# Why does the CMB conform to black-body formula for 2.71 degrees, when it's the stretched emission at a far hotter temp?

Physicist Chad Orzel in "Einstein's Legacy" discusses Planck's Black-Body formula, stating that it fits perfectly to everything we see, from toasters to stars. Fine. Then he says it perfectly describes the radiation curve of the CMB as belonging to a black body at 2.71 degrees. I'm confused, since the CMB was not EMITTED at that temp, but at one with a very different shaped curve. How would stretching those original frequencies out by a factor or a thousand (or whatever it is) result in a curve representative of 2.7 degrees?

The Planck spectrum of a blackbody is given by $$B(\nu, T) = \frac{2h\nu^3}{c^2} \left( \exp[h\nu/k_BT] - 1\right)^{-1} \ ,$$ where $$\nu$$ is the frequency, $$T$$ is the temperature and $$h$$ and $$k_B$$ are the usual Planck and Boltzmann constants and $$c$$ is the speed of light in vacuum.

If you transform the frequencies of light by applying a redshift $$z$$, then the frequencies decrease by a factor of $$1+z$$ $$\nu' = \nu/(1+z)\ .$$

In which case, the Planck function just becomes $$B' = \frac{2h\nu'^3(1+z)^3}{c^2} \left( \exp[h\nu'(1+z)/k_BT] - 1\right)^{-1} \ ,$$ which can just be rewritten as $$B' = (1+z)^3\ \frac{2h\nu'^3}{c^2} \left( \exp[h\nu'/k_BT'] - 1\right)^{-1} = (1+z)^3 B(\nu', T') \ ,$$ where $$T' = T/(1+z)$$.

i.e. This has just been transformed into a new Planck function at a lower temperature and with a different normalisation.

• Impressive equations, but am still struggling. Forget the CMB. The graph of intensity vs wavelength for 6000 K is very steep, whereas the graph at 3000 K has a low profile. If 6000 K black-body EM had wavelengths stretched 2x, how could the graph become like 3000 K's graph? Jan 28, 2023 at 19:21
• @GulbenkianD not sure what you mean by low profile. Or what you mean by the graph of intensity versus wavelength is steep. If a 6000K blackbody spectrum had all its wavelengths stretched by 2 it would have exactly the same shape as a 3000K blackbody. Jan 28, 2023 at 19:57
• If you are worried about the intensity, remember that the energy density is also being decreased by the expansion of the universe. @GulbenkianD Jan 28, 2023 at 20:00
• @GulbenkianD at long wavelengths the intensity goes to zero - there's no one value you can point to and say that the maximum is 10x greater. You're comparing apples to oranges. This answer shows that the graph does have the same shape, but squeezed in the vertical direction and stretched in the horizontal one. Jan 29, 2023 at 13:43
• @GulbenkianD Your impression that the shape changes is only an optical illusion when you plot the Planck curve in a linear fashion. If you plot it logarithmically, you can see that a temperature change only results in a upward/downward and sideways displacement of the curve. See commons.wikimedia.org/wiki/File:Planck_law_log_log_scale.png . This also follows algebraically from the Planck formula as shown in my answer below. Jan 29, 2023 at 16:04

According to the Hubble law all radiation is progressively red-shifted with distance traveled, with the wavelength changing from $$\lambda$$ to $$\lambda(1+z)$$ (with $$z>0$$). Now the formula for the Planck black body curve is

$$B_\lambda=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/(\lambda kT)}-1}$$

You can see that if you replace $$\lambda$$ with $$\lambda(1+z)$$ in the exponential factor, the original shape is retained by replacing $$T$$ with $$T/(1+z)$$. So the original temperature appears to have been reduced by a factor $$1/(1+z)$$ by the red-shifting of the spectrum. In addition, the absolute intensity has dropped because you also get a factor $$1/(1+z)^5$$ due to the red-shift.

From the double-logarithmic plot below you can see that a temperature change (be it real or apparent due to the redshift) does not change the shape of the Planck curve but only displaces it upwards/downwards and sideways. Your impression that the shape changes is only an optical illusion when you plot the curve linearly (as it is usually done).