Collected short fiction.., p.179
Collected Short Fiction of Greg Egan, page 179
δu = –R(u,n,u)
But u doesn’t change relative to the geodesics as it’s parallel-transported along them, between Q and R and between S and P — that’s the definition of geodesics — so we can attribute this entire discrepancy, δu, to the difference in direction of the geodesics at S and R. Since the two geodesics start out parallel, the first rate of change of their separation n is zero. But since they nonetheless manage to acquire a relative “tilt” of δu, after we follow them a unit distance in the u direction, the second rate of change of their separation is δu, which is –R(u,n,u). In other words:
∇u∇un = –R(u,n,u) (16)
To compute the second rate of change in the volume between the geodesics of a whole cluster of particles (which we’ll assume for simplicity to have an initial volume of 1), we need to take the second rate of change of the distance between them in each of the three dimensions perpendicular to u, and add up the results. But we might just as well do this over all four coordinate directions instead, because any contribution parallel to u will always be zero. We can write this most succinctly by defining a new tensor, known as the Ricci tensor, extracting the second rate of change of distance in each of the coordinate directions by feeding a coordinate 1-form into the very first slot of R (the one that we usually leave “unfed” in order to get a vector, rather than a number, as the final result) while setting n, the initial separation between geodesics, to the corresponding coordinate vector.
Ricci(v,w) = R(dx,v,∂x,w) + R(dy,v,∂y,w) +
R(dz,v,∂z,w) + R(dt,v,∂t,w) (17)
(∂u∂uV)/V = –Ricci(u,u)
A tensor defined this way — by slotting coordinate 1-forms and vectors into another tensor and adding up over all the coordinate directions — is called a contraction of the original tensor. We say that the Ricci tensor is “the contraction of the Riemann tensor on its first and third slots.” You can form a contraction over any two slots of a tensor, but if they both take vectors or both take 1-forms, you must lower or raise one index first, so you can feed coordinate vectors to one, and coordinate 1-forms to the other. If you don’t, the result isn’t coordinate independent.
The negative of the Ricci tensor gives the proportional second rate of change of the volume between geodesics, which we’d like to relate somehow to the stress-energy tensor T. In analogy to Equation (15), a reasonable first guess would be:
Ricci = 4πκ T (maybe?)
There’s a problem, though: if you calculate div Ricci, the divergence of the Ricci tensor, it’s not zero. This means the equation we’ve just written is incompatible with div T = 0, the conservation of 4-momentum!
Luckily, it turns out that we can use the Ricci tensor to construct another tensor that is divergence free. First, define a contraction known as the Ricci scalar, which is normally written as R (not in bold face, since it’s a number, not a tensor). Because the Ricci tensor as we initially defined it had rank (2,0), we have to perform the contraction on a version which has had one index lowered, to become rank (1,1).
R = Ricci(dx,∂x) + Ricci(dy,∂y) +
Ricci(dz,∂z) + Ricci(dt,∂t) (18)
There’s a certain combination of the Ricci tensor, the metric g, and the Ricci scalar that’s divergence free. This is known as the Einstein tensor, and it’s always written as G.
G = Ricci – (R/2)g (19)
In the next section we’ll say a bit about why this tensor is divergence free, but before doing that let’s write the equation connecting G to the stress-energy tensor. First, note that in Minkowskian coordinates:
G(∂t,∂t) = Ricci(∂t,∂t) – (R/2)g(∂t,∂t)
= –(∂t∂tV)/V + (R/2)
using Equation (17), and the fact that the Minkowskian metric gives g(∂t,∂t)=–1. Now, in spacetime that isn’t very strongly curved, the Ricci scalar, R, turns out to be “dominated” by the last term in Equation (18), Ricci(dt,∂t). Because we’re using Minkowskian coordinates, the equivalent expression for the (0,2) tensor is –Ricci(∂t,∂t), which in turn is equal to (∂t∂tV)/V. So G(∂t,∂t) is approximately equal to (–∂t∂tV)/2V — half the value we’d get from the Ricci tensor — and to be compatible with Equation (15), we must have:
G = 8πκ T (20)
This, at last, is the Einstein equation, linking spacetime curvature with the density of matter and energy!
