general relativity - Why doesn't $ds^2 = 0$ imply two distinct points $p$ and $p'$ on a manifold are the same point?

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?

Answer

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 $0$.