# Finding a constant in an expanding universe

1. May 24, 2014

### utku

Finding a constant in an axpanding univarse

Hi,
I want to obtain a constant that proportional to total matter density of universe.

Let ρ be matter densitiy and ρr be radiation desity.

In an expanding universe volume times total density must be constant. It is wrong?

let me show it is not correct because,

a^3(ρ + ρr) = ρ0 + ρr0/a

where ρ0 is matter density at present and similarly ρr0 is radiation density at present.And "a" is scale factor that function of time coordinate.
As you can see volume times total matter density not constant.

And energy of radiation decays like 1/a due to the expansion, total energy is conserved but energy of radition is not conserved. How this energy transform?

thanks.

2. May 24, 2014

### Simon Bridge

Probably - depends on the Universe.

In a closed universe that is expanding so that total volume increases with time, then the total matter density is M/V(t) ... this decreases with time unless more matter comes from someplace.

Your calculation was for energy density ... including only matter and light.
Did you miss out some forms of energy?
You also failed to justify the equality.

If you assume that the matter density must be a constant, and you want to find a constant proportional to the matter density, then why not just use the matter density?

3. May 25, 2014

### yogi

Total energy may be zero - for matter the negative energy of each local G field summed over the Hubble sphere must be considered - radiation and other forms of energy also correspond to an mc^2 energy, and consequently may be cancelled by an associated negative potential.

4. May 25, 2014

### Simon Bridge

... and so on: i.e. it depends on the Universe.
@utku: any of this of any use?

5. May 25, 2014

### Chalnoth

Only for matter that has no pressure on cosmological scales, i.e. normal matter and dark matter (normal matter experiences pressure in compact objects, but not at very large scales, as galaxies don't apply pressure to one another, except when they are actively colliding).

Total energy is not conserved in all situations in General Relativity. In particular, it isn't conserved in an expanding universe. For a more in-depth analysis, see here:
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

6. May 26, 2014

### utku

Thanks for your reply Simon Bridge and Yogi,Chalnoth.

I will look the article that Chaltoth send.
Some Energy of the radiation loses due to the expansion. What is this energy transform to?

7. May 26, 2014

### Chalnoth

It doesn't transform into anything. Energy simply isn't conserved in an expanding universe.

8. May 26, 2014

### George Jones

Staff Emeritus
You forgot the $pdV$ term in the first law of thermodynamics as applied to cosmology.

\begin{align} 0 &= dE + pdV\\ &= \mathrm{d}\left(a^3 \rho \right) + p \mathrm{d}\left(a^3 \right) \end{align}

If what you wrote above were true, then the first term in my last equation would be zero; this isn't correct, it is the sum of the two terms that is zero.

Try to use the photon gas equation of state

$$p = \frac{1}{3} \rho_r$$

and what I wrote above to derive the dependence of $\rho_r$ on $a$.

Last edited: May 26, 2014
9. May 27, 2014

### utku

Thanks to all.
I search the forum and find very usefull answers to many questions about cosmology. I think before make new reply ı must search the forum:)

2-
yes, i calculate George Jones, and it must depend a^(-4).