Deriving the Simplest Geometry

In Chapter 12 of Incandescence, the Splinterites succeed in deriving a possible geometry for the spacetime they inhabit. They come up with the simplest possible geometry that conforms to Zak’s principle (that the sum of the three perpendicular “weights”, or tidal accelerations, is zero, after the effects of spin have been removed) while explaining the fact that the ratio of the garm-sard weight to the shomal-junub weight is less than the apparent historical value of three.

The spacetime geometry that they discover is what we would call the Schwarzschild spacetime, which is the geometry of the vacuum around any spherically symmetrical body whose spin and electrical charge are zero, such as a non-rotating black hole. In the novel there is a rough sketch of how they performed their calculations; this page gives the details. Nothing here will require prior knowledge of general relativity, and anyone should be able to get the gist of it, but you’ll need a bit of high-school level algebra, trigonometry and calculus if you want to verify every calculation along the way.

Zak’s principle is really just a formulation of Einstein’s equation for spacetime curvature in a vacuum. (The rock of the Splinter is far from being a vacuum, of course, but the Splinter is so small that its own gravitational field is negligible, and what dominates the local spacetime curvature is the shape that the Hub has given to the surrounding vacuum.) The more general case, including the effects of matter, is discussed in this wonderful exposition: “The Meaning of Einstein’s Equation” by John C. Baez and Emory F. Bunn.

[A technical note for anyone who’s read my Foundations articles on general relativity: those articles use a coordinate vector basis for all their calculations, whereas the calculations on this page always use an orthonormal vector basis, because that’s how the Splinterites do their differential geometry. Consequently, some of the results here will appear to be different from those given in the Foundations articles, but in fact they’re both useful special cases of general results that encompass both choices.]


The basic mathematical tool which the Splinterites use to deal with the curved nature of spacetime is known as a connection. We will start by illustrating this concept with the example of the curved surface of the Earth (where we will ignore all bumps and irregularities, and pretend it is a perfect sphere).

Everywhere on Earth, with the exception of the poles, we can define two perpendicular directions, east and north. (The other main compass points, west and south, are just the opposite directions to these two, so mathematically it’s simpler to treat them as “negative east” and “negative north” respectively.) Because the surface of the Earth is curved, over large distances these directions behave in ways quite different from the perpendicular x and y directions on a sheet of flat graph paper. On a sheet of graph paper, we’d say that the x direction at any point is parallel to the x direction at any other point, but whether we picture the Earth from space, or whether we carry out various tests that are confined to its two-dimensional surface, it seems clear that we can’t say that the direction east in London is parallel to the direction east in Havana.

On the surface of the Earth, the closest thing we have to a straight line is known as a great circle, or geodesic. The geodesic between two cities is the route that planes would fly in a perfect world (ignoring various logistical complications), because it always comprises the shortest total distance. In a three-dimensional view of the Earth, the geodesic that joins city A with city B is an arc of the circle with a radius as large as that of the Earth itself, that passes through both cities, and whose centre lies at the centre of the Earth. But we don’t need to take that Earth-from-space view; we could equally well define the geodesic from A to B by requiring that everywhere along its length, it acts locally like the straightest possible line: as we walk along it, we won’t detect any swerving to the left or right.

Parallel Transport

The sense in which the direction of a geodesic “stays true” is the closest thing to the notion of directions being “parallel” that we can have on a curved surface: if you travel the great circle route from London to Havana, then it makes a certain amount of sense to claim that you were travelling “in the same direction” from start to finish. And as well as the direction you were travelling, you could claim just as reasonably that the direction to your left, or to your right, or at any fixed angle from “straight ahead” was as close to “the same direction” all along as it could possibly be. We will say that a direction is parallel-transported along a geodesic if the direction makes a fixed angle with the geodesic everywhere. But it’s important to note that this still doesn’t let us claim that two directions at different locations are (or aren’t) parallel in any absolute sense. If you travel a triangular route, each side of which is a geodesic, and parallel-transport a direction at the start all the way around the triangle, it won’t generally arrive back pointing in the same direction! In the diagram on the left, the blue arrow transported from P to Q to R and then back to P ends up finally as the red arrow at P, quite different from the original.

Connection Example

To describe the behaviour of parallel transport mathematically, we first recall that we have two perpendicular directions, east and north, defined everywhere (except the poles). Suppose, at town A, we head along a geodesic, a great circle that runs due east through A. Using parallel transport, we can take the directions east and north at A and carry them a small distance, d, along the geodesic. Then we can compare these “transported versions” (the blue arrows in the diagram on the right) with the directions east and north that already existed at our destination (green arrows). What we will find, in general, is that the two disagree slightly: the local directions will be rotated by a small angle compared to the ones we carried from town A.

For small distances d, the vector we get by subtracting the transported east from the local east will be proportional to d; here we’re supposing that we draw the two vectors as each being one unit long, for whatever unit we’re using to measure the distance d, and then subtracting one from the other by drawing the vector that joins their tips. So if we divide the difference in the vectors by d, we’ll get a vector representing the change per distance travelled that is very close to constant: that is, the ratio is independent of d. With a little trigonometry, it’s not hard to show that:

(local east – transported east) / (distance travelled east) = (tan θ) / RE north
(local north – transported north) / (distance travelled east) = – (tan θ) / RE east

where θ is the latitude of our town A, and RE is the radius of the Earth. For example, consider a town that lies at latitude 45 degrees north, as shown in the diagram above right. The great circle that runs due east through the town will have its highest northern latitude there, and will “turn” southwards as you go east. Since the transported east will pick up a southwards component, subtracting it from the local east will give a difference that points north. Similarly, the transported north will pick up an eastwards component, so subtracting that from the local north will give a difference that points west, or “negative east”.

