In my differential geometry course I have been presented the following metric, to do basic computations with: $$g=-a^2 (dt\otimes dt)-ae^y(dt\otimes dx)-ae^y(dx\otimes dt)-\frac{e^{2y}}{2}(dx\otimes dx)+(dy\otimes dy).$$ Another physics professor, who teaches classical electrodynamics, asked me to come up with a physical interpretation of the metric when he saw it in my notes.

Now, I haven't taken any courses on general relativity, but by the look of it I think it could be interpreted as the metric of spacetime with an inhomogeneous mass distribution in the $y$ axis.

Is this interpretation correct? Is the mass distibution obvious from the expression of the metric, of at least straighforward to compute? What exactly is going on with the coupling between the $x$ and $y$ coordinates?

I'm tagging as general relativity out of my assumption but can utag if it turns out to be wrong.

  • 1
    $\begingroup$ I think you should remove the sign in front of the first term. Also, if $a$ is a constant, you should absorb it into the definition of the time variable $t$. I suspect the sign of the last term is also wrong, so $$g= (dt\otimes dt)-e^y(dt\otimes dx)-e^y(dx\otimes dt)-\frac{e^{2y}}{2}(dx\otimes dx)-(dy\otimes dy).$$ $\endgroup$
    – Cham
    Nov 5, 2019 at 0:50
  • $\begingroup$ Im afraid I can't change that sign without perturbating the signature of the matrix, which would make it non semi-Remannian.So is it with the other sign. I could just multiply by -1 if thats a more comftable signature choice but it won't change much. I'll also not absorb $a$ since it clearly represents $c$ (velocity of light in the vacuum) and allows me to stay wothin SI units, but i can rename it $c$. $\endgroup$
    – Mario
    Nov 5, 2019 at 8:40

1 Answer 1


I would not say that this is a metric in generall reletivity due to the sign of $dt^2$ and $dx^2$. In GR you would get different signs infront of the time and the space. Which one is positive and which one is negative depends on your convention.

So I whould not think about it that abstract (In four dimensions with curved spacetime). But much simpler just as a curved threedimensional space. Since there is no curvature in the $y-direction$ you might even look at the projection, and only consider a curved surface.

But I don't know what object this surface will form.

Have you calculated the curvature?

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    $\begingroup$ The three eigenvalues of the matrix formed by $g$ (when written as a quadratic form), are $1$ and $\left( -2a^2 -e^{2y} \pm \sqrt{ 4 a^4 + 12a^2 e^{2y} + 4 e^{4y} } \right)/4$. The only negative eigenvalue is $\left( -2a^2 -e^{2y} - \sqrt{ 4 a^4 + 12a^2 e^{2y} + 4 e^{4y} } \right)/4$. This means that the signature is $(-,+,+)$ for this metric and so I disagree with your first paragraph. $\endgroup$ Nov 4, 2019 at 18:03
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    $\begingroup$ Another easy check for the signature of this metric is to compute the determinant, which is negative. This means it has to have the right signature. $\endgroup$ Nov 4, 2019 at 18:05
  • $\begingroup$ But only if you are interpreting y as time right? $\endgroup$
    – tomtom1-4
    Nov 4, 2019 at 18:09
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    $\begingroup$ Your first paragraph is not correct, different signs are only required if the metric is diagonal. What it matters is that the signature is (-,+,+) which, as pointed out above, it is. The curvature and other calculations are straightforwardly derived from those of the Gödel metric (en.wikipedia.org/wiki/G%C3%B6del_metric) of which I have found this is a variant. $\endgroup$
    – Mario
    Nov 5, 2019 at 0:16

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