Are there actual correct representations of curved spacetime?

Question

2d spacetime diagrams correctly visualize how space and time relate, and change when you change referential.

The problem is that, when we see representations of curved spacetime, it only shows a curvature in 2 space dimensions, and I'm not even sure the curved shape showed actually has the same curvature (scalar, Christoffel symbols, etc) as what relativity predicts.

So is there actual representations, of curved spacetime (a 2d curved shape, with one direction being is time), with the exact chrystofell symbols and curvature as what relativity predicts?

In a way, it would be a "correct" visualisation of general relativity in a 2d spacetime.

Answer 1

In any area of mathematics or physics, we can introduce diagrams as a way of exactly representing certain facts or properties, if we carefully define how the diagram is to be interpreted. In this respect such precisely-defined diagrams are like algebraic symbols. Indeed, some areas of maths are handled by developing a notation where the symbols are like little diagrams.

In the case of spacetime we are dealing with manifolds. The most natural way to allow a diagram to give information about a manifold depends on the type of manifold. Here are examples.

1. Euclidean manifold in 2D. This is the easiest: an ordinary flat diagram on a paper or screen maps naturally to such a manifold. No special interpretation is needed; the interpretation is obvious.

2. Minkowskian manifold in 2D. This is the kind of manifold you encounter in Special Relativity. It is flat (the Riemann curvature is everywhere zero) but it is not quite the same as a flat piece of paper, because the latter has Euclidean geometry. In the Minkowski space the invariant is $-\Delta t^2 + \Delta x^2$ but in the Euclidean space the invariant is $+\Delta z^2 + \Delta x^2$. This means that when the Euclidean diagram is used to visualize and Minkowski spacetime, some interpretation is needed. For example, the length of a line in some arbitrary direction on the diagram is not related in any simple way to the length of the corresponding spacetime interval (i.e. $(\Delta z^2 + \Delta x^2)^{1/2}$ does not behave like $(-\Delta t^2 + \Delta x^2)^{1/2}$). Lines that are orthogonal in the spacetime sense are mostly not at right angles to one another on the diagram. However the situation is not too bad: lines forming a pair whose angle is bisected by a photon worldline are spacetime-orthogonal, and distances parallel to any given time- or distance-axis are proportional to time intervals and spatial intervals in the corresponding reference frame, with the proportionality constant indicated by a hyperbola of constant interval from the origin.

3. Curved 2D space with positive signature. This is the kind of manifold which can represent space (not spacetime) around a gravitating object such as a star or a black hole, especially when there is spherical symmetry so we can use a 2D slice to tell us about the whole space. In this case it is often possible to use an embedding diagram. Here we go up one dimension, to 3D, such that the 3D space is Euclidean. Then we draw a surface in this 3D space. The surface is 2D, but it need not be flat. For many cases it is possible to construct this surface in such a way that distances along the surface embedded in Euclidean 3D space (so the distances we naturally see when we look at the diagram) exactly match the distances in the curved 2D space. This kind of diagram is widely used to help us get intuition about the Schwarzschild-Droste spacetime. We treat the region outside the horizon, and take a slice through spacetime at some given time. We then have a 3D space which with spherical symmetry. We take a slice through that space, thus getting a 2D manifold with positive signature. The embedding diagram looks like this

This is a precise diagram; it is called Flamm's paraboloid. Here are a few things you can deduce. (i) Each small region has the shape of a saddle, therefore the Gaussian curvature is negative. (ii) If you obtain the Gaussian curvature at each point, the value matches that of the Schwarzschild-Droste space. (iii) The distance from the horizon to any point is finite. (iv) As you walk outwards, increasing your distance from the horizon by $\Delta r$, the circumference of the circle about $z$ grows by less than $2\pi\Delta r$, and furthermore the diagram indicates exactly by how much it grows.