In contrast, if we travel due north rather than east, we’ll find that the local directions agree exactly with the transported directions. And if we travel in a direction that lies somewhere in between, for small distances we can just take the east and north components of the direction we’re travelling, and add their respective contributions. For example, if we travelled at a bearing that meant we moved 0.8 metres east and 0.6 metres north for every total metre we travelled, we would get:

(local east – transported east) / (total distance travelled) = 0.8 (tan θ) / RE north
(local north – transported north) / (distance travelled east) = – 0.8 (tan θ) / RE east

If the changes due to travelling north weren’t zero, they would be added in too, multiplied by 0.6.

We can summarise the way the local east and north directions change, compared to parallel transport, as follows:

changeeasteast = (tan θ) / RE north
changeeastnorth = – (tan θ) / RE east
changenortheast = 0
changenortheast = 0

This is just a matter of defining a convenient shorthand. For example, the second line is saying that as you travel a short distance d to the east (because east is the subscript to the word change), the local north changes compared to the parallel-transported north by having – d (tan θ) / RE east added to it. [Anyone who’s studied differential geometry is free to mentally substitute an upside-down triangle for the word change in this notation.]

A collection of equations like the four above, listing the way a complete set of directions change compared to parallel transport, comprise what’s known as a connection. Using this notation, the defining equation of a geodesic becomes:

changetangenttangent = 0

where “tangent” is the direction tangent to the geodesic. Of course, we defined parallel transport in the first place in terms of maintaining a constant angle with a geodesic; this just puts that definition into an equation with our “change” notation. The geodesic equation can be used in either direction: if you happen to know the geodesics on a surface, it will let you work out the connection, but conversely, if you’re starting from the connection, it will let you find the geodesics.


Commutator Example

Two smugglers plan to rendezvous in the jungle for a clandestine transaction. They don’t want to risk being seen travelling in the same direction, so starting from the same small town, one smuggler travels 300 kilometres east then 500 kilometres north, while the other travels 500 kilometres north then 300 kilometres east. Will their plan work?

The answer is: no! These actions, moving x kilometres north and y kilometres east, do not commute: they don’t give the same result when combined in different orders. Travelling x kilometres north always changes your latitude by the same amount, but travelling y kilometres east corresponds to a different change in longitude depending on your latitude. (If the smugglers had chosen to travel a certain number of degrees of latitude and longitude, they would have met up successfully; vectors based on coordinate systems always commute.)

In the limit of small distances x and y, the discrepancy between the two destinations reached by following the instructions in different orders turns out to be proportional to x times y, and we can define the commutator of the directions east and north, which we will write as [east, north], to be the ratio:

[east, north] = (vector from north, east destination to east, north destination) / (x y)

With a little trigonometry and calculus, it’s not hard to show that:

[east, north] = – (tan θ) / RE east

where θ is latitude and RE is the radius of the Earth. If this expression looks familiar, it’s not your imagination, and it’s not a coincidence either. Though we won’t prove it, the commutator between two directions can be derived from the connection, with the following formula:

[a, b] = changeabchangeba

What this means is that, on a curved surface, if you want to form a closed loop by going a distance x in direction a and a distance y in direction b, you can’t generally do so with a four-sided figure, a quadrilateral; you need to add in a fifth step where you go a distance x y in the direction [a, b]. In the example we’ve shown, the extra step just backtracks and cancels out an overshoot in one direction, but in general [a, b] can point in any direction, and need not be a multiple of a or b.

The Geodesic Squeeze

Suppose that two nearby great circles on the Earth are initially parallel. (We’ve stressed that the notion of directions being “parallel” isn’t defined on curved surfaces, but if the two points where the directions are located are close enough we can compare the directions with parallel transport along any short route, and the choice of route won’t make a significant difference.) As we move along these great circles, we know that they will grow closer together; for example, two meridians separated by a kilometre at the equator will grow ever closer until they meet at the north pole. What we’d like to do is quantify the rate at which they are accelerating towards each other.

First, we’ll do this with elementary trigonometry and calculus. Suppose the distance separating two meridians at the equator is x0. Then the difference in their longitude will be an angle of x0 / RE in radians. That angle will remain constant as we move northwards from the equator, but at any given latitude θ, where the radius of the circle of latitude is RE cos(θ), it will correspond to a distance of x = x0 cos(θ). Now the latitude θ will be related to the distance we’ve travelled north, call this y, by θ = y / RE. So we have:

x = x0 cos(y / RE)
dx/dy = – x0 sin(y / RE) / RE
d2x/dy2 = – x0 cos(y / RE) / RE2
(1 / x0) d2x/dy2 |y=0 = – 1 / RE2

So the rate at which the curvature of the Earth squeezes two parallel geodesics together can be summed up by the number - 1 / RE2; this is the rate at which the geodesics’ separation “accelerates” as you move along them, if they are initially separated by one unit of distance. For greater or smaller separations, the acceleration changes proportionately.

Geodesic Deviation Example

Now, in order to learn how to do the same calculation for other geometries that aren’t as easy to analyse directly as the surface of a sphere, we want to see how to get this result from our connection for the geometry.

In the diagram on the left, the points A and B lie on the equator, and we’ll assume that they’re separated by a distance of 1. By following the meridians that pass through A and B a distance of 1 to the north, we reach points E and C.

The tangents to the meridians at A and B (blue arrows) agree when compared by parallel transport along the path AB, and because the meridians themselves are geodesics, they parallel-transport their own tangents to the blue arrows at E and C. All four of these arrows simply agree with the local north. However, if we parallel transport the blue arrow at C over to E, we get the red arrow at E. The difference between the red and blue arrows at E reflects the fact that the geodesics AE and BC are accelerating towards each other.

