Collected short fiction.., p.172
Collected Short Fiction of Greg Egan, page 172
x2 = g[(x1,y1),(cos A, sin A)]
= x1cos A + y1sin A (10a)
y2 = g[(x1,y1),(–sin A, cos A)]
= y1cos A – x1sin A (10b)
This is the standard way of expressing the change of coordinates for a rotation in space. There’s another way, though, which is worth writing down because of its similarity with the most common form of the equivalent spacetime equations. Put
s = tan A
= sin A / cos A
Then s is just the slope of the x2 axis as a line in the (x1,y1) reference frame, and the vector (1,s) points along the x2 axis, while the vector (–s,1) points along the y2 axis. These are not unit vectors, but we can apply Eqn (7) and divide out their lengths:
x2 = g[(x1,y1),(1,s)] / |(1,s)|
= (x1 + sy1) / √(1 + s2) (11a)
y2 = g[(x1,y1),(–s,1)] / |(–s,1)|
= (y1 – sx1) / √(1 + s2) (11b)
Spacetime Geometry
The fact that spacetime has four dimensions, as opposed to the three of space alone, is important, but it’s far from being the distinguishing feature of spacetime geometry. Our simple problem in interstellar travel involves only one dimension of space and one of time — a two-dimensional “slice” through four-dimensional spacetime — but the geometry that applies to that slice is not the same as the geometry of the familiar two-dimensional Euclidean plane.
In Euclidean geometry, given two fixed points, the (x,y) coordinates you give those points will depend on the reference frame you choose. The coordinates you give P and Q in Figure 5 depend on where you stand, and the direction you’re facing. But the distance between the points, PQ, which can be calculated with Pythagoras’s theorem, must always be the same. A quantity like this, which every observer agrees on, is called an invariant. That the distance between points is an invariant in Euclidean geometry seems almost too obvious to mention, but it’s worth checking that both Eqns (10) and (11) yield the result:
x22 + y22 = x12 + y12
Points in spacetime are usually called events, to distinguish them from points in space. Events can be specified by giving their space and time coordinates, (x,t), according to a particular observer. For example, in Figure 1, if the event of the spacecraft’s launch is (0, 0), the event of its arrival would be (8.7, 14.5).
The question is, is there a “spacetime distance” between these two events, which everyone can agree on? Is there an invariant in spacetime geometry equivalent to the Euclidean concept of distance?
In the 1880s, Michelson and Morley carried out a series of experiments which established that the speed of light in a vacuum is always the same, regardless of any motion of either the source of the light or the observer making the measurement. Light travels along paths that cut across spacetime in such a way that everyone agrees on its speed. This is the fact that Einstein used to uncover the geometry of spacetime.
Imagine the set of world lines traced out in spacetime by all the pulses of light, travelling in every possible direction, that could pass through a given event. This is known as the light cone for that event. If we’re only dealing with one dimension of space, “every possible direction” means either left-to-right or right-to-left, as illustrated by the two dashed 45° lines in Figure 6, but if you imagine spinning this diagram around the t-axis, you’ll see what the case for two spatial dimensions looks like, and why the term “light cone” is used.
In this diagram, as in Figure 1, we’ve chosen units where the speed of light (normally referred to as c) is one. In everyday units, c = 300,000 km/sec, but using light-years and years (or any similar choice, like light minutes and minutes) conveniently makes c = 1. Be warned, though: if you plug distances and times in metres and seconds into any of the formulas we’re about to derive, they won’t work.
Given a velocity of one, the equations for a pulse of light travelling left-to-right or right-to-left through the event (0,0) are:
x = t
x = –t
These two cases can be encompassed by a single equation for the whole light cone:
x2 – t2 = 0 (12)
The Michelson-Morley experiments showed that, no matter what your own velocity is, you will agree that this is the equation for the light cone. So for any two events whose separation in spacetime is such that a pulse of light can travel between them, such as O and P in Figure 6, the quantity x2–t2 must be zero — whoever calculates it, and no matter what individual x and t values they measure.
