This is a cache of https://www.gregegan.net/ALLSKIES/01/Gravity.html. It is a snapshot of the page at 2021-09-09T02:43:34.642+0000.
Gravity in Multiply Connected Space — Greg Egan

# The Book of All Skies

## Gravity in Multiply Connected Space

### Multiply Connected Space

The ordinary space we inhabit is usually assumed to be simply connected. This means, roughly, that in empty space if you take a string that connects two points, A and B, and someone else does the same, then by moving the strings around — without detaching them from their fixed endpoints — and stretching or shortening them as necessary, you could make the two of them coincide along their entire lengths. In a certain sense, if we don’t care about the exact shapes of different paths, there is only one way to get from A to B.

Infinite Euclidean space is simply connected. So is the surface of a sphere. But the surface of a torus, or doughnut, is not: if you wrap one string around the torus the long way once, and wrap another one twice, then even if they start and end at the same points, you will never be able to change the number of times they wrap around the hole, so they can never be made identical.

A space that is connected (in the sense that there is always a path between any two points), but not simply connected, is known as a multiply connected space, because there are multiple, fundamentally different ways to get from A to B.

If we choose A and B to be the same point, then we can always get from A to A by doing nothing: following a “path” of zero length that goes nowhere but A. Then the difference between a simply connected space and a multiply connected space boils down to whether or not there are any other ways of getting from A to A: any loops through the space that can’t be contracted down to a single point. In The Book of All Skies, the protagonist lives on a world that has formed in the presence of two structures known as the Hoops, which are disks that connect different “skies”: different expanses of otherwise ordinary space. If we ignore one of the Hoops and just consider the effects of a single one, the multiply connected space it gives access to comprises a certain number N of copies of ordinary space, numbered 1, 2, 3, ..., N, which have been spliced together as follows: if you pass through the Hoop in space i in one direction, you emerge from the Hoop in space i+1, and if you pass through the Hoop in space i in the other direction, you emerge from the Hoop in space i–1. Whenever this recipe suggests you would go higher than space N or lower than space 1, you simply wrap around to space 1 or space N respectively.

To give a simple illustration, in the image on the right there are just three copies of ordinary space, with the Hoop that joins them in black. The precise way they are connected is illustrated by the coloured curves, which together comprise a single loop. If you join up these curves by matching the colour at their endpoints, you can see that moving clockwise around the red/green curve at the bottom takes you to the green/purple curve in the middle, and from there to the purple/red curve at the top, which connects back to the start of the red/green curve at the bottom.

This loop can’t be shrunk down to a point, because if you tried to shrink it, it would have to cut across the edges of the Hoop, which are taken to be singularities that are not actually part of the space.

The purpose of this page is to discuss the way that Newtonian gravity can be adapted to a multiply connected space like this. Of course, General Relativity is a more accurate theory of gravity, and that theory is necessary to fully understand all the possibilities for spaces with exotic topologies. But if we treat the Hoops as we have described them as a fait accompli that splice together regions of (approximately) flat space, containing bodies of a similar size and composition to the planets in our own solar system, then Newtonian gravity is a reasonable approximation for how we would expect gravity to work under those circumstances.

Now, in much of what follows, we will actually be talking about the way that electrostatics — the study of electric fields due to motionless electric charges — would work in the same kind of flat, multiply connected space. The underlying mathematics is essentially the same: in ordinary space, the gravitational attraction between particles with mass, and the electrostatic attraction between oppositely charged particles, both follow an inverse square law, with the force of attraction inversely proportional to the square of the distance between the particles. But why bother switching to electrostatics as a proxy for gravity, if the mathematics is identical, and gravity is what we are really interested in?

There are two reasons to do this. The first is that some of the mathematical tricks we will use make the most intuitive sense if we describe them in terms of familiar electrical phenomena, like conductive metals and oppositely charged particles. The gravitational analogues of these things are either wildly exotic, or flat-out physically impossible, which would make it distracting to talk in those terms, even though the final results we end up with still apply to the gravitational field.

The other reason is that the theory of electrostatics in multiply connected spaces actually has real-world applications in ordinary space, and those applications motivated most of the mathematics that we’ll use to solve the problems that interest us. And even in the imaginary world of the novel, where the gravitational field in a multiply connected space is accessible to observation and experiment, some of the electrostatic analogues are still extremely useful, since they are accessible to simpler, smaller-scale experiments.

### The Potential, the Field and the Flux

First, a quick primer on the mathematical ingredients we will be working with. In both electrostatics and gravity, there is a quantity known as the potential associated with the gravitational or electric field produced by a massive body or a charged body respectively. We will define it as the potential energy per unit of mass or charge that a second object possesses by virtue of being at a particular location in space.

To give a simple example, in a uniform gravitational field (which is how we often approximate the gravity on the surface of the Earth over medium distances), the gravitational potential φ is:

φ(z) = g z

where z is the height above the ground in metres and g ≈ 9.8 metres per second squared is the local gravitational acceleration. What this is saying is that an object at height z possesses this much potential energy for each kilogram of its mass. If we let it drop from a height z1 to z2, the potential energy it loses will have to end up in some other form (e.g. kinetic energy, because it accelerates, and/or heat energy due to friction, if it moves through air).

