# A positively curved visualization of gravity?

1. Dec 5, 2013

### Karl Coryat

Huffingtonpost ran a story heaping praise on a guy for demonstrating GR with the old "rubber sheet" analogy. This demo bothers me for several reasons, not least of which is that the surface is negatively curved, on which parallel geodesics diverge -- the exact opposite of what is supposedly being demonstrated.

But, it made me wonder: Is there a comparable 2D visualization of warped space, that extends to infinity like the rubber sheet, with positive curvature?

If one imagines a basketball pressed against the rubber sheet, the portion of the sheet in contact with the ball would be positively curved, but the rest would (as before) be negatively curved, one assumes.

I also assume (due to converging parallel geodesics) that a massive object creates positive curvature, and that this curvature extends to infinity. Yet, I am unable to imagine a 2D representation of this; every example of positive curvature I've seen is on a closed surface (i.e., a sphere). Are my assumptions about curvature wrong, or is this a property of 3D surfaces that cannot be reproduced on 2D surfaces (or both)? Thank you.

2. Dec 5, 2013

### WannabeNewton

2D representations of space-times are embedding diagrams which take space-times having sufficiently high degrees of symmetry (such as a time translation symmetry and rotational symmetry), suppress the symmetry directions, and embed the resulting 2 dimensional submanifold into an ambient 3D space (such as $\mathbb{R}^{3}$). These are not depicting space-time itself; the converging geodesic congruences that you mention are associated with space-time and not these embedded submanifolds i.e. the curves on these surfaces do not represent the worldlines of massive particles; furthermore the curvature you visualize with regards to these embeddings is extrinsic curvature i.e. how these embeddings bend relative to the ambient space (such as $\mathbb{R}^{3}$). The curvature that causes geodesic congruences to converge due to gravity is intrinsic curvature (of space-time!) and is an entirely different notion.

You cannot hope to depict space-time itself as a 2D realization; this is futile. Embedding diagrams help visualize various aspects of the geometry of space-time but you must be careful in interpreting exactly what kind of geometry is being described by these diagrams. Take for example Schwarzschild space-time. We can slice up Schwarzschild space-time into spatial slices of constant time $t = \text{const.}$ and choose any single one of these because of the time translation symmetry; we have now narrowed things down to 3 dimensions. Next, the spherical symmetry allows us to restrict ourselves to a $\theta = \text{const.}$ plane and this brings us down to 2 dimensions. We can now embed this 2 dimensional submanifold into $\mathbb{R}^{3}$ using the leftover cylindrical symmetry and the result is a 2D depiction of the spatial geometry of Schwarzschild space-time (relative to the above slicing) and is known as Flamm's paraboloid: http://en.wikipedia.org/wiki/Flamm's_paraboloid#Flamm.27s_paraboloid

3. Dec 6, 2013

### A.T.

I guess you mean the one discussed here:
Mods might want to merge these threads.

The sheet represents the spatial curvature outside the big mass, which indeed has locally negative curvature, similar to the sheet:
http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid

1) As WN noted, free faller follow geodesics in space-time, not in the space that is represented by the sheet.

2) Whether two geodesics diverge or converge depends on the situation. Two horizontally arranged objects falling vertically will get closer together, so their geodesics will converge. Two vertically arranged objects falling vertically will be pulled apart by tidial effects, so their geodesics will diverge.

Yes, this is qualitatively like a spatial slice of the Schwarzschild metric:
http://commons.wikimedia.org/wiki/File:Schwarzschild_interior.jpg[ [Broken]

No, some geodesics diverge. But again: They are geodesics in space-time, not in the space represented by the sheet. This is actually the biggest problem of this analogy. To describe gravitational attraction geometrically you have to include the time dimension:

The interior part has positive curvature, but exterior curvature is negative:

Last edited by a moderator: May 6, 2017
4. Dec 6, 2013

### Naty1

Karl..... As noted, a visualization of gravitational curvature involves space and time. And it's non linear which doesn't aid clarity. Also, some spacetime curvature is not gravitational. edit: and there is no single measure of curvature in General Relativity.

