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a.k.a., Space 11: How to do Interstellar Navigation

Various antecedents you may want to have peered at first:

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: B (Bob's care package) A (Alice) (Carol) C a ~ 12/13 b ~ 3/5 c ~ ?

  1. Carol lives at South Pole Station (a cool place where I think I will be hosting all of my geometry problems from now on).

  2. 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 …)

  3. 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…)

  4. 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.

  5. 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.

  6. 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.

  7. 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).
    So, …

    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:

Bob's package Bob's package's "1 hour from now" events Alice Alice's "1 hour from now" events Carol Carol's "1 hour from now" eventsn Package Launch (Now)

(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:

B (Bob's care package) A (Alice) (Carol) C a ~ 12/13 b ~ 3/5 c ~ ?

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 an angle, which means 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 b
and 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,

B (Bob's care package) A (Alice) (Carol) C 30° a ~ 12/13 b ~ 3/5 c ~ ?

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.

So,... picking up where we left off, I, the intrepid hero, am sailing off to your right into the sunset, with my incredibly reliable gerbil keeping time for me. You, the diligent historian, will eventually reconstruct everything I'm seeing from all of the reports you'll get — from the cloud of NSA bugs that I'm flying through — into a big, happy space-time diagram in which:

  • My time axis, everything that is happening "Here" according to me, is — as everyone would reasonably expect because I'm moving — slanted away to the right from your own natural, obvious, and vertical notion of "Here",

  • My space axis, everything that is happening "Now" in my direction of motion, according to me, is, — as nobody expected prior to 1905, — slanted up from your own natural, obvious, and horizontal notion of "Now" and by the same angle,

that second item being what makes all the difference, ruins Galactic Empire stories, and happens to be the only thing you really have to remember about Relativity because it's enough to derive all of the other wacky effects you hear about.

Here's how:

Introduce a 3rd player, call her Alice, moving away from you to the left at exactly the same speed that I'm moving to the right.

Let's also give Alice a gerbil clock that is identical to mine. Same laws of physics applying to an incredibly reliable identical-twin gerbil running on its own identical gerbil-wheel. Thus far, everything Alice does mirrors what I'm doing, as far as you're concerned. Therefore she's got the same tilt to her time axis, her space axis, same spacing on all of her ticks, etc.

Rewind everything to when Alice and I were in the same place. Call that moment "here" and "now" for both of us. And then we zoom in on what happens in that first tick afterwards:

Look at all the events I consider simultaneous with my first tick. One of them is on Alice's ship, and it's somewhat before when she gets to her first tick. Meaning her first tick isn't happening fast enough, as far as I'm concerned. Which means I'm "seeing" her clock running slow — and a quick look at that grayed right triangle shows that it's by a factor of √1−v² — or, if you're one of those weirdos who insists on using stupid units that entail the speed of light being some c≠1, then it's √1−(v²/c²), whatever.

For you, this should be no mystery:  my definition of "simultaneous" is fucked up in not being horizontal. And if we flipped things around and look at where Alice thinks my first tick should be, she can just as easily conclude that it's my clock that's running slow.

And if I tell you that second gray triangle to the right is just the first one rotated by 90°, then it's not too hard to see that the distances Alice and I are measuring in the direction of motion will likewise have to be fucked up and by exactly the same factor.

If you're getting the idea that my deciding that a bunch of events are all happening "now" is a pretty arbitrary thing, you wouldn't be far wrong. Apart from the moment we meet, I'm not actually there on Alice's ship, so one might reasonably conclude it doesn't actually matter what I think about her clock. In fact, we're never really going to be able to compare notes because if we stay on our ships and never fire our engines, we never see each other again.

It's also a fair bet that other observers will have yet other ideas as to which event on Alice's ship is one tick from now according to them. And then we're stuck in this rhetorical black hole where all opinions are equally valid as to where "one tick from now" actually is, leading us to conclude that the notion of "one tick from now" must actually be nonsense, therefore time doesn't exist, and everything decays into this heap of moral relativism, Satan wins, etc.

We can still salvage something. There may not be a universally agreed, absolute notion of when "one tick from now" is. But, everyone still has to agree on which event on Alice's ship Alice thinks is one tick from now and that Alice and all of her passengers, just like her gerbil, must be experiencing one tick's worth of physics in that whole time.

And if we arrange for one of her passengers, Dave, to jump ship at the point that I think of as being one tick from now, everyone (including me) can and must still agree that Dave could only have experienced one √1−v²th of a tick while aboard Alice's ship.

If we then have Dave immediately catch a ride with Bob, who is coming towards me with the same velocity as Alice is moving away, so that Dave can arrive back where I am just in time for my second tick, we can all similarly calculate that Dave only experiences another √1−v²th a tick, thus arriving back at my place somewhat younger than expected,…

…and that is what rubs it in our faces that time is not this Absolute Thing, i.e., the way Newton and Gallileo thought it was.

Einstein probably figured he was done at this point, but I'm sure he hadn't reckoned with the tenacity of 20th century SF writers, so I'm going to go a bit further with this and extend the diagram to the left (a whole lot).

Let's suppose for the sake of argument everything thus far is all happening out in space 500 light-years from Earth, and let's suppose there exists some magical Tachyon or Subspace Transmitter Ansible Thing that allows communicating with Earth in real time. Remember, Sinclair was able to talk to Geneva on Gold Channel and get immediate responses.

Do we need the super-genius Vulcan working out the Intermix Formula that's likely to make the ship go up in the biggest fireball since the last sun in these parts exploded, which is why nobody'd ever thought to try it before?

Do we need freak sunspot activity interfering in just the right way with the Unobtanium-Powered Stargate?

Do we need Mysterious Tech from the Ancient, Dead Civilization that takes up the inside of an entire planet and can only be used for just the one episode?

Or… maybe,… just maybe,… we can look at where Bob's and Alice's "Now" axes are.

If we know how to make a subspace transmitter capable of sending messages instantaneously across 500 light-years, then we can make two of them: One for Bob, one for Alice. The physics of it, whatever it is, should work just as well on their ships as on mine, so all I have to do is give my message ("pls KILL DAVE right now. kthnxbai") to Bob when we meet just as Dave is returning from his journey, Bob sends it to Earth, Earth relays it to Alice, it arrives when Dave is first setting out, and voilà: trivially easy time loop.

For extra fun, let's see just how incredibly fast Bob and Alice have to be going in order to send a message back in time, say, one whole day. That ought to be enough to wreak havok, right?

Stretch the diagram so that the [Earth ⟷ Me] distance is 500 light-years, and then we make each of the ticks 12 hours. 500 years to 12 hours is a ratio of 365,000 to 1. Which makes all of those grey triangles really, really thin. Which means it's enough if Alice and Bob are going 1/365,000th of the speed of light, which is,… wait for it,…

900 meters per second.

This is so ridiculously easy we don't even need spaceships. Alice, Bob, and their respective correspondents on Earth who are relaying the messages can all be flying SR-71 Blackbirds, and that's enough for it to be game over for causality right there.

