# How to measure the curvature of the space-time?

I know G.R. change our vision of space and time as a unique surface than can bend. We can associate the curvature of the space-time as the gravity created by the mass of planets, stars... But how can we measure the curvature of the space-time for calculate, as an example, the position of a star?

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## 5 Answers

If you want a direct, physical measurement of curvature, here's a plan that would take lots of money and decades, possibly centuries to set up. Perfect for physics!

What you need are three satellites equipped with lasers, light detectors, precision aiming capabilities, and radio communication. These three satellites are launched into space and position themselves far away from each other so that they form the points of a very large triangle. The satellites then turn on two lasers, aiming each one at the other two. Each satellite reports to the others when it is receiving the laser light. Once the satellites are all reporting that they see the laser light from the others, they measure the angle between their own two laser beams. Each satellite transmits this angle back to headquarters on Earth.

The overall curvature of space can be determined from these angles. If the sum is 180 degrees, like you learned in geometry class, then the space around the satellites is flat.

If the sum is more than 180 degrees, then space has positive curvature there, like the surface of a sphere. You can picture the situation on Earth by drawing a line from the North Pole to the equator, continuing a quarter way around the world along the equator, then heading back to the North Pole. You've just drawn a triangle with three 90 degree angles for a sum of 270.

If the sum of the angles is less than 180, the region of space has negative curvature like a saddle.

Let's say the satellites are surrounding a star. Since light bends towards masses, the satellites will have to aim away from the star so that the light will bend around the star and hit the other satellites. This means that the angles of the resulting triangle will be larger than normal (i.e., flat space), meaning the sum will be greater than 180 degrees. Thus, we can conclude that space has positive curvature near a mass.

The exact relation between the sum of the angles of the triangle and the total curvature inside that triangle is given by $$\sum\limits^3_{i=1} \theta_i = \pi + \iint_T K dA$$ where $\theta_i$ is the angle measured at each satellite (measured in radians), $T$ is the 2D triangular surface defined by the three satellites being integrated over, $K$ is the Gaussian curvature at each point in the triangle, and $dA$ is the infinitesimal area with curvature $K$. For a region of space with zero total curvature, the angles will sum to $\pi$ radians (180$^\circ$). Positive curvature leads to a sum larger than $\pi$, negative curvature to a sum smaller than $\pi$.

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Mark H: "Once the satellites are all reporting that they see the laser light from the others, they measure the angle between their own two laser beams." -- What, exactly, do you mean by "measur[ing] the angle between ... two ...", please ? Something similar to what's described in this answer to the question "What is the notion of a spatial angle in general relativity?" ?? – user12262 Apr 24 '14 at 20:48
user12262: I mean measuring with something like a protractor. The laser beams are two lines in space that are heading off in different directions. I want the angle between them. A goniometer (en.wikipedia.org/wiki/Goniometer) would probably be more appropriate for the satellites. – Mark H Apr 25 '14 at 2:51
Mark H: "I mean measuring with something like a protractor [... or] goniometer" -- Do you mean that some kind of protractor might be of any use before it was measured whether/how the spacetime region was curved which cointained the various separate parts of the given kind of protractor (i.e. several distinct "markings on the rim" and the separate "center of the protractor") ? IOW: How do you suppose that specific number values ought to be attached to specific "markings on the rim" before having finished the curvature measurement? – user12262 Apr 25 '14 at 5:03
Might be worth mentioning that Gauss did a version of this experiment; we was immersed in the study of non-Euclidean geometry at the time and he may have been seeking evidence for spatial curvature in doing the measurement described here – WetSavannaAnimal aka Rod Vance Jun 17 '15 at 11:28

Measuring Curvature

Curvature can be quantified by many tensors, and their various contractions give rise to a plethora of scalars describing curvature. In general relativity the most common are,

• Riemann curvature tensor, $R^{a}_{bcd}$ which measures to what extent the metric is not isometric to flat Euclidean space. In another manner, it measures the failure of parallel transportation.
• Ricci tensor, $R_{ab}=R^{c}_{acb}$ which appears directly in the field equations of general relativity.
• Ricci scalar, $R=g^{ab}R_{ab}$, which, informaly, when positive at a particular point suggests a ball around the point has a smaller volume that a ball of equal radius in Euclidean space.

