Collected short fiction.., p.173
Collected Short Fiction of Greg Egan, page 173
Figure 11 shows the world lines of two astronauts, a man spacewalking in Earth orbit, and a woman tethered to a ship cruising past with velocity v. At event A, they pass within waving distance and synchronise their watches.
Despite appearances to the contrary — due to the distortions of drawing spacetime on a piece of paper — the time that elapses for the man from A to C is exactly the same as the time that elapses for the woman from A to E. The proper time, √(t2–x2), is the same in both cases. And while CD is obviously perpendicular to the man’s world line, BE is equally perpendicular to the woman’s world line, since it’s parallel to the x2 axis.
Suppose the man’s watch records 10 seconds elapsing from A to C, and the woman’s watch records an equal time elapsing from A to E. At C, the man considers the woman to have reached only event D, so less time must have passed for her: her watch must be “running slow”. But conversely, at E the woman considers the man to have reached only event B, so less time must have passed for him, and his watch must be running slow. Again, the situation is perfectly symmetrical.
Of course, neither can see the other’s watch immediately — assuming they can see it at all, with binoculars or whatever — so in practice they could only figure out these relationships after the fact: by keeping a record of their observations and then subtracting out the time it took the light to reach them. But time dilation has nothing to do with these time lags; the effect involves the situation deduced by each astronaut after taking account of time lags.
Eqn (18b) is:
t2 = (t1 – vx1) / √(1 – v2)
Given that the man measures no separation in space, x1, between A and B, this becomes:
AE = AB / √(1 – v2)
AB = AE √(1 – v2)
For example, if AE is 10 seconds, and v = 0.6 (60% of lightspeed), then:
AB = 10 √(1 – (0.6)2)
= 10 √(1 – 0.36)
= 10 √(0.64)
= 8 seconds
AB is the time the woman concludes has elapsed for the man, when she’s reached event E and 10 seconds has passed for her. By symmetry, AD (the time the man concludes has elapsed for the woman, when 10 seconds has passed for him) is also 8 seconds. Both astronauts are correct, they’re just talking about different things.
With all this symmetry, you might be starting to wonder how space travellers can ever return home younger than their twins. What makes this possible is the fact that a two-way voyage is not symmetrical; however you look at it, the space travelling twin takes a detour from the straight-line path of the twin who stays at home.
The analogy in space is completely familiar. In Figure 12, the shortest path from A to C is the straight line that happens to run due north. If a man and a woman set off from A, and the man travels due north, but the woman travels north-east for the first half of the trip and north-west for the second half, it’s obvious that when the two meet up at C, the woman will have travelled farther. AG and GC are both longer than AH and HC:
AG = √(AH2 + HG2)
> AH
GC = √(HC2 + HG2)
> HC
In Figure 13, when the astronaut tethered to the ship reaches G, the ship reverses and heads back towards the Earth. (In reality, a perfectly sharp turn like this would require infinite acceleration, but to simplify the analysis we’ll ignore any actual rounding of the corner at G.) The situation looks exactly the same as Figure 12, but now AG and GC are shorter than AH and HC:
AG = √(AH2 – HG2)
< AH
GC = √(HC2 – HG2)
< HC
We can quantify this, by noting that if the ship’s velocity is v for both stages:
HG = v AH
AG = √(AH2 – v2 AH2)
= AH √(1–v2)
HG = v HC
GC = √(HC2 – v2 HC2)
= HC √(1–v2)
(AG + GC) = AC √(1–v2)
After synchronising watches at A, when the woman passes the man for the second time, at event C, her watch shows less elapsed time than the man’s. In spacetime, a straight world line is the longest path between two events, not the shortest.
