# Light bending in the hot plate model of curvature

1. May 1, 2014

### Sonderval

Light bending in the "hot plate" model of curvature

In the Feynman lectures, feynman describes the hot plate model of space curvature and shows that light is bent around the center of the plate, see Fig. 42-6
http://www.feynmanlectures.caltech.edu/II_42.html#Ch42-S1

However, the hot plate corresponds to a region of positive curvature, if I understand things correctly. Outside of a mass, the spacetime curvature should be negative.

I've tried to match the Feynman picture with the Schwarzschild metric. The space part of ds² is (approximately)
$ds^2 = (1+2GM/r) dr^2$
(where I use a -+++ metric because I'm only interested in space components right now.)
If I understand this formula correctly, it says that the "rulers" in feynman's picture get shorter the further I am away from the central mass, which agrees with the negative curvature interpretation.

From this model alone I would expect light to bend away from a mass - which is of course wrong. Is this due to the time dilation near the mass that affects the geodesics? Or am I making another stupid mistake?

2. May 1, 2014

### Staff: Mentor

Only in certain directions. For the vacuum region around a massive body, the curvature, roughly speaking, is negative in the radial direction but positive in the angular directions. See further comments below.

Curvature is not described by the metric; it's described by the Riemann curvature tensor. The picture Feynman is giving is not really a good description of spacetime curvature; at best it's a description of the spatial curvature of a slice of constant Schwarzschild coordinate time, but the resulting model has a lot of problems when you try to use it to analyze actual physics.

No, that's not what it says. What it says is that a constant coordinate interval $dr$ corresponds to a smaller radial distance the further away you are from the central mass. But "smaller radial distance" just means fewer rulers fit into a given coordinate interval $dr$; it doesn't mean the lengths of the rulers themselves change. But again, all this is not a description of spacetime curvature; at best, it's a description of spatial curvature of a slice of constant Schwarzschild coordinate time.

No, it's due to using the wrong mathematical object, the metric instead of the Riemann curvature tensor. Or, for a simpler way of viewing what the curvature tensor is telling you, think of tidal gravity, since spacetime curvature and tidal gravity are really the same thing. For freely falilng objects in the vacuum region around a massive body, tidal gravity tends to increase their radial separation (negative curvature) and decrease their angular separation (positive curvature). If you compute the appropriate Riemann tensor components, you get radial components that look like $- 2 M / r^3$, and angular components that look like $M / r^3$.

3. May 2, 2014

### Sonderval

Thanks for the answer. I think it pushed me in the right direction and I finally realised that I have confused two things - the shortest distance in space and a geodesic. Here is my current understanding, I would be grateful if you could tell me whether it is o.k. or whether I'm still confusing things.

If I connect two fixed points in vacuum with a rubber band, it will form a straight line. (This is the situation I implicitly assumed and - as you said - which only looks at the spatial curvature at constant time.) If I now approach the system with a mass from one side, the rubber band will bend towards the mass, exactly as expected from the Feynman hot plate model with negative "curvature" (i.e. with a temperature that decreases with increasing radius). Feynman's picture 42-6 is correct, but it is not analoguous to light bending at all.

If I connect two points in space-time with a light-ray, the ray follows a geodesic that bends around the mass, similar to a particle in free fall that would also take a path that bends away from the mass.

So whereas in flat space a straight line as constructed with a rubber band and a straight line as followed by a light ray coincide, this is no longer true in curved space-time.

Mathematically, in the rubber band case I am minimising the spatial part of ds² only, whereas for a photon I have ds²=0 and thus get an equation that connects dt² and dr² (as done by Lerner in this paper http://astronomia.udea.edu.co/sitio...lares/Relatividad/2010-1/Papers/AJP001194.pdf )

Is this a correct understanding of the situation?

4. May 2, 2014

### A.T.

If I understand the hot plate model correctly, the sign of curvature is not the sign of the first derivative of the temperature. The sign of curvature rather depends on how the second derivatives of temperature in different directions relate.

5. May 2, 2014

### pervect

Staff Emeritus
I think you have the right idea, but I'd word it differently.