That makes x2–t2 a good candidate to take the place in spacetime of x2+y2 in space. A small change to Eqns (1) and (3) can accommodate this:
g[(x,t),(u,w)] = xu – tw (13)
|(x,t)|2 = g[(x,t),(x,t)]
= x2–t2 (14a, if x2–t2 > 0)
|(x,t)|2 = –g[(x,t),(x,t)]
= t2–x2 (14b, if x2–t2 < 0)
|(x,t)|2 = 0 (14c, if x2–t2 = 0)
The new formula for g given in Eqn (13) is known as the Minkowskian, or “flat spacetime”, metric, as opposed to the Euclidean metric of Eqn (1). This metric meets the same conditions of symmetry and linearity as the Euclidean metric, spelt out in Eqns (2).
From Figure 6, it’s clear that some vectors in spacetime, such as OR, can have negative values for x2–t2, so the new equivalent of Pythagoras’s theorem, Eqns (14), need to take account of this possibility. Although it might seem a shame to have to divide vectors in spacetime into different classes and treat them each somewhat differently, the three possibilities involve very real physical distinctions, so it’s a good idea not to try to gloss over the differences.
If x2–t2 > 0, the vector (x,t) is called spacelike. A spacelike vector slopes away from the time axis at a greater angle (and hence a greater velocity) than a light ray. No object’s world line can point in a spacelike direction. In Figure 6, not even a pulse of light at event O can travel fast enough to reach event Q. For events with spacelike separation, |(x,t)| is called the proper distance between them; an observer who judges them to have happened simultaneously measures t = 0, so |(x,t)| = x.
If x2–t2 < 0, the vector (x,t) is called timelike. A timelike vector slopes away from the time axis at a smaller angle than a light ray. Ordinary objects’ world lines point in timelike directions. In Figure 6, there’s nothing to stop a spacecraft cruising from event O to event R — and an observer on the spacecraft would consider that the two events were separated only in time, not space. For events with timelike separation, |(x,t)| is called the proper time between them; an observer who judges them to have happened at the same place measures x = 0, so |(x,t)| = t.
If x2–t2 = 0, the vector (x,t) is called lightlike, or null — because the vector’s length is zero, however large its individual x and t components are. Only photons, the particles of light (and other massless particles) have world lines pointing in null directions.
The light cone is a real, physical structure, not an artifact of the coordinates you happen to choose. And since it marks the division between timelike vectors and spacelike ones, every observer will assign a given vector to the same class. Motionless versus stationary is a matter of opinion. Timelike versus spacelike is not.
To convert between spacetime coordinates for different observers, we’ll need to be able to project one spacetime vector onto another. It would be nice to be able to use the equations we’ve established for vectors in space, like Eqn (7), and simply substitute the new metric. But rather than doing that blindly, we need to take a closer look at what the idea of projection really means, in spacetime.
Suppose we want to determine the x-coordinate of the event P in Figure 7. We want to pin down its location in space, by marking it on the x-axis. In space alone, we’d do this simply by drawing a line through P at right angles to the x-axis — but is there any justification for doing this on a spacetime diagram?
DP and EP are the world lines of two pulses of light, aimed at each other, which collide precisely at event P. Assuming that the two pulses leave “at the same time”, they must travel equal distances in order to arrive together. So the event Q, mid-way between D and E, marks P’s location on the x-axis. Because the light rays DP and EP both make 45° angles with the x-axis, QP must be perpendicular to the axis.
But now suppose we use the same method to project OP onto some arbitrary spacelike vector OG, instead of the x-axis. The same two light rays intersect OG at B and C, and the event S is mid-way between them. So according to an observer who considers that B and C (rather than D and E) happen “at the same time”, S and P (rather than Q and P) happen “at the same place”. The two concepts are bound together, just like “left-right” and “forwards-backwards”; exactly what you mean by each one depends on what you mean by the other.
OS is the projection of OP onto OG, in exactly the same sense as OQ is the projection of OP onto the x-axis. Does the spacetime metric, Eqn (13), give us the length of OS? First, since SP isn’t perpendicular to OG, it will help to know what its direction is. By drawing in another 45° line, SH, parallel to BP, we can make a triangle SHC, the same shape as BPC but half the size (since BS and SC are equal). That means H must bisect PC, and the angles either side of it will be equal.
