Writing Metric in Matrix Form: Method?

[ChrisJ]

In $c=1$ units, from my SR courses I was told for example, that the Minkowski metric $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$ can be written in matrix form as the below..

\eta = <br /> \begin{pmatrix}<br /> -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 <br /> \end{pmatrix}<br />

And it was just kind of given to me, but now as I am trying to learn GR and practise more with weird and unusual metrics I find that I do not know a formalism for turning a given metric of the form $ds^2 =$.. into a matrix form $g =$ .

Am I correct in thinking that the following metric $ds^2 = \frac{1}{y^2} dx^2 + \frac{1}{y^2}dy^2$ is just simply..

g = <br /> \begin{pmatrix}<br /> y^{-2} &amp; 0 \\<br /> 0 &amp; y^{-2}<br /> \end{pmatrix}<br />

If so, what about weirder ones with cross terms (i.e. values in the matrix that are not just along the diagonal ).

Is there a standard formalism for doing this? I have tried searching but not sure I am using the correct terms to get the results I want, or if I do find stuff it uses a lot of notation that I am unfamiliar with.

 

[Orodruin]

Am I correct in thinking that the following metric $ds^2 = \frac{1}{y^2} dx^2 + \frac{1}{y^2}dy^2$ is just simply..

g =<br /> \begin{pmatrix}<br /> y^{-2} &amp; 0 \\<br /> 0 &amp; y^{-2}<br /> \end{pmatrix}<br />

Yes. What you are doing is really writing a matrix representation of the metric.

If so, what about weirder ones with cross terms (i.e. values in the matrix that are not just along the diagonal ).

Is there a standard formalism for doing this? I have tried searching but not sure I am using the correct terms to get the results I want, or if I do find stuff it uses a lot of notation that I am unfamiliar with.

In general, the line element is given by
$$
ds^2 = g_{ab} dx^a dx^b.
$$
If you have the line element, just write out the sum and start identifying components (taking into account that the metric is symmetric so that $g_{ab} = g_{ba}$. The matrix representation of the metric has the metric components $g_{ab}$ as its elements.

Edit: For example, consider the coordinates $\xi = x-t$ and $\eta = x+t$ in 2D Minkowski space (those are called light-cone coordinates. You would obtain that $x = (\xi + \eta)/2$ and $t = (\eta-\xi)/2$ and therefore
$$
ds^2 = -dt^2 + dx^2 = \frac{1}{4}[(d\xi + d\eta)^2 - (d\eta - d\xi)^2] = \frac{1}{2} d\xi \,d\eta
= g_{\xi\xi} d\xi^2 + 2 g_{\xi \eta} d\xi\, d\eta + g_{\eta\eta} d\eta^2.
$$
Identification directly gives $g_{\xi\eta} = 1/4$ and $g_{\xi\xi} = g_{\eta\eta} = 0$.

 

[Ibix]

In tensor notation, $ds^2=g_{ij}dx^idx^j$. If you want to use matrix notation for it (careful! Tensors are not matrices and the rules for multiplication are not the same), it's $ds^2=\vec{dx}^T\mathbf{g}\vec{dx}$.

So your example is correct. Essentially, the coefficient of $dx^idx^j$ goes in the i,j position of the matrix representation of the tensor. The only trap for the unwary is that $dx^idx^j=dx^jdx^i$, so for off-diagonal elements if you have $ds^2=\ldots+2Adx^idx^j+\ldots$ then you put $A$ in the position i,j and also A in j,i.

 

[ChrisJ]

In tensor notation, $ds^2=g_{ij}dx^idx^j$. If you want to use matrix notation for it (careful! Tensors are not matrices and the rules for multiplication are not the same), it's $ds^2=\vec{dx}^T\mathbf{g}\vec{dx}$.

So your example is correct. Essentially, the coefficient of $dx^idx^j$ goes in the i,j position of the matrix representation of the tensor. The only trap for the unwary is that $dx^idx^j=dx^jdx^i$, so for off-diagonal elements if you have $ds^2=\ldots+2Adx^idx^j+\ldots$ then you put $A$ in the position i,j and also A in j,i.

Ok thanks both,

So, if I am understanding you both correct, something like $ds^2 = -xdv^2 + 2dvdx$ would be

<br /> g = \begin{pmatrix} -x &amp; 1 \\ 1 &amp; 0 \end{pmatrix}

 

[Ibix]

Yes.