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Bjoern
2007-Jan-25, 09:00 PM
One often reads that the dark energy (cosmological constant) has an equation of state of the form
p = -rho,
where rho is the energy density, and p is the pressure. However, I don't understand entirely how it is derived. In the following, I will be referring specifically to the following paper:

The Cosmological Constant and Dark Energy, P. J. E. Peebles and Bharat Ratra, astro-ph/0207347

(It's a really nice paper, explaining both the basic concepts and the available evidence; I'd recommend it to anyone interested in DE.)

The crucial equations in that paper are (13), (14), (15), (18), and (19); see also the bottom half of page 8 for more details.

(19) says that the energy-momentum tensor for dark energy is given by
T^Lambda_munu = rho_Lambda g_munu
From (13) and (15), it is easy to conclude that g_munu (the well-known Robertson-Walker metric), is a diagonal matrix with the diagonal elements (for the simplest case of a flat universe, K=0):
(1, -a^2(t), -a^2(t), -a^2(t))
where a(t) is the scale factor. We thus should have that T^Lambda_munu also is a diagonal matrix with the diagonal elements (rho_Lambda, -a^2(t) rho_Lambda, -a^2(t) rho_Lambda, -a^2(t) rho_Lambda).

Comparing this to the usual form of the energy-momentum tensor for homogenously distributed, isotropic stuff, equation (18), we get:
rho = rho_Lambda (duh)
p = -a^2(t) rho_Lambda
Thus, the equation of state seems to be not p = -rho, but p = - a^2(t) rho!

I'm really confused by this; what am I doing wrong? The paper says that if one combines equations (14), (18) and (19), where (14) is the usual Minkowski metric, one gets p = -rho. That's obviously right - but why am I allowed in using the Minkowski metric instead of the Robertson-Walker metric here???

I suspect it has something to do with the remarks made directly below (14), i. e. that a freely falling inertial observer can always choose locally Minkowski coordinates. But after all, we are not asking what the energy-momentum tensor (and hence the equation of state for dark energy) look like locally, but we want to know that globally! (or don't we???)

Please help me out......... :(

publius
2007-Jan-26, 04:25 AM
That's exactly what it is, I'm sure (well, as sure my knowledge can be, which isn't much in the grand scheme!). That cosmological constant T reduces to p = -rho locally for a free-falling observer, which I think would be a "co-moving" observer here.

The stress-energy tensor's component vary with the coordinates, anyway. Consider a spherical rotating mass like earth or a star. The T distribution you see there is frame dependent, but the EFE always holds in those coordinates.

For example, if we're in a "stationary frame", we see a spinning ball of mass-energy. If we're in a free-falling frame, we see a spinning ball that is moving towards us. And if we go into a rotating frame but radially stationary, we see a stationary ball of mass energy.

Those transforms are complex to say the least, but how much pressure/momentum transport, along with energy/energy current terms you see is frame dependent.

There are things there, like conservation laws that apply only locally in the most general sense.

And second, note that with expanding space, the more space you get, the more total dark energy you have. The density remains constant, but you get more volume. That's what the scale factor terms are reflecting, I believe. If you transform to a co-moving/geodesic following/whatever frame, you see that p = -rho, but the total volume is increasing with time, so you're getting more dark total dark energy globally integrated. Or something.


-Richard

Bjoern
2007-Jan-26, 01:58 PM
That's exactly what it is, I'm sure (well, as sure my knowledge can be, which isn't much in the grand scheme!). That cosmological constant T reduces to p = -rho locally for a free-falling observer, which I think would be a "co-moving" observer here.

The stress-energy tensor's component vary with the coordinates, anyway. Consider a spherical rotating mass like earth or a star. The T distribution you see there is frame dependent, but the EFE always holds in those coordinates.

For example, if we're in a "stationary frame", we see a spinning ball of mass-energy. If we're in a free-falling frame, we see a spinning ball that is moving towards us. And if we go into a rotating frame but radially stationary, we see a stationary ball of mass energy.

Those transforms are complex to say the least, but how much pressure/momentum transport, along with energy/energy current terms you see is frame dependent.

There are things there, like conservation laws that apply only locally in the most general sense.

And second, note that with expanding space, the more space you get, the more total dark energy you have. The density remains constant, but you get more volume. That's what the scale factor terms are reflecting, I believe. If you transform to a co-moving/geodesic following/whatever frame, you see that p = -rho, but the total volume is increasing with time, so you're getting more dark total dark energy globally integrated. Or something.


Thanks. That's about what I suspected...

But I still don't understand why we are allowed to make such a coordinate system change that in this case. After all, for arriving at the Friedmann-Lemaitre equations, one needs the Robertson-Walker metric, not the Minkowski metric. So how can one use for the dark energy p=-rho in the FL equations, if this relation in reality does not hold for the metric from which these equations were derived?

publius
2007-Jan-27, 06:16 AM
Thanks. That's about what I suspected...

But I still don't understand why we are allowed to make such a coordinate system change that in this case. After all, for arriving at the Friedmann-Lemaitre equations, one needs the Robertson-Walker metric, not the Minkowski metric. So how can one use for the dark energy p=-rho in the FL equations, if this relation in reality does not hold for the metric from which these equations were derived?

I wish I knew enough to say something, but my general guess is it's all a matter of cooridinates. The cosmological constant is a "feature" of the field equation, not really an independent source term. You move it over to the right and it looks like an additional source, but that will depend on the coordinates. Choose a local Minkwoski looking frame and it looks like p = -rho locally.


-Richard

Ken G
2007-Jan-27, 06:34 AM
It might help to recognize that phyically, the EOS of dark energy cannot depend on a(t). This treatment of dark energy has it depend only on space, and the properties of space are not changing with age, there's just more of it. The EOS is quintessentially local, it makes no difference at all what the rest of the universe is doing. Perhaps another way to say this is, if you consider the EOS "now", a(t)=1, and the distinction doesn't matter. So your question comes down to, if p=-rho now, could it have been different when the universe had a different scale? If you trace back the assumptions about this dark energy, perhaps you can find exactly where the assumption is inserted that it does not depend explicitly on age.