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So recently I had a problem with a given metric $g(T, X) = \begin{bmatrix} -X^2 & 0\\0 & 1\end{bmatrix}$ which had a singularity at X = 0. For a given coordinate transformation $(T,X) \rightarrow (V,Z)$, we are asked to show that the singularity was a coordinate singularity.

So, my goal was to show that the new metric had no singularities. I found that: $$ g(V,Z) = \begin{bmatrix} -Z & B\\B & \frac{C}{Z}\end{bmatrix}$$, where B and C are just some nonzero constants.

Now the determinant of this metric comes out to be a nonzero constant because the Z's cancel out. This suggests that there are no zero eigenvalues and thus no singularity.

However, If I plug in Z=0 (which corresonds to X=0 for this problem) I get an undefined term of $\frac{C}{0}$. This to me seems to suggest a singularity, which seems to be contradictory to my previous determinant calculation.

Is it possible for a metric to have an undefined term (and still be invertible/non-singular at that point) and if so what would this correspond to physically?

For reference, the given transformation was $V = T + log(X)$ and $Z =X^2$.

Edit: So turns out for this problem $log(X)$ was intended to be a natural log and so the &g_{zz} term didn't have a z dependence. But my question remains. If this were a base 10 log and we acquired the metric shown, how could we have an undefined term when the overall determinant of the matrix is a constant?

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  • $\begingroup$ Note that to answer the given question, you only need to show that $g(V,Z)$ is not singular on the image of the locus $X=0$. $\endgroup$
    – mmeent
    Apr 21 '20 at 18:33
  • $\begingroup$ I'm a little new to differential geometry. What do you mean by "the image of the locus at X=0". $\endgroup$
    – ZacharyC
    Apr 21 '20 at 18:35
  • $\begingroup$ The points that the singularity of the original metric get mapped to. (You do not really care if the new metric has a coordinate singularity somewhere else.) $\endgroup$
    – mmeent
    Apr 21 '20 at 18:42
  • $\begingroup$ Given the exercise, I strongly suspect you made a mistake applying the coordinate transformation. $\endgroup$
    – mmeent
    Apr 21 '20 at 18:43
  • $\begingroup$ I found the new metric from the given transformation two ways: one using the transformation law of metric tensors and the other by just computing the differentials directly (so like dZ = 2XdX). Both gave me the same answer which makes me think I got it right. $\endgroup$
    – ZacharyC
    Apr 21 '20 at 19:02
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Is it possible for a metric to have an undefined term (and still be invertible/non-singular at that point) and if so what would this correspond to physically?

how could we have an undefined term when the overall determinant of the matrix is a constant?

This is actually something that happens a lot, since not all the coordinate system you can choose to describe locally a physical system are globally well defined.

The maybe simplest but most relevant case is the Schwarzschild solution, which describe a spherically symmetric and static vacuum solution of the Einstein field equations. In unit G_N = 1, it can be written in a diagonal form as:

ds^2 = -f(r)dt^2 + f(r)^-1 dr^2 + r^2d\Omega^2,

where \Omega is the solid angle and f(r) = (1-2M/r).

Clearly the determinant is not divergent for any positive r, since in it appears the product f(r)*f(r)^-1. It is also not invertible everywhere, since in r = 0 the determinant is 0. On the other hand, there is a coordinate singularity in r = 2m, which has many interesting physical conseguences, even if its not a real singularity.

To prove that the singularity is only of the coordinate system it is enough to find a coordinate transformation in which the singularity is not present. For the case of Schwarzschild these are, for example, the Kruscal coordinates.

However in most cases it is possible to evaluate the nature of a singularity by studying scalars built on from its curvature. One of the most common and relatively simple to compute is the Kretschmann scalar, which is the contraction of the Rienmann tensor with itself. For the Schwarzschild example, it is not divergent at r=2m, which then is surely a coordinate singularity, and divergent at 0 which instead it is a true singularity.

About the physical interpretation of a coordinate singularity, it strongly depends on the class of observer you are trying to describe with those coordinates. For the case of Schwarzschild, the fact that a metric component blows at r = 2M implies a peculiar behavior for observers at the spatial infinity (r \rightarrow \infty). Locally speaking, an observer A crossing the sphere r=2M will not experience nothing particular, i.e. standard physics will apply for any experiment he will try to perform locally. On the other hand, an observer B at spatial infinity will never see the outcome of any of those experiments because it would require an infinite amount of time for him to see A actually crossing the r = 2M radius.

This example is to explain that the law of physics are still the same at a coordinate singularity, but it is not possible to describe within the same set of coordinate physics on the global space-time.

Edit: For the Schwarzschild example, it is not divergent at r=2m, which then is surely a coordinate singularity, and divergent at 0 which instead it is a singularity.

As pointed out by @AndrewSteane the above sentence is wrong, in particular the word surely, since a non-diverging curvature invariant does not imply that a singularity is not true. Only the opposite applies, a diverging curvature invariant implies the existence of a true singularity. For the Scharzschild case, however, the existence of a coordinate system (the kruskal one) in which the singularity r=2m does not show implies that indeed it is a curvature singularity, and thats why its Kretschmann scalar does not diverge.

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    $\begingroup$ careful: diverging curvature invariant does imply curvature singularity, but finite curvature invariant does not necessarily imply there is no curvature singularity $\endgroup$ Apr 21 '20 at 22:36
  • $\begingroup$ @AndrewSteane True, but the singularity should show up in at least one of the curvature scalars, not necessarily in the Ricci scalar. $\endgroup$
    – Bruce Lee
    Apr 22 '20 at 7:14

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