You say, "Fine, so we don't do real-time communication. But surely, we could still have something where the message to Star Fleet Command takes 2 weeks to get there?" This changes nothing. Rerun the previous problem and ask how fast Alice and Bob have to be going to get a message back in time 4 weeks plus one day. The answer comes out to around 24,000 m/s. Yes, that's a bit harder. It's roughly how fast the Earth moves in its orbit around the sun. I think we'll be able to manage that.

If the communication delay is anything less than 500 years (minus 12 hours), Alice and Bob can go fast enough to make up for it and get the message delivered in time (a day ago).

The problem with "Meanwhile, back on Earth,…" is this:  If you are 500 light-years away from Earth, then "meanwhile" can, depending on how fast you're going and in what direction, refer to anything between the cord-cutting ceremony for a shiny, new Star Fleet Academy building in Marin City, California, and the first Spanish Explorers landing at Point Reyez 1,000 years earlier.

That much is not a problem so long as you and the Meanwhiles cannot actually talk to each other.

But if you're relying on a Meanwhile for story purposes, then, chances are, you are assuming that they can. Why else would you be bothering with the happenings on Earth if it weren't going to affect your characters less than 500 years in their future?.

And that right there is where you're going wrong.

Because once there's any kind of conversation along the "Now" lines, that means they all can talk to each other. Easily.

From there it's a very short and easy step to having the rogue Star Fleet cadets teaching the Ohlone tribes how to make phasers, and would-be conquering Spaniards and causality all get toasted extra crispy.

Note that this is not an argument for FTL travel/communication being impossible. Just that if it were possible, in the sort of arbitrary and ubiquitous way that you would need for most Galactic Empire stories, then we're in Bill and Ted Land. Which we could actually be, for all we know, though, if so, I'd like to think we'd have noticed this by now. In any case, all of your stories then end up being unintentional time-travel stories, which will then consequently officially suck, because with two (2) pairs of stargates, a couple of non-FTL-but-really-fast ships, and a big enough fuel supply, you can fix anything…

…and then we spend the rest of the movie chasing down Biff's Sports Almanac.

This entry was originally posted at https://wrog.dreamwidth.org/61845.html. Please comment there using OpenID.
(Yes, more Space! Part 1 is here but you don't need to go back that far to follow this)

Having at least covered (see Part 8) the question of why the speed of light is constant, the next order of business is the weird and wacky consequences. I'll skip straight to the one that nobody seems to get:

If you're writing your Galactic Empire story and you find yourself needing to say, "Meanwhile, back on Earth,…" that, right there, means you are doing it all wrong.

This can be explained, but first, we need some building blocks:

A Digression on Events and Light-Rays

An event is some instantaneous thing that happens somewhere.

  • On April 11, 1945, just off Okinawa,
    the USS Enterprise gets hit by a kamikaze.

  • On 287 พשּׁ₽ủ⿓, 389284th Year of the BL🐡R🐢G,
    somewhere near Gamma Leporis (~30 light-years from Earth)
    construction is completed on The Giant Commemorative Mirror.

  • On June ▒, 200▒, in a particular hospital room in Bellevue,
    my son is born.

I could do this all day. The entire history of the universe is basically one huge grab-bag of events.

We typically identify events by saying when and where they happened, but that is perhaps a bit misleading since actual numbers for time and place turn out to be negotiable given that observers are free to set up their own coordinate systems. Surveying your back yard, you can calculate latitude and longitude for all of your fence posts, but a prequisite and key piece of that puzzle is the direction you decided to call "North" (South Pole Station inhabitants can have particular fun with this). Change that and the latitude/longitude numbers all change,… but that doesn't cause the fence posts to move.

There's a similar choice with time. If you, sitting in your spaceship observing things, fire your engines, then you're changing the direction that you're calling "The Future". A series of events that were originally all going to take place near your ship is instead going to occur at progressively farther distances from you. But, again, despite that, the events themselves, The Things That Happen, our fence posts, remain wherever/whenever they are.

In particular, if the Things That Happen send messages to each other, then everyone observing them has to agree on who's sending what to whom. Because light rays carry information, whatever happens at the source has consequences at the destination, and no switching of viewpoints should be able to affect that.

So if the kamikaze blowing up on the deck of the Enterprise sends light out into space, which travels for 30 light-years, then encounters the Gamma Leporis mirror, and bounces back to the aforementioned hospital room, where yours truly had the foresight to set up a telescope pointed in exactly the right direction, then we can be in the birthing room watching/experiencing the battle of Okinawa in real time, with all that implies

(..."Oh cool! Enterprise control tower just got blown up. Okay, hon, time for another push…" and I'm pretty sure this would have had consequences if I'd actually done this.)

Building a Map

What you might not have expected is that just this much, i.e., having the universe be a grab-bag of events tied together by light-rays, is, by itself, enough for any single observer to map out everything in his/her vicinity.

Here's one way to do it:

  • We have bugs everywhere, helpfully planted by the NSA. I suppose if we want to be slightly more realistic, it'll suffice that there be a bug present at just the events where we need them, and they just get there somehow (and exactly how, we probably don't want to know).

  • I have a clock — a device based on some well-understood physical process, e.g., an incredibly reliable gerbil running on his little gerbilwheel specially contrived to make a "tick" every so often in a uniform way and sufficiently quickly that I can get whatever timing accuracy I need.

  • I have a transmitter that the bugs are all tuned to listen to. Every time my clock ticks, my current timestamp (= number of ticks since I started my clock) heads out into the universe at the speed of light.

  • The bugs have two jobs:
    1. Remembering what time it is, or, rather, what timestamp they most recently received from me. Since I'm constantly sending these out, they'll constantly be getting new ones. That way, they don't need clocks of their own.
    2. Reporting back to mommy. Every time a bug notices Something Interesting in its immediate vicinity, it compiles a (brief) report ("ship go boom" or "baby born") with my latest timestamp, and then broadcasts it in all directions.

  • Finally, I can have a very sensitive and very directional receiver.

So I sit there for a while collecting bug reports. The trumpet sounds and Time comes to an End, or maybe I just get bored and decide I'm done with all this. And then I can compile all of the reports I've received from everywhere and plot when and where all events occur onto one big, happy space-time diagram:

The way you read this is Time goes bottom to top. The black vertical axis in the middle is me and all of my clock ticks, evenly spaced because my gerbil is so incredibly reliable. Easy.

The horizontal axis is one of the spatial dimensions, call it "X", and there's some choice here about which way to point it. For the sake of argument I'll use some event that I care about and point it that way, meaning when the event finally occurs, it'll be on that line somewhere.

The speed of light being constant, once we send a signal, we know exactly how far away it'll be at any given time. So we can draw, in blue, lines representing all of the timestamp signals heading outwards, and they all have to have the same slope (because the slope is how fast they're going and that number is always the same).

I've taken the liberty of bolding one of the lines. All of the events where there are bugs receiving that particular timestamp have to be on that particular blue line.

Similarly, in red, I've drawn the possible paths of incoming signals from the bugs, and, again, helpfully bolded one of them. If I receive a signal at a particular time from a particular direction, then it had to have come from some event along that red line.

