Coordinates for diagonal metric tensors

  • #1
Ibix
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In the recent thread about the gravitational field of an infinite flat wall PeterDonis posted (indirectly) a link to a mathpages analysis of the scenario. That page (http://www.mathpages.com/home/kmath530/kmath530.htm) produces an ansatz for the metric as follows (I had to re-type the LaTeX - any typos are mine):
mathpages said:
We can restrict the range of possible metrics based on symmetry considerations. We seek a stationary coordinate system, so the metric coefficients must be independent of ##t##. In addition, the field is clearly independent of the transverse coordinates ##y## and ##z##, so we consider a diagonal metric of the form$$d\tau^2=g_{tt}(x)dt^2-g_{xx}(x)dx^2-g_{yy}(x)dy^2-g_{zz}(x)dz^2$$where ##g_{yy}(x)=g_{zz}(x)##
OK - I follow almost all of that. Stating it in painful detail:
  • We choose the wall to be at rest and pick constant spatial coordinates to be at rest wrt the wall. The wall isn't moving or changing in these coordinates, so the metric can't depend on the time coordinate.
  • We pick two spatial directions to lie in the planes at constant distance from the wall. The wall is symmetric under translation and rotation in this plane, so the metric can't depend on these coordinates.
  • That leaves perpendicular to the plane of the wall as our remaining direction, and the only possible coordinate the metric can depend on.
But what symmetry is reflected by setting the off-diagonal elements to zero? I can see why ##g_{yz}=g_{zy}=0##, because this enforces rotational symmetry in the y-z plane on the assumption that the y and z directions are perpendicular. But why shouldn't (e.g.) ##g_{xy}## or ##g_{tx}## be non-zero? We don't expect rotational symmetry in these planes (and, indeed, the on-diagonal elements aren't expected to be equal), so I'm not sure what we're enforcing here. I know we can always pick coordinates such that the off-diagonal elements are zero, but haven't we already used up our freedoms to pick coordinates with the bullets above?

There's clearly something basic I'm missing here, but hopefully I won't have to kick myself too hard when I get an answer...
 

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  • #2
PeterDonis
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what symmetry is reflected by setting the off-diagonal elements to zero?
There are two types of off-diagonal elements to consider (if we eliminate, as you say, ##g_{yz}## and ##g_{zy}## based on symmetry in the y-z plane). (Note that the same symmetry should allow one to set ##g_{yy} = g_{zz}## without loss of generality.)

Setting the off-diagonal elements with ##t## in them equal to zero is equivalent to assuming that the spacetime is static and choosing the spacelike surfaces of constant coordinate time to be the appropriate static surfaces. I don't see any issue with that for this scenario.

Setting the off-diagonal elements ##g_{xa}## to zero, where ##a = y, z##, is equivalent to assuming that the surfaces of constant "height" ##x## are orthogonal to the "vertical" vector ##\partial / \partial x## (or, to put it another way, to the proper acceleration of an object at rest at constant height). This seems reasonable since any non-orthogonality would pick out a particular direction in the y-z plane, which we have already agreed should be rotationally symmetric.

I know we can always pick coordinates such that the off-diagonal elements are zero,
You can always do this locally (i.e,. within the confines of a local inertial frame), but you can't always do it globally. A simple example is Kerr spacetime, which does not admit any globally diagonal coordinate chart.
 
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  • #3
Ibix
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Thanks Peter.
Setting the off-diagonal elements with ##t## in them equal to zero is equivalent to assuming that the spacetime is static and choosing the spacelike surfaces of constant coordinate time to be the appropriate static surfaces. I don't see any issue with that for this scenario.

Setting the off-diagonal elements ##g_{xa}## to zero, where ##a = y, z##, is equivalent to assuming that the surfaces of constant "height" ##x## are orthogonal to the "vertical" vector ##\partial / \partial x## (or, to put it another way, to the proper acceleration of an object at rest at constant height). This seems reasonable since any non-orthogonality would pick out a particular direction in the y-z plane, which we have already agreed should be rotationally symmetric.
Of course - so a non-zero off-diagonal element implies that the relevant coordinate basis vectors aren't orthogonal. I think I may need to kick myself a little bit over that, since it's obvious that the inner product of two different basis vectors is not generally zero when there are off-diagonal elements in the metric.
You can always do this locally (i.e,. within the confines of a local inertial frame), but you can't always do it globally. A simple example is Kerr spacetime, which does not admit any globally diagonal coordinate chart.
Hm. Can you not just take the metric tensor in some coordinates, treat it as a matrix and diagonalise it? (Maxima at least gave me four distinct eigenvalues when I tried it. The phone version gave up trying to do the eigenvectors, but having seen the eigenvalues I assumed that was algebraic complexity getting out of hand.) Presumably that's naive in some way. Am I going to run into something like trying to Einstein synchronise clocks in a rotating frame, where you get a kind of screw discontinuity in the plane of simultaneity?

Actually this was part of my confusion. I can easily introduce off-diagonal elements in even a trivial metric by picking complicated coordinates. So off-diagonal elements can just mean "didn't choose straightforward coordinates". But then what is it about Kerr spacetime that means you can't globally diagonalise the metric?
 
