Let's suppose I have a spacetime manifold $M$. Let $p$ be a point on my manifold. Now I move from $p$ to some other point $p'$. Presumably I should have moved some "distance" right? How can I speak of notions of space and time if I have no conception of distance?

But now consider light moving through spacetime. Suppose my light starts at $p = (0,0,0,0)$ and travels to $p' = (1,1,0,0)$. By the definition of the spacetime interval $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$, this should mean $ds^2 = (1)^2 - (1)^2 - 0 - 0 = 0$. So $ds^2=0$.

Yet I have moved from point $p$ to $p'$. So I clearly have moved along some path along the curve, but the length of this path is zero. Shouldn't that mean $p$ and $p'$ are the same point?

Note: I think I may be suffering from an overly Euclidean mindset and my brain hasn't adapted yet enough to the non-Euclidean logic of semi-Riemannian manifolds.

My Question

Can someone resolve this contradiction?


4 Answers 4


Let's separate out some definitions:

metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ such that the following axioms hold for all $x,y,z \in X$:

  • $d(x,y) \geq 0$,
  • $d(x,y) = 0 \Leftrightarrow x = y$,
  • $d(x,y) = d(y,x)$, and
  • $d(x,z) \leq d(x,y) + d(y,z)$.

pseudo-metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ such that the following axioms hold for all $x,y,z \in X$:

  • $d(x,x) = 0$,
  • $d(x,y) = d(y,x)$, and
  • $d(x,z) \leq d(x,y) + d(y,z)$.

metric(2): (aka "inner product") Given a vector space $V$ over a field $F$, which is either $\mathbb{R}$ or $\mathbb{C}$, a function $g : V \times V \to F$ such that the following axioms hold for all $x,y,z \in V$ and $a \in F$:

  • $g(x,y) = \overline{g(y,x)}$;
  • $g(ax,y) = a g(x,y)$,
  • $g(x+y,z) = g(x,z) + g(y,z)$,
  • $g(x,x) \geq 0$, and
  • $g(x,x) = 0 \rightarrow x = 0$.

pseudo-metric(2): (aka "pseudo inner product") Given a vector space $V$ over a field $F$, which is either $\mathbb{R}$ or $\mathbb{C}$, a function $g : V \times V \to F$ such that the following axioms hold for all $x,y,z \in V$ and $a \in F$:

  • $g(x,y) = \overline{g(y,x)}$;
  • $g(ax,y) = a g(x,y)$,
  • $g(x+y,z) = g(x,z) + g(y,z)$, and
  • $\exists\ v \in V : g(x,v) \neq 0$.

Now you want to define a distance between points on a manifold. You are intuitively looking for a (pseudo-)metric(1) here, a distance function on a set without any extra structure. The problem is all you are given is a (pseudo-)metric(2) on the tangent space at each point. Your (pseudo-)metric(2) can only give you magnitudes of tangent vectors at points. Intuitively, these are "infinitesimal distances." You need to integrate such magnitudes along a path in order to get distances between points.

But this is the crux of the issue: What path do you choose? Even for a nice manifold like the surface of a 2-sphere (that is, something with a real metric(2), not just a pseudo-metric(2), on its tangent bundle), the distance between points is path dependent. You could fly directly from New York to London along a great circle (geodesic), or you could stop by in Beijing.

If you have positive-definiteness working for you, you could take the infimum over all paths from one point to another. Consider curves of the form \begin{align} \gamma : [0,1] & \to M \\ \lambda & \mapsto p \\ 0,1 & \mapsto p_1,p_2. \end{align} Then $$ d(p_1,p_2) = \inf_\gamma \int_0^1 \left(g_p \left(\frac{\mathrm{d}p}{\mathrm{d}\lambda}, \frac{\mathrm{d}p}{\mathrm{d}\lambda}\right)\right)^{1/2} \, \mathrm{d}\lambda $$ defines a distance function in the metric(1) sense as long as $g_p$ is an honest metric(2) inner product at each $p$.

Unfortunately, when you try this with a Lorentzian manifold equipped with a pseudo-metric(2), the construction fails to produce anything useful. Even taking an absolute value before the square root, there will always be a piecewise differentiable null path between any two points. Thus there will be differentiable curves of length arbitrarily close to $0$, and so the pseudo-metric(1) you induce is trivial: all distances are $0$.

  • $\begingroup$ I usually see the last definition as $d(..)=..\int \sqrt{g(..)}$. Everything should work even without the square root, the only thing I am not sure about is the triangle inequality. $\endgroup$
    – Void
    Commented Mar 4, 2015 at 8:15
  • $\begingroup$ Actually I am now sure that it will violate the triangle inequality - it can be seen in terms of infinitesimals: take the vector $XZ$ representing an infinitesimal path to have a norm $1$. Now you can take the point $Y$ halfway between $X$ and $Z$ and the prescription without the square root will tell you that the path $XY + YZ$ has half the length of $XZ$. $\endgroup$
    – Void
    Commented Mar 4, 2015 at 8:25
  • $\begingroup$ @Void You're quite right. I just forgot the square root. $\endgroup$
    – user10851
    Commented Mar 4, 2015 at 17:32

I think it might help to think about the spacetime interval $\text{d}s^2$ as a measure of movement in spacetime relative to the speed of light. Let's say that you want to move from a point $p=(0,0,0,0)$ to another point $p'=(t,x,0,0)$. The quantity $\text{d}s^2 = c^2\text{d}t^2-\text{d}x^2$ is then:

  • Positive if $x<ct$, which means that you traversed the distance slower than the speed of light;
  • Zero if $x = ct$, which means that you traversed it exactly at the speed of light;
  • Negative if $x>ct$, which means that you traversed the distance faster than the speed of light.

