# Incandescence

## Kem’s Results on Symmetry

In Chapter 20 of Incandescence, Kem stumbles upon an invaluable insight into the consequences of symmetry in curved geometry:

“I’ve been thinking about symmetries,” Kem said. “If you look at the relationship between the direction of a natural path and a motion of symmetry, it should be the same all along the path.”

What the Splinterites have discovered here is an incredibly beautiful and powerful result, which is known to us as Killing’s theorem.

The basic idea is very simple. Suppose that you have a way of moving figures around in your geometry that perfectly preserves their shapes, keeping all lengths and angles the same. In flat space, an example of this would be rotating things around some axis. Now, think of the vectors that describe the velocities of all the points on an object as you move it this way. Every point on the object will have a vector associated with it; in fact, because you might be moving an object that lies anywhere at all, or one that fills the entire space, there will be a vector associated with every single point in the space. This set of vectors, one for each point, comprise what is known as a vector field. We will use the notation wA for the vector at point A. How does the vector at some other point B compare to that at point A? In flat space, we can say this: the difference between the two, wBwA, must be perpendicular to the vector sAB that points from A to B. Why? Because the motion we’re considering preserves all lengths and angles, it must preserve the distance between A and B: the two points can’t be growing closer or moving apart. Since point A is moving with velocity wA and point B is moving with velocity wB, the vector that joins them, sAB, will be undergoing a change of wBwA. But since the length of sAB can’t be changing, if it’s changing at all it can only be changing direction. And when a vector is changing direction, its rate of change must be perpendicular to the vector itself. (For example, consider the second hand of a clock. Because the hand remains fixed in length, the velocity of its tip will always be perpendicular to the direction of the hand.)

In curved geometry, essentially the same result holds for infinitesimally small regions. If a vector field w is associated with a motion that preserves all lengths and angles, and we look at the change in w over a small distance in some direction u, we find that the change will be perpendicular to u, for whatever choice of u we make. (The concept of “change” for a vector in curved space is made a bit more precise here.) What’s the value of this? Suppose we have a geodesic, the closest thing to a straight line in our curved geometry (this is what the Splinterites call a “natural path” when discussing the motion of objects in spacetime). Call the tangent vectors to this geodesic u. Starting at some point A, consider the length of the projection of wA onto the tangent uA. In other words, draw wA as an arrow at A and join the tip of that arrow to the geodesic by means of a line perpendicular to uA, and see how far it is from A to the foot of that line. We will call this distance wA·uA, or the “dot product” between wA and uA.

How does the length of the projection change, as we move from point A to some other point B? The answer is: it doesn’t! Because the curve is a geodesic, by our definition of “change” the direction of the curve, u, isn’t changing. Any change in w is entirely perpendicular to u, so that has no effect on the length of the projection. So what we have is a simple quantity, w·u, related to the direction of the geodesic and to our shape-preserving motion, that will remain fixed along the length of the geodesic.

This result turns out to have widespread implications. For a start, if we take the case where the geodesic is the world line of a particle in free fall, then the tangent u is what is known as the “4-velocity” of the particle, the vector that takes the role of velocity in special relativity. If we multiply u by the mass of the particle, we get its “4-momentum” or “energy-momentum vector”, which combines the classical notions of a momentum vector and kinetic energy. Now, if we’re in flat spacetime, and we choose the vector field w to be associated with any of the symmetries of that spacetime:

1. translations in time (e.g. moving everything forwards in time one second)
2. translations in space (e.g. shifting everything one metre in a chosen direction), or

then the quantities we get from w·u that remain fixed along the particle’s world line are, respectively:

1. its energy
2. its linear momentum (specifically, the component of its momentum in the direction we chose to move everything), and
3. its angular momentum about our chosen axis.

In other words, we have just derived the laws of conservation of energy, momentum and angular momentum for a free-falling particle from the geometry of flat spacetime. (Note that w·u gives these quantities per unit of mass; if we want the total quantities, we just multiply by the particle’s mass, m.)

In curved spacetime there will generally be less symmetry, but wherever there is some symmetry we can exploit it in the same way to get quantities that are conserved for our free-falling particle. For example, in the Schwarzschild spacetime around a non-rotating black hole, translations in time and rotations around the black hole are both symmetries, and we can use these symmetries to find two quantities that are conserved for a particle orbiting a black hole, or a ray of light deflected by the hole. (Some calculations that make use of this method are given here.) We will finish with three examples that are very closely related: an ant crawling on a vase, a plane flying across the surface of the Earth, and the cosmological redshift.

Firstly, consider a vase with a complicated shape that is nevertheless rotationally symmetric. That is, its cross section might be a fantastically elaborate curve, but the surface itself is created by spinning that curve around an axis. Assume that we know the function R(z) that tells us the radius of the vase, R, at any height from the bottom, z.

