Special relativity: components of a metric tensor

[Uku]

Homework Statement

An interval in Minkovski space is given in spheric coordinates as;

ds^{2}=c^{2}dt^{2}-dr^{2}-r^{2}d\theta^{2}-r^{2}sin^{2}\theta d\phi^{2}

Now I have to find the covariant and contravariant components of the metric tensor.

Homework Equations

General expression of a metric tensor is:
G=g_{\mu\nu}dx^{\mu}dx^{\nu} which also equals ds^{2}

The Attempt at a Solution

I have some messy answer written down from the lecture, but can't get it clear.
Seems that I have written down the covariant components of the metric as
g_{00}=1
g_{11}=-1
g_{22}=-r^{2}
g_{33}=-r^{2}sin^{2}\theta
Are these the covariant components that the problem asks me to find? How are they found?

I see a correlation in the Schwarzschild metric

G=[PLAIN]http://rqgravity.net/images/gravitation/Gravitation-92.gif

After this it seems that the lecturer has written down that

g_{\mu\sigma} g^{\mu\rho}=\delta^{\rho}_{\sigma}

Is this correct?

So I get a matrix on which all the components on the main diagonal are 1. From there I can derive that the contra-variant components of the metric are simply inverses of the above covariant metric.

 

[vela]

If you expand the implied summations in ds^2 = g_{\mu\nu} dx^\mu dx^\nu[/tex], you get<br /> <br /> ds^2 = g_{00} dx^0 dx^0 + g_{01}dx^0dx^1 + \cdots + g_{32}dx^3dx^2 + g_{33}dx^3dx^3<br /> <br /> Keeping in mind that x^0 = t, x^1 = r, x^2 = \theta, and x^3 = \phi, compare that expression to <br /> <br /> ds^2 = dt^2 - dr^2 - r^2 d\theta^2 - r^2\sin^2\theta\,d\phi^2<br /> <br /> You can just read off the components of g_{\mu\nu}.<br /> <br /> Your notes are correct in that g_{\mu\sigma}g^{\mu\rho}=\delta^\rho_\sigma. Consider<br /> <br /> x_\rho = g_{\rho\sigma} x^\sigma=g_{\rho\sigma}g^{\sigma\mu}x_\mu<br /> <br /> You can see that all g_{\rho\sigma}g^{\sigma\mu} does is change the label of the index on x_\mu. In other words, it&#039;s just the identity.

 

[Uku]

Thanks!

A well, actually I have a question still.
The bit:

ds^{2}=g_{\mu\nu}dx^{\mu} dx^{\nu}

indicates summation over 16 (4x4) values, yes? With only the main diagonal being significant on the g.

EDIT: yes, that is what happens.