So with the metric convention that you use, special relativity dictates that any massive particle can only traverse positive spacetime intervals, and any massless particle can only traverse zero spacetime intervals, since all the distances are measured relative to how a photon would move between the points.

  • $\begingroup$ So are you saying that, if I am at a point $p$ and I move to a point $p'$, that the rules for traveling on the manifold depend if the particle is massive or not? $\endgroup$ Commented Mar 4, 2015 at 2:36
  • $\begingroup$ Manifolds in general don't come with a notion of "traveling on" them, but yes, massive particles have speeds less than $c$ and massless particles travel at speed $c$ unless you include phenomena like, say, glass. $\endgroup$
    – CR Drost
    Commented Mar 4, 2015 at 2:39
  • 2
    $\begingroup$ @StanShunpike, what I mean is that according to special relativity, a massless particle can only move from $p$ to $p'$ if $(p'-p)^2=0$, a massive particle can only move from $p$ to $p'$ if $(p'-p)^2>0$, and no particle can move from $p$ to $p'$ if $(p'-p)^2<0$. So the spacetime distance $\text{d}s^2$ between the two points determines which kinds of particles are allowed to move from $p$ to $p'$ and not. $\endgroup$
    – jabirali
    Commented Mar 4, 2015 at 2:50
  • $\begingroup$ @jabirali Ohhh...I'm still not sure that completely answers my question but +1 because I didn't realize that. So you are saying that massless particles travel exclusively on light-like intervals, whereas massive particles travel exclusively on space-like and no particles (besides hypothetical ones) travel along space-like. Wow! Somehow this seems obvious in retrospect but I never thought about it like that. Cool! $\endgroup$ Commented Mar 4, 2015 at 2:57
  • $\begingroup$ Specifically, the part that I don't think is answered is....okay, let's say then that the only particles that travel along light-like intervals are massless particles...that still leaves me with my original problem of traveling some distance between two points and the interval being zero. Why doesn't that imply they are the same point? $\endgroup$ Commented Mar 4, 2015 at 2:59

Yeah, you've not yet adapted. That's OK. Let me take you through it.

In this conventional world of classical physics we have separate notions of distance and time, with the idea that either two events happen at the same time and therefore have an objective distance between them, or two events happen at different times and therefore have an objective time between them. It is always one or the other, not both, not neither: if there is a time gap between two events, then there is some reference frame which sees them both at distance L for any L you'd like; otherwise if the time gap is zero then everyone agrees on the distance between the two events. This allows you to freeze a moment in time and speak of distances.

In relativity, we make things a little more complicated, but also more realistic. It's almost the same story but not quite. Think about, say, a supernova explosion -- what we see as a bright flash in the starry sky would, if you were to "look down upon it", look like an event in spacetime with a horizon of light announcing the event expanding outward at the speed of light. That "expanding bubble" is important. In relativity we call that bubble a "light cone".

Imagine two expanding-at-speed-$c$ bubbles. Topologically, either one bubble is inside the other, or they both intersect on a ring (when they're large enough to intersect at all!), or they both intersect at just one point where they "kiss" each other. Those are your three possibilities: time-separated, space-separated, and null-separated. They correspond to a positive metric, negative metric, and 0 metric. The Lorentz transforms preserve this topology because they preserve the metric in general; the light cones get mapped to other light cones with the same structure relative to each other.

Given time-separated, you still have that two things are not objectively space-separated, as there is a reference frame for a spaceship which passes through both events in spacetime. Given space-separated, you also have that they are not objectively time-separated; that's a little more subtle, but imagine someone on the expanding "ring" where the bubbles "intersect": they see both events simultaneous with their local apparatus. The claim is, given the right velocity, they would draw the distances back to the original events both as L, so that they think that both events were simultaneous, proving that there is no objective time-ordering.

Null-separated is the new beast that you'll have to accept, lying between those. It's objectively "time separated" because up to a point, one light cone is "inside" the other one, so you can say that one comes "before" the other. However, in the limit as you go faster and faster, trying to be at both events at the same time, you literally see no time elapse between the two events. Similarly it's objectively "space separated" but there are reference frames which make the distance between the two events arbitrarily small.

Those are the metric values and what they mean. "Distance" is relative to this new notion of "basically at the same place, basically at the same time" and can go negative for proper distances or positive for proper times.

You can view this, too, if you like, as broadening the notion of "present" up from a plane in classical physics to the space between two light cones. One light cone represents all the events in the "past" which "have been seen" by this spacetime point (light from them has had an opportunity to reach the point); another light code represents all the events in the "future" which "have seen" this spacetime point; the spacelike-separated stuff is all in a "relativistic present" relative to this point; different Lorentz transforms choose different planes through the point as "present planes" but this choice is more arbitrary.


While the others have provided more technical answers, I hope to offer a more straight-forward explanation. The interval $ds^2$ is:

$ds^2 = g_{\mu\nu}dx^\mu dx^\nu = 0$

But this is a sum and it does not imply that the $dx^\mu dx^\nu$ are zero.

  • $\begingroup$ But zero sum would imply the vectors were zero if $g$ were a positive definite billinear form, which condition would make it a "real" metric (in the mathematician's sense). It's the nontrivial signature that allows nonzero null vectors. $\endgroup$ Commented Jun 19, 2017 at 13:17

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