What’s going on with point D? In the previous section, we discussed the fact that on a curved surface, you can’t generally construct a closed quadrilateral having equal lengths for opposite sides; to close the figure, there is an extra piece needed, derived from the commutator of the directions for the main sides. When we come up with a connection-based expression for the difference in the arrows at E, we will need to take that into account, because the connection formulas will give us changes in vectors per unit distance travelled: that is, they will give us the change along ED rather than EC.

To quantify the difference at E, we could write out 5 expressions for the changes in north along the five legs of its journey in a loop leading back to E, but we’re really just interested in the rate at which things change in the limit of small distances. So instead of writing separate expressions for the changes along ED and BA, we’ll write the net result of those two eastwards changes at different latitudes as:


And instead of writing separate expressions for the changes along AE and CB, we’ll write the net result of those two northwards changes at different longitudes as:

- changeeast(changenorthnorth)

Then finally, we need to include the change due to the extra side, DC, which has the direction [east, north]. The overall result is:

relative acc = changenorth(changeeastnorth) – changeeast(changenorthnorth) + change[east, north]north (1)

Some of the terms above happen to be zero in this particular case, but we’ve written out the full equation for reference. With arbitrary directions a and b in place of east and north, this gives the rate at which geodesics accelerate relative to each other in any curved geometry.

In this specific case, making use of the formulas we’ve previously obtained for the connection and the commutator on the surface of a sphere, we have:

relative acc = changenorth(changeeastnorth) – changeeast(changenorthnorth) + change[east, north]north  
  = changenorth(–(tan θ) / RE east) + change– (tan θ) / RE eastnorth  
  = – d[(tan θ) / RE] / dsnorth east – (tan θ) / RE changeeastnorth  
  = – (dθ/dsnorth) d[(tan θ) / RE] / dθ east + (tan θ)2 / RE2 east  
  = – (1 / RE) (sec θ)2 / RE east + (tan θ)2 / RE2 east  
  = [(tan θ)2 – (sec θ)2] / RE2 east  
  = – 1 / RE2 east  

In going from the second line to the third, where we needed changenorth of a direction, east, multiplied by a function, - (tan θ) / RE, that itself changes as we move north, we used a rule that’s really the same as the usual product rule for derivatives: (fg)’ = f’g + fg’. That is to say, we found changenorth for each factor separately while leaving the other unchanged, and add the resulting terms. What is changenorth of a function of latitude? It is the rate of change of that function with respect to actual distance travelled north, snorth, which we then turned into a derivative with respect to θ by noting that dθ/dsnorth = 1 / RE.

The final result here tells us that for each unit of separation to the east between our north-going geodesics, they will have a relative acceleration of – 1 / RE2, in agreement with our direct calculation.

Geodesics in Spacetime

In flat spacetime — that is, spacetime far from any massive body — objects that are subject to no force move in straight lines at a uniform velocity. This means that their world lines, the lines they trace out over time in the four dimensions of spacetime, are also straight lines.

The basic idea of general relativity is that gravity is accounted for, not by a force that acts on any particular object, but by a change from a flat geometry to a curved one, such that all the world lines of all objects in free fall are geodesics of the curved spacetime geometry.

So, an object in free fall will have a world line whose tangent obeys the geodesic equation. What is the tangent to an object’s world line? It’s simply the direction in spacetime that someone fixed to the object considers to be the pure “time” direction. To advance in time rather than move through space simply means to stay where you are, and since the free-falling object itself is the observer’s idea of a fixed location, the direction that is tangent to its world line (the direction that points along with the world line, rather than away from it) will be the observer’s idea of the direction “time”. If we call that direction u, we have:

changeuu = 0

As well as telling us about free-falling objects, a spacetime connection will tell us the degree to which an object fails to be in free fall. If an object’s world line is not a geodesic, because it is subject to a non-gravitational force, then we define its acceleration vector by the formula:

acc = changeuu

The Connection for a Spherically Symmetrical Spacetime

A spacetime connection will look very similar to the kind we’ve seen for a two-dimensional surface, with one obvious exception: there will be four perpendicular directions to consider, for the three dimensions of space and one of time, rather than just two. But there is also a subtler distinction. A connection describes the gradual rotation of the set of perpendicular directions compared to parallel transport. When those directions are all directions in space, the fact that we’re talking about a rotation, rather than some more general kind of change, is encapsulated in a symmetry of the connection:

If changeab = F c,
then changeac = – F b.

The first two equations of our connection for the surface of the Earth clearly obey this rule. (If the connection is a bit more complicated than this, and there are more than two terms on the right-hand side, then this symmetry applies to all the individual terms; we’ll see an example of this shortly.)

In spacetime we want the same general result, but the useful definition of a “rotation” (as the Splinterites eventually realise) is slightly different. In space, the length of a vector v squared is vx2 + vy2 + vz2, where vx, vy and vz are components measured in three perpendicular directions, and this quantity is unchanged if we rotate the vector. In spacetime, however, the corresponding quantity that needs to remain unchanged for a 4-dimensional vector u is – ut2 + ux2 + uy2 + uz2 (where we’re measuring things in units such that the speed of light is equal to 1, so time and space are in identical units). The presence of that first minus sign wends its way into all the algebra of spacetime “rotations”, but the consequences for a spacetime connection can be summed up very simply:

If changeat = F c,
then changeac = F t.

In other words, when time is one of the directions involved in the rotation, the original minus sign goes away.

