Dark energy and the Cosmological Constant

1. Mar 15, 2015

Quds Akbar

So the Universe expanded very rapidly in its very first moments (inflation). The Universe then slowed down and is speeding up again, and Dark energy is supposed to be responsible for this accelerated expansion.

The cosmological constant might as well be dark energy, but why is it still being considered when the Universe's expansion's rate has changed over time? And the cosmological constant is constant, so it should remain constant since it existed, right?

So I'm guessing my question is, why do physicists still consider the cosmological constant if the Universe's expansion rate has changed over time? Or was dark energy simply not responsible for inflation?

2. Mar 15, 2015

phinds

Dark Energy is not known to have been responsible for inflation, but we don't know what was responsible, so it's hard to say.

Expansion and acceleration of the expansion are not the same thing. Dark energy is only responsible for the acceleration.

3. Mar 15, 2015

Quds Akbar

But why do we believe in the cosmological constant if the expansion since the beginning of the universe was not constant?

4. Mar 15, 2015

Garth

Because the Cosmological Constant, aka DE, has only become dominant in the latter history of the universe.

If it is DE, not $\Lambda$, then it might evolve over time and indeed it might possibly be the inflaton field at an earlier, higher energy stage. The trick will be getting the idea to work!

Garth

Last edited: Mar 16, 2015
5. Mar 15, 2015

Bandersnatch

It kinda works like this:
You've got three 'things' that influence expansion: matter, radiation and the cosmological constant (which is like a property of space). As the universe expands (there's more space), matter and radiation get diluted: on average there's less of them per unit volume of space (radiation gets diluted faster due to being additionally redshifted). The cosmological constant doesn't get diluted - it remains the same per unit volume of space.

So, way back when everything was closer together, the matter and radiation density were dominant (i.e. pulled everything together slowing down the rate of expansion), but with time they got diluted so much that the influence of the cosmological constant took over and became dominant - driving the accelerated expansion.

The cosmological constant is (or at least appears to be) constant, because how much a unit of space 'pushes outwards' never changed. It's the other influences that went down.

6. Mar 15, 2015

phinds

Yeah, what he said

7. Mar 15, 2015

marcus

Hi, Quds, I agree with Phinds that Bandersnatch's explanation was especially good---as verbal explanations go. There is a limit to how clearly you can think about the standard cosmic model if you think purely in words without ever considering definite measured quantities like RATES of distance expansion, numerically expressed. Words can get in the way of understanding because they come with a lot of other associations. Like the word "acceleration" is automatically associated with driving a car and brings up a whole batch of unconscious expectations.

It could be that you are content with the verbal understanding and don't want to look at the actual expansion RATES, expressed numerically. It's only optional. Only read what I'm writing if you are curious and like numbers. There are two important rates of distance expansion to learn: the present rate H0 and the longterm constant future rate H which is being gradually approached over time.

To keep this post simple I will only describe the present rate and leave the other for another post if you say you want a more numerical understanding.

The distances we are talking about are long-range cosmic scale distances between disconnected things, things not bound together by gravity, e.g. in orbits. We aren't talking about distances within our own galaxy, or between us and our nearest neighbor galaxies (to which ours is gravitationally bound).

Those long-range distances are currently increasing by 1/144 of one percent per million years.

Another way to say the same thing, if you like scientific notation (powers of ten) is to say that the current fractional growth rate H0 is 2.20 x 10-18 per second.

Last edited: Mar 15, 2015
8. Mar 16, 2015

Quds Akbar

What is the "A"?

9. Mar 16, 2015

Bandersnatch

It's a greek letter 'lambda', used to denote the cosmological constant. What Garth means there is that there are other than Λ theoretical possibilities for dark energy, where it would no be constant. The quoted statement can be a bit misleading, though, as Λ still is DE.

10. Mar 16, 2015

Garth

I disagree, there is a difference between $\Lambda$ (on the left hand side of Einstein's Field Equation) and DE (on the right hand side of EFE) - see the discussion here.

Garth

Last edited: Mar 16, 2015
11. Mar 16, 2015

Bandersnatch

Fair enough, but... He'll still find the the one referred to as the other in the less rigorous sources if he continues reading about the subject. Even your earlier post from which the statement was quoted can be confusing in this way, as you both equate the two and differentiate between them.

12. Mar 16, 2015

wabbit

I sometimes wish this kind of units were used more often instead of the like of "km/s/Mpc", which combines units from at least two different systems and can make quick order-of-magnitude estimates tricky.

My favorite ? "About 7 % per billion years".

Last edited: Mar 16, 2015
13. Mar 16, 2015

Staff: Mentor

If I understand your argument in that other thread correctly, you are saying that if $\Lambda$ is on the LHS of the EFE, it must be a constant (not varying in either space or time), because otherwise it would violate the Bianchi identities; whereas if it is on the RHS, it can vary? That's not a valid argument; the RHS of the EFE has to obey the Bianchi identities just like the LHS does (since the EFE is just an equality between the LHS and RHS), so if $\Lambda$ must be a constant if it's on the LHS, it must be a constant if it's on the RHS as well.

