Why the factor $(1+z)^4$?

I am reading through Gravitation by Misner, Thorne and Wheeler. In the exercice 29.5 (page 794) it is asked to calculate the flux of energy $$S$$ for a distant source in a flat spacetime. The end result is $$S=\dfrac{L}{4\pi r^2 (1+z)^4}$$ where $$L$$ is the luminosity of the source and $$z$$ is the redshift.

Why is there the fourth power of the redshift instead of the general relativistic $$(1+z)^2$$?

Note I am asking for the physical reason, the mathematical one can be found in the cited rerefence by Ellis.

• Could you clarify what $r$ represents. May 4 '20 at 19:26
• @RobJeffries It is the distance from the source May 4 '20 at 19:37
• Which distance?! May 4 '20 at 20:49

The physical reason for this is most transparent when thinking about an energy flux of photons.(However the same can be derived when considering purely classical fields.) There are three physical effects at play:

1. Due to the geometry, the photons get spread over an area that is larger by a factor of $$(1+z)^2$$, reducing the flux by the same factor.
2. Due to time dilation the source appears to be emitting less photons per unit of time. This reduces the photon flux by another factor of $$1+z$$.
3. Again due to time dilation the source appears to be emitting photons of a lower frequency and therefore lower energy. Consequently, the energy flux is decreased by another factor $$1+z$$.

Together these three effects lead to the total energy flux being reduced by a factor $$(1+z)^4$$. This gives a physical (somewhat handywavy) reason for this factor which of course be check by a rigorous calculation (which the OP already did).

• But if $r$ is supposed to be the co-moving distance, then the equation is incorrect isn't it and the factor should be just $(1+z)^2$, which I think deals with #2 and #3. The effect of #1 only applies if $r$ is the angular diameter distance? May 5 '20 at 11:08
• @RobJeffries According to the question in MTW $r$ is the distance between the source and the observer at the time of emission. So indeed there should the factor under #1 due to expansion of the sphere at $r$. May 5 '20 at 12:38

It's all to do with the various forms of distance in cosmology.

The flux is given by the luminosity divided by $$4\pi d_L^2$$, where $$d_L$$ is the "luminosity" distance. In fact, this defines what you mean by the luminosity distance.

In turn, the luminosity distance is related to the "proper distance" $$d_p$$, which I am guessing is what your $$r$$ represents (you don't say and I don't have MTW to hand), in a flat universe by $$d_L = (1+z)^2 d_p \tag*{(1)}$$ So mystery solved, if $$d_p \equiv r$$.

As for why (1) is true, the answer by @mmeent is as good as any.

• Sorry, but this is not giving an answer to physical reason why. May 5 '20 at 12:39
• I agree, but we need to get to the bottom of what$r$ is supposed to be. @mmeent May 5 '20 at 13:23
• MTW is quite clear. $r$ is the distance between source and observer on a time slice at the time of emission. May 5 '20 at 17:55
• @mmeent Fine. As I said in my answer, I don't have MTW at home and despite me asking for clarification, it isn't defined in the question and another answer claims it isn't clear in MTW. May 5 '20 at 17:57

Could it be that the redshift lowers the temperature by $$T\to T/(1+z)$$ and the total emitted radiation is $$\propto T^4$$?

The question, as expressed in MTW, is indeed as clear as mud, since they really do not say (imv) which distance measure they mean by $$r$$. They ask in part b) to compare with 29.27, in which $$R$$ is again poorly defined (imv) but is in fact comoving distance. It seems that they want to define $$r$$ according to local distances in a Lorentz frame. Of course, this definition makes little sense in an expanding cosmology, but we do have $$R = (1+z) r$$ locally. The same formula defines angular diameter distance as described by Rob Jeffries. If you make the substitution, you get the result.