Once I receive a report and extract the timestamp, that, along with the time direction of receipt is enough to nail down exactly where the event is on this diagram.

(… It is a bit unfortunate that this 2-dimensional picture can only represent happenings in a 1-dimensional universe, but that's really all we need for now. If you want, you can imagine a Y axis pointing straight into the page/screen. The bent blue line becomes a blue cone, while the red line remains a single ray intersecting it at exactly one point, so this all still works. If you're totally insane, you can try putting back the Z axis as well, coming orthogonally out of reality, though if you were already able to do that you probably didn't need me to be explaining this stuff…)

What's interesting here is where the left-right axis comes from. Since I'm not moving — yeah, I know, hold that thought,— light rays will cover the same distance going out to an event and back. Even if the event is happening on some moving object, or perhaps the bug doing the report is itself moving around, or both, we simply don't care, because the bug receiving the timestamp, the event itself, and the bug's reporting of it, is all over and done with in such a short span of time that any errors resulting from the relative movements will be teentsy.

Therefore if the timestamp received at event E was 1006 and the corresponding report came back at 1026, then, obviously, the event E
  • happens midway between the two times, at 1016, and

  • at a distance of 10 ticks away from me, because if the signal is taking 20 ticks to get back to me, it must have taken 10 ticks to go out and another 10 ticks to come back.

Yes, I'm using "tick" as a unit of distance:  it's how far light travels in a clock tick. The speed of light is then 1, i.e., one tick per tick, and that's one less multiplication we have to do, so yay. This is also why we have the blue and red lnes being sloped 45°, in case you were wondering; life is so much more convenient when you use the right units.

We can also have other events happening at the same time (1016) at different distances away. But in order to infer that something happened at time 1016, the timestamp has to be just as much earlier as the receipt time is afterwards. Which then dictates that all of those events are on the same horizontal line as you'd expect.

Notice how I'm using my clock to measure distance. This is in fact the only good way to measure distance; actual, physical rulers are far too prone to being stretched, bent, or broken. Yes, things can happen to light beams, too, but we'll generally know what's happening and be able to compensate for it.

But the real point is so that we can do all of our measurements without having to go anywhere. All measurements come to us if we live long enough.

Now for the fun part…

Another Point of View

… in which Somebody Else, say, you, tries to reconstruct what I've done.

In what follows, you are somewhere off the screen, with your own clock built in its own peculiar way (you have an extraordinarily dependable hamster) ticking away at whatever rate along a vertical line, broadcasting your own timestamps, none of which I'll be bothering to show.

Your own "tick" is whatever it is, but for for the sake of clarity I'll assume your distance unit likewise matches up with your time unit so that you, too, can have light rays sloped at 45°.

What remains is to modify the bugs ever-so-slightly so that they're smart enough to distinguish your timestamp signals from mine and nice enough to include both timestamps in whatever reports they send. And then I ignore your timestamps, you ignore mine, and all proceeds as before.

At some point the trumpet sounds, or you get bored, and now it's your turn to reassemble the puzzle pieces.

There being NSA bugs in my clock, that's as good a place as any to start and you'll have all the information you need to plot where and when all of my clock ticks are taking place. You thus discover that I'm moving at some particular velocity (because different people can have different velocities and my future will be in a different place from yours; who knew?)

The velocity is constant because we're all just coasting along with engines turned off, nobody is spinning, and we're out in deep space where there's no gravity to worry about. And whatever direction it is I'm moving, we'll take that to be your X direction so that everything can stay on the page.

Once that's done, given that we have to agree on which events are joined by light rays, then for any other event, like E, no matter what place time numbers you assign to it, it remains the case that a (blue) timestamp message travels from me to E and a (red) bug report gets sent back; you have to have those light-rays linking E to the same ticks on my clock as I do. Which means once you've nailed down all of my clock ticks, E is nailed down as well.

The same goes for all of the other events I know about, including all of the ones on that now-not-so-horizontal line where I have t=1016.

Recall that the only assumption that went into this is that I have a constant nonzero velocity in your X direction, which necessarily tilts my time axis from your vertical. The rest of my grid of light-rays and events folds up sideways like an empty wine-bottle carton being stepped on, and there's only one way it can fold once you know how fast I'm going. Which means there's no way around concluding that my X axis, i.e., all of the events I consider to be simultaneous with my own here now, must be tilted from your own horizontal one…

and, it's not too hard to show, by the same angle. Meaning if, according to your grid, my 1026 clock tick is 500 light-years away from my 1016 clock tick — the sort of scale you need for a Galactic Empire story, — then, for you, E is happening 500 years later than my 1016 clock tick (rather than simultaneously, as it is for me).

It also turns out to be fairly simple to find an observer for whom E is happening 500 years earlier than my 1016 clock tick.

We don't even need to know exactly where you are or how fast your clock is ticking; which is why I haven't bothered to plot any of that.

Everything thus far is an unavoidable consequence of events being connected by light-rays and the speed of light being constant…

…and this much is enough to screw the pooch for any kind of Faster Than Light travel/communication.

(though exactly how will need to wait for Part 10)

This entry was originally posted at http://wrog.dreamwidth.org/61663.html. Please comment there using OpenID.
8th-May-2017 10:15 pm - Exodus
So, wow, it's been 15 years. And yours truly is Not Actually Illegal in the Russian Federation (at least, not so far as I know and not yet).

But I'm still joining the Exodus to Dreamwidth because, well, fuck every last bit of that noise (also, fuck the ads, and maybe also fuck brad, while I'm at it...).

In other news, I learned enough of the API to fix the cross references in my journal to all point to Dreamwidth so that you can painlessly follow my various n-part series which might possibly be continuing; we'll see. (I may be sufficiently annoyed to change all of cross references on the LJ side as well, whee...)

Now I just need to turn off commenting on the LJ side (just c'mon over — or if you don't want to do the wholesale move, just create an account and claim your LJ OpenID (note that some of you who've already moved here still need to do that...)

This entry was originally posted at http://wrog.dreamwidth.org/61213.html. Please comment there using OpenID.

(Still waffling on whether this should be Space Travel part 8 or Relativity part 1; we'll see...
If you want to go back to part 1, that's here though none of the prior material really matters for this one.)

It's weird to me how everybody knows there's a problem.

How, if you are, say, Issac Asimov or George Lucas or Gene Roddenberry trying to write your Galactic Empire/Federation/Whatever story, you've got to do some kind of handwave about Relativity. We know this because we were taught at some early age that trying to go faster than lightspeed is Somehow Bad, even if nobody ever explains the details.

It's particularly annoying to me because many of the details actually could be explained without going beyond 6th grade math. If you get how right triangles work, that's pretty much all you really need for Special Relativity, except nobody seems to bother trying.

