Collected short fiction.., p.176
Collected Short Fiction of Greg Egan, page 176
|δv| = ε2 √[(E cos y)2 (–vy)2 + E2 (cos y)4 (vx)2]
|v| = √[(E cos y)2 (vx)2 + E2 (vy)2]
|δv| / |v| = ε2 cos y
which is proportional to the area of the loop, once you take into account the fact that ε units of longitude comprise a shorter distance at higher latitudes, by a factor of cos y. For a given loop, δv is perpendicular to v and proportional to it in length, so this confirms that the parallel-transported version of v has simply been rotated (if δv wasn’t perpendicular, that would imply some expansion or contraction as well).
Equation (17) can be rewritten as:
R = ∂x⊗dy – (cos y)2 ∂y⊗dx (18a)
δv = –ε2 R(v) (18b)
where R is a tensor of rank (1,1). From our earlier description of tensors, a tensor of rank (1,1) would be fed a 1-form and a vector, to produce a number. But you can also leave the vector part “unfed” — combined with nothing — and use the tensor to produce one vector from another. In other words, Equations (18a) and (18b) are the same as Equation (17) because (∂x⊗dy)(v)=vy∂x and (∂y⊗dx)(v)=vx∂y, with the 1-forms dy and dx combining with the vector v, and the vectors ∂x and ∂y left untouched as vectors.
The tensor R is almost the Riemann curvature tensor — a famous geometrical object containing all the information about the curvature of a given space or spacetime. There’s one slight omission, though; we haven’t considered the fact that in general, you need to specify what plane the loop lies in. There’s only one possibility in two dimensions, but in three or four you have to specify a choice. The full Riemann tensor takes this into account: as well as feeding it the original vector v, you have to feed it two other vectors, say w and z, which together single out the plane of the loop. What’s more, you can move the factor of ε into those vectors, so they specify the size of the loop as well.
We can modify Equations (18a) and (18b) to work like this:
R = (∂x⊗dy – (cos y)2 ∂y⊗dx) ⊗ (dx⊗dy–dy⊗dx) (19a)
δv = –R(v, ε ∂x, ε ∂y) (19b)
Here, the (dx⊗dy–dy⊗dx) part of R simply converts the last two vectors fed to R into the number ε2. But it would cope just as well with whatever vectors you fed it; for example, if you reversed the roles of ∂x and ∂y and traversed the standard loop in the opposite direction, it would give –ε2 instead.
In the next article, we’ll see how the curvature of spacetime can be linked to the distribution of matter and energy, through Einstein’s equation — and then we’ll look at one solution of that equation in detail: a black hole.
3: Black Holes
Mass
Velocity and Acceleration
Energy and Momentum
The Stress-Energy Tensor
Conservation of 4-momentum in Curved Spacetime
The Einstein Tensor
The Bianchi Identity
The Schwarzschild Solution
· · · · ·
The previous article in this series began building the framework of ideas needed for general relativity by describing the geometry of manifolds — mathematical spaces without any notion of distance or angle — and then showing how it was possible to add a metric that defined these things in a very general way. The idea of parallel transport of a vector was introduced: moving along any path, you can carry a kind of “reference copy” of a vector from your starting point with you. A path is called a geodesic if it continues to follow the parallel-transported copy of its initial direction, never swerving away from its original bearing. Parallel transport of a vector around a closed loop can produce a reference copy back at the starting point that fails to match the original vector, and this effect is used to quantify the curvature of space (or spacetime), via the Riemann curvature tensor.
Einstein’s equation links the curvature of spacetime with the presence of matter and energy. We haven’t quite said all that we need to about curvature, but this article will begin by attacking the other side of the equation. This will give us some insight into why the equation takes the form it does, before we reach the final goal: examining one solution of the equation, the Schwarzschild solution, which describes a black hole.
