# Is it always possible to have a (local) time coordinate in GR?

Apologies for the confusing title, it is late here. I'm wondering exactly what meaning the "time coordinate" has in General Relativity. We always write the line element as $$\tag{1} ds^2=g_{00}(dx^{0})^2+2g_{0i}dx^0dx^i+g_{ij}dx^idx^j,$$ with the assumption that $x^0$ is some kind of time coordinate, and $x^i$ are spatial coordinates. However, as light cone coordinates show, if we pick a random coordinate system, the metric will not be in the form (1). Is the existence of coordinate systems as in (1) an axiom or can it be derived somehow? By a time coordinate I mean that the time coordinate lines should have timelike tangent vectors, and similarly with spacelike coordinates.

• Comment to the post (v2): Consider to include your definition of 'local' and 'time coordinate' for clarity. – Qmechanic Jan 7 '17 at 8:21
• Consider studying the ADM formulation of General Relativity. It is always possible to choose the diffeomorphism gauge such that the degrees of freedom of General Relativity are described by the spatial 3d metric $q_{ab}$. – Prof. Legolasov Jan 7 '17 at 12:39
• @Solenodon Paradoxus you have to assume that the spacetime is globally hyperbolic to prove the existence of the foliation you mention. The answer provided by Ocelo7 below is completely local and it does not need global hypotheses, it is valid also for pathological spacetimes where no global spacelike foliations exist. – Valter Moretti Jan 7 '17 at 13:25
• @ValterMoretti I disagree, you can always consider a foliation of some local coordinate patch. – Prof. Legolasov Jan 7 '17 at 15:11
• Yes you are right, but the proof is then a refinement of the one outlined below. – Valter Moretti Jan 7 '17 at 15:27

Let $(M^{n+1},g)$ be a Lorentzian manifold. Given $p\in M$, we will show that there is a coordinate system $(x^\mu)$ defined on an open set $p\in U\subset M$ such that $\partial_0$ is a timelike vector field, and $\partial_i$ are spacelike vector fields for $i=1,\dotsc,n$.
Let $(x^\mu)$ be an arbitrary chart defined on $U\ni p$. It is known that $T_pM$ is the span of $\{\partial_0,\partial_1,\dotsc,\partial_n\}$. As $g_p$ has signature $(-,+,\dotsc,+)$, we may find linearly independent vectors $v_\mu$, $\mu=0,1,\dotsc,n$, such that $g_p(v_0,v_0)=-1,$ $g_p(v_i,v_i)=+1$. These vectors are linear combinations of $\{\partial_0,\partial_1,\dotsc,\partial_n\}$. By a linear change of coordinates, we can find a coordinate system $(y^\mu)$ such that $\partial/\partial y^\mu=\partial_\mu'=v_\mu$ at $p$. By continuity, there is a neighborhood $V_0\subset U$ of $p$ such that $g(\partial_0',\partial_0')<0$, i.e., $\partial_0'$ is timelike on $V_0$. Similarly, there exist neighborhoods $V_i$ such that $g(\partial_i',\partial_i')>0$ on $V_i$. We take $V=V_0\cap\cdots \cap V_n$, which is a neighborhood of $p$. By changing each coordinate value by a constant, we can adjust the origin without changing the aforementioned vector fields. Then $(y^\mu)$ is the desired coordinate system on $V$.
This is a common misconception, which I will illustrate using the usual $r$-coordinate in Schwarzschild-Droste spacetime. In Schwarzschild-Droste coordinates, the $r$-coordinate vector has components $(0,1,0,0)$, which you can show is spacelike for $r>2M$ and timelike for $r<2M$. In Gullstrand-Painleve coordinates the $r$-coordinate vector has the components $(0,1,0,0)$ in these different coordinates, and is spacelike everywhere!! But this is not the same vector as before: just apply the usual transformation law for vector components. This is despite the fact that the $r$-coordinate itself is identical in both cases; what I mean by identical, at least in this sentence, is you can think of $r$ as a scalar on the manifold (ignoring $r=2M$ if you like), and they're the same scalar.
Hence, the technical definition is in terms of hypersurfaces $r=\textrm{const}$. One can show it comes down to the sign of the component $g^{rr}$ of the inverse metric, which is $1-2M/r$ in all cases. Hence $r$ is spacelike for $r>2M$, timelike for $r<2M$, and null for $r=2M$ (the latter is not defined in the case of Schwarzschild-Droste coordinates). If the metric is diagonal in some given coordinate system, then $g^{rr}=g_{rr}^{-1}$ so the nature of coordinate vectors coincide, so it is clear where the misconception comes from.