This equation is not unique in meeting the requirement that div T = 0. Because of the compatibility of the metric with parallel transport, all covariant derivatives of the metric are zero, and hence the divergence of any constant multiple of the metric is also zero. So there’s no fundamental reason why the true equation for spacetime curvature might not be:
G + Λg = 8πκ T (21)
The symbol Λ (this is a Greek letter, the capital lambda) stands for a number called the cosmological constant, and its value is still very much a matter of debate. A negative Λ would cause empty spacetime to be curved as if it contained energy; a positive Λ would cause it to be curved as if it contained “negative energy,” in the sense that it would cause geodesics to move apart rather than come together. When Einstein first developed general relativity, he chose a small positive value for Λ that would balance the curvature caused by the overall density of matter in the universe, keeping everything static, because at the time there was little observational evidence to support what is now common knowledge: the universe is expanding. When Einstein learnt of this, he declared the cosmological constant to be the greatest mistake of his life, and decided that the true value was exactly zero. However, recent astronomical observations suggest a positive value, sufficient not only to overcome the mutual attraction of matter, but to cause the universe to expand ever more rapidly in the future. Whether or not this is the final verdict, there’s still plenty of scope for quantum mechanical treatments of the vacuum, and of gravity itself, to shed more light on the issue of why Λ takes whatever value it actually has.
Although Λ is immensely important in cosmology, on any “small” scale — at least up to the size of clusters of galaxies! — it’s definitely insignificant, and for the remainder of this article we’ll simply assume that Λ=0, and use Equation (20).
The Bianchi Identity
Figure 8 shows a path that leads from a point, P, around a small cube whose edges are all one unit long, and point in the directions u, v and w. This path traverses every face of the cube exactly once, but it traverses every edge an even number of times, backwards as many times as forwards.
If you parallel-transport a vector b around this path, it will come back unchanged, because every step you travel along an edge in one direction, you eventually travel again in reverse, undoing the effect. However, we can write this overall lack of change as a sum of the changes we get from parallel transport around six simple loops: in each of three planes defined by pairs of the three vectors (e.g. u and v), we do one loop for the face of the cube that’s closest to P, and another for the opposite face, which is displaced one unit in the direction of the remaining vector (e.g. w). For the loop around the opposite face we have to get there and back from P along an edge of the cube, but since we use the same edge for both trips, the effect of that part of the path cancels out.
We move around opposite faces in opposite directions; for example, as we travel around the closest face to P in the u-v plane, the change in b is δb=–R(b,u,v), but for the opposite face it’s δb=–R(b,v,u)=R(b,u,v). However, these two terms might not cancel each other out, because R can be different on the two faces. Different by how much? By the length of the distance between the faces, which is one unit, times the rate of change of R in the direction w, which is ∇wR. So the change in b due to these two loops is ∇wR(b,u,v). Combining this for all three planes, and equating it to the overall result of zero change that we know we must get, yields:
0 = ∇wR(b,u,v)+∇uR(b,v,w)+∇vR(b,w,u) (22)
This equation is known as the Bianchi identity, and it’s the reason that G is divergence free. We won’t go through the proof that div G = 0, but basically it consists of a bit of algebraic rearrangement of Equation (22). So you can ultimately trace the fact that div G = 0 back to Figure 8, and what it says about the way changes in curvature must fit together over any volume of spacetime.
There are two ways to interpret this. One is to take div G = 0 as merely a handy clue that G is the correct choice of tensor to equate with T, since we already know that div T = 0. Another is to consider Einstein’s equation as explaining conservation of 4-momentum. Given Einstein’s equation, 4-momentum must be conserved, because div G = 0 isn’t an additional, physical hypothesis that might or might not hold, it’s a geometrical tautology: the undeniable fact that every edge in the cube in Figure 8 is traversed in opposite directions an equal number of times.
The Schwarzschild Solution
In empty space, T=0, so Einstein’s equation becomes G=0, and since most of the universe is near enough to vacuum, metrics whose curvature satisfies the “vacuum Einstein equation” are enormously important. One obvious vacuum solution is flat Minkowskian spacetime: if the Riemann curvature tensor R is zero, Ricci and G are also zero. This is a pretty good description of small regions of interstellar and intergalactic space — though not of the galaxy, or the universe, as a whole.
