In Greg Egan’s remarkable Orthogonal Trilogy, physics differs from ours in one crucial respect: the Lorentz metric is replaced with the Riemannian metric , putting time on the same footing as space. As one would expect from this author, the consequences are worked out in great detail, and are integral to the plot. This comes at some predictable expense to characterisation and versimilitude: still I’ve found the first two books sufficiently interesting that I can (almost) ignore that much of the exposition consists of the characters giving each other physics tutorials, conducting experiments (which remarkably always work, even if the results often come as a surprise) or listening to each other give lectures and seminars. That said, the physics is leavened by some set-piece ethical dilemmas, and there is also a well-developed biological theme, concentrated on the unique (but not completely implausible) morphology of the Orthogonal race.
The purpose of this post is to do one of the exercises left implicitly to the reader in the second book, The Eternal Flame, in which it is observed that when a photon is deflected by a stationary particle of ordinary matter (called a `luxagen’ in the trilogy), it may glance off at two different angles. Moreover, there is a maximum deflection angle, which is independent of the energy of the photon.
Here I’ll show that this follows from the relation between energy and momentum in Orthogonal physics, and the two conservation laws, and implies that Orthogonal photons are heavier than luxagens. I’ll also show that the same behaviour holds in our universe, whenever the moving particle is heavier than the particle it strikes. My solution uses piles and piles of algebra to arrive at a surprisingly simple answer. Egan’s book hints at a simple trigonometric solution: made more plausible because his characters are of course brilliant at Euclidean geometry, thanks to the double motivation from their physics and our shared mathematics. Or maybe there is a better way to organize the algebra, perhaps starting `We take the Lagrangian …’, or maybe `By working in the reference frame in which the centre of momentum is fixed …’. Either way, I would like to know it.
Much of my interest arises from how Egan (and his characters) arrive at this setup: for more background see the supplementary material on Egan’s website, or read the first book in the trilogy, The Clockwork Rocket. At the time of writing I still have the third book to look forward to.
Space-time and the energy-momentum relation
There is -dimensional space-time. In our universe, when we measure a time interval by the distance that light travels in this interval, space-time has the Lorentz metric . For example, a photon may move from to in one unit of time, and, corresponding to the fact that no time passes for the photon, the Lorentz distance between these space-time points is zero.
In the Orthogonal universe space-time has the Riemannian metric , where is a constant with dimension distance/time; by choosing units appropriately (thus sweeping under the carpet the first half of the first book on Yalda’s experiments on Mount Peerless, the beautiful dual Pythagorean Theorem, and the fact that in Orthogonal physics, the speed of light is not constant), we may assume that the numerical value of is . We take the following (and only the following, for the time being) as further axioms:
- Physical laws are invariant under the group of transformations preserving the Lorentz/Riemannian metric.
- Every particle has a symmetry invariant tangent vector to its world line called its –velocity with units distance/time; in our universe and in the Orthogonal universe . A stationary observer measures the particle’s velocity as the -vector , as illustrated in the diagram below (click for a larger pdf).
- Multiplying the -velocity, normalized to have length (our universe) or (the Orthogonal universe), by the mass of a particle gives its -momentum , where is the -vector momentum.
Note that (1) only makes sense because we measure time and distance in compatible units: thus is dimensionless and whenever a time appears, such as in the diagram above, it is multiplied by . Therefore the symmetry group of space-time can reasonably mix them up.
The way I imagine (2), half-remembered from when I did special relativity 20 years ago (my reference frame), is that the -plane has clocks every metre, connected by rigid rods, which record the time of passing of every particle. The particle is at position (which conveniently enough happens to have a clock) at time , and at position (similarly conveniently placed) at time . The stationary observer can therefore collect the measurements from the clocks (simply by travelling to each one), and approximate the particle’s velocity as and so . Everyday objects in our universe, such as cars, trains and coronavirus, move at a negligible fraction of the speed of light, so is typically huge compared to , and corresponding the velocity measured by the observer is a tiny fraction of the speed of light.
(Note this is not the velocity that an observer travelling with the particle would measure: this requires the notion of proper time; this is the usual way to define the -vector velocity, but except for the informal diagram and the motivation in the previous paragraph, we’re not going to use it here.)
In general, writing for the -velocity, and as usual in physics, for the square of its -vector Euclidean norm
axiom (2) implies that . By symmetry invariance, the Lorentzian norm squared of the tangent vector is , and the Euclidean norm squared is (using the assumption that ). Now since has units distance/time, and we care only about the direction of the tangent vector, not its magnitude, it is most convenient (and usual in physics, although I’ve never seen it clearly explained why) to require, as in axiom (3), that the Lorentzian length is .
Thus in our universe , and so , where is the usual dimensionless Lorentz factor. In the Orthogonal universe ; this is again dimensionless because the in the numerator implicitly has units distance/time.
Therefore the -vector velocity is
in our universe and
in the Orthogonal universe.
According to (3) the familiar -vector momentum is obtained by multiplying the spatial component of each of these -vector velocities by the rest mass . In our universe, this agrees with special relativity: a particle moving at -vector velocity is heavier according to a stationary observer by the factor . In the Orthogonal universe, we find instead that the -vector momentum is , so fast moving particles are lighter according to a stationary observer.