Now, we want to start with a general description of a connection for a spacetime that is (a) spherically symmetrical about some point, and (b) unchanging over time. The geometry of any imaginary sphere centred on the point of symmetry will be the same as that of any normal sphere, such as the surface of the Earth; however, because spacetime is curved, we can’t expect the distance from the centre of the sphere to be related to its surface area by the usual formula, A = 4 π r2 (any more than we expect the circumference of a circle of latitude on the Earth to be related to its distance from the north pole by C = 2 π r). So what we will do is define a coordinate r = √(A / [4 π]). This r won’t measure the distance to the centre of each sphere, but it will give us the sphere’s total area with the usual formula. And because each individual sphere’s geometry is perfectly conventional, the length of its equator, or any other great circle on it, will be 2 π r for the r we have defined this way.

On each of our imaginary spheres, we will use standard latitude and longitude coordinates, which we’ll call θ and φ. We will supplement the directions east and north from our previous example with a third spatial direction, out, which points outwards away from the centre of the spheres.

How does time enter into the picture? Someone who remains at a fixed distance from the centre of symmetry, at a fixed latitude and longitude, will define a direction we will call time: this is simply the direction through spacetime they are “travelling” by staying still. We won’t define a time coordinate at this stage; choosing a direction will be enough.

Now that we’ve defined our four perpendicular directions, east, north, out and time, we have to consider all the ways these four directions can change, compared to parallel transport, in each of the four directions. In principle, that threatens to give us 16 equations defining our connection, with up to four terms on the right hand side of each, but fortunately the symmetries that are present in any connection, along with the extra symmetries of our particular situation, mean that many of the terms are zero, and the others are interdependent in a way that will simplify our calculations considerably. The non-zero equations are:

changetimetime = G(r) out (2a)
changetimeout = G(r) time (2b)
changenorthout = H(r) north (2c)
changenorthnorth = H(r) out (2d)
changeeastout = H(r) east (2e)
changeeasteast = H(r) out + (tan θ) / r north (2f)
changeeastnorth = – (tan θ) / r east (2g)

Here G(r) and H(r) are unknown functions of r that we will need to determine.

First of all, note that this connection satisfies the symmetries we discussed earlier; for example, the first two equations, where both terms have a factor of G(r), describe a “rotation” in a spacetime plane involving the out and time directions. All the other terms involve purely spatial rotations, and match up in pairs with opposite signs.

Given that we have defined time to be the time direction for someone who remains at a fixed distance r and a fixed latitude and longitude, the first equation here is telling us that someone under those conditions will be experiencing an acceleration of G(r) in the out direction. It might seem odd to describe someone in a fixed location as accelerating, but the central insight of general relativity is that free fall, the condition of weightlessness, is motion without acceleration, whereas any object that resists free fall is being acted on by a force, and hence is accelerating. From this perspective, acceleration is not a matter of whether an object happens to be moving or not moving relative to something else; it is a matter of whether an object’s world line is as straight as possible (a spacetime geodesic), or not. To remain at a fixed distance from a massive object means that something else, other than gravity is imposing a force on you to keep you in place (for example, a rocket engine). That force causes you to accelerate — to fail to be in free fall — and you certainly feel the acceleration, since you experience a weight that you would not experience in free fall.

The second equation is required by the symmetry of the connection; the time and out vectors must be rotated together by parallel transport.

The third and fourth equations describe the fact that, as you move north across one of our imaginary spheres, the direction that points out from the centre of the sphere will rotate compared to its previous position. In flat space, we would simply have H(r)=1/r; the difference between the local vertical directions on a sphere, separated by a unit distance, is inversely proportional to the radius of the sphere. But here we can’t assume the geometry is the same, so H(r) is to be determined.

Finally, the fifth, sixth and seventh equations mix two things: they restate what we’ve just said for the case when you move east rather than north; by symmetry that effect must be the same in either direction, so the same function H(r) appears again. And the terms with (tan θ) / r simply repeat the relationship between east and north that we worked out for the surface of the Earth; since the two-dimensional geometry of our imaginary spheres themselves (as opposed to the way they fit in with the other dimensions) is identical to that of the Earth’s surface, these terms are exactly the same, with r in place of RE.

Imposing Zak’s principle

To complete our description of the spacetime geometry, we need to find the two mysterious functions of r, G(r) and H(r), that appear in the connection. The way we will do this is to impose Zak’s principle on two different moving bodies, giving us two equations that we can combine to solve for these unknown functions. We derived a general formula for computing tidal acceleration — the relative acceleration between geodesics — in equation (1), but it will turn out that we only need to use that heavy machinery once, because in all the other cases we can employ some simple tricks to find the result.

The first moving body we will consider is the Splinter itself, which is rotating around “the Hub” in a tide-locked orbit: that is, keeping one face oriented towards the centre of the orbit. Each point on the Splinter that lies in the plane of the orbit will move on the equator of one of our imaginary spheres, remaining at some fixed r coordinate, and at the fixed latitude θ=0. The world line of each such point will form a helix in spacetime: advancing in time while circling the Hub. Accordingly, the tangent to the world line will take the form:

u(r) = C(r) east + √(C(r)2 + 1) time (3)

where C(r) is a function of r that we need to determine. Why is the second term √(C(r)2 + 1)? Recall our formula – ut2 + ux2 + uy2 + uz2 for the squared length of a 4-dimensional vector in terms of its components in four perpendicular directions. Just as the directions east, north and out are vectors of squared length 1, time (with ut=1, and all other components zero) has squared length –1, and we want any world line’s tangent to be the same. The coefficients we’ve given here will combine to give u(r) a squared length of –1, no matter what C(r) is.

If we use our connection to compute the acceleration of any point of the Splinter in the plane of the orbit as a function of its r coordinate, we find:

acc(r) = changeu(r)u(r)  
  = changeC(r) east + √(C(r)2 + 1) time[ C(r) east + √(C(r)2 + 1) time ]  
  = C(r)2 changeeasteast + (C(r)2 + 1) changetimetime  
  = [ C(r)2 (G(r) – H(r)) + G(r) ] out (4)

where we’ve made use of the fact that we’re on the equator, θ=0, to discard the (tan θ) / r term from changeeasteast.

