a.k.a., Space 11: How to do Interstellar Navigation
Various antecedents you may want to have peered at first:
- the Spherical Trigonometry lesson
- Space 10 on consequences of Relativity (why clocks slow down, lengths change, and how FTL breaks things), or
- Space 9 if you need to brush up on spacetime diagrams and why simultaneity gets screwed up
Today's post is about Hyperbolic Geometry, wherein you learn what those "Warning, Evil, Don't Look" columns are about. It's now safe to look; well okay, no it isn't, but too late! AHAHAHAHAHAHAHA.
Hyperbolic geometry is basically Geometry On Drugs and we know that's never going lead anywhere good.
To be fair, Spherical Geometry is arguably also on drugs, but at least it's easier to explain in that, having had lots of experience with basketballs and whatnot, you already know what a sphere is. Having a concrete place for the "points" to live, I can then tell you
- what "lines" are (great circles, or planes slicing the sphere through the origin / center of the sphere),
- how to measure "distance" along a "line" segment (measure angle between endpoints from the center of the sphere),
- how to measure "angles" between "lines" (the planes will intersect; there's an angle there; done), and
- what "circles" are (they're um, circles, … or, if you like, planes that don't necessarily go through the origin, or cones coming out of the origin; whatever works for you),
and then you're basically good to go, ready to do all of the geometry/trigonometry you could ever want, once you've heeded my warnings that Certain Things Will Be Different (no such thing as "parallel", triangles add up to 180 plus area instead of just 180, do not feed them after midnight, etc…).
Unfortunately, the place where we're Doing Geometry today is this inside-out Hyperboloid Sheet Thing with a fucked up metric, … and if you've actually seen one of those in real life, I will be very surprised, especially since it's not something that can exist in ordinary 3D space. Oddly enough, it will end up relating to something you do have day-to-day experience with, namely (cue reverb and James Earl Jones voice)… Your Future,… but I'm not sure how much help that's going be in visualizing it.
But before I get to how that works, let me just, like last time, plunge in and give you a triangle to solve, which will make zero sense until I give you the obligatory contrived word problem that motivates it:
Carol lives at South Pole Station (a cool place where I think I will be hosting all of my geometry problems from now on).
Carol has a care package she wants to send to Bob, who is on an expedition towards Sigma Octans, which for the purposes of this problem will be the southern pole star, what Polaris is to us northern hemisphere folks, in other words always directly overhead at South Pole Station, even if it might not quite be in real life. This is convenient because Carol has this nice radio dish carved into the ice (suffice it to say, when it's 40 below outside, you really want your equipment to have as few moving parts as possible). And because of her unique situation, if she can launch her package straight up fast enough, it'll be going the right direction so that she can use her dish to track it every step of the way there. (Psst, don't tell her that glaciers move around …)
Unfortunately, launching of interstellar rockets directly from South Pole Station would violate numerous international treaties and environmental regulations. However, Carol has friends. One of her friends, Alice has her own spaceship. Alice, in fact, will shortly be passing directly over Carol's position, flying low enough that, for our purposes, she may as well be plowing straight through South Pole Station, even though that would be a particularly egregious violation of international treaties and environmental regulations …
… well, all right, maybe not that low. Pick an altitude low enough for Carol to reach with the Giant Trebuchet but not so low that it disturbs the penguins and everybody will be happy. Alice gets the package and we don't really care how this happens. (yeah, I know,… SPS is too cold/dry for penguins; sigh…)
Now, Alice, like Bob, is also an interstellar traveller. In fact, she's just getting back from Beta Virgo, (very briefly) passing through the neighborhood to pick up some odds and ends, and will thence be headed off towards Lambda Pisces — which for the purposes of this problem will be taken to be the vernal equinox point in the sky — or 90° away from where Bob is going.
Alice is going 3/5 lightspeed. And since, as we know, course changes at that speed are freakishly expensive, she's going to need to do all this without slowing down or otherwise changing her trajectory in any way.
Carol has figured out that if she could launch Bob's package herself at 12/13 lightspeed towards Bob at that exact moment that Alice will be passing overhead, then it will reach him when he'll be celebrating his next birthday, which is what she wants.