I provided this answer in another discussion which may aid your thinking [based on many inputs from experts here] :

In another discussion DrGreg explained things this way [my synopsis]:

You can picture world line [paths] curves in SR as you would curves on a flat graph paper. When gravitational curvature is involved, as in GR, the graph paper itself on which the curved worldlines are drawn is itself curved.
[This is as close a 'visualization' as I have gotten]

PeterDonis provided me the criteria for distinguishing gravitational curvature:

Here is another tip on identifying gravitational spacetime curvature:

Someone else posted an explanation:

interpretation: a fast bullet and a slow bullet follow different trajectories [the fast bullet travels further for example] but the spacetime curvature for each [sourced from the moving frame of each] is the same. On the other hand, if one bullet is rotating really fast, it will exhibit additional SET based rotational KE [in the frame of the moving bullet] so its gravitational curvature will differ from that of a slower moving bullet.

Also, there is no one measure of gravitational curvature that encapsulates 'gravity'.

There are a number of 'curvature' measurements; here are two well known:

[mostly from Wikipedia...]

The explanation immediately above is for 'tidal gravity'.

So I've come to the conclusion I cannot visualize spacetime curvature. [edit, add:..only particular aspects of curvature..as described..]

Last edited: Dec 6, 2013
5. Dec 6, 2013

### Staff: Mentor

??

If you're referring to "curvature due to acceleration", that is *path* curvature (curvature of a worldline), not spacetime curvature. You can't curve spacetime by accelerating.

6. Dec 6, 2013

### A.T.

Yes, all of this intrinsic curvature stuff is related to 'tidal gravity'. And that is also what the OP asks about (converging/diverging geodesics).

But local 'gravity' in the sense of the attraction is eqivalent to an accelerating frame, which doesn't involve intrinsic curvature, just curvilinear coordinates as shown in the animation with a apple above.

For a radial line, you can. It looks something like shown here:
http://www.relativitet.se/spacetime1.html

But to show orbits, you need 2 space dimensions, so together with time 3. Curved 3D is difficult to visualize.

7. Dec 6, 2013

### Naty1

PeterDonis:
I just looked at my notes and you have properly corrected me on this once before!!....don't give up on me, I'll eventually get it right!!

I like the distinction AT provided:

Last edited: Dec 6, 2013
8. Dec 6, 2013

### Staff: Mentor

I agree, he put it very succinctly and well.

9. Dec 7, 2013

### Karl Coryat

Thank you for the enlightenment. The situation is more interesting and subtle than I had realized, and the "rubber sheet" even farther from a real demonstration than I thought.

A.T. mentioned that the curvature beyond the Earth is negative, but that at some radius (I assume it's within the Earth's radius), the curvature inflects to positive. This makes me wonder: Suppose we had a hollow chamber inside the Earth's core, say 100km from the center of mass. We let two massive objects freefall. From our fixed perspective in the chamber, would we view the objects' paths as diverging?

Last edited: Dec 7, 2013
10. Dec 7, 2013

### Staff: Mentor

It's positive everywhere inside the Earth, assuming the Earth is solid. See below.

No; at least not if the chamber is spherically symmetric around the Earth's center of mass. (Technically, we also have to assume that the Earth itself is spherical.) If that is the case, then spacetime inside the chamber will be flat, and freely falling objects inside will not converge or diverge.

[Edit: After seeing A.T.'s post, I should clarify that I assumed the two objects inside the chamber did not have enough mass to create any detectable spacetime curvature on their own.]

