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My goal is to visualize somehow the curvature of time, as opposed to the curvature of space. I know that we generally talk about spacetime curvature altogether; however, the fact that spacetime has signature (+, +, +, -) indeed shows that time is a "special" component --> it is the only one which carries the minus sign. Hence, I expect that even geometrically the time curvature should be peculiar with respect to space curvature.

Now, if we focus only on $x$ spatial axis and time $t$, then we could depict the simplified-curved-spacetime as a 2D pseudo-Riemannian manifold (signature (+,-) ). Compare it to a 2D proper-Riemannian manifold as a sphere (signature (+,+) ).

Of course I know how to visualize a sphere; but is there a way to visualize a pseudo-Riemannian 2d manifold? Can I visualize the difference with respect to a proper-Riemannian 2d manifold (like the sphere)?

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    $\begingroup$ You should take the line element $s^2=x^2-t^2=0$ for light or equal -1 for particle, so you get t over x $\endgroup$
    – Eli
    Sep 9 '19 at 16:34
  • $\begingroup$ @Eli is somewhat right, but consider a smooth 1D submanifold of 3D (i.e. a smooth Euclidean path). As in that case, visualising the full nature of the path requires us to embed the curve into $\mathbb{R}^3$, to fully see how a Minkowski curve behaves we would need to embed it into a Minkowski space. Although investigating curvature is independent from that. Also, Ricci curvature? $\endgroup$
    – acarturk
    Sep 9 '19 at 18:31
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    $\begingroup$ One approach to "visualization" is to isometrically embed the spacetime $M$ of interest into a higher-dimensional flat spacetime $N$ such that the metric on $M$ is naturally inherited from the flat metric on $N$. The abstract of arXiv:1109.4211 says: "any simply-connected smooth complete surface with curvature bounded above by a negative constant admits a smooth isometric embedding into [three-dimensional Minkowski space]..." (Not sure if this is a global embedding, though.) For a more complete list of results of this type, you could try posting a question on Math SE. $\endgroup$
    – Chiral Anomaly
    Sep 10 '19 at 1:01
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    $\begingroup$ By the way, the question might attract more positive attention if you don't use the phrases "time curvature" and "curvature of time." If I understand what you're asking, those phrases don't mean what you want, if they mean anything at all. $\endgroup$
    – Chiral Anomaly
    Sep 10 '19 at 1:10
  • $\begingroup$ In addition to the comments by @ChiralAnomaly, I think you're also mixing concepts of (local) differential geometry with topology. The sphere is 2D manifold with intrinsic curvature, but it's also topologically closed, and those are not equivalent facts. So it's unclear to me what you mean by this question other than taking a $t$-$x$ slice, but I think you've rejected that. $\endgroup$
    – Brick
    Sep 13 '19 at 14:43
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The answer is no, there is no way to visualize curved spacetime of gravity in a simplified way such as a 2D-manifold. In popular science there are many attempts to provide us with an idea what fourdimensional curvature of spacetime might look like, but they are providing only a very vague orientation, and up to now no kind of simplified representation of curved spacetime has been found.

However, the gravity of the Schwarzschild metric may not only be represented by curved spacetime but alternatively also as gravitational time dilation in uncurved, flat space:

For this purpose we consider the Schwarzschild metric

$$ ds^2 = -(1 - \frac{2GM}{c^2 r}) c^2 dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} } dr^2 + r^2 (d\Theta^2 + sin^2 \Theta d\Phi^2)$$

and the gravitational time dilation from the point of view of a far-away observer

$$ C = \frac{\tau}{t} = \sqrt{1 - \frac{2GM}{c^2 r}}$$

As you may observe, we can insert the second equation into the first one, and we get: $$ ds^2 = -c^2 (Cdt)^2 + {(\frac {dr}{C})}^2 + r^2 (d\Theta^2 + sin^2 \Theta d\Phi^2)$$

This is still the Schwarzschild metrics of curved spacetime. Now we compare this equation with the Minkowski metrics of flat space:

$$ ds^2 = - c^2 dt^2 + dr^2 + r^2 (d\Theta^2 + sin^2 \Theta d\Phi^2)$$

The result, both equations are very similar, and C (= the gravitational time dilation) is the only difference. That means that gravity may be described entirely by gravitational time dilation in uncurved, flat space such that there is no kind of "curved time".

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