We will call the r coordinate of the Splinter’s centre of mass r0; the centre of mass will be in free fall, so acc(r0)=0. This will only be true if:

C(r0) = √(G(r0) / [H(r0) – G(r0)]) (5)

which in turn gives us the tangent to the world line of the Splinter’s centre of mass:

u(r0) = √(G(r0) / [H(r0) – G(r0)]) east + √(H(r0) / [H(r0) – G(r0)]) time (6)

From the east component of this tangent vector, we can deduce the rate at which the Splinter is orbiting the Hub. Because east is defined to be a unit vector in terms of distances measured by a stationary observer, we only have to multiply it by (1 / r0) to get an angular velocity, since by definition of the r coordinate the full circumference of the equator is 2 π r0, with all angles and distances related by the same factor of r0. So the angular velocity of the Splinter — by its own clocks, since u(r0) points in the Splinter’s own time direction — will be:

dφ/dτSplinter = (1 / r0) √(G(r0) / [H(r0) – G(r0)]) (7)

Because we are assuming a spherically symmetrical spacetime geometry, this equation will also hold for an object orbiting at a slight incline to the Splinter’s orbit, such as a stone displaced a small distance from the Null Line in the shomal or junub direction. Such a stone will cycle back and forth across the plane of the Splinter’s orbit, completing one cycle with each orbit. (See the second animation on the Tidal Accelerations page for an illustration of this). The tidal acceleration in this direction, which we will call ashomal, will be equal to minus the square of the angular frequency of the motion. Why? Because the mathematics here is the same as that of an object bouncing on the end of a spring. If a spring that is stretched by a distance x causes its tip to accelerate at a rate of –ω2x, then it will oscillate with an angular frequency ω. Like the force on a stretched spring, tidal acceleration is also simply a multiple of the distance from the point of equilibrium, so the oscillations it creates obey exactly the same equations.

So we have a value for ashomal in terms of G and H:

ashomal = – (1 / r02) (G(r0) / [H(r0) – G(r0)]) (8)

Next, we will determine the rate at which the Splinter is spinning. As we move along the Splinter’s world line, how does the out direction (to which the Splinter is fixed, because it’s tidally locked to keep the same face towards the Hub) change with respect to parallel transport?

changeu(r0)out = √(G(r0) / [H(r0) – G(r0)]) changeeastout + √(H(r0) / [H(r0) – G(r0)]) changetimeout  
  = √(G(r0) / [H(r0) – G(r0)]) H(r0) east + √(H(r0) / [H(r0) – G(r0)]) G(r0) time  
  = √(G(r0) H(r0)) rarb  
where rarb = √(H(r0) / [H(r0) – G(r0)]) east + √(G(r0) / [H(r0) – G(r0)]) time  

The vector rarb here is a unit vector perpendicular in spacetime to u(r0), so it is a purely spatial direction as far as the Splinter is concerned. (Two vectors in space are perpendicular when the sum of the products of their components, their “dot product”, is zero; in spacetime, this formula is changed by subtracting, rather than adding, the product of the time components.) The vector rarb is also perpendicular to out. So this result is telling us that, compared to parallel transport, the direction out is rotating in the rarb direction with an angular velocity of √(G(r0) H(r0)). It follows that the tidal acceleration ararb that precisely balances the centrifugal force due to the Splinter’s rotation must be:

ararb = G(r0) H(r0) (9)

The third tidal acceleration we need is in the garm/sard, or out direction. We start by noting that every point on the Splinter is orbiting the Hub with the same angular velocity, and the argument we used to link that velocity to the east component of u(r) works just as well for values of r other than r0. It then follows from equation (3) that:

C(r) / r = C(r0) / r0  
C(r) = C(r0) r / r0  
  = (r / r0) √(G(r0) / [H(r0) – G(r0)])  

[This formula is not exactly right, because it gives a constant angular velocity across the Splinter with respect to time measured by clocks on the Splinter itself, whereas what we really need for perfectly rigid motion is a constant rate of change of the longitudinal coordinate, φ, with respect to the Schwarzschild time coordinate, t. However, close to the Splinter’s centre the distinction becomes ever smaller, and it turns out that for the purpose of computing the rate of change of acceleration at the centre it makes no difference at all.]

Substituting this into equation (4) gives us:

acc(r) = [ (r / r0)2 (G(r) – H(r)) (G(r0) / [H(r0) – G(r0)]) + G(r) ] out (10)

We want to know how this acceleration (which is zero at r=r0) changes as we move further from, or closer to, the Hub than the Splinter’s centre of mass. However, we can’t quite get this simply by taking the derivative with respect to r, because the coordinate r doesn’t measure actual distances. So first we need to know the ratio between a change in r and an actual distance in the out direction.

Parallel Transport

We can find this ratio by considering how much the length of a small arc of a circle will change if we add a layer of actual width 1 to the circle. In the diagram on the left, we make use of equation (2e),which tells us how the direction out compares to its parallel-transported version when transported by a distance r φ to the east: the difference is a vector of length r φ H(r).