Last edited: Dec 7, 2013
11. Dec 7, 2013

### A.T.

Do you really mean two massive objects, with non-negligible gravity on their own? They will obviously converge, with or without the Earth around them. Now you may ask why two massive objects in space converge, if both create negative curvature outside of them. Here you have to differentiate between local curvature at some point (which is negative outside of the masses), and net curvature over a larger area that includes both masses (which is positive).

But if you mean test particles with negligible mass, it depends on the shape and position of the cavity. A spherical cavity in a uniform density sphere should contain uniform gravitational fields, where space-time has no intrinsic curvature. So the test particles will neither diverge, nor converge, just move at constant velocities (if cavity is centered) or accelerate at the same rate (if cavity is off center).

12. Dec 8, 2013

### Karl Coryat

Yes, I neglected to say ignore any curvature due to the objects themselves.

I am not talking about a chamber at the center of a spherical Earth, where spacetime would be clearly flat. I mean, a chamber at some distance from the center (e.g., 100km). Presumably spacetime is not flat there -- PeterDonis notes that it is positive, if I understand correctly. Does that mean that two "dropped" test objects would diverge as they (slowly) approach the center of mass?

Common sense suggests that all free-falling objects converge toward a center of mass, regardless of their location. So I'm asking whether the curvature inflection from negative to positive means freefalling paths "behave" differently close to a center of mass. That seems paradoxical, to simultaneously fall toward and diverge away from the center...hence my confusion. Thank you for your patience!

13. Dec 8, 2013

### A.T.

Just forget the cavity as it makes it more complicated. Consider test particles that don't interact with the matter of the big mass (except for gravity), like neutrinos but even less interacting.

No, positve curvature means that intially parallel geodesics converge, no matter their alignment. The tidal effect inside the mass is compressive from all sides. So starting at the same velocity they will initially converge no matter how they where arranged inside the mass.

Outside the mass tides stretch radially, and compresses tangentially, so the converging / diverging depends on the alignment of the test particles.

14. Dec 8, 2013

### yuiop

Ironically, the reviled rubber sheet analogy gives the correct answer here. Initially stationary objects would accelerate straight towards the centre (converging).

15. Dec 8, 2013

### yuiop

Imagine a perfectly spherical non rotating steel ball the size of the Earth. If we rotate this steel ball so that it has the same angular velocity as the Earth, we will find that clocks at the equator will be running slower relative to clocks at the poles. If we now add enough water to the surface, the liquid surface will naturally form an oblate spheroid. Clocks at sea level on this liquid surface will now all run at the same rate. This suggests that the time dilation factor is related to the gravitational potential. Stationary objects tend to be accelerated from regions where clocks run fast to where clocks run slower. Assuming some validity to this observation, a plot of the time dilation factor has some relation to the plot of the gravitational potential. This diagram shows a plot of $d\tau/dt$ on the vertical axis versus radius r on the horizontal axis, for the interior and exterior Schwarzschild metrics with $dr=d\Omega=0$.

The magenta curve is the interior metric which represents the potential inside the gravitational object and the blue curve is the exterior metric which represents the region outside the gravitational body. The vertical red line represents the surface of the gravitational body. If the curves are rotated around the r=0 axis, an embedding diagram is generated which has similarities to a conventional gravitational well diagram (and to the rubber sheet analogy). I should mention that one limitation of the graph I have plotted is that interior Schwarzschild metric assumes constant density which is not realistic for a real gravitational body.

Like all analogies, the rubber sheet model has its limitations and it is not supposed to be demonstration of what causes gravity. It is merely a easily visualised and simple demonstration for school children of Wheeler's idea that "'Matter tells space how to curve. Space tells matter how to move" and it does not do a bad job of demonstrating that. Any objects placed on the sheet deform the sheet. Any deformations of the sheet alter how objects move along it. Any objects moving across this landscape respond to the geometry of the landscape and dynamically alter the landscape.

Another limitation is that the effective potential is dependent on the motion of the test bodies, so no single embedding diagram can represent the potential 'landscape' around a given massive body for test objects with different velocities.

