I Euclidean geometry and gravity

[raagamuffin]

I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté !

If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees?
How about if I'm looking at this paper outside of the (reasonable) gravitational sphere of influence ? Does it matter where it's being observed from?

Or is this question nonsensical? For example, presumably the measuring apparatus (protractor) would also be warped by an equal amount?

 

[PeterDonis]

If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees?

It depends on how the paper is oriented and what its state of motion is. But for a piece of paper of ordinary size and a black hole of stellar mass or larger, any effects of non-Euclidean geometry will be too small to observe.

A better way of probing the spatial geometry around a black hole is this: suppose I construct two spherical shells around the hole, one inside the other. I carefully measure the surface area of each shell. Then I measure the radial distance between the shells. How will that distance compare to the radial distance between two concentric spherical shells as predicted by Euclidean geometry?

The answer to that is that the measured radial distance will be larger than the Euclidean prediction.

How about if I'm looking at this paper outside of the (reasonable) gravitational sphere of influence ? Does it matter where it's being observed from?

The observations I described above, for the two concentric spherical shells, depend on local measurements. Trying to make those observations from a distance brings in other effects, such as the behavior of light in curved spacetime, that make it much more difficult to interpret the results.

presumably the measuring apparatus (protractor) would also be warped by an equal amount?

No, it wouldn't. It's true that objects being held at rest in a gravitational field, so they feel weight, don't behave exactly the same as objects in free fall; internal stresses in the objects can change their shape. But for measurements like the ones I described above with the spherical shells, you can make the measurements using, for example, very small rulers, so such effects are negligible.

 

[raagamuffin]

Thanks, the sphere example is a nice illustration.
Please allow me to probe into your answers a bit.
(If you wouldn't mind sticking to the 2D world, it just helps me visualize things a bit better.)

Let's extend the example, to say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?

Trying to make those observations from a distance brings in other effects, such as the behavior of light in curved spacetime, that make it much more difficult to interpret the results.

I understand this issue with observation.Theoretically speaking then, should the observer expect a different answer every time, or is there a predictable answer (or set of answers) ?

 

[PeterDonis]

say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?

If the paper were oriented radially, then yes, you would be able to draw a triangle on it whose angles would not add up to 180 degrees. In effect, it would be as if you were drawing the triangle on a Flamm paraboloid:

https://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm's_paraboloid

Please note that the Flamm paraboloid visualization is very limited, and should be used with caution. But it does capture the specific case we're discussing.

 

[PeterDonis]

Theoretically speaking then, should the observer expect a different answer every time

Not if the conditions are the same. But what someone very distant observes won't be easily interpreted as "I'm seeing non-Euclidean geometry".

 

[raagamuffin]

thanks, this is good information. I'll work through the Flamm paraboloid writeup.

Not if the conditions are the same. But what someone very distant observes won't be easily interpreted as "I'm seeing non-Euclidean geometry".

One last question, if we can't interpret the data well enough that comes from curved space time, would it also be the case that we can't get usable data (about spacetime curvature) from light that comes as a result of gravitational lensing ? or infer the geometry of space from the light that comes from black hole boundaries (instead relying on theoretical data)?
(Perhaps this is a cosmology thread question.)

 

[Ibix]

we can't get usable data (about spacetime curvature) from light that comes as a result of gravitational lensing ?

Sure we can get usable data. But you have to take into account the effect of curvature in the spacetime the light travels through, as well as the curvature at the area you're looking at. So what you actually see doesn't have a straightforward interpretation in terms of the local geometry. You can still deduce what's there.

It's not radically different from looking at something through a bottle. The image is distorted so what you see isn't what's there, but if you know how to do the maths you can correct for the distortion. The maths is a bit harder in GR than normal optics, that's all.

 

[PeterDonis]

if we can't interpret the data well enough that comes from curved space time

That's not what I said. What I said is that, if you're observing from a distance objects that are at rest fairly close to a black hole's horizon, there won't be a simple way to interpret your observations as showing you the non-Euclidean geometry of space in the region you're observing. That's a much narrower statement than the one you made in the quote above.

would it also be the case that we can't get usable data (about spacetime curvature) from light that comes as a result of gravitational lensing ?

