# Energy density questions

1. Aug 19, 2012

### spriag

Hi,

Been doing some cosmology revision and have a few questions owing to my terrible notes.

One of the past papers asks the question
"The density parameter of the Universe today has the value Ω = 0.25.
Using the value of Hubble’s constant from (b), calculate the mean density of matter today. Estimate the matter density at the epoch of recombination (which occurred at a redshift z  1000)."

In b) got value of H at 69.3km/s/mpc. Using pcrit=3H^2/8piG got the value 8.949*10^-27, then using Ω=ρ/ρcrit got the value 2.235*10&-27kg/m^3.

The second part of the question slightly confuses me. I know εb=ε0b(1+z)^3 so would it just be 2.235*10^-27 *c^2 to give εb, and sub in z=1000. Would i be looking to solve for εb or ε0b, the notation is confusing me a bit.

Second question is "Use the conservation of mass and photon number, and the scaling of
photon energy, hν, with redshift, z, to calculate how the energy densities
of matter and of radiation scale with (1 + z)." My notes don't cover this so any help would be appreciated.

Sorry for the first post for help situation, thanks in advance.

2. Aug 19, 2012

### marcus

My help may not be worth much, read at your own risk. I'll tell you my personal non-academic way of visualizing, which might NOT be what you're looking for and might merely confuse you. But you can choose to ignore this, I'll just toss an intuitive image out FWIW (for what it's worth).

You are wondering about how matter dens. and radiation dens. change with 1+z.

I picture a box with matter in it, going back in time. At redshift z, the volume is less by a factor of (1+z)3 so the matter density is greater by that same factor (1+z)3.

I picture a box with photons in it, going back in time. At redshift z, the volume is less by a factor of (1+z)3 so the NUMBER density of the photons is greater by that same factor. However the energy of each photon is also bigger now, by a factor of 1+z. So the energy density of the radiation in the box is now greater by a factor of (1+z)4.

Basically it seems to me that this is the kind of thing you are addressing, except you are supposed to be putting in numbers with powers of ten and units, like kilograms, joules, meters etc. etc. You are on your own as far as that goes. Putting numbers and units into the picture is your department. There is a "homework help" section here (up in the "Science Education" menu) for that kind of thing.
https://www.physicsforums.com/forumdisplay.php?f=152
It's very good. You should really be going there for coaching and suggestions. However, welcome to PF!

Last edited: Aug 19, 2012
3. Aug 22, 2012

### cepheid

Staff Emeritus
ε is the energy density in the past, at the epoch corresponding to redshift z, whereas ε0 is the energy density today, at z = 0. What makes it obvious which one is which is that ε > ε0 by the factor of (1+z)3, and you know that the density goes up with increasing z (i.e. the farther you go into the past). So this means that ε must be the past value and ε0 the present value.

The "b" subscript typically indicates that this is the energy density associated with baryons, but I don't think that's what you're looking for here. You're finding the density corresponding to the the density parameter (Ω) for all matter, not just for baryons. So you'd be better off using a subscript "m" for matter or something.

Just to say what marcus said again, maybe in a slightly different way. The total number of matter particles is conserved, which means that the number in any given "box" in the universe is constant. So as the volume of that box increases, the density decreases, since density is inversely proportional to volume (the same number of particles is now spread out over a larger volume). The volume is proportional to the cube of the scale factor, which means that the mass density is proportional to the inverse of the cube of the scale factor (and so is the energy density, since the energy here is just the rest energy associated with the mass of the particles). The inverse of the scale factor is (1 + z). That's how we get the (1 + z)3 dependence. Matter just dilutes as a result of the expansion of the universe exactly as you would expect.

For photons, the argument is the same, except that not only do the photons dilute as their "box" expands, but also the energy per photon is decreased due to the increase in wavelength. So, figure out how wavelength changes with scale factor, and you should be able to get the result for how the energy density of photons depends on z. EDIT: esp. since marcus gave the answer.

4. Aug 22, 2012

### Kraflyn

Hi.

Yes, everything written so far is correct. But the author requires some hard equations I guess. So here's the deal with energy densities.

In following notes:

$\rho$ denotes $T_{00}$ component of energy density tensor: energy density itself;
$p$ denotes $T_{ii}$ component of energy density tensor: impulse density;
$z$ denotes redshift.