SP, then, must make exactly the same angle with the t-axis as OG does with the x-axis. Since OG is the vector (u,w), and SP has coordinates (TQ, UR):
TQ/UR = tan A
= w/u
TQ u = UR w
TQ u – UR w = 0
g[(TQ, UR),(u,w)] = 0
Applying the Euclidean metric to perpendicular vectors in Euclidean space gives zero. By the same criterion, this shows that SP and OG really are perpendicular, in spacetime, even though the lines we draw for them on paper are not. In a perspective drawing of a room where a certain wall is viewed face-on, right angles on that wall will be right angles in the drawing — but right angles on the floor, the ceiling, and other walls will not. Don’t take this analogy too seriously — the details in each case are quite different — but when you’re drawing a spacetime diagram it would be a great surprise if you could show everything without distortion.
Now, to find OS, we make use of Eqn (14a), the new version of Pythagoras’s theorem:
OS2 = OT2 – OU2
OS = OT (OT/OS) – OU (OU/OS)
= g[(OT, OU),(OT/OS, OU/OS)]
and note that (OT/OS, OU/OS) is a unit vector:
|(OT/OS, OU/OS)| = (OT/OS)2 – (OU/OS)2
= (OT2 – OU2) / OS2
= 1
As in Euclidean space, we can find the unit vector in the direction of OG by dividing (u,w) by its length:
(OT/OS, OU/OS) = (u,w) / |(u,w)|
and so:
OS = g[(OT, OU),(u,w)] / |(u,w)|
Now, we know that:
(x,y) = (OQ, OR)
= (OT, OU) + (TQ, UR)
Making use of Eqn (2c):
g[(x,y),(u,w)] / |(u,w)| = g[(OT, OU),(u,w)] / |(u,w)|
+ g[(TQ, UR),(u,w)] / |(u,w)|
= OS + 0
= OS
We’ve shown that the spacetime metric gives the correct length of the projection OS, where S is the point on OG mid-way between two light rays that converge on P. But we’ve assumed that OG is spacelike. What about projections onto the t-axis, and other timelike vectors? Two pulses of light that leave the t-axis at different moments will never even meet — the second one will never catch up with the first. For timelike vectors, we need to take a slightly different approach.
Suppose that instead of arranging for two pulses of light to collide at P, we imagine bouncing a radar pulse off some object at P, and waiting for the reflection to come back. The time the reflection occurs will then be exactly halfway between the time the pulse was sent out and the time it returns. In Figure 8, a pulse leaves the t-axis at D, and returns at E, so R, mid-way between D and E, must mark P’s time coordinate.
As with the projection onto the x-axis, RP is at right angles to the t-axis. But if we apply the same method to another timelike vector, OG, the same radar pulses would be observed leaving and returning at B and C. B and C happen “at the same place”, according to an observer whose world line points in the direction OG, so the event S, mid-way between B and C, must happen “at the same time” as P according to that observer — just as R happens “at the same time” as P for any observer whose world line is parallel to the t-axis.
Again, SP isn’t perpendicular to OG in the normal Euclidean sense, when we draw it on paper. But the same construction shows that it’s perpendicular in the spacetime sense. The only difference comes when we calculate the length of OS; because OG and OS are timelike vectors, we have to use Eqn (14b) instead of (14a):
OS2 = OU2 – OT2
OS = OU (OU / OS) – OT (OT / OS)
= –g[(OT, OU),(OT/OS, OU/OS)]
= –g[(OT, OU),(u,w)] / |(u,w)|
This means there are two slightly different equations for the length of the projection of (x,t) onto (u,w), depending on whether (u,w) is spacelike or timelike. (If (u,w) is lightlike, then the length of the projection is always just zero.)
length of projection = g[(x,t),(u,w)] / |(u,w)| (15a, if u2–w2 > 0)
length of projection = –g[(x,t),(u,w)] / |(u,w)| (15b, if u2–w2 < 0)
Rotations in Spacetime
Now, imagine a ship that happens to be cruising past the Earth with velocity v. The event of the ship’s closest approach to the Earth is taken to be the origin of spacetime coordinates, for both an observer on Earth and an observer on the ship. But they differ on the direction of the time and space axes, just as the directions “left-right” and “forwards-backwards” are different for the two coordinate systems in Figure 4.
