wrog | Entries tagged with math

My explanation of Special Relativity, trying to keep it almost completely geometric, continued from part 2 wherein I describe

what the Moving People spacetime grid looks like when mapped out by the Stationary People,
how all we have left to do is figure out their unit spacing, and
how there can exist Intermediate People who see the Moving and the Stationary People moving in opposite directions at the same speed.

We now take a moment to introduce, out of left field, a new definition that will turn out to be very useful:

Give a pair of events, separated by a distance $Δz$ and elapsed time $Δt$ , the interval between them is defined as the quantity ${Δz}^{2} - {Δt}^{2}$

which looks like a squared distance except for an annoying minus sign,
which seems to depend on whose coordinates we're using (Stationary People vs. Moving People vs. somebody else), but we will now show that:

For any given pair of distinct events, the interval between them is an invariant

… by which we mean every observer that we care about who calculates this in their own coordinates will get the same number.

We'll start by showing that Moving and Anti-Moving (née Stationary) People have to agree on this.

Recall where these numbers ( $Δz, Δt$ ) come from. We have one event pick out a history line while the other picks out a snapshot, these have to intersect at some common point/event, counting units along the history line gives you the elapsed time while counting units along the snapshot gives you the spacing. And you'd still get the same numbers if you switch which event picks out the history vs. the snapshot (since you'd just be going around the other side of the rectangle/parallelogram).

So the Moving People get ( ${Δz}_{m}$ , ${Δt}_{m}$ ) and the Anti-Moving People get ( ${Δz}_{s}$ , ${Δt}_{s}$ ).

Normally, with two arbitrary Peoples moving at arbitrary velocities, we'd be stuck at this point, but we're actually doing two sneaky things:

We are using the point of view of the Intermediate People, using their spacetime chart, where the velocities are equal and opposite. The angles that the Anti-Moving People's history and snapshot lines make vs. a downward light ray are the same as for Moving People's lines vs. an upward light ray.
We have the Moving People follow history from one event while the Anti-Moving people follow history from the other event.

Which gives us two extra right angles that you, perhaps, weren't expecting to see.

Which means we have two right triangles sharing the same hypotenuse (dotted red line between the two common points).

Which means we have two different ways to invoke the Pythagorean Theorem to get that length,

Which means we have two different sums of squares that have to be equal.

The one remaining potential problem is that what we're squaring are segment lengths on a page that we'd then have to divide by the (as yet unknown) time/distance unit spacings to get the corresponding Δz and Δt values.

But, fortunately, in this viewpoint, since the Moving and Anti-Moving People are carrying identical clocks at the same speed in opposite directions, all of these unit spacings have to be the same. So we can cancel them out, leaving us with

{Δz}_{s}^{2} + {Δt}_{m}^{2} = {Δz}_{m}^{2} + {Δt}_{s}^{2}

Subtracting ${Δt}_{s}^{2} + {Δt}_{m}^{2}$ from both sides gets what we want

{Δz}_{s}^{2} - {Δt}_{s}^{2} = {Δz}_{m}^{2} - {Δt}_{m}^{2}

This being purely a statement about Moving and Stationary People measurements, we do not have to care what the Intermediate People measurements actually are. For this to work, it suffices that the Intermediate People merely exist.

And for that. we only need the Moving vs. Stationary People relative velocity to be STL.

And, since there's nothing special about this particular choice of Moving People, that means everyone going STL according to The Stationary People will agree with The Stationary People on the interval value for this pair of events.

Or any given pair of events since there was nothing special about this particular pair.

One easy consequence of this:

Since agreeing on the value of ${Δz}^{2} - {Δt}^{2}$ includes agreeing on its sign, that means agreeing on whether $| Δz |$ is less than, equal to, or greater than $| Δt |$ and hence whether the trajectory between those events is STL, lightspeed, or FTL.

Therefore, everyone going STL according to The Stationary People must see each other's trajectories as STL and agree that all other trajectories are not. Which, among other things, means it doesn't matter which of them was originally picked to be the Stationary People; we still unambiguously get the same group of People Who Are Going STL with respect to each other, which is then the group of observers we care about.

Why FTL Sucks, part 2

If you want to postulate FTL People existing and imagine that one of them might accidentally get chosen as the initial Stationary People, then it will suffice to make a small addition to our list of requirements (i.e., to be in the set of observers I care about) to immediately disqualify such folks: "Must be able to measure the length of your own arms using light rays and reflectors," or maybe, "Must be receiving light from all directions." (Yes, there need to be distant stars everywhere for this argument to work; this is not a problem in our universe.)

And, yes, having all STL people see each other as moving STL is very weird.

For example, if you have people going opposite directions, each at 2/3 lightspeed, their relative velocity still has to be STL and therefore cannot be 4/3 lightspeed. It's actually 12/13 lightspeed, if you must know. Once again:

Velocities do not combine the way you'd think they do

In fact, it turns out, the only time they do add/subtract the way you'd think is either when one of the velocities is zero or when they're equal and opposite (and thus sum to zero).

About proper time

The proper time is the time lapse between two events that are occuring in the same place. There's a sense in which this kind of time lapse is more "real" than other kinds of Δt, since in this case you or whichever of your friends happens to be there is sitting right there directly experiencing, up close and personal, both events and All of The Physics happening between them.

But this concept also makes sense for any two events that have a STL (i.e., negative) interval between them, because that means we can have somebody right there, coasting at that velocity, who then sees both events occurring in the same place.

Same place means Δz is zero for that observer, so the proper time squared is minus the interval and so this is a number all observers will agree on. Note that this includes agreement on the sign since everybody agrees on which time direction is the future and hence which event is happening first. (If you have trouble with this, imagine everybody has both a clock and a steam kettle with escaping steam.)

This (finally) gives us the Moving People time/distance unit spacing, i.e., if you find two events on a Moving People trajectory whose interval is 1 (calculated in your own or anybody else's coordinates), then you have two events that are 1 unit apart in proper time and therefore 1 unit of elapsed time according to the Moving People (because they're seeing them happening in the same place).

Which means that when a Moving Person advances forward in time by 1 unit according to us Stationary People, they're also moving a positive distance $v < 1$ , which then makes that interval $- (1 - v^{2})$ , meaning the proper time is less than 1, and therefore the point at which they will have experienced a full unit of time is farther into the future. And the closer they are to lightspeed (1), the farther it gets. This is time dilation at work: For anything that we see as moving, its clocks ticks more slowly than ours (according to us).

Except, since there's nothing special about us, it must also be that, according to them, it's our clocks that are ticking more slowly. Which may seem like a contradiction. But it's not, because we're not agreeing on what's simultaneous and we're counting time in different directions.

