To study curvature, though, we’re initially going to use examples where we don’t have to solve equations in general relativity to find the metric. Rather, we’ll look at curved surfaces in flat space, where the metric is “inherited” from the three-dimensional space in which the surface is embedded. Spacetime curvature doesn’t arise like that — the metric is not inherited from some larger, flat “hyperspace” — but we can still define measures of curvature that are equally applicable in both cases.

  Figure 5 shows the surface of the Earth, covered with a familiar coordinate grid: x and y are just longitude and latitude. As mentioned earlier, there are problems with this grid at the poles and the 180° meridian, but we’ll stick to a region that avoids those mathematical trouble spots. We will make one slight change from the standard way of measuring latitude and longitude: we’ll measure these angles in radians, rather than degrees. If you haven’t come across this before, it’s an extremely simple idea; 360° is equal to 2 π radians — the circumference of a circle with a radius of one — and any smaller angle’s equivalent in radians is just equal to the length of a proportionately smaller arc.

  As before, imagine a train in motion across this surface. But now that the notion of “distance” is allowed, we can ask: what’s the relationship between the train’s speed, in kilometres per hour, and vx and vy, the rate at which its x- and y-coordinates are changing? If the train was heading purely in the y-direction, due north along a meridian, its speed would be equal to Evy, where E is the radius of the Earth. Why? Because vy is just ∂ty, the rate of change of the train’s angle from the equator. Multiplying by the radius of the meridian the train is travelling on — which is just the radius of the Earth — converts an angle (measured in radians) into an arc length; equally, it converts a rate of change with time of that angle into the rate of change of the arc length.

  If the train was heading purely in the x-direction, along a line of latitude, almost the same argument would apply. However, the radius of each circle of latitude obviously depends on the latitude, and it’s not hard to see that the value is E cos y. So the train’s speed would be (E cos y) vx.

  Given that the train is actually heading partly in each of these two directions, we can use Pythagoras’s Theorem to find its speed. This is where we make use of the plain old Euclidean metric of flat, three-dimensional space. The two directions, ∂x and ∂y, are at right angles to each other, and from a satellite’s point of view they can be treated like vectors in a Euclidean plane. The speed of the train is:

  |v| = √[(E cos y)2 (vx)2 + E2 (vy)2] (11)

  What metric for the curved surface of the Earth will agree with this result? A metric applied to the same vector twice is supposed to yield the square of the length of that vector, |v|2=g(v,v), so:

  g(v,w) = (E cos y)2 vxwx + E2 vywy (12)

  is clearly compatible with Equation (11), since this gives:

  g(v,v) = (E cos y)2 (vx)2 + E2 (vy)2

  We can use Equation (12) to find the train’s speed from the rates at which its longitude and latitude are changing, anywhere on the surface of the Earth (except for the poles and the 180° meridian). However, we can’t substitute, say, the x- and y-coordinates of the point P in Figure 5 in place of vx and vy, and expect Equation (12) to give us the distance from O to P. Why not? Because the metric’s relationship to the coordinates changes as you change latitude; that “cos y” in the formula means you can’t just add up the length of all the steps you took as you walked from O to P, and expect them all to bear an identical relationship to the amounts by which they increased your latitude and longitude. In Euclidean space, if you head off along a straight line, equal strides always involve the same increments in your x- and y-coordinates. In general, this isn’t true. What you can always do is use calculus to add up the distance along any path, taking account of the varying relationship of distance to the coordinates. That’s not difficult, but we won’t go into the details here.