The shortest curve in space would be (modulo a few rather highly technical points that don't apply to GR) a spatial geodesic. The shortest curve in space is not a geodesic in space-time, but it's a geodesic in the spatial hyperslice of space-time. To define the hyperslice, you need a way of slicing space-time into space+time.

This "shortest curve in space" is, as you point out, different from a null geodesic, the path that light follows.

It would be better to call it a geodesic (in the given hyperslice) rather than a straight line, I think. A moderately tricky point is that we assume all parts of the rubber band "hold station", and this notion of the rubber band "holding station" implicityl defines the necessary time-slice.

But your main point is correct - the spatial geodesic followed by the rubberband is not the same curve as the null geodesic followed by the light beam.

Something I feel I should add:

Feynman may not have gone through all the math rigorously, but the rubberband model (with idealized rubber bands whose mass approaches zero) would in fact have a curved geometry, using the "sum-of-angles-in-the-triangle" test.

Real rubberband would respond to both the tidal forces and the underlying spatial curvature. You need to idealize the rubberbands to eliminate their response to the tidal force to get the purely spatial part of the geometry.

The part of the Riemann that describes the underlying spatial curvature is the topogravitic part when it's decomposed using a technique using the Bel decomposition.

Last edited: May 2, 2014
6. May 2, 2014

### Sonderval

@A.T.
I don't think so - as long as the temperature increases from center to the outside, a path bending outwards as in Fig. 42-6 is always shorter than a "straight" path.

@pervect
Thanks for the clarificaton of the nomenclature. I think now I understand it.

7. May 2, 2014

### A.T.

The sign of curvature doesn't tell you whether a path bends outwards or inwards. It tells you how neighboring, initially parallel paths behave relative to each other.

A light ray inside the planet (positive curvature) will bend in the same direction (relative to the center) as a light ray outside of the planet (negative curvature). But two neighboring tangential initially parallel rays will come closer inside the planet (positive curvature) and deviate apart outside of the planet (negative curvature).

Last edited: May 2, 2014
8. May 2, 2014

### Staff: Mentor

Ok so far (but it should be noted that the "straight line" is a straight line in space--the space of the vacuum at some constant time--not spacetime).

If you don't do anything to hold it in place, it will *fall* towards the mass. If you do hold it in place, whether or not it bends depends on how you hold it.

No, both the particle in free fall and the light ray bend *towards* the mass, not away from it. Gravity is attractive.

Also, note that the "bending" you are imagining here is still not "bending" of a path in spacetime; it's bending of a path in space which is a projection, into a surface of constant time, of the spacetime path the light ray (or free-falling particle) follows. The paths that the light ray and the particle follow in spacetime are not bent at all; they are geodesics, which means they are as straight as they can possibly be.

Once again, you are confusing "straight lines" in space with "straight lines" in spacetime. The two are not the same.

Which means you're not describing any path in spacetime, only in space. Comparing paths in space with paths in spacetime is comparing apples and oranges.

9. May 2, 2014

### Staff: Mentor

One clarification I think is important: the rubber band does not "follow" a spatial geodesic; its points follow a family of timelike geodesics (at least, they do if we assume it is in free fall--but if it is in free fall, it won't bend when gravity is present, it will just fall, as I noted in my previous post). The spatial geodesic is obtained by taking the intersection of the family of timelike geodesics with some surface of constant time. But no object actually "follows" that spacelike geodesic.

The reason I think this is important is that, as I said in my previous post, it's important not to confuse paths through space with paths through spacetime. Comparing the two is comparing apples and oranges, and IMO does not help with understanding.

10. May 2, 2014

### Sonderval

@A.T.
Seems I misunderstood you the first time because I fully agree with your last post. Sorry for that.

@PeterDonis
Thanks - yes, the confusion between "shortest distance in space" and "shortest distance in spacetime" was exactly my problem.

Concerning the rubber band, I'm imagining a situation where I have a contraption that fixes the two end points in some way and have the rubber band free in between - the band would then bend in exactly the opposite way from Feynman's Fig. 42-6. I agree that this is a very artificial situation (as your last post makes very clear) - but this was exactly the source of my confusion, which seems to be cleared up now. Thanks again.

