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Why is the Minkowski metric a diagonal in a 4x4 matrix? What does the diagonal do?

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    $\begingroup$ What does "What does the diagonal do" mean? Are you asking what the significance of it being globally diagonalisable is? $\endgroup$ – Nihar Karve Feb 9 at 15:32
  • $\begingroup$ yes why diagonalisable $\endgroup$ – user288531 Feb 9 at 15:49
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For a given spacetime the metric tensor may be written in both (globally) diagonal and non-diagonal forms depending on what coordinates we choose. For example for flat spacetime one diagonal form (not the only diagonal form) is the Minkowski metric:

$$ ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 $$

But we could choose to write the metric using rotating cylindrical coordinates, and we would get:

$$ ds^2 = -\Big(1 - \frac{r^2\Omega^2}{c^2}\Big)dt^2 + r^2d\theta^2 + 2r^2\frac{\Omega}{c} d\theta dt + dr^2 + dz^2 $$

This is not diagonal because we have the term $2r^2\frac{\Omega}{c} d\theta dt$, but it's the same spacetime, and the metric is only non-diagonal because we have made a poor choice of coordinates.

So the interesting questions are:

  1. what determines whether for a particular geometry there is a choice of coordinates for which the metric is (globally) diagonal?

  2. why do we want the metric to be diagonal anyway?

(1) is hard. Technically we require: In d dimensions, you can write the metric in diagonal form if and only if your manifold can be foliated by d mutually orthogonal families of surfaces. But good luck with that. There is no simple way to take some arbitrary geometry and find coordinates in which the metric is diagonal. Not all metrics can be written in a diagonal form - for example the Kerr metric cannot, regardless of how inventive we are in choosing coordinates.

As for (2): having the metric in diagonal form is computationally convenient. It makes its determinant easy to calculate so it is easy to invert. It also reduces the amount of work needed to calculate the Christoffel symbols.

As an addendum to this, during a discussion of this issue in the chat room ACuriousMind pointed out that for specific cases the "hypersurface orthogonality" condition is equivalent to the "Frobenius condition" (also just called "integrable" by mathematicians) $\xi \wedge \mathrm{d}\xi = 0$ for the one-form $\xi$ dual to the Killing vector. For more on this see this article on the Schwarzschild metric (the link is a PDF).

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  • $\begingroup$ It might be worth noting (1) that "the metric" is distinct from "the components of the metric in a particular basis", and more importantly (2) that "a particular basis" is distinct from "a particular coordinate basis". By my reading, nothing in the question requires even considering coordinates. $\endgroup$ – Mike Feb 10 at 16:56
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Two facts are sufficient to make Minkowski tensor globally diagonal:

  • Symmetry of the metric tensor $g_{ij} = g_{ji}$, the simple reason for this can be found in the answer to this question Why is the metric tensor symmetric?. For every symmetric matrix one can switch to the basis of eigenvectors, where it is diagonal. Therefore, it can be diagonalized at least locally.
  • Homogeneity of the space-time, $g_{ij}$ does not depend on the coordinates. Diagonalizing it in one point, we make it diagonal everywhere.
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Being diagonal is a coordinate-dependent concept: the components of the matrix associated to the metric tensor depend on the coordinate system you use. Thus a very simple example of a non-diagonal metric is the standard Euclidean metric $\delta = dx^2 + dy^2$ on $\mathbb R^2$ in the coordinate system $(x,z) = (x, x+y)$, where it has the coordinate expression $$\delta = dx^2 + d(z-x)^2 = 2dx^2 + dz^2 - 2 dx dz.$$

In fact, there's some very famous solutions that have non-diagonal metrics. Such as the Kerr metric for a rotating black hole in General Relativity.

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