Cosmological constant times the metric tensor

[Isaac0427]

In the EFE, what does adding Λg_μν mean and why is it not included in the Einstein tensor?

 

[PeterDonis]

The $\Lambda g_{\mu \nu}$ term is the cosmological constant term. It's not included in the Einstein tensor because the Einstein tensor is derived from the Riemann tensor (via the Ricci tensor), which involves derivatives of the metric. The $\Lambda g_{\mu \nu}$ term, as is obvious from looking at it, involves no derivatives of the metric; it's just a constant times the metric.

Physically, the Einstein tensor describes spacetime curvature; that's why it involves derivatives of the metric. The $\Lambda$ term doesn't describe spacetime curvature; it just describes a constant "something" that is present everywhere. The "something" has various names; in cosmology, it is called "dark energy", and is what causes the expansion of the universe to accelerate.

 

[Isaac0427]

The $\Lambda g_{\mu \nu}$ term is the cosmological constant term. It's not included in the Einstein tensor because the Einstein tensor is derived from the Riemann tensor (via the Ricci tensor), which involves derivatives of the metric. The $\Lambda g_{\mu \nu}$ term, as is obvious from looking at it, involves no derivatives of the metric; it's just a constant times the metric.

Physically, the Einstein tensor describes spacetime curvature; that's why it involves derivatives of the metric. The $\Lambda$ term doesn't describe spacetime curvature; it just describes a constant "something" that is present everywhere. The "something" has various names; in cosmology, it is called "dark energy", and is what causes the expansion of the universe to accelerate.

So is the metric times the cosmological constant is constant? I thought the metric could vary. R and R_μν can vary, right?

 

[PeterDonis]

So is the metric times the cosmological constant is constant? I thought the metric could very. R and Rμν can very, right?

You're mixing up different kinds of "variation". Let's look at the vacuum Einstein Field Equation (we'll leave out the stress-energy tensor here for simplicity):

$$
R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = 0
$$

What does this equation mean, in general? The Ricci tensor and Ricci scalar are expressions involving the first and second derivatives of the metric tensor. So the above equation is really a set of ten second-order differential equations for the components of the metric tensor. That is, it's a set of ten second-order differential equations for ten functions $g_{00}$, $g_{01}$, etc.

What does the $\Lambda$ term do? It just adds a constant times the metric tensor to the differential equation. In other words, in each of the ten differential equations for the ten functions $g_{00}$, $g_{01}$, etc., it adds a constant $\Lambda$ times that function. So the differential equation for the function $g_{00}$ has a term $\Lambda g_{00}$ in it, the differential equation for the function $g_{01}$ has a term $\Lambda g_{01}$ in it, etc. None of these terms add any derivatives of the functions; they only add a constant times the functions. That's why $\Lambda$ is called a "constant", but the Einstein tensor is not.

Now it's true that the functions themselves, $g_{00}$, $g_{01}$, etc., can vary from point to point in spacetime. If they didn't, if all their derivatives were zero, the EFE would be trivial (it would just describe the flat Minkowski spacetime of SR). But in a differential equation, there is a big difference between terms involving derivatives of a function (the Einstein tensor) and terms involving only a constant times the function itself (the $\Lambda$ term).

 

[Isaac0427]

You're mixing up different kinds of "variation". Let's look at the vacuum Einstein Field Equation (we'll leave out the stress-energy tensor here for simplicity):

$$
R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = 0
$$

What does this equation mean, in general? The Ricci tensor and Ricci scalar are expressions involving the first and second derivatives of the metric tensor. So the above equation is really a set of ten second-order differential equations for the components of the metric tensor. That is, it's a set of ten second-order differential equations for ten functions $g_{00}$, $g_{01}$, etc.

What does the $\Lambda$ term do? It just adds a constant times the metric tensor to the differential equation. In other words, in each of the ten differential equations for the ten functions $g_{00}$, $g_{01}$, etc., it adds a constant $\Lambda$ times that function. So the differential equation for the function $g_{00}$ has a term $\Lambda g_{00}$ in it, the differential equation for the function $g_{01}$ has a term $\Lambda g_{01}$ in it, etc. None of these terms add any derivatives of the functions; they only add a constant times the functions. That's why $\Lambda$ is called a "constant", but the Einstein tensor is not.