11. May 2, 2014

### pervect

Staff Emeritus
I would expect that the idealized rubber band would curve in the same manner as particle orbits and light orbits do. I haven't calculated anything in detail (recently), but the reason I think this should happen is that if you look at PPN theory, http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism, of the two main parameters $\beta$ and $\gamma$, light deflection is only sensitive to the later parameter, which represents "How much space curvature is produced by a unit rest mass".

And if we set $\gamma$ equal to zero, the spatial curvature of the geometry goes to zero, and the deflection halves, which comes across in popularizations as "half the deflection of light is due to spatial curvature". So it's distinctly odd to see the rubber band curve the other way, I'd expect it to curve in the same way.

12. May 3, 2014

### pervect

Staff Emeritus
OK, doing a quick calculation and using some results from https://www.physicsforums.com/showthread.php?t=686147#post4354272

If we assume we have a purely spatial metric

$ds^2 = g(r)\, dr^2 + h(r)\, d\phi^2$

Then we can look for the parameterized curve $r(\tau)$, $\phi(\tau)$ that's a geodesic using the geodesic equation (sorry to used advanced concepts, but I'm trying to pursue a harder solution than my previous hand-waving). Note that $\tau$ is really an affine parameter and not really time, though you can envision tracing out the spatial geodesic as a function of $\tau$if you remember that the end result is just a curve in space.

$$\frac{d^2r}{d \tau^2} + \Gamma^r{}_{rr} \left( \frac{d r}{d \tau} \right)^2 + \Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$

and we can rewrite the normalization condition of unit velocity as

$$g(r) \, \left( \frac{dr}{d\tau} \right)^2 + h(r) \, \left( \frac{d \phi}{d\tau} \right)^2 = 1$$

Though a unity normalization condition is traditionally, any constant value would do. The form of the geodesic equation used DOES require that the right hand side of the above be constant. We will use unity for ease of calculation from here on.

All we really need to calculate for our purposes is the value of $\frac{d^2r}{d \tau^2}$ at the point of closest approach.

At the point of closest approach $dr / d\tau$ will be equal to zero, which gives us the simple relationship:

$$\frac{d^2r}{d \tau^2} + \Gamma^r{}_{\phi \phi} \left( \frac{d \phi }{d \tau} \right)^2 = 0$$

Substituting for the Christoffel symbols from https://www.physicsforums.com/showthread.php?t=686147#post4354272

$$\Gamma^r{}_{\phi \phi} = - \frac{(\frac{dh}{dr})}{ 2g}$$

and defining $\omega = \frac{d\phi}{d\tau}$

we get

$$\frac{d^2r}{d \tau^2} = \left( \frac{\frac{dh}{dr}} { 2 g h} \right) \omega^2$$

For a flat space metric h(r) = r^2 and g=1 , which gives us the expected relation for a spatial geodesic of

$\frac{d^2r}{d \tau^2} = 1/r$ given that $r^2 \omega^2 = 1$, i.e. this is like v=1 and a radial acceleration of v^2 / r = 1/r if we think of a particle moving with unit velocity along the geodesic

For the schwarzschild metric h will also be equal to r^2, but g will be $g = 1 / (1 -\frac{2 G M}{c^2 r})$. The only difference from the previous result is the factor of 1/g, and since 1/g < 1, we get a lesser value for $\frac{d^2r}{d \tau^2}$, which means that in the Schwarzschild metric, at the point of closest approach, $\frac{d^2r}{d \tau^2}$ is still outwards but lower than it is for flat space , implying that relatively speaking, it "curves outwards" less or "curves inwards" more than a straight line in flat space would.

This is rather hasty, but I think it's an OK derivation.

13. May 3, 2014

### Sonderval

@pervect
Although I think I can (at least roughly) follow your calculation, I'm not sure what you are really showing. That a rubber band does indeed bend inwards when calculating with a Schwarzschild metric? Is this a correction or a corroboration of your previous post? Sorry - seems I'm a bit confused today.

Last edited: May 3, 2014
14. May 3, 2014

### A.T.

Maybe you should define "bending inwards" vs. "bending outwards", ideally with a sketch.

15. May 3, 2014

### Sonderval

@A.T.
I was going to simply point you to Feynman's Fig. 42-6 - but then I realised that things are not that simple if space is curved.