4. Curved spacetime. [corrected after helpful comment by peek-a-boo] The case of a curved manifold with a Minkowski signature is more difficult. There is, in principle, a Lorentzian (not Euclidean) space of higher dimension in which the manifold can be embedded, but our diagrams have to be drawn in Euclidean space and there is no easily interpreted (i.e. isometric) embedding into Euclidean space.

Finally, another remark on case (3), the curved 2D manifold with positive signature. Another example occurs in cosmology. Space on a very large scale may be flat, or it may have constant non-zero curvature. If the curvature is positive then you get a 3-sphere, and a cross-section through it gives a 2-sphere. That is a 2D manifold whose curvature and topology exactly matches that of the surface of an ordinary sphere in Euclidean space. So the widely used 'balloon model' of the cosmos, where an expanding balloon represents the expanding space of the cosmos, is a correct and exact model if it is interpreted correctly. Note especially that it is the surface of the sphere that represents a slice through the space of the cosmos.

Answer 2

You are right to be skeptical about the curvature being different when illustrations go to lower dimensions. Differential geometry is unfortunately very sensitive to the number of dimensions of the manifold. Just to illustrate, suppose we simplified it down to please the flat-Earth folks and pretended that the Earth was two-dimensional. We then set out to solve Einstein's field equations for orbit. Since orbit is a vacuum, this amounts to setting the Ricci tensor equal to zero: $$R_{uv}=0$$

Spoiler alert: the solution, in two-dimensions of space, is zero gravitational field. Spacetime in less than three dimensions of space is just not rich enough to have nonzero curvature but with zero Ricci tensor. Gravity in one dimension of space and one dimension of time is even more sickly.

So, the most accurate picture of gravity in general relativity is just the full picture, and most visuals (especially the horrid trampoline analogy) don't do it justice. As you've already gathered, this is annoying because we didn't evolve to imagine four dimensions of spacetime. That being said, there are some intuitive ways to get around this. Some previous replies have already given a few of these.

I'll discuss one I've used for a while now that I keep returning to. In the weak field approximation, the overwhelming majority of the gravitation we experience is due to the $g_{00}(x,y,z)$ component of the spacetime metric. This controls time dilation as a function of location, rather than spatial curvature. One way to leverage this is to imagine space flowing, like a river, towards gravitational wells. This is known as the river model. A great paper by Braeck and Grøn explain this in mathematical detail, while the YouTube channel ScienceClic English has a fantastic video illustrating this point of view with animations that show why it works so well.

This is a good enough way to think that you can easily derive the equation for gravitational time dilation in orbit by just imagining someone falling from infinity and then invoking the equivalence principle before they land. They will land, by energy conservation, at the planet's escape velocity $$v_e=\sqrt{\frac{2GM}{r}}$$ But when they land, by the equivalence principle, they are no longer moving in the original inertial freefall frame they were once in. In a sense, they are now moving against that frame at a constant speed of $v_e$ which means the good old fashioned time dilation formula from special relativity can be used: $$\Delta \tau=\sqrt{1-\beta^2}\Delta t=\sqrt{1-\frac{2GM}{c^2 r}}\Delta t=\sqrt{1-\frac{r_s}{r}}\Delta t$$ Including length contraction follows immediately by the same logic, and the solution in orbit $$\Delta \tau=\sqrt{1-\dfrac{3}{2}\dfrac{r_s}{r}}\Delta t$$ is also easy using a little vector argument. This is remarkably elegant for solving what is essentially the full blown Einstein field equations for the Schwarzschild metric, which is no easy task from the point of view of differential calculus. It's as if space itself is falling inwards towards the Earth at a speed of the escape velocity.

That answer might not be exactly what you were looking for, but hopefully it gives you another tool for thinking about general relativity without the heavy mathematical machinery.

Answer 3

As an addition to Andrew Steane's answer and a bit off from the question itself: There are Penrose diagrams. They don't visualize the local properties of the curved space-time exactly, but they do represent the global causal structure of the space-time faithfully.