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http://en.wikipedia.org/wiki/Tolman_surface_brightness_test

It says:

In a simple (static and flat) universe, the light received from an object drops inversely with the square of its distance, but the apparent area of the object also drops inversely with the square of the distance, so the surface brightness would be independent of the distance. In an expanding universe, however, there are two effects that reduce the power detected coming from distant objects. First, the rate at which photons are received is reduced because each photon has to travel a little farther than the one before. Second, the energy of each photon observed is reduced by the redshift. At the same time, distant objects appear larger than they really are because the photons observed were emitted at a time when the object was closer. Adding these effects together, the surface brightness in a simple expanding universe (flat geometry and uniform expansion over the range of redshifts observed) should decrease with the fourth power of (1+z). To date, the best investigation of the relationship between surface brightness and redshift was carried out using the 10m Keck telescope to measure nearly a thousand galaxies' redshifts and the 2.4m Hubble Space Telescope to measure those galaxies' surface brightness.[1] The exponent found is not 4 as expected in the simplest expanding model, but 2.6 or 3.4, depending on the frequency band.

I'm trying to understand where $(1 + z)^4$ comes in.

Specifically, it says:

there are two effects that reduce the power detected coming from distant objects.

Question

Do the two effects each contribute by $(1 + z)^2$, which are then multiplied together?

It also says

First, the rate at which photons are received is reduced because each photon has to travel a little farther than the one before.

Let's say just this factor was removed. Would the predicted brightness then be reduced by $(1 + z)^2$?

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The surface brightness is defined as the flux per solid angle $$S=\frac{f}{\Omega}$$. The solid angle is $$\Omega=\pi\bigg(\frac{D}{d_A}\bigg)^2$$ where $d_A$ is the angular diameter distance. This gives us $$S=\frac{f d_A^2}{\pi D^2}$$ where the angular diameter distance is related to the luminosity distance (see http://en.wikipedia.org/wiki/Luminosity_distance) by $$d_A=\frac{d_L}{(1+z)^2}.$$ You can arrive at this result for the angular diameter distance by combining the two equations for angular diameter distance and luminosity distance (see the wikipedia page for distance measurements in cosmology http://en.wikipedia.org/wiki/Distance_measures_(cosmology)). Combining this relation with that of the surface brightness gives you $$S=\frac{f d_L^2}{\pi D^2 (1+z)^4}$$ which is decreasing with the 4th power of (1+z) as you stated above. The actual exponent for surface brightness will depend on the form of the luminosity distance from the cosmological model used so you only get the $(1+z)^{-4}$ relation if luminosity distance is independent of z. It is also interesting to note that in the stationary case where $d_A=d_L$ there is no (1+z) dependence. Hope this helps.

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