Collected short fiction.., p.171

Collected Short Fiction of Greg Egan, page 171

 

Collected Short Fiction of Greg Egan
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  Latifa didn’t over-think the puzzle, and in twenty minutes she’d made her choice. She clicked the button beside her selection and confirmed it, satisfied that she’d done her best. After three years in the game she’d proved to be a born chemical match-maker, but she didn’t want it going to her head. Whatever lay behind her well-judged guesses, it could only be a matter of time before the software itself learned to codify all the same rules. The truth was, the more successful she became, the faster she’d be heading for obsolescence. She needed to make the most of her talent while it still counted.

  * * *

  COMING TO THIS EBOOK IN 2024

  or … purchase Instantiation from Amazon Kindle or The Best of Greg Egan from Subterranean Press

  Foundations

  by Greg Egan

  From the author's website at — http://www.gregegan.net/FOUNDATIONS/

  Foundations / created Thursday, 4 December 1997 / revised Saturday, 22 July 2000

  Copyright © Greg Egan, 1997-2000. All rights reserved.

  * * *

  Foundations is a series of articles, first published in the magazine Eidolon, on some of the theories of twentieth-century physics that have most influenced modern science fiction. However, these are not essays on the history or philosophy of science; their aim is to show how the central idea of each theory leads to detailed, quantitative predictions of real physical effects. For example, the article on special relativity derives formulas for time dilation, Doppler shift, and aberration.

  These articles are for the interested lay reader. No prior knowledge of mathematics beyond high school algebra and geometry is needed.

  * * *

  Foundations Table of Contents

  1: Special Relativity Spacetime

  Rotations in Space

  Spacetime Geometry

  Rotations in Spacetime

  Time Dilation

  The Flight to Sirius

  Doppler Shift and Aberration

  2: From Special to General

  Gravity as Spacetime Curvature

  Manifolds

  Curved Geometry

  Parallel Transport

  The Curvature Tensor

  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

  4: Quantum Mechanics

  The Birth of Quantum Mechanics

  Complex Numbers

  Wave Mechanics

  Matrix Mechanics

  The Uncertainty Principle

  The Action Principle

  1: Special Relativity

  Spacetime

  Rotations in Space

  Spacetime Geometry

  Rotations in Spacetime

  Time Dilation

  The Flight to Sirius

  Doppler Shift and Aberration

  · · · · ·

  Anyone who reads science fiction will be familiar with some of the remarkable predictions of twentieth-century physics. Time dilation, black holes, and the uncertainty principle have all been part of the SF lexicon for decades. In this series of articles I’m going to describe in detail how these phenomena arise, and along the way I hope to shed some light on the theories that underpin them: special relativity, general relativity, and quantum mechanics. The foundations of modern physics.

  These articles are meant for the interested lay reader. If you can follow high school algebra and geometry, and aren’t afraid to take in a few new concepts — which is the whole point, after all — nothing here should faze you.

  Spacetime

  The idea that we inhabit a four-dimensional spacetime is a very natural and intuitive one. It’s only because we take the duration of objects so much for granted that we tend to gloss over it and refer to them as three-dimensional. Since most of the Earth’s landscape changes slowly, factoring out time from our mental models and paper maps is a very pragmatic thing to do, but it’s this unchanging space that we imagine for convenience that’s the abstract mental construct, not spacetime. Spacetime is simply what we live in, all four dimensions of it.

  Drawing a diagram of spacetime comes almost as naturally as making any other kind of map; every historical timeline is halfway there, and placing a timeline for Germany next to one for France, then sketching in the movement of armies between the two, is as good a spacetime diagram as anything you’ll find in particle physics. Of course, a spacetime diagram in ink on paper has only two useful dimensions, so it generally only shows time plus one dimension of space (though one more can be added, using the standard techniques for drawing three-dimensional objects). Fortunately, many problems in special relativity involve only one dimension of space; for example, a spacecraft flying from here to Sirius would almost certainly travel along a straight line.

  Figure 1 is a spacetime diagram for such a flight. Distances are in light-years and times are in years. For the sake of simplicity, the slight “proper motion” of Sirius relative to the Sun, and any orbital manoeuvres and planetary take-offs and landings by the spacecraft, are ignored. The spacecraft accelerates at the start of the journey, shuts off its engines and cruises for the middle stage, then decelerates at the end, bringing it to a halt just as it arrives. (There’s no special reason for all three stages to cover equal distances; this is just one possible flight plan of many.) Given that the distance to Sirius is almost nine light-years, it’s reasonable to treat stars and spacecraft alike as mere specks, tracing out one-dimensional world lines, rather than worrying about the fact that they’re really solid objects whose histories in spacetime are four-dimensional “world hypercylinders”.

  When you draw a map, you have to choose a compass direction to point “up” on the page. North is often convenient, but it’s a completely arbitrary choice, and on a house plan, say, it might be more useful to align the map so that the street frontage is horizontal. Similarly, to draw a spacetime diagram you have to choose a reference frame: you have to pick some object, such as the Sun, and treat it as fixed. The chosen object’s world line will then be vertical — it will be “moving” only in time, not space — as will the world line of any other object at a constant distance from it. So the world lines for the Sun and Sirius are vertical here because the diagram was drawn that way; it’s a matter of convenience, not a statement that the Sun is “truly motionless”, any more than north is “truly up”.

