# Inverse proportional to cosmological scale factor

How can I show using calculation that the temperature of the universe is inversely proportional to the cosmological scale factor?

I am just curious, as my textbook states this fact but does not show the calculations, and so I was thinking how one could show this?

I would appreciate the help.

EDIT: so I know for a fact that the equation:

$$z+1=\frac{R}{R_0}$$

Where $$z$$ is the Doppler shift and $$R$$ is the cosmological scale factor in the present and the $$R_0$$ is the scale factor at $$t_0$$. but we also know that:

$$\frac{R}{R_0}=\frac{\lambda}{\lambda_0}$$

But we know from Wien's Law that: $$\lambda T=2.9\times 10^{-3}$$

Hence we have: $$\frac{R}{R_0}=\frac{\lambda}{\lambda_0}=\frac{T_0}{T}=z+1$$

This is all I can extract at this point the relationship between temperature and the scale factor, I do not know if I am on the right track, or if this is correct at all? I would appreciate if someone can help me check and help me go further from here to show that the relationship between $$T$$ and $$R$$ is inversely proportional.

• Fact 1: Early universe was radiation dominated. From statistical considerations, (Stefan-Boltzmann's law) $$\rho_{rad}\propto T^4$$, where $$\rho_{rad}$$ is the energy density for radiation and $$T$$ is temperature.
• Fact 2: Radiation has the following equation of state $$p=\rho/3$$, $$p$$ being pressure. Plugging that into Friedmann's equations you will get $$\rho_{rad}\propto a^{-4}$$, where $$a$$ is the scale factor.
Combining both you easily obtain the known relation, $$a\propto \frac{1}{T}$$.
• You already have the answer in your last equation, $R = (R_0*T_0) /T$. However notice this is an approximation and Redshift is generally more complicated, and model dependent. Also notice that the content of the universe also affects the relation. But for this level I guess is a decent short computation. – ohneVal Mar 25 at 10:24
• Oh, can I just state that: $$\frac{R}{R_0}=\frac{T_0}{T}$$ therefore we have, $R = \frac{k}{T}?$ – Aurora Borealis Mar 25 at 10:26