I have a question as follows, which I hope can be related to the thought/calculation process for general relativity, please:

We suppose that a man and a woman start walking due North at the same time. The man starts walking North from London, and the woman starts walking North from Paris. They carry sophisticated "distance meters" with them that can tell them the how far apart they are.

The distance meter reads 300 km as they begin their long journey to the North. Puzzlingly, some hours later the meter reads only 290 km. But they have both been walking completely North, so how can this be? After some more hours, the distance meter again reads a smaller number 280 km.

Then, the astute couple begins to define between themselves a new mysterious "attractive" force between them as follows: Rate of change of distance meter readout = Cumulative distance traveled from starting city + (speed of walking * time till next reading) * some-proportionality-constant.

My question is, where is the Ricci curvature hidden in this primitive "theory"? Is it absolutely indispensable to first derive the metric tensor for the sphere of Earth radius, followed by the Christoffel symbols, followed by the Riemann curvature tensor, followed by the Ricci curvature in order to have a predictive theory for their distance meter readings as they each travel North?

More seriously, what are the indispensable insights gained from each of the following steps for this particular example: Metric Tensor -> Christoffel Symbols -> Riemann Curvature -> Ricci Curvature in order to make useful predictions in this case?

The real metric tensor in this case (combining the London Bridge, the Eiffel Tower, asphalt roads, the Thames, houses, swimming pools, skyscrapers, etc) can probably never be solved analytically, is this correct?

How can we solve this particular situation from the metric tensor to get the size of the "attractive" force between the man and woman as they travel North?

Thank you.


2 Answers 2


Your example is basically the motion of test particles constrained to a 2D surface (in this case, the surface of the Earth) in flat spacetime. If the surface in question is sufficiently smooth, then we can define a spatial metric $\gamma$ which is inherited from the embedding of the surface in flat 3D space.

Here's how this works for the 2-sphere. We can coordinatize points on the 2-sphere using the angular coordinates $(\theta,\phi)$ such that $$\pmatrix{x(\theta,\phi)\\y(\theta,\phi)\\z(\theta,\phi)} = \pmatrix{R\sin(\theta)\cos(\phi)\\ R\sin(\theta)\sin(\phi) \\ R \cos(\theta)}$$ The 2-sphere is defined by the equation $x^2+y^2+z^2 = R^2$. The line element in flat 3D space is given by $\mathrm d\sigma^2=\mathrm dx^2+\mathrm dy^2+\mathrm dz^2$. Plugging in the coordinate expressions above and doing some algebra yields $$\mathrm d\sigma^2 = R^2 \mathrm d\theta^2 + R^2\sin^2(\theta) \mathrm d\phi^2$$ which is the induced line element on the 2-sphere. The components of the corresponding induced metric $\gamma$ can be read off as $$\gamma_{ij} = \pmatrix{R^2 & 0 \\ 0 & R^2\sin^2(\theta)}$$

The full spacetime line element and metric in the coordinates $(t,\theta,\phi)$ are then simply $$\mathrm ds^2 = -c^2\mathrm dt^2 + R^2 \mathrm d\theta^2 + R^2\sin^2(\theta) \mathrm d\phi^2$$ $$g_{ij} = \pmatrix{-c^2 & 0 & 0 \\ 0 & R^2 & 0 \\ 0& 0 & R^2\sin^2(\theta)}$$ and the inverse metric components are $$g^{ij} = \pmatrix{-\frac{1}{c^2} & 0 &0 \\ 0 & \frac{1}{R^2} & 0 \\ 0 & 0 & \frac{1}{R^2\sin^2(\theta)}}$$ From here we can compute the Christoffel symbols, which is a straightforward exercise (the only non-constant component of the metric tensor is $g_{\phi\phi}$, so almost all of them vanish). That's all we need for the geodesic equation, so if we want to understand the motion of test particles then we're basically done.

What are the indispensable insights gained from each of the following steps for this particular example: Metric Tensor -> Christoffel Symbols -> Riemann Curvature -> Ricci Curvature in order to make useful predictions in this case?

Once we have the Christoffel symbols, we don't need the Riemann or Ricci curvatures to predict how particles will move. As in the example above, if you are handed the metric (e.g. if you are considering constrained motion in flat spacetime) then by taking a few derivatives you can compute the $\Gamma$'s, and therefore have the equations of motion for test particles.

The point you may be missing is that in GR, we generally don't have the metric; instead, we start with the stress-energy tensor $T_{\mu\nu}$ which tells us the distribution of energy and momentum in the spacetime we are considering. From there, note that the $\Gamma$'s involve the first derivatives of the (unknown!) metric, while the Riemann tensor involves its second derivatives. The Ricci and Einstein tensors are just algebraic combinations of the above, so the expression $$G_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$ is an elaborate system of second-order, nonlinear, coupled PDE's for the components $g_{ij}$ of the unknown metric.

