# Question about the metric tensor

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?

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.

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

stevendaryl
Staff Emeritus
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.