If we move out to a wider scale appropriate for spaceflight and astronomy, the gravitational potential due to a spherical body of mass M in ordinary space is:

φ(r) = –G M / r

where G ≈ 6.67 × 10–11 newtons metres squared per kilogram squared, r is the distance from the centre of the spherical body, and the formula only applies when r is greater than the radius of the body. The minus sign here might be confusing at first glance, but since 1/r decreases as r increases, the minus sign means that the potential increases with increasing altitude (as a negative number that becomes smaller in magnitude), just as it did with z. How can both formulas be correct, though, if the first gives a positive value and the second a negative value? Potentials are not defined in an absolute sense; only the difference between the potential at two different points has any physical meaning. It’s convenient to set the potential at ground level to zero in the first formula, while it’s convenient to set the potential infinitely far from a planet or star to zero in the second formula. But we could add any constant value we liked to either formula, and it would still give the same predictions for how a falling object behaves.

The electrostatic potential outside a uniformly charged spherical body with total charge Q obeys the same kind of law:

φ(r) = Q / (4 π ε0 r)

Here 1/(4 π ε0) ≈ 9 × 109 newtons metres squared per Coulomb squared; a Coulomb is just a unit of charge. Given a field due to the charge Q, this tells us the potential energy per unit of charge of some other charged object. The difference in sign with the gravitational potential reflects the fact that, while masses are always positive and give rise to a gravitational attraction, electrical charges, which can be positive or negative, obey the rule that opposite charges attract and like charges repel. So if Q is positive, for each unit of positive charge another object possesses, its potential energy will increase the closer it comes to the body with charge Q, just as if it were climbing a hill. Given the chance, it will roll down the hill: that is, it will be repelled. But if Q is positive and we place a negatively charged object in the same electric field, the potential energy will decrease the closer the object comes to the body with charge Q, so that object will be attracted.

In both the gravitational and electrostatic cases, the potential is often the easiest quantity to work with to describe the field. But given the potential (a single number at each point in space), we can also derive a three-dimensional vector (at each point in space) that gives a more direct measure of how the field acts on objects. In the gravitational case, what we will call the gravitational field is just the acceleration due to gravity at any point, g. This is a vector, because we don’t just give the amount of acceleration, but also the direction.

In the electrostatic case, we describe the electric field, usually written as E, as a vector that gives us the force per unit of charge on an object. The acceleration that force produces will then depend on the mass of the object. The gravitational field can equally well be described as the force per unit of mass on an object ... which will be equal to its acceleration. In this case, the two roles the mass of the object plays cancel out: it determines how strong the gravitational force is, for a given field, but then it also determines how much force is needed to accelerate the object.

How, exactly, do we get from the potential to the field? If we use ordinary Cartesian coordinates x, y, z, the components of the field vector are just the opposite of the rate of change of the potential in each coordinate direction:

E = (–∂φ/∂x, –∂φ/∂y, –∂φ/∂z)

This kind of vector is known to mathematicians and physicists as (the opposite of) the gradient of the potential, and it can also be written without reference to any coordinate system as:

E = –∇dφ

We won’t get into a detailed discussion of why the opposite of the gradient of potential energy [per unit of mass or charge] should yield a force [per unit of mass or charge], but as an intuitive picture, just imagine the potential φ as describing the varying height of an “energy landscape” where a force drives objects to “roll downhill”.

It is the field, not the potential, that obeys the famous inverse square law of gravity and electrostatics. For electrostatics, this is known as Coulomb’s Law:

E(r) = Q er / (4 π ε0 r2)

where er is a vector of length one that points away from the charged body. In gravitational physics, the equivalent is Newton’s Law of Gravity:

g(r) = –G M er / r2  The diagram on the top right shows the contour lines of constant potential around a positive (red) and negative (blue) point charge adjacent to each other; these lines are for equally spaced values of φ, with φ=0 along the black line in the middle. We have not drawn contour lines closer than a certain distance around each charge, as they would be too closely spaced to convey much information. The same diagram also shows the electric field E drawn as an arrow, at a few points; this illustrates the force on a small positive “test charge” (a charge too small to affect the surrounding field significantly) if it was located at the base of the arrow.

We can see a couple of things on this diagram. One is that the field vector is always perpendicular to the contour lines for the potential. The other is that the size of the field vector depends on how closely spaced the contours are, at the base of the arrow. Both of these properties follow directly from the mathematical definition of the field as proportional to the gradient of the potential.

We can get a less cluttered sense of the field if, instead of trying to draw a suitably sized arrow at a multitude of points, we draw a set of curves known as flux lines; these are the green curves in the diagram on the bottom right. Flux lines are drawn so that the tangent to the curve points in the direction of the field at each point it passes through, but a single flux lines does not convey any information about the strength of the field. However, we can still infer the strength, in either of two ways. If we also draw contours for the potential, then the spacing of those contours indicates the strength of the field. Independently, we can also choose to draw flux lines that are spaced so that the number of them passing through a given area (in a three-dimensional picture) is proportional to the strength of the field. However, if we draw a two-dimensional picture of flux lines because we are only showing a two-dimensional slice through the field — as we are doing here — the spacing can only give us a rough, qualitative sense of the field strength. It is only in three dimensions that the geometry and mathematics can work together to make the density of the flux lines per unit area an accurate measure of the field strength.

