Question about Metric Tensor: Can g_{rr} be Functions of Coordinate Variables?

[kkz23691]

Hello

Say, the metric tensor is diagonal, $g=\mbox{diag}(g_{11}, g_{22},...,g_{NN})$. The (null) geodesic equations are

$\frac{d}{ds}(2g_{ri} \frac{dx^{i}}{ds})-\frac{\partial g_{jk}}{\partial x^{r}}\frac{dx^{j}}{ds}\frac{dx^{k}}{ds} = 0$

These are $N$ equations containing $N$ partial derivatives $\frac{\partial g_{rr}}{\partial x^{l}}$.

The question is - does this mean $g_{rr}$ (a total of $N$ of them) can be functions of up to one coordinate variable each?
Say, in cyl. coordinates $ds^2=g_{11}(r)dr^2+g_{22}(\theta)d\theta^2+g_{33}(z)dz^2+g_{44}(t)dt^2$
What is your understanding - can say, $g_{22}$ be a function of $t$? Or could $g_{11}$ be a function of $z$?

It just seems that if in the most general case $g_{rr}=g_{rr}(x^1,x^2,...,x^N)$ the geodesic equations should be at least $N^2$, to carry the information for all possible partial derivatives...

Any thoughts?

 

[Mentz114]

I'm not sure if I understand the question. I think you've confused the coordinate r with the index r in the equation.

Here it is with m replacing r as the surviving index.

$\frac{d}{ds}(2g_{mi} \frac{dx^{i}}{ds})-\frac{\partial g_{jk}}{\partial x^{m}}\frac{dx^{j}}{ds}\frac{dx^{k}}{ds} = 0$

There is no reason why the metric coefficients should not be functions of all or any of the coordinates.

 

[kkz23691]

Then we would have $N^2$ nonzero partial derivatives $\frac{\partial g_{rr}}{\partial x^l}$, while the geodesic equations are only $N$.

 

[stevendaryl]

Then we would have $N^2$ nonzero partial derivatives $\frac{\partial g_{rr}}{\partial x^l}$, while the geodesic equations are only $N$.

The geodesic equation is for determining x^\mu(\tau) given g_{\mu \nu}. It's not for determining g_{\mu \nu}. There are 4 equations and four unknowns:

\frac{d^2 x^\mu}{d\tau^2} =...

 

[kkz23691]

I think you've confused the coordinate r with the index r in the equation.

My bad, should have used something else instead of r.

There are 4 equations and four unknowns

Ah, I see. There are $N$ coefficients of the metric tensor which, through the $N$ geodesic equations output $N$ parametric equations for the geodesic. This makes sense I shouldn't have looked at the number of derivatives.