  Equation (12) can be rewritten in a way that makes it clear that the metric is a geometrical object, independent of the coordinate system. Equations (3a) and (9) allow us to rewrite the coordinates of the vectors in terms of inner products with the coordinate 1-forms: vx=, vy=, and similarly for the coordinates of w.

  g(v,w) = (E cos y)2 + E2

  This in turn can be expressed with a more compact notation:

  g = (E cos y)2 dx⊗dx + E2 dy⊗dy (13)

  The symbol ⊗ is called the tensor product. Just as dx and dy can be combined with a single vector to yield a number, dx⊗dx and dy⊗dy can be combined with pairs of vectors to yield numbers. You can multiply as many 1-forms as you like in this fashion, and they need not be the coordinate 1-forms. For example, if m, n and p are 1-forms and u, v and w are vectors:

  m⊗n⊗p(u,v,w) =

  You can also combine vectors in the same fashion, or even a mixture of vectors and 1-forms:

  u⊗v(m,n) =

  m⊗n⊗w(u,v,p) =

  The tensor product of r vectors and s 1-forms is called a tensor of rank (r,s). Since each vector can create a number if combined with a 1-form, and each 1-form can create a number if combined with a vector, you can “feed” r 1-forms and s vectors to such a tensor, and it will give you a number. Because this number is produced by multiplying a string of inner products together, and each inner product is linear in both the 1-form and the vector, tensors are completely linear: feed them anything twice as big, and the number they produce from it will be doubled.

  Equation (13) shows that the metric for the surface of a sphere is a tensor of rank (0,2): it can be fed two vectors, and from them it produces a number. In the previous article, we showed that both the Euclidean and Minkowskian metrics are linear, so this aspect should come as no surprise. We also showed that those metrics were symmetric, i.e. g(v,w)=g(w,v), which is also clearly true of Equation (13), since it’s the sum of two tensors which combine identical 1-forms with the first and second vectors fed to them. The most general form a two-dimensional metric can take is:

  g = gxxdx⊗dx + gxydx⊗dy + gyxdy⊗dx + gyydy⊗dy (14)

  where each of the numbers gxx, gxy, gyx, gyy can vary from point to point in the manifold, but gxy must equal gyx everywhere, to ensure that the metric is symmetric. These numbers are called the coordinates of the metric, and like those of vectors and 1-forms they depend on the particular coordinate system being used. But Equation (14) as a whole is independent of the coordinate system. To take a simple example, if we switched from x and y to new coordinates u and v, where u=x/2 and v=y/2, we’d make the substitutions du=dx/2, dv=dy/2 (i.e. dx=2du, dy=2dv):

  g = 4gxxdu⊗du + 4gxydu⊗dv + 4gyxdv⊗du + 4gyydv⊗dv

  = guudu⊗du + guvdu⊗dv + gvudv⊗du + gvvdv⊗dv

  So the metric’s new coordinate guu is equal to 4gxx, but that’s balanced by the fact that du⊗du yields numbers one-quarter the size of those dx⊗dx yields from the same vectors. The same thing holds for guv, gvu, and gvv. The metric’s coordinates have changed, but the thing itself is unaltered, just as a train travelling down a railway track is unaltered by the coordinates used to measure its velocity.

  Parallel Transport

  It’s common knowledge that the “straightest possible” line — the technical name for this is a geodesic — between two points on the surface of a sphere is an arc of a great circle, a circle whose radius is equal to the radius of the sphere. If you travel along a great circle, there’s a sense in which your velocity vector is always pointing in “the same” direction, as opposed to swerving from side to side. But what exactly does this mean? It’s easy to say when two vectors at different points are parallel in Euclidean space — in a rectangular coordinate system the two vectors will have identical coordinates — but in curved space, the issue is a great deal more subtle.

  Figure 6 shows the velocity vector ∂t at two different points for a ship travelling at a constant speed along a great circle. From a satellite’s view, the velocity is clearly different from P to Q, but both vectors lie in the same plane: the plane of the great circle. As the ship moves around the Earth, it can’t head off along a truly straight line; that would take it up into space! Instead, its velocity vector has to rotate downwards, in order to stay horizontal as the Earth curves away. This rotation is entirely perpendicular to the surface of the Earth; if it was partly sideways, the ship would swerve off course.

  That’s simple enough, but how can we characterise this from the ship’s point of view? The coordinates of its velocity clearly don’t stay constant: at P the ship is heading north-east, and at Q it’s heading south-east. In general, vx and vy are going to change in a complicated manner, even as the ship maintains a constant speed along a geodesic.

