# When can a metric be put in diagonal form?

1. Feb 17, 2015

### binbagsss

I'm looking at deriving the Schwarzschild metric in 'Lecture Notes on General Relativity, Sean M. Carroll, 1997'
and the comment under eq. 7.8, where he seeks a diagnoal form of the metric...

- Is it always possible to put a metric in diagonal form or are certain symmetries required?
- What exactly is the motivation for diagnoal form - is it because computing christoffel symbols etc is simpler or?

2. Feb 17, 2015

### Ibix

I believe you can always diagonalise the metric at any point, and that the first-derivatives of the metric can always be made to vanish at that point. Second derivatives and above will not, in general, vanish. This is the formal definition of "manifold looks locally Lorentzian". Riemann Normal Coordinates come out of this, too.

Have a look at equation 2.34 et seq in Carroll's notes.

3. Feb 17, 2015

### Staff: Mentor

Yes, but only at that point. If you want to be able to find a coordinate chart in which the metric is diagonal over a finite region, that won't always be possible. See below.

(Note that "diagonal" is a property of the metric as expressed in a particular coordinate chart; it's not an intrinsic property of the geometry of spacetime itself. Even for spacetimes where the metric is diagonal in some charts, such as Schwarzschild spacetime in the Schwarzschild chart, there will be other charts where it is not diagonal, such as Schwarzschild spacetime in the Painleve chart.)

I'm not sure if the exact conditions for the existence of a coordinate chart in which the metric is diagonal over a finite region of a manifold are known. (I think I know at least one, however--see below.) However, I know of at least one example of a metric that is not diagonalizable: the Kerr metric. There is no coordinate chart on the Kerr metric that is diagonal over a finite region.

The example of the Kerr metric suggests at least one possible condition for there to be a chart in which the metric is diagonal: there must be a family of timelike worldlines that fills the spacetime and is hypersurface orthogonal, i.e., one can "slice" the spacetime into 3-dimensional spacelike slices such that each worldline in the family intersects each slice exactly once, and is orthogonal to each slice when they intersect.

There are probably a number of them, but one that I think is important is that it makes it easier to invert the metric.

4. Feb 17, 2015

### PAllen

From a counting argument, I suspect you need two extra identities to ensure you can find coordinates that are globally orthogonal (= globally diagonal). The argument is that out of 10 algebraically independent components of the metric, you want 6 to vanish. Gauge invariance suggests you can make 4 vanish. For the other two, you need something more, and two killing vector fields are not enough: the Kerr vaccuum has axial and timelike killing fields, yet cannot be brought into globally orthogonal form. Of course, the Schwarzschild geometry can be put in such a form. In any case, you certainly need very special conditions to be able to have a globally orthogonal metric in GR.

[Edit: cross posted with Peter, who made a similar point about Kerr metric.]

5. Feb 17, 2015

### Staff: Mentor

Actually, on thinking it over, this won't work. The ZAMO congruence in Kerr spacetime is orthogonal to surfaces of constant coordinate time in Boyer-Lindquist coordinates. There may be a more elaborate condition involving hypersurface orthogonality that will work, but just the existence of a hypersurface orthogonal timelike congruence is not enough by itself.

Hmm. In FRW spacetime, there is no timelike KVF, but there are at least three spacelike KVFs. So perhaps three spacelike KVFs is the magic number. Or perhaps a combination of three spacelike KVFs and a hypersurface orthogonal timelike congruence. (Note that the congruence need not be a family of integral curves of a timelike KVF; it isn't in FRW spacetime, as noted.)

6. Feb 18, 2015

### Ben Niehoff

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.

In other words, you must have d twist-free congruences, all mutually orthogonal.

7. Feb 18, 2015

### Staff: Mentor

Ah, ok, so I was not that far off.

So, just to belabor this a bit, any spherically symmetric spacetime automatically has three mutually orthogonal twist-free congruences, correct? (Because it has three mutually orthogonal spacelike KVFs, and each one generates a twist-free congruence.) So any spherically symmetric spacetime that has a fourth twist-free congruence orthogonal to the other three (which Schwarzschild and FRW both do) will have a diagonalizable metric.

