Change of basis for a metric

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I would like to know how, given a metric with non-zero off-diagonal components [tex]g_{mn}, m \neq n[/tex], one can find if another (orthonormal or null) frame exists in which [tex]g_{\mu \nu}=constant[/tex] for all components of the metric. And if it exists how to compute this basis. Thanks!
 

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  • #2
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It's often possible to find a coordinate transformation to remove the diagonal element. Given,

[tex]ds^2=A^2dt^2+B^2dtdr-K^2dr^2[/tex]

one can look for a transformation to T, R

[tex]dT=fdt+hdr[/tex]
[tex]dR=kdt+pdr[/tex]

so the [itex]dtdr[/itex] terms cancel. I'm not sure if this answers your question.
 
  • #3
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Thanks, this is indeed what I am looking for. Do you know maybe if it is possible with the Kerr metric (given, e.g. the metric in the Boyer-Lindquist coordinates) ?
 
  • #4
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I don't think it's possible to have a physical frame ( ie corresponding to an observer) which loses the diagonal element because it represents actual angular momentum. The Wiki article on the Kerr spacetime also has metric for the co-rotating observer which has the diagonal element.
 
  • #5
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[tex]dT=fdt+hdr[/tex]
[tex]dR=kdt+pdr[/tex]
These equations are not gaurenteed to give you a coordinate change as the 1-form dT may not be exact in order to actually find the quantity T = T(t,r) , thenecessary co-ordinate change. I will come up with an example if necessary, it is similar to thecase of holonomic and non-holonomic constraints in classical mechanics
 
  • #6
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I guess though you are considering local considersations only so you will be able to find some domain where this is possible
 
  • #7
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Hi nughret,
yes, I have some doubts about this process and it's not clear whether a general method exists.

M
 
  • #8
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In the case of a Riemannian manifold I am aware of a beautiful proof which gives a positive answer to the case of finding on orthonormal base. My memory of the details are a bit hazy and we use the principle frame bundle (or some prolongation of this) and then consider whether we can find a subgroup of the natural group which acts on this bundle. For the Riemannian case we get, using polar decompostion, the necessity of a global symmetric non-degenerate tensor, i.e. the Reimannian metric.
Now obviously for a Lorentz manifold such a proof will need to be altered. I shall try to fully recall the proof in the R case and then apply it to the L case and let you know what I find.
 
  • #9
Hurkyl
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In the case of a Riemannian manifold I am aware of a beautiful proof which gives a positive answer to the case of finding on orthonormal base.
That doesn't sound right. Spheres, for example, don't have any (global) basis, let alone an orthonormal one, because every vector field must have a zero.
 
  • #10
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That doesn't sound right. Spheres, for example, don't have any (global) basis, let alone an orthonormal one, because every vector field must have a zero.
Of course. I was referring to the problem of being able to find smooth local orthonormal bases through any point. The proof is used in the more general setting of vector bundles and as i said i will try to recall it, we have to consider the orthogonal subgroup of the general linear group and you can then see where the polar decomposition comes in. For the Lorentz case we will need to consider the Lorentz orthogonal subgroup of the General linear group etc..
 

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