  The idea that we can take a vector at P and move it along some path to Q — without “really” changing it, even though its coordinates might change — is known as parallel transport. It’s this idea that defines a geodesic: parallel transport tells you how to carry a kind of “reference copy” of your initial velocity with you as you go. If your actual velocity agrees with the reference copy all the way, you’re moving along a geodesic; if it doesn’t, you’re not.

  Physically, you could do this by setting a gyroscope spinning with its axis in the direction of your initial velocity (and subtracting out the gradual tilt of the axis towards the vertical, as the local definition of “horizontal” changed), but what we need is a mathematical recipe to predict the difference between the gyroscope’s direction and a fixed compass bearing. We’ll assume that parallel transport is a linear process for the vectors being “transported”: adding two vectors or multiplying a vector by some number at the starting point, and then transporting the result, ends up giving you the same final vector as doing the transporting first and the adding or multiplying later. We’ll also assume that over very short distances the difference between gyroscope and compass bearing is linear for the vector describing your motion; this is really just saying that the recipe for parallel transport, though it will vary from point to point (like the metric), doesn’t undergo any wild jumps that would stop you from treating it as fixed over a small enough region. Then if we know what happens to ∂x when we move a short distance ε in the x-direction, what happens to it if we move in the y-direction, and what happens to ∂y after the same two moves, all these assumptions of linearity let us work out what happens to any vector, moved (a short distance) in any direction.

  It’s standard notation to write the effects of these four possible moves as:

  ∇x∂x = Γxxx ∂x + Γyxx ∂y (15a)

  ∇y∂x = Γxxy ∂x + Γyxy ∂y (15b)

  ∇x∂y = Γxyx ∂x + Γyyx ∂y (15c)

  ∇y∂y = Γxyy ∂x + Γyyy ∂y (15d)

  Don’t be put off by the unfamiliar symbols; as ever, this is just shorthand for things we’ve just discussed. After increasing our x-coordinate by a small amount ε, “the difference between the ∂x the coordinate grid gives us and the parallel-transported reference copy of the ∂x we started off with, divided by ε” is what we’ll call ∇x∂x, with the subscript x in ∇x indicating the direction of the move. We divide out the particular change in x, ε, because the effect is linear, and what we really care about is the rate per coordinate unit. This is a vector, with coordinates Γxxx and Γyxx. Once you include the two coordinate vectors and the two directions you can consider moving in, there are eight of these numbers at every point, and together they characterise a particular way of doing parallel transport. They’re known as the connection coefficients for the geometry, or sometimes the Christoffel symbols.

  This gives us the language in which to talk about parallel transport, but we still don’t know what recipe to use in any given case — what the values of the connection coefficients are, in terms of the metric. It turns out that there are two equivalent ways to pin this down. You can assume either that: (1) parallel transport must yield geodesics that give the shortest possible distance between nearby points (or in spacetime, the longest distance), or (2) parallel transport of vectors changes neither their length nor the angle between them, and what’s more ∇y∂x=∇x∂y: the x coordinate vector changes when you move in the y-direction in exactly the same way as the y coordinate vector changes when you move in the x-direction.

  It’s beyond the scope of this article to prove that these requirements are identical, but it’s worth knowing both. Condition (1) is very simple to talk about, and makes great intuitive sense. In Euclidean space, straight lines are the shortest paths between points, so it’s only reasonable that the “straightest” path between points in curved space should be the shortest. In Minkowskian spacetime, straight lines are the longest paths, so the same should apply to geodesics in curved spacetime. In fact, the fundamental reason for all of this lies in quantum mechanics, but that’s a topic for a later article. Condition (2) is a little more complicated, but the first part sounds pretty reasonable: if the reference copy of our ship’s velocity changed length as we moved, what sort of standard would it be? And if we carried a reference copy of “starboard” (to the right) that didn’t continue to lie at 90° to the reference copy of our velocity, we wouldn’t know which to steer by, since the two would contradict each other.