Probably the easiest way to see this is to look at it from the Intermediate point of view.

We meet. We each tick off a unit of time. But once we've done that, we're not in the same place anymore, and each of our snapshots is tilted back into the other person's past. So this actually does work:

For extra fun, if someone were to hitch a ride upwards with Moving Person at the meeting point, then jump off right at the moment where the Stationary People think 1 unit of time has elapsed, with sufficient velocity downwards so as return to the meeting point (i.e., where the Stationary People consider the meeting point to be) promptly at the stroke of 2 (units of Stationary People proper time), then we can see that:

Both of the traveler's upwards and downwards journeys have the same interval, i.e., on both legs, a particular distance (length of the upwards blue arrow) gets traversed in 1 unit of time, at least according to the Stationary People, but, then, everybody else's calculation of the interval has to match what the Stationary People get.
The traveler is experiencing (much) less than 1 unit of proper time on each leg.
The traveler will be (much) less than 2 units older when they return to their starting point.

This is the resolution of the famous "Twin Paradox", in which one member of a pair of identical twins journeys and the other stays behind, and it's only a paradox if you think it's somehow against the rules for identical twins to age differently, but by now it ought to be clear that Time is not this Absolute Thing anymore.

If you need actual numbers: The figure is to scale if both the Moving and Stationary People have a relative velocity of 12/13 lightspeed and the Intermediate People are seeing them each go at 2/3 lightspeed in opposite directions.

For the sake of not writing more fractions than we have to, let's have the distance/time unit be 13 years. Then, the jumping off point is, for the Moving People, 5 years after the initial meeting, but, for the Stationary People, it'll be 13 years after and 12 (light-)years away from (above) the initial meeting (13²−12²=5², so it's still 5 years proper time when the Stationary people calculate it). The traveler's return journey is 12/13 lightspeed downwards according to the Stationary People and 62/63 lightspeed downwards according to the Intermediate people (so even though the return jouney looks a whole lot longer on the Intermediate People chart, there's a whole lot more time dilation).

Upon returning (bottom event), the traveler will be 10 years older, as opposed to the Stationary People being 26 years (2 units) older.

About proper distance

Two distinct events that are simultaneous have a proper distance between them. One can debate exactly how "real" this is, but, being the positive square root of the (positive) interval (i.e., |Δz| for whoever sees Δt as zero), it is, at least, a number that everyone must agree on.

But this concept also makes sense for any two events that have a FTL (i.e., positive) interval between them, because we can invert that FTL velocity to get something STL that is how fast someone needs to be coasting in order to see those events being simultaneous.

Yes this is different from proper time in that we make no attempt to get agreement on signs, hence why we take the trouble to say positive square root and say $| Δz | instead of Δz . That's because, even though our 1-dimensional universe does have a universally agreed "up" and so we$ could define a signed proper distance if we wanted, we would then lose in higher dimensions. That is, adding back the x and y directions makes Δz into a vector and then one can do purely spatial rotations, making nonsense of any notion of spatial "sign".

Why FTL Sucks, part 3

Any FTL pair of events will be seen as simultaneous by somebody. Also people going slightly slower or slightly faster than the velocity you need to see them as simultaneous will disagree about which event comes "first", a distinction which won't matter if there's no way to communicate or send physical objects between the events,…

…but if you can, then we have a Problem. Whatever technobabble method we the Stationary People can use to get stuff from one event to another across an FTL interval should be doable in the same way by any of the STL Moving People to similarly get stuff moved across any other FTL interval that has the same proper distance including the backards version of that first interval (i.e. swapping origin with destination), or any number of other choices that connect the destination to somewhere/when before its origin and voilà! instant, easy time loop.

Or to put it another way, the moment you can take the stargate / ansible / warp drive / thing that can make jump points in arbitrary empty space / whatever and put it on a ship going at any STL velocity in any direction and have it still work the way it did in the Stationary Place — which it should because there's a whole lot of Physics that depends on this — that's all you need to wreck causality.

And yes, nothing says the universe can't be acausal (it's mainly humans that like causality), it still means anyone who's developing FTL travel/communication (or putting it in a story) is inevitably going to need a plan for dealing with an acausal / Bill&Ted universe.

Why FTL Sucks, Stargate SG-1/Atlantis Edition

The stargates make instantaneous connections across hundreds of light-years, routinely. So, e.g., the Stationary people can have a pair of stargates separated by 99.999999 (light-)years, one dials the other, you can go out, then close the connection, immediately dial back the other way, and return with virtually no time having passed at the starting place.

You could, if you want, imagine that the far-end events are all offset in time by some amount, but then that would mean for one of the directions you'd have to be going backwards in time, which is not supposed to be possible without the Stupid Solar Flare Plot Device games they play when they actually want to do time travel. So we have to assume the opening and closing connection events at both ends are synchronized/simultaneous (not that this actually saves us).

Now take the same pair of gates, reduce the separation to 99.99999700000002 years (the difference is roughly Earth-Jupiter, so no one will care) and have them set in motion going exactly 0.02/100.000001≈0.0002 lightspeed or roughly 60 km/s in a particular direction that we will call "up".

Meaning we now have Moving People in charge of the gates.

Their snapshots will be slanted futurewards (slope = 100.000001/0.02 = 5000.00005) if you follow them upwards, or pastwards if you follow them downwards.
If you follow a snapshot up, it'll take 100.000001 years (Stationary People distance) to reach the other gate's trajectory (it's moving away, remember), and you're also going forward in (Stationary People) time by 0.02 years ≈ 175 hours
(Upper gate's position (z) being 99.99999700000002 + 0.02×5000.00005 = 100.000001, so this works out. I could have just made you solve for where the snapshot intersects the upper gate's trajectory, but I figured I'd just give it to you. You're welcome.).
The interval is then (100.000001)²−(0.02)²=(99.999999)², so the proper distance, i.e., the distance as measured by the Moving People between the two (stationary in their view) gates, is (surprise!) 99.999999 years.

Which means those gates can operate exactly as before, even though they're moving, because the gates themselves can't tell that they're moving. So, if you go in one end, you come out the other end 99.999999 (light-)years away at exactly the same (Moving People) time, but according to the Stationary People time, you're coming out 100.000001 years up and forwards in time by 175 hours.

Or those same distances/times backwards when you do the return/downward transit.

Now arrange for a second pair of Anti-Moving gates, i.e., with all of the same parameters but moving downwards, timing/placing them so that someone coming out of an upwards journey through the Anti-Moving gates can immediately dive into the upper Moving gate and arrive back where they started 350 hours (a little under 15 days) in the past.

