# Ricci scalar and General Relativity cartoon pictures

In many cartoon pictures trying to explain general relativity, there is usually a massive object embedded in a "spacetime mattress" and heavier the object, deeper it sinks.

However, what is actually plotted?

I guess it might be the Ricci scalar, or scalar curvature $R$.

Ricci curvature describes the change of volume of a geodesic ball. When approaching a massive star, does the volume of geodesic ball increase or decrease?

• – Qmechanic Aug 12 '16 at 21:10
• I would give good odds that the grid was drawn straight from their imagination, and possibly analogs of trampolines, and not from one tiny, minuscule, infinitesimaly small part of math :). Once it got going, the other TV companies etc, just copied it. – user108787 Aug 12 '16 at 21:18
• Better to think of spacetime curvature in the 3-d, at least, like this. – heather Aug 12 '16 at 21:23
• The example with "the spacetime matress" is wrong. I never liked it. I never use it. Spacetime is more complex than it could be visualized by a plane or a matress. – Physics Guy Aug 12 '16 at 22:05

These are called embedding diagrams. Usually, if done properly, they are surfaces with the same intrinsic metric as some simple slice through the spacetime, like the equatorial plane.

For example, on this site about relativity, black holes, and related ideas, you'll find this page, where you'll see a similar image for a neutron star. Disclosure: I wrote most of the site and created the graphics in question. I flipped the image upside down, because I think it makes more sense to people who've never seen such a thing. But it's the same idea as the more common sinking-into-a-mattress picture. There's another with a black hole on the following page. In either case, if you look at the metric intrinsic to these surfaces as embedded in flat three-dimensional space, they're the same as the metrics of the slices through the middle of the spacetimes.

While the surface's shape itself can't account for the time components of the metric, I used colors to indicate the time-time components. Embedding diagrams like this also necessarily suppress a spatial dimension, but they are actually precisely "correct" in a certain sense. That is, when done properly they're not just "drawn straight from [the] imagination" as count_to_10 suggests, or "wrong" as Physics Guy suggests. They do have their limitations, but if you know what they mean, they can be useful.

To be more explicit, for the neutron star picture, I solved the TOV equation using a tabulated equation of state. (I forget precisely which EOS.) This gives me the metric of the spacetime in and around the neutron star. Then, I evaluated the metric on the $x$-$y$ plane (so, constant $t$ and $z$ coordinates), and extracted the $g_{xx}$, $g_{xy}=g_{yx}$, and $g_{yy}$ components. Better yet, use polar coordinates $(r, \phi)$. Then the line element in this slice is \begin{equation} ds^2 = \left[ 1 - \frac{2m(r)}{r} \right]^{-1} dr^2 + r^2 d\phi^2, \end{equation} where $m(r)$ is the mass enclosed by the sphere of radius $r$.

I then created a surface in three-dimensional flat space, using coordinates $(x', y', z')$. Assuming it was rotationally symmetric (like the spacetime), I just chose the height $h$ of the surface at each point above the $x'$-$y'$ plane. Now switch to polar coordinates $(r', \phi')$, and the line element of the surface is \begin{equation} ds'^2 = \left[ 1 + \left( \frac{dh} {dr'} \right)^2 \right] dr'^2 + r'^2 d\phi'^2, \end{equation} where $h(r')$ is some function that I can solve for just by equating the terms in the two line elements. This will involve a little integration of $m(r)$, but it's not too hard numerically. The result is a surface in flat 3-d space with the same intrinsic metric as the slice through the center of the warped spacetime.

Outside the star (or black hole), we have $m(r) = M$, the (constant) total mass of the star, and we can solve to find \begin{equation} h(r') = \sqrt{8M(r-2M)} + \mathrm{const.} \end{equation} This is the same whether you're dealing with a planet, white dwarf, neutron star, or black hole (as long as we can ignore the spin). Inside the star, things get trickier. For non-black holes, the surface just smoothly connects at the center. You'd hardly be able to see it for white dwarfs. As shown on that web site, you can start to see the smooth but non-flat shape for a neutron star. And for a black hole, inside the horizon $r<2m$, you'd get an imaginary number for $h$, so you just cut off the diagram.

There's a good discussion of this in MTW — around pages 613 and 837. Aslo, while I haven't read it, it looks like this paper contains a good discussion of embedding diagrams, and it's free.

• I much prefer to see these diagrams drawn that way up; it discourages attempting to explain gravity using gravity. – m4r35n357 Aug 13 '16 at 14:58
• Thanks @m4r35n357 . I've gotten a lot of pushback on that from other relativists because they think that way is unfamiliar to the public, but I stand by my choice as a more sensible one. And I like the analogy of a hill as something people would understand more immediately. – Mike Aug 13 '16 at 17:42
• Hi @Mike I understand the part where you set $t$, $z$ to constant, and the metric now a 2x2 matrix and depends on $(x,y)$. Is it true that you let the line element on the Euclidean embedded surface equal the line element on the constant-$(t,z)$ surface? And further assume $x'=y'=z'=f(x,y)$ then solve $f(x,y)$, correct? – Machine Aug 15 '16 at 17:46
• @ChenChao Yeah, that's the idea. I've updated my answer to add a little more detail. – Mike Aug 15 '16 at 19:04

Ricci curvature is 0 for the external Schwarszchild solution, e.g., outside a black hole or an (ideal) spherical star. To first order in the change of radius of the ball, the ball remains the same size as it is proportional to R, The zero valued Ricci scalar. There are more terms of higher order.

In that case the relevant scalar curvature is the Kretschmann curvature K, an invariant of the fully contracted Weyl tensor. See https://en.m.wikipedia.org/wiki/Kretschmann_scalar.

It is proportional to mass square, so yes, the more massive the higher that scalar Weyl curvature. What happens to the ball's size isn't depicted much in those cartoons, more the depth of the mattress as the massive body sinks into it.