Thinking a bit more, I think I can now re-phrase my original thought more precisely:

In flat space-time, the geodesic connecting two points A and B followed by a massive free falling object, by a light ray and the shortest distance between two points all coincide.
If I now approach the system with a mass, the geodesic (or rather, its spatial projection) of a massive particle lies outwards (further away from the mass) than that of a lightray, and the line of shortest distance lies inwards (towards the mass).
In other words, if the distance between A and B is timelike, the geodesic's projection in space lies on one side of a light ray, if it is spacelike, the shortest distance path lies on the other side of the light ray.

Is this correct?

16. May 3, 2014

### Staff: Mentor

No, they don't. You are confusing curves in space with curves in spacetime again. In spacetime, all three of these curves are different: the first is timelike, the second is null, and the third is spacelike. You are obviously aware of this distinction, since you refer to it later in your post; but I don't think you've fully considered its implications. See below.

The correct statement is that the *spatial projections* of the spacetime curves followed by a massive free-falling object traveling between two spatial points, and by a light ray traveling between those same two spatial points, coincide with the spatial geodesic representing the shortest distance between those two points in a surface of constant time.

No, this is not correct as it stands, because there are many, many different possible geodesics that a massive particle and a light ray could follow. You need to impose some constraints so we know which particular geodesics you are talking about. That's why the distinction between curves in space and curves in spacetime is important.

My guess is that what you are thinking is something like this: pick two spatial points very far away from the mass, but on opposite (or nearly opposite) sides of it. Call these spatial points A and B. (We assume that A and B are both at the same--very large--radius from the center of the gravitating body.) Then look at the spatial projections of (1) the spacetime geodesic followed by a massive particle that is on a free-fall orbit that will take it from A to B; (2) the spacetime geodesic followed by a light ray that travels from A to B. Compare these two spatial projections with (3) the spacelike geodesic in a surface of constant time that represents the shortest distance from A to B.

If the above is what you are thinking, then I *think* your statement is correct, in the sense that the spatial point of closest approach to the center of the gravitating body is larger for #1 than for #2, and larger for #2 than for #3. However, I'm less sure of the latter comparison than I am of the former.

17. May 4, 2014

### A.T.

And what does this figure show? "Bending inwards" or "bending outwards"?

18. May 5, 2014

### Sonderval

@PeterDonis
You nailed my question exactly, sorry for being so imprecise in my language to obfuscate matters.
More precise follow-up question below.

@A.T.
As I said, I realised that it is not so simple, that's why I did not really point to that figure...
But I think I can make it more specific:
Consider the two points A and B on an isoscles triangle with the mass sitting in the third corner C - where the distance betweem A, B and c is very large (think for example of A being earth, B being a far-away galaxy and C being a galaxy in the middle between the two but not exactly on their onnecting line, similar to a gravitational lensing effect).

If there were no mass, the middle point M on the egde A-B is closest to the third corner and its distance L to that corner can be calculated by simple trigonometry. Now I add a mass at point M.

The closest distance L of the space projection of a light ray geodesic is then smaller than that of the geodesic of a massive particle (and L becomes larger the slower the particle is). The closest distance of a spatial shortest-distance line (so that the spacetime points A and B at the start and end of the line have a spacelike separation) is still smaller than that of a light ray. (This is the situation Peter describes.)

If we assume - for simplicity - that A, B and C are at rest to each other (if the distance between them is sufficiently great, this should not be too problematic) and if we choose a coordinate system at rest wrt the three points, we now can construct the line with the smallest spacetime distance $S^2=\int ds^2$ as a function of the time separation $\delta t$ of the starting and the end point in space time. (If $S^2=0$, we connect the points with a light ray, if it is larger, we connect with a geodesic of a massive particle, if it is smaller, the end points are separated by a spacelike distance).

We look at the spatial projection of all these paths and at their smallest distance L to the mass at C.
The questions then are:
1. What is the function $L(\delta t)$ ?
2. Is there any path for which $L(\delta t)$ is smaller that the shortest distance calculated from pure geometry without a mass at C? This is a path that "curves inwards" compared to the case of a flat spacetime.