The typical way to solve these equations is to use symmetry arguments to constrain the form of the unknown metric. For example, assuming a static and spherically symmetric vacuum spacetime leads to coordinates $(t,r,\theta,\phi)$ in which the line element takes the form $$\mathrm ds^2 = -A(r)\mathrm dt^2 + B(r)\mathrm dr^2 + r^2(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2)$$

So the problem of finding the unknown metric has already been reduced to finding two unknown functions of the single coordinate $r$. This is vastly simpler than the general case, in which you have 10 independent components $g_{ij}$, each of which is a function of all 4 coordinates. Plugging this in to Einstein's equations and turning the crank is still a tedious exercise (you can find it worked out in full detail here), but it ultimately yields very simple equations for $A$ and $B$: $$A(r) B(r) = c^2$$ $$\frac{d}{dr}\left(r^2 A'(r)\right) = 0$$ From there, one can demand that this reproduce Newtonian gravity in the large-$r$ limit; the result is the Swarzschild metric. Once you have that, you can compute the Christoffel symbols and obtain the equations of motion for test particles.

So in summary, the entire point of solving Einstein's equations - at least in terms of predicting the motion of test particles - is to obtain the components of the metric, and thereby the Christoffel symbols which appear in the particles' equations of motion. If you already have the metric/$\Gamma$'s, the Riemann/Einstein tensors don't give you any additional insight into how particles will move.

  • $\begingroup$ @ J Murray thank you very much! I think am starting to understand the bigger picture... So, getting from the metric tensor to the Christoffel symbols and then the geodesic equations of motion is really the "trivial" part of GR, and can even possibly be done by symbolic computer, and the main difficulty is actually in the precursor step, of getting from the reference-frame-invariant Stress Energy tensor to get at the metric in the first place, is this correct? $\endgroup$
    – James
    Oct 15, 2021 at 11:59
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    $\begingroup$ @James Yes, and indeed it can easily be done by computer - there are plenty of Mathematica notebooks floating around which include that functionality. Note also that there are plenty of interesting question in GR - such as the propagation of gravitational waves, the nature of singularities and black holes, etc - which are somewhat adjacent to the operational procedure I have, but those topics are more advanced. $\endgroup$
    – J. Murray
    Oct 15, 2021 at 12:23
  • $\begingroup$ @ J Murray thank you again, the answer was really instructive and clears up what the early textbook chapters on GR are about in the bigger context. $\endgroup$
    – James
    Oct 15, 2021 at 12:36

Only the line element on the surface of a sphere is needed which is

$ds^2 = A^2 ( d\theta^2 + \sin^2 \theta d\varphi^2)$

I used for the radius of the sphere the symbol $A$ to make clear that this value is a constant that will not be changed in this problem. If both persons walk exactly towards the North at different coordinates $\varphi_1$ and $\varphi_2$ and same $\theta_S$ ($d\theta=0$ )the distance between them is

$ds^2 = A^2 \sin^2 \theta_S \Delta\varphi^2$

After walking a bit along $\theta$-lines the $\varphi$ coordinate distance $\Delta\varphi$ does not change, but $\theta_S$ changes to $\theta_E$ (we also assume that both walkers walk the same "length" in $\theta$, so that $d\theta=0$):

$ds^2 = A^2 \sin^2 \theta_E \Delta\varphi^2$

We can now compare both distances:

$$\Delta(ds) = A (\sin\theta_S -\sin \theta_E) \Delta\varphi$$

Assuming that both persons start in the northern hemisphere the angles $\theta_S$ and $\theta_E$ are both significantly below $90^\circ$ (which should be the case in Paris and London) we can use linear approximation for $\sin\theta \approx \theta$ we get (actually if the linear approximation is not valid for large $\theta$ the property of the sinus that it is a monotonously increasing function for $\theta < 90^\circ$ can be used to get to the result $\Delta(ds)>0$ ):

$$\Delta(ds) = A (\theta_S -\theta_E) \Delta\varphi$$

As $\theta_S > \theta_E$ we see that $\Delta(ds)>0$, i.e. the initial distance between both walkers is larger than the final one. That's the correct answer.

A force obtained from the curvature is not needed. The use of curvilinear coordinates may lead sometimes to surprising results because we are used to cartesian coordinates. But curvilinear coordinates can have any form one can imagine as long as they are differentiable.

  • $\begingroup$ thank you! "A force obtained from the curvature is not needed," Yes, I thought the relation here should have been quite simple too, though reading through the early chapters on general relativity, one would have thought it's an accounting course for indices... have Einstein heard of python? :) In any case, suppose if one were to do the whole-nine-yards calculations to get curvature first from the line element, and then calculate the distance on manifold between the two as a function of time and so forth, would the final expression for the answer turn out to be the same? Thank you. $\endgroup$
    – James
    Oct 15, 2021 at 9:32
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    $\begingroup$ @James Actually I was a bit too fast. It is not really correct, but the result is the same. $\endgroup$ Oct 15, 2021 at 9:48
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    $\begingroup$ @James I elaborated the answer and corrected the first post which was posted too swiftly --- sorry for that. But the final answer is the same. $\endgroup$ Oct 15, 2021 at 10:14
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    $\begingroup$ @James Simple questions have in most cases simple answers (BTW also an insight of A. Einstein). $\endgroup$ Oct 15, 2021 at 10:16
  • $\begingroup$ @ Frederic yes, I'm just learning how to calculate with these "aparatus" of general relativity and how to get concrete results using them... $\endgroup$
    – James
    Oct 15, 2021 at 10:31

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