Again by (3), the energy of the particle is given by in our universe, and in the Orthogonal universe. Substituting in the -vectors, we find that in our universe has Lorentzian length , and in the Orthogonal universe, has Euclidean length . Equivalently, energy and momentum are related in our universe by
and in the Orthogonal universe by
The former is one statement of the energy-momentum relation between the energy and the -momentum . Observe that the Orthogonal energy-momentum relation can be stated very neatly as and , where . The analogue for our universe, using the identity , is and . We show both by the triangles below, noting that only the Orthogonal triangle is a genuine geometric representation, rather than a helpful aide-memoire.
We now need one final pair of axioms
- In any system of particles, the sum of all -vector momenta is conserved;
- In any system of particles, the sum of all energies (as recorded in the time component of -vector momenta) is conserved.
The diagrams below show a stationary particle with mass hit by a heavy particle with mass moving in the direction (and also in time, of course). Again click on the diagram for a larger pdf.
We can suppose that after the collision motion is contained in -space (of which the diagram above shows only the -plane), so all -coordinates are zero: let (oriented clockwise) and (oriented anticlockwise) be the deflection angles of the lighter and heavier particle. The -momenta conservation equations are, in our universe,
Notice that these are homogeneous in . We can therefore divide through by and obtain the equations for the Orthogonal universe by replacing hyperbolic functions with trigonometric functions throughout. As a further small simplification we introduce and observe that for our universe
and for the Orthogonal universe,
Therefore a unified system of equations is
where are in our universe and in the Orthogonal universe. The expression is positive in our universe and negative for the Orthogonal universe; it has the three constants , and (all with units of mass) which we regard as determined by the experimental setup. The unknowns are , , and , appearing only on the right-hand sides.
It is routine to solve these equations numerically using
NSolve in Mathematica. As an example, we suppose that the heavier particle has three times the mass of the lighter, so . The graph below show the post-collision velocity of the heavier particle for each of the two possible deflection angles , as the energy of the heavier particle increases from least (red) to greatest (violet).
This agrees with the experiment conducted in Orthogonal where the heavier particle is a photon, and the characters observe the deflection angle by visual observation of a system of mirrors. The graph below the has the same axes, comparing the post-collision velocity in the Orthogonal universe (solid) and our universe (dashed), choosing the energies to get comparable shapes. Notice that the maximum deflection angle does not depend on the energy of the heavier particle, or even on which universe we are in: we show below that its sine is the ratio of the masses. Thus in these graphs it is .
The graph below is the equivalent of the first graph showing the final energy of the heavier particle. If you are surprised that red is now at the top, note that in Orthogonal physics, the energy is , so is lower for faster moving particles.
Tangent space equation
Differentiating the momentum conservation equations using that , we obtain
At an extremum for , we have . Multiplying the top equation by and the bottom equation by and subtracting, we obtain
Differentiating the equation from energy conservation, using that and , so the same sign appears both times, we get
Comparing these relations between the independent differentials and , we see that the coefficients must be in the same ratio: that is
where is in our universe and in the Orthogonal universe.
Solution for maximum deflection
First step: solving for and
The sum of the squares of the two equations for momentum is
By the equation from the tangent space, at maximum deflection . Substituting and using
in which the only unknowns are and . The only equation we have not yet used is the energy conservation equation
It implies that
and, by squaring this relation and subtracting , that
where the sign is for our universe and for the Orthogonal universe. Using these two relations to eliminate and and recalling that is positive for our universe and negative for the Orthogonal universe, we obtain
with the same conventions for the universe dependent signs. Now only appears. After multiplying through by we can simplify the right-hand side as follows
It seems somewhat remarkable to me that all the signs cancel, and we are left with such a simple equation. Perhaps this is a sign that there is some more direct derivation that I am missing. Anyway, cancelling the factors of on each side and rearranging, we get the final equation
which clearly has unique solution
Using the relation from energy conservation we immediately get
Final step: finding and
Substituting these expressions for and into the equation from the tangent space equation we obtain
Again there is a surprising simplification: the numerator of the left-hand fraction becomes if , and so vanishes if . (Of course only the specialization is physically meaningful.) The quotient by is . Hence we have
Similarly one can show that the denominator factorizes as
Therefore the expression above for simplifies to
Using the momentum conservation equation
and the same factorizations to simplify we get
Rewriting the equation for so it uses only sines and then substituting for using the equation above gives
Hence writing for we get by a rearrangement and squaring the following quadratic equation in :
It is routine to check that the difference between the two sides is
and so the unique solution is . (This is one of several places where we can see that there is a maximum deflection angle if and only if .) Since is positive, we have , and, as claimed some time ago, the sine of the maximum deflection angle is simply the ratio of the masses.
The equation for above now implies that
A somewhat remarkable feature of the unique solution for the maximum deflection
where is the total energy (up to the factor in our universe), is that it depends on which universe we are working in only by a change from the hyperbolic function (our universe) to the trigonometric function (Orthogonal universe). Moreover, the maximum deflection angle does not depend on the universe at all.