In fact, the number of flux lines that cross a given surface is of interest beyond its role in interpreting pictures representing the field. We can define a quantity called the flux, for a given field and a given surface S, as:

flux(S) = ∫S E · n dA

That is, we integrate the dot product of the field vector and n, a unit-length normal to S, across the whole surface, where the integral is weighted by the area dA of each infinitesimal portion of the surface. The dot product here is a continuous, precise version of the quantity we get approximately by simply counting how many flux lines pass through the surface, if we space them out so their density is porportional to the field strength.

For both gravity and electrostatics, there is a wonderful result known as Gauss’s Law, which relates the flux through any closed surface S, like a sphere, with the mass or charge inside the surface. For the gravitational acceleration, we have:

S g · n dA = –4 π G M

where M is the total mass enclosed by the surface S. For the electrostatic field:

S E · n dA = Q / ε0

where Q is the total charge enclosed by S. For a single, spherical massive or charged body, surrounded by a concentric spherical surface S, these results can easily be obtained directly from our previous formulas for the fields in question. But the marvelous thing is that they apply in any situation, no matter how complicated.

### Laplace’s Equation

So far, we have simply stated formulas for the gravitational and electrostatic fields around spherical bodies in ordinary space — that is, simply connected Euclidean space. But our aim is to understand these fields in multiply connected spaces. If we can no longer apply the usual inverse-square-law formulas, what criterion should we be using to decide what the correct solution will be?

The answer to that boils down to a combination of Gauss’s Law, which we have just described, and an equation that the potential obeys in empty space (that is, space containing no significant mass, or charge). This equation is known as Laplace’s equation, and it states that:

2φ/∂x2 + ∂2φ/∂y2 + ∂2φ/∂z2 = 0

That is, if we take the second derivative of the potential with respect to each of the Cartesian coordinates x, y, z, the sum of them all is zero.

This can also be written without reference to any coordinate system as:

2φ = 0

If we write r in terms of Cartesian coordinates:

r = √(x2 + y2 + z2) = (x2 + y2 + z2)1/2

then we have:

∂(1/r)/∂x = –x/r3
2(1/r)/∂x2 = –1/r3 + 3x2/r5

and similar expressions for y and z. Adding them all up, we arrive at:

2(1/r) = –3/r3 + 3(x2 + y2 + z2)/r5 = 0

That is enough to tell us that the formulas we’ve seen for the gravitational and electrostatic potentials outside spherical bodies, which are proportional to 1/r, satisfy Laplace’s equation.

By making use of the divergence theorem, a result in advanced calculus, it’s possible to show that Gauss’s Law, applied to surfaces that enclose no mass or charge, actually requires the potential to satisfy Laplace’s equation in empty space. Taken together, these two results have a nice interpretation in terms of flux lines:

New flux lines can only emerge from regions in space that contain non-zero mass, or a net charge. The number of new flux lines that emerge from such a region is proportional to the mass, or charge, within. If a region is empty, every flux line that flows into it somewhere will emerge from it somewhere else.

So, our aim will be to find solutions for the potential, φ, in multiply connected space that obey Laplace’s equation everywhere in empty space, and also obey Gauss’s Law on surfaces that enclose some mass, or net charge.

### Conducting Disks and Image Charges

If an object is made of an electrically conductive material, such as a metal, then if its electrical environment isn’t changing, to a fair approximation it will end up with an electric potential that is the same across the whole object.

Why? In a conductor, there are electric charges that are free to move about. If the potential within the object is different from place to place, then those charges will be free to move from locations where their potential energy is high to new locations where it is lower. As they do so, their own electric fields will alter the potential, reshaping the energy landscape. The only way the motion of charge will stop is if we run out of mobile charges, or the energy landscape throughout the conductor becomes perfectly flat: that is, the potential is the same everywhere. And since we are interested more in the mathematical ideal of a conductor than the real thing, we will assume that it is always the latter.

What does this have to do with multiply connected space? Suppose we take a very simple example of the kind of space that we are interested in, where there is a single “Hoop” that joins just two “skies”. That is, there are two regions of ordinary Euclidean space, but they are spliced together along a disk, bounded by the Hoop. By passing through the Hoop you can cross back and forth between the two regions.

Now, suppose you place a particle with charge Q somewhere in the first region, and a particle with equal and opposite charge, –Q, at a point in the second region that is the mirror image of the first particle, with respect to a mirror in the plane of the Hoop. What can we say about the electric field in those circumstances?

As our earlier image of a pair of equal and opposite charges in ordinary space shows, in the plane that lies midway between them the electric field is perpendicular to that plane, and the potential is constant. We don’t yet have any formulas for an electric field in our multiply connected space, but because the situation is so symmetrical, we can still say something about the field as it passes through the Hoop: any force there must be perpendicular to the plane of the Hoop.

Why? Suppose the force had a component that lay in the plane of the Hoop. If we reflected everything in the plane of the Hoop and also swapped the two regions, this force would be unchanged ... but that would also swap the locations of the positive and negative charges, which amounts to multiplying Q by –1, which in turn will multiply the electric field everywhere by –1. The only vector that can be both unchanged and multiplied by –1 is the zero vector. So there is no component of the electric field in the plane of the Hoop.

Since the field is the gradient of the potential, this also means that the potential must be constant over the entire disk enclosed by the Hoop. That is: we have the same situation as we have for a conductor!  Away from that disk, we have empty space, where the potential must obey Laplace’s equation, and the point charges, which require any surface that encloses them to obey Gauss’s Law. But all of these aspects of the field, in either of the two regions, will also be true for ordinary space if the disk of the Hoop is instead occupied by a perfectly conducting metal disk, and a single charge is present in the same location.

