Metric tensor and spacetime

1. Oct 7, 2008

student85

Hi guys. I'm taking a GR course right now, my first one. I was reading the textbook and I was wondering if you guys could help me out just to make sure I'm getting things straight here. I'm reading about the metric tensor, and I'm pretty sure I am expected to know what the metric tensor for a given spacetime is, if the line element if given to me. So, say we have this line element:

ds2=ev(r)dt2-e$$\lambda$$(r)dr2-r2(d$$\theta$$2+sin2$$\theta$$d$$\phi$$2)
Is the metric tensor simply the coefficients of each differential arranged in the diagonal of a matrix, with the rest of the elements in it equal to zero? Or am I way off here?

NOTE: The Thetas and the phi in the expression are not supposed to be exponentials, I don't know why they came out that way.

Last edited: Oct 7, 2008
2. Oct 7, 2008

cristo

Staff Emeritus
The line element is related to the metric tensor as follows:

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

So, yes, for this line element, the metric tensor is diagonal. Note that, in general, the metric tensor can have off diagonal terms.

Inline latex is obtained by using [itex] brackets.

3. Oct 7, 2008

tiny-tim

Hi student85!

(type "itex" instead of "tex", and it gives you in-line tex, which fits better. )

Yes … the coefficients of dsomething2 go into the diagoanl positions in the matrix.

And if the line element contained a mixed term, for example, dxdt, then half of the coefficent would go into each of the two corresponding off-diagonal positions.

4. Oct 7, 2008

student85

Thanks guys!
I love this forum!