# Einstein equivalence principle cannot entirely predict gravitational time dilation

The Einstein equivalence principle can be used to derive the gravitational redshift of photons, but it does so in an unusual way. The derivation is as follows. Consider an source of photons on the ground and a receiver at the top of a tower of height $H$. At the time when a photon is emitted we move into the freely falling frame and after a time $t \approx H/c$ the photon will have reached the top of the tower. During this time interval the receiver has accelerated to a velocity $v=gt = gH/c$ away from the photon and therefore will detect the photon as redshifted with frequency, $$\nu_r = \nu_e\left(1+\frac{v}{c}\right) = \nu_e\left(1-\frac{gH}{c^2}\right).$$

When performing the proper derivation we rely on two events separated by a time $\Delta t$, which may be two distinct photon emissions. Under this derivation we see that not only will individual photons be redshifted, but that a sequence of photons will arrive later by the same redshift factor. This is the true effect of gravitational time dilation and as far as I can tell, it cannot be predicted by using only the Einstein equivalence principle.

Perhaps what I find most interesting is the fact that the equivalence principle derivation hinges on the fact the a photon is a quantum object which is emitted and absorbed at a single instant. It is almost as if the predictions of the EEP rely on this aspect of quantum theory.

Is it possible to construct a thought experiment using the Einstein equivalence principle alone, that can predict both gravitational redshift of $n$ photons along with the delay in their arrival times?

You can derive gravitational time dilation from the equivalence principle without any quantum mechanics of any type whatsoever. And I daresay that is the common method.

You can consider a frame at rest with the top of a tower at a particular time. It will have been rushing upwards at the bottom so that it can accelerate downwards and finally come to rest right at the top. According to its own proper time it sees a frequency of classical light at the top. And it sees a frequency of classical light at the bottom back when it left. Compared with the frame that was at rest when-where the light left, this frame sees it arrive red shifted.

And you never have to bring up a quantum anything. You don't even have to use light, its just that a freely falling frame that is at rest at a moment up high used to be rushing upwards when it was down lower. And then you use regular special relativity time dilation for moving frames and get the result you need.

Keeping in mind that the inertial frames should be infinitesimal in size and duration.

• Yes. I understand this to be the standard derivation, but it assumes that the light is emitted and received instantaneously. If you try to reformulate this argument using two events, each separated by one period of the lights frequency, then you will find that the time between these two events is unchanged. This is in fact the very same method by which ones derives gravitational time dilation using the metric (a single event cannot be used). – Hedra Nov 1 '15 at 6:09
• I can even rephrase this again, two photons emitted $\Delta t$ seconds apart at the bottom, will be received $\Delta t (1 + \frac{gH}{c^2})$ seconds apart at the top, but as far as I can tell, this cannot be shown with the equivalence principle. – Hedra Nov 1 '15 at 6:09
• @Hedra You aren't rephrasing, you claim I make an assumption that I don't make, then claim you can't get a result that you can get. I don't know why you say the things you say. – Timaeus Nov 1 '15 at 6:22
• I'm sorry that we are not understanding each other and I don't intend to frustrate or offend. I'll try to give your answer more thought. – Hedra Nov 1 '15 at 6:35
• I see the distinction in your scenario. I am happy with the infinitesimal approach using time dilation. My issues appear to be with the derivations which make use of a Doppler-shift effect. – Hedra Nov 1 '15 at 7:18