In other words, if we could determine the electric field when ordinary space contains a single point charge and a metal disk, that would also tell us the electric field when our multiply connected space contains two equal and opposite point charges in mirror-image positions on either side of the Hoop.

In the real world, of course, we are more interested in going in the other direction, and the main reason mathematical physicists have sought solutions for Laplace’s equation in this kind of exotic space has been to yield a precise formula for the potential in ordinary space containing a point charge and a metal disk. But in the world of The Book of All Skies, the parallel between the two situations also means that the ability to measure the field that arises in that kind of table-top experiment can offer some hints about the potential in multiply connected space — though not a complete answer, of course, since it only describes the potential in situations where there are pairs of mirror-image charges.

The two diagrams on the right can be interpreted either way. They can be seen as showing the contours of the potential (in red and blue) and the flux lines (in green) for a multiply connected space with two regions joined by a Hoop (black), with one region containing a positive charge above the Hoop, and the other containing a negative charge below the Hoop.

Or, they can be seen as showing two slightly different scenarios in ordinary space. In both cases, the black line represents a metal disk seen edge-on, and there is a single point charge that is either positive or negative, positioned either above or below the metal disk.

Note that the metal disk, in either scenario, does not have a net charge of zero; rather, as witnessed by the flux lines that are either all pointing in towards it, or all emerging from it, it has a charge that is the opposite sign to that of the point charge (though not as large). To account for this, we need to consider the metal disk to have been grounded: connected by a narrow wire to the Earth, or some other very large object with which it can exchange charge, allowing it to reach a potential of zero. If we didn’t do this, and kept the disk isolated with a net charge of zero, then although it would still be at a constant [non-zero] potential, the corresponding setup in the multiply connected space would no longer have the same symmetry.

So, we’ve shown a connection between a simple electrostatic experiment in ordinary space, and a special kind of situation involving pairs of charges in a multiply connected space. But what we really want is a way to calculate the potential due to a single charge in a multiply connected space.

### Charge Layers and Dipole Layers

If you pass through the Hoop into the adjoining region, the potential and the electric field change smoothly and continuously; nothing special happens as you cross the plane of the Hoop. However, from the point of view of someone who remains in a single region, these quantities undergo abrupt changes, in the sense that if you compare corresponding points by travelling all the way around the Hoop and looking at the other side, the values will generally not match up.

There are also discontinuities like this that can arise in ordinary space: in the previous section, we looked at the example of a metal disk with a charge distribution induced by a nearby point charge. But in order to deal with more general situations, we need to consider other possibilities.

It turns out that the kinds of discontinuities we need can be generated with a combination of two ingredients. The first is a charge layer: a two-dimensional distribution of electric charge. But unlike the case with a metal disk, we will not restrict things by requiring the charge distribution to create a constant potential.

In general, the potential as you cross from one side of a charge layer to the other is continuous, but the component of the electric field perpendicular to the layer undergoes a jump that is proportional to the local charge density of the layer. One way to see this is with Gauss’s Law: if we enclose a small, square-shaped piece of a layer with charge density σ and area A in a rectangular prism of negligible height, then the total charge enclosed is Q = σA. Suppose the component of the electric field pointing in a direction n perpendicular to the layer is E1 on one side of the layer, and E2 on the other, where n points from the first side to the second side. Ignoring the negligibly-sized sides of the prism, the two square sides will contribute a flux of (E2E1) A. So we have:

Total flux = Q / ε0 [Gauss’s Law]
(E2E1) A = σA / ε0
E2E1 = σ / ε0

So, we can account for any jump in the perpendicular component of the electric field across the plane of the Hoop by pretending that we have a charge layer there, whose density σ is chosen to match the jump. But to deal with a jump in the potential, we need to use the second ingredient: a dipole layer.

An electric dipole, in general, is just a pair of equal and opposite charges, but the sense in which we will use the word is an idealisation of that notion. Suppose you have charges –q and q separated by a distance dx, in a direction given by the unit-length vector u. We can compute the potential of this configuration with Coulomb’s Law, just by adding up the contributions from the two charges. But we can obtain a simpler result if we suppose that the separation dx grows smaller and smaller while the value of q grows larger, in a way that keeps the product of the two, q dx, equal to some constant P. Then in the limit where dx → 0 and q = P/dx → ∞, the potential and field of the dipole turn out to be finite (at least away from the point occupied by the charges), because the infinitude of the individual charges is countered by their ability to cancel each other’s field more effectively as they come closer together. In fact, the potential in this case is just equal to P times the opposite of the derivative, in the direction of the separation vector u, of the ordinary Coulomb potential of a single unit charge, because the process of taking a derivative is the same kind of limit as the process of forming an idealised dipole.

We won’t write down the potential of a dipole, because we are more interested in a dipole layer: a two-dimensional distribution of dipoles. And specifically, we are interested in the case where the dipoles all point in a direction perpendicular to the layer. In that situation, not only do we have a connection between the (opposite of the) derivative of the potential of a single charge and the potential of a dipole, we have the fact that the perpendicular component of the electric field is the opposite of the derivative of the potential in that component’s direction. Combining the two, we see that the potential of a dipole layer like this is just the perpendicular component of the electric field of a charge layer — which, as we have seen, has a discontinuity proportional to its charge density.