Instant, cheap time machine. Look Ma! No solar flares! And way, way less work than that McKay-Carter Intergalactic Gate Bridge they built to get from Milky Way to Pegasus that spanned 3 million light years and used up hundreds of gates.

Are we done yet?

Next up: The Storrow Drive Theorem or How We Put Back the Other Dimensions

My revamped explanation of Special Relativity, trying to keep it completely geometric and not invoking Walls of Math, continued from here where I have now completely beaten to death the concept of "Stationary" and it's time to

Meet the Moving People

Actually, we're going to be particular about who we associate with. We want Moving People who will likewise be able to say that they don't think of themselves as moving, meaning they can't be accelerating or spinning, either. Which leaves having them coast at some constant velocity that is non-zero (because otherwise they'd be stationary and One of Us) and upwards (just to pick a direction).

We also want them moving slower than lightspeed (STL), for Reasons.

Why FTL Sucks, part 1

At this point we are not saying that faster-than-lightspeed (FTL) People cannot exist. It's just that we have no idea how they'll be measuring distances. They outrun all light signals, so the whole business of setting up a screen to reflect off of will never work; no reflections will ever reach them.

Not that they can't have some New Method for measuring distance, but whatever it is, it'll be Different, they won't be able to not know it's different, and, beyond that, there's little we can say about it, so I'm not going to try. So let's just focus on the folks we know do exist.

Also, if, later on (spoiler alert), it turns out we don't actually need FTL people for anything, so much the better.

The Stationary People are Not Special

If the Moving Person and their friends do everything we did as above to establish their own coordinates for everything that happens, they should get a consistent world view out of it, even if it turns out to be different from ours in some unexpected way.

In other words, feel free to imagine them building an entire Moving John Hancock Tower, complete with its own (moving) floors, someone on every floor who the original Moving Person likewise thinks of as stationary, everybody with their own clocks, set in such a away that they think they're all synchronized.

Which means they've got their own spacetime grid, complete with its own floor histories and snapshots. And we just have to plot what that has to look like on our own (stationary) grid.

The "horizontal" grid lines are easy: They're just the trajectories of the various people on the moving floors.

Constant velocity trajectories are linear, so we can say that at least of the original Moving Person.
Moving clocks repeat some physical process which should keep working the same way everywhen along a trajectory, so their tick events have to be evenly spaced.
Having repeated distance measurements turn out the same — taking a particular shape made out of events and being able to repeat it arbitrarily later along a trajectory — means all of the trajectories are linear, parallel, and have the same tick spacings.

Note that slope (= rise over run) and velocity (= distance over time) are the same thing in this representation (one reason to prefer time going left-to-right). STL means they're less steep than the (45°) light rays. And parallel means all Moving People velocities are the same.

Or, to summarize,

The Moving People's trajectories / floor histories are all lines with the same slope $v$ (velocity)

Their "vertical" snapshot grid lines are where shit gets weird.

Watch what happens when Moving People try to measure/verify distances. Someone aims their projector, either upwards or downwards or both (two projectors, yay), light ray (yellow) heads out, bounces off of a screen, Upper Bounce and/or Lower Bounce, depending, then comes back, and now they, too, have a clock image they can compare with what's sitting in their lap.

The three events Sent, Upper (or Lower) Bounce, Received must form three corners of a rectangle since the light rays are all 45°.

The Moving People don't think they're moving, so they have no reason to think of the light as traveling different distances going out and back. Which means the time the Moving People infer for the bounce events must be whatever their clock is saying halfway between Sent and Received, the event at the center of the rectangle.

The yellow rectangle, being a rectangle, has axes of symmetry, light rays coming out of its center that I have helpfully drawn in magenta. That is, reflecting everything through one of them (doesn't matter which) maps the rectangle onto itself while swapping the diagonals, therefore each diagonal makes the same angle with that light ray; it just swings the other way.

Or, equivalently, the slopes of the diagonals must be inverses. Recall slope = rise over run, and mirroring everything through 45° swaps rises with runs, so if one diagonal is sloped $v$ , the other must be sloped $1 / v$ .

No matter how big or small the rectangle is, if it's centered on the Halfway Tick, then that other diagonal has to coincide with the $1 / v$ -sloped (blue) line through Halfway Tick, so that must be where all events deemed simultaneous with Halfway Tick end up. Same goes for all other clock ticks and all other Moving People trajectories. Therefore,

The Moving People's snapshots are all lines with slope $1 / v$

In particular, these lines cannot be vertical — as Newton and everyone else had assumed — unless the velocity is zero. In other words, the Stationary People and Moving People will disagree on which pairs of events are simultaneous.

You can also see that compressing the (left-right) time scale by a factor of a hundred million in order to make ordinary, everyday velocities like one meter per second visibly non-horizontal will also take a snapshot line that was already off from vertical by only one part in a hundred million and reduces that to one part in ten quadrillion; it's no wonder Newton et al thought it was vertical.

In other news:

The Moving People's distance units in their snapshots are spaced the same as the time units in their histories

This follows from we the Stationary People seeing both diagonals of the rectangle having the same length while the Moving People infer the same number of (distance/time) units along each diagonal (because they also take speed of light to be 1).

Notice that if we had made Newton's assumption that their snapshots are vertical like ours, then their clock ticks would have to be spaced farther apart to line up with ours because their trajectories are sloped. So we won't take anything for granted here. Their unit spacing is whatever it is; we'll work it out later.

Thus, now we know their grid (plotted onto the our Stationary People spacetime grid), up to some scaling factor, looks like this:

(a bunch of rhombuses) which by itself is enough to prove all manner of fun facts. For example,

The Moving People and The Stationary People agree on what their relative velocity is.

(When you see what goes "wrong" later, you'll be glad we checked this. Also if you're thinking this is obvious, try coming up with your own argument before looking at mine.)

Start at some rendezvous between a Stationary and a Moving Person. The Stationary People pick some amount of time to wait ( ${Δt}_{s}$ ), doesn't matter how long, then see how far the Moving Person has gone ( ${Δz}_{s}$ ) during that time, then do the division ${Δz}_{s} / {Δt}_{s}$ . Easy.

The Moving People do likewise, but using their own grid lines. If they pick the right ${Δt}_{m}$ , then we get this diagram:

We then notice that the two gray triangles have the same angles, hence are similar, hence lengths of corresponding sides have the same ratio.

The triangle on the right is smaller than the upper triangle by some factor v, and we get that same number. whether we divide the two short sides, the two medium sides, or the two hypotenoi^*.

^*Yeah, I know this is not the actual plural of 'hypotenuse'; in a just universe it would be.

