Does space time cone live in spacetime manifold or in its tangent space?

Given that it is defined by g(v, v) >= 0 I think it operates on velocity vectors which means space time cone is a structure defined on tangent space at every point.

However wikpedia says "In special and general relativity, a light cone is the path that a flash of light, emanating from a single event (localized to a single point in space and a single moment in time) and traveling in all directions, would take through spacetime."

This would imply the cone is in spacetime (manifold) itself.

I still think cone is defined in tangent space. Is this correct?

  • 1
    $\begingroup$ Well, I'm not sure there's a "correct" answer, because each author is free to define words however he wants to, provided he uses them consistently. But I certainly agree that it is much more natural and satisfying to define the light cone in the tangent space, not in the underlying manifold --- and I am willing to bet that this is what most careful authors do. $\endgroup$
    – WillO
    Nov 29, 2017 at 2:44
  • $\begingroup$ It depends on how you define the null geodesics. Your idea would be correct for a general differentiable manifold where we must construct the geodesics based on the vector field of velocities defined on the tangent space. However, for the metric space, such as pseudo-Riemannian, we can define the null geodesics based on solving the Lagrange equations (or equivalently by using the Christoffel symbols). This way we have the null geodesics defined on the metric manifold. $\endgroup$
    – safesphere
    Nov 29, 2017 at 2:55

1 Answer 1


It mostly depends on exactly how you define "light cone," but in the most common sense of the term, it lies in the spacetime manifold itself. In special relativity, the standard definition of the light cone emanating from a spacetime point $x_0$ is $$\{ x_0 + \Delta x | g(\Delta x, \Delta x) = 0\}.$$

In general relativity the concept is more complicated: it's the set of points in spacetime that can be connected to the point $x_0$ by a null geodesic. (It's not generally a "cone" in the sense that it's not necessarily "flat," at least not in the naive sense of "flat".) But in general relativity, where we are often concerned with local physics, we often consider "infinitesimal" light cones of the form $$\{ x_0 + dx | g_{\mu \nu}\, dx^\mu\, dx^\nu = 0\},$$ where the four-vectors $dx$ are infinitesimal. We can meaningfully add the vectors if we are considering a small enough region that we can neglect the curvature and treat is as locally flat. In this case, the vectors $dx$ live in the tangent space of the manifold at $x_0$.

  • $\begingroup$ doesn't g operate on tangent vectors, in this case dx or delta(x)? $\endgroup$
    – Fakrudeen
    Nov 29, 2017 at 19:58
  • $\begingroup$ @Fakrudeen In general relativity, yes (which agrees with what I said), but in special relativity it operates on non-infinitesimal four-vectors, which are best thought of as lying inside the Minkowski spacetime manifold, not in a tangent space. $\endgroup$
    – tparker
    Nov 29, 2017 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.