So a dipole layer has a discontinuity in the potential, as you cross from one side to the other, that is proportional to the density of dipoles in the layer. This means we can account for any discontinuity in potential across the plane of the Hoop by pretending that there’s a suitable dipole layer present.

What about discontinuities in the components of the electric field that lie in the plane of the Hoop, rather than perpendicular to it? The fact that the electric field is the opposite of the gradient of the potential means that any such jump is completely determined by the behaviour of the potential, so if we get the potential right, those components of the electric field will also be taken care of.

Now, suppose we have a multiply connected space with just two regions, as in the previous section, but this time, instead of a carefully positioned pair of charges, we have a single charge, located at some arbitrary point. We might as well suppose it is in the first of the two regions, as there is no fundamental difference between the two.

Seen by someone confined to either region, there will be jumps in the potential and the electric field across the disk enclosed by the Hoop. But anyone travelling through the Hoop will find these quantities to vary continuously, and to be completely unchanged at any given point as they cross from one region to the other. We can accommodate all of these requirements by imagining that there are both charge layers and dipole layers positioned above and below the plane of the Hoop, in both regions, for a total of four layers of each kind (as in the diagram on the right). As someone approaches the Hoop from above in the first region, they encounter layers with charge density σa and dipole density δa, but then, when they pass through the Hoop and emerge in the second region, they immediately encounter layers with the opposite densities, so that (in the idealisation where the separation between the different layers goes to zero) there is effectively just empty space. Similarly, if someone approaches the Hoop from above in the second region, they encounter layers with charge density σb and dipole density δb, but when they pass through the Hoop into the first region, they again encounter layers with the opposite densities.

However, anyone confined to the first region will see the Hoop containing a total charge density of σa–σb and dipole density of δa–δb, while anyone in the second region would see the opposite values. These quantities need not be zero. Note that we are allowing all these densities to vary across the Hoop from point to point; they do not need to be constant.

Our goal is to find the potential anywhere in the two regions, given that there is an actual point charge somewhere in the first region, along with these fictitious charge and dipole layers. To do this, we need to determine exactly what the total densities σa–σb and δa–δb should be in order to make everything consistent.

We will write φ(σa) and φ(σb) for the contributions to the potential due solely to the charge layers with densities σa and σb respectively, at a point infinitesimally close to the layers themselves. Since the potential is continuous across charge layers, we don’t have to worry about which side of the layer we are on. And we’ll write Ea) and Eb) for the upwards component of the electric field due to those charge layers, at a point infinitesimally close to them, and just above them. In this case, because of the discontinuity in E, it will take the opposite value just below the same layers.

Similarly, we’ll write φ(δa) and φ(δb) for the contributions to the potential from the dipole layers with densities δa and δb, at a point infinitesimally close to them, and just above them. And we will write Ea) and Eb) for the upwards component of the electric field due to those dipole layers, at any point infinitesimally close to them. (Though we haven’t proved it, the perpendicular component of the electric field is continuous across a dipole layer.)

And finally, we will write φQ and EQ for the potential and the upwards component of the electric field due to a point charge Q somewhere in region 1.

Putting all these contributions together, we can obtain two different expressions for the potential at some arbitrary point within the Hoop, due to the charges and dipoles we have placed in each region. In this view, the two regions are entirely separate, and the fields that are produced by these sources do not leak from one region into another. But in order for this picture to agree with one in which the regions are actually joined along the disk enclosed by the Hoop, the two potentials at any point on that disk must agree. We can solve the resulting equation to obtain the potential that the two charge layers must be producing:

φQ + φ(σa) – φ(σb) – φ(δa) – φ(δb) = φ(σb) – φ(σa) – φ(δb) – φ(δa)
φ(σa) – φ(σb) = –½φQ

Similarly, the upwards components of the electric field must agree, and we can solve the equation to obtain the field that the two dipole layers must be producing:

EQ + Ea) – Eb) – Ea) – Eb) = Eb) – Ea) – Eb) – Ea)
Ea) – Eb) = –½EQ

So, we now have the potential and electric field due to the charge layers and dipole layers, but only within those layers. What we need to do now is work backwards from this to find the total densities of the layers, σa–σb and δa–δb. Once we have those densities, we can use them to compute the potential in either region. In region 1, we will add the potential from the layers to the potential produced by the point charge, while in region 2 we will only have the (opposite) contribution from the layers themselves.

Unfortunately, while the perpendicular component of the field, E(σ), is determined solely by the charge density σ in the layer at that point, the potential φ(σ) depends on the entire distribution of charge across the whole layer. Similarly, for a dipole layer, the potential is determined entirely by the local density, but E depends on the whole distribution.

In order to make progress, then, we will work numerically, approximating the distribution of charges and dipoles across the disk of the Hoop with a finite number of samples. If we can compute matrices that convert these sets of charge/dipole densities into potentials/field strengths, then we can use the inverse of those matrices to work backwards, and find the densities we need.  We can approximate the disk by chopping it up into a number of small squares, then use a formula for the potential φ(x, y, z) due to a square charge layer of constant surface density σ, with half-side-length h, centred at the origin in the x-y plane:

f(a, b, c, z) = a log([c + √(a2 + c2 + z2)] / [b + √(a2 + b2 + z2)]) + z arctan(a b / [z √(a2 + b2 + z2)])

x± = x – (±h)
y± = y – (±h)

φ(x, y, z) = [σ / (4 π ε0)] [ f(y, x+, x, z) + f(y+, x, x+, z) – f(x, y, y+, z) – f(x+, y+, y, z) ]

This formula can be obtained by integrating the Coulomb potential. The potential within the plane z = 0 follows from a simple substitution, with the z in the denominator of the arctan’s argument no problem since the arctan itself is bounded, and the whole final term in f(a, b, c, 0) vanishes.