But Alice has to do the launch. And it's a fair bet that Alice tossing the package out sideways (at 90° from her current course) at 12/13 lightspeed is simply not going to work. Not only does she have to compensate for her own velocity, but there's also Relativity to contend with (…ooo, bet you didn't see that coming).
How fast should Alice launch the package? And in what direction?
We start with Carol, Alice, and Bob's package, all in (roughly) the same place at the same time. Call that launch event Here and Now. Then everybody goes their separate ways. The triangle is a kind of snapshot of the future, essentially, where everybody is, say, an hour from now.
But it's a weird kind of snapshot. We have to be careful in defining the shape we care about because it's not actually the configuration that will be observed by any one person; those shapes will all differ. The fundamental problem here is that, as you'll recall, Carol and Alice (and Bob's package, for that matter) will not, in general, be agreeing on which events are "an hour from now".
What everyone will agree on will be, say, the particular event that's on Alice's ship and is an hour from now according to Alice. Even if another observer, like Carol, thinks Alice's clock is running too slow, she and all other observers will at least agree that Alice sees that event as being an hour from now.
More generally, any pair of events that can be visited by the same person has a proper time that elapses between them. That is, if they can both be visited at all, then they can be visited by someone who's just coasting (constant velocity), who can just sit there with a clock while the two events just happen right there, actually experiencing the time lapse first-hand. And everybody else can find out what this number is by taking the time lapse Δt and spatial separation Δx that they themselves see and doing √Δt²−Δx². If you're not yet convinced of this, go stare at this diagram (upper gray triangle) some more.
So we take every possible observer that could be passing through our launch event in every possible direction/velocity and plot all of their where-I-am-an-hour-from-now events on a space-time diagram. You get this shape:
(except this needs to be folded along the vertical line where Carol is so that Alice is coming out of the page, since there are 2 spatial dimensions we care about here and this stupid page can only show one at a time). If we were only doing one spatial dimension this magic shape would be a hyperbola (t² − x² = 1 hour). 2 dimensions makes it a hyperboloid of revolution (t² − x² − y² = 1 hour), and analogously for the 3 dimensional hyperboloid of spherolution-or-whatever that you can't visualize anyway.
Here, we happen to be using Carol's coordinate system, i.e., she's the one who's stationary in the middle, but the hyperboloid shape we get is always going to be the same no matter who's point of view we use.
The next fun thing is we can measure proper distances along paths within the hyperboloid. The story on proper distance is much like the story on proper time: For any pair of events where you'd need to go faster-than-light to get from one to the other, there will exist an observer who sees those events as simultaneous. The proper distance between those events is whatever distance that observer measures. And everybody else can find out what this number is by taking the distance Δx and the time lapse Δt that they themselves see and doing √Δx²−Δt². If you're not yet convinced of this, go stare at this diagram some more (the righthand gray triangle this time).
The one added wrinkle is that the hyperboloid-thing is an annoyingly curved surface, but we can deal with that by breaking it up into lots of patches small enough that we can pretend they're flat. For every patch there's somebody passing through who sees everything in the patch as simultaneous, so we'll just let them be the authority on what the distances and angles are there. Then we glue all of these measurements together and do the Usual Calculus Thing of taking the limit as the patch size goes to zero. It helps that each of the "stuff that SoandSo thinks is an hour from now" planes is tangent to the hyperboloid right at SoandSo's hour-from-now event, so SoandSo is, in fact, that observer that we want for the neighborhood of SoandSo's hour-from-now event.
Or we just let Carol do √</strong>dx²−dt² everywhere to get the same numbers. (and it's dx,dt instead of Δx,Δt because Calculus!).
Bottom line is, given any smooth path through a bunch of events on the hyperboloid, it has a proper length that everyone will agree on. Meaning the transformation to someone else's coordinates, e.g., Alice's, not only leaves the hyperboloid in place, but it preserves distances along paths and the angles between them; it's an isometry, a way to "roll" the hyperboloid rigidly.
So if you're asking what the shortest path is between two events on the hyperboloid, then just pick one of them. It'll be Somebody's hour-from-now event. Transform to Somebody's point of view, i.e., roll the hyperboloid so that this event goes to the middle (where Carol's hour-from-now is above). See where the other event ends up and take the radial path out to it (yeah, you still have to prove this is shortest but this is way easier than the general case). Notice how this path is what you get when you take a plane through the origin (launch event) and the two events and see where else it slices the hyperboloid.