  It’s possible to use Condition (2) to write a completely general formula for the connection coefficients in terms of the metric, but we’ll just state the results for the surface of a sphere:

  ∇x∂x = (sin y cos y) ∂y (16a)

  ∇y∂x = (–tan y) ∂x (16b)

  ∇x∂y = (–tan y) ∂x (16c)

  ∇y∂y = 0 (16d)

  Equation (16a) says that ∂x (east) tilts in the y-direction (north) when you travel in the x-direction (east), to an extent that’s zero either on the equator or on the poles (since either sin y or cos y is zero in those cases), and negative in the southern hemisphere. To see this, find the point on Figure 6 where the great circle is tangent to a circle of latitude, just west of point Q. The two curves are initially parallel, but then the circle of latitude — the definition of “east” — veers north relative to the great circle. Equation (16b) says that ∂x decreases in the x-direction (i.e. shrinks) as you travel in the y-direction: as you travel north, a velocity measured in degrees or radians per hour east comes to mean less and less in terms of kilometres per hour, as the circles of latitude shrink. Equation (16c) says that as you watch ∂y while moving in the x-direction (i.e. keep an eye on what happens to “north”, relative to your gyroscope, as you travel east), it tilts in the negative x-direction (west) by an amount that’s greater at greater latitudes (tan y starts off equal to zero at y=0, and grows with latitude). That makes sense: the north pole lies precisely on your left when you start out east, but if you travelled along a great circle like the one in Figure 6, it would swing to the rear of you, which is west, as you turned to the south.

  The Curvature Tensor

  What happens when you parallel-transport a vector around a closed loop? Figure 7 shows this happening on the surface of the Earth: a vector v is transported along each of the three sides of a triangle PQR built from geodesic arcs. The angle between the vector and each geodesic it moves along remains unchanged, and to make the example even simpler, v is chosen initially to be a tangent vector to one side of the triangle, PQ, at P. After being transported from P to Q this is still true — the angle between the vector and the curve remains zero. Transport from Q to R is almost as easy: the vector just ends up making the same angle with QR at R and at Q … and then the same can be said about its angle with RP at R and at P.

  At the end of the process, back at P, v no longer points in the same direction. This couldn’t happen in Euclidean space, where the vector would remain parallel to its initial position throughout. In curved space, though, there is no absolute criterion for saying one vector is parallel to another vector at another point. All you can do is transport a copy of the first vector over to the second one, by a particular route, and see if they agree — and whether they do or don’t will generally depend on the path you take.

  The failure of the vector to return to its initial direction can be linked to a well-known property of triangles in curved space: the failure of their angles to add up to 180°. The angle the vector v makes with successive sides of the triangle jumps at each vertex, by 180° minus the angle at the vertex; in total, this means it ends up rotated by 540° minus the sum of the angles. If that sum was 180°, the net rotation would be 360° and the transported vector would match the original. On a sphere, the angles of a triangle always add up to something more than 180°, so the transported vector in Figure 7 is rotated by less than 360° and fails to line up with the original.

  It’s clear that this discrepancy, δv, between the original vector and the transported version must shrink as the triangle (or any other loop) gets smaller, because on a small enough region of the Earth’s surface the geometry is indistinguishable from Euclidean geometry. However, we can still ask exactly how rapidly δv shrinks, and it turns out that for small enough loops it’s proportional to the area of the loop, and unaffected by its shape. So we can create a kind of standardised small loop, a square which consists of travelling ε units in the x-direction, ε units in the y-direction, –ε units in the x-direction, and –ε units in the y-direction, where ε is a small number whose precise value doesn’t matter. All that’s needed to compute the discrepancy in any vector transported around this loop are the connection coefficients, and the rates at which they’re changing in the x- and y-directions.

  The result for the surface of a sphere turns out to be quite simple:

  δv = ε2 (–vy ∂x + vx (cos y)2 ∂y) (17)

  Note that this is perpendicular to the original vector:

  g(v,δv) = ε2 ( (E cos y)2 vx(–vy) + E2 vyvx (cos y)2)

  = 0

  and its size relative to the original vector is:

 

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