as I' = gamma^2 x I. $\endgroup$ – S. McGrew Dec 30 '18 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.