I suppose part of the problem here is the mystique of General Relativity. True story:

Arthur Eddington is at a meeting of Royal Society in 1919 where Sir J.J. Thomson (President) concludes a talk saying nobody's ever really stated in clear language what Relativity is about. The meeting disperses, Ludwig Silberstein (the author of one of the early books on relativity) comes up to Eddington and essentially says, "So you must one of three people in the world who actually get this stuff." Eddington demurs, but Silberstein pushes further, "C'mon, don't be modest," at which point Eddington replies, "Actually, I'm trying to think who #3 is..."

Then again, given the sheer number of people I've encountered who have Misner, Thorne, and Wheeler's Gravitation sitting on their bookshelves, that's clearly not the case anymore.

But if even Albert Fucking Einstein had to take five years off to get up to speed on Differential Geometry — which to some extent is just the Partial Differentiation Chain Rule on Steroids and working through all of the various consequences, but if you didn't get past high school calculus, even that much is going to be a bit hard to take — what hope is there for the rest of us ordinary mortals?

But it so happens that, for Special Relativity, you don't need differential geometry. So let's get started:

Why is the speed of light constant?

How did they even get this idea that the speed of light is always the same? That's the part They never explain. They figure there's no way anybody's going to get that without having the full-ass physics course, so they don't bother trying.

The short answer is that they've done the experiments and that's the way it actually is. Since that's unsatisfying, I'm going to try giving you the half-assed physics course instead:

Read more...Collapse )
26th-Jun-2014 01:57 am - On Caucus Math
party politics
(... The curse of being a math geek living in a state where they have caucuses instead of primaries (not to mention having spent some time observing party rules committees) is I end up thinking about this stuff...)

So here's a fun issue with caucuses:

First, a quick review of the basic rule for awarding delegates at caucuses.

Your precinct is allocated some number of delegates, D, to elect. Some number of people attend your caucus and each declares a preference for a particular candidate. Twenty minutes later, once the blood has been mopped off of the floor, the battle lines have hardened, and everyone who might have been inclined to change his/her mind has been talked to death, you then compute for each candidate the following number
(# of votes for that candidate)
—————————————  × D
(# attendees)
I'll note first of all that every attendee will in fact be included here, because if you don't ever actually declare a preference, that's treated as equivalent to declaring a preference for "Uncommitted," this extra fake candidate that's always added to the mix. So it's guaranteed that all of these "delegate-share" numbers will indeed add up to D.

The next step is to split each delegate-share into a whole number plus a fractional part. For each candidate, the whole number gets awarded directly, and, if those numbers by themselves don't sum to all of the available delegates, you then rank the fractional parts and distribute any remaining delegates, one each, to the highest candidates in that fractional ranking.

(...And yes, for those of you who know about this, I'm skipping the 15% threshold rule, which some states apply at the precinct level. Thankfully, in Washington, we got rid of that 8 years ago, since it's a complete waste of time at the precinct level [also has any number of bad effects, but that's a whole 'nother discussion].)

So first, I'll present Survival Trick Number One, so that you can survive in a chaotic caucus environment without having to do long division in your head. It goes like this: we rewrite the formula above as follows:
(# of votes for that candidate)
((# attendees)/D)
i.e., just divide numerator and denominator by D, which works because multiplication is commutative (...except that the DNC stupidly ruins the commutativity by including a 3-decimal rounding rule in the process, but, as it happens, this doesn't affect things very often at the precinct level, and in any case this still works fine as a rule of thumb so that you can wrap your head around what's going on...)

Bottom line is there's a certain magic number of votes ((# attendees)/D) that you need to get a "whole delgate".

Meaning you can take your caucus, divide it up into blocs of that many people who are all voting for the same candidate. For each such bloc, that candidate gets a "whole delegate", and then whatever votes you have left over, you rank those, and the candidates that are highest on that ranking get the remaining delegates. The advantage of doing things this way is that you're just counting votes without having to do any long division in your head.

Concrete example:

You're in a precinct that's been allocated two delegates.
Twenty people show up.
14 are Kerry supporters.
6 are Dean supporters.
If you follow the worksheet, then it's (14/20)*2 = 1.4 vs. (6/20)*2 = 0.6.
Kerry gets 1 whole delegate and then, because 0.6 beats 0.4, Dean gets the other one.

Or you can do it my way, seeing that (20 attendees)/(2 delegates) = 10 votes needed to get a whole delegate. Thus, Kerry's 14 votes produce one whole delegate (10 votes) with 4 left over that then lose to Dean's 6 leftovers, and so Dean gets the other delegate out of the "fractional ranking", never mind that we're not having to rank fractions any more.

Now for the problem.
It turns out that the number of votes that you need to get a whole delegate is NOT the same as the number of votes you need to win a delegate out of the fractional ranking.
In fact, if your candidate is getting awarded any whole delegates at all, there's a fair argument that some of your votes are being wasted.

What do I mean by this?

Back to our example: Thus far, it's 14 to 6 with each candidate ending up with one delegate. But then the Kerry folks wonder if they can do better. And it turns out, they can!

After a brief strategy session, 7 of the Kerry voters change their preference to Uncommitted. Which now means the totals are 7 Kerry, 7 Uncommitted, and 6 Dean. Since you need 10 to get a whole delegate, there are now zero whole delegates, the fractional ranking then has to award two and they go to Kerry and Uncommitted.

Except,... since the "Uncommitted" folks are really all Kerry supporters, it's a good bet that "Uncommitted" delegate will be signing in for Kerry at the next caucus level.

Which means Kerry has just effectively cleaned up and claimed both delegates.

WTFF? How did that happen?

On the other hand, he did have more than 2/3 of the vote in that precinct, so there's some argument that this isn't actually a totally unfair outcome and perhaps the real question is why should the Kerry supporters have to jump through this extra hoop to get the delegates that are rightfully theirs?

The problem is that, while the number of votes you need to get a whole delegate is
(# attendees)/D,
the number you need to guarantee one out of the fractional ranking is actually
(# attendees)/(D+1),
which, in our example is 20/3 = 6+2/3.

... or, more precisely, if your candidate has at least (# attendees)/(D+1) and there is at least one candidate with strictly more than (# attendees)/(D+1) (even if it's only the slightest ε more), then you are guaranteed a delegate out of the fractional ranking (since in that case there cannot be more than D groups with (# attendees)/(D+1) votes, and you're one of them, so you win)

Dean with 6 votes isn't quite there, and that makes all the difference in the world.

Meanwhile back in the first scenario, where the Kerry supporters are spending 10 votes to get a "whole delegate", this can now be seen as a ripoff, spending 10 when they only needed to spend 6+2/3, thus wasting 3+1/3 of their votes, which costs them a delegate.

More generally, (# attendees)/(D+1) will always be less than (# attendees)/D and, if you're in a close race, chances are you're going to care about that difference.

And apparently, they've even thought about this in Iowa, or, rather, it's the only way I can account for Iowa's version of the threshold rule which not only makes things way more complicated, but also introduces the nastier features of thresholds into the lower-delegate caucuses where they weren't originally a problem.

My fix, which will most likely never be adopted, is much simpler:

Instead of multiplying by D, multiply by (D+1).

