# Cosmological constant term and metric tensor

1. May 22, 2015

### exponent137

Why cosmological constant term $\Lambda g_{uv}$ in Einstein equation is proportional to $g_{uv}$. Why it is even proportional to $g_{uv}$ in spacetime of MInkowski?

Last edited: May 22, 2015
2. May 22, 2015

### bcrowell

Staff Emeritus
I have a detailed discussion of this in section 8.1.4 of my GR book, http://www.lightandmatter.com/genrel/ , but a quick and dirty argument is that if we have in mind something that's a form of stress-energy built in to the structure of space itself, then its stress-energy tensor acts like a special tensor that's built in to the structure of spacetime. One way of stating the equivalence principle is that the only such built-in tensor is the metric itself.

Minkowski space isn't a solution of the Einstein field equations with a nonzero cosmological constant.

3. May 22, 2015

### Mentz114

In his paper Inside Gravity* Prof Padmanabhan refers to the fact that the Einstein-Hilbert action can be split in a bulk part and a surface part. If one discards the bulk part and extremizes the action then the EFE come out unchanged and the cosmological constant emerges as a constant of integration. So it can be put in or appear naturally with much the same result.

More importantly, if any metric is multiplied by a constant this is the same as putting a constant curvature everywhere. ( As ben says, I think)

* available online free
Gen Relativ Gravit (2008) 40:2031–2036
DOI 10.1007/s10714-008-0669-6

Last edited: May 22, 2015
4. May 22, 2015

### bcrowell

Staff Emeritus
Huh? I don't understand what you mean by this.

5. May 23, 2015

### exponent137

Benjamin Crowell, you wrote in the book:
"
Because of these issues, Einstein decided to try to patch up his field equation so that it would allow a static universe. Looking back
over the considerations that led us to this form of the equation, we see that it is very nearly uniquely determined by the following
criteria:
1. It should be consistent with experimental evidence for local conservation of energy-momentum.
2. It should satisfy the equivalence principle.
3. It should be coordinate-independent.
4. It should be equivalent to Newtonian gravity or \plain" general
relativity in the appropriate limit.
5. It should not be overdetermined.
This is not meant to be a rigorous proof, just a general observation that it's not easy to tinker with the theory without breaking it.

A failed attempt at tinkering Example:
As an example of the lack of “wiggle room” in the structure of the field equations, suppose we construct the scalar Tcc
a , the trace of
the stress-energy tensor, and try to insert it into the field equations as a further source term. The first problem is that the field
equation involves rank-2 tensors, so we can’t just add a scalar. To get around this, suppose we multiply by the metric. We then
have something like Gab = c1Tab + c2gabTcc , where the two constants c1 and c2 would be constrained by the requirement that the theory agree with Newtonian gravity in the classical limit.
To see why this attempt fails, note that the stress-energy tensor of an electromagnetic field is traceless, Tc c = 0. Therefore the beam of light’s coupling to gravity in the c2 term is zero. As discussed onpp. 273-276, empirical tests of conservation of momentum would therefore constrain c2 to be . 10-8.
One way in which we can change the eld equation without violating any of these requirements is to add a term g_ab, giving

"
What can be still, more simple explained: maybe in Minkowski spacetime limit this part should be proportional to (-1, 1, 1, 1) because of equivalence principle (EP), beside other? Is this the most visualized aspect? Why otherwise, EP is not valid? The most key property of vacuum energy is that density analogy is negative, where pressure analogies are expectedly positive. Why EP causes that "density" of vacuum energy in negative?

You give first and important information about this, Baez, Bunn, and Carroll did not explain this problem in their one paper, and one book, but can you a little more visualize?

Last edited: May 23, 2015
6. May 23, 2015

### Mentz114

Sorry. My misunderstanding. I was thinking of de Sitter space where the solution of the EFE is $G_{\mu\nu}=-\Lambda g_{\mu\nu}$. This spacetime has no matter but has constant curvature everywhere ( that is how I interpret "four dimensional spaces of constant and isotropic curvature").

It doesn't address the question 'why is lambda multiplied by the metric'.