If we take the opposite of the derivative of this potential with respect to z, that immediately gives us the potential for a dipole layer. If we take the second derivative, it gives us the upwards component of the electric field for a dipole layer, for which the final result at z = 0 simplifies to:

g(a, b) = [√(a2 + b2)] / [a b]

E(x, y) = [δ / (4 π ε0)] [ g(x+, y+) + g(x, y) – g(x, y+) – g(x+, y) ]

We now have all the ingredients we need to construct our two matrices, apply their inverses to the required potential/field values to obtain a finite set of charge/dipole densities, and use those densities to compute the potential and field anywhere in the two regions.

The diagrams on the right show the results for the two regions, where we have approximated the disk with 1961 small squares, each with half-side-length h equal to one fiftieth the radius of the Hoop. Note that in region 2, we have used a finer set of contours for the potential, in between those used in region 1, because the gradient is much lower.

The potential will obey Laplace’s equation almost everywhere by virtue of the way we have constructed it, and will give results in accord with Gauss’s Law for surfaces that enclose both the charge and the Hoop in region 1 and the Hoop in region 2, since all the layers we add at the Hoop in region 1 are cancelled out by those in region 2. The only flaw is that a finite number of squares cannot make the splice in the potential between the two regions perfectly smooth as we cross through the Hoop, but the deviation here is quite small.

The sharp bends in the contour lines near the edge of the Hoop are not artifacts of the approximation; they are present even in the exact, closed-form solution that we will derive in the next section.

We can use the same approach for multiply connected spaces with any number of regions. In general, if there are n regions, there will be n–1 independent functions σi for the total charge density at the Hoop, and another n–1 independent functions δi for the total dipole densities, with the counts reduced from n because the sum of these densities across all regions must be zero; this happens automatically if we require any traveller through the Hoop to pass through charge and dipole layers of opposite sign. And we have a total of 2(n–1) equations that arise from matching potentials and fields across the splices between the regions; in our model, we simply have equality between the corresponding points within the Hoop in all n regions.

We will write Mφσ for the matrix that takes the charge densities at a set of points on the disk and returns the potentials at the same points, and MEδ for the matrix that takes the dipole densities at those points and returns the upwards component of the electric field. And we will write λ as an abbreviation for the factor 1/(2 ε0), which converts a charge density to the upwards component of the electric field just above the layer at that point, and also converts a dipole density for an upwards-pointing dipole layer to the potential just above the layer.

If we take the potential and field at a point within the Hoop in region 1, and subtract the values for each of the other regions in turn, we end up with the following system of linear equations:

φQ + Mφσ1 – σr) + λ (δ1 + 2 (δ2 + ... + δr–1) + δr) = 0   for r = 2, ..., n–1
φQ + Mφσ (2 σ1 + σ2 + ... + σn–1) + λ (δ2 + ... + δn–1) = 0
EQ + MEδ1 – δr) + λ (σ1 + 2 (σ2 + ... + σr–1) + σr) = 0   for r = 2, ..., n–1
EQ + MEδ (2 δ1 + δ2 + ... + δn–1) + λ (σ2 + ... + σn–1) = 0

For n=2, this is simply:

φQ + 2 Mφσ σ1 = 0
EQ + 2 MEδ δ1 = 0

with the solution:

σ1 = –½(Mφσ)–1 φQ
σ2 = –σ1
δ1 = –½(MEδ)–1 EQ
δ2 = –δ1

For n=3, the equations are:

φQ + Mφσ1 – σ2) + λ (δ1 + δ2) = 0
φQ + Mφσ (2 σ1 + σ2) + λ δ2 = 0
EQ + MEδ1 – δ2) + λ (σ1 + σ2) = 0
EQ + MEδ (2 δ1 + δ2) + λ σ2 = 0

with the solution:

σ1 = –2MEδ (3MφσMEδ + λ2I)–1 φQ
σ2 = MEδ (3MφσMEδ + λ2I)–1 φQ – λ (3MEδMφσ + λ2I)–1 EQ
σ3 = –σ1–σ2
δ1 = –2Mφσ (3MEδMφσ + λ2I)–1 EQ
δ2 = Mφσ (3MEδMφσ + λ2I)–1 EQ – λ (3MφσMEδ + λ2I)–1 φQ
δ3 = –δ1–δ2

### Sommerfeld’s Method

The method of charge and dipole layers yields reasonably accurate results if we include enough sample points, but applying it is a bit unwieldy. Fortunately, there is another approach that involves less brute-force computation, and can give us an exact formula for the potential for the case of two regions.