(Having said all that, the diagrams and animations by A.T. are more informative and accurate representations of the effects of time and space on the motion of bodies in GR.)

#### Attached Files:

• ###### embed.jpg
File size:
9.8 KB
Views:
493
Last edited: Dec 8, 2013
16. Dec 8, 2013

### A.T.

The problem of the rubber sheet is not being an analogy of gravity in general, where the rubber sheet could for example represent the gravitational potential (gravity well). The problem of the rubber sheet is that it has nothing to do with how attraction results from General Relativity. See:
http://en.wikipedia.org/wiki/Gravity_well#Gravity_wells_and_general_relativity

Why would anything that is initially stationary on the rubber sheet start moving along the rubber sheet? Because gravity pulls it down the dent? That is the circular kind of argument that just confuses people. See here and countless similar threads.:

In the context of GR the rubber sheet represents just a geometry, a distortion of distances. It could just as well have a hill instead of a dent and still represent the same curved space. But in the logic of the rolling balls on the sheet that would imply repulsion, which is of course nonsense.

17. Dec 8, 2013

### Karl Coryat

I think I got it

Thank you for the insights and the excellent resources. I also found this tool extremely helpful. If I may put it in my own words, please correct me if I'm off:

- The rubber sheet is generally a geometrically inaccurate representation of the curvature of spacetime due to a planet-like mass, but nevertheless, under the conditions of Earth's gravity, the demonstration does happen to crudely reproduce some physical effects of spacetime curvature (at the expense of providing a circular explanation).

- Some actual models of spacetime curvature (e.g. Flamm's paraboloid, which is a visualization of Schwartzschild spacetime, a solution of Einstein's equations relevant to a planet-like object) do superficially resemble the rubber sheet, but (a) this is coincidental and (b) they represent a momentary slice in coordinate time, and no particle could actually move across such a surface, as all intervals on the surface are spacelike.

- If we are to discuss positive vs. negative curvature affecting the geodesics of test particles, this has nothing to do with the geometry of Flamm's paraboloid, let alone the rubber sheet. The direction of these curvatures only affects moving particles in models where time is assigned to one of the dimensions (as the model is embedded in our observers' world with our three familiar dimensions). And, these models have little in common geometrically with the rubber sheet.

18. Dec 9, 2013

### A.T.

I would summarize it this way:

- Rolling balls on a rubber sheet can be used as a qualitative analogy for the gravity well (gravitational potential). That's why it gives the correct qualitative result. But that has nothing to do with explaining General Relativity and curved space-time, because it applies equally to Newtonian Gravity.

- The indented rubber sheet can be used as a qualitative visualization of the space (not space-time) distortion in General Relativity (Flamm's paraboloid). But that has has nothing to do with explaining how masses attract each other in General Relativity, which requires including the time dimension (space-time distortion). Flamm's paraboloid represents a distortion of spatial distances between coordinates, and could just as well be shown with the funnel upwards, so the rolling balls would give a wrong result. Therefore rolling some balls on the curved surface representing Flamm's paraboloid makes no sense.

- The local intrinsic curvature of space-time you are asking about is primarily related to tidal-effects, or gravity gradient.

Last edited: Dec 9, 2013
19. Dec 9, 2013

### WannabeNewton

You summarized it perfectly in my opinion but do take into account the things that A.T. added in his summary. You have to keep in mind that the most important flaw in the so called "rubber sheet analogy" is that it doesn't take into account space-time structure (the spatial geometry embedding diagrams also fail to do this but they don't claim to describe space-time, they are meant to describe spatial geometries).

Some use of differential geometry actually makes it explicit why the "rubber sheet analogy" is flawed as far as depicting of space-time geometry goes. Take some arbitrary space-time with metric tensor $g_{\mu\nu}$ and imagine a very small spherical ball of massive test particles that are all initially at rest with respect to one another (you may be asking yourself what it even means to have a spherical ball in curved space-time and this is a valid concern; we assume the particles are infinitesimally separated from one another in such a way so that they form an infinitesimal spherical ball in the usual vector space sense).

