Why is Minkowski metric diagonal?

Question

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

Answer 1

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).

Answer 2

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.

Answer 3

There are several ways a metric can be diagonal.

1. "Frameable"
The Kerr-Newman solution has a metric given by a line element of the form: $$\left(h^1\right)^2 + \left(h^2\right)^2 + \left(h^3\right)^2 - \left(h^4\right)^2,\tag{1}\label{1}$$ where $$ h^1 = \frac{Bdr}{C}, \quad h^2 = Brdθ, \quad h^3 = \sin{θ}\frac{A^2 r dφ - Sdt}{B}, \quad h^4 = \frac{cC}{B}\left(dt - \frac{S \sin^2{θ}}{c^2}r dφ\right),\\ A = \sqrt{1 + \frac{S^2}{c^2}}, \quad B = \sqrt{1 + \frac{S^2 \cos^2{θ}}{c^2}}, \quad C = \sqrt{1 + \frac{2V + S^2}{c^2}}, \quad S = \frac{J}{Mr}, $$ and $$V = -\frac{GM}{r} + \frac{μ_0}{8π}\frac{Q^2}{r^2}\tag{2}\label{2},$$ for a rotating object with mass $M$, electric charge $Q$ and angular momentum $J$.

I don't believe there is any single coordinate system, however, that it can be made diagonal in such a way that you have no mixing going on, as you do with $h^3$ and $h^4$ between the longitude coordinate $φ$ and the time $t$; only the radius $r$ and co-latitude $θ$ do, with $dr$ going only with $h^1$ and $dθ$ only with $h^2$. That's the essence of what makes it rotating. You can see that directly: if you shut off the rotation, setting $J = 0$, which sets $S = 0$, then $h^3$ and $h^4$ become separated, $dφ$ goes only with $h^3$ and $dt$ goes only with $h^4$.

There is a condition required for a chrono-geometry to be uniformly diagonal, like this, with respect to a co-frame $\left(h^1, h^2, h^3, h^4\right)$. I can't find the cite, or name the theorem, but I think it has to be globally hyperbolic. For instance, if the space-time is not time-orientable, then you might have a case where your future-pointing direction is along the positive $t$ axis, but then as you go along a path (a "time-travel" closed time-like path) you end up back at the same place and time, with the future-pointing direction now going along the negative $t$ axis. No global co-frame exists for such a chrono-geometry.

All Riemann-Cartan chrono-geometries, by construction, are "frameable" in this way, since the co-frame is already part of its defining infrastructure.

2. "Scaleable"
If you turn off the rotation for the Kerr-Newman solution, the result is the Reissner-Nordstroem solution has the same form ($\ref{1}$), and the same potential $V$ as given by ($\ref{2}$), with $$h^1 = \frac{dr}{D}, \quad h^2 = r dθ, \quad h^3 = r \sin{θ} dφ, \quad h^4 = c D dt, \quad D = \sqrt{1 + \frac{2V}{c^2}}. $$ Then, the metric reduces to scaling coefficients on the coordinates.

Again, I don't know the exact conditions for this to be possible. Only a subset of the "Frameable" cases have this property.

3. "Integrable"
If you turn off the charge, setting $Q = 0$, and the mass, setting $M = 0$, then $V = 0$, $C = 0$ and the result is ($\ref{1}$), with $$h^1 = dr, \quad h^2 = r dθ, \quad h^3 = r \sin{θ} dφ, \quad h^4 = c dt.$$

As presented, the metric is still "Scaleable", but it can be integrated to: $$x = r \cos{θ}\cos{φ}, \quad y = r \cos{θ}\sin{φ}, \quad z = r \sin{θ},$$ with the result being that ($\ref{1}$) becomes: $$dx^2 + dy^2 + dz^2 - (c dt)^2.$$

This is an extension of Euclidean geometry, whose line element is $$dx^2 + dy^2 + dz^2$$ to a Minkowski chrono-geometry. So, you're essentially asking why the Euclidean line element is diagonal, too. But, that's the essence of what it means for the geometry to be flat. It means you can uniformly lay out a coordinate grid that gives you something analogous to the Pythagorean Theorem.

Contrast that to the sphere, or a similar surface like the Earth. If you lay out a grid on it, like is done with the range lines in the old Northwest Territory in the United States, then it warps and buckles, because of the Earth's surface being curved. There are numerous defects in the grid on account of this - places where you have kinks or other abrupt adjustments. The laying out happened between 1784 and 1787. I don't know how much Washington, himself, was involved in this. He was a land surveyor by profession, and he had large holdings in the area, but it's all still there today.

Answer 4

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.