14. Mar 17, 2015

Garth

Hi Peter,

Well it first depends on whether you want to stay within GR or not. If we are to stay within GR then $\Lambda$ must be constant, otherwise we start again with some other non-metric theory. Of course it might just be that an observation of '$\Lambda$/DE' varying may force such consideration in future.

However the GR Field Equation (EFE) is given by: $R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} = 8\pi T_{\mu\nu}$

The whole construction of GR and the Equivalence Principle revolves around the 'covariant conservation' of both sides of the EFE, see MHW chapters 15 (Bianchi Identities and the boundary of a boundary (is zero)), 17.2 (Automatic conservation of the source - a dynamic necessity), 17.3 (Cosmological Constant).

If we move $\Lambda$ over to the RHS to become DE then we have: $R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} = 8\pi [T_{\mu\nu} + E_{\mu\nu}]$, where $E_{\mu\nu}= -(\Lambda/(8\pi) )g_{\mu\nu}$

If the LHS of the EFE is conserved then the whole of the RHS: G, the stress-energy tensor of 'matter' $T_{\mu\nu}$and DE $E_{\mu\nu}$, must be 'covariantly conserved': 8$\pi (G[T^{\mu}_{\nu} + E^{\mu}_{\nu}]);_{\mu} = 0$. G may be allowed to vary in theories such as Brans Dicke and https://en.wikipedia.org/wiki/Self-creation_cosmology [Broken], however in general G is considered constant.

In which case we must have: $(T^{\mu}_{\nu} + E^{\mu}_{\nu});_{\mu} = 0$ and any variation in the DE stress-energy tensor $E_{\mu\nu}$ must be 'mopped up' by that of matter $T_{\mu\nu}$, this will violate the equivalence principle to some degree.

So if we allow variation of $E_{\mu\nu}$ i.e. DE but not $\Lambda$ then it is possible to consider an evolving DE and still stay within GR.

However, if we start varying $\Lambda$ then it would be very easy to stray into very strange territory.

Garth

Last edited by a moderator: May 7, 2017
15. Mar 17, 2015

Staff: Mentor

These two statements are inconsistent. "Evolving DE" is just another name for "$\Lambda$ is not constant". It's either one or the other.

Let me try and clarify a bit. If we put $\Lambda$ on the LHS of the EFE, we have

$$G_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}$$

Taking the covariant derivative of both sides (not assuming $\Lambda$ constant) gives

$$\nabla^{\mu} G_{\mu \nu} + \Lambda \nabla^{\mu} g_{\mu \nu} + g_{\mu \nu} \nabla^{\mu} \Lambda = 8 \pi \nabla^{\mu} T_{\mu \nu}$$

Now, if we want to say that the covariant derivative of the LHS must be zero, then we must have $\nabla^{\mu} \Lambda = 0$, since every other term on the LHS already has zero covariant derivative as an identity. This then makes the covariant derivative of the RHS equal to zero, which means that the energy and momentum of ordinary matter and radiation are locally conserved (see further comments below).

However, we could also shift the $\nabla^{\mu} \Lambda$ term to the RHS and write

$$\nabla^{\mu} G_{\mu \nu} + \Lambda \nabla^{\mu} g_{\mu \nu} = 8 \pi \left( \nabla^{\mu} T_{\mu \nu} - \frac{1}{8 \pi} g_{\mu \nu} \nabla^{\mu} \Lambda \right)$$

Now the LHS is identically zero, so the RHS must be as well, hence the two terms inside the parentheses must cancel, but neither one needs to vanish by itself. But that would still be true if we left the $\Lambda$ term on the LHS; the difference is just algebra, not anything physical. Just shifting a term from one side of an equation to the other doesn't somehow force the term to be constant, or allow it to vary.

The physical question is whether you want to allow the covariant derivative of $T_{\mu \nu}$, exclusive of any $\Lambda$ term, to be nonzero; in other words, whether you want to allow energy and momentum to be exchanged, locally, between ordinary matter and radiation, and "dark energy" (or "scalar field", or whatever you want to call the $\Lambda$ term now that it can vary in space and time). Both possible answers, "no" and "yes", can be consistently modeled within GR. Whether both kinds of models are physically reasonable is a different question, one on which I'm not sure there is a consensus among physicists.

16. Mar 17, 2015

Garth

Hi Peter,

Well I have already covered that point in my last post.

I would agree that $\Lambda$ or DE are most likely 'fixed', as you suggest, as its the simplest solution and there is no substantial evidence to say otherwise.

However that does not prevent others from investigating an evolving DE and I was just showing how that could happen within the EFE, so long as it is DE and not $\Lambda$ that you allow to 'vary'.

In your FE with $\Lambda$ moved across (I think it needs editing as you have left it on the LHS as well) you need to include G as well as that might vary is some Brans Dicke type modification of GR and the covariant differential operator $\nabla^{\mu}$ can operate on the whole of the RHS.
$$\nabla^{\mu}G_{\mu\nu} = 8\pi\nabla^{\mu}(G(T_{\mu\nu} - \frac{1}{8\pi}g_{\mu\nu}\Lambda))$$

However as $\Lambda$ is constant by definition there is not much point in doing this. Onthe other hand if we call it DE we can give it a more general energy-momentum tensor that approximates to $E_{\mu\nu}= -(\Lambda/(8\pi) )g_{\mu\nu}$ then it may be allowed to evolve as I describe in my post above #14.

