When can a metric be put in diagonal form?

[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?

Thanks in advance.

 

[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.

 

[PeterDonis]

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.

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.)

Is it always possible to put a metric in diagonal form or are certain symmetries required?

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.

What exactly is the motivation for diagnoal form

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

 

[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 vacuum 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.]

 

[PeterDonis]

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.

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.

or the other two, you need something more, and two killing vector fields are not enough

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.)

 

[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.

 

[PeterDonis]

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

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.

 

[Ben Niehoff]

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.

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).

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.

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?).

 

[PeterDonis]

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

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.)

the twist-free congruences do not have to be Killing (and they do not have to be geodesic)

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.)

a Killing congruence is not necessarily twist-free

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).

 

[Ben Niehoff]

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.)

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

 

[PeterDonis]

Look up orthogonal coordinates in $\mathbb{R}^3$.

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

 

[bcrowell]

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

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.

 

[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.