Conversely, an axisymmetric spacetime that is not spherically symmetric only has one twist-free congruence from axisymmetry, so even if it has another one orthogonal to the first (as Kerr spacetime does, the ZAMO congruence), that's not sufficient for diagonalizability.

Last edited: Feb 18, 2015
8. Feb 18, 2015

### Ben Niehoff

In fact, I think Birkhoff's theorem tells you that the 4th Killing vector must exist, and it happens to be orthogonal to the first three.

However, keep in mind: In order to write a diagonal metric, the twist-free congruences do not have to be Killing (and they do not have to be geodesic). Also, a Killing congruence is not necessarily twist-free (consider that generated by $\partial_t + \partial_\phi$ in Minkowski space).

Yes.

However, I don't know if there is a simple way to look at a metric and decide if it can be written in diagonal form, in general. I'm not aware of any obstruction that is easy to calculate (but maybe one exists?).

9. Feb 18, 2015

### Staff: Mentor

Yes. (The orthogonality comes out as part of the proof of the theorem, so I'm not sure I would say it "happens to be" orthogonal.)

Agreed. In FRW spacetime, the congruence of comoving worldlines is not Killing. (It is geodesic, though; I can't think of an example offhand of a congruence that is both not Killing and not geodesic, leading to a diagonalizable metric.)

Yes, agreed.

In fact, in an axisymmetric but not spherically symmetric spacetime, I think a timelike Killing congruence cannot be twist-free (because, heuristically, it has to combine $\partial_t$ with zero or more spacelike Killing vectors with constant coefficients, and in order for the congruence to be twist-free the coefficients would have to vary with "distance from the axis", as they do for the ZAMO congruence in Kerr spacetime).

10. Feb 18, 2015

### Ben Niehoff

There are loads of examples which are neither Killing nor geodesic. Look up orthogonal coordinates in $\mathbb{R}^3$.

11. Feb 18, 2015

### Staff: Mentor

I was thinking specifically of "well known" spacetimes ("well known" meaning ones that have appeared in the literature).

12. Feb 18, 2015

### bcrowell

Staff Emeritus
Some possibly relevant information:

Grant and Vickers, "Block diagonalisation of four-dimensional metrics," http://arxiv.org/abs/0809.3327

De Turck and Yang, "Existence of elastic deformations with prescribed principal strains and triply orthogonal systems," http://zmath.sub.uni-goettingen.de/ZMATH/serials/en/search/zmath/?an=0544.53012 [Broken]

Tod, "On choosing coordinates to diagonalize the metric," http://iopscience.iop.org/0264-9381/9/7/005

In 3 dimensions, diagonalization is always locally possible, but in 4 it isn't in general (Grant and Vickers). This seems to be plausible based on counting degrees of freedom as in PAllen's #4, but it seems that a rigorous proof was nontrivial? I assume that "locally" means diagonalizing it exactly in some finite region, with the only obstruction to extending it globally being some kind of topological issue. (Globally, we don't even necessarily expect to be able to cover the whole manifold with one chart.) I assume they *don't* mean "locally" in the more trivial sense, which follows from the equivalence principle.

As others seem to have stated, I think staticity suffices (locally) in 3+1 dimensions. (I guess this is because staticity allows us to write the spacetime as a direct product of a time dimension with a 3-space, and diagonalization is possible a 3 dimensions.) More generally, the abstract of Tod says, "It is shown that diagonalizability of the metric generically imposes restrictions on the third derivative of the Weyl tensor when n=4[...]" (I don't have access to the whole paper.)

Because of the topological issues, I doubt that any amount of symmetry suffices to guarantee a metric that's diagonalizable globally. For example, take the direct product of a 3-sphere with a line of time. It has a very high degree of symmetry, but I don't think we can find coordinates that diagonalize the metric globally, simply because we can't cover the whole thing with one chart.

Last edited by a moderator: May 7, 2017
13. Feb 19, 2015

### Ben Niehoff

In the mathematical literature, "local" usually means "within a given chart". On the unit circle, for example, $d \theta$ is locally the gradient of a function, but not globally so. Likewise, all 2-dimensional manifolds are locally conformally flat---this just implies the existence of isothermal coordinates.