And yes, for the Moving People, the segment lengths aren't actually ${Δt}_{m}$ and ${Δz}_{m}$ because of the as-yet-unknown (distance/time) unit spacings, but, fortunately, since the spacings are the same, they cancel out here, and so the Moving People get the same answer doing ${Δz}_{m} / {Δt}_{m}$ .

Now that you know how to do that, you can probably figure out how to show that

Moving people also see light rays going at velocity 1

(exercise!)

Let's try something slightly more complicated:

There is an intermediate trajectory that looks the same to both the Moving and Stationary People

…by which I mean there can be Intermediate People moving in the same direction as the Moving People, but enough slower so that the Moving People see them moving downwards at the same speed that we the Stationary People see them moving upwards.

You'd think this would be obvious. Except, as it turns out, …

The intermediate velocity is not actually half of the Moving People velocity.

We can, from the previous diagram, consider the range of possible intermediate trajectories, starting with our own (Stationary, horizontal) one, and then gradually increasing the slope until it matches the Moving People's velocity (v).

Stationary People will care about where the trajectory crosses the (vertical) blue arrow to get their Δz and then divide, thus getting a velocity $v_{s}$ , ranging from 0 to v.

Moving People do likewise except, because their notion of what's simultaneous is different, they will be concerned with where the trajectory intersects the slanty downward blue arrow. The velocity they see, $v_{m}$ , will correspondingly range from $v$ to 0.

Subtracting $v_{s} - v_{m}$ gives us something that goes from $- v$ to $+ v$ and crosses through zero exactly once somewhere. That crossing is the trajectory we want.

Also, the trajectories where either of us is seeing a velocity of $v / 2$ (crossing one of the blue arrows exactly half way up the triangle) are clearly not the same, thus neither is the one we want, and that the actual Intermediate velocity generally needs to be larger, giving us our first warning that

Velocities do not combine the way you'd think they do

Presumably, we'd already know this because slopes do not add. (Hint: 45° is slope 1; combining two 45° angles gives you 90° which is infinite slope, whereas slope 2 is around 63.435°).

Yes, this isn't entirely constructive since we haven't told you what the intermediate velocity is. It's complicated — well okay, it's not that hard, but you need to solve an equation and it's quadratic (exercise!). Fortunately, for what follows, all we need is that the trajectory exists and the velocity is no faster than what the Moving People were originally doing and is therefore STL.

We then invoke our previous result — that every pair of observers has to agree on their mutual relative velocity — to get:

The Intermediate People see the Moving and Stationary People going the same speed in opposite directions

So we now have a viewpoint where The Moving People are slower and The Stationary People become the The Anti-Moving People. We can now take advantage of symmetry to show …

… something completely and totally cool that I will now defer to the next installment

(yeah, ok, I don't actually know Latin)

So, here's another go at explaining Special Relativity. I remain annoyed at how few people really get it, even amongst avid SF consumers, entirely too accustomed to generations of SF writers papering over FTL issues with technobabble, 'cause we need that galactic empire, don'tcha know.

Also, it's long past time to dump poorly motivated 1930s pedagogy that real physicists abandoned long ago (e.g.,. "Your mass increases as you go faster", "What? why?", "Fuck you, it just does" [spoiler alert: just No; forget you ever heard that. And if, in 2026, anyone is still trying to teach it that way, someone needs to sit them down for A Talk]).

All we need is basic geometry you knew or could have learned about in 6th grade plus a bit of algebra (up to Pythagorean Theorem). I think I can get by without using a single square-root sign.

But there will be fewer handwaves this time. Because we have to be clear why things have to Not Be The Way You Expected and not leave wiggle room. Here goes:

The speed of light as a constant

Shine a flashlight off of a moving boat. How fast does the light go?

It could be like throwing a baseball, where speed of the boat matters.
Or it could be like the waves that spread out when you jostle the boat or have it move, where replacing the boat with a possibly-moving helicopter tossing out stationary gravel makes zero difference—the wave fronts have a velocity with respect to the water and only the current/lack-thereof in the river/lake matters.

This is part of why they spent so much time from the 1600s onward arguing over whether light was a "particle" or a "wave" and they couldn't make headway because light goes so fast its speed was impossible to measure…

…until the 19th century, when the tech got good enough and the theory advanced to being able to say what light is (thanks, Maxwell), and we (sort of) get an answer: "It's like the waves in the water, but (surprise) there is no water". What we have is a number for "speed of light in a total vacuum", derived completely from constants, leaving no way to have it depend on any boat or water-current velocity, never mind there's nothing there that can have any kind of current in it.

Or, rather, the moment you try to imagine there being something there, you should then be able to infer the speed of any current in it by looking at the wave patterns you get from sources moving in various directions, even if this "water" is otherwise totally undetectable.

Except, they tried this. Result: The speed of the "current" is always zero (thanks, Michaelson and Morley).

Which then gets extra-special weird if you now arrange for there to be two boats in the same place, one stationary and one moving in some direction and somehow the magic "water" is moving in sync with both boats at the same time. How does this even work?

Answer: It can't. This is the point where Isaac Newton's head explodes.

The only way out is if one or both of the boats is having their distance and time measurements fucked with. But to get a handle on that, we need to go back to first principles on what it means to measure distance and time.

There's also the question of what it means for a boat to be "moving" when you don't have water anymore. Actually, let's start there:

I am Stationary

Let's make our lives as simple as possible by being somewhere out in interstellar space, far away from any distinguishing stars, planets, asteroids, or other landmarks of any kind, and where there's no gravity to speak of. Turn the engines off so that there's nothing accelerating us. Do shit with gyroscopes to make sure we are not spinning — gets rid of all of the weird (centrifugal, etc.) forces that would cause. (I vaguely recall Einstein spending multiple pages on what an "inertial reference frame" is, but this is all you get from me)

What we're left with is Newton's First Law: No forces on us means we are either "at rest" or moving at constant velocity in some direction. Except we can't tell the difference if all we have to look at is a bunch of distant stars that are all moving randomly. There would have to be Somebody Else local looking at us to say whether/how we're moving with respect to them. And even given that, we could still say, "No, we are stationary and you are the one who is moving."

If we can't tell the difference, perhaps it shouldn't matter. So given all of this (i.e., that I feel no forces, therefore none of the scenarios where I would know that I'm moving are in effect), I'll just declare myself Stationary and go from there.

Secondly, if I only trust measurements made by Other People Who Are Also Stationary, then I will not be relying on assumptions about how moving around affects measurements; i.e., if something is dicking with those numbers, I don't have to know in advance how that works. So, for Other People, it suffices to be able to tell whether they are also stationary; if not, then we just ignore them for now.

Measuring Time

Time is easy. Use your clock.

