# Can General Relativity Metric Tensor be independent of a particular co-ordinate index in a local area?

For example in a particular local area, can the metric tensor be totally independent of $z$ co-ordinate in $(t,x,y,z)$ co-ordinate system? This way the distance function will not contain $z$ co-ordinate. Probably this is same as asking if a diagonal element be zero in metric tensor?

Or this leads to degeneracy of the metric?

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The signature of the metric, i.e., the number of positive and negative eigenvalues, is coordinate-independent, so if you already require your metric to have a certain signature, that's not going to change in any coordinates.

Since the context of your question is general relativity, where we require spacetime to be Lorentzian and the metric nondegenerate, the answer is "no".

However, if you have a coordinate singularity, it can look that way. For example, take the Minkowski spacetime in the usual Minkowski coordinates and substitute $z = \frac{2}{3}Z^{3/2}$: $$\mathrm{d}s^2 = -\mathrm{d}T^2 + (\mathrm{d}x^2 + \mathrm{d}y^2 + Z\,\mathrm{d}Z^2)\text{.}$$ At $Z = 0$, the contribution of $\mathrm{d}Z$ completely disappears. However, what's actually going on is that our new coordinate chart fails to cover that region of spacetime.

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The metric in GR must be pointwise diagonalizable1 to that of Minkowski, $\mathrm{diag}(-1, 1, 1, 1)$. GR is a generalization of SR to arbitrary smooth coordinates, and at every point there exists a change of coordinates that puts you into a free-falling frame. This is basically the mathematical version of the Equivalence Principle.

More abstractly, the tangent space at a point would have dimension less than $4$ if the metric did not have full rank there, and so you wouldn't be in the 4D Lorentzian manifold that is spacetime. In a sense, you couldn't distinguish the $z$-direction as existing. Even though Lorentzian manifolds loosen the positive definiteness requirement for the inner product that Riemannian manifolds have, they still require some degree of nondegeneracy such that

for all non-zero $x$ there exists some $y$ such that $\langle x, y \rangle \neq 0$.

But in the case you are proposing, the unit vector in the $z$-direction, $\hat{e}_z$, would have $g_{\mu\nu} e_z^\mu v^\nu = 0$ for any vector $\vec{v}$.

Note of course that the components $g_{\mu\nu}$ can all be independent of a coordinate. Your metric can be independent of where you are in spacetime, but it cannot be completely ignorant of which direction you are pointing.

1 As fqq points out, this is not the same as having all nonzero diagonal elements.

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"The metric in GR must be diagonalizable to that of Minkowski" Um, no, that's not quite true. Only LOCALLY is it Minkowskian, but not globally (e.g. the Schwarzschild solution cannot be globally transformed to a diagonal constant matrix). I think the property you wanted to invoke was non-degeneracy... –  Alex Nelson Dec 22 '13 at 23:11
No, it is not. You're not asking if a diagonal element of the matrix representing the metric for some particular coordinates can be zero; what you want for the distance to be indipendent of some direction is that all the elements in the corrisponding column and row of the metric be zero (as in the coordinate singularity of Stan Liou's answer), otherwise you can get a dependence through cross-terms (e.g. you don't have $\mathrm{d}z^2$, but you have $\mathrm{d}x \mathrm{d}z$).
Since the metric has indefinite signature, you can choose coordinates such that some diagonal element is zero. If $g=\mathrm{diag}(1,-1,-1,-1)$ in the $(t,x,y,z)$ system, you can simply change to lightcone coordinates $x_\pm=t\pm x$ and get $$g'=\begin{pmatrix} 0 & 1/2 & 0 & 0 \\ 1/2 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}$$
$$\mathrm{d}s^2=\mathrm{d}t^2-\mathrm{d}x^2-\mathrm{d}y^2-\mathrm{d}z^2=\mathrm{d}x_+\mathrm{d}x_--\mathrm{d}y^2-\mathrm{d}z^2$$