That is, for each candidate you instead compute
(# of votes for that candidate)
—————————————  × (D+1)
(# total number of voters)
and then proceed as before, awarding the whole numbers, and again if the total number of delegates awarded in this way is different from D, use the ranking of the fractions to fix it. The only difference now is the (remote) possibility that there will be too many (i.e., D+1) delegates awarded via the whole numbers, in which case, instead of giving out delegates to whoever is at the top of the ranking, we're taking a delegate away from whoever is at the bottom (except that if all of the whole numbers are indeed adding up to (D+1), that means all of the fractional parts will necessarily be zero, so we don't even have to look at any ranking; you just pick somebody at random to lose one).

So in our example, for Kerry the magic number is 14*3/20 = 42/20 = 2.1 and for Dean it's 6*3/20 = 18/20 = 0.9, Kerry gets 2 and we are done; we don't even need to consult the fractional ranking at all.

Or, calculating things my preferred way, you need 20/3 = 6+2/3 votes to get a whole delegate, Kerry has 2 such blocs, Dean doesn't have any, and again we're done. At which point it should be blindingly obvious that there's absolutely no advantage to be had by splitting your voters up over multiple fake candidates; you'll get the same result either way.

Which is the way caucus rules should be (i.e., they just give you an answer and no amount of gameplaying changes it).

But, even though they're never going to adopt this rule, you can still use it, the point of it being that if you're ever in a situation where the Multiply By (D+1) rule is giving you a different answer than the actual Multiply By D rule, that's where you have to watch out.

This effect is most pronounced in the 2 delegate caucuses but it can show up in higher-delegate caucuses as well.

Example 2

4 delegate caucus
20 people show up
17 for Kerry
3 for LaRouche

So now it takes 20/4 = 5 votes to get a whole delegate. Thus, Kerry supporters can spend 15 to get 3 of the 4 delegates. Unfortunately, that means they only have 2 votes left over, which lose to LaRouche's 3 in the fractional ranking and so we get one LaRouche delegate.

What does the Multiply By (D+1) rule say? In this case (D+1)=5, meaning you only need 20/5 = 4 votes to get a delegate out of the fractional ranking, and, with 17 votes, Kerry supporters can produce four such blocs which will all beat LaRouche's 3 votes.

... the only problem being that those blocs need to be for four different candidates. But this is actually doable:

5 for Kerry
4 for Uncommitted (actually Kerry)
4 for Sharpton (actually Kerry)
4 for Kucinich (actually Kerry)
3 for LaRouche

Now Kerry only gets 1 whole delegate while the fractional ranking awards the rest to Uncommitted, Sharpton, and Kucinich, who all change their votes to Kerry at the next caucus, and thus Kerry cleans up all four delegates.
8th-Jan-2014 12:21 am - Today's game theory intro
William has learned about Rock, Paper, Scissors. In honor of that, a puzzle:

Same game: Rock breaks Scissors, Paper covers Rock, Scissors cut Paper. So far so good. Now we add a couple twists:
  • I do research and find myself a Better Rock, a chunk of pure New England granite that rules, absolutely crushes all lesser Rocks. Even though it still loses to Paper I'm happy with it.
  • Meanwhile you've been doing your own research; being of a technological bent you know there are better ways to do Scissors; carbon steel with a diamond edge; not only cuts Paper but completely destroys other Scissors as well. Granted, even a diamond edge is no match for an any actual Rock, let alone mine, but you still win against everything else, so you're happy.
So... my Rock beats your Rock, your Scissors beat my Scissors, and Paper vs. Paper is the only draw possibility left.

What's my strategy? What's your strategy?

And how does this change if I go out and get really good, battle-ready Paper as well, e.g., some of that Tyvek stuff that they use to insulate houses; something that will entirely shred your Paper, even if your high-tech Scissors will still make short work of it. Meaning that not only is there no longer any possibility of a draw, but out of the 9 possible scenarios, I'm winning in five of them.

Unfair, right?


(This is Part 7. There are previous installments; this one is a digression from something I said in Part 6, though you can also start from the beginning at Part 1)

In Part 6, I said:

The theoretical absolute best we can do with rockets is if we can get the exhaust velocity up to the speed of light. This means our exhaust will be pure radiation, that we are somehow powering a huge-ass laser with 100% efficiency, since that's the only way we get all of the exhaust going the same direction. And, boy howdy, do you not want to be following along directly behind,…

Which leads rather directly to this:

The best possible rocket engine and the best possible directed energy weapon are exactly the same thing.

Remember this the next time you're watching Star Wars, Babylon 5, Battlestar Galactica, etc. All of those little fighter ships where the engine is distinct from the guns? Those scenes where they're accelerating forward, closing with the enemy, firing forward with everything they've got?

Wrong. Wrong. Wrong. No military contractor worth its salt is going to waste resources mounting a second gun on a ship when there's already this totally effective and somewhat expensive first gun.

Conversely, if you've got a phaser/laser/gamma-ray-laser that can do real damage from a distance, then most likely that is your engine. If you're in a universe where rockets are your only form of propulsion, you are definitely not going to be wasting resources on a 2nd engine. It's going to be correspondingly expensive to fire, too. Nor will you get off that many shots before you're hurtling away.

What you need to do is arrange to be headed towards your target with as much velocity as you can manage. Then, when you're really close, you flip around and shove the throttle to maximum. It'll look exactly like you're landing on your target (modulo the small matter that you'll want to not be too predictable about it, see below). Best if, once you've killed all of your relative velocity, you can whip out some (really strong) tethers to attach yourself with before continuing to fire, so that you can be expending as much energy as possible on your target vs. propelling yourself away.

Of course, if you actually can get that close you're probably still better off with a burrowing torpedo that can blow up your target from the inside.

In fact, I'm rather having trouble shaking the conclusion that directed energy weapons aren't at least as stupid as rockets. On the other hand, if you're stuck in a universe where rockets are all you have, then so be it. Swords, planted bombs, and bioweapons are all very nice if you can get close enough to use them, but sometimes you just can't.

Also, to be sure, planet and asteroid-based gamma-ray-laser cannons will be a different story. They'll have room for arbitrarily huge reserves of antimatter compared with what you'll have available in your fighter ship, and the momentum consequences of firing off huge blasts will be negligible for them.

Suffice it to say, you'll want to stay well out of range of those. Except for the small problems that,

  1. being lasers, they'll have lots and lots of range, and
  2. the moment you stop firing your own engine for any length of time, your trajectory becomes immediately predictable; figure by the time we have practical antimatter distilleries, we'll have the software for this worked out just fine, too
  3. given the stupidity of rockets, you won't be able to be constantly firing your engine for any length of time before running out of fuel
So, good luck with that.

Granted, if I were going up against an entire planet, I'd probably want to arrange for a dinosaur-killing asteroid to do the dirty work for me. Hide an armada behind it to take out anyone who tries to come near to divert it.

Which then means that any sensible planet is going to have an entire inventory of asteroids of various sizes lined up at its Lagrange points to be able to deal with any such threat, at which point I'd then concentrate my efforts on subverting the folks in charge of the asteroid inventory.