Mass
If we want to quantify the amount of matter and energy in a region of spacetime, a good place to start is the idea of mass. According to Newtonian physics, when we weigh an object we’re measuring the gravitational force that the Earth exerts upon it, and this force is taken to be proportional to the object’s mass. Mass is usually defined quite differently, though, through the property of inertia: in the absence of complications like friction, when you apply a certain force to an object its rate of acceleration will be inversely proportional to its mass. Imagine pushing two items of furniture on frictionless pallets across a level surface; even though you’re not opposing gravity, the same push will accelerate a 100-kilogram sofa half as much as a 50-kilogram bookcase.
Are these two ways of measuring mass — gravitational and inertial — necessarily equivalent? If they are, then neglecting the effects of atmospheric drag, a truck and a pebble should both fall off a cliff with the same acceleration all the way: however much harder it is to accelerate the truck, the gravitational force on it is proportionately greater. In a vacuum, all objects should fall to Earth at exactly the same rate, whatever their mass, and whatever they’re made of. Centuries of experiments have confirmed that they do, so this is no surprise to anyone at this point in history, but from a Newtonian perspective it’s quite baffling that there are no known exceptions to this rule. No other force works like this. The electrostatic force between two objects depends on their electric charges; a proton and a positron have identical positive charges, but very different masses, so although they’ll experience the same electrostatic force — the same push — if placed in the same electric field, they won’t accelerate identically like the truck and the pebble. What’s so special about the gravitational force that it’s always perfectly matched to an object’s inertial mass?
Einstein’s answer is that gravity isn’t a force at all. Rather, in the absence of forces, any object — whatever its mass and composition — simply follows a geodesic in spacetime: it takes the straightest possible world line in the direction it happens to be heading. In the curved spacetime near the Earth, the geodesic of an object that started out stationary would carry it straight to the centre of the planet if nothing got in its way. The only reason a pebble and a truck sitting motionless on the edge of a cliff aren’t following such paths is because the cliff pushes up on them, with an electrostatic force between the electrons of the atoms at the surfaces making contact. The different forces required by the pebble and the truck to keep them from falling aren’t really opposing two different “gravitational forces.” If you define an object’s acceleration in curved spacetime as the degree to which its world line fails to be a geodesic — by analogy with the case in flat spacetime, where having a constant velocity means having a perfectly straight world line — the cliff is simply applying different forces to produce the same acceleration in two different masses.
If the idea that a motionless object can be accelerating strikes you as bizarre, imagine swinging a weight on the end of a rope: once it’s swinging in a fixed circle, you still need to apply a constant force to accelerate it towards you, just to keep it from getting further away. What you’re doing is curving a path that would otherwise be straight: cut the rope and the weight will fly off in a straight line. Letting the rope hang vertically is similar: the force you’re applying to keep the weight motionless is still keeping its world line from being the straightest possible path through spacetime, a path that would carry it towards the Earth. Being “motionless in space” (relative to some massive object like the Earth) generally doesn’t produce the straightest possible world line in curved spacetime. Compare this to a ship travelling east at a fixed latitude, say 45° S. The ship is “motionless” in the dimension of latitude — it’s not drawing closer to either the south pole or the equator — but it can only do this if its engines are constantly applying a south-directed force to keep it from heading north along a great circle, the geodesic it would otherwise naturally follow if merely propelled forward.
So, your inertial mass tells you how much force must be provided (by the ground, or the floor, or the chair you’re sitting in) to accelerate you sufficiently to keep you motionless with respect to the surface of the Earth, in exactly the same way as it tells you how much force must be provided to accelerate you into motion. The idea of a “gravitational mass” that determines your response to a gravitational field is illusory. There is only one kind of mass: inertial mass.
However, as we’ll see shortly, matter isn’t the only thing with inertia.
Velocity and Acceleration
To provide a full description of matter and energy as the source of spacetime curvature, we need to introduce the relativistic versions of some simple ideas from classical physics. The ordinary velocity vector, v, of an object in three dimensions tells you how fast the object is travelling in each of three directions — the velocity’s coordinates vx, vy and vz describe how fast the object’s x, y and z coordinates are changing with time — and the length of v is the speed of the object, how fast it’s moving overall.