Energy and impulse are related in FLRW model by equation
$\dot{ \rho} = -3 \frac{\dot {a(t)}}{a(t)} \left( \rho + p \right)$
Here $a(t)$ is scale factor proportional to a distance traversed by a particle emitted at the very moment of big-bang-bong. This equation comes from Einstein equations for FLRW metric $\delta s^2 =\delta t^2 -a^2(t) \left[ \delta r^2 + r^2 \delta \Omega^2 \right]$.

There are basically 3 possibilities for energy source: matter, radiation and cosmological constant.

Matter: it refers to non-relativistic matter. It is characterized by the fact that it has rest energy, so there is energy densitiy. However, being non-relativistic, one can approximate it by saying it's impulse is negligible compared to rest energy. So matter has impulse 0. Thus, $\rho_M$ exists, of course, but $p_M =0$.

Radiation: all relativistic matter, massive or massless alike. Has energy of course, and impulse is 1/3 of the energy. So, $p_R = \rho_R /3$. Why $1/3$? It comes from statistical mechanics. Of all particles present in a box, $1/3$ of them will move in Your direction in 3D world.

Cosmological constant: it acts anti-gravitationally. Seriously: $p_\Lambda = - \rho_\Lambda$.

So, let us employ Einstein equation now. First: matter! $p_M = 0$, and hence
$\dot{ \rho_M} = -3 \frac{\dot {a(t)}}{a(t)} \rho_M$
In other words,
$\frac{\dot{ \rho_M}}{\rho_M} = -3 \frac{\dot {a(t)}}{a(t)}$
So $\rho_M = 1/a^3(t)$. I chose constant of integration as 1 for simplicity. So You see, matter density behaves like ordinary density we are accustomed to. It preserves overall rest energy, here taken to be equal to 1. One may take this as conservation of particle number, obviously.

Next in line is: radiation! $p_R = \rho_R /3$, and hence
$\dot{ \rho_R} = -3 \frac{\dot {a(t)}}{a(t)} \left( \rho_R + \frac{\rho_R }{3} \right)$
In other words,
$\frac{\dot{ \rho_R}}{\rho_R} = -4 \frac{\dot {a(t)}}{a(t)}$
So $\rho_R = 1/a^4(t)$. I chose constant of integration as 1 for simplicity. So You see, radiation density does not behave like ordinary density. As space-time expands, radiation energy is not conserved... Well, it is conserved, but not in the way one might expect at first. The other Einstein equation makes sure it is so.

So, how come photon energy drops with universe expanding? As universe expands, it stretches. Just notice stretching factor $a(t)$ in FLRW metric. So, photon's wavelength also stretches, and this way photon loses energy just by being in a flow in universe. Yeah, that means Earth and every object in universe stretches too! No, no, no, no!!!!! You see, there is something fishy here, because we don't stretch. Just photon's wavelength. Yes, I know, it's a fairytale. There is an explanation for it, but You won't find it with standard explanations. So just take it for granted: $\rho_R = 1/a^4(t)$.

Finally, here comes the king: cosmological constant! Well, as name suggests - it's constant!

So, what is redshift $z$ here? How does it relate to the only free parameter in our theory: $a(t)$? Well, if photon had wavelength $a(t)$ and then universe flows a while and now photon's wavelength today is $a_0$, then redshift $z$ is by definition
$1+z=\frac{a_0}{a(t)}$
In other words: wavelength observed today / wavelength originally emitted.

So, today $\rho_{M0} = 1/a_0^3$. Yesterday it was $\rho_M = 1/a^3 (t)$. So, $\rho_{M0} a_0^3 = \rho_M a^3 (t)$. In other words,
$\rho_M = \rho_{M0} \frac{a_0^3}{a^3 (t)} = \rho_{M0} (1+z)^3$

This explains notation You mentioned.

Finally, we all wonder: "so how is photon's energy then conserved...?"
OK, Einstein equations, and there are 2 of them, because we work with time and space both, also say that first law of thermodynamics is in action here. Universe expands adiabatically, so thermodynamics says
$p \; dV = -d \left( \rho V \right)$
This is just conservation law. This is exactly the second Einstein equation. When You put photon numbers in, You get identity. So everything is fine in the end.

I hope this helped a bit. Please do ask if You happen to have some further trouble with it.

Cheers.

Last edited: Aug 22, 2012