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I have a problem understanding the ideas behind a basic assumption of cosmology. The Friedmann equations follow from Newtonian mechanics and conservation of Energy-momentum $(E_{kin}+E_{pot}=E_{tot})$ or equally from Einsteins field equation with a Friedmann-Lemaitre-Robertson-Walker metric. In a radiation dominated, flat universe, standard cosmology uses the result from electrodynamics for radiation pressure $P_{rad}=\frac{1}{3}\rho_{rad}$, where $\rho_{rad}$ is the energy density of radiation in the universe and $P_{rad}$ is the associated pressure. It then puts the second Friedmann equation into the first to derive the standard result $\rho_{rad} \propto a^{-4}$, where $a$ is the scale factor on the metric. Putting this result into the first Friedmann equation yields $$a\propto\sqrt{t}$$ where $t$ is the proper time.

Therefore we used the standard radiation pressure of classical electrodynamics to derive an expression for the expansion of the universe. My problem is in understanding why this is justified. Is the picture that photons crash into the walls of the universe to drive the expansion really valid? Certainly not (what would the walls of the universe be, and what are they made of ;)?), but at least this is how the radiation equation of state is derived. Is there any further justification for this? Be aware of that I'm talking here about a radiation dominated universe, so matter and dark energy can be neglected. Therefore, can we not derive a certain rate of expansion for the universe without anything mysterious like Dark Energy?

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The Friedmann equations that describe our Universe do not follow from Newtonian mechanics! It also seems you are confusing some results for matter with those for radiation. $\rho_{\text{matter}}\sim a^{-3}$, while $\rho_{\text{rad}}\sim a^{-4}$. However, this doesn't have a huge bearing on your question per se. – Danu Mar 18 '14 at 12:25
Thanks for your comment, i corrected the mistake. The first and second Friedmann equation follow, besides from the cosmological constant directly form Newtonian mechanics. See for example: – JakobH Mar 18 '14 at 12:33

Your initial thought process is flawless; in a radiation dominated universe, $a\propto\sqrt t$. That said, it is not correct to interpret this as the photons exerting some sort of pressure that drives the expansion. In modern FRW cosmology, a positive pressure ($\frac{1}{3}\rho_{rad}$ as you pointed out) corresponds to an energy density that decelerates the expansion of the universe, which I don't need to tell you is the opposite of driving the expansion. Another thing that must be understood is that most interpretations assume an already existing, positive $\dot a(t)$. In fact, if you look at the first derivative of the scale factor in this case, you'll find $\dot a\propto t^{-1/2}$, which shows that the expansion of the universe slows down over time during radiation domination. This is can be interpreted quite the opposite to what you have done. All of the radiation is gravitationally attracted together, which tends to slow the rate of expansion over time. Take a look at the following plot:

Effective potential from Friedmann equations. Plot taken from note for "Dark Energy" course, York University, Matthew Johnson authored.

From the Friedmann equations, you can show that $\frac{\dot a^2}{2}+V(a)=0$. The yellow line corresponds to the potential, $V(a)$, that you would get from radiation, the reddish-purple line is for matter, and the blue line is for a cosmological constant; all for a flat universe. From this, you can see that if we assume the universe is expanding initially, it will keep expanding but the expansion slows quickly under radiation domination. If the universe were closed in a radiation-only universe, then at some point $\dot a$ would reach zero; the universe would then start to collapse and (as the graph shows) the rate of collapse would accelerate ($a$ would be dropping again, remember). This reinforces the idea that radiation slows the expansion and tries to make the universe collapse again.

As for your second question that referred to not needing dark energy to explain the expansion. We do not need dark energy to explain the expansion of the universe; you are correct in stating that. However, as you can see, in a matter or radiation dominated universe, the rate of expansion, $\dot a$, would be constantly decreasing. This is in sharp contradiction with our observations, which is that the rate of expansion is accelerating. Therefore, we need dark energy as a driving pressure to accelerate expansion. I mentioned earlier that a positive pressure decelerates the expansion. When we use a cosmological constant, we find that the pressure of dark energy is $P_{DE}=-\rho_{DE}$, it has a negative pressure. We also can easily show that only fluids with an equation of state parameter (the coefficient in front of the $\rho$) that is less than $-\frac{1}{3}$ can cause the expansion to accelerate. Again referring to the plot, you can see that using a cosmological constant as our dark energy leads invariably to a dark energy dominated universe where the rate of expansion grows exponentially. This is, in fact, in very good agreement with observations. So yes, even a purely radiation-filled universe does not required dark energy to explain expansion, but we require it to explain accelerated expansion. Radiation, like matter, decelerates expansion as its gravitational attraction tends to try to re-collapse the universe.

I hope that clears things up.