The Flight to Sirius
To calculate how much time passes on board the spacecraft travelling to Sirius in Figure 1, consider each stage of the journey separately. The middle stage is easy:
ship time (cruising) = √((3.5)2 – (2.9)2)
= 1.96 years
To analyse the acceleration stage, we need to know the exact shape of the curved world line. To determine this with complete mathematical rigour takes a bit of work, but there’s a nice intuitive way to reach the same answer. In classical mechanics, the distance travelled in time t by a uniformly accelerating object starting from rest is:
x = at2 / 2
(Why? Because its final velocity is v = at, but it started out with v = 0, so its average speed is halfway in between, v = at/2.) Given x and t, the rate of acceleration is:
a = 2x / t2
The same relationship holds in relativistic mechanics, but t is replaced by the proper time, √(t2–x2), measured along a straight line in spacetime between the endpoints of the curved world line:
a = 2x1 / (t12 – x12) (20a)
Acceleration here must be measured in units consistent with the speed of light being one, such as light-years/year2. In those units, one Earth gravity is:
9.8 metres/sec2 = 9.8 x 3600 x 24 x 365 / 300,000,000 light-years/year2
= 1.03 light-years/year2
Unfortunately, there’s no easy shortcut for computing the proper time along the curved world line, which is the elapsed ship time. This is given by:
ship time (accelerating) = ln (1 + a (x1 + t1)) / a (20b)
In Figure 1, the craft travels 2.9 light-years in 5.5 years, so Eqn (20a) gives:
a = 2 x 2.9 / ((5.5)2 – (2.9)2)
= 0.266 light-years/year2
and Eqn (20b) gives:
ship time (accelerating) = ln (1 + 0.266 x (2.9 + 5.5)) / 0.266
= 4.41 years
By symmetry, the deceleration stage takes exactly the same time, so the total shipboard time for the journey is 4.41 + 1.96 + 4.41 = 10.78 years, compared to 14.5 years Earth or Sirius time.
Doppler Shift and Aberration
A beam of light can be thought of as consisting of a series of evenly spaced wavefronts in spacetime. (This doesn’t begin to do justice to the physics of electromagnetic waves, but it captures the essential geometry of the situation.) In four-dimensional spacetime, these wavefronts are three-dimensional hyperplanes, but since we’re only going to tackle problems involving one or two dimensions of space, wavefronts will appear either as lines in two-dimensional spacetime (Figure 14) or planes in three-dimensional spacetime (Figure 15).
Figure 14 shows light travelling right-to-left through one-dimensional space; the wavefronts are all inclined at 45° in spacetime. If the t-axis was the world line for your eye, it would be struck by a wavefront at regular intervals; the number of times this happens per second is called the frequency of the light, and the time between wavefronts is called the period. The distance between wavefronts in space is called the wavelength; in units such that the speed of light is one, the wavelength and the period will be equal.
The direction and spacing of a series of wavefronts like this can be described by a single spacetime vector, called the propagation vector, parallel to the wavefronts’ world lines. Both the space and time components of the propagation vector are given a length of 1/L, where L is the wavelength of the light, and the space component points in the direction the light is travelling. In the one-dimensional example of Figure 14, that just means the x coordinate has a minus sign to indicate right-to-left, so the propagation vector is (–1/L,1/L).
The propagation vector as a whole has length zero, of course — whatever its individual space and time components — since it points in a lightlike direction in spacetime.
|(–1/L,1/L)|2 = 1/L2 – 1/L2
= 0
If you draw a vector from the origin to any one of the wavefronts, the propagation vector provides a simple way to determine, mathematically, which wavefront each vector is touching. The lines for the wavefronts in Figure 14 all take the form:
x + t = nL
where n is an integer. So events on these wavefronts have (x,t) coordinates (x, nL–x). The spacetime metric operating on the propagation vector and one of these vectors gives:
g[(–1/L, 1/L),(x, nL–x)] = –x/L – n + x/L
= –n
which identifies the wavefront with a number that’s completely independent of the particular vector you chose. So in Figure 14, without going to the trouble of working out the coordinates of points A to E, we know that:
g[P, A] = 1
g[P, B] = 1
g[P, C] = 1
g[P, D] = –2
g[P, E] = –3
Figure 15 shows a series of planar wavefronts, moving through the coordinate systems of two observers in relative motion. The two planes defined by either the x1 axis or the x2 axis and the shared y axis are the slices through three-dimensional spacetime that Earth-based and shipboard observers consider to be “space, at time zero.” Because these spacelike slices cut through the wavefronts at different angles, both the angle of approach and the spacing between the wavefronts — the wavelength of the light — will be different for the two observers.