  However, some reference frames are different from others. Orienting a map so that a given straight road runs vertically is one thing; arranging for a meandering river to appear as a straight line is a much harder task. If we chose the spacecraft to be the fixed point, everything we did would be complicated by the need to straighten out the curved sections of its world line when it accelerates and decelerates. To avoid this kind of complication, special relativity deals only with inertial reference frames, which take as their fixed point an object that is not accelerating. Unlike the idea of being motionless (motionless compared to what?) this condition is easily defined in the middle of interstellar space: if you’re not firing your engines, and everything in the ship is weightless, then you’re not accelerating.

  Of course, we can imagine a hypothetical second spacecraft which never accelerates, but conveniently happens to match the first spacecraft’s velocity for some part of the journey — such as the entire middle stage, when the engines are shut off, or even just for an instant during the acceleration or deceleration stages. That way, we can analyse the first spacecraft’s viewpoint at any given moment, without adopting a reference frame in which it appears motionless from start to finish.

  In a reference frame fixed to the Sun, the world line for the spacecraft starts out being vertical, tips over as it accelerates, has a constant slope in the cruising stage, then comes back towards vertical again as it decelerates. The world line for a pulse of light that leaves the solar system at the same time as the spacecraft is shown for comparison; it has a constant angle (45° in this diagram), because it travels at a constant velocity all the way.

  To be in motion relative to the Sun means tracing out a world line at an angle to the Sun’s world line. That might sound like nothing but a novel way to describe the situation, but it’s the key to all the relativistic effects of space travel. Two people facing different directions in ordinary space see the same objects differently. Two people driving between the same two towns will travel different distances, if one takes the most direct route while the other takes a detour. In spacetime, the effects are analogous, but not quite identical, because the geometry of spacetime is not quite the same as the geometry of space.

  Rotations in Space

  Despite the differences, analysing the effects of rotating your angle of view in ordinary space makes a useful rehearsal for tackling the problem in spacetime. It’s easier to deal with ordinary space, where we can rely on everyday geometrical intuition, and then the results can be carried over to spacetime with only a few small changes.

  First, a quick review of the geometry we’ll need. Pythagoras’s theorem says that the square of the hypotenuse (OP in Figure 2) of any right-angle triangle equals the sum of the squares of the other two sides (OQ and PQ).

  OP2 = OQ2 + PQ2

  The sine of the angle marked A is equal to the ratio between the side opposite it (PQ) and the hypotenuse (OP).

  sin A = PQ / OP

  PQ = OP sin A

  The cosine of A is equal to the ratio between the side adjacent to it (OQ) and the hypotenuse (OP).

  cos A = OQ / OP

  OQ = OP cos A

  The tangent of A is equal to the ratio between the side opposite it (PQ), and the side adjacent to it (OQ).

  tan A = PQ / OQ

  = (OP sin A) / (OP cos A)

  = sin A / cos A

  There’s a simple relationship between the sine and cosine of an angle, which comes straight from the definitions and Pythagoras’s theorem:

  (cos A)2 + (sin A)2 = (OQ / OP)2 + (PQ / OP)2

  = (OQ2 + PQ2) / OP2

  = OP2 / OP2

  = 1

  The notation “(x,y)” beside the point P is a reminder that points can be referred to by their x- and y-coordinates, written as an ordered pair. The arrow drawn from O to P is a reminder that every point can be thought of as defining a vector from the origin to the point. The advantage of dealing with vectors, rather than just points in space, is that the same geometry can then be applied to other vectors, like velocity and acceleration.

  To make it easier to carry things over from Euclidean geometry to spacetime geometry, it will help to restate some of these familiar ideas in slightly different language. In both Euclidean and spacetime geometry, there’s a formula for taking two vectors and calculating a number from them which depends on the length of the vectors and the angle between them. This formula is known as the metric for the geometry. (You might also have come across it as the “dot product” of two vectors.) It’s usually written as g:

  g[(x,y),(u,w)] = xu + yw (1)

  Eqn (1) defines the Euclidean metric. Eqns (2a)-(2c) demonstrate some of its properties: it’s symmetric (swapping the two vectors leaves the value unchanged), and it’s linear (its value is simply multiplied and added as shown, if you apply it to a vector that’s been multiplied by a factor, or had another vector added to it).

  g[(u,w),(x,y)] = ux + wy

  = xu + yw

  = g[(x,y),(u,w)] (2a)

  g[a(x,y),(u,w)] = g[(ax,ay),(u,w)]

  = axu + ayw

  = a (xu + yw)

  = a g[(x,y),(u,w)] (2b)

  g[(x,y)+(p,q),(u,w)] = g[(x+p,y+q),(u,w)]

  = (x+p)u + (y+q)w

  = (xu + yw) + (pu + qw)

  = g[(x,y),(u,w)] + g[(p,q),(u,w)] (2c)