By definition, the r coordinate is equal to the arc length divided by the angle, so the change in the r coordinate divided by the actual distance increment (which is 1 in this case) is:

dr/ds = r H(r) (11)

Now that we have this ratio, we can compute the total acceleration per unit distance in the garm/sard direction. We take the opposite of the derivative of equation (10) with respect to r, multiply by dr/ds, and set r equal to r0. The result is:

Asard = r0 H(r0) ( [G(r0) (H'(r0) – G'(r0))] / [H(r0) – G(r0)] + 2 G(r0) / r0G'(r0) ) (12)

We have used the notation Asard here as a reminder that this acceleration is a total which comprises both the pure tidal acceleration, asard, and an extra acceleration due to centrifugal force. Because the centrifugal force is balanced exactly by the tidal acceleration along the direction of the orbit, ararb, we have asard = Asard + ararb, and hence Zak’s principle tells us:

ashomal + ararb + asard = ashomal + 2 ararb + Asard  
  = 0 (13)

With a little rearrangement, and putting r in place of r0 for simplicity, equations (8), (9), (12) and (13) yield:

r3 H(r) [ H(r) G'(r) – H'(r) G(r) ] + G(r) = 0 (14)
Dust cloud free-falls towards Earth

To obtain a second equation, we will now consider a second free-falling body: a cloud of dust that starts from rest and free-falls towards the Hub (see animation on the right).

We will define the quantity b(r1) to be the second rate of change with time of the r coordinate of a speck of dust that starts from rest at r=r1, evaluated at the instant the speck begins to fall; the time here is measured by an observer at rest at r=r1. Note that because the r coordinate is not an actual distance, this second derivative is not a physical acceleration.

Suppose that two specks of dust are initially at rest at the same r coordinate, r1, and are separated by a small angle α; to be concrete, suppose α is the difference in their latitude coordinates. As they fall, their r coordinates will remain equal, and the angle between them will remain fixed at α. The distance between them, in the limit of small α, will equal the length of the arc that joins them, which by definition of the r coordinate is α r. So the second rate of change with time of that distance, as they begin their fall from r1, will simply be α b(r1). If we then divide that by their initial separation, α r1, we find that the tidal acceleration in the transverse direction for our dust cloud as it begins to fall is:

atransverse = b(r1) / r1  

How can we link this to the functions G and H? From equation (2a), G(r1) out is the acceleration experienced by a body that remains fixed at r1, so if instead of being fixed that body is allowed to fall, – G(r1) will be the second rate of change with time of the actual distance fallen. If we call that distance s, we have d2s/dτ2 = – G(r1), and:

b(r1) = d2r/dτ2|r=r1  
  = d[dr/dτ]/dτ|r=r1  
  = d[(ds/dτ)(dr/ds)]/dτ|r=r1  
  = (d2s/dτ2)(dr/ds) + (ds/dτ) d[(dr/ds)]/dτ|r=r1  
  = G(r1) r1 H(r1)  

where in the last step we’ve made use of the fact that ds/dτ=0 at r=r1 (i.e. the dust is falling from rest at r1), and we have substituted the value for dr/ds from equation (11). So we have the transverse tidal acceleration in terms of G and H:

atransverse = G(r1) H(r1) (15)

In order to find the radial tidal acceleration, we need to bring out the heavy machinery of equation (1), substituting out for east and time for north, then making use of our particular spacetime connection in terms of G and H, and the general formula for a commutator in terms of the connection, [a, b] = changeabchangeba:

aradial out = changetime(changeouttime) – changeout(changetimetime) + change[out, time]time  
  = changeout(G(r) out) + changeG(r) timetime  
  = – dG(r)/ds outG(r)2 out  
  = – (dr/ds) dG(r)/dr outG(r)2 out  
  = – (r H(r) G'(r) + G(r)2) out (16)

Commutator between time and out

These calculations show that an approach that might have been tempting, simply to take the derivative – dG(r)/ds, would not have been correct. Because out and time don’t commute, there is an extra term. But what does it mean, physically, that [out, time] = – G(r) time, rather than zero? If you go forwards in time one second, then outwards one metre, you won’t arrive at the same spacetime event as going outwards one metre, then forwards in time one second; after taking the second path you’d need to wait an extra G(r) seconds in order to meet up with the endpoint of the first path.

Essentially, what we’ve encountered here is “gravitational time dilation”: different intervals of time passing at different distances from the Hub. More precisely, if we imagine slicing up spacetime into three-dimensional hypersurfaces that are considered to be purely spatial by all stationary observers, then less time passes between the same two hypersurfaces when you’re closer to the Hub than when you’re further away.

Because the transverse acceleration applies in two of the perpendicular directions, Zak’s principle gives us:

2 atransverse + aradial = 0 (17)

which in turn means that:

2 G(r) H(r) + r H(r) G'(r) + G(r)2 = 0 (18)

We also have equation (14):

r3 H(r) [ H(r) G'(r) – H'(r) G(r) ] + G(r) = 0 (14)

So between the two, we ought to be able to solve for G(r) and H(r). As the first step to doing this, we will define two new functions:

P(r) = G(r) H(r) (19a)
Q(r) = H(r) / G(r) (19b)

Solving these for G(r) and H(r), we get:

G(r)2 = P(r) / Q(r) (20a)
H(r)2 = P(r) Q(r) (20b)

Taking derivatives of equations (19) yields:

P'(r) = G'(r) H(r) + G(r) H'(r) (21a)
Q'(r) = [H'(r) G(r) – H(r) G'(r)] / G(r)2 (21b)

If we multiply equation (14) through by G(r), substitute from (19a) and (20a), then multiply through by Q(r) / P(r), we get:

r3 Q(r) [ H(r) G'(r) – H'(r) G(r) ] + 1 = 0 (22)

Solving for the expression in brackets from (21b), and again substituting (20a) for G(r)2, gives:

r3 P(r) Q'(r) + 1 = 0  
P(r) = 1 / [r3 Q'(r)] (23a)
P'(r) = – [r Q''(r) + 3 Q'(r)] / [r4 Q'(r)2] (23b)

Combining (20a), (21a) and (21b), we find:

H(r) G'(r) = (1/2) [P'(r) – P(r) Q'(r) / Q(r)] (24)

