# Where did radiation energy go?

1. Apr 26, 2007

### aludwig

I am a newbie to cosmology, and have just finished reading "The first three minutes" by Weinberg. On page 76 he relates that in the early universe almost all of the energy was in radiation (photons), and relatively little in matter (the masses of the nuclear particles). But now most of the energy is in matter. I assume the amount of energy in matter remained fairly constant. Is this correct, and if so, again assuming energy is conserved, where did all of the energy in the early radiation go to?

2. Apr 26, 2007

### mathman

A lot of it went into creating matter.

3. Apr 26, 2007

### Wallace

A couple of points to note. The first is that according to current theory, most of the energy in the Universe is not in fact in the form of nuclear particles. In fact 96% of the total energy is in the exotic forms we call dark matter and dark energy.

That aside, it is still true that there is far more energy in the form of familiar matter (neutron,protons,electrons mainly) than radiation today. The reason for this is that the energy of photons diminishes as the universe expands due to being redshifted. Formally the energy density of matter evolves with the size of the Universe (which we denoted by the 'scale factor' $$a(t)$$) as:

$$\rho_m \sim a^{-3}$$

If you think about this for a moment this is simply what you get by keeping the energy per particle constant, total number of particles constant but increasing the volume. So if we note the $$V \sim a^{3}$$ then the above expression becomes

$$\rho_m \sim \frac{1}{V}$$.

Now, for photons we also have the redshift. This means that the energy per particle does not stay constant, but changes as

$$E = E_0\frac{a_0}{a}$$

this means that the energy density of radiation evolves as

$$\rho_rad \sim a^{-4}$$

Since radiation energy density decays more rapidly with $$a(t)$$ than matter it means that even though initially there was more energy in radiation, the expansion of the Universe has caused there to now be much more energy density in matter.

4. Apr 27, 2007

### aludwig

Thanks for the replies. I am familiar with the red-shift, which Weinberg also discusses, and why this would effect density. But Weinberg refers to total energy, not density. If the energy per photon decreases, either the total energy in all photons decreases, or the number of photons increases. If it is the former, then I still don't see how energy is conserved. If it is going into new matter, what particles are being created, and what is the process?

5. Apr 27, 2007

### Garth

Energy is not generally conserved in GR; GR conserves energy-momentum which is significantly different.

Garth

6. Apr 27, 2007

### Wallace

Yep 'Energy' is a 3D quantity. We shouldn't bee too surprised to find that it is not conserved in GR. So as the Universe expands the total energy does indeed decrease.

7. Apr 29, 2007

### George Jones

Staff Emeritus
The last result in post #3 by Wallace can be derived by "conservation of energy" in the form of the first law of thermodynamics; the photons lose energy because of the work done expanding space.

The work done is given by $dW = p dV = 1/3 \rho dV,$ where $\rho$ is the energy density of the photons and the second equality follows from the radiation equation of state $p = \rho / 3.$ Then, for the photons,

$$\begin{equation*} \begin{split} 0 &= dE + dW\\ &= \frac{dE}{dt} + \frac{dW}{dt}\\ &= \frac{d \left(\rho V \right)}{dt} + \frac{1}{3} \rho \frac{dV}{dt}\\ &= \frac{4}{3} \rho \frac{dV}{dt} + \frac{d \rho}{dt} V\\ &= \frac{4}{3} \rho \frac{d}{dt} a^3 + \frac{d \rho}{dt} a^3\\ &= 4 \rho a^2 \dot{a} + \dot{\rho} a^3. \end{split} \end{equation*}$$

Then,

$$\int^t_{t_0} \frac{\dot{\rho}}{\rho} dt' = - 4 \int^t_{t_0} \frac{\dot{a}}{a} dt'$$

gives

$$\rho = \rho_0 \left( \frac{a_0}{a} \right)^4.$$

Last edited: Apr 29, 2007
8. Apr 30, 2007

### aludwig

Suppose the universe mass is large enough that it eventually starts to collapse towards the big crunch. All of the photons will then be blue-shifted, with the greatest shift from the most distant stars, and it looks to me like the above equations just un-wind. So if this is correct, in this case energy really is conserved, in the form of a kind of potential energy.