# Change of basis for a metric

## Main Question or Discussion Point

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

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

$$ds^2=A^2dt^2+B^2dtdr-K^2dr^2$$

one can look for a transformation to T, R

$$dT=fdt+hdr$$
$$dR=kdt+pdr$$

so the $dtdr$ terms cancel. I'm not sure if this answers your question.

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) ?

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.

$$dT=fdt+hdr$$
$$dR=kdt+pdr$$
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

I guess though you are considering local considersations only so you will be able to find some domain where this is possible

Hi nughret,
yes, I have some doubts about this process and it's not clear whether a general method exists.

M

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.

Hurkyl
Staff Emeritus