# Gravitational entropy in the early universe

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1. Apr 2, 2015

### Khashishi

Penrose wrote in the Road to Reality that gravitational clumping increases the entropy of the universe. The early universe was very low in entropy because it was very smooth, with very little clumping.

So, is it accurate to say that the early universe was high in entropy except for the gravitational contribution? Almost like the universe was allowed to equilibrate without gravity and then gravity was suddenly turned on? If gravity was turned off for the beginning of the universe, would this explain the low entropy of the initial universe?

2. Apr 3, 2015

### Staff: Mentor

No, because there isn't a separate gravitational contribution. Penrose was not saying that there are two kinds of entropy, gravitational and non-gravitational, and you just add them up. He was saying that entropy in the presence of gravity works fundamentally differently from entropy in the absence of gravity. In the absence of gravity, a clumped state has lower entropy than a smooth state; but in the presence of gravity, it's the other way around.

No, because you can't "turn gravity off". More precisely, turning gravity off would be equivalent to spacetime being flat, and we know the spacetime of the early universe was not flat. In fact, it was more curved then than it is now.

3. Apr 3, 2015

### Khashishi

How do we know the early universe was not flat?

4. Apr 3, 2015

### Staff: Mentor

Because it was expanding rapidly, much more rapidly than now. A higher rate of expansion means a greater spacetime curvature. (More precisely, it was expanding much more rapidly, but there was also a higher rate of change of the expansion rate. Both of those things are manifestations of higher spacetime curvature.)

From the standpoint of the Einstein Field Equation, the reason for the greater curvature is that the universe was a lot denser then than it is now, so the source term on the RHS of the EFE was much larger, hence the curvature on the LHS was also much larger. (Also, the early universe was radiation-dominated rather than matter-dominated, so pressure was a significant source of gravity, increasing the RHS of the EFE still further.)

5. Apr 3, 2015

### Khashishi

But what if something is missing from the model? The reason I say that is because as far as I know, current measurements show that the universe is flat. I know that inflation explains why the universe is almost flat, but another explanation is that maybe the universe really is perfectly flat. Then it must have always been flat. Maybe flatness was part of the initial conditions of the universe.

I know you can't answer a question like "what if the model is wrong", but you can address how well grounded the model really is. I know dark energy is supposed to be negligible in the early universe, but as far as I know, dark energy is just a fudge factor to make Einstein's field equations predict an accelerating universe.

6. Apr 3, 2015

### Khashishi

A follow up question. Was the curvature in the early universe positive or negative?

7. Apr 3, 2015

### Staff: Mentor

Current measurements show that the universe is spatially flat. We are talking about spacetime curvature, not space curvature. The only way to have a flat spacetime is to have zero matter or energy anywhere. So I could have answered your earlier question by saying that we know the early universe was curved because we know there was matter and energy present. (And as I said in a previous post, because the density of matter/energy was much larger in the early universe, the curvature was much greater as well.)

It has always been positive--more precisely, the Ricci scalar has always been positive.

8. Apr 3, 2015

### Staff: Mentor

This is incorrect in two ways. First, a cosmological constant is not a "fudge factor" added to the EFE; it belongs there. This is easiest to see if you derive the EFE from a Lagrangian; the most general Lagrangian that satisfies the desired criteria (no more than second derivatives of the metric) is not $R \sqrt{-g}$ but $\left( R + \Lambda \right) \sqrt{-g}$.

Second, a cosmological constant is not the only way to get accelerating expansion. Any equation of state with $w < - 1/3$ (i.e., with $p < - (1/3) \rho$ ) will give accelerating expansion. A scalar field can have such an equation of state. So our current model, which has $w = -1$ (i.e., a cosmological constant as opposed to some other kind of field producing accelerating expansion), is not the only possible model that gives accelerating expansion; it is our best current model because $w = -1$ best matches the data.