Now it's true that the functions themselves, $g_{00}$, $g_{01}$, etc., can vary from point to point in spacetime. If they didn't, if all their derivatives were zero, the EFE would be trivial (it would just describe the flat Minkowski spacetime of SR). But in a differential equation, there is a big difference between terms involving derivatives of a function (the Einstein tensor) and terms involving only a constant times the function itself (the $\Lambda$ term).

Thank you. This is one of the the most helpful answers I have ever gotten on this site.

 

[PeterDonis]

Thank you. This is one of the the most helpful answers I have ever gotten on this site.

You're welcome!

 

[Matterwave]

The "something" has various names; in cosmology, it is called "dark energy", and is what causes the expansion of the universe to accelerate.

I don't think it's fair to conflate dark energy with the cosmological constant. The cosmological constant is one candidate "model" for dark energy, but the dark energy might come from a dynamical field or some other such models (e.g. quintessence).

 

[Isaac0427]

The Ricci tensor and Ricci scalar are expressions involving the first and second derivatives of the metric tensor.

What is the exact relationship between the Ricci tensor and the metric, and the Ricci scalar and the metric?

 

[vanhees71]

In the EFE, what does adding Λg_μν mean and why is it not included in the Einstein tensor?

This is a very clever question, and to my knowledge nobody knows the answer. It depends on how to physically interpret the term with the cosmological constant in the Einstein field equations.

The concept is the following: The most straightforward "derivation" of the Einstein field equation is to assume the equivalence principle, which leads to the idea to formulate generally covariant field equations for the pseudometric tensor of a pseudo-Riemannian space-time manifold. The observable gravitational effects are then due to the curvature of this space-time manifold.

Further it is in a way natural to assume that the field equations should be of 2nd order in the partial derivatives (and thus also in the covariant derivatives) of the pseudo-metric tensor. It is most easy to use Hamilton's principle to derive field equations under the constraints of a symmetry (here general covariance) by finding the possible actions that are invariant under the symmetry given the field-degrees of freedom (i.e., here the pseudo-metric tensor). To get 2nd-order PDE's the action should be formulated as a functional of the field components and its first derivatives modulo a total divergence. Now the general covariance demands that the Lagrangian should be a scalar function under general coordinate (diffeomorphism transformations) of the fields and its derivatives. Now there is no such thing with only first-order derivatives only, but the curvature scalare $R$, which has 2nd derivatives of the pseudo-metric components but these appear only linear with coefficients that contain no derivatives, which means that they can be rewritten as expressions with only first-order derivatives plus a total four-divergence. It turns out that this is the only such invariant you can build from the pseudo-metric. The only other possibility is just a constant term in the Lagrangian. Now, because in the four-volume measure $\sqrt{-g} \mathrm{d}^4 q$ you have the $\sqrt{-g}$ term, this is not a trivial contribution and indeed leads to the cosmological constant term.

From this point of view, one would conclude the cosmological constant is part of the gravitational field and belongs to the left-hand side of the Einstein equations. The right-hand side is derived from the same principles introducing other fields (or classical particle densities) discribing matter and radiation into the action. It turns out that the coupling between the matter-degrees of freedom must be to the energy-momentum tensor of the matter Lagrangian (which is a kind of "minimal-coupling argument" due to the gauge nature of the general covariance with respect to the metric tensor), which has to appear with a universal gravitational coupling constant. This gives the right-hand side, i.e., in some sense the "sources" of the gravitational field.

Now there's also an ambiguity in the matter action, as long as you only consider the special-relativistic case, where a constant addition to the Lagrangian doesn's play any role. On the other hand it means a constant contribution to the total energy density in the Hamiltonian. Now, when using the equivalence principle to generalize the usual actions for matter (e.g., the Lagrangian for the electromagnetic field and charges which is used to derive Maxwell's equations from the action principle), such a constant is precisely like a contribution to the cosmological constant.