What is a clock? Some physical process — a totally reliable ISO Standard Gerbil In A Cage, or a spinning cesium-137 nucleus in a particular mode, we don't care, just pick something — takes a known amount of time to do a thing, making a "tick". Add some standardized mechanism to count ticks that everyone can build. As long as basic physical laws are the same everywhere (a not unreasonable assumption that may not actually be true but has sure worked really, really, really well so far), this should work. So I can now time everything happening where I am.

Measuring Distance

The trick here is to get as much mileage (pun intended) as we can out of the speed of light being this totally universal constant, which we now assume because that's what the experiments seem to be telling us..

If it's a constant, then it's like a conversion factor:

299,792,458 meters is one second;

9,460,536,207,068,016 meters is one (Gregorian) year.

Most authors would trouble to say "light-second" or "light-year" here, but (I think) that's actually less helpful. Light goes 1 second per second, 1 year per year, or 1 meter per meter, or Just 1, Period (or "Full Stop", as my Brit friends would say). If it's already clear from context what we're measuring, the "light-" prefix is just unnecessary extra noise.

(I don't want to go so far as to say time and distance are the same thing; they're really not, even if it is natural to use the same units. Just like we can naturally measure distances on the surface of a sphere in degrees or radians, but we wouldn't want to say that distance and angle are the same thing; that would be confusing.)

Seems like we should now be able to use clocks and light to measure distance:

I have (1) a clock in my lap, (2) a video projector that can project an image of that clock onto a screen some distance away from me, and (3) sufficiently good eyesight to read what's on the screen

Light needs time to get there and back, so the screen image will be delayed. If the clock image I see is, say, exactly 2 microseconds behind the clock in my lap, then the light must have traveled a total distance of 2 microseconds (≈ 2000 feet), one microsecond going out and one microsecond coming back.
Even if the screen is moving, we can still know where it was at the exact moment that the light-ray bounced off of it: at exactly one microsecond (distance) away from me at exactly one microsecond ago.

Oh look, I just measured a distance. Go me.

What "Stationary" means

If I can now look at the screen over some period of time and see that distance does not change, i.e., that clock on the screen steadfastly remains 2 microseconds behind the clock in my lap, then I can know for certain that the screen cannot be moving. At least not in the radial (towards/away-from me) direction.

But since I'm not spinning, then transverse motions are also covered, i.e., if, I'm not having to be move my eyes or turn my head to track it, it can't be moving in those directions either, and we're done. I now have somebody located Elsewhere who I can trust.

My new friend can now put her own clock — identical construction, operating principles, and time units — next to her screen where I can see it, and I, in turn, can set up a screen next to me and return the favor.

She has to see the same delay — light rays repeatedly covering the same distance back and forth, take the same amount of time, — so we have to agree on how far apart we are.
If one of our clocks were to tick faster than the other, we'd have to agree on whose it is. Except that if we each have identical equipment and there's nothing else around us to distinguish where we are (e.g., if I were next to a black hole and she weren't, that would definitely be A Problem, so let's suppose there's nothing like that happening), meaning if flipping the entire universe 180° around the midpoint between us does nothing important except exchanging our identities, then how can swapping our names change the clock speeds? So this can't happen. Yay, symmetry. Which then means…
The difference between our clock times being fixed, we can now synchronize our clocks. I suppose at this point we could have a political problem where we can't come to an agreement, but there'd still be a fixed difference, and so I will know what to subtract from her time readings to translate them to the time standard I want to use, so we may as well stipulate/assume her clock is synchronized with mine and be done with it.

Now repeat this for all of my other stationary friends. Now imagine an entire network of stationary friends located in all of the places I care about. I now know All Distances, to everyone and also between everyone because we talk to each other (over radios or whatever). This gives us a fishnet of spatial measurements, enough to derive some kind of coordinate system from (or I could take a WW2 Tokyo approach and use people's names to identify locations; whatever works).

And all of those in-synch clocks can time everything happening elsewhere on my network.

Finally, one last (temporary) simplification:

1-Dimensional Universe

Hi, we all live in the John Hancock Tower, where you never have to leave the building, where going anywhere means riding the elevator or going up and down stairs if that's more to your liking (since we're out in interstellar space with no gravity, stairs shouldn't be quite so onerous). Hey, it's got condos, offices, restaurants, movie theatres, everything you could ever want, right? (Yes, once upon a time, it really was considered Cool and Futuristic to construct buildings around the idea that people would want to live like this.)

One dimensional — we can add back the other dimensions later — means the floor number (z) is the only position coordinate we need. The aforementioned fishnet is now just an up-and-down line of stationary people, one on every floor, each of whom has a clock synched with mine.

Which then allows us to fit everything that happens on a 2-dimensional page, a spacetime grid with space going up-and-down and time scrolling left-to-right (yes this is flipped from how I've have previously done this and how most physics books do this; cope). Every horizontal line is a history of all of the events that happen on a particular floor of The Tower. Every vertical line is a snapshot of the universe/Tower at a particular time.

Some set of events will happen. I will record what happens on (or near) my floor with the time it occurs. All of my friends will do likewise. Everything goes onto the grid. And, at the End of Time, we get a picture of What Happened as far as We The Stationary People are concerned.

In the figure, we can see what a typical distance measurement looks like. I point my projector up (or down), send out a light ray at 'Sent', the light bounces off of a screen at 'Upper (Lower) Bounce' at some particular distance (n) above (below) me, and comes back to me at 'Received'. Both bounces and 'Halfway Tick' are all in the same snapshot since they're all taking place at the same time. Easy.

What Nice Units We Have

One consequence of using the same units for distance and time — whether it's feet (= nanoseconds), microseconds (= kilofeet), seconds, etc. — is that all light rays, whether traveling upwards or downwards, will be going at 45° (traversing one distance unit per time unit) on our grid.

Why care about this? Because so much becomes more obvious when you scale time vs. distance the right way.

E.g., imagine surveying your backyard measuring north-south distances in meters and east-west distances in Astronomical Units (1AU=150 million km), i.e., compressing the east-west scale by a factor of 150 billion for basically no reason. That would be stupid, right?

But that's essentially what we were doing before 1905 and why it took 200+ years to figure out Newton had screwed something up.

And now it's time to…

Meet the Moving People

(to be continued)

(why I should not be allowed to write textbooks, part 342)

This is mainly because I wanted to have all of the formulas in one place. Also curious to see how much I can compress the derivation and still have it vaguely make sense. Also I wanted to learn more MathML.

The 2 body problem

We have two bodies with mass. Gravity attracts them to each other. How do they move?