To quantify this, suppose the shipboard observer sees starlight with wavelength L2, at an angle A2 from the direction of travel — anything from 0° for a star that’s straight ahead to 180° for a star directly behind the ship. (The second angle needed to pin down most stars, the angle measured around the ship’s axis, makes no difference to the analysis.) The propagation vector according to the shipboard observer will be:
(x2,y,t2) = (–cos A2, –sin A2, 1) / L2 (21)
since the space component of this, (x2,y), points in the right direction — towards the ship, at an angle A2 — and both the space and time components taken individually have length 1/L2. (The length of the space component is just |(x2,y)|, in the usual Euclidean sense.)
Using Eqns (19) — leaving the y coordinate unchanged — we can transform this vector into Earth-based coordinates:
x1 = (v – cos A2) / (L2 √(1–v2)) (22a)
y = –sin A2 / L2 (22b)
t1 = (1 – v cos A2) / (L2 √(1–v2)) (22c)
But the time and space components of the propagation vector in Earth coordinates must both have length 1/L1, where L1 is the wavelength measured by an Earth observer. The single time component is easiest to deal with:
1/L1 = (1 – v cos A2) / (L2 √(1–v2))
L2/L1 = (1 – v cos A2) / √(1–v2) (23)
Eqn (23) describes the Doppler shift, the difference in the wavelength of light as measured by two observers in relative motion. For A2 = 0° and 180°, respectively:
L2/L1 = (1–v) / √(1–v2)
= √((1–v)/(1+v))
L2/L1 = (1+v) / √(1–v2)
= √((1+v)/(1–v))
Assuming v is positive, starlight from straight ahead always has a shorter wavelength — a blue shift — and starlight from behind always has a longer wavelength — a red shift. The angle at which the crossover occurs — the direction in which the wavelength is unaltered — depends on the velocity:
1 = (1 – v cos A2) / √(1–v2)
1 – v cos A2 = √(1–v2)
cos A2 = (1 – √(1–v2)) / v
For a ship travelling at 90% of lightspeed, the wavelength of starlight from straight ahead is less than a quarter the usual value. This is enough to shift all visible light into the ultraviolet (though it also shifts a band of infrared wavelengths into the visible spectrum, so stars won’t necessarily vanish from sight). Looking back, the effect is reversed: wavelengths are multiplied by a factor of more than four, turning all visible light into infrared (and rendering some UV visible).
For angles in between, the Doppler shift varies smoothly, ringing the spacecraft with bands of stars shifted by different amounts — the famous “starbow” of interstellar travel. Individual stars are different colours anyway, so all the stars at a given angle won’t look identical, but the average colour of the sky will be graded like a circular rainbow, positioned roughly at the crossover angle, 50° in this case.
Eqns (22) for the propagation vector also show that the direction the wavefronts move within the (x1,y) plane is:
(x1,y) = ((v – cos A2) / (L2 √(1–v2)), –sin A2 / L2)
If the wavefront is seen by an Earth-based observer to approach at an angle A1 from the x1 axis, the ratio of y- to x1-coordinates for this vector must equal tan A1:
tan A1 = sin A2 √(1–v2) / (cos A2 – v) (24)
Eqn (24) describes an effect known as aberration. For travellers moving at a substantial fraction of lightspeed, the familiar constellations will appear to have been pushed forward in the sky, crowded around the direction in which the ship is travelling.