Or maybe just taking a trip down to the planet itself, sneaking in, and detonating the huge antimatter reserve where the phaser cannon is located.

Of course, if everybody has sufficient resources to be distilling out the insane quantities of antimatter needed to be fighting these battles, I'd have to wonder what the hell they're fighting over. Not that this would be the first time in human history where a war got started for completely stupid reasons (cf. WWI)

And round and round we go.

(and there's a Part 8 now, where we move on to Something Completely Different)


(This is Part 6. There are previous installments, though if you only made it as far as Part 3: Rockets Are Stupid, that's good enough for this one.)

Rockets are Even More Stupid Than You Thought

Meanwhile, back on the launch pad, staring at the 2500 tons of Saturn V that I've just told you how to make smaller, we can ponder what's going to be possible once our technology gets Really Good.

The theoretical absolute best we can do with rockets is if we can get the exhaust velocity up to the speed of light. This means our exhaust will be pure radiation, that we are somehow powering a huge-ass laser with 100% efficiency, since that's the only way we get all of the exhaust going the same direction.

And, boy howdy, do you not want to be following along directly behind,…
… which inspires an observation about fighting space battles, which I'm going to defer to Part 7.

Anyway, this Best Possible Rocket brings the fuel cost for getting your Winnebago-sized Command Module to the moon and back again down to a mere 327 grams.

The catch is that half of that 327 grams will need to be antimatter. This also assumes you've solved the problem of storing it in a reasonable way — and if so, the Fusion Power People would really like to hear from you; and no, they won't necessarily be obsolete, because antimatter is merely a storage medium; you still have to extract the energy from somewhere. It should be noted that the amount of energy needed to make that amount of antimatter — and what you get back when you let it recombine — is roughly that of an 8 megaton bomb.

So, if you're imagining this to be the family car, where you can just hop in and fly to the moon for a week when the kids are off school, guess again. Unless you like the idea of each of your neighbors having an 8 megaton bomb in the garage and DUI being about much more than just the occasional lamp post or pedestrian. Hell, let's just have every auto repair garage, bus station, and airport be a terrorist candy shop, where the stakes in question are not just single office buildings but whole continents and planetary habitability.

Suffice it to say, there's a whole range of social and political problems we're going to need to have completely solved before we get to any kind of ubiquitous space travel regime.

Never mind that said problems will have bitten us in the ass long before we have practical antimatter distilleries. No matter how many countries we can get to sign the nuclear non-proliferation treaty, if you're Joe Sixpack sitting at L1 with a 6 ton rock and the right software, that's an 8 megaton bomb you can drop anywhere on earth, no nuclear tech needed.

To The Stars?

And all of that insane energy expense is just to get to the fucking moon. You want to go to the stars?

Using this theoretically perfect rocket which will never actually exist, accelerating at one g (earth gravity) for a day costs you 0.3% of your ship. That may not sound like a lot but you'll need to keep that up continuously for a year to get to 3/4 the speed of light. At which point 2/3 of your ship is now gone (which means at least 1/3 of it was antimatter to start with).

That may be good enough to reach Alpha Centauri, but to get anywhere real, you need to keep this up. Four more years (proper time) of accelerating at 1g — meaning you'll have to spend 242/243 of your original ship; picture launching one third of a Saturn V half-filled with antimatter — puts you 75 light-years out from earth, with finally enough time dilation (100 to 1) so that you can coast across some reasonable fraction of the galaxy (thousands of light-years) in a single lifetime.

Confined to a Winnebago.

And you thought space exploration was going to be fun and exciting.

Also, you better hope you picked a good destination, because, unless you happen to have a perfectly placed neutron star or black hole in your path — at which point you will then also be needing enough shielding to cope with all of the crap likely to be in the vicinity of any such object, because going at any significant fraction of light-speed means you'll need to get really, really close to make any kind of tight turn (also, good luck with the tides) — course changes will be essentially impossible once you get going fast enough.

Nor will you be able to stop anywhere along the way or even slow down at the end of the trip, unless you've arranged to be able to spend another 242/243 of your ship and allowed for another 5 years (proper time) to do it. Meaning we're now launching something 73 times the size of the Saturn V, half-filled with antimatter, and stuck in the Winnebago for a minimum of ten years, in order to be able to do any kind of interstellar travel beyond 150 light-years.

Rockets just suck.

Why I Shouldn't be Allowed to Write for SF Television

This, by the way, is something else that Star Trek and similar shows get wrong.

((Update: It seems I've done Roddenbury an injustice; he apparently did make a pronouncement about impulse engines early on. So you'll have to read this as, "What Star Trek would have been like if impulse engines were rockets." In any case, it doesn't let any of the other shows off the hook: BSG, I'm lookin' at you.))

It's not that I'm going to fault them for postulating the existence of something like Warp Drive, which you just need if you're going to do Galactic Empire stories. Nor am I going to fault them for not dealing with Relativity properly, because the sad fact is that most SF authors only understand Newtonian Universes anyway, and I'd just as soon they stay in their comfort zone and tell stories that make sense on their own terms, rather than attempt to include Relativity and make an utter, complete hash of it. The other fun thing is that if you try to have Warp Drive and Relativity in the same story, then that generally means you have a time travel story even if you don't realize it, at which point JWZ's Law probably applies.

No, where Star Trek — along with everybody elseactually screws up is with the impulse engines, whether they're called that or "thrusters" or "reaction engines" as on some other shows, they are clearly intended to be rockets of some sort. And then they get used in every episode as a completely routine means of puttering around a planetary system.

To which I say, "Wrong." Firing a rocket is cannibalizing your spacecraft; it uses exponential amounts of fuel; you never want to do it if you have any alternative available. Rockets are the propulsion method of last resort.

Meaning if you actually had anything at all like Warp Drive, you would contrive a way to use it for everything you possibly could. You'd use it in-system, you'd install it on the shuttlecraft, you'd use it in the space dock, you'd use it for going to the grocery store. You would use it everywhere that you didn't have some other reasonable alternative (like space elevators, solar sails, tethers, whatever).


The only proper scene in which the impulse engines would even be brought up would be something like the Battle Aftermath Scene. In which the ship has been wrecked by Commodore Decker's planet killer or some such. They've barely eked out a victory, but half the crew is dead. Bodies are scattered everywhere. Shit is on fire. You can hardly see through the smoke. Sulu and Chekov have big nasty burns. Kirk has his shirt ripped off and is bleeding in a dozen places. McCoy and Chapel are buzzing around doing triage. Warp drive is trashed. Dilithium crystals are hopelessly fused, etc.

And now it transpires that they're spinning out of control into a planet or something. At which point we have a dramatic pause and musical cue as Kirk calls down to Engineering.

"Scotty," he says, "Ready the impulse engines."

Some of the younger bridge crew startle at this. It comes up in the training sessions but you never imagine that you'll hear it for real. Because if you do, it usually means you're about to die.

And then we have the long anguished close-up on Scotty. He looks around at the debris in the engine room and realizes the captain is probably making the right call. And finally...

"Aye, captain."

