I Measuring Space Curvature w/ Swarzschild Coordinate System: Q&A

[andrewkirk]

I have some questions about the curvature of space (NB not of spacetime) near a planet like Earth. Unambiguously defining space curvature requires choice of a coordinate system, so I choose the Swarzschild system. Here are my questions:

1. Would constant-time hypersurfaces under the Swarzschild metric exhibit spatial curvature near Earth?
2. Given current equipment accuracy, how large a triangle would be needed to reliably indicate the existence of such curvature by observing the angle excess of the triangle? If the triangle needs to be so large as to not fit within the Earth's atmosphere, I imagine using three laser beams making the sides of the triangle, fired from space stations that are at the triangle's vertices.
3. Has such an experiment been conducted, and did it verify the spatial curvature of the hypersurfaces?
4. If the answer to 1 is Yes, is there any coordinate system S that produces a foliation under which there is a nonzero open interval $I = (t1,t2)$ of time and a nonzero open volume $V$ of space such that, for all $t\in(t1,t2)$ the region $\{t\}\times V$ is spatially flat? In short, is there a coordinate system that can flatten out the spatial curvature within a nonzero volume and a finite time interval?
5. Is there a more appropriate coordinate system than the Swarzschild for this question? If so, what is it and what would the answers to 1-3 be for that?

I expect 1 and 4 could be answered by doing some extended calculations from Einstein's equation, but I think they'd be long and I'm hoping they may have already been done somewhere.

Thank you very much for any answers.

Andrew

 

[PeterDonis]

Would constant-time hypersurfaces under the Swarzschild metric exhibit spatial curvature near Earth?

Yes.

Given current equipment accuracy, how large a triangle would be needed to reliably indicate the existence of such curvature by observing the angle excess of the triangle?

A rough estimate of the fractional variation due to spatial curvature is $GM / c^2 R$, where $M$ is the mass of the Earth and $R$ is the distance from the center at which the measurements are being made. For measurements near the Earth's surface this works out to about 1 part in a billion.

Has such an experiment been conducted, and did it verify the spatial curvature of the hypersurfaces?

Not to my knowledge. The effect is very small so it would be hard to measure directly.

As for indirect measurements, the obvious one is the bending of light by a planet or star. Roughly speaking, half of the bending predicted by GR is due to space curvature, so if space curvature were not present, the observed bending would only be half of the GR prediction. The GR prediction has certainly been confirmed for the Sun, to enough accuracy to rule out alternative theories that only predict half the bending. I don't know if it's been confirmed for the Earth.

is there a coordinate system that can flatten out the spatial curvature

Yes, Painleve coordinates will do it.

Is there a more appropriate coordinate system than the Swarzschild for this question?

It depends on what you mean by "appropriate". What picks out Schwarzschild coordinates is that the surfaces of constant time are orthogonal to the worldlines of static observers (observers who stay at the same altitude above the gravitating mass), and those worldlines are the integral curves of the timelike Killing vector field that makes the spacetime itself static. So those surfaces of constant time are the "natural" ones for the spacetime, and static observers are the "natural" observers.

The surfaces of constant time in Painleve coordinates are orthogonal to the worldlines of observers who are freely falling from rest at infinity. So those surfaces are the "natural" ones for such observers. It's rare for such observers to be considered when looking at a planet or a star, but when we talk about black holes things change, because at and inside a black hole's horizon there are no static observers.

 

[pervect]

I would estimate the angular excess of a triangle as being something proportional to $\frac{rs}{R^3}A$ where $r_s$ is the Schwarzscild radius of the object, and A is the area of the triangle. For the Earth, $r_s$ is about 9 millimeters.

The minimum value of R would be the radius of the Earth, 6.3 million meters, a the formula won't work inside the Earth.

This could be wrong, but that'd be my first guess. It's motivated from the angular excess of a spherical triangle, <<link>>, which is the sum of the angles of a spherical triangle minus pi. This is proportional to the area of the spherical triangle multiplied by some constant. At this point, I'm not worried about the value of this constant.

$\frac{r_s}{r^3}$ is the Gaussian curvature, so I'm basically estimating that we multiply the Gaussian curvature by the area of the triangle and some constant to get the angular excess, assuming the Gaussian curvature stays fairly constant inside our traingle.

The problem with doing this experimentally is drawing a spatial geodesic, the shortest distance between two points in the curved space. I'm pretty sure the spatial projection of a light beam will not be such a spatial geodesic.

If we use light beams rather than difficult-to-experimentally realize spatial geodesics, we are basically doing some variant of the deflection of light experiment. Motivationally, I'd expect we'd see the spatial curvature effects of the same order of magnitude as so-called "Newtonian" gravitational deflection of light, as we see in other light deflection experiments where GR basically doubles the deflection. See the usual arguments for the meaning of what "Newtonian" deflection really means in this context.

Note that this is really off the top of my head and not a careful analysis. I may take some time to think it over more if I have time.