A more interesting vacuum solution is that which allows the moon to orbit the Earth, and planets to orbit the sun. To analyse the spacetime geometry around a star or a planet, we’ll assume that the geometry is spherically symmetrical. It turns out that there’s only one possible “class” of solutions that meet this criterion, all with the same general shape. The sole freedom left is to plug in a number that lets you set the scale — and by comparison with Newtonian gravity it’s easy to identify that number with the mass of the star or planet that lies at the centre of the vacuum geometry.
This class of solutions is known collectively as the Schwarzschild solution, and the metric is given by Equation (23). M here stands for the mass of the star, and we’ve chosen units where not only is the speed of light, c, equal to 1, but the gravitational constant κ is also 1. This makes all the algebra much simpler, and though it’s a pain to convert to and from conventional units, the less cluttered equations in between are generally worth it. In geometric units, as this system is called, everything is measured in distances — we’ll use metres. Time is measured in metres (the time it takes light to travel 1 metre, 3.3 nanoseconds), and mass is measured in metres (the mass that Newtonian gravity predicts would cause an acceleration, at a distance of 1 metre, of 1 metre per metre squared; this is 1.35 x 1027 kilograms, making the mass of the sun, 2 x 1030 kilograms, equivalent to about 1480 metres).
g = –(1–2M/r) dt⊗dt + 1/(1–2M/r) dr⊗dr +
r2(cos θ)2 dφ⊗dφ + r2 dθ⊗dθ (23)
The spacetime coordinates used for the Schwarzschild metric are called r, φ, θ and t. If you picture a sphere centred on the star, φ can be thought of as the longitude and θ the latitude of any point on the surface of that sphere. (It doesn’t matter where you put the “equatorial plane” and which hemisphere you call “north,” because the geometry is spherically symmetrical.) If you compare the part of the metric involving φ and θ with the metric we derived in the previous article for the surface of the Earth, you’ll see that it’s identical; we’ve just changed the names of the coordinates from x and y, and the radius of the sphere from E to r.
So we can imagine the star surrounded by spheres like onion layers, each with a different r coordinate, and each with the same geometry as the surface of a sphere in Euclidean space with a radius of r. The surface area of each onion layer is 4πr2, and since you can measure this without going any nearer to the star, this offers the simplest way to interpret r. But is the r coordinate actually the distance to the centre of each sphere? No. Distance is defined by the metric, and assuming that you’re stationary relative to the star, so that ∂r is your idea of a purely spatial direction, |∂r|=√g(∂r,∂r) is equal to 1/√(1–2M/r). For r greater than 2M, this will be greater than 1, which means that distances measured radially are going to be greater than changes in the r coordinate. There are “more onion layers” packed in here than there would be in Euclidean space.
That tells us a bit about the geometry of space according to stationary observers, but what about the passage of time? It’s sometimes said that “clocks run slow” in a strong gravitational field, and there are a number of works of science fiction where the protagonists deliberately travel close to a massive object (such as a black hole, of which we’ll have more to say shortly) in order to experience additional time dilation, ageing even less compared to Earth-bound people than they would from the effects of travelling through flat spacetime at the same velocity. This effect is certainly real, but the statement about clocks “running slow” needs to be treated as cautiously as the same statement about moving clocks. No clock ever truly runs slow unless it’s broken — and blaming the “flow of time” is as misleading as blaming the “flow of distance” if you happen to travel from one town to another by a longer route than someone else. Some paths through spacetime from A to B are simply shorter than others, and while curvature complicates this whole business, clocks are no more “slowed down” by gravity than your odometer is “sped up” when you drive over a mountain and register more kilometres from one side to the other than someone who took a road tunnel instead.
