Cosmo-logical constant

[kleinwolf]

Is it allowed to write :

 

[math] G_{\mu\nu}(g)=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\Lambda g_{\mu\nu}+\Lambda^1_{\mu\nu}=\kappa T_{\mu\nu} [/math]

 

where [math]\Lambda^1_{\mu\nu}[/math] is a constant symmetric tensor (depending neither on space nor time)...and other symbols well-known in GR.

 

We find, since in vacuum the Minkowski metric is a particular solution in cartesian coordinates, that [math]\Lambda^1_{\mu\nu}=-\Lambda \textrm{diag}(1,-1,-1,-1)[/math].

 

Else, could someone give a comprehensive summary of the conditions on [math]G_{\mu\nu}[/math] forbidding similar constant (of integration?) ?

[kleinwolf]

An error occured in my question :

 

In fact [math]\Lambda_1[/math] should be understood as independent of "g" (the unknown, i.e. the metric), but can depend on space-time, since for example in spherical coordinates, [math]\Lambda^1_{\mu\nu}=-\Lambda \textrm{ diag}(1,-1,-r^2,-r^2\sin(\theta)^2)[/math]

[timo]

I remember being told that the original term is the only possibility. A quick reminder how the geometry-side of the Einstein equation can be [roughly] justified:

1) The matter-side, the energy-momentum tensor is divergence-free (=> local conservation of energy and momentum).

2) So should be the term representing the geometry side if you want to have a differential equation.

3) The geometry side shall somehow include the 2nd derivatives; the curvature. The Ricci tensor contains those (dunno to what extent using the Ricci tensor is required) but is not divergence-free.

4) Explicitly subtracting the divergence of the Ricci tensor then makes the geometry side a divergence-free function of the 2nd derivatives.

5) But: Since the metric tensor itself is divergence-free, adding any multiple of the metric will not spoil divergence-freeness of the geometry term while thought-wise it can still be considered part of the geometry.

 

So what would that mean for your Lambda:

- It should be a divergence-free tensor of rank 2.

- It should not be some kind of particle content or otherwise it would logically belong to the matter part T. Substracting parts of T from both sides of the Einstein equation and calling that -Lambda probably isn't what you were after.

- It should, by your constraint, not depend on the metric - whatever that exactly means.

 

I would assume that "should not be some particle content" might be the killer constraint for you.