Not at all. There are certainly ways of getting useful data about spacetime curvature from a distance.

But the non-Euclidean geometry of space close to a black hole is not the same thing as spacetime curvature. It is a consequence of the geometry of spacetime around a black hole, which is curved, but it has no simple relationship to spacetime curvature.

or infer the geometry of space from the light that comes from black hole boundaries

"The geometry of space" is not even an invariant. Remember that GR is a theory of spacetime, not space. The geometry of spacetime is an invariant. The non-Euclidean "geometry of space" around a black hole that we've been discussing depends on splitting up spacetime into "space" and "time" in a particular way. That way is the one that makes the most sense for an observer at rest relative to the black hole, but it's still a choice that ultimately has nothing to do with the physics; it's just a convenient way of organizing information so we humans can work with it. The geometry of spacetime is the fundamental thing.

 

[PAllen]

If the paper were oriented radially, then yes, you would be able to draw a triangle on it whose angles would not add up to 180 degrees. In effect, it would be as if you were drawing the triangle on a Flamm paraboloid:

https://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm's_paraboloid

Please note that the Flamm paraboloid visualization is very limited, and should be used with caution. But it does capture the specific case we're discussing.

Some additional cautions may be stated:

A paper so oriented is not inertial, so you have to imagine some force holding it in place, that has no effect on the spacetime. The force must be different at different points on the paper.

Also, if I choose a different set of events (using a different possible time coordinate) on each stationary paper world line to be simultaneous, I can discover a strictly Euclidean triangle - the Gullestrand Panlieve foliation.

 

[A.T.]

Let's extend the example, to say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?

If you manufacture a large enough sheet of paper in flat space, and then move it into curved space, it will be torn apart, no matter if/how you apply forces to try preventing that.

 

[PeterDonis]

If you manufacture a large enough sheet of paper in flat space, and then move it into curved space, it will be torn apart, no matter if/how you apply forces to try preventing that.

First, you should say flat spacetime and curved spacetime. Those are the invariants.

Second, while paper made out of the ordinary materials we usually make paper out of probably would be torn apart, yes, we could imagine making paper out of thin sheets of Kevlar, or carbon nanotubes, or something that would be strong enough to hold itself together against the stresses it would undergo while being moved. But such an object, while it would not be torn apart, would change its shape. That's the crucial point.

 

[Jaime Rudas]

A better way of probing the spatial geometry around a black hole is this: suppose I construct two spherical shells around the hole, one inside the other. I carefully measure the surface area of each shell. Then I measure the radial distance between the shells. How will that distance compare to the radial distance between two concentric spherical shells as predicted by Euclidean geometry?

The answer to that is that the measured radial distance will be larger than the Euclidean prediction.

From this, I understand that for a sphere of area $A$ around a very massive and dense star, its radius $R$ will be greater than $\frac12 \sqrt{\frac A \pi}$. If this is correct, for the case of the sphere around the black hole, can its radius be calculated, or is it undefined?

 

[PeterDonis]

From this, I understand that for a sphere of area $A$ around a very massive and dense star, its radius $R$ will be greater than $\frac12 \sqrt{\frac A \pi}$.

That's correct. Note, though, that when you have a star in the center instead of a black hole, the spacetime geometry inside the star is different from the geometry outside, and that has to be taken into account.

If this is correct, for the case of the sphere around the black hole, can its radius be calculated, or is it undefined?

It's undefined. The black hole has no center in the usual geometric sense. The locus $r = 0$, the singularity, is not a place in space at the center of the hole; it's a moment of time that's to the future of all events inside the hole.

 

[PAllen]

First, you should say flat spacetime and curved spacetime. Those are the invariants.

Second, while paper made out of the ordinary materials we usually make paper out of probably would be torn apart, yes, we could imagine making paper out of thin sheets of Kevlar, or carbon nanotubes, or something that would be strong enough to hold itself together against the stresses it would undergo while being moved. But such an object, while it would not be torn apart, would change its shape. That's the crucial point.

