# The Cosmological (constant)

1. Jun 22, 2011

### jfy4

Hi,

I am wondering if the cosmological constant is a constant in the sense that it can only have one value, ie some constant element of the reals, or if it can be a scalar function too dependent on the coordinate variables, eg $\Lambda(r,t)$.

2. Jun 22, 2011

### atyy

Usually it's constant to maintain that the divergence of the stress-energy tensor is zero.

I think there are ways of adding it not to the field equations, but to the action, and varying with respect to it, to also maintain energy conservation. http://arxiv.org/abs/gr-qc/0505128

Last edited: Jun 22, 2011
3. Jun 22, 2011

### jfy4

That makes sense. If I may, would the satisfaction of
$$\nabla_{\beta}\left( T^{\alpha\beta}-g^{\alpha\beta}\Lambda\right)=0$$
justify the inclusion of a cosmological constant that was a scalar function?

Last edited: Jun 22, 2011
4. Jun 22, 2011

### atyy

Do you mean something like the potential of a scalar field forming part of the stress-energy tensor of matter (http://ned.ipac.caltech.edu/level5/Carroll2/Carroll1_3.html" [Broken])?

Last edited by a moderator: May 5, 2017
5. Jun 22, 2011

### jfy4

I think this answers my question.
This sounds like it is ok to include a cosmological constant of the form
$$\Lambda(x_{\alpha})=\Lambda_0+V\,[\phi(x_{\alpha})]$$
that consists of an initial cosmological constant, $\Lambda_0$, summed with a scalar function $V$. Have I interpreted this correctly?

Last edited by a moderator: May 5, 2017
6. Jun 23, 2011

### haushofer

The LHS of the Einstein equation should be divergenceless, because the right hand side is (energy momentum conservation). This brings one to the addition of a term

$$\Lambda g^{\mu\nu} \ \ \rightarrow \nabla_{\mu}(\Lambda g^{\mu\nu}) = \nabla^{\nu}\Lambda = 0$$

So,

$$\partial_{\mu} \Lambda = 0$$

Hence, lambda must be a constant.