His children are about to be murdered and there's nothing he can do about it. He motions two of his surviving lieutenants over and together they remove the cover and break the seal on the impulse controls. There's a huge lever there that takes three men to move.

Back on the bridge, Kirk flicks a switch on his armrest, "All hands! Jettison Stations! Level Three Emergency! Unnecessary mass into the tubes! Repeat! Level Three Jettison Emergency! Unnecessary mass into the tubes!"

Cut to scenes around the ship of surviving crew members, rummaging through every room, grabbing everything not nailed down — wreckage, equipment, random belongings — and stuffing it all into chutes specifically designed for this purpose.

"Sulu, what's our time factor?"

"We need 5000 ΔV in the next 30 minutes or we're dead."

Flick. "Engineering?"

"Impulse engines ready, Captain. You'll have 195 seconds of burn time."

"Thank you, Scotty. We'll make it count. Spock, how's our mass situation?"

"Down 23% We need to lose another 5. Another 10 minutes, 42 seconds if we can keep the jettison rate up."

"That's cutting it too close." Flick. "All hands! Level TWO Jettison Emergency! Repeat! Level TWO!" Cut to more scenes around the ship. Now they're gathering the dead bodies and stuffing them into the tubes. Cut to exterior view of the ship with expanding cloud of debris and bodies and crap.

… and so on.

Really. Rockets just suck.

Up next, Part 7: Space Battles


(This is Part 5. The previous installments are Part 1: Another Anniversary, Part 2: Climbing the Wall, Part 3: Rockets Are Stupid, Part 4: L1 Rendezvous )

Sisyphus Revisited

So, before I get on with confessing my sins and explain what I've been lying about, I'm realizing I need to say a bit more about instabilities and why that horribly weird spiral trajectory — which I'm going to guess would have been especially horrifying for those Mercury and Gemini astronauts who all got their start as experimental aircraft test pilots; just mention "spiral dive" to any pilots you know and see what they have to say about it (hint: it's not something one usually lives to tell about) — and about why said trajectory is something to be embraced rather than feared.

Imagine a ball bearing rolling back and forth in a parabolic valley.

This scenario, the Simple Harmonic Oscillator, shows up all over the place, pendulum clock, weight on spring, child on a swing. It's like 90% of physics is about making as many situations as possible look like simple harmonic oscillators. Not suprising since this one of the easiest things to solve; we have this hammer; if we can make everything look like a nail, so much the better.

Qualitatively, there's only one solution:

x ∝ cos(ωt)

The ball just goes back and forth and back and forth and back and forth. Forever and ever; the epitome of stability. Once you know the frequency ω, you've got the whole story.

(I use the wonky "is proportional to" (∝) symbol so as not to have to be writing out lots of pointless constants. If, in what follows, you also want to imagine (t-t0) wherever you see t, feel free. So really what we're saying is x=x0cos(ω(t-t0)), the most general solution. Or you can just ignore the math altogether.)

And now we sneak in and make a little, teentsy sign change. What harm could it do?

The next morning, the security guards wake up and find that the parabolic valley has been turned upside-down/inside-out/whatever and we now have a parabolic hilltop (... along with Spock getting a goatee, Federation → Empire, Cat → Dog, etc...).

Whereas before, we had one simple solution, there are now nine (9) and they're all different qualitatively.

Instead of running away screaming, I'm going to make a table listing them all, so that you can get a sense of what we're up against. You may find it easiest to read them in a clockwise or counter-clockwise order.

x ∝ −e−ωt
The ball has always been rolling in from the left; someday it may reach the top but we won't live to see it.
x ∝ +sinh(ωt)
The ball approaches from the left,
makes it over the top,
and rolls down to the right
x ∝ +e+ωt
The ball has always been rolling away to the right; if it was ever at the top, long ago, nobody remembers that far back.
x ∝ −cosh(ωt)
The ball approaches from the left,
fails to get to the top,
rolls back to the left
x = 0
The ball was, is, and ever shall be perfectly balanced at the top of the hill. Amen.
x ∝ +cosh(ωt)
The ball approaches from the right,
fails to get to the top,
rolls back to the right
x ∝ −e+ωt
The ball has always been rolling away to the left; if it was ever at the top, long ago, nobody remembers that far back.
x ∝ −sinh(ωt)
The ball approaches from the right,
makes it over the top,
and rolls down to the left
x ∝ +e−ωt
The ball has always been rolling in from the right; someday it may reach the top but we won't live to see it.

Tell me where you are and how fast you're going, and I'll tell you which box you're in.

The colorings are energy levels. All of the gray boxes have the same energy. Meaning if, when you're at some particular place and you have just the right velocity, you'll be in one of the gray boxes. If you're going faster, then you're in one of the blue boxes; if you're going slower, you're in one of the red boxes.

The gray boxes are essentially boundaries, drawing the fine line between success (blue boxes) and failure to cross over (red boxes). And you can skirt as close to them as you dare, so if, say, you're in one of the blue boxes, by reducing your energy you can make your trajectory be arbitrarily close to the trajectory in the gray box on either side. The closer you get, the more time it takes, so if you're going over the hill, you can arrange to take exactly as much time as you want by picking the right velocity/energy level.

In the center is chaos. An infinitesimal change to an x=0 scenario has eight possible outcomes. Rounding errors will ruin your day if you're not sufficiently clever.

Now, as I mentioned earlier, L1 is actually a saddle point. That is, assuming we orient our axes the right way, it's a parabolic hilltop in the x direction but in both of the y and z directions it's a parabolic valley. Parabolic valleys are places where we can park arbitrary amounts of energy while we're passing through (well okay, there are limits). In other words, if we're going through L1 from the Earth to the Moon or vice versa, we can make the transit take however long we want by changing the size of the spiral.

And since the various frequencies/periods stay roughly the same if we don't go too far out, that means we can spiral around as many times as we want while going through. But also, since we can mess with the y and z directions independently, that gives us even more choices re what direction we're going once we're out of the neighborhood of L1.

What we need is to build a map. Essentially, you can think of there being a (4-dimensional but never mind that) sphere of possible ways to park energy in the y and z directions. Imagine that sphere as being the center box in the table above (i.e., what the orbits would be if we weren't moving at all in the x direction).

Then you have the upper-right and lower left "Rolling Away" boxes which are now (5-dimensional) tubes leading from the sphere to elsewhere, and then the pair of "Rolling in" boxes, which are tubes from elsewhere back to the sphere. These bound the set of possible useful transit trajectories (the blue stuff) that take us from the elsewheres on the left (in the big hole where Earth is) to the elsewheres on the right (the moon and everything outside), which are what we want out of this.

At which point our agenda is simple (hahahahaha): Solve for where the sphere is. Figure out where the tubes go. Once we know where the tubes go, we know what our choices are.

The Actual Lay of the Land

Something is indeed rotten in Denmark and the core of it is that we're not actually doing the happy two-body problem that Newton solved, where angular momentum is conserved, everything has to move on conic-section-shaped trajectories, and ellipses are forever. Counting on our fingers, we see that Earth is one, Moon is two, and spacecraft makes three (3) bodies there, at which point there are no closed-form solutions, Newton gave up, Lagrange figured out a few things and then gave up.

