Where did radiation energy go?

In summary, according to Weinberg, the amount of energy in matter remained fairly constant, but the energy in radiation (photons) decreased due to the expansion of the universe. Most of the energy in matter now is in the form of familiar matter (neutron,protons,electrons mainly), while radiation has a decreasing amount of energy.
  • #1
aludwig
3
0
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?
 
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  • #2
A lot of it went into creating matter.
 
  • #3
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' [tex]a(t)[/tex]) as:

[tex]\rho_m \sim a^{-3}[/tex]

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 [tex] V \sim a^{3}[/tex] then the above expression becomes

[tex]\rho_m \sim \frac{1}{V}[/tex].

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

[tex] E = E_0\frac{a_0}{a} [/tex]

this means that the energy density of radiation evolves as

[tex] \rho_rad \sim a^{-4} [/tex]

Since radiation energy density decays more rapidly with [tex]a(t)[/tex] 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
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
Energy is not generally conserved in GR; GR conserves energy-momentum which is significantly different.

Garth
 
  • #6
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
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 [itex]dW = p dV = 1/3 \rho dV,[/itex] where [itex]\rho[/itex] is the energy density of the photons and the second equality follows from the radiation equation of state [itex]p = \rho / 3.[/itex] Then, for the photons,

[tex]
\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*}
[/tex]

Then,

[tex]
\int^t_{t_0} \frac{\dot{\rho}}{\rho} dt' = - 4 \int^t_{t_0} \frac{\dot{a}}{a} dt'
[/tex]

gives

[tex]\rho = \rho_0 \left( \frac{a_0}{a} \right)^4.[/tex]
 
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  • #8
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.
 

1. Where does radiation energy go when it is absorbed?

When radiation energy is absorbed, it is converted into heat energy. This heat energy can be used for various purposes, such as warming the surrounding environment or powering machinery.

2. Does all radiation energy get absorbed?

No, not all radiation energy gets absorbed. Some of it may be reflected or scattered by objects, while some may pass through without being absorbed. The amount of radiation energy that gets absorbed depends on the type of radiation, the material it is passing through, and the distance it travels.

3. Can radiation energy be stored?

Yes, radiation energy can be stored in certain materials, such as radioactive elements. These materials release energy slowly over time, which can be harnessed for various purposes.

4. What happens to radiation energy that is not absorbed?

Radiation energy that is not absorbed can continue to travel through space until it is either absorbed or scattered by other objects. Some types of radiation, such as gamma rays, can travel long distances without being absorbed.

5. Is radiation energy dangerous?

Radiation energy can be dangerous to living organisms if they are exposed to high levels for extended periods of time. However, small amounts of radiation energy are present in our everyday environment and are not harmful. It is important to protect ourselves from excessive exposure to radiation energy through proper safety measures.

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