A rotation of the vase will move each part of it at a speed that is proportional to its distance from the axis, R(z). So if we describe the unit vector field (a vector field where all the vectors have length 1) that points horizontally around the vase as east, then the vector field associated with a rotation will be w=R(z) east.

Now suppose that an ant crawls across the surface of the vase, following a geodesic of that surface: the closest thing there is to a straight line. Calculating the shape of these geodesics might be difficult, but it turns out that if the ant is travelling at a constant speed we can say something very simple about its motion. Because the rotation of the vase is a symmetry, the projection of w onto the tangent to the ant’s path will be constant along the entire path. That projection will be proportional to the component of the ant’s velocity that points around the vase, or “eastwards”; if we call this veast, it will obey the simple rule:

veast(z) R(z) = constant

If, at some point where the radius of the vase is R1, the ant’s eastward speed is v1, then its eastward speed at any other point will be:

veast(z) = v1 R1 / R(z)

In other words, the ant’s speed around the vase will be inversely proportional to the radius of the vase: the narrower the vase, the greater the speed in that direction. Of course, the total speed of the ant is assumed to be constant, so what this really means is that the narrower the vase, the more the geodesic will veer towards the east, and the wider the vase, the more it will point towards the “north”.

These calculations will also hold for much simpler shapes than the vase, of course; for example, they also describe the motion of a plane travelling a great-circle route above the surface of the Earth. In that case, with a little trigonometry it’s easy to verify our formula directly. In place of z it’s convenient to use the plane’s latitude, θ. We then have:

R(θ) = RE cos θ

where RE is the radius of the Earth. If the plane’s longitude is φ, its location in three-dimensional Cartesian coordinates will be:

(x,y,z) = (RE cos θ cos φ, RE cos θ sin φ, RE sin θ)

Suppose the plane’s great circle route crosses the equator at longitude φ=0, i.e. at (RE, 0, 0), and reaches a maximum northern latitude of M at φ=π/2, i.e. at (0, RE cos M, RE sin M). Its general position after it has travelled an angle of α away from the equatorial crossing will then be:

(x,y,z) = cos α (RE, 0, 0) + sin α (0, RE cos M, RE sin M) = (RE cos α, RE sin α cos M, RE sin α sin M)

We can find the plane’s velocity vector v by taking the derivative of its position with respect to α and then, assuming a constant speed of v, multiplying through by v / RE.

v = (–v sin α, v cos α cos M, v cos α sin M)

At any point (x,y,z) that lies on the surface of the Earth, if we also know its latitude θ, the unit vector pointing in an easterly direction is:

east = (–y, x, 0) / [RE cos θ]

For our plane’s coordinates this becomes:

east = (–sin α cos M, cos α, 0) / cos θ

The easterly component of the plane’s velocity is v·east, the dot product of the velocity vector and the unit vector in the easterly direction:

veast(θ) = (v sin2 α cos M + v cos2 α cos M) / cos θ = v cos M / cos θ

This makes sense, because when θ = M, at the route’s highest latitude, the plane will be travelling due east and its eastwards velocity will just equal its total speed v. At the equator, θ = 0, we have an eastwards velocity of v cos M, because the route makes an angle of M with the equator where it crosses it. This formula corresponds exactly to the one we gave for the ant on the vase if we put R1 = RE, the radius of the equator, and v1 = v cos M, the eastwards speed of the plane when it’s on the equator.

The vase scenario also turns out to be closely analogous to the phenomenon of the cosmological redshift: the fact that we observe light that was emitted in the distant past to have less energy and momentum, and hence a longer (redder) wavelength than when it was emitted. If we assume as a useful approximation that the universe is homogeneous and isotropic (that is, at any given moment it is the same everywhere, and looks the same in all directions), then its geometry at time t can be summed up simply by an overall length scale, R(t). What’s more, because the universe is homogeneous, translating things in any spatial direction will be a symmetry. (It takes a bit of work to make this precise, but if the universe is finite then its shape at time t is the higher-dimensional equivalent of a sphere of radius R(t), and we can rotate all the spheres for different values of t by the same angle, in the same direction. If the universe is infinite, there is a similar operation that we can perform.)

Now, suppose a photon free-falls through the universe, and we measure its momentum at two different times. We can choose the spatial direction that the photon is moving in as the direction for our symmetry, so we end up with a vector field w that always points in that direction. If we call a unit vector in that direction ex, then we will have w=R(t) ex, because, just like the vase, we need to move the spacetime through larger distances when it is spatially bigger in order to “rotate” the whole thing rigidly.

The projection of the photon’s 4-momentum onto ex is just its linear momentum, p, so we have:

p(t) R(t) = constant

This in turn means that if the photon’s momentum is p1 when the length scale of the universe is R1, at any other time it will be:

p(t) = p1 R1 / R(t)

So if the photon is emitted when R=R1 and then the universe expands, R(t) increases and the photon’s momentum decreases proportionately. That’s the cosmological redshift!  