The first order of business is to arrange our own seating so that their center of mass is stationary from our point of view. Once we do that, what we see is

a big mass $M$ at a small distance $r$ away from the center, moving at small velocity $v$ and
a small mass $m$ at a big distance $R$ away moving at big velocity $V$ ;

M r = m R

;

M v = m V

; and they're always opposite each other, mirroring each other's motions. Thank you, conservation of momentum.

We also notice that, at any given moment, you can always draw a line through the origin and both bodies. The velocities will usually not be parallel to the line, but there's always a plane that contains the line and one of the velocities, and since the other velocity is always the exact opposite direction from the first, it, too will be in the plane. And then there's nothing that's ever trying to pull either of the bodies out of that plane, we realize that plane doesn't move, they stay in it forever and so we can just rotate our coordinates so that it's the x-y plane, forget about z, and pretend we're in a 2-dimensional universe from now on.

And now for The Trick.

Whether you're following Hamilton's or Langrange's way of doing things (skipping how that works for now), the energies, as functions of positions and velocities, give you everything you need to know to solve for how things are going to move. So let's just write them down. Total kinetic energy will (because I hate dividing by 2) be half of

m V^{2} + M v^{2} = (m {(\frac{M}{M + m})}^{2} + M {(\frac{m}{M + m})}^{2}) {(V + v)}^{2} = \frac{M m}{M + m} {(V + v)}^{2}

while the potential energy, which will be all due to gravity here, is

- (\frac{G M m = G (M + m) (\frac{M m}{M + m})}{R + r})

What's interesting/weird here is how we have managed to recast everything in terms of just the relative distance $R + r$ and relative velocity $V + v$ and that these are the same formulas we'd get if we were trying to figure out what happens with a single object of "reduced" mass $μ = M m / (M + m)$ bouncing around a gravity well generated by something at the origin with mass $M + m$ (that has been magically nailed in place so that it doesn't move). Any solution we get for the latter problem will readily translate back.

One possibly-slight-surprise here is that, despite my talk of "small" and "big", we are, at no point ever actually depending on, say, $m ≪ M$ , or even $m < M$ . m could even be bigger and this will all still work; you just have to be sure to remember to do that last step (i.e., translate back to a 2-body problem and not be surprised when it turns out the "big" object is moving around, too).

So, scratch one body. Yay.

Also, we will not actually be mentioning the reduced mass μ ever again because from now on, we can just do everything in terms of energy/whatever per unit mass (of the one object).

Alsoalso, we can introduce a new constant $k ≔ G (M + m)$ so as not to have to look at that ever again, either. As it happens, in real life, k tends to be way more accurately known than any of G, M, or m, individually, so good riddance.

One body inverse-square force problem

Now that we're down to one body, in a $- k / r$ gravity well generated by (SHUT UP), what happens?

If the body's position in polar coordinates is $⟨ r, θ ⟩$ , then we have radial and angular velocities being $\overset{•}{r}$ and $r \overset{•}{θ}$ , respectively, where the • is time derivative, i.e., $\overset{•}{r} = d r / d t$ , and so on. (Yes, we're doing the Dot Thing.)

All forces come out of the center, so angular momentum per unit mass

L ≔ r^{2} \overset{•}{θ}

will be constant. Which, first of all, means $θ$ will be always increasing or always decreasing as time goes on (good to know!).

Also, if you can imagine lots of really thin triangles with base $r dθ$ and height $r$ (and therefore area $= ½ r^{2} dθ$ ), we see that $L$ is is twice the area being swept out per unit time, and therefore if we eventually get an orbit that's periodic with period $T$ (spoilers!), then the total area enclosed will be $½ L T$ (save for future reference).

For any $L$ , there will be a corresponding circular orbit radius $r_{0} = L^{2} / k$ , which I am introducing now solely for my own convenience and may or may not have anything to do with where the object is actually spending its time.

The second constant of motion will be the energy per unit mass, $E$ , expressed thusly (again multiplying by 2, just because I can):

2 E = {\overset{•}{r}}^{2} + (r^{2} {\overset{•}{θ}}^{2} = \frac{(L^{2} = k r_{0})}{r^{2}}) - \frac{2 k}{r}

Now if what we want is to get the shape of the orbit/trajectory, then we don't care quite so much about $\overset{•}{r}$ or $\overset{•}{θ}$ as we do about their ratio $dr / dθ$ , which inspires the following truly awesome and sneaky maneuver:

\overset{•}{r} = \frac{dr}{dt} = (\frac{dθ}{dt} = \overset{•}{θ} = \frac{L}{r^{2}}) \frac{dr}{dθ} = L \frac{d}{dθ} (\frac{−1}{r})

which we can square and substitute into the previous equation. Then, dividing both sides by $L^{2} = k r_{0}$ we get

\frac{2 E}{k r_{0}} = {(\frac{d}{dθ} (\frac{−1}{r}))}^{2} + \frac{1}{r^{2}} - \frac{2}{r r_{0}}

where we now see two terms on the right as a square crying out to be completed, so we add $1 / {r_{0}}^{2}$ to both sides, while at the same time noticing that we can also add arbitrary constant crap to the thing we're differentiating in the middle, hence

\frac{1}{r_{0}^{2}} (ε^{2} ≔ 1 + \frac{2 E r_{0}}{k}) = {(\frac{d}{dθ} (\frac{1}{r_{0}} - \frac{1}{r}))}^{2} + {(\frac{1}{r_{0}} - \frac{1}{r})}^{2}

where on the left we have also consolidated yet more constant crap into a new constant, the eccentricity (but introducing it as $ε^{2}$ because the stuff on the right can't be negative).

And now we stop to ask the class, "So, does anyone know of some $f (θ)$ that does $f^{2} + {(df / dθ)}^{2} = {f_{0}}^{2}$ (a constant)?"

Inevitably, some nerd in the front who did the assigned reading or maybe they really are able to figure out on the fly that you can rearrange this into $\pm df / \sqrt{{f_{0}}^{2} - f^{2}} = dθ$ and then of course you want $f = f_{0} sin-or-cos (something)$ which both the df and the square root expression change to cos-or-sin()s which cancel leaving you with $d (something) = \pm dθ$ and then the something has to be ±θ plus a constant will be all, "Oo, pick me!" and then we get our answer (this is actually how all differential equations get solved, in case you didn't know):

\frac{1}{r_{0}} - \frac{1}{r} = - \frac{ε}{r_{0}} \cos (θ - θ_{0})

where the particular choice to have it be −cos() rather than +sin() makes no difference — because of that extra $θ_{0}$ which covers all of the bases — but happens to be easy to rearrange to get the version more people are familiar with:

r_{0} = r (1 + ε \cos θ)

And yes, I quietly rotated the coordinate system to make the $θ_{0}$ go away again because I was pretty sure I would get tired of looking at it. Good catch.