Since the ship is cruising past the Earth with velocity v, an observer on Earth would consider the ship’s world line to be:
x1 = vt1 (16a)
while an observer on the ship itself considers it to be stationary:
x2 = 0 (16b)
This is just the t2 axis. So on a spacetime diagram drawn by the Earth observer, such as Figure 9, the t2 axis is given by Eqn (16a). What’s more, since (x1,t1) = (v,1) solves Eqn (16a), (v,1) is a vector pointing along the t2 axis.
The x2 axis must be perpendicular to the t2 axis, in the spacetime sense. It’s easy to see that (x1,t1) = (1,v) points in the right direction:
g[(1,v),(v,1)] = v – v
= 0
which means the x2 axis, in Earth and ship coordinates, is:
t1 = vx1 (17a)
t2 = 0 (17b)
The shipboard observer’s idea of the things that are happening “right now” at time zero isn’t the same as the Earth observer’s: the two x-axes in Figure 9 don’t coincide, and the further you move away from the origin, the further apart they become. But this is no different from the situation where two people, standing in the same location but facing different directions, fail to agree as to which objects are “precisely to the right”, and their disagreement is greater for objects a kilometre away than a metre away. Our bodies carry the definitions of left-right, forwards-backwards, and up-down with them. What Einstein showed is that we also carry our own definition of the direction “future-past”, perpendicular to the other three. Relative motion means that two peoples’ future-past axes point in different directions in spacetime, and it’s as unreasonable to expect their idea of “to my left, but no earlier or later” to match up under those circumstances as it is to expect two people facing north and north-east to mean the same thing by the phrase “to my left, but not forwards or backwards”.
We now have everything we need to write down the equations for the change of coordinates between the Earth’s reference frame and the ship’s. Using Eqns (15a) and (15b) in turn to project the vector with Earth coordinates (x1,t1) onto, first, the spacelike vector (1,v) which points along the x2 axis, and then the timelike vector (v,1) which points along the t2 axis:
x2 = g[(x1,t1),(1,v)] / |(1,v)|
= (x1 – vt1) / √(1 – v2) (18a)
t2 = –g[(x1,t1),(v,1)] / |(v,1)|
= (t1 – vx1) / √(1 – v2) (18b)
These equations are called a Lorentz transformation, or to be more specific, a boost. (Lorentz transformations also include completely general rotations in four-dimensional spacetime, which might include some rotation in space.)
The reverse transformation, from ship coordinates (x2,t2) to Earth coordinates (x1,t1), is identical to Eqns (18), except that v is replaced by –v, since from the ship’s point of view the Earth appears to be travelling in the opposite direction.
x1 = g[(x2,t2),(1,–v)] / |(1,–v)|
= (x2 + vt2) / √(1 – v2) (19a)
t1 = –g[(x2,t2),(–v,1)] / |(–v,1)|
= (t2 + vx2) / √(1 – v2) (19b)
As you’d expect, it follows from either Eqns (18) or (19) that:
x22 – t22 = x12 – t12
Whatever the Earth-based and shipboard observers disagree on, for a given event they always calculate the same value for x2–t2.
Time Dilation
Suppose a man and a woman are walking on the surface of the Earth (over short enough distances for curvature to be negligible). They start out together, but the man walks due north, and the woman walks north-east, as in Figure 10.
By the time the man reaches point B, the woman would have to have travelled much farther, all the way to point E, to have reached the same latitude and be precisely to the man’s right. But … so what? By the time the woman reached point D, the man would have to have travelled all the way to point C to have gone as far in the north-east direction as the woman, and be precisely to her left. Each might think that the other has to “run fast” in order to “keep up”, but their situation is completely symmetrical.
Time dilation is the spacetime equivalent of this scenario. The length of a path in spacetime is the time that has elapsed along that path, and because the spacetime version of Pythagoras’s theorem has a minus sign, the time that elapses along what looks like the longer path is actually less, not more.
