# Equation of state

1. May 5, 2010

### Ranku

Cosmological constant dark energy has equation of state parameter = -1. What is the equation of state parameter of gravitational energy?

2. May 6, 2010

### Chalnoth

There isn't one. In the standard formulation of General Relativity, there is no such thing as gravitational energy.

3. May 11, 2010

### Ranku

Since dark energy density is related to negative pressure, is there no way to construct a relation between 'positive pressure' and 'gravitational energy density', even though gravitational energy does not occur in general relativity?

4. May 11, 2010

### Chalnoth

No, I don't think so.

One way you could do it, though, is through the conservation of the stress-energy tensor. The stress-energy tensor is a 2nd-rank tensor (sort of like a matrix) that includes components such as energy, momentum, pressure, and shear (shear includes things like twisting forces). This tensor is always conserved in a very particular way: its so-called covariant derivative is zero.

If you examine this conservation law in flat space-time, you get an energy transport equation: the time rate of change of energy in a region of space is equal to the flow of energy into/out of that region.

However, in curved space-time things get a little bit more complicated. What you could do is make up a new "total energy density" which is always conserved: as in the flat space-time example, the total energy density within a region of space-time only changes if it flows from one place to another. You could do this by adding an extra term to the equations that exactly cancels the extra terms you get from curved space-time, and call this your gravitational potential energy.

5. May 11, 2010

### Ranku

Hmm...interesting. Well it's good to know that a positive pressure:gravitational energy relation could at least be considered, whether it works or not.
Thanks.

6. May 11, 2010

### AWA

Would't that leave you with a completely new stress-energy tensor with two components(simplifying a lot): one negative "dark" energy (pressure) and one "gravitational" positive energy pressure?
It's interesting, if a bit radical. In the past there have been attempts along those lines with negative and positive gravity or with inertia versus gravity(de Sitter) but they were all turned down because they seemed to bring up unphysical concepts like "negative energy".Since 1998 with the accelerated expansion surprise and "dark energy" there is more going in that direction.

7. May 11, 2010

### Chalnoth

Hmm, that's not what I was talking about. The basic idea is this. If we have massless particles in flat space-time, we can write energy conservation as:

$$\frac{d\rho}{dt} = -\bigtriangledown \vec{p}c$$

That is, for the energy density in a region of space can only change if some of that energy flows into or out of the region. When we have curved space-time, we instead have:

$$\frac{d\rho}{dt} = -\bigtriangledown \vec{p}c +$$curvature-related terms

I won't bother to look up the precise form of the curvature-related terms on the right hand side. But suffice it to say it's always possible to write energy conservation in this way. So I could simply make up "gravitational potential energy" and set it to the negative of those curvature-related terms. Then I would have conservation of total energy again.

This has nothing to do with making up a new tensor, just making up a new quantity.

Er, dark energy must have a positive energy density to produce an accelerated expansion.

8. May 11, 2010

### AWA

Ok, and wouldn't adding that new made up quantity change anything in the tensor,at least conceptually?
About dark energy sign, I know itis supposed to have conventionally positive density to attain
accelaration, but as the OP spoke in terms of negative pressure and we were calling "gravitational energy" positive I switched the signs,but I coul have made the gravitational energy negative and dark energy positive since the point seeemed to be that they are opposite and cancel each other out. Let's not get hung up about the signs here.

9. May 11, 2010

### Chalnoth

No, the tensor remains exactly the same. This would just be a way of talking about gravitational potential energy.