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"All of the radiation is gravitationally attracted together, which tends to slow the rate of expansion over time." - this would happen in any case whether the matter in form of radiation or not. – Anixx Apr 18 '14 at 21:53
@Anixx true, but the question is specifically about radiation. I was illustrating how radiation does not drive expansion by putting pressure on the universe, but it slows expansion due to gravity instead – Jim Apr 18 '14 at 22:04
any photon travelling over expanding space looses energy. As such it becomess less massive, and consequently less contributes to the expansion's slowdown, effectively increasing the expansion. – Anixx Apr 18 '14 at 22:08
@Anixx contributing less to the slowdown implies the slowdown continues, it does not effectively increase the rate of expansion. Furthermore, the question specifies it is the radiation dominated epoch; as can clearly be seen by the friedmann potential, as the universe expands, the rate of expansion falls off like $1/a$ during radiation domination. It does not ever increase the rate of expansion – Jim Apr 18 '14 at 22:11
@Anixx Additionally, by your own logic, as the expansion slows, the amount that the photons are redshifted decreases, which extends the duration of the impact on the expansion rate that they have. AND even if the photons red-shifted into oblivion, that would mean that even in a universe populated by only radiation, the expansion would remain constant. At no point would it "effectively increase" – Jim Apr 18 '14 at 22:14

The fallacy here is assuming conservation of energy. See, Noether's theorem ensures the energy is conserved if the hamiltonian of the system is time invariant. But the universe is expanding, so the very frame in which you describe it is changing over time, therefore you cannot apply energy conservation.

It is, however, rotationally invariant, so you know angular momentum is conserved. It is also shift invariant, so linear momentum is also conserved.

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All observations up to date indicate that we live in an Euclidean universe, i.e. k=0, i.e. $E_{tot}=0$. Therefore the Energy is conserved and furthermore it is not the Friedmann equations which are in question here. The result $a \propto \sqrt{t}$ is a standard result, which can be found in every textbook about cosmology. I want to understand the thought process behind the derivation. Nevertheless, thanks for your answer. – JakobH Mar 18 '14 at 18:52
@Jakobh: Your understanding of the topic isn't quite right, energy is not conserved in our universe's expansion (…). Also, according to standard inflationary paradigm, curvature is not equal to exactly 0 and it is changing with time (very rapidly so early on!) – Benjamin Horowitz Apr 17 '14 at 15:24

It is not necessary to assume that the universe has walls in order for the matter content of the universe to have nonzero pressure. The standard assumption is that the matter/radiation content of the universe is infinite, but at a finite volume density. Also note that there could be some sort of "end of matter" at some radius beyond the cosmological horizon that would be, in principle, completely undetectable, which would enable you to get rid of the "infinite matter" assumption.

Irrespectively, it's the initial state of the universe that produces the pressure, not some equilibrium state with an external wall. Also, note that most of the point of the thermodynamic machinery involving walls or reservoirs is to remove the dependance of the final result on these walls or reservoirs.

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There are the walls of the universe, at finite distance from here. But we cannot reach them because of length contraction: the closer we to the wall the more we contracted radially. It is the cosmic horizon. For eternal de Sitter (expanding) universe cosmic horizon is just de Sitter event horizon.

Any particle approaching the horizon looses its speed due to length contraction, similarly to how it happens with a black hole. In limit all particles that reached the horizon contribute to its area, but then emitted back as de Sitter radiation (analog of Hawking radiation).

As such, it exerts pressure on the boundary.

If our universe is anti-de Sitter (contracting), then the boundary even has the property of a mirror: everything which goes in the direction of the boundary will be reflected back, so the photon gas slows down the contraction of the universe while becoming hotter.

enter image description here

See here for details: Can Hubble red shift be interpreted as time dilation?

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Explain the downvote, please. – Anixx Apr 18 '14 at 22:16
The information here is more or less fine, but it applies to de Sitter universes, when the question asked about specifically FRW universes – Jim Apr 18 '14 at 22:16
@Jim this does not very much differ as de Sitter is the future limit of FRW. FRW is different in that it has the boundary at even closer dictance from the observer than the de Sitter event horizon. But in the infinite future the FRW horizon and de Sitter horizon will coincide. – Anixx Apr 18 '14 at 22:20
again, I do not disagree with that statement; but I am once again forced to point out something explicitly stated in the question; this question is only considering the radiation domination era. This does not include the far future of the FRW universe where the horizons coincide – Jim Apr 18 '14 at 22:30
the Hubble radius was smaller, but there was no "boundary" that caused things to length contract as they approached, nor one the siphoned off speed from a particle as they drew nearer. Comoving velocity decreases everywhere in the universe by the same amount, not specifically at some boundary. And the hubble horizon is not a barrier that pressure can be applied to; it is a description of the approximate limits of causal interactions on an observer at its center and a description of the rate of expansion of the universe – Jim Apr 18 '14 at 22:39

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