Then you undo the roll, i.e., roll everything back to where it was. Since the roll is linear — takes straight things to straight things; lines to lines, planes to planes — we will still have a plane containing the origin and the two events and its intersection with the hyperboloid will be the path we want, a "great hyperbola" if you like, and that path will indeed be shortest, even in the general case …
… just like with the sphere, where, if you, say, know all of the shortest paths coming out of the south pole (the meridians), and you know how to roll the sphere rigidly so that any point on earth can be rolled to the south pole, then you know what the shortest paths are everywhere (great circles).
So let's have a look at that triangle again:
We are looking "down" from the far future onto the hyperboloid at where Alice, Carol, and Bob's package will be one hour from now.
Or one year from now. Just as in spherical geometry, where we don't care what the radius of the sphere is, here we likewise don't care how far in the future we're going — the velocities are constant so the only thing that changes is the scale.
The sides of the triangle a, b, c are great hyperbola segments that are expressing velocity angles.
WTF is a velocity angle, you ask? It's like this:
Velocities are slopes in space-time. Slopes are annoying because they don't add the way you'd expect — well okay, they do sort of if they're Really Small, and this worked out fine for Gallileo and Newton, but beyond a certain point they really don't. E.g., if you watch me going down the road at 3/4 lightspeed and I toss something out in front of me at 3/4 lightspeed, which I'm perfectly allowed to do, then due to the wacky time and distance effects shifting back to your point of view, you're going to see the thing I tossed going the road at 0.96 lightspeed, and 0.96 ≠ 0.75 + 0.75.
Velocity angles, on the other hand, are just path lengths on the unit hyperboloid. If you have two paths that are both in the same plane [i.e., same great hyperbola slice], you can add or subtract their lengths just fine. (And if they're not, that's why we have trigonometry.) It's so much easier to work with velocity angles.
And if we know the angles but badly need to know the slopes/velocities, recall from circle/sphere-land that if you have an angle you get the corresponding slope by taking the tangent (tan a = sin a / cos a). And it's the same story here except we need to instead use the almost but not quite entirely different hyperbolic trigonometric functions (tanh a = sinh a / cosh a).
So to summarize, a is how fast Carol sees the package going, expressed as a velocity angle, where the actual velocity is tanh a = 12/13. Likewise, tanh b = 3/5, and tanh c = (mystery!).
So how do we solve this? Taking our cue from trig class, it does seem like we have a side-angle-side situation brewing here. Solve for c to get "how fast?" and the angle at A to get the direction.
By the way, the angle at B is meaningful, too, since if Carol wants to be able to talk to the package, then we'll need to arrange that its dish antenna is pointed in the correct direction (i.e., at Carol), and for that, we might want to know what direction that actually is.
Note that if you were to simply assume that B is 90 minus A, you will lose, because we're not in Kansas anymore. Just to emphasize this, I've drawn a curved line for side c, not because Bob's package is going to be following a curved trajectory but because the page you're looking at, like Kansas, is annoyingly flat and thus incapable of representing this triangle faithfully. Recall that we have the same problem depicting spherical triangles (hint: How do you draw the Ecuador-Congo-NorthPole triangle on a flat page without stretching or tearing it? Give up, now.)
Still, we can count our blessings. This is a right triangle. Things could be worse. But now, I'll take pity on you and make it easier.
Here, have a Pythagorean Theorem:
cosh c = cosh a cosh b
(... You may be thinking, "Um, that doesn't look much like a Pythagorean Theorem," but, really, it is. It may help if you recall that cosh() has its own power series expansion good for small arguments (…except that since it has an infinite radius of convergence, it's good for arbitrarily large values of small, but never mind that…)
|cosh c||= 1 + c²/2 + [ignore this shit]|
|cosh a cosh b||= (1 + a²/2 + [ignore this shit])(1 + b²/2 + [ignore this shit])|
|= 1 + a²/2 + b²/2 + [ignore this shit]|
Subtract out the 1s, multiply through by 2, ignore the [ignore this shit], and then you should be seeing something familiar. Just like for spherical geometry where the Pythagorean Theorem is
cos c = cos a cos band cos c expands to (1 − c²/2 + [ignore this shit]); same story except for a teentsy sign change. Meaning both spherical and hyperbolic geometry look just like plane geometry if you're using it for Really Really Small triangles.