Alternatively, make it out of something like rubber, and it will change shape without great stress or tearing.

 

[PAllen]

Claim: There is no difficulty defining a congruence of world lines representing a piece of paper far from a BH, that then has each worldline accelerate and decelerate such that you have a representation of a sheet of paper stationary near the BH. The kinematic decomposition of this congruence will necessarily show expansion and compression. However whether a material approximation to this math breaks or not depends on material properties. Unless one is talking about getting close to the horizon of stellar ( rather than supermassive) BH, the stress would not be extreme.

 

[pervect]

Thanks, the sphere example is a nice illustration.
Please allow me to probe into your answers a bit.
(If you wouldn't mind sticking to the 2D world, it just helps me visualize things a bit better.)

Let's extend the example, to say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?


I understand this issue with observation.Theoretically speaking then, should the observer expect a different answer every time, or is there a predictable answer (or set of answers) ?

It's unclear to me what mathematics models a "large sheet of paper" in the current context. So I don't have anything rigorous. But I'll talk about the "small piece of paper" case.

A "small piece of paper" will have a sectional curvature that can be computed from the Riemann tensor, wiki has the formula https://en.wikipedia.org/wiki/Sectional_curvature

The easiest thing to do is to define the "small paper" by two unit vectors that are orthogonal, then the wiki formula for the sectional curvature is

$$R_{abcd} X^a Y^b Y^c X^d = -R_{abcd} X^a Y^b X^c Y^d$$

Sorry for the tensor notion, but I don't know how to avoid it. You may be familiar with matrices. In tensor notation they would be two index tensors, like say $Q_{ab}$. If you imagine a sort of matrix that's 4 dimensional rather than 2 dimensional, you'll have some vague idea of what the Riemann tensor is about. Space time is 4 dimensional, so the riemann tensor has 4x4x4x4=256 components, but most of them are zero or the same. There's only about 20 or so unique values of it in a 4d space-time. I forget how many unique componets are in 3d space, and in 2d there is only one non-zero component.

What we are relying on here is the idea that 2 dimensional surfaces have only one non-vanishing curvature component, which is given by the sectional curvature formula Then the wiki formula above tells you how to get this value at any point on your piece of large paper.

But the point is that the sectional curvature is varying with position. Thus, If you have a large sheet of paper, the sectional curvature will vary with the position on the paper, because parts of the paper will be far away from the black hole and other parts will be close. The "small paper" works when the Riemann tensor components don't change much in the paper.

You will find formula for the sum of the angles of a spherical triangle being greater than 180, and the amount the sum is greater than 180 will depend on the area of the triangle. This pretty much carries over to the case of interest.

Wiki covers this briefly in https://en.wikipedia.org/wiki/Spherical_geometry. Here is a brief summary of some (not all) of the points wiki makes about spherical triangles and the sum of angles of spherical triangles.

The angle sum of a triangle is greater than 180° and less than 540°.

The area of a triangle is proportional to the excess of its angle sum over 180°.

Two triangles with the same angle sum are equal in area.

So this method lets us get the sum of angles of a "small" triangle, where the Riemann curvature tensors components are constant within the triangle. For Schwarzschild geometries, I believe the components are proportional to 1/r^3, so you want r not to vary much.

I don't have anything rigorous for "large sheets of paper", as I said. But a model comes to mind.

Imagine have a piece of paper that's flat at infinity with a "bump" in the middle. The "bump" represents the curved sections, where the geometry isn't euclidean, and in the envision model it's basically spherical. (You could also have a saddle surface which needs hyperbolic geometry, but I'm not imaginig that case).

Using this model, we can see that probably a large triangle can be oriented so most of its edges are in the "flat" portion, and the angles of the large triangle will sum to 180, but you can change this by having parts of the triangle pass through the section with a high curvature. You can break the large triangle up into a lot of small triangles and use the small triangle formula to get the sum-of-angles of the small triangles, but I'm not sure how to get anything useful out of that at the moment.