The Weyl tensor $C_{abcd}$ measures the tidal forces experienced along a geodesic. In addition, in dimensions $d\geq 4$, the vanishing of the Weyl tensor indicates the metric is conformally flat, i.e. a conformal transformation which changes the metric by an overall factor,

$$g_{ab}\to\Omega^2(x)g_{ab}$$

can be used to make the metric flat. Furthermore, the tensor governs whether radiation may propagate through spacetime with no matter content. Finally, the Ricci decomposition can be used to express the Riemann curvature in terms of various other curvature forms, which includes a Weyl piece.

Matter and Curvature

The Einstein field equations relate the curvature of a spacetime manifold to the matter content:

$$R_{ab}-\frac{1}{2}g_{ab}R + \Lambda g_{ab}=8\pi G T_{ab}$$

where $\Lambda$ is the cosmological constant, and $T_{ab}$ is the stress-energy tensor describing the matter content, which is derived from the Lagrangian of the matter. As an example, suppose we had some matter with a particular energy, then one of the entries would be $T_{00}=\epsilon$, where $\epsilon$ is the energy density. A general decomposition of the tensor is given here. Often we are not able to solve the equations exactly for a particular system. In this case we resort to perturbation theory, where we linearly perturb a known background, i.e.

$$g_{ab}\to g_{ab}+h_{ab} +\mathcal{O}(h^2)$$

Movement of Test Particles

The movement of a test particle on a particular spacetime is governed by the geodesic equation,

$$\frac{\mathrm{d}^2 x^\alpha}{\mathrm{d}s^2} + \Gamma^\alpha_{\beta \gamma} \frac{\mathrm{d}x^\beta}{\mathrm{ds}}\frac{\mathrm{d}x^\gamma}{\mathrm{d}s} = 0$$

where $\Gamma^{a}_{bc}$ are the Christoffel symbols, i.e. the Levi-Civita connection expressed in the coordinate basis, with the assumption that the metric tensor is torsionless. The general concept is that geodesics are the natural paths test particles will follow along a manifold. For example, for flat Minkowski space, the connection vanishes, and hence we obtain $x(s)$ is a linear function as expected, as $x''(s)=0$.

Physical Example

The Schwarzschild metric describes a spherical matter content, which is static, i.e. with a $\partial_t$ Killing vector and $SO(3)$ symmetry; e.g. it could describe the Earth, or a black hole:

$$\mathrm{d}s^2 = \left( 1- {2GM \over r}\right) \mathrm{d}t^2 - \left( 1- {2GM \over r}\right)^{-1}\mathrm{d}r^2 - \underbrace{r^2 \mathrm{d}\theta^2 -r^2 \sin^2 \theta \mathrm{d}\phi^2}_{\text{metric on the sphere}}$$

Simply by inspection, we see it is already in agreement with a classical understanding of gravitation as it possesses a singularity at $r=0$, as does the classical relation $F \propto r^{-2}$. The singularity at $r=2GM$ reflects that we have chosen an inappropriate coordinate system. It is not a curvature nor conical singularity, the curvature forms can be shown to be non-singular at $r=2GM$.

Evidence for General Relativity

The recent discovery of gravitational waves provides empirical evidence supporting the theory which predicted their existence. A gravitational wave may travel at the speed $c$, but also below depending on the amplitude. Essentially, it employs spacetime itself as a medium. A particular wave metric:

$$\mathrm{d}s^2 = \mathrm{d}t^2 -\mathrm{d}r^2 + H(t-r,x^1,x^2)(\mathrm{d}t-\mathrm{d}r)^2 - \mathrm{d}(x^1)^2 -\mathrm{d}(x^2)^2$$

where conditions on $H$ are determined by the demand that the stress-energy tensor vanishes to ensure the waves are truly purely gravitational, rather than, for example, electromagnetic. Specifically, $\Delta H = 0$, where $\Delta$ is the Laplacian with coordinates $x^1,x^2$.