It’s straightforward in principle to use the metric of Equation (23) to find the proper time along any world line, but the detailed calculations for a complete journey to and from the vicinity of a massive object are a bit too messy to present here. Fortunately, there’s a much easier way to quantify gravitational “time dilation” that also tells us something about the view of the stars from near such an object. Suppose you follow the world line of a photon, as it travels from a point in space far from the object and strikes the eye of someone who is stationary relative to the object. That is, someone whose world line is a line of constant r, φ and θ, and hence whose 4-velocity will be pointing solely in the direction of ∂t. Everyone’s 4-velocity u must satisfy g(u,u)=–1, so if u=ut∂t:
–1 = g(u,u)
= (ut)2 g(∂t,∂t)
= –(ut)2 (1–2M/r)
ut = 1/√(1–2M/r)
u = 1/√(1–2M/r) ∂t (24)
This tells us, incidentally, that the t coordinate isn’t a measure of proper time for our observer, any more than the r coordinate is a measure of proper distance. The proper time that elapses along this observer’s world line will be less than any change in the t coordinate, because ∂τt — that is, the rate of change of t with respect to proper time τ — is equal to u(t)=1/√(1–2M/r), which is greater than 1.
The t coordinate is useful, though: because it doesn’t appear directly in the metric, Equation (23), the geometry of spacetime is independent of the value of t. You can think of the whole of Schwarzschild spacetime as being made up of lots of slices with different values for t, all piled one on top of the other, with the pile stretching from the past into the future. Unlike the onion layers of different r coordinates, which each have the geometry of a different-sized sphere, all these t-slices are identical. In fact, you can take any shape “drawn” on spacetime and increase the t-coordinate of every point by the same amount, and the new version will be identical to the original.
Ways of moving things that preserve their size and shape in this way are called isometries (Greek for “same distance”), and the vectors that produce them, such as ∂t, are known as Killing vectors (after the mathematician Wilhelm Killing). For example, adding thirty degrees to the longitude of every point on the coastline of Africa would just rotate the continent around the Earth, leaving its size and shape unchanged, whereas adding thirty degrees to the latitude of every point would distort the shape enormously. The longitude coordinate vector is a Killing vector, the latitude coordinate vector isn’t.
Though we won’t prove it, the projection of a Killing vector onto the tangent to a geodesic is the same everywhere along that geodesic. (If you want to test this claim with a simple example, consider the projection of the longitude coordinate vector onto the tangent to a great circle.) In the Schwarzschild geometry, since ∂t is a Killing vector and the world line of an astronaut in free fall is a geodesic, g(∂t,w) is constant for the astronaut’s 4-velocity w. A photon’s world line is also a geodesic, but in that case we have to use the photon’s 4-momentum, P, as the tangent. (The 4-velocity of a photon is a meaningless idea, because the 4-velocity must have a length of 1, but any lightlike vector has a length of zero.) The energy that an observer with 4-velocity u measures for a photon is g(u,P), so using the value of u from Equation (24) we have:
E = g(u,P)
= g(1/√(1–2M/r) ∂t, P)
= 1/√(1–2M/r) g(∂t, P) (25)
Since P is the tangent to a geodesic, and ∂t is a Killing vector, g(∂t,P) must be constant along the photon’s entire path. Let’s call this constant value E∞, since for very large values of r, 1/√(1–2M/r) gets so close to 1 that it might as well be 1, and hence the energy someone far away would measure for the photon is just g(∂t,P). This lets us write:
E = 1/√(1–2M/r) E∞ (26)
This equation is known as the gravitational blue shift, since it describes how the energy of a photon looks greater — pushing it towards the blue end of the spectrum — to someone deeper in a gravitational field. For example, at a distance of r=5.55M, E=1.25E∞, so an observer would see all the stars in the sky as being 25% “bluer” than someone far away in space.
Because the energy of light is proportional to its frequency — the number of complete oscillations the light wave performs in a second — this immediately tells us something about time as well. By measuring a greater energy for the photon, our observer is also using the light as a signal to compare his or her local clock with a clock far away, and by this method, local time seems to be “running slower” by 25%. This is not to say that the frequency of distant stars represents some kind of absolute standard for time. Like the comparison between two clocks in relative motion that we made in the article on special relativity, this is just a way of drawing a connection between two different observers — both of whom are correctly measuring proper time along their respective world lines.
