Coordinate transformation and metric tensor

[archipatelin]

General four-dimensional (symmetric) metric tensor has 10 algebraic independent components.

But transformation of coordinates allows choose four components of metric tensor almost arbitrarily.

My question is how much freedom is in choose this components?

Do exist for most general metric any coordinates transformation
g_{\mu\nu}(x)\rightarrow\tilde{g}_{\mu\nu}(\tilde{x}) which transforming matric to new form?

<br /> \[ \left( \begin{array}{cccc}<br /> -1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \tilde{g}11 &amp; \tilde{g}12 &amp; \tilde{g}13\\<br /> 0 &amp; \tilde{g}12 &amp; \tilde{g}22 &amp; \tilde{g}23\\<br /> 0 &amp; \tilde{g}13 &amp; \tilde{g}23 &amp; \tilde{g}33 \end{array} \right)\]<br />

Or​

<br /> \[ \left( \begin{array}{cccc}<br /> -1 &amp; \tilde{g}01 &amp; \tilde{g}02 &amp; \tilde{g}02 \\<br /> \tilde{g}01 &amp; \tilde{g}11 &amp; 0 &amp; 0\\<br /> \tilde{g}02 &amp; 0 &amp; \tilde{g}22 &amp; 0\\<br /> \tilde{g}03 &amp; 0 &amp; 0 &amp; \tilde{g}33 \end{array} \right)\]<br />
​

 

[bcrowell]

Do you want it to have this form at a particular point, or globally?

If all you want is a particular point, then you can always put it in the form diag(-1,1,1,1). This is guaranteed by the equivalence principle.

If you want this to happen globally, then I'm pretty sure it's impossible, for an arbitrary given metric. For example, I think the first form you gave has no rotation, but spacetimes like the Godel universe have rotation that you can't get rid of by a change of coordinates.

 

[archipatelin]

Do you want it to have this form at a particular point, or globally?

I mean globally transformation.

I know that the general two-dimensional metric can be transformed into the orthogonal traceless form

<br /> \[ \left( \begin{array}{cc}<br /> g_{00}(t,x) &amp; g_{01}(t,x) \\<br /> g_{01}(t,x) &amp; g_{11}(t,x) \end{array} \right)\rightarrow<br /> \left( \begin{array}{cc}<br /> h(\tau,\chi) &amp; 0 \\<br /> 0 &amp; -h(\tau,\chi) \end{array} \right).<br /> \] <br />​

What can we say about the transformation coordinates for general four-dimensional metric?

 

[bcrowell]

I mean globally transformation.

I know that the general two-dimensional metric can be transformed into the orthogonal traceless form

<br /> \[ \left( \begin{array}{cc}<br /> g_{00}(t,x) &amp; g_{01}(t,x) \\<br /> g_{01}(t,x) &amp; g_{11}(t,x) \end{array} \right)\rightarrow<br /> \left( \begin{array}{cc}<br /> h(\tau,\chi) &amp; 0 \\<br /> 0 &amp; -h(\tau,\chi) \end{array} \right).<br /> \] <br />​

What can we say about the transformation coordinates for general four-dimensional metric?

Then I think the answer is what I said in #2: there's not much you can do in general.

Keep in mind that you typically can't write down the metric for an entire spacetime in a single coordinate chart, so not only can you not put the metric in a special form, you can't put it in *any* form.

Maybe there is something more that can be said about special forms of the metric on an open neighborhood of any point. Certainly there are special forms in an open neighborhood if you have some kind of symmetry (static, stationary, ...).