Well okay, maybe that didn't help at all. Onward,…)
So b is Alice's velocity angle. And Alice's velocity is tanh b. And we can actually get by without calculating b because we have Happy Trignometic Identities:
1 + sinh² = cosh² (analogous to cos² + sin² = 1)
sech² + tanh² = 1 (analogous to 1 + tan² = sec²)
the second line being the first line divided by cosh² (or multiplied by sech², sech being 1/cosh, because for some reason they had to have a different name for that).
So if Alice's velocity (tanh) vs. Carol is 3/5, then her sech is 4/5, and her cosh is 5/4.
Similarly, for a,for Bob's package vs. Carol, the relative velocity (tanh) is 12/13, the sech is 5/13, and so the cosh is 13/5.
Then, for the package vs. Alice, we multiply to get the cosh, (13/5)×(5/4) = 13/4, which means the sech is 4/13 and so that velocity (tanh) then has to be (√153)/13 = 0.951 lightspeed.
What about the direction, the angle A? SOHCAHTOA, baby! That is, since we now know all 3 sides of our right triangle, any of the those relations will do, e.g., let's do Opposite over Adjacent:
tan A = tanh a / sinh b
It's not just a/b because Not Plane Geometry (although, again, when a and b get really small — meaning the velocities are small enough that relativity barely does anything — it really does start looking like plane geometry, as you would expect). Note that it's tan A not tanh A because A is an honest-to-god ordinary angle, not one of these velocity-angle-bullshit things.
Running the numbers, the only new thing we need is sinh b; the cosh is 5/4 so the sinh must be 3/4. And then it's (12/13) / (3/4) = 16/13, take the arctangent, and we find that Alice needs to toss out the package out at 50.91° from her exhaust plume in the Plane of Bob. To get the other angle, the tangent will be (3/5) / (12/5) = 1/4 which means that Alice needs to fix the dish so that it's not facing her but rather pointed 14.04° away in Carol's direction when she launches the package.
And we're done and probably a lot more easily than the method they taught you in Physics 101. (And wow do those angles not add up to 90°.)
If we were to make the problem slightly harder by pointing Alice in a different direction, say towards Alpha Centauri, which is at declination −60° and therefore in a direction 30° from where Bob is headed,
then we no longer have a right triangle and so the Hyperbolic Pythagorean Theorem won't work anymore; we will need the Hyperbolic Law of Cosines instead
coshc = cosha coshb − sinha sinhb cosC
= (13/5)(5/4) − (12/5)(3/4)(√3/2) =
and now Alice needs to toss out the package at 0.806 lightspeed. To get the angle, while there is a Hyperbolic Law of Sines, I'd rather use this nameless beast
tan A = sin C / (sinh b / tanh a − cosh b cos C)which you'll notice reduces to the previously quoted opposite-over-adjacent rule in the event that C = 90°. And then you turn the crank to find that A = 118.37° and B = 15.96°.
But you probably didn't care quite so much about that actual numbers so much as the general idea of being prepared for the possibility of space pirates kidnapping you and forcing you to do their interstellar navigation for them because they will have somehow achieved space travel without inventing computers or understanding math or anything, a situation which you will now survive now that you're learned my handy math tricks, which makes total sense because we're basically talking about people stealing shit and then confining themselves to Winnebagos for decades at a time, as if it will ever make sense to travel interstellarly with rockets because rockets are so fucking stupid or if some other kind of propulsion technology even existed or will ever exist which it won't and what was I saying again?
Oh, right. Practical. Totally.
Just in case you were wondering, you are now qualified to sit in Pavel Chekov's seat. Yay.Or, at least, you will be once you learn how to deal with gravity. Though, honestly the situations where gravity matters are so exceedingly rare in interstellar flights I'm not sure why you'd want to bother. Seriously, getting out of earth's and sun's gravity wells is going to slow Bob's package by a factor of 1 part in 120 million, which means it'll be a quarter second late for every year of travel time. Carol is going to be so pissed. This entry was originally posted at https://wrog.dreamwidth.org/62027.html. Please comment there using OpenID.