The setup I'm imagining is that $\frac{r_s}{r^3}$ is essentially constant over some path. The not-off-the-top-of-the-head way to calculate it is to use the parallel transport equations to parallel transport a vector around the triangle. For this to work, the angular excess needs to be the same as the net rotation of a vector transported around the closed curve of the triangle. I think this is true, I don't have a textbook reference for it though.

The necessary Riemann tensor components are basically multiples of $\frac{r_s}{r^3}$. If we set up an orthonormal basis for the Schwarzschild metric, basically $R_{\hat{x}\hat{y}\hat{x}\hat{y}} = R_{\hat{t}\hat{z}\hat{t}\hat{z}}$

[add]. A bit of an afterthought, but I should mention that I'm using geometric units with c=1. Without using geometric units, there would be some extra factors of $c^n$ in here.

The "hats" are because I'm using an orthonormal basis, not a coordinate basis.

 

[pervect]

Some more comments. The mathematical aspects of splitting space-time into a spatial part can be described in various ways, but one of the most fundamental is by a projection operator that maps the 4-d space-time into a 3-d space.

For the case in question, the projection operator can be made particularly simple. If we take the Schwarzschild metric:

$$ds^2 = -(1-2m/r)dt^2 + \frac{dr^2}{1-2m/r} + r^2 d\theta^2 + r^2 \sin^2 \theta \, d\phi^2$$

Then if we consider surfaces of constant time, $dt^2$=0, and the projected 3-metric is just:

$$ds^2 = \frac{dr^2}{1-2m/r} + r^2 d\theta^2 + r^2 \sin^2 \theta \, d\phi^2$$

This projection operator choice map maps (t,r,theta,phi) into (r,theta,phi) - it basically eliminates time, leaving the spatial part of the metric. This is not the only possible way to split space-time into space+time, but it's a natural way of splitting it for a static observer.

Using this 3-metric, we can compute the Riemann components.

What do we do with them when we get them? From the analysis using the easier to describe spherical trignometry, rather than the Riemanian approach, we expect that the angular excess of a triangle will be proportional to the area of the triangle, and the constant of proportionality will be related to the Gaussian curvature of the surface.

How do we relate angular excess of a triangle to the rotation of a vector? Let's work with spherical triangle - or more generally, a spherical polygon - in two dimensions as an easy case to analyze. We construct a polygon from geodesic segments. If we consider some starting segment, and take it's tangent vector, when we parallel transport this tangent vector along that geodesic segment, it doesn't rotate, as all geodesic parallel transports their own tangent vectors. At every vertex of the polygon, we rotate the vector to point along the next segment, and add the amount of rotation to a sum we accumulate. This sums adds together the exterior angles of the polygon, not the interior angles. The relationship is that the exterior angle is 180 degrees minus the interior angle.

When we transport the vector completely around the closed polygon, if we had no curvature the vector would be rotated through 360 degrees. In the presence of curvature, the vector will not be rotated 360 degrees, and the difference will be the angular excess or debit (I'm not quite sure of the sign here).

How do we tie this back into the Riemann formula? If we have two unit, orthogonal vectors, $\hat{x}$ and $\hat{y}$ so that $g_{ab} \hat{x}^a\hat{x}^b = g_{ab} \hat{y}^a \hat{y}^b = 1 \quad g_{ab}\hat{x}^a\,\hat{y}^b = 0$ then, regardless of our choice of basis, the amount of rotation of a vector transported around a unit area loop in the plane defined by our two vectors $\hat{x}$ and $\hat{y}$ will be

$$R_{abcd} \, \hat{x}^a \hat{y}^b \hat{x}^c \hat{y}^d$$

There's probably a much more elegant way of doing this with bivectors and two-forms, but I think the above approach is more accessible. Written this way, we don't have to specify a particular choice of basis, leaving us open to using any desired choice of basis, including a coordinate basis if desired or computationally convenient. Note though that we do need scale $\hat{x}$ and $\hat{y}$ so that they are of unit legnth in whatever basis we choose.

That sums up the mathematical aspects. For the physical aspects, I suppose that I like to consider a stretchy string as a physical mechanism that will find the curve of shortest length between two points in space. The string has a higher energy when it's stretched, and it adjusts its configuration to the lowest energy state which is the shortest path. However, any such stretchy string that we can actually make will have mass, which means that gravitational fields or tidal gravity will spoil this length-minimizing property. Mathematically, we can construct a frame-dependent idealization of what one might call a "massless string" that would follow the desired geodesic path, but I don't think there's any direct physical way of generating our geodesics. To physically measure the angular excess or debit, we'd need to be able to physically construct the geodesic paths somehow.

This leads us back to methods that measure space curvature from light deflection. One of the common treatments uses PPN theory, where "space curvature" is encoded as the PPN parameter beta. Several experements are sensitive only to this particular PPN parameter, so I believe it's reasonable to regard such experiments as "measuring space curvature".