This tells you everything you need to know about an object’s motion, but there’s a way of “re-packaging” the same information that’s more useful in relativity. People using different coordinate systems might disagree about every aspect of the three-dimensional vector v: not just its individual coordinates, but even its overall length, the speed of the object. But what happens if we extend the vector into four dimensions? Let’s define a vector u called the 4-velocity of the object, with coordinates ux, uy, uz and ut that describe how all four spacetime coordinates are changing for the object with time. Whose time? We want the 4-velocity to be independent of any coordinate system, so we define u as the rate of change with respect to the time shown by a clock carried along with the object itself: this is known as proper time, and it’s usually referred to by the Greek letter tau, τ. We’re defining the 4-velocity u as being ∂τ, the rate of change of things with respect to τ. For example, ux=∂τx: the x coordinate of u is just the rate of change of the object’s x coordinate, with respect to a clock moving alongside the object.
Consider a spaceship moving past the Earth with a constant speed of v, a situation where we only need to worry about one space coordinate, plus time. Call coordinates in which the Earth is stationary x and t, and coordinates in which the ship is stationary λ and τ. It’s easy to describe the ship’s 4-velocity u in its own coordinates, because we’ve defined u as ∂τ. So uλ=∂τλ=0 (the ship is motionless in its own coordinates) and uτ=∂ττ=1 (the ship’s clock keeps perfect time with respect to itself). Assuming that we’ve chosen coordinates for the ship in which the metric g is just the Minkowskian metric, we then have:
g(u,u) = (uλ)2 – (uτ)2
= 02 – 12
= –1 (1)
The negative sign for g(u,u) tells us that u is a timelike vector, as you’d expect for the direction of an object’s world line, and its length is the square root of –g(u,u), which is just 1. To describe u in Earth coordinates, we use the Lorentz transformation that we derived in the article on special relativity, rewritten slightly to apply to coordinate vectors rather than coordinates themselves:
∂λ = (∂x + v∂t) / √(1 – v2) (2a)
∂τ = (v∂x + ∂t) / √(1 – v2) (2b)
As in previous articles, we’re making life simple by using units where the speed of light is equal to 1. Since u=∂τ, this immediately tells us:
u = (v∂x + ∂t) / √(1 – v2) (3a)
ux = v / √(1 – v2) (3b)
ut = 1 / √(1 – v2) (3c)
If the ship’s speed v increases, both of the individual coordinates of u grow larger, but due to the nature of the spacetime metric the effects on the overall length of u cancel each other out. If we compute this with the Minkowskian metric in Earth coordinates:
g(u,u) = (ux)2 – (ut)2
= v2/(1–v2) – 1/(1–v2)
= –1 (4)
The agreement with Equation (1) should come as no surprise: the length of a spacetime vector is completely independent of the coordinates used. And since we can pick Minkowskian coordinates like λ and τ that are stationary with respect to any object — even in curved spacetime this is possible over a small region around the object at a given moment, just as we can always pick Euclidean coordinates over a small region of the Earth’s curved surface — it’s always going to be true that g(u,u)=–1. The 4-velocity is always a unit timelike vector, a vector with a length of 1 that points along an object’s world line. You can recover the object’s ordinary velocity v in a given coordinate system by taking the space coordinates of u and dividing them by the time coordinate, e.g. for the example we’ve just given, in Earth coordinates, vx=ux/ut =v.
Just as the acceleration of an object is defined in classical physics as the rate of change of its velocity with time, its 4-acceleration vector, a, is defined in relativistic physics as the rate of change of its 4-velocity with proper time. How can u change, if its length must always be 1? Only by the object’s world line changing direction in spacetime, which is what it means to change your ordinary velocity.
But how should we judge a “change of direction” when spacetime is curved? The physical evidence that your 4-velocity isn’t “changing direction” is simply that you’re weightless, because no force needs to act on you in order for you to follow your world line. If you’re in a spaceship that’s (a) orbiting the Earth, (b) falling straight towards a planet (without atmospheric drag), or (c) cruising through interstellar space, in all three cases you’ll be weightless. In all three cases, you’re following a geodesic. So acceleration means moving along a world line that is not a geodesic. This is true in either flat or curved spacetime, but to compute acceleration in curved spacetime you need to work out the change in an object’s 4-velocity from moment to moment by using parallel transport to carry its earlier 4-velocity forward along its world line for comparison with the later value. This is known as taking the covariant derivative of the vector u, in the direction u,which we write as ∇uu. So a=∇uu, and for a geodesic ∇uu=0.