In the 1890s, the physicist Arnold Sommerfeld found a way to compute the potential of a point charge in the presence of a half-infinite conducting plane (a plane that extends to infinity on one side of a single, infinitely long edge). He also attempted to extend his approach to the case of an infinite conducting strip, bounded by two edges. There was a flaw in his solution for the second scenario, but seventy years later, two researchers at the Ford Motor Company, L. C. Davis and J. R. Reitz, showed how his method could be used to compute the potential in the presence of a conducting disk. We plotted the final result of those calculations earlier, but in fact, using Sommerfeld’s method to solve that problem starts with the potential in precisely the kind of multiply connected space that interests us. To proceed, we first need to set up a coordinate system that is better suited to the problem than the usual Cartesian coordinates in three dimensions. Bipolar coordinates in two dimensions serve as a starting point. We choose two points, A and B, as the poles of the system, and the coordinates given to any point P are ρ = log(AP/BP) and θ = ∠APB. The curves of constant ρ and θ are generally circles, though they became straight lines for some special values.

We can turn this into a three-dimensional coordinate system by rotating the plane by a third coordinate, γ, around the line ρ=0, which is the perpendicular bisector of AB. As γ ranges from –π/2 to π/2, A and B each sweep out half of a circle, and we arrange for that circle to coincide with the edge of the Hoop.

Inside the circle, θ = –π just above the plane of the Hoop, and θ = π just below it, while in the same plane, but outside the circle, θ = 0. This allows us to easily extend the range of θ to other regions, simply by adding multiples of 2π. For example, if we had a multiply connected space with three skies, θ could lie between –π and π in the first, between π and 3π in the second, and between 3π and 5π in the third, with θ = 5π being equivalent to θ = –π.

Now, suppose we have a point charge Q with coordinates (ρQ, θQ, γQ), and we wish to compute the potential at some arbitrary point (ρ, θ, γ), where we are working in a multiply connected space with n regions. We start by considering the function R that measures the ordinary Euclidean distance between these points, if we ignore whatever multiples of 2π have been added to the θ coordinates and just treat them both as lying in the first region.

R = a √[2 (cosh(ρ–ρQ) – cos(θ–θQ) + sinh ρ sinh ρQ (1–cos(γ–γQ))) / ((cosh ρ – cos θ) (cosh ρQ – cos θQ))]

Here a is the radius of the Hoop. We know that K/R for any constant K is a solution of Laplace’s equation ... but in our multiply connected space, what that would give us, with the usual choice of K, would be the potential due to n point charges, with one in every region, all in matching locations.

The next step is to take the θ coordinate of the point charge, θQ, and replace it with a complex parameter α, which is construed to have a real part in the same range as θ (i.e. it will span n multiples of 2π), and any imaginary values we like. We will write R(α) for the resulting formula; this is still the same function of six quantities as it was originally, but now one of them is α in place of θQ.

Next, we consider the function:

F(α) = (i / n) √[(cosh ρQ – cos θQ) / (cosh ρQ – cos α)] exp(iα/n) / [exp(iα/n) – exp(iθQ/n)]

As a function of the complex parameter α, this is a finite, differentiable function everywhere, except when α = θQ, where it goes to infinity. But we have chosen it so that it does so like the function 1/(α – θQ). In the terminology of complex analaysis, we say that F(α) has a simple pole at α = θQ with a residue of 1.

It then follows from Cauchy’s Residue Theorem that if we integrate K/(2πi) F(α)/R(α) around any simple closed curve in the complex plane that encloses the point α = θQ, the integral we get will just be K/RQ).

If you haven’t studied complex analysis this might all sound a bit daunting, but don’t be put off. All it’s really saying, for our purposes, is that we can replace n point charges as the source of a K/R potential with a new mathematical fiction: a “line charge” whose θ coordinate is no longer a single real value, but now consists of any suitable loop in the complex plane. The role of F(α) here is to help specify the charge density along that loop that will give the correct result. And though the coordinates of this line charge, and the charge density, will both take on complex values, when everything is integrated over the loop the result is exactly the same potential as the n point charges in the original locations.

Of course, we don’t want the potential from n point charges, we only want a single charge. But having turned the n points into a loop, maybe there is some way we can divide up that loop, and single out the piece we want. The diagram on the left shows one loop in the complex plane that encloses the point θQ, for the case n=3. The points of the form θ + 2 π k ± α1 i, where:

cosh α1 = cosh(ρ–ρQ) + sinh ρ sinh ρQ (1–cos(γ–γQ))

are points where R(α) = 0. Because the function R(α) involves a square root, it must have branch cuts: lines in the complex plane where it jumps abruptly from one square root to the other. The usual branch cut for the square root function itself is the negative half of the real axis, where, for example, we jump from using i as √(–1) to using –i. But we can align the branch cuts any way we like, so long as they start from the points where R(α) = 0, and the dashed lines show a convenient choice. The path of integration here moves back and forth along part of each branch cut, on either side, and we can bring these parts of the loop arbitrarily close to the cut, but they don’t cancel each other out, because the function takes different values on the different sides of the cut.

The two vertical sides of the rectangle here, where the real part of α is –π or 5π, do cancel each other out, because the quantity we are integrating is periodic: adding 2 π n to α leaves the value of F(α)/R(α) unchanged. And the integral along the horizontal parts of the loop goes to zero as the height of the rectangle goes to infinity; our choice of F(α) helps with this, by guaranteeing that the integrand goes to zero as the imaginary part of α goes to positive or negative infinity.

