On the back page of the CPEP poster “The History and Fate of the Universe” a table gives distance scale (size), temperature, and energy (radiation) for seven expansion times (10-43 sec to now). http://www.cpepphysics.org/main_universe/history-chart.html

I have a reference for calculating temperatures in a radiation and matter dominated space but not for distance and energy. Appreciate some help.

  • $\begingroup$ Note that the step from era 1 to era 2 is speculation. We don't know by how much the universe exponentially expanded during the inflationary epoch. $\endgroup$ – Rob Jeffries Aug 26 '15 at 22:35
  • $\begingroup$ ^ what he said. All we know is the minimum it would have had to expand $\endgroup$ – Jim Aug 27 '15 at 15:47
  • $\begingroup$ Related physics.stackexchange.com/q/68493 $\endgroup$ – Rob Jeffries Aug 27 '15 at 18:53

If I understand correctly, you are just asking about the relation between energy and distances in both radiation and matter (and cosmological constant) dominated eras of the expansion of the universe.

Consider the Einstein equation

$$ G_{\mu\nu} = 8\pi G T_{\mu\nu} \ ,$$

where $G$ is Newton's constant. In a FLRW unverse $G_{\mu\nu}$ is diagonal and using standard cosmology (homogeneity and isotropy), the energy-momentum tensor takes the form $T_{\mu\nu}=\text{diag}(\rho,-p,-p,-p)$, where $\rho$ is the energy density and $p$ the pressure. The $00$ component of the Einstein equation is sum of inverse squared spatial curvature $R_c=\pm \frac{a}{\sqrt{|k|}}$ and inverse squared Hubble radius $R_H$

$$ G_{00} = 3 \left[\frac{1}{R_c^2} + \frac{1}{R_H^2} \right]=3\left[\frac{k}{a^2} + \left(\frac{\dot{a}}{a}\right)^2\right]=8\pi G\rho \ , $$

where $k$ is the (constant) curvature, $a$ the scaling factor and the factor $3$ comes from the three spatial dimensions. The ratio $\frac{\dot{a}}{a}=H$ is called the Hubble constant (it is not really a constant, since $a=a(t)$). We rewrite

$$ H^2= \frac{8\pi G}{3} \rho - \frac{k}{a^2} \ . $$

From the Friedmann equations we get the energy conservation equation

$$\dot{\rho}=-3\frac{\dot{a}}{a}(\rho+p) \ . $$

Now we solve this equation in two scenarios:

Radiation domination
The system behaves like ultrarelativistic matter, where we know from statistical thermodynamics, that $p=\frac{\rho}{3}$. Solving

$$ \dot{\rho} = -3\frac{\dot{a}}{a}\left(1+\frac{1}{3}\right)\rho = -4 \frac{\dot{a}}{a}\rho $$

we see that $\rho \sim a^{-4}$ (the energy density scales with the scaling factor $a^{-4}$). And since $\rho \sim H^2$ we get $a(t) \sim t^{1/2}$.

Matter domination
The system consits of nonrelativistic matter, where the negligible kinetic energy is expressed as $p=0$. The solution to

$$ \dot{\rho} = -3 \frac{\dot{a}}{a}\rho $$

is $\rho \sim a^{-3}$ and thus $a(t) \sim t^{2/3}$.

$\Lambda$ domination
In a cosmological constant, or $\Lambda$ dominated universe, the Einstein equations take the form

$$ G_{\mu\nu} + \Lambda g_{\mu\nu} = 8 \pi G T_{\mu\nu} \ , $$

where $g_{\mu\nu}$ is the metric. We follow the same steps as above and get

$$ H^2= \frac{8\pi G}{3}\rho - \frac{k}{a^2} + \frac{\Lambda}{3} \ . $$

If the last term dominates, we have

$$ \frac{\dot{a}}{a} \sim \text{constant} \ , $$

which means that $\dot{\rho}\sim \rho$ and thus $a(t) \sim \exp\left[ \frac{\Lambda t}{3}\right]$.

The last thing you probably want to know is how to calculate energy into temperature and vice versa. Energy is often measured in electronvolt, where $1\ eV= 1.602 \cdot 10^{-19}\ J$. Using the Boltzmann constant and $1 \frac{eV}{k_B} = 11'602\ K$ we can now calculate temperatures from energies end vice verca.

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  • $\begingroup$ I see how you interpreted the question. If you were correct, you have my +1 for this answer. However, you'll have to forgive me if I hold off on that until the OP's intention for the question in verified. $\endgroup$ – Jim Aug 27 '15 at 15:50
  • $\begingroup$ But the present universe is not matter dominated. So you had better add a solution that includes dark energy. This is also appropriate (perhaps) for the inflationary epoch. $\endgroup$ – Rob Jeffries Aug 27 '15 at 18:48
  • $\begingroup$ Clever, thanks for response. Will respond tomorrow. $\endgroup$ – Jim Johnson Aug 27 '15 at 23:01
  • $\begingroup$ @RobJeffries: Thanks for the comment, I added the third scenario. $\endgroup$ – Clever Aug 28 '15 at 6:50
  • $\begingroup$ After reading your response and rereading Hyperphysics, I was able to compute the values in the table referenced. $\endgroup$ – Jim Johnson Aug 28 '15 at 21:45

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