Metric Elements and Einstein Equations in a Variable-Dependent Formulation

[eljose]

I have two questions...let's suppose we have a metric in the form:

$$ds^2 =f(t)dt^2 +g(x)dx^2 +H(y)dy^2$$

So every element of the metric only depend on a variable..my question is..does this mean that the Einstein Equations (vaccuum) are of the form:

$$R_ii =0$$ i=t,x,y ?..

-And the second question is i know that $$det(g_ab )=f(t)g(x)H(y)$$ but ..what's the form of the Lagrangian?..i guess:

$$L= \int_ V dVf(t)g(x)H(y)(f(t)R_00 +g(x)R_11+ H(y)R_22 )$$

 

[pervect]

I have two questions...let's suppose we have a metric in the form:

$$ds^2 =f(t)dt^2 +g(x)dx^2 +H(y)dy^2$$

So every element of the metric only depend on a variable..my question is..does this mean that the Einstein Equations (vaccuum) are of the form:

$$R_ii =0$$ i=t,x,y ?..

Yep. I threw this line element (with a k(z) dz^2 term added) into GrTensor II, the Riemann was zero, as well as the Ricci and the Einstein.

 

[eljose]

And a "Mixed" one?.. $$ds^2 = f(t)dt^2 + g_ij dx^i dx^j$$ Where Einstein summation is assumed...

 

[Parlyne]

This line element is just a simple diffeomorphism of normal Minkowski space. Consider changing the coordinates to T, X, and Y, such that:

$$dT = \sqrt{-f(t)} dt$$
$$dX = \sqrt{g(x)} dx$$
$$dY = \sqrt{H(y)} dy$$

In these coordinates, the line element becomes:
$$ds^2 = - dT^2 + dX^2 + dY^2$$