I'm about to be taking an undergrad course in GR. I'm trying to get ahead of the game, and I've hit a problem.

The metric for flat Minkowski space (using the $-+++$ signature and $c=1$) is:

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

which in polar coordinates is: $$ds^2 = -dt^2+dr^2 +r^2(d\theta^2 +d\phi^2\sin^2\theta)$$

Firstly, the $r^2(d\theta^2 +d\phi^2\sin^2\theta)$ part of the metric is identical to the metric for a 2-sphere, which to me suggests that the Minkowski metric above should signify curved spacetime - why is this part of the metric there if the space the metric describes is flat? In my mind, the metric containing "something curved" should mean that the space it describes is curved.

Also, why does simply doing this:

$$ds^2 = -dt^2+dr^2 +(r^2+b^2)(d\theta^2 +d\phi^2\sin^2\theta)$$

to the metric produce a wormhole geometry ($b$ a constant - the size of the wormhole's throat)?

I suppose my questions are summarised by: what does a 2-sphere have to do with flat Minkowski spacetime or the geometry of a wormhole?

  • $\begingroup$ A similar situation is a 'flat' FRW spacetime with metric $ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)$. Here the spatial part is flat but overall spacetime is curved. You can't just look at part of the metric. $\endgroup$
    – jacob1729
    Commented Sep 10, 2019 at 16:17
  • 2
    $\begingroup$ Ordinary Euclidean 3-space contains a sphere, which is curved. Does that mean Euclidean space is curved? $\endgroup$
    – Javier
    Commented Sep 10, 2019 at 16:25
  • $\begingroup$ So just because the metric contains something that looks curved doesn't mean the spacetime is necessarily curved. Thanks! $\endgroup$
    – Ali
    Commented Sep 10, 2019 at 16:41
  • 1
    $\begingroup$ You can’t conclude anything about the curvature of a space by looking at the curvature of one of its lower-dimensional subspaces, as far as I know. $\endgroup$
    – G. Smith
    Commented Sep 10, 2019 at 16:43

1 Answer 1


The metric is telling you how to measure distances for the given coordinate system on the given space time. You don't get curvature by changing coordinates, which is what you did between your first two expressions. Curvature is given by the Riemann tensor, usually as contracted into the Ricci tensor or Ricci scalar, depending a bit on exactly what you want to do. You'll get those in your course. The Ricci scalar for Minkowski space is invariantly 0 regardless of which of the coordinate expressions you use.

In the spherical coordinate form, you're basically getting the (completely ordinary, not-specific-to-relativity) fact that arc length depends on both radius and angle.

The wormhole part is really a different question. You should ask it separately.

  • $\begingroup$ So it's just a coincidence that the metric for a 2-sphere appears in the metric for flat spacetime? $\endgroup$
    – Ali
    Commented Sep 10, 2019 at 16:28
  • $\begingroup$ Not a coincidence. The spatial part of Minkowski is a "regular" space, and that's the metric for such a space in spherical coordinates. $\endgroup$
    – Brick
    Commented Sep 10, 2019 at 16:34

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.