In General Relativity spacetime is a four-dimensional oriented and time oriented Lorentzian manifold $M$. Consider then one coordinate system $(U,x)$ on the open set $U\subset M$.

This determines the coordinate functions $x^\mu : U\to \mathbb{R}$ so that

$$x(\mathscr{A})=(x^0(\mathscr{A}),x^1(\mathscr{A}),x^2(\mathscr{A}),x^3(\mathscr{A})), \quad \forall \mathscr{A}\in U.$$

My question is: when we have a coordinate system, when does $x^0$ represents time and $x^i$ for $i=1,2,3$ represents spatial coordinates?

How can we know if a coordinate system is like this?

In the way I have learned, an observer is a timelike future pointing worldline $\gamma : I\subset \mathbb{R}\to M$ together with four vector fields along $\gamma$, $e_\mu : I\to TM$, with $e_0 = \gamma'$ and $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. In that setting one defines the time interval between the events $\gamma(0)$ and $\gamma(1)$ to be:

$$\tau=\int_0^1 \sqrt{g_{\gamma(\lambda)}(\gamma'(\lambda),\gamma'(\lambda))}d\lambda$$

In the way I've been learning GR this is the only situation in which time appeared. No more has been said about time, nor space, separately.

So my question is: given a chart, and given this notion of observer and of time interval, when does $x^0$ represents time and when $x^i$ represents space?

  • 2
    $\begingroup$ What if it's neither? I.e., what if you have two null coordinates? $\endgroup$ Mar 18, 2017 at 17:59
  • $\begingroup$ That's exactly my point. I imagine coordinate systems like this where $x^\mu$ is neither spacelike nor timelike do exist, but I'd like to know when $x^0$ does mean time. I mean, is there some criteria for that? $\endgroup$
    – Gold
    Mar 18, 2017 at 18:01
  • $\begingroup$ I'm not sure that this is enough but, if you have a metric tensor with a signature where all except one diagonal components have one sign, isn't the component with different sign the one associated with time? Maybe you'd like an answer in the stage of manifold + chart, without introducing yet the metric structure. $\endgroup$
    – anonymous
    Mar 18, 2017 at 18:02
  • 6
    $\begingroup$ It represents a time coordinate when the basis vector of the tangent space associated with that coordinate is timelike. $\endgroup$
    – Slereah
    Mar 18, 2017 at 18:27
  • 1
    $\begingroup$ Very relevant: physics.stackexchange.com/q/303471 $\endgroup$
    – Ryan Unger
    Mar 19, 2017 at 22:51

2 Answers 2


Following @Slereah's comment I think I've arrived in a quite convincing argument that I decided to post. A coordinate system $(U,x)$ such that

$$g\left(\dfrac{\partial}{\partial x^0},\dfrac{\partial}{\partial x^0}\right)>0$$

in other words, on which $\partial_0$ is time-like has the property that $x^0$ represents time.

In order to justify this, consider an observer $\gamma : I\subset \mathbb{R}\to M$ such that at least a portion of its worldline lies in $U$. Consider that in $U$ we can write


In particular it is obvious that $\gamma'(\tau)= \frac{\partial}{\partial x^0}|_{\gamma(\tau)}$. Assume furthermore, that $\partial_0$ is normalized to unity, since this is possible because it is timelike and requires merely a rescale by its length.

Now we recall the so-called standard clocks postulate:

There are standard clocks that can be carried by observers along their worldlines and such that those clocks register between events $\gamma(a)$ and $\gamma(b)$ the time interval $\Delta \tau$

$$\Delta \tau = \int_a^b \sqrt{g_{\gamma(\lambda)}(\gamma'(\lambda),\gamma'(\lambda))}d\lambda$$

This postulate allows us to parametrize an observer's worldline by proper time.

In that sense, the parameter of the worldline, according to the postulate, has the meaning of the time registered by the clock carried by the observer considering the clock has started at the event $\gamma(0)$.

But it is clear from the definition made in the postulate that the condition that $\gamma$ is parametrized by proper time is that $g_{\gamma(\tau)}(\gamma'(\tau),\gamma'(\tau))=1$. This obviously happen in our situation.

But the parameter of the worldline is exactly the coordinate $x^0$! In that sense, directly from the postulate, $x^0$ represents the time registered by one observer's clock whose worldline is $\gamma$.

  • $\begingroup$ Note that when you "[a]ssume that $\partial_0$ is normalized to unity", this effectively redefines $x^0$ as well. Specifically, you've gone from coordinates $(x^0, x^i)$ to coordinates $(\tilde{x}^0(x^0), x^i)$; and the argument that $\tilde{x}^0$ is a time coordinate for some observer then holds. But an arbitrary coordinate $x^0$ for which $\partial_0$ is timelike need not actually correspond to the proper time for any observer; for example, the coordinate $t$ in the Schwarzschild solution does not exactly correspond to the proper time measured by any observer, inertial or otherwise. $\endgroup$ Jul 18, 2019 at 13:36

This is really a question about SR, not GR. SR can be done with any coordinates.

There doesn't have to be any such thing as a time coordinate. For example, in 1+1 dimensions, suppose that $(t,x)$ are Minkowski coordinates. You could instead use coordinates $(u,v)=(t+x,t-x)$. Neither $u$ nor $v$ is timelike. You can also make a coordinate system in which both coordinates are timelike. All that matters is that the vectors $\partial_\mu $ are linearly independent, so that they span the tangent space.


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