tom.stoer said:
I think we should try to be more exact.
There is a Doppler shift ω → ω' = βω;
There is a trf. of Ω → Ω' = β-2Ω;
I'm not sure what you mean. What' trf. an abbreviation for, and what does the variable Ω represent?
Since we're trying to be exact, the relativistic doppler shift is
z =\sqrt{\frac{1+\beta}{1-\beta}} = \frac{1+\beta}{\sqrt{1-\beta^2}} = (1+\beta)\gamma
where ##\beta = v/c##
The frequency and wavelength of a pulse of light is shifted by a factor of z. As a consequence, the length of a light pulse is shifted by a factor of 1/z.
The energy density (usually denoted ##T^{00}## in GR) scales as z^2, a factor of z comes from the shift in frequency (giving each individual photon more energy) and another factor of z comes from an increased photon arrival rate. I worked out the details of the transformation of the stress-energy tensor T in the previous thread.
Wikki gives, for a black body
<br />
I = \left( \frac{2\,h\,\nu^3}{c^2}\right) \left(\frac{1}{e^{\frac{h\,\nu}{k\,T}}-1} \right)<br />
where I is the energy/unit time, or the power. Given the energy /unit volume ##\rho##, I would be equal to ##\rho## * (volume) / (time) == ##\rho## * (area) * (length)/(time) = ##\rho## * (area) * c. So I and ##\rho## are proportional , I = ##\rho A c##
Neither A or c should be affected by the lorentz transform, so I should transform in the same manner as ##\rho##.
Thus we can say that for a black body (I'm not sure how good an approxiation the sun is of one)
<br />
I = (1+\beta)^2 \gamma^2 \left( \frac{2\,h\,\nu_0^3}{c^2}\right) \left(\frac{1}{e^{\frac{h\,\nu_0}{k\,T}}-1} \right)<br />
where ##\nu_0## is the frequency of emission.
Substituting ##\nu_0 = \frac{\nu}{\gamma \, \left(1+\beta\right)}## where ##\nu## is the frequency of reception should give the received intensity law as a function of received frequency, if I haven't made any errors.
This gives a black-body like spectrum, I think - but multiplied by a diferent scale factor.