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  • #4
PeterDonis
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Can you not just take the metric tensor in some coordinates, treat it as a matrix and diagonalise it?
Am I going to run into something like trying to Einstein synchronise clocks in a rotating frame, where you get a kind of screw discontinuity in the plane of simultaneity?
Yes, that's basically it: the chart can't be global, because the coordinate surfaces will have these kinds of discontinuities so the same event will map to multiple coordinate values.

what is it about Kerr spacetime that means you can't globally diagonalise the metric?
First, what I'm going to say assumes that we are talking about charts with one timelike and three spacelike coordinates. There is probably a way to extend what I'm saying to other cases, but it would probably take someone slicker at math than I am to be able to describe such an extension.

In Kerr spacetime, the "cross term" that can't be gotten rid of is a "time-space" cross term. The reason you can't get rid of this term is that there does not exist any hypersurface orthogonal congruence of timelike worldlines. In other words, there is no family of timelike curves in the spacetime that (a) fill the entire spacetime, and (b) are everywhere orthogonal to a set of spacelike 3-surfaces that foliate the spacetime.

(Note that the above is often described as Kerr spacetime being stationary but not static. However, there are non-static spacetimes that do have a globally diagonal chart, and hence do have a family of worldlines and surfaces that meet the requirements above. The best known example is FRW spacetime.)

In Kerr spacetime, you can find charts that have no "space-space" cross terms. I'm not aware of any commonly known spacetime that does not admit such a chart. It might well be that there is no such spacetime, if we restrict to charts with three spacelike coordinates. The reason I think that is that the three spacelike coordinates are coordinates on surfaces of constant time (so we have already assumed that we have a spacetime that can be foliated into such surfaces, not necessarily all with the same geometry, since the geometry could be a function of time), and those surfaces, being everywhere spacelike, have Riemannian geometry. I'm not sure it's possible to have the kind of "screw discontinuity" you refer to in a manifold with Riemannian (as opposed to Lorentzian) geometry. But I'm not familiar enough with this area to know for sure.
 
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  • #5
DrGreg
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Can you not just take the metric tensor in some coordinates, treat it as a matrix and diagonalise it?
The metric tensor matrix doesn't act directly on coordinates, it acts on differentials of coordinates. You can always diagonalise at a single event, but to find coordinates that diagonalise the metric matrix everywhere you have to solve differential equations, and sometimes the equations have no solutions (or at least none that are compatible with the constraints you are putting on your coordinates, e.g. that certain objects are at rest).
 
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  • #6
PeterDonis
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to find coordinates that diagonalise the metric matrix everywhere you have to solve differential equations, and sometimes the equations have no solutions
Do you know of an example of a Riemannian manifold (i.e., spacelike everywhere) that does not admit a global diagonal coordinate chart?
 
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DrGreg
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Do you know of an example of a Riemannian manifold (i.e., spacelike everywhere) that does not admit a global diagonal coordinate chart?
Sorry, I don't know whether such a thing exists or not.
 
  • #8
Ibix
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The metric tensor matrix doesn't act directly on coordinates, it acts on differentials of coordinates. You can always diagonalise at a single event, but to find coordinates that diagonalise the metric matrix everywhere you have to solve differential equations, and sometimes the equations have no solutions (or at least none that are compatible with the constraints you are putting on your coordinates, e.g. that certain objects are at rest).
Right - so the naive matrix diagonalisation approach tells me how to get a local Minkowski frame (give or take scale factors of the eigenvalues) at any event. But to fit those local frames together requires solving those PDEs (presumably something like extending the Minkowski frame axes as space-like geodesics and rotating the "next frame along" to match up), and that can't necessarily be done without assigning multiple coordinates to one event.
In Kerr spacetime, the "cross term" that can't be gotten rid of is a "time-space" cross term. The reason you can't get rid of this term is that there does not exist any hypersurface orthogonal congruence of timelike worldlines. In other words, there is no family of timelike curves in the spacetime that (a) fill the entire spacetime, and (b) are everywhere orthogonal to a set of spacelike 3-surfaces that foliate the spacetime.
And this is exactly what happens with the naive approach to a rotating frame in Minkowski spacetime, albeit by choice. There, I chose a family of worldlines that don't satisfy (b). In the Kerr spacetime, I cannot find a set of worldlines that do.

Can I pick such a family if I promise to stay outside the ergosphere? I can have observers hovering at constant coordinates indefinitely. Or does "cutting a hole" in the patch of spacetime I'm covering cause trouble?
 
  • #9
PeterDonis
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Can I pick such a family if I promise to stay outside the ergosphere?
No. Kerr spacetime is not hypersurface orthogonal anywhere.
 
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  • #10
Ibix
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No. Kerr spacetime is not hypersurface orthogonal anywhere.
Thanks. Is there a way to decide this for an arbitrary spacetime?
 
  • #11
martinbn
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Do you know of an example of a Riemannian manifold (i.e., spacelike everywhere) that does not admit a global diagonal coordinate chart?
Some of them don't admit global charts at all.
 
  • #12
PeterDonis
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Some of them don't admit global charts at all.
If you are referring to cases like a sphere not admitting a single global chart, that's true, but it's not the sort of thing I'm looking for. Let me rephrase my question: is there an example of a Riemannian manifold for which no atlas of charts exists that contains any diagonal charts?
 
  • #13
martinbn
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In dimension three it is true that every point has a neighborhood with local coordinates on which the metric is diagonal. In higher dimensions it isn't true in general.
 

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