Einstein Equation with Cosmological Constant & Variable x^u

[alejandrito29]

with the action $$\int (k(R- 2 \Lambda)+L_m) \sqrt{g}$$ the Einstein equation is:

$$R_{uv}-\frac{1}{2}R g_{uv}- \Lambda g_{uv} = k' T_{uv}$$

How is the Einstein equation if $$\Lambda=\Lambda(x^u)$$? with $$x^u$$ a coordinate

 

[haushofer]

The same. You derive the Einstein equations with a variation with respect to the metric, not the coordinates. So you obtain the Einstein equations

$$R_{uv}-\frac{1}{2}R g_{uv}- \Lambda(x) g_{uv} = ksingle-quote T_{uv}$$

However, you're in serieus trouble with energy conservation. A constant CC is allowed, because then

$$\nabla_{\rho}(\Lambda g_{\mu\nu}) = \Lambda \nabla_{\rho}g_{\mu\nu}=0$$

If Lambda is general, energy conservation is spoiled.