First note that each particle traces out some worldline in space-time so that the entire ball of particles trace out a family of worldlines that "knit together" to form a vector field $\xi^{\mu}$ which represents the 4-velocity field of this ball of particles. Now we set up our system so that the particles making up the spherical ball are all initially at rest with respect to one another. We want to know how space-time curvature then causes these particles to start converging at that very instant once we let them go into free fall; put a different way, in what way does space-time curvature cause the worldlines of the particles (which are initially at rest relative to one another) to start converging towards one another? From here on out, we go to the instantaneous rest frame of any one particle in the ball and calculate things from the perspective of an observer $O$ at rest in this instantaneous rest frame.

I won't derive the following relation here but if we assume the infinitesimal ball of particles has a volume $V$ then we can show that the divergence of $\xi^{\mu}$ is related to the rate of change of $V$ relative to $O$ by $\nabla_{\mu}\xi^{\mu} = \frac{1}{V}\xi^{\mu}\nabla_{\mu}V = \frac{1}{V}\frac{d V}{d\tau}$ where $\tau$ is the proper time read by $O$'s clock. So if we want to see how space-time curvature causes the initially stationary ball of particles to start converging, which is equivalent to looking at the second order rate of change of the volume of the ball (second order because space-time curvature is second order in $\nabla_{\mu}$) then we want to focus on the rate of change of $\theta\equiv \nabla_{\mu}\xi^{\mu}$ (often called the expansion scalar) relative to $O$ i.e. we want to focus on $\xi^{\mu}\nabla_{\mu}\theta = \frac{\mathrm{d} \theta}{\mathrm{d} \tau}$; keep in mind all of this is at the very instant that we let the initially stationary ball of particles go into free fall.

I also won't' derive the following expression (if you want to see where it come's from, read this: http://en.wikipedia.org/wiki/Raychaudhuri_equation) but it can be shown that for a spherical ball of particles that are all initially at rest relative to one another and allowed to go into free fall, $\frac{\mathrm{d} \theta}{\mathrm{d} \tau}$ at that instant is given by $\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -R_{\mu\nu}\xi^{\mu}\xi^{\nu}$ where $R_{\mu\nu}$ is the Ricci curvature tensor. Now in the instantaneous rest frame that we are working in, $\xi^{\mu} = (1,0,0,0) = e_{\hat{t}}$ where $e_{\hat{t}}$ is the unit vector pointing along the time axis of this instantaneous rest frame so $\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -R_{\hat{t}\hat{t}}$.

Now Einstein's equation $R_{\mu\nu} = 8\pi(T_{\mu\nu} - \frac{1}{2}Tg_{\mu\nu})$ combined with the strong energy condition (http://en.wikipedia.org/wiki/Strong_energy_condition#Strong_energy_condition) implies that $R_{\mu\nu}\xi^{\mu}\xi^{\nu} = R_{\hat{t}\hat{t}} > 0$ if we are not in vacuum (which is the case we are considering here); this is basically the statement that observers always see positive space-time curvature, as far as the Ricci curvature tensor goes. This implies that $\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -R_{\hat{t}\hat{t}} < 0$ i.e. the initially stationary particles composing the spherical ball start converging at the very instant we let them go into free fall.

The important point is that it was $R_{\hat{t}\hat{t}}$ (i.e. the time-time component of the Ricci curvature tensor in the instantaneous rest frame of $O$) that caused the initially stationary particles to start converging towards one another. If we ignore the "time" part of space-time then we can't even explain why gravity as a manifestation of space-time geometry causes these particles to start converging! This is why the "rubber sheet analogy" is a completely incorrect representation of space-time geometry and the kinematics of freely falling particles in general relativity.

Last edited: Dec 9, 2013