  Eqn (3) is just a restatement of Pythagoras’s theorem; the notation |(x,y)| means the length of the vector (x,y) — also referred to as its magnitude — or if you prefer to think in terms of the coordinates of a point, |(x,y)| is the distance from the origin (0,0) to the point (x,y).

  |(x,y)|2 = g[(x,y),(x,y)]

  = x2 + y2 (3)

  Eqn (4) states that the cosine of the angle between two vectors (x,y) and (u,w) is equal to the metric function applied to the two vectors, divided by both their lengths. Eqn (4) will take a bit of work to prove, but in doing so we’ll solve the whole problem of rotations in space.

  cos B = g[(x,y),(u,w)] / (|(x,y)||(u,w)|) (4)

  where B is the angle between (x,y) and (u,w).

  If you want to know the x-coordinate of a point like P in Figure 3, you draw a line through P at right angles to the x-axis, and see where it hits the axis. In the process, the vector OP is shown to be the sum of two vectors: OQ, which is parallel to the x-axis, and QP, which is perpendicular to it.

  The same thing can be done with any other vector in place of the x-axis. If a line from P to OG meets OG at a right angle, at point S, then OS is called the projection of OP onto OG. And again, OP is shown to be the sum of two vectors: OS, which is parallel to OG, and SP, which is perpendicular.

  How long is OS? If the angle between OP and OG is B:

  OS = OP cos B

  = |(x,y)| cos B (5)

  What if we don’t know B? Suppose all we know are the coordinates of P, (x,y), and the angle A that OG makes with the x-axis. Projecting OS onto the coordinate axes to make OT and OU, and projecting OT back onto OS to make OV:

  OS = OV + VS

  = OT cos A + OU sin A

  = g[(OT, OU),(cos A, sin A)]

  We don’t know (OT, OU), but we do know that it’s the part of the vector (x,y) parallel to OG, when (x,y) is written as a sum of parallel and perpendicular components:

  (x,y) = (OQ, OR)

  = (OT, OU) + (–QT, UR)

  Making use of Eqn (2c):

  g[(x,y),(cos A, sin A)] = g[(OT, OU),(cos A, sin A)]

  + g[(–QT, UR),(cos A, sin A)]

  = OS + (–QT cos A + UR sin A)

  = OS + (–PS sin A cos A + PS cos A sin A)

  = OS

  OS = g[(x,y),(cos A, sin A)] (6)

  What’s this vector (cos A, sin A)? It points in the same direction as the vector OG, but from what we know about sine and cosine:

  |(cos A, sin A)| = √((cos A)2 + (sin A)2)

  = 1

  A vector like this, with a magnitude of one, is known as a unit vector. Eqn (6) tells us that to calculate the length of the projection of (x,y) onto OG, we just apply the metric function to (x,y) and the unit vector in the direction of OG.

  What if we don’t know the angle A, but only the coordinates of G, (u,w)? We can still compute the unit vector in this direction, just by dividing (u,w) by its own length, automatically re-scaling it to a length of one.

  (cos A, sin A) = (u,w) / |(u,w)|

  OS = g[(x,y),(cos A, sin A)]

  = g[(x,y),(u,w)/|(u,w)|]

  = g[(x,y),(u,w)] / |(u,w)| (7)

  Here we’ve made use of Eqn (2b) to shift the factor 1/|(u,w)| outside the metric.

  Equating the two formulas for OS from Eqns (5) & (7) gives:

  |(x,y)| cos B = g[(x,y),(u,w)] / |(u,w)|

  cos B = g[(x,y),(u,w)] / (|(x,y)||(u,w)|) (8a)

  g[(x,y),(u,w)] = |(x,y)||(u,w)| cos B (8b)

  Eqn (8a) is identical to Eqn (4), which is what we set out to prove.

  Setting B equal to 90°, Eqn (8b) shows that the metric function for two perpendicular vectors is zero, since the cosine of 90° is zero. That’s why the metric is able to “pick out” the part of one vector that’s parallel to another; the part that’s perpendicular simply yields zero.

  Calculating the way a point’s coordinates change when the reference frame is rotated in space is easy, now. In Figure 4, imagine you’re standing at the origin, O, looking straight ahead in the direction of the axis marked y1, with x1 pointing directly to your right. P marks some fixed object in front of you, say a tree. OQ measures how far to the right of you the tree is, and OR measures how far ahead it is. (Negative numbers would be used if it was to the left, or behind you.)

  Now suppose you turn your entire body through the angle A, so you’re looking in the direction y2, and x2 points directly to your right. The new coordinates you’d give P are OS and OT. We already have OS for exactly this situation, Eqn (6), and all that’s needed to work out OT are the coordinates of a unit vector that points along the y2 axis. From Figure 4 it’s clear that (–sin A, cos A) does the job, so:

  OT = g[(OQ, OR),(–sin A, cos A)] (9)

  Writing (x1,y1) for the coordinates of any point in the original reference frame, and (x2,y2) for the coordinates of the same point in the rotated reference frame, Eqns (6) and (9) become:

 

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