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I feel that this probably goes way over OP's head, but nice job! – Danu Apr 23 '14 at 20:02
One problem that I do have with this answer is that the question by OP: "How do we actually measure curvature?" was not really addressed... I think there should be a section on experimental methods/evidence (e.g. from the CMB?) as well as mathematical background. – Danu Apr 23 '14 at 20:04
@Danu: Oops, I thought when he said 'measure curvature,' he meant in the mathematical sense. Will add a note on experiments shortly. – JamalS Apr 23 '14 at 20:10
@Danu: Added gravitational waves as an example of empirical verification. – JamalS Apr 23 '14 at 20:26
This is false: "For example, for flat space, the connection vanishes, and hence we obtain $x(s)$ is a linear function as expected, as $x′′(s)=0$." Perhaps you might specify Minkowski/Cartesian coordinates. Your Schwarzschild metric is wrong and the discussion following it makes no sense. In the static weak-field limit, reproducing Newtonian low-velocity orbits requires the coefficient of $\mathrm{d}t^2$ to be $1+2\Phi$ to first order in Newtonian potential $\Phi$, and so not $\propto r^{-2}$. On a more minor note, Schwarzschild geometry in isotropic coordinates would work better for this. – Stan Liou Apr 23 '14 at 20:47

Following Synge's description of a "five-point curvature detector" 1 and its generalization (e.g. as indicated in [2]), the curvature of five participants ($A$, $B$, $N$, $P$, $Q$) who were and remained (chrono-geometrically) rigid to each other (i.e. finding constant ping duration ratios) is measured as the real number value $\kappa_5$ for which the corresponding Gram determinant vanishes; i.e. as solution of

0 = $\Tiny{ \begin{array}{|ccccc|} 1 & \text{Cos[} \sqrt{ \kappa_5 \frac{A \, B}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{A \, N}{A \, B} \frac{A \, N}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{A \, P}{A \, B} \frac{A \, P}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{A \, Q}{A \, B} \frac{A \, Q}{B \, A} } \text{]} \, \, \\ \, \, \text{Cos[} \sqrt{ \kappa_5 \frac{B \, A}{A \, B} } \text{]}& 1 & \text{Cos[} \sqrt{ \kappa_5 \frac{B \, N}{A \, B} \frac{B \, N}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{B \, P}{A \, B} \frac{B \, P}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{B \, Q}{A \, B} \frac{B \, Q}{B \, A} } \text{]} \, \, \\ \, \, \text{Cos[} \sqrt{ \kappa_5 \frac{N \, A}{A \, B} \frac{N \, A}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{N \, B}{A \, B} \frac{N \, B}{B \, A} } \text{]} & 1 & \text{Cos[} \sqrt{ \kappa_5 \frac{N \, P}{A \, B} \frac{N \, P}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{N \, Q}{A \, B} \frac{N \, Q}{B \, A} } \text{]} \, \, \\ \, \, \text{Cos[} \sqrt{ \kappa_5 \frac{P \, A}{A \, B} \frac{P \, A}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{P \, B}{A \, B} \frac{P \, B}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{P \, N}{A \, B} \frac{P \, N}{B \, A} } \text{]} & 1 & \text{Cos[} \sqrt{ \kappa_5 \frac{P \, Q}{A \, B} \frac{P \, Q}{B \, A} } \text{]} \, \, \\ \, \, \text{Cos[} \sqrt{ \kappa_5 \frac{Q \, A}{A \, B} \frac{Q \, A}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{Q \, B}{A \, B} \frac{Q \, B}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{Q \, N}{A \, B} \frac{Q \, N}{B \, A} } \text{]} & \text{Cos[} \sqrt{ \kappa_5 \frac{Q \, P}{A \, B} \frac{Q \, P}{B \, A} } \text{]} & 1 \, \, \end{array} }$,

where the chronometrically measured ping duration ratio $\frac{A \, B}{B \, A}$ is the ratio between $A$'s duration from having stated a signal indication until having observed the corresponding reflection (or "ping echo") from $B$ and $B$'s duration from having stated a signal indication until having observed the corresponding reflection from $A$,
the chronometrically measured ping duration ratio $\frac{A \, N}{A \, B}$ is the ratio between $A$'s duration from having stated a signal indication until having observed the corresponding reflection from $N$ and $A$'s duration from having stated a signal indication until having observed the corresponding reflection from $B$,
the chronometrically measured ping duration ratio $\frac{A \, N}{B \, A}$ is the ratio between $A$'s duration from having stated a signal indication until having observed the corresponding reflection from $N$ and $B$'s duration from having stated a signal indication until having observed the corresponding reflection from $A$, with $\frac{A \, N}{B \, A} := \frac{A \, N}{A \, B} \frac{A \, B}{B \, A}$, and so on.