Using (19a), (20a), (23a), (23b) and (24), we can transform equation (18) into:

Q'(r) [ Q(r) – r Q'(r) + 2 ] – r Q(r) Q''(r) = 0 (25)

Equation (25) is solved by a linear function:

Q(r) = r / M – 2  

which in turn gives us:

P(r) = M / r3  
G(r) = (M / r2) (1 / √(1 – 2M / r)) (26)
H(r) = (1 / r) √(1 – 2M / r) (27)

Having found G(r) and H(r), we now know the connection, equations (2a-g), for this spacetime geometry! Furthermore, we can substitute these expressions for G(r) and H(r) back into our various intermediate results. One thing we find is that:

b(r) = –r G(r) H(r) = – M / r2

which is telling us that the second rate of change of the r coordinate of a body falling from rest, with respect to time for a stationary observer, is just the same as the Newtonian value. Of course r is not quite equal to distance; we also have:

dr/ds = r H(r)  
  = √(1 – 2M / r) (28)

Looking at the tidal accelerations in the Splinter, we have:

ashomal = M / [r2 (r – 3M)] (29a)
ararb = M / r3 (29b)
Asard = 3M (r – 2M) / [r3 (r – 3M)] (29c)
Asard / ashomal = – 3 (1 – 2M/r) (29d)

So the magnitude of the “ratio of weights” inside the Splinter will be less than 3, and the observed value for the ratio, around two-and-a-quarter, can be explained if r = 8M.

The Schwarzschild Metric

Readers who have studied some general relativity might be wondering how close we’ve come to presenting the Schwarzschild geometry in its traditional form, as a metric in terms of the Schwarzschild r and t coordinates:

ds2 = – (1 – 2M / r) dt2 + 1 / (1 – 2M / r) dr2 + r2 (dθ2 + (cos θ)22) (30)

The reciprocal of equation (28), squared, gives us the term 1 / (1 – 2M /r) dr2, and the terms involving dθ and dφ are just the metric on any sphere of radius r, but so far we haven’t made any explicit mention of the Schwarzschild t coordinate. This coordinate doesn’t measure proper time; rather it is defined to have a useful geometric property. If we carve up spacetime into three-dimensional slices that are perpendicular everywhere to time, the time direction for a stationary observer, then we will assign the same value of the Schwarzschild t coordinate to each slice.

To find the relationship between t and proper time for a stationary observer, we note that in the commutator diagram just after equation (16), the two out vectors lie in surfaces separated by a fixed increment of the Schwarzschild t coordinate (which corresponds, at the left of the diagram, to 1 unit of proper time, as represented by the time vector.) If we abbreviate dt/dτ (the amount of Schwarzschild time corresponding to 1 unit of proper time) as T(r), then we have from the diagram that T(r) is changed to T(r)(1 – G(r)) as we move outwards one unit of proper distance, or dr/ds units of r. In other words:

T'(r) = T(r) G(r) / (dr/ds)  
  = T(r) M / [r (r – 2M)] (31)

It’s not hard to check that this is solved by:

T(r) = 1 / √(1 – 2M / r)  
dτ/dt = √(1 – 2M / r) (32)

In fact, (32) is not the unique solution to (31), but it is unique if we require dτ/dt to approach 1 as r goes to infinity. So we can single out the Schwarzschild t coordinate by the condition that, very far from the black hole, it simply advances at the same rate as proper time for a stationary observer. (There is also an arbitrary choice of the origin, t=0, but that has no effect on the metric.)

The Wind Past the Splinter

When computing the garm-sard acceleration, we made the assumption that every point on the Splinter circled the Hub with the same angular velocity as measured by a clock tied to the Splinter at that point. However, that wasn’t quite correct! The Schwarzschild coordinates t and φ both give symmetries of the spacetime geometry: if you take any set of events in spacetime and increase the t coordinates of all of them by the same amount, the geometry of that set will be unchanged, and the same applies to their φ coordinates. So if the Splinter is to circle the Hub with no change whatsoever in its shape or dimensions, the t and φ coordinates need to change in step. In other words, it is dφ/dt — with t the Schwarzschild time coordinate, not time measured by a clock tied to the Splinter — that must be the same throughout the Splinter.

To demonstrate that the simplification we used doesn’t actually change anything, we’ll sketch a nice consistency check that we can perform on our results. Using the new, more rigorous assumption that dφ/dt is constant throughout the Splinter, we will calculate the velocity of the orbiting “wind” of the accretion disk relative to the Splinter, and show that the Coriolis acceleration (due to the wind’s motion in the Splinter’s spinning frame of reference) is perfectly matched by the garm-sard acceleration we calculated previously. This match-up explains, from the Splinter’s point of view, why the wind can blow past in a straight line, rather than having its path bent by the weight that stationary objects experience.

We start by finding dφ/dt for any object freely orbiting the Hub. We will abbreviate this as ωc(r), the “coordinate angular velocity”. We find this by first taking the ratio between the east and time components of u(r) from (6), which gives us the linear velocity of the orbiting object in a stationary reference frame, then dividing through by r to get an angular velocity, and finally multiplying by dτ/dt from (32) to convert from stationary proper time to coordinate time. The result is very simple:

ωc(r) = √(M / r3) (33)

This is exactly the same form as the Newtonian orbital angular velocity, though of course the meaning of r here is slightly different.