To take two examples, for stars that appear to a stationary observer to be 90° away from the ship’s destination:
cos A2 = v
and for stars that appear to a moving observer to be 90° away from the same point:
tan A1 = –√(1–v2) / v
For a ship travelling at 90% of lightspeed, the constellations that would normally have occupied the entire forward hemisphere are squashed together into a circle 25° in radius. Looking backwards, the stars that would have been confined to a 25° circle in the stationary view are spread out to fill half the sky.
2: From Special to General
Gravity as Spacetime Curvature
Manifolds
Curved Geometry
Parallel Transport
The Curvature Tensor
· · · · ·
The first article in this series described some of the ways in which the geometry of spacetime affects travellers moving (relative to their destinations, or each other) at a substantial fraction of the speed of light. By generalising from the Euclidean metric, which captures such familiar aspects of geometry as Pythagoras’s Theorem, to the Minkowskian metric suggested by the fact that the speed of light in a vacuum is the same for everyone, we analysed the “rotated” view of spacetime that two observers in relative motion have with respect to each other, and derived formulas for time dilation, Doppler shift and aberration.
This article and the next will build the framework needed to provide a similar account of the strange effects that have been predicted to take place in the vicinity of a black hole. To do this, we need to generalise yet again: from flat geometry, to curved.
Gravity as Spacetime Curvature
The basic premise of general relativity is simple: the correct way to account for the acceleration of objects due to gravity is to consider spacetime to be curved in the presence of matter and energy. How does curvature explain acceleration? If two explorers set off from different points on the Earth’s equator, and both head north, their paths will grow steadily closer together, despite the fact that they started out in the same direction. In spacetime, if two nearby stars start out being motionless with respect to each other, their world lines will draw closer together, despite the fact that those world lines were initially pointing in the same direction. We could say that the force of gravity is pulling the stars together … but we don’t say there’s a “force” acting on the explorers, do we? Of course, the two-dimensional surface of the Earth is a visibly curved object embedded in a larger (and more or less flat) space, but we have no reason to believe that spacetime is embedded in anything larger. Rather, general relativity assumes that whatever gives rise to spacetime geometry in the first place is tied up with the presence of matter and energy in such a way that the resulting geometry is sometimes curved.
Manifolds
Before exploring curved geometry, it will be useful to take a look at a kind of geometry that’s neither flat nor curved: geometry without any metric at all. Essentially, this is like asking what you can say about lines drawn on a sheet of rubber that remains true however much you stretch or squeeze the sheet: distances and angles lose all meaning, but you can still talk about such things as whether or not two lines intersect. Why is this relevant to general relativity, which does assign a metric to every part of spacetime? Firstly, everything that’s true without reference to a metric can be safely carried over to regions of spacetime where the metric varies from place to place. Secondly, it reflects the situation you find yourself in when you begin to solve a problem in general relativity: initially, you have no idea what the metric is, since that’s the very thing the equations are supposed to tell you.
Let’s start with a familiar situation — two-dimensional space, with flat geometry — and see what concepts can survive the loss of the metric. Choose a city’s central post office as the origin for coordinates, measure distances north and east of it, and ignore the curvature of the Earth. Every building in the city can be identified by a pair of numbers, (x,y), specifying distances east and north of the post office. Vectors associated with objects in the city can also be given coordinates in much the same way. For example, a train’s velocity can be assigned two coordinates — call them vx and vy — stating how fast the train is travelling east, and how fast it’s travelling north.
Now, imagine that this city lies, not on solid rock, but on a vast sheet of rubber. The railway track and all the buildings are made of equally flexible material, so the whole city can be stretched and squeezed without anything being disrupted. What’s more, the imaginary grid lines that initially measured distances north and east of the post office have been painted onto the ground, so they too can flex with it. Then a giant hand descends from the sky and gives one edge of the city a mighty tug.
Figure 2 shows the result. But the cosmic intervention doesn’t stop here; further stretching and compression is applied, at random, 24 hours a day. The city dwellers simply have to adapt to the fact that streets no longer meet at the same angle from hour to hour, and buildings are no longer separated by fixed distances. These concepts are soon discarded as irrelevant.
