# Special relativity: components of a metric tensor

1. Jun 25, 2010

### Uku

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution
I have some messy answer written down from the lecture, but cant 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 Schwarzchild metric

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

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.

Last edited by a moderator: May 4, 2017
2. Jun 25, 2010

### vela

Staff Emeritus
If you expand the implied summations in $ds^2 = g_{\mu\nu} dx^\mu dx^\nu[/tex], you get $$ds^2 = g_{00} dx^0 dx^0 + g_{01}dx^0dx^1 + \cdots + g_{32}dx^3dx^2 + g_{33}dx^3dx^3$$ Keeping in mind that [itex]x^0 = t$, $x^1 = r$, $x^2 = \theta$, and $x^3 = \phi$, compare that expression to

$$ds^2 = dt^2 - dr^2 - r^2 d\theta^2 - r^2\sin^2\theta\,d\phi^2$$

You can just read off the components of $g_{\mu\nu}$.

Your notes are correct in that $g_{\mu\sigma}g^{\mu\rho}=\delta^\rho_\sigma$. Consider

$$x_\rho = g_{\rho\sigma} x^\sigma=g_{\rho\sigma}g^{\sigma\mu}x_\mu$$

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's just the identity.

Last edited: Jun 25, 2010
3. Jun 25, 2010

### 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.

Last edited: Jun 25, 2010