# Signature of a metric

1. Dec 5, 2009

### wglmb

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

I have a metric and I need to find the signature.

2. Relevant equations

$ds^{2} = -(1-e^{-x^{2}})\ dt^{2} + 6x\ dy^{2} + 9\ dx\ dy + y^{2}\ dx^{2}$

3. The attempt at a solution

In matrix form, the metric is
$\begin{pmatrix} -(1-e^{-x^{2}}) & 0 & 0 \\ 0 & y^{2} & 9 \\ 0 & 9 & 6x \end{pmatrix}$

Now, I'm pretty sure my lecturer hasn't told us what a signature is, but after a bit of wikipedia-ing I see that I need to diagonalise this matrix.
What I want to know is, is that really what I need to do? 'cos it's a really horrible equation I'd have to solve to find the eigenvalues...
$\lambda ^{3} + (-6x -y^{2} + 1 - e^{-x^{2}})\lambda ^{2} + (6xy^{2} + 9^{2} - 6x(1-e^{-x^2}}) - y^{2} (1-e^{-x^{2}}))\lambda + (1 - e^{-x^{2}})(6xy^{2} + 9^{2}) = 0$

2. Dec 5, 2009

### Hurkyl

Staff Emeritus
The signature is a constant....

That said, that doesn't look like a metric. Are there restrictions on x,y that you haven't told us about?

3. Dec 5, 2009

### wglmb

No, but a later question is "Is the metric well-defined for all $0<x<\infty$?"
Haven't though about it, but I assume the answer is no, and that suitable restrictions will make it OK.

yeeees, but how do I find it?

The wikipedia way is to diagonalise the matrix (by finding eigenvalues, then eigenvectors, and then doing a bit of mtx multiplication) and count the number of +ve and -ve entries. Is that really what I need to do? I see no easy way of solving the (cubic!) characteristic eigenvalue equation.

4. Dec 5, 2009

### Hurkyl

Staff Emeritus
You aren't being precise enough! I think you have made two implicit assumptions:
* You want the general solution
* You want an exact solution
But you don't need either of those to answer the question you're really interested in, do you?

5. Dec 5, 2009

### wglmb

uh, sorry, I can't see what you're getting at... are you saying I don't need to find the eigenvalues at all?

6. Dec 5, 2009

### Hurkyl

Staff Emeritus
No, I'm saying that the solution you need doesn't have to be valid for all t,x,y, nor does it need to be an exact one.

7. Dec 5, 2009

### wglmb

Ah, I think I see! (I hope)
You mean that, since the signature is constant, I can take any x,y,t that I like and calculate it for them?

8. Dec 6, 2009

### wglmb

Thanks for your help - I think I've got it.

As $x \rightarrow \infty$ the signature is 0
As $x \rightarrow 0$ the signature is 1

So this implies that the metric is not well defined for all $0<\infty$, since the signature should be constant.

Last edited: Dec 6, 2009