There are lots of weird nooks and crannies, and we're in the process of stumbling onto one of them. Indeed the very fact that we can even have saddle points where chaotic things are happening is all part of why there can't ever be a closed form solution.

But now we need the Big Picture.

L1 is at minus 170km but "the top" is not at 0km. Why? Because, this whole time, I've actually been using a rotating reference frame where the earth and moon are fixed. Which means, among other things, there's centrifugal force to contend with, which gets stronger the farther out you go.

Meaning that 6000 km deep hole where the Earth is is not in the middle of a plane, but rather at the center of this parabola-shaped hill (well okay, parabola-of-revolution-shaped hill), which turns out to be a volcano with a 6000-km deep crater at the top of it. The circular rim of the crater is where the moon's orbit is; everything slopes downwards in all directions outside of it.

Things are further messed up because the moon is sufficiently big to put the earth off-center. That is, since earth and moon actually revolve around their common center of mass, the earth is displaced somewhat (4600km) in the direction opposite to the moon. Which then tilts that aforementioned circular crater rim; rather than being a constant −160½km altitude, the point opposite the moon on the rim (the L3 Lagrange point) is a bit lower (−161km) because it's closer to the earth.

Now if L3 is the low point on the rim, you might be thinking the place opposite it, where the moon is, should be the high point, but

  1. the moon is there, and
  2. the moon has its own gravity, which we have to add back (450km deep hole, remember?)
So, no.

Where is the high point? Follow the rim 120° from L3 in the direction of the moon's orbit and you get to L5 and O'Neill's space colony. If you'd gone 120° the other way you'd have gotten to L4 instead. L5 and L4 are both at −160km and are the real (twin) hilltops. They are as high as you can go and there is no place that's 0km after all.

Things are actually further messed up because of the Coriolis force, which I haven't told you about, which happens to be crucial for understanding why L4 and L5 are stable even though they are hilltops which should otherwise be completely disastrous from a stability point of view. Fortunately, for L1 and L2, the Coriolis force only messes with the frequencies and tilts the various axes a bit; it doesn't change the overall qualitative picture, so I can skip that part.

(Nor was I never clear on why O'Neill preferred L5 to L4. Everything you can do with L5, trajectoriwise, you can do with L4; it's all symmetric, see. I'm also now wondering if the hilltop genuinely is the best place to be; it's actually the hardest place to get to in the Earth-moon system. There are so many tasks you need to do to maintain a space colony and keep everybody alive; station-keeping was never even remotely the biggest problem. Stability also means it's harder to leave, which will suck if you ever want to move the colony somewhere else. Though I suppose being at hilltop may reduce the probability that random rocks will arrive from infinity and ruin your day. That, to me, would be a much better selling point than stability — if it's actually true; haven't done the math on that one yet...)

So to get out from L1, instead of having to climb 170km as you might have originally thought, it's looking like, depending which direction we go, we only have to climb 10km at the most.

But it gets better.

The presence of the moon actually cuts a huge notch in the crater rim. If we continue our hike along the rim from L3 past the peak at L5 we'll find ourselves headed decisively downwards. Then the rim wall splits, going around either side of the big hole where the moon is. Directly across the moon from where they split, the walls rejoin on the far side and the rim continues around up to L4. L1 is the saddle point on the inner wall; L2 (you knew there had to be an L2) is the saddle point on the outer wall. And that is the last of the flat spots; Lagrange proved that there could only be five and this is where he left things 200 years ago.

L2, at −169km, is a measly one (1) kilometer higher than L1. As long as you have at least 140 m/s (313 mph) of velocity when you get to L1, you'll have enough energy to get to L2. And everything I've said about L1 (i.e., that it's a saddle point, that there are tubes, etc...) is true of L2 as well.

So if you're stationary at L1, you just need to put on 140 m/s of ΔV. But it's actually easier than that. The moon is right there. L1 has two outgoing tubes, one headed back towards Earth, the other outward. L2 likewise has two incoming tubes. See where the outward bound L1 tube intersects L2's from-inwards tube. Find the pair of intersecting orbits that comes closest to the moon. That is where you want to do your burn and chances are it'll be a lot smaller than 140 m/s (because the deeper you are, the faster you're moving and the faster you're moving, the less ΔV you need to achieve a particular energy change, i.e., to gain that last kilometer).

Once you are at L2, you are definitively outside the crater.

At which point we switch to the Earth-Sun rotating frame, where there is an entirely different set of Lagrange points. As it happens, the Earth-Sun L1 and L2 points are each about 1.5 million kilometers from Earth, your being at the Earth-Moon L2 point means you're now moving in a 444,000 km radius circle around the earth — exactly where depending on the time of the month — at something like 1200 m/s which is 300 m/s faster than what you'd ordinarily need to stay in circular orbit around Earth at that distance.

Which means, once you get sufficiently beyond L2 and away from the moon's influence, you're being flung away. Depending on how you timed things — and you can hang out in the halo orbits as long as you need to in order to time things just right — you can arrange to get flung away in any direction you want; and you'll be left with enough energy to both get away from the moon and get up another 63km worth of wall (this "wall" now being the wall around the solar crater whose rim is where the Earth's orbit is). Which is good because the Earth-Sun L1 and L2 points are both only about 50km higher from where you are now.

Or you can view everything from a completely non-rotating frame and see that the Moon just gave you a big gravitational assist. And when you get to Earth-Sun L2, the Earth is going to give you one, too, if you've played your cards right.

Except that, once you know where the tubes are, it's no longer a matter of chance. That is, you know where the outgoing tube from Earth-moon L2 is and where the incoming tube for Earth-sun L2 is and thus where they intersect. You then have a bunch of trajectories you can use.

And from Earth-sun L1 or L2, we can similarly go all sorts of other places, Sun-Mars L1, Sun-Mars L2, Sun-Jupiter L1, Jupiter-Ganymede L2, and on, and on. All pairs of co-orbiting bodies in the solar system, sun-planet, planet-moon, etc. each have their own L1 and L2 points guarding the entrances to their respective craters. Since everything is time-reversible in classical mechanics, you have trajectories going both ways, i.e., for every weird spiral trajectory that sends you away from L1 or L2 off to wherever, there's a corresponding one that brings you back in. You can string these trajectories together playing mix and match with them, giving you a way to visit any planet or moon that you like — admittedly, these low-fuel trajectories tend to be really slow, but if you're an unmanned satellite, you don't care.

This is the essence of the Interplanetary Transportation Network.

(which might seem to be a conclusion, but I actually have more to say about rockets in Part 6)

Random update:
I, of course, forgot to mention the really cool part of Farquhar's thesis, which was the proposal that Collins be put in a halo orbit at L2 behind the Moon — which, as noted above, is a measly 1km worth of additional energy/effort beyond my have-him-orbit-L1 plan.

It then so happens you can make the radius big enough so as to remain visible from Earth at all times.

And then we do Far Side landings with no gaps in communication.

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