Consider a redshift experiment in the Earth’s gravitational field and assume that a special relativistic theory of gravity exists, which need not be further specified here. For such an experiment we may neglect all masses other than that of the Earth and regard the Earth as being at rest relative to some inertial system. In a spacetime diagram (height $z$ above the Earth’s surface versus time), the Earth’s surface, the emitter and the absorber all move along world lines of constant z. he transmitter is supposed to emit at a fixed frequency from $S_1$ to $S_2$. The photons registered by the absorber move along world lines $\gamma_1$ and $\gamma_2$, that are not necessarily straight lines at an angle of 45◦, due to a possible interaction with the gravitational field, but must be parallel, since we are dealing with a static situation. Thus, if the flat Minkowski geometry holds and the time measurement is given by (2.5), it follows that the time difference between $S_1$ and $S_2$ must be equal to the time difference between $A_1$ and $A_2$. Thus, there would be no redshift. This shows that at the very least (2.5) is no longer valid. The argument does not exclude the possibility that the metric $g_{\mu\nu}$ might be proportional to $\eta_{\mu \nu}$.

How do they conclude that? Why the fact that the proper times are the same imply that there is no redshift?

experiment proper time

SOURCE : (Graduate texts in physics) Norbert Straumann -General relativity-Springer (2013)


1 Answer 1


Emitter says: I emitted $N$ wave crests during time $t$.

Absorber says: I absorbed $N'$ wave crests during time $t'$.

We know that $N'= N$ and the text says that $t'= t$

Well, for me that above looks like the defintion of "no redshift". Absorber received what emitter emitted, no change, no shift of any kind.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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