In the previous article, we used the symbol ∇ to write the changes in coordinate vectors relative to their parallel-transported versions, e.g. on the surface of the Earth, using longitude and latitude as x and y coordinates, ∇x∂x=(sin y cos y) ∂y. This means that as you travel east (take a covariant derivative in the x-direction, ∇x) in the northern hemisphere (where sin y cos y is positive), the local direction east (∂x) “veers north” (in the direction of ∂y) relative to a gyroscope bearing or a great-circle geodesic that was pointing east when you first set out. But you can take covariant derivatives in any direction, not just coordinate directions, and you can take covariant derivatives of any vector, not just the coordinate vectors. All you have to do is ask how the vector changes relative to a parallel-transported copy of its initial value, as you travel in the specified direction.
Energy and Momentum
Another powerful concept from classical physics is the momentum vector for an object, which is just its velocity vector multiplied by its mass: p=mv. This quantifies the intuitive notion that a 1-gram bullet travelling at 1 kilometre per second, and a 1-kilogram bowling ball travelling at 1 metre per second, have something in common. Since force is defined as mass times acceleration, and acceleration is the rate of change of velocity, force can equally well be defined as the rate of change of momentum. This tells us just what it is that the bullet and the bowling ball have in common: to bring them to a halt in one second, to reduce their momentum to zero, you’d need to apply exactly the same force, 1 Newton, in either case.
Momentum turns out to be conserved: for a collection of objects — maybe interacting among themselves, but subject to no external force — the total momentum never changes. Why not? When the objects aren’t interacting, they’re subject to no forces at all, so they’ll simply keep moving with whatever constant velocities they happened to possess. When two of the objects do interact, they’ll exert equal and opposite forces on each other, and whatever change in momentum one of them experiences as a result, the other will experience an equal and opposite change. The total momentum vector remains constant.
A closely related idea is that of kinetic energy, K, which is a number rather than a vector. Energy in general can be defined as the capacity to “do work,” in the technical sense of moving a load some distance against a resisting force — it’s no coincidence that this idea developed most rapidly in the age of steam engines. Suppose you extract energy from a moving object of mass m and speed v by making it drive a piston that resists its motion with a constant force, bringing it to rest in a time t. The object’s average velocity over that period will be v/2, so it will travel a distance of vt/2. Its deceleration will be v/t, and the force needed to produce this will be mv/t. So the “work done” by the object will be the force applied, mv/t, times the distance moved, vt/2, which comes to mv2/2. This is the classical formula for kinetic energy: K=mv2/2. Although the bullet and the bowling ball mentioned earlier have the same momentum, the kinetic energy of the bullet is a thousand times greater: both can be stopped in 1 second by a force of 1 Newton, but the bullet will travel 500 metres in that time (averaging half its initial speed of 1 km/sec, as the force gradually decelerates it), the bowling ball a mere half a metre.
Energy in general turns out to be conserved, like momentum, but kinetic energy is often converted into other forms when objects interact. Sometimes these forms are really just kinetic energy “in disguise”: the frictional heating or sound produced by most objects colliding is mainly just a transfer of kinetic energy from the colliding objects to individual molecules. But kinetic energy can also be converted into various kinds of potential energy: when you release a plucked guitar string, its energy cycles back and forth between the kinetic energy of motion and the potential energy stored by the material of the string when it’s stretched — though of course it all eventually leaks away as sound, and a tiny amount of heat. Like kinetic energy, changes in potential energy can sometimes be “disguised” because they’re happening down at the level of individual molecules. When a meteor hits the Earth, most of its kinetic energy ends up as heat, some of which goes to drive chemical reactions in the surrounding rock — rearrangements of atoms which change their overall electrostatic potential energy.