The first thing to notice is that if $ε = 0$ then this is a circular orbit and now, hopefully, the mysterious name I gave for $r_{0}$ should start making sense.

The second thing to notice is that if $ε > 1$ there's a range of angles where the stuff in parentheses can go negative, i.e., everything around 180° between the two magic angles where $\cos θ = −1 / ε$ is actually forbidden. Which would then mean we have something coming in from infinity at one of the magic angles, swooping around to do a close flyby at $θ = 0°$ , and then fucking off back to infinity at the other magic angle … what you might call a hyperbolic trajectory.

If instead $ε < 1$ , then θ goes all of the way around without any problems, … must, in fact, do so and keep doing so (thank you, angular momentum), which then means we are indeed stuck in something periodic, with r varying between a closest approach $r_{0} / (1 + ε)$ , at $θ = 0°$ , to a maximum distance $r_{0} / (1 - ε)$ , at $θ = 180°$ . Adding these and dividing by 2 gives us the semi-major axis

a ≔ r_{0} / (1 - ε^{2}) = - k / 2 E

the distance to the actual geometric center of this thing, which usually will be some distance away ( $a ε$ ) from where all of the gravity is coming from. (Also, don't freak about the minus sign; $ε < 1$ means $E$ has to be negative.)

If you're having trouble figuring out what this shape is, it may be more recognizable in Cartesian coordinates. However, life will be easier if we also shift the origin, so we set $⟨ r \cos θ, r \sin θ ⟩ = ⟨ x - a ε, y ⟩$ , at which point the orbit becomes

(r_{0} = a (1 - ε^{2})) = r + ε (x - a ε)

or, canceling the $a ε^{2}$ s, getting $r$ by itself on one side, and squaring everything:

{(a - ε x)}^{2} = r^{2} = {(x - a ε)}^{2} + y^{2}

or, eventually

1 = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{(a^{2} (1 - ε^{2}) = a r_{0})}

which, as long as $ε < 1$ , is taking a unit circle and stretching it, — in the $x$ direction by $a$ and in the $y$ direction by $b ≔ a \sqrt{1 - ε^{2}} = \sqrt{a r_{0}}$ — which is what an ellipse is.

Since the unit circle has area $π$ , our ellipse has to have area $π a b = ½ L T$ , per that thing we saved for future reference earlier, which, after we square it and cancel an $r_{0}$ from both sides, rearranges into

{(\frac{2 π}{T})}^{2} a^{3} = k

which is that square-cube law that's awfully useful for getting orbital periods from orbit sizes and vice versa.

If $ε > 1$ , then $a$ is negative, which is weird but not a showstopper, and we end up with the $y^{2}$ term being negative, which means we have a hyperbola instead of an ellipse (surprise!). And then the stuff after that (defining $b$ as the square root of a negative number, etc) fails and you don't get a period out of this (again, surprise! it's not periodic anymore).

We will leave the $ε = 1$ case — where the energy $E$ is zero and the object is always travelling at exactly the escape velocity, no more, no less — as an exercise.

Well okay, I guess there is a way to get Trilorne to hover.

If we're willing to toss the usual definition of "North" (rotation axis points that way), we can put the sun over the equator, and then we can have it be tide-locked. Meaning this planet is basically Mercury but farther away so that only the stuff directly underneath the sun is getting fried to shit.

And maybe having an actual atmosphere will help, too, in various ways, though I can't imagine there not being freaky weather patterns, e.g., some kind of permanent cyclone storm around the solar-maximum point wherever it happens to be at the moment, but that won't damage the story too much because nobody ever goes to the Fire Lands anyway.

The Wall then runs around ( surprise )

Crossposts: https://wrog.livejournal.com/70683.html

So there's this Arthur C. Clarke short story, "The Wall of Darkness" (1949). I read it as a kid and found it really haunting. Clarke does Haunting really well.

If you haven't read it already and want to go read it before I completely and totally ruin it, feel free.

(I found the whole thing by searching for "Trilorne" in google, which then gave me a google books hit; we'll see how much longer The Algorithm lets people do that).

But you've already had 70 years, so… onward…

( let the ruining begin... )

Crossposts: https://wrog.livejournal.com/70528.html

Continued from Part 3, what happens when there is no "parallel", the rules for circles aren't what you thought they were, and so on.

Napier's Rules

So how does trigonometry work in this world?

See, I belatedly realized that spewing walls of equations like this is not actually going to be much use when you're stuck in a rowboat in the middle of the North Atlantic having to navigate by the stars with no cell phone and no GPS. Because, chances are, you also have No Internet, and then my blog entries with their handy tables go to waste.

It would be far better if I can teach you how to derive these relationships instead, i.e., in a way that you might actually be able to vaguely remember while sitting in a boat in the middle of the North Atlantic.

But first I'm going to introduce a bit of gratuitous extra notation. Write

ᶜᵒθ

— pronounce it "co-theta" if you want — to mean (90° − θ). I do this because:

I can,
it's less typing,
it's way less degree vs. radian waffling, which I already do too much of,
but also,
you get all of the following useful and amusing equivalences (no, really; read them aloud):

sin ᶜᵒθ = cos θ
cos ᶜᵒθ = sin θ
tan ᶜᵒθ = cot θ	(= 1/tan θ, in case you've forgotten)
cot ᶜᵒθ = tan θ
csc ᶜᵒθ = sec θ	(= 1/cos θ, and no, I don't know why reciprocals get these special names)
sec ᶜᵒθ = csc θ	(= 1/sin θ, because, seriously, WTFF?)

You'd almost think they planned it this way.

Hopefully, it goes without saying that ᶜᵒ(ᶜᵒθ) = θ, except I had to go and say it, didn't I? (Damn.)

And now let's start with a right triangle, with vertices/angles and sides/lengths labeled a,b,c,A,B, the way you usually see it in trigonometry class, ( and then derive stuff about it )

Crossposts: https://wrog.livejournal.com/70126.html

Continued from Part 2, exploring the benighted universe where "parallel" is Not a Thing.

How circles work

So, to review the weird things we've seen so far:

When we have a circle radius of 90°, otherwise known as a straight line, and we're traversing the circumference, i.e., measuring the total length along it as we sweep out 360° from the pole in the middle, we get 360° worth of path (phrasing it this way so that if this turns out we're on a projective plane rather than a sphere and what we're really doing is traversing the same 180° path twice, I won't have been lying to you), which, being 4 times the radius, is slightly less than one might have expected (2π being roughly 6.28).
If we attempt a circle of radius of 180°, we stay firmly nailed to the antipode of the center, our circumference traversal goes nowhere and thus we get a circumference of zero.