Next, we need to know the unit tangent to the world line, or 4-velocity, for an object that circles the Hub at a given radial coordinate, r, and with a given coordinate angular velocity ωc that might or might not be the orbital value at r. To find this, we more or less work backwards from what we just did; given an ωc we convert it to an angular velocity wrt stationary proper time by dividing through by dτ/dt from (32). We multiply that by r to convert to a linear velocity, which must then be the ratio of the east and time components of our 4-velocity u. It’s then just a matter of “normalising” u, dividing it through by a factor that ensures that its squared length is –1. The result is:

uc) = [√(r – 2M) time + ωc r3/2 east] / √((r – 2M) – ωc2r3) (34)

We can now apply this formula to the two objects that are of interest to us. Firstly, we want to know u for a particle of wind freely orbiting the Hub, so we simply substitute (33) into (34), yielding:

uwind = [√(r – 2M) time + √M east] / √(r – 3M) (35)

Secondly, we want u for a particle of rock fixed to the Splinter at r, but whose coordinate angular velocity is that of the centre of the Splinter, ωc(r0). This is:

urock = [√(r – 2M) time + √(M (r / r0)3) east] / √((r – 2M) – M (r / r0)3) (36)

This 4-velocity urock gives us the direction of time for the rock of the Splinter, and from that it’s easy to find the eastwards direction of space for the rock, which we will call rarbrock; we simply switch the east and time components, yielding an eastwards pointing vector that is orthogonal to urock:

rarbrock = [√(M (r / r0)3) time + √(r – 2M) east] / √((r – 2M) – M (r / r0)3) (37)

To find the linear velocity of the particle of wind relative to the rock, we project the vector uwind into the reference frame of the rock, taking its dot product with rarbrock and urock. Then we take the ratio of those space and time components:

vwind = uwind·rarbrock / uwind·urock  
  = √[M (r – 2M)] (r03/2r3/2) / ( M r3/2 – (r – 2M) r03/2 ) (38)

We need to know how this wind speed varies with distance from the centre of the Splinter, so we take the derivative of it with respect to r, multiply by dr/ds from (28), and then set r0 equal to r (because the Splinter is so small that we can treat this rate of variation as constant). This gives us:

d(vwind)/ds = (3/2) (√M) (r – 2M) / [ r3/2 (r – 3M) ] (39)

Finally, the Coriolis acceleration due to a velocity in the plane of rotation of a reference frame is double the velocity times the rate of rotation. The rate of rotation of the Splinter’s reference frame can be seen from (29b) to be √(M / r3), so we have:

d(aCoriolis)/ds = 3 M (r – 2M) / [ r3 (r – 3M) ] (40)

This is in perfect agreement with Asard from (29c)! In other words, the garm-sard “weight” is precisely matched to the Coriolis acceleration, so the wind is seen from the Splinter to be subject to no net force, allowing it to move in a straight line.

We’re now in a position to see exactly why the original simplification we made caused no problems. Equation (32) tells us that the time on a stationary clock runs slower than the Schwarzschild t coordinate, but less so with increasing r. However, though the time on a clock that moves with the Splinter will, in turn, run slower than a stationary clock, with increasing r it will be moving faster, and hence run even slower. So in comparing a clock tied to the Splinter with the Schwarzschild t coordinate, any change in r will have two opposing effects, and at the centre of the Splinter they will balance out. That’s not to say that Splinter time agrees with the t coordinate there; it certainly doesn’t run at the same rate. But the ratio between the two rates is, to first order, unchanged as you move across the Splinter.

We can quantify this by combining dτ/dt from (32) with dτ/dτrock, which is just the time component of urock from (36), to obtain:

rock / dt = dτ/dt / (dτ/dτrock)  
  = √(1 – 2M / r) √((r – 2M) – M (r / r0)3) / √(r – 2M)  
  = √(1 – 2M / rM r2 / r03) (41)

We then take the derivative with respect to r:

d[dτrock/dt]/dr = M (1 / r2r /r03) / √(1 – 2M / rM r2 / r03) (42)

At r=r0 this derivative is zero. So the distinction between the world lines of the Splinter having a constant value of dφ/dt and a constant value of dφ/dτrock is a second-order effect that makes no difference to any derivatives we compute at the centre of the Splinter, such as the one that gave us the garm-sard acceleration.

The Splinterites’ False Starts

On this page we’ve described the first spacetime geometry found by the Splinterites that satisfies Zak’s principle and agrees with the empirical fact that the “ratio of weights” is always less than three. However, before finding this geometry the Splinterites found two others which satisfied Zak’s principle, but gave incorrect values for the ratio of weights.

The first geometry was based on the assumption that there was a universal notion of time, which did not get mixed up with the directions of space. By searching for a connection that respected that assumption, the Splinterites found a geometry where the ratio of weights was always precisely three. And that geometry, of course, is a spacetime that describes Newtonian gravitational physics around a central mass. We’re not used to thinking of Newtonian physics as relating to “spacetime geometry” at all, but the brilliant French mathematician Élie Cartan put Newtonian gravity in precisely these terms as part of his work to elucidate Einstein’s theory of general relativity. The calculations the Splinterites would have performed are given in this companion page on Newtonian Spacetime Geometry.

The second geometry the Splinterites found was based on the assumption that time could be treated completely identically to the spatial directions; this led to a ratio of weights that was always greater than three. We’ll leave it as an exercise for the reader to go through the calculations on this page and make the necessary changes to derive that result.

The correct ratio of weights only arises from a third alternative, where the time dimension is distinguished by having a different sign associated with it in the spacetime length, –ut2 + ux2 + uy2 + uz2, that is invariant for all observers. This also implies that there must be a certain speed that is agreed upon by all observers, which turns out to be the speed of light. In our history, this realisation arose from a long process of studying electrostatic and magnetic phenomena, leading up to Maxwell’s equations of electromagnetism, which describe light itself. But what we’ve shown here is that in the right environment, it could also be revealed by measurements of tidal effects, and reasoning involving nothing but gravitational physics.

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Incandescence / Deriving the Simplest Geometry / created Saturday, 5 January 2008 / revised Friday, 3 April 2009
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