What is the signature of the given metric?

[wglmb]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

In matrix form, the metric is
\begin{pmatrix}<br /> -(1-e^{-x^{2}}) &amp; 0 &amp; 0 \\<br /> 0 &amp; y^{2} &amp; 9 \\<br /> 0 &amp; 9 &amp; 6x<br /> \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

 

[Hurkyl]

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?

 

[wglmb]

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

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

The signature is a constant...

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.

 

[Hurkyl]

I see no easy way of solving the (cubic!) characteristic eigenvalue equation.

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?

 

[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?

 

[Hurkyl]

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.

 

[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?

 

[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&lt;\infty, since the signature should be constant.