Garth

17. Mar 17, 2015

Staff: Mentor

And my point is that the distinction you are drawing between "DE" and $\Lambda$, which appears to be based on which side of the EFE the term is written on, is not valid. Physically it doesn't matter which side of the equation you write the $\Lambda$ term on, or whether you call it "DE". What matters physically is whether or not energy and momentum can be exchanged between ordinary matter/radiation and "dark energy" or whatever you want to call it. If the answer is "yes", then $\Lambda$ can vary. If the answer is "no", then it can't. Which side of the equation things are written on has nothing to do with that.

That's not the EFE; it's the covariant derivative of the EFE. The term in $\Lambda$ that I left on the LHS is identically zero (because $\nabla^{\mu} g_{\mu \nu}$ is identically zero). I left it on the LHS to underscore that fact.

Also, once again, getting hung up on which side of the equation a term is on is pointless; the physics doesn't depend on that.

I was only considering standard GR, not modifications of it; my point was to show that both $\Lambda$ constant and $\Lambda$ varying in space and time can be modeled in standard GR.

It's worth noting, though, that when considering the possibility of $G$ varying, it's important to separate, conceptually, two different functions that $G$ serves in standard GR. First, it provides a conversion factor between geometric units (curvature) and stress-energy units (energy density). Second, it tells how much curvature is produced by a given amount of stress-energy. It's only the second function that can vary in Brans-Dicke type theories.

18. Mar 17, 2015

Garth

If the cosmological constant (LHS) and DE (RHS) are the same thing then they cannot vary - the cosmological constant is constant by the Bianchi identities, and consequently DE cannot evolve (except by it simply varying in its effect because of the decreasing densities of other contents of the universe.)

However the identical mathematical structure ($p = -\rho c^2$) belies the difference between CC and DE. On the LHS the cosmological constant is simply a modification of space-time curvature and therefore an addition to the behaviour of gravitation - that is repulsive at cosmological ranges and attractive at short ones. On the RHS the DE is seen as an extra content of the universe, a mysterious substance, and a source of gravitation.

I think we are saying the same thing, it is just that when others talk about DE evolving and its EoS departing from $\omega = -1$, I prefer to keep it on the RHS where it can exchange energy and momentum between ordinary matter/radiation, and because I like to reserve the LHS for a constant $\Lambda$.

Okay, I understand what you are doing with the $\nabla^{\mu}g_{\mu\nu}$, sorry for any confusion.

You make an interesting point about the two functions of G in the EFE - my question would be how do you measure geometric units, derived from 'stress-energy' units, apart from the amount of curvature produced by a given amount of stress energy? A similar question in the Post Newtonian approximation of the One Body Problem is whether the Robertson parameter $\alpha$ has to be unity? It does as it is a consequence of the empirical definition of mass M.

Garth

Last edited: Mar 18, 2015
19. Mar 17, 2015

Staff: Mentor

No, this is not correct. You can have a varying $\Lambda$ term on the LHS or the RHS, it's just a matter of algebra. Whether or not the LHS or RHS of the equation has zero covariant divergence is a red herring; the physical question is whether $T_{\mu \nu}$, the stress-energy tensor exclusive of any "dark energy", has zero covariant divergence.

This is really a matter of interpretation, not physics. If $\Lambda$ is variable and you put it on the LHS, you're just saying that the behavior of gravitation is variable in space and time, instead of saying that "dark energy" is variable in space and time. You'll get the same predictions for observables either way.

I have no problem with this as a preference; I tend to have the same preference. But you were making (or appearing to make) a much stronger (and false) claim: that $\Lambda$ can only be variable if it's put on the RHS of the equation.

20. Mar 17, 2015

marcus

If I understand correctly, Garth, I think I agree with you. For me, here's the important thing that distinguishes. We do not know that ALL spacetime curvature is caused by mass-energy density, pressure and associated material quantities.
To me, it seems important to acknowledge this incompleteness of our knowledge.
What we have observed and measured is a residual curvature, a property of spacetime geometry.

We cannot blithely move Lambda over to the other side, the "matter" side, because doing so involves mythology. When we do that we mythologize an imaginary energy density which "causes" the curvature in the same way we are used to having other energy density cause curvature.

It involves assuming something that we do not know to be the case, that all curvature arises from some material property. This is a superstition encouraged by past experience, but it is not science. We have to acknowledge that there may be no actual energy in "dark energy". Lambda may simply be a small constant intrinsic curvature---just the way spacetime geometry IS, before any material effects on geometry register.

Anyway, that's my point of view. It's encouraged by the fact that there is no evidence that Lambda varies, and so far the measurements of w are very close to -1, consistent with Lambda as a curvature constant. If it were to turn out after finer and finer measurements that w = -1 is excluded and that w < -1 then that would be evidence for an energy interpretation, and reason for me to change my viewpoint.