Similarly, curvature values $\kappa_n$ may be determined for any $n$ mutually rigid participants (sensibly for $n \ge 4$).

Whether and how such measured "discrete" curvature values of $\kappa_n$ in general and of $\kappa_5$ in particular may be related to curvature tensors (or corresponding scalars) which are considered in differential geometry has been asked here already.

EDIT (in response to comments):

As an illustration of the simplest case, the value $\kappa_4$ expressing "discrete" curvature for 4 given participants who are mutually rigid (as generally required) and who are even at rest to each other, i.e. such that $\frac{A \, B}{B \, A} = 1$, $\frac{A \, C}{C \, A} = 1$, and so on,
consider the "spherical quadrangle with vertices $A_1$, $B_1$, $I_{20}$, $I_{10}$" in this picture:

If the required values of ping duration ratios between those four participants are equal to the ratios of great circle arclengths between those vertices,
then the value $\kappa_4$ for which the corresponding 4th order Gram determinant vanishes equals

$\kappa_4 = \left( \frac{A_1 \, B_1}{ R/c } \right)^2$,

where $R$ is the radius of the sphere, and $c$ the "speed of light".

References:

[1.] J.L.Synge, "Gravitation. The general theory"; ch. XI, §8: "A five-point curvature detector".

[2.] S.L.Kokkendorff, "Gram matrix analysis of finite distancé spaces in constant curvature" (Discrete and Computational Geometrie 31, 515, 2004).

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Don't take this the wrong way, but this seems like it was written in the most technically complicated way to answer the question without any regard to the ability of the OP or of other interested persons to understand it. Consider writing a summary of your answer beneath it that outlines the details of your answer in a technical level more appropriate for the probable audience. – Jim Apr 24 '14 at 0:23
@Jim: "[...] without any regard to the ability of the OP or of other interested persons" -- Do you possibly feel overtaxed by my referring to the notion of duration? Or (even simpler!) the notion of "duration ratios" (i.e. plainly measured real numbers, which in the case considered were even supposed to be constant)? Do you presume that the OP and the audience at large couldn't handle some arithmetic with those numbers written as Gram determinants ? ... [to be continued] – user12262 Apr 24 '14 at 4:51
Jim: "Consider writing a summary of your answer beneath it that outlines the details [...]" -- I am considering doing that; adding to the references and links given in my answer already. Now, please don't take it the wrong way, but are you going to submit similar requests to people/answers who/which, unashamedly on a website dedicated to physics, addressing a question about measurement, referred to notions of differential geometry, thereby seemingly taking for granted comprehension of "manifolds", "tangent spaces", "metric tensors", etc. ?? I might! (Effectively, I am already.) – user12262 Apr 24 '14 at 4:52
Great answer in terms of actual measurements. Actually what I miss is a definition of which curvature it measures - Ricci scalar? – Keith Apr 24 '14 at 5:04
@user12262 I asked not for my benefit but for those who would come here looking for a simpler answer. And yes, I do intend to ask that of other users – Jim Apr 24 '14 at 13:24

There are measurable effects of the space-time curvature due to a star - the precession of the perihelion of Mercury and the bending of light as it passes near the sun, for instance. Indeed these were the first effects of GR to be experimentally confirmed.

There are even measurable effects of the space-time curvature due to the Earth: the redshifting of light as it climbs out of the gravity well, time dilation of ground-based clocks relative to ones in orbit (which GPS has to correct for), and the frame-dragging recently measured by Gravity Probe B.

All of these things can act as measures of the degree of space-time curvature near an object.

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You don't even need to go to orbit to measure time dilation: putting an atomic clock on an international airline flight or even driving one up a mountain is more than enough. – Matt Nordhoff Apr 25 '14 at 2:26

∆A+∆B+∆C = 180 in flat space. However, a triangle which was measured around the earth would have angles that sum up to an answer which is less than 180º. This is because space is bent downwards like a saddle by mass.

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While a true statement, I don't think that this really answers the question of how curvature is actually measured. – Kyle Kanos Dec 1 '14 at 3:32