The integrals from each pair of branch cuts with the same real part for α can be combined. So we end up with:

K/RQ) = sum of n integrals, where Re[α] = θ + 2 π k for k = 0, ... , n–1

But if we take just the first of these n integrals, the one along the pair of branch cuts where Re[α] = θ, we get precisely the solution of Laplace’s equation that we need!

Though the setup required to reach this conclusion has been quite elaborate, the final claim is not too hard to check. For a start, the integral of K/(2πi) F(α)/R(α) along any curve will satisfy Laplace’s equation, because it represents a distribution of charged points, all of them individually producing a 1/r potential. That there are complex coordinates and charge densities involved makes no difference, and the final potential we get from integrating around each pair of branch cuts is real-valued.

The only other thing we need to check is that we measure the correct amount of flux passing through any surface that encloses the point charge and all the Hoops. But as we go from region to region, integrating flux over n different surfaces, that amounts to changing θ by various multiples of 2π, which is the same as summing the result over all n pairs of branch-cut integrals. And we know that the potential from that sum is just K/R. If we had used K/R itself we would get an answer n times too large, but by splitting up K/R into these n pieces, we get exactly the right result.

After combining the integrals for the two branch cuts where Re[α] = θ, we end up with the following expression for the potential, φ, due to a point source in a multiply connected space with n regions:

φ = K/(π a) √[½ (cosh ρQ – cos θQ)(cosh ρ – cos θ)] ∫σ 1/[(x – τ) √(cosh(n arccosh(x)) – cosh(n arccosh(σ)))] dx

where:
τ = cos((θ–θQ)/n)
σ = cosh(α1/n)
cosh α1 = cosh(ρ–ρQ) + sinh ρ sinh ρQ (1–cos(γ–γQ))
a is the radius of the Hoop

K = Q / (4 π ε0) for an electrostatic point charge Q
K = –G M for a gravitational point mass M

Part of the integrand here can be expressed in terms of Chebshev polynomials:

cosh(n arccosh(x)) – cosh(n arccosh(σ)) = Tn(x) – Tn(σ)

For the case n=2, where Tn(x) = 2x2–1, we have:

φ = K/(2 π a) √[(cosh ρQ – cos θQ)(cosh ρ – cos θ)] ∫σ 1/[(x – τ) √(x2–σ2)] dx
= K/(2 π a) √[(cosh ρQ – cos θQ)(cosh ρ – cos θ) / (σ2–τ2)] arccos(–τ/σ)
= K/(π a) √[½ (cosh ρQ – cos θQ)(cosh ρ – cos θ) / (cosh(ρ–ρQ) – cos(θ–θQ) + sinh ρ sinh ρQ (1–cos(γ–γQ)))] arccos(–τ/σ)
= K/(π R) arccos(–τ/σ)

R can be expressed in terms of (ρ, θ, γ) coordinates with the complicated formula given previously, but it is still just the ordinary Euclidean distance between the two points if they are treated as being in the same region.

If you read the paper by Davis and Reitz, you will see that their equation (30) for the same potential presents it with 2 arctan(√((σ+τ)/(σ–τ))) where we have written arccos(–τ/σ), but these can be shown to be identical, and the arccos form seems simpler.

This formula makes it easy to check that the sum of potentials across the two regions is K/R. Adding 2π to θ changes the sign of τ while leaving σ unchanged, so it changes arccos(–τ/σ) to π – arccos(–τ/σ). This makes the sum of the arccos terms π, and the sum of the potentials K/R, as claimed.

### Gravity at the Nubs

In the previous section, we saw that the sum of the potential of a point source over all the different regions is just K/R, where K is an appropriate constant. In The Book of All Skies, the protagonist lives on a world where the ball of rock beneath her feet is more or less the same over many neighbouring regions. So, the gravitational field she experiences under those conditions will be very close to the normal K/R potential that gives rise to ordinary Newtonian gravity. However, these balls of rock eventually dwindle away, giving rise to places known as “nubs” where there is a substantial mass in one region, but none at all in the neighbouring regions. Here, the geometry of the gravitational field becomes much stranger.

The images on the right show a simulated land mass in a multiply connected space with six regions, where one region has a full sphere of rock, and its two immediate neighbours have nubs. This is a much simpler situation than the world of the novel — and this is not a map of the terrain at the nub, in the world of the novel! — but it can still act as a reasonable qualitative guide to how the gravitational field will behave when the matter responsible for it is no longer evenly distributed between the regions.

The Hoop here is much larger than the body of rock; only part of it is shown, with the disk in red, and the edge in blue. The top image shows the one full sphere, while the bottom image shows a nub, where some land protrudes into the region, but does not continue all the way around the Hoop. The black spikes indicate the local vertical.

The images below show contours of constant gravitational potential in all six regions of the model; region 1 contains the full sphere of rock, regions 2 and 6 contain nubs, and the other regions contain no solid matter. These are for slices that bisect the Hoop; a small part of the Hoop is shown edge-on, in black. Moving counter-clockwise through the plane of the Hoop takes you to a region with a higher number, until you reach region 6, where it takes you back to region 1.      ### References

 “Solution to Potential Problems near a Conducting Semi-Infinite Sheet or Conducting Disk” by L. C. Davis and J. R. Reitz, American Journal of Physics, 39, 1255 (1971). Online at Semantic Scholar.  The Book of All Skies / Gravity in Multiply Connected Space / created Monday, 23 August 2021