Construction of metric from tensor products of vectors

[gerald V]

1.
The metric $g_{\mu \nu}$ of spacetime shall be constructed from tensor products of vectors (relevant are the unit vectors in the respective directions). One such vector shall be called $A$.

Homework Equations

$g_{\mu \nu} = \lambda \frac{A_\mu A_\nu}{g^{\alpha \beta} A_\alpha A_\beta} + ...$, where neither $\lambda$ nor the further terms shall involve $A$. Now $g$ shall be differentiated w.r.t. $A$, this is $\frac{\partial g_{\mu\nu}}{\partial A_\tau}$ .[/B]

The Attempt at a Solution

Everything works fine except one term originating from differentiating the denominator. It contains a factor $A_\alpha A_\beta \frac{\partial g^{\alpha\beta}}{\partial A_\tau}$. I have no idea what to do with this expression. Is it zero? Or is it equal to to $A^\alpha A^\beta \frac{\partial g_{\alpha\beta}}{\partial A_\tau}$? Or what else?

Many thanks for any advice.

 

[George Jones]

It contains a factor $A_\alpha A_\beta \frac{\partial g^{\alpha\beta}}{\partial A_\tau}$. I have no idea what to do with this expression.

To find an expression for this factor, contact $g_{\mu \nu} = \lambda \frac{A_\mu A_\nu}{g^{\alpha \beta} A_\alpha A_\beta} + ...$ with $A^\mu A ^\nu$, and take the derivative with respect to $A_\tau$.

 

[George Jones]

It contains a factor $A_\alpha A_\beta \frac{\partial g^{\alpha\beta}}{\partial A_\tau}$. I have no idea what to do with this expression. Is it zero? Or is it equal to to $A^\alpha A^\beta \frac{\partial g_{\alpha\beta}}{\partial A_\tau}$? Or what else?

Actually, it might be better to look at
$$0 = \frac{\partial}{\partial A_\tau} \left( \delta^\alpha_\beta \right) .$$