In quantum field theory you must renormalize the total energy density of the quanta described by them. In special relativity it's an unobservable total energy, and is simply subtracted to make the "vacuum state" of energy 0, which leads just to the level above which all energies are defined. In GR it contributes to the cosmological constant. Now renormalization implies the introduction of an energy-momentum renormalization scale, and renormalization depends on the choice of this scale. So does the absolute value of the total energy! And renormalizing it at a low scale implies large changes at higher scales, most prominently at the humongous Planck scale. To fit the value measured by investigations of the fluctuations of the cosmic background radiation (with COBE, WMAP, and PLANCK a high-precision science today!) you need a fine-tuning of parameters to an accuracy of 120 orders of magnitude, when using the Standard Model of Elementary Particles as the model to describe matter. The main culprit is the mass-renormalization of the Higgs boson, which is introduced into the theory as an elementary scalar particle. This (among other things) makes the so successful standard model so ugly that physicists urgently look for "physics beyond the Standard Model".

In that sense the cosmological constant or "dark energy" (depending on which of the two possible interpretations you use) is the most puzzling enigma of contemporary physics. There's a famous review article by Weinberg about the subject:

Steven Weinberg. The cosmological constant problem. Rev. Mod. Phys., 61:1, 1989
http://link.aps.org/abstract/RMP/V61/P1

 

[PeterDonis]

I don't think it's fair to conflate dark energy with the cosmological constant.

I agree that a cosmological constant is not the only possible model that could account for the accelerated expansion of the universe. As far as I know, it is the leading candidate, and it is certainly the simplest. All of the other models involve some kind of nonzero stress-energy tensor (i.e., something other than a constant times the metric). For this discussion, I've left out all such possibilities for simplicity, but you're right that they are possibilities.

 

[vanhees71]

Sure, there are various models for inflation and other ideas about the cosmological constant or things that act like one (inflaton fields, quintessence,...). Perhaps somebody here is an expert on these ideas and can elaborate that further.

 

[PeterDonis]

What is the exact relationship between the Ricci tensor and the metric, and the Ricci scalar and the metric?

The Ricci tensor is a contraction of the Riemann tensor, which is the tensor that describes all aspects of spacetime curvature, and is built out of first and second derivatives of the metric; more precisely, it is built out of derivatives of and products of the Christoffel symbols, which are built out of first derivatives of the metric. The Ricci scalar is the contraction of the Ricci tensor.

More details on this Wikipedia page and the links given there:

https://en.wikipedia.org/wiki/Introduction_to_the_mathematics_of_general_relativity#Curvature_tensor

 

[Nugatory]

What is the exact relationship between the Ricci tensor and the metric, and the Ricci scalar and the metric?

You won't find a gentler and more friendly introduction than http://preposterousuniverse.com/grnotes/grtinypdf.pdf (but take a look at http://preposterousuniverse.com/grnotes/ for more context).

The Christoffel symbols are defined in terms of the metric in equation 36.
These in turn define the components of the Riemann tensor in a given coordinate system, using equation 44.
The components of the Ricci tensor and the Ricci scalar are defined in terms of the components of the Riemann tensor using equations 45 and 46.
So we get from the metric to the Ricci tensor and scalar in multiple steps.

Some things to watch out for:
1) The symbol $R$ is used for the Riemann tensor, the Ricci tensor, and the Ricci constant; you tell them apart from the number of indices.
2) The Einstein summation convention is used throughout, so repeated indices are summed over: $g_{\mu\nu}A^\mu$ is a shorthand notation for $g_{0\nu}A^0+g_{1\nu}A^1+g_{2\nu}A^2+g_{3\nu}A^3$. This gets pretty complicated pretty quickly when there are multiple indices being summed over.
3) The notations $A^2$ and $A^3$ don't mean "A-squared" and "A-cubed" respectively; they mean the second and third components of $A$ (which because it has a single raised index is a vector).