I am a little confused about proving that the temperature of a redshifted blackbody differs from the intrinsic temperature by the factor (where z is the reshift)$$T=\frac{T_{intrinsic}}{(1+z)}$$ So I've been looking in various sources finding all sorts of proofs that at a certain point seem to contradict each other.I will be listing below 3 proofs that I found:
FIRST PROOF:
This one is usually used in cosmology and uses the first law of thermodynamics and the fact that in its expansion, the universe evolves adiabatically: $$dQ=dU+dL$$ $$dQ=0 \implies dU=-dL$$ The energy density of a universe filled with radiation is $$u=\frac{4\sigma}{c}T^4$$ And the radiation pressure is $$p=\frac{u}{3}$$ $$ dL=pdV$$ and $$dU=d(uV)$$ and $$V=\frac{4\pi R_0^3 a^3}{3}$$ where where "a" is the scale factor and "R0" is the present radius of the universe.Substituting everywhere we obtain that $$\frac{dT}{T}=\frac{da}{a}$$ and knowing that $$a=\frac{constant}{1+z} \implies T_0 =\frac{T}{1+z} $$ where "T0" is the present temperature of the universe
SECOND PROOF:
This is the proof with which I am the most pleased ( also used in cosmology ). We can express the luminosity of a black body as: $$L=\frac{dN}{dt} \cdot \frac{hc}{\lambda}$$ where dN/dt represents photons emited per unit time.Because with the expansion of the universe all distances strech we know that : $$ dt=\frac{constant}{a} ; dN =constant ; \lambda=constant\cdot a$$ where a is the scale factor. We also know that the distance between us and the body increases with 1+z so the observed flux(watts per unit area) will therefor be :
$$F=\frac{L_{intrinsic}}{4 \pi r_{initial}^2 (1+z)^4}$$ This is actually what we observe and this is the definiton of the luminosity distance $$d_L=r_{initial} (1+z)^2$$
THIRD PROOF (no more expansion of the universe)
The Planck distribution law is: $$ B_\lambda=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$$ $$B_{\lambda_0}=\frac{2hc^2}{\lambda_0^5}\frac{1}{e^{hc/\lambda_0 kT_0}-1}$$
If we take a redshifted blackbody we have : $$\lambda=(1+z)\lambda_0$$
The integrated intensities(over the entire spectrum) are : $$\int_{0}^{\infty} B_{\lambda_0} d\lambda_0 = \int_{0}^{\infty}\frac{2hc^2}{\lambda_0^5}\frac{1}{e^{hc/\lambda_0 kT_0}-1}d\lambda_0$$ $$\int_{0}^{\infty} B_{\lambda} d\lambda= \int_{0}^{\infty}\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}d\lambda = \frac{1}{(1+z)^4}\int_{0}^{\infty}\frac{2hc^2}{\lambda_0^5}\frac{1}{e^{hc/\lambda_0 (1+z) kT}-1}d\lambda_0$$ Now if we assume $$T=\frac{T_0}{1+z}$$ And after integrating we obtain the new luminosity(watts) of the blackbody: $$L=\frac{L_0}{(1+z)^4}$$ proving the assumption right: $$T=\frac{T_0}{1+z}$$
My question is why in the second proof the luminosity is less by a factor of $(1+z)^2$ and not $(1+z)^4$ like in the third one. ( in the second proof $(1+z)^2$ also comes from the distance expansion)
Thank you!