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So how does quintessence evade this? Doesn't quintessence act like a varying lambda?

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- Thread starter bcrowell
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- #1

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So how does quintessence evade this? Doesn't quintessence act like a varying lambda?

- #2

Matterwave

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That's about all I can say...=/

- #3

PAllen

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Section 4.5 of the following, which I found for another thread, discusses quintessence:

http://relativity.livingreviews.org/Articles/lrr-2001-1/fulltext.html [Broken]

http://relativity.livingreviews.org/Articles/lrr-2001-1/fulltext.html [Broken]

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As far as verifying that a varying lambda violates conservation of stress-energy, here's what I have:

[tex]G_{ab}=8\pi T_{ab}+\Lambda g_{ab}[/tex]

The divergence is:

[tex]\nabla^a G_{ab}=8\pi \nabla^a T_{ab}+\nabla^a (\Lambda g_{ab})[/tex]

[itex]\nabla^a G_{ab}[/itex] is supposed to vanish identically. (I haven't checked this myself.) Applying the product rule to the final term, we get two terms. The term involving the covariant derivative of the metric vanishes identically, because that's basically the defining property of the covariant derivative. So the result is:

[tex]0=8\pi \nabla^a T_{ab}+\nabla_b \Lambda[/tex]

So if we have forms of matter that would otherwise have given a vanishing divergence for T, but then we add in a non-constant lambda, we get a contradiction. I don't see how it matters whether or not the varying lambda is lumped in with T or kept separate. Still mystified.

The review article by Carrol basically says that we'd like energy conditions to be satisfied, because otherwise energy can flow faster than c. w<-1 (giving a big rip) violates this. But to me, it seems far easier to accept violation of an energy condition than to accept a nonvanishing divergence of the stress-energy tensor, which would mean local nonconservation of energy.

This thread seems relevant, but doesn't seem to resolve the issue: https://www.physicsforums.com/showthread.php?t=508923

[tex]G_{ab}=8\pi T_{ab}+\Lambda g_{ab}[/tex]

The divergence is:

[tex]\nabla^a G_{ab}=8\pi \nabla^a T_{ab}+\nabla^a (\Lambda g_{ab})[/tex]

[itex]\nabla^a G_{ab}[/itex] is supposed to vanish identically. (I haven't checked this myself.) Applying the product rule to the final term, we get two terms. The term involving the covariant derivative of the metric vanishes identically, because that's basically the defining property of the covariant derivative. So the result is:

[tex]0=8\pi \nabla^a T_{ab}+\nabla_b \Lambda[/tex]

So if we have forms of matter that would otherwise have given a vanishing divergence for T, but then we add in a non-constant lambda, we get a contradiction. I don't see how it matters whether or not the varying lambda is lumped in with T or kept separate. Still mystified.

The review article by Carrol basically says that we'd like energy conditions to be satisfied, because otherwise energy can flow faster than c. w<-1 (giving a big rip) violates this. But to me, it seems far easier to accept violation of an energy condition than to accept a nonvanishing divergence of the stress-energy tensor, which would mean local nonconservation of energy.

This thread seems relevant, but doesn't seem to resolve the issue: https://www.physicsforums.com/showthread.php?t=508923

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George Jones

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Look at the expressions for density and pressure between equations (62) and (63) from Carroll's article. Zero phi dot gives a constant cosmological constant. Phi dot non-zero means that density is not the negative of pressure, and so the source is not a scalar times the metric (in an orthonormal frame). Phi non-zero but small means that the is source is approximately (in some sense) a scalar times the metric.

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PAllen

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http://arxiv.org/abs/astro-ph/9711102

http://arxiv.org/abs/astro-ph/9708069

From the first, it seems claimed that the scalar field just contributes to T, and that divergence of T would still be expected to be zero. But, maybe I misunderstand, I just skimmed it...

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Look at the expressions for density and pressure between equations (62) and (63) from Carroll's article. Zero phi dot gives a constant cosmological constant. Phi dot non-zero means that density is not the negative of pressure, and so the source is not a scalar times the metric (in an orthonormal frame). Phi non-zero but small means that the is source is approximately (in some sense) a scalar times the metric.

Maybe it's just because it's getting late at night in California and I've already had two beers, but... Your first paragraph hints at some quantitative reasoning that I'm just not finding in your second paragraph, which never refers to the stress-energy tensor or its divergence...?

One of the Friedmann equations is [itex]\ddot{a}/a = \frac{1}{3}\Lambda - \frac{4\pi}{3}(\rho+3P)[/itex]. If [itex]\rho[/itex] and P are negligible, we can observe the l.h.s. and infer lambda. If, at some later cosmological epoch, we observe a different value for the l.h.s, we infer a different value for lambda. (Of course we don't actually wait billions of years and then observe again, but conceptually we do this through the "lookback time" of supernova surveys, for example.) If observation allows us to infer one value of lambda for one epoch and then another value for another epoch, how is this different from a varying lambda?

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George Jones

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Maybe it's just because it's getting late at night in California

Getting late for me too (I'm in the same timezone, but at 54N), so what I write will still be quite cryptic. I'm very busy tomorrow, so I don't know when I can write a decent answer.

Your first paragraph hints at some quantitative reasoning that I'm just not finding in your second paragraph, which never refers to the stress-energy tensor or its divergence...?

Granted I don't calculate the divergence, but my second paragraph does explicitly refer to the stress-energy tensor. In a comoving orthonormal frame, the stress-energy tensor of a perfect fluid is T = diag{rho, p, p, p}. My second paragraph references Carroll's explicit expressions for rho and p. For quintessence, p =/= -rho, so rho cannot be factored out to give rho g (since I'm near the west coast, I use the east coast metric ). Consequently, the source for quintessence is not a scalar times the metric, so its divergence cannot be calculated as if it were. If, however, phi dot is small, p almost equals -rho.

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Thanks, George, for your #8 -- I'll have to take some time to parse it carefully.

One other thought that may be relevant:

If we want to test empirically whether stress-energy is conserved, we can do it using laboratory experiments, or we can do it using cosmological observations. But if we're going to do it using cosmological observations, we can't just consider different solutions to the Friedmann equations and see which ones match the data the best, because all such solutions conserve stress-energy. We would actually have to look for violations of the Friedmann equations. But that would be tough to do, because we can't measure the P and rho of our quintessence or whatever, and it's not analogous to dust or radiation, for which we can confirm the equation of state using laboratory experiments. What we're actually measuring is a(t), which tells us about the metric. But that only allows us to calculate the divergence of the Einstein tensor, which vanishes identically.

One other thought that may be relevant:

If we want to test empirically whether stress-energy is conserved, we can do it using laboratory experiments, or we can do it using cosmological observations. But if we're going to do it using cosmological observations, we can't just consider different solutions to the Friedmann equations and see which ones match the data the best, because all such solutions conserve stress-energy. We would actually have to look for violations of the Friedmann equations. But that would be tough to do, because we can't measure the P and rho of our quintessence or whatever, and it's not analogous to dust or radiation, for which we can confirm the equation of state using laboratory experiments. What we're actually measuring is a(t), which tells us about the metric. But that only allows us to calculate the divergence of the Einstein tensor, which vanishes identically.

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