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I ran across this problem on an exam a couple days ago:

Consider the transformation to the new coordinates $(r,u,z,\bar{z})$ $$r=\sqrt{(x^1)^2+(x^2)^2+(x^3)^2} \qquad u = x^0-r \qquad z=\frac{x^1+ix^2}{x^3+r} \qquad \bar{z}=\frac{x^1-ix^2}{x^3+r}$$ Find the metric tensor $g_{\mu \nu}$

At first glance, it seamed easier to just solve for the contravariant metric and invert it later $$g^{\mu\nu} = \epsilon^{\nu} \epsilon^\mu = \frac{\partial q^\nu}{\partial x^j} \hat{e}^j \frac{\partial q^\mu}{\partial x^i} \hat{e}^i = \frac{\partial q^\nu}{\partial x^j} \frac{\partial q^\mu}{\partial x^i} \eta^{ij} $$

However, when I computed the elements, I found it to be singular, with only three entries non-zero.

When I tried the other way around and computed $x^i = f(q^j)$ to get the covariant metric directly $$g_{\mu\nu} = \epsilon_{\nu} \epsilon_\mu = \frac{\partial x^i}{\partial q^\nu} \hat{e}_i \frac{\partial x^j}{\partial q^\mu} \hat{e}_j = \frac{\partial x^i}{\partial q^\nu} \frac{\partial x^j}{\partial q^\mu} \eta_{ij} $$

I got a non-singular metric with clearly non-zero elements.

I've read a few posts to understand the significance of a singular metric tensor ($det(g)=0$), but I'm yet unclear as to why one representation (covariant) of the metric was non-singular, while the other (contra) was singular when they should be describing the same thing.

Aren't they related by $G_{co}= \frac{1}{G_{contra}}$?

How is this possible if only one of them is singular?

One last note: this paper(On Degenerate Metrics and Electromagnetism, T.P Searight) seemed to acknowledge this problem, but I got lost in the heavy algebra because I didn't take a GR class yet

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  • $\begingroup$ What kind of metric do you start from? Is it Minkowski? Also I think you are confused with notation, the transformation rule to express $g_{\mu\nu}$ in the new coordinates should be the one pointed out by @frederic-thomas in his answer. The way you contract the $j$ index is not correct. Also, what do you mean by capital $G$? Do you mean that the components of the covariant metric can be found by taking the inverse of the respective component of the contravariant metric? That is only true for diagonal metrics. $\endgroup$
    – Urb
    Jun 18, 2020 at 20:15
  • $\begingroup$ @Urb Sorry, I fixed my equations. Yes, I did use the Minkowski metric. By $G$ I meant the matrix formed of the elements $g_{ij}$. The covariant matrix is the inverse of the contravariant metric, is that correct? $\endgroup$ Jun 18, 2020 at 22:05
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    $\begingroup$ Minor comment to the post (v4): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Jun 19, 2020 at 10:45

1 Answer 1

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Due to the invariance of the line element $ds^2$ we can use the following relation: (here we assume that the coordinates $x^\mu = (x^0, x^1, x^2, x^3)$ are from a cartesian Minkowski metric $\eta_{\mu\nu}$):

$$ g_{ij} dq^i dq^j =ds^2= dx^\mu dx^\nu \eta_{\mu\nu} = \frac{\partial x^\mu}{dq^i} dq^i \frac{\partial x^\nu}{dq^j} dq^j \eta_{\mu\nu} \equiv \frac{\partial x^\mu}{dq^i} \frac{\partial x^\nu}{dq^j}\eta_{\mu\nu} dq^i dq^j$$

Einsteinian summation convention is assumed. From this it should be plausible that one has the following transformation law:

$$g_{ij} = \frac{\partial x^\mu}{dq^i} \frac{\partial x^\nu}{dq^j}\eta_{\mu\nu} $$

The advantage of this particular case is that one only needs to compute the products of differential quotients for $(\mu,\nu) = (0,0;\, 1,1;\, 2,2;\, 3,3)$. This should allow to compute the components of the metric tensor $g_{ij}$ as it was required. I did not check on the singular character of the matrix $(g_{ij})$, but I guess as it was an exam question that the result should be kind of reasonable.

EDIT: Contravariant and covariant metric tensor components are related by

$$ g^{ij}g_{jk} = \delta^i_k$$

i.e. indeed there are reciprocal to each other. By the way, for particular values of the coordinates the metric tensor can form a singular matrix (i.e. without inverse) or even not be defined, as it is the case for instance for the Schwarzschild metric:

$$ds^2 = \left(1-\frac{GM}{r c^2}\right) dt^2 - \left(1-\frac{GM}{r c^2}\right)^{-1} dr^2 -r^2 d\Omega^2$$

which is not defined for $r_s=GM/c^2$, the Schwarzschild radius. Then in this area $r_s$ other coordinates have to be chosen, for instance Kruskal-coordinates.

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  • $\begingroup$ Sorry, there was an error with my latex, I fixed the equations. These are the actual ones I used. $\endgroup$ Jun 18, 2020 at 22:08
  • $\begingroup$ In response to your edit: I see, but is it possible for one coordinate system to have a singular covariant metric, but a non-singular contravariant (or vice-versa)? If yes, what does this imply? $\endgroup$ Jun 20, 2020 at 0:37
  • $\begingroup$ Let's take polar coordinates in flat space: $ds^2=dr^2+r^2 d\phi^2$. The covariant metric tensor is $g_{ik} =diag(1,r^2)$, the contravariant one $g^{ik} = diag(1,1/r^2)$. The covariant metric has no inverse at r=0, the contravariant one diverges at $r=0$, so it's not defined. I guess, this is what you mean with singular. If it is not defined, it cannot be inverted, how to invert something that is not defined ? Of course one can study it for $r<<1$ and even invert it if $r\neq 0$. Manifolds have coordinates only locally, that's part of their definition. Polar coordinates aren't defined at r=0. $\endgroup$ Jun 20, 2020 at 9:51

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