In GR the spacetime manifold is equipped with a metric which makes it a Lorentzian manifold. It is the metric that is doing the separation of space and time (so that we end up with three dimensions of space and one of time for each spacetime event covered by the metric). Is that correct?

If so: I am trying to understand what kind of curve in spacetime stands as a world line. I know of world lines as timelike curves. But since the metric is needed to tell apart null, spacelike and timelike curves, defining a world line this way does not seem straightforward. Is there a metric-independent way to define what a world line is?

For example which geodesics are world lines and which are not?

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    $\begingroup$ Comment to the question (v2): The notion of a world line as a timelike curve (which does not have to be a geodesic if non-gravitational forces act) is defined on a Lorentzian manifold $(M,g)$, not on a bare manifold $M$ without a metric. $\endgroup$
    – Qmechanic
    Mar 16, 2016 at 10:23
  • $\begingroup$ @Qmechanic. Thanks. I have read this Wikipedia page a dozen times trying to figure that out but was left confused. I am also confused about the physical meaning of the bare manifold; maybe this is for another question though. $\endgroup$ Mar 16, 2016 at 10:39
  • $\begingroup$ @StéphaneRollandin The bare manifold is basically a way to label physical events with points on a 4d manifold such that different physical events get different 4d points. There isn't much physics to do until you start to put fields (such as a metric tensor field and a stress-energy tensor field) on it. You don't even have a Lorentzian manifold until you put at least a metric on it. $\endgroup$
    – Timaeus
    Mar 16, 2016 at 19:08
  • $\begingroup$ @Timaeus. I had the feeling that the bare manifold was significant because of questions like physics.stackexchange.com/q/1787 but it seems I have been confusing curvature and manifold topology. $\endgroup$ Mar 16, 2016 at 23:12

2 Answers 2


Yes the metric does the separation of space and time. Without the metric you just have a 4d manifold with 4d points called events and a topology (a sense of small regions of spacetime). You can even have curves and they can have tangents.

But once you have the metric then a 4d curve can have tangents that can be identified as spacelike or timelike or lightlike. And now you can ask whether your manifold is Lorentzian or not.

It is the metric that is doing the separation of space and time (so that we end up with three dimensions of space and one of time for each spacetime event covered by the metric). Is that correct?


Is there a metric-independent way to define what a world line is?

Absolutely not. Consider the manifold $$M=\{(a,b,c,d):a,b,c,d\in\mathbb R\},$$ and the curve $t\mapsto (t,0,0,0)$.

Either of the two metrics $$\mathrm ds^2=\mathrm da^2-\mathrm db^2-\mathrm dc^2-\mathrm dd^2,$$ and $$\mathrm ds^2=\mathrm db^2-\mathrm da^2-\mathrm dc^2-\mathrm dd^2,$$ make it a Lorentzian manifold. But the curve $t\mapsto (t,0,0,0)$ is a worldline (has all tangents be timelike) with the first metric and the curve $t\mapsto (t,0,0,0)$ is not a worldline (has nontimelike tangents) in the second metric.

Whether or not something is a worldline depends on the metric. That's life.

For example which geodesics are world lines and which are not?

The curves (differentiable maps from parameters into the manifold) that have a timelike tangent at every 4d point (event) along the curve are the worldlines. The other ones are not worldlines.

But even being a geodesic depends on the metric. So if you had two metrics on the same manifold they could disagree on whether a curve is a geodesic.

That said, when someone hands you a Lorentzian manifold they hand you a manifold and a metric. So you always have a metric.


I wonder if you are mixing up manifolds, metric and coordinate systems.

Mathematicians have a precise definition of a manifold, but for our purposes it's just something that sets the dimensionality. We're normally only interested in the manifold $\mathbb{R}^4$, though I guess other manifolds would creep in if we were dealing with spacetimes with non-trivial topologies.

The metric defines distance on the manifold. It allows us to measure the distance along a curve between any two points. The metric is coordinate independent, so it's wrong to say:

It is the metric that is doing the separation of space and time

The metric will have a signature that specifies how many timelike and spacelike dimensions there are, and we are normally only interested in metrics with the signature (3,1), or (1,3) depending on your sign convention. These have one timelike and three spacelike dimensions, but how we split spacetime up into timelike and spacelike parts is done by choosing a coordinate system.

And finally, the coordinate system is how we choose to measure out space and time. The most intuitive example is to use three orthogonal spatial axes $x$, $y$ and $z$ and a time axis $t$, but this is far from the only choice. For example in describing the spacetime geometry round a black hole we may use the obvious simple coordinates (usually as polar coordinates), but we might choose Gullstrand-Painleve, Eddington-Finkelstein or Kruskal-Szekeres coordinates and probably lots of others that I can't think of. These split up time and space in different ways.

The point of all this is that if you take any curve in spacetime you can divide it up into a series of infinitesimal straight lines $\mathbf{ds}$ then calculate the length of $\mathbf{ds}$ using:

$$ ds^2 = g_{\alpha\beta}dx^\alpha dx^\beta $$

where $g_{\alpha\beta}$ is the metric tensor and the $dx^\alpha$ are our four coordinates. For example in flat spacetime $g$ is the Minkowski metric and we can write the equation as:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

The values of $g$ and the $dx$s will depend on what coordinate system we have chosen, however the key point is that no matter what coordinates we choose the value of $ds^2$ we end up with is always the same.

So regardless of what coordinates we use we can always tell if a curve is timelike just by writing down an expression for $ds^2$ anywhere along its length and then seeing if $ds^2$ is positive, negative or zero. And we can use whatever coordinate system is most convenient because the answer doesn't depend on what coordinates we choose.

So we can tell if any random curve can be the world line of a massive object just by showing that $ds^2$ is everwhere negative along the curve. Similarly we can show the curve is the trajectory of a massless particle by showing that $ds^2$ is everywhere zero on the curve.

However, although we can choose whatever bizarre coordinates we want, we cannot do the calculation unless we have a metric. That's because without a metric the curve is just a set of spacetime points with no concept of distance between points on the curve.

I appear to have ranted on a bit: I wasn't sure exactly what you were asking so I just wrote down anything that seemed relevant. If I've misunderstood what you are asking please say so ina comment and I'll edit the answer accordingly.

  • $\begingroup$ Yes I definitely have been mixing up manifolds, metric and coordinate systems. Your answer does not read like a rant at all, it is very clear and informative. The only thing is that when I said "the metric is doing the separation of space and time", I meant it the way @Timaeus understood it in its answer, that is in the sense that without the metric signature the four dimensions of the manifold are undifferentiated in terms of space and time. But now I see that the physics is not in the bare manifold so this consideration may not have much sense anyway. $\endgroup$ Mar 16, 2016 at 23:10

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