Meaning if we have to explain to the residents what "π" is, we're going to lose horribly. Best we can do is, "So: Circumference to radius? That's a ratio. It's literally all over the map. But as radius approaches zero, once you're under 90°, you'll notice the ratio is always getting bigger. If you work at it, you can prove that it's bounded and it converges to this weird transcendental number like e. And, no, don't ask us how we came up with this…"

We need to understand better how curved paths work. ( and so ye shall... )

Crossposts: https://wrog.livejournal.com/69769.html

Continued from Part 1, in which we discover at least one consequence to doing away with the concept of "parallel lines".

Let's talk about Area

Having noticed that isoceles right triangles give us a natural way to define/measure distances, we see that we can do area this way as well. That is, the area of ΔAPX is clearly the angle at P times some constant, which we may as well just take to be 1 if we haven't defined a unit of area yet. ( so let's do that ... )

Crossposts: https://wrog.livejournal.com/69460.html

So, as part of my possibly-continuing "Geometry on Drugs" series, here is a prequel to my post on spherical geometry, which was more of a "hey, this is useful" post in which much there's a whole lot you're expected to take on faith. It was really more intended for the hardcore engineering type who needs to see that use case up front.

This version is going back to first principles, where we do the axiom wanking and you (hopefully) get a sense of why things turn out the way they do.

Also, this is the practice run before I launch into the Essence of Hyperbolic Geometry, so, … Onward …

The Geometry Axiom Everybody Hates

Start with this diagram and the inevitable question that comes up:

Start with a line ℓ and a point A not on it. How do you put a line through A that doesn't intersect ℓ?

(In other news, I am now convinced that the Unicode committee contained at least one disgruntled geometry teacher. How else to explain why there's this isolated script ℓ code point?)

We can drop a perpendicular from A meeting ℓ at some point X, and then it's obvious that the line you want (dotted) is the one perpendicular to XA. If you tilt it even slightly away from 90°, then it simply must intersect ℓ somewhere.

Proof by diagram. We're allowed to do that, right? ( Read more... )

Crossposts: https://wrog.livejournal.com/68946.html

So here's a puzzler

Brief review of the physics of wind instruments

Your standard wind instrument is a pipe attached to some kind of sound source (lip or reed). Being of a certain length it resonates at particular frequencies and suppresses the others, ulimately being capable of producing tones that are multiples of a certain fundamental frequency f, 2f being an octave higher than f, 3f being a fifth above that and so on.

The main things to keep in mind about the multiples of f (the "harmonic series"):

going up by a particular interval means multiplying the frequency (or dividing the wavelength) by a particular ratio (2 ↔ octave, 3/2 ↔ perfect fifth, 5/4 ↔ major third, etc.…), and
intervalwise, the notes are getting closer together the higher you go.

Eventually, the notes get so close to each other that you can do scales of a sort -- this is how the natural (i.e., valveless) French horn works; the tube is so damned long that you're always playing in the 8f-16f range and then you just fudge the few notes that are out of tune with lip and hand-in-the-bell tricks. ... and likewise for why baroque trumpet parts are so insanely high-pitched.

But if you want to do anything useful down in the lower register, you have no choice but to mess with the length of the tube somehow, the two most popular methods being

(the woodwind solution) poke holes in the tube (and cover them with removable keys)
(the brass solution) insert valves or slides in the tube to change its actual length

We'll focus on the second solution since I really have no freaking clue what's going on with woodwinds. Oddly enough, the brass solution is simpler but it didn't come along until fairly late in the game. Not until the 19th century did metalworking technology finally get good enough that Heinrich Stölzel was able to construct the first actual valved instrument in 1814. Once this happened it didn't take long to catch on -- hence the 19th century explosion in the use of brass instruments in orchestral works (cf. Wagner, Mahler...).

So how do valves actually get used?

Let's first get clear what a valve is: ( Yes, we are going deep into the weeds. You knew this was going to happen )

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.

( bring on the drugs... )

Crossposts: https://wrog.livejournal.com/68636.html

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?

Discuss.

So, I remember this demo.

It was something that, back in grade school, the engineer father of one of my best friends really liked to put on -- did it on several occasions that I remember. It was one of these Physics is Cool demos.

Start with a room about 40-50 feet long, almost exactly like this one, in fact if we can trust Google's scale marker, that's pretty much what it was (yes, the room takes up the whole length of the building; yes, the Internet is really scary these days...). I remember the ceiling as being really high, but since I was smaller back then, that recollection is suspect. I also distinctly recall someone accidentally sticking the end of a flagpole into the ceiling (we had our Boy Scout troop meetings in that room), so it couldn't have been that high. From the photo, 12 feet seems likely.

There's a stepladder right at the back wall of the room.

Somewhere towards the middle of the room, there's a hook in the ceiling.

A piece of twine is tied to the hook, long enough to almost reach the floor; at the other end of the twine, a Rather Heavy Object of some sort, ... shotput or bowling ball or somesuch, I forget.

Up at the front of the room is a chair with a 1-2 foot diameter basket on it.

He pulls two volunteers out of the audience. Volunteer #1 is somebody's little sister. He positions her near the middle of the room, between the basket and the hook, and hands her this big-ass butcher's knife. "This is extremely sharp. I want you to hold this right here, just like that." Blade is pointed towards the back of the room, angled downward.

Volunteer #2 takes the bowling ball up the stepladder. "Yeah, that's about high enough. Just let go of it."

Ball swings forward, right down to the floor and up again, reaches the knife, the twine parts pretty much instantaneously, the ball sails through the air, and lands right in the middle of the basket.

And there was much rejoicing.

What's a bit frustrating now is that, since I was in 6th or 7th grade at the time, this was all before I'd had any Actual Physics. Meaning "Physics is Cool" was pretty much all of the content that I got out of this. It was mentioned that the ball needed to be heavy and the knife needed to be sharp, but beyond that I had no idea which other details of the setup were crucial and which didn't matter at all. Nor was there really much of an explanation of why it worked, what principle was being demonstrated -- or maybe there was one, and I just didn't have the background to appreciate it and so it all just went kind of went whoosh.

Actually, now that I think about it, there may indeed have done some kind of brief "inertial foo gravitational mass mumble mumble", but I'm sure he knew up front that when you've got something that works mainly because it just drops out of the math and you've got an audience that doesn't know a whole lot of math, you're just going to lose them if you try to explain too much, so you just skip the boring part and get on with the show.

To be sure, it's fairly cool to be able to set up a scenario, calculate out in advance where things are going to go, then pull the trigger and have them Actually Go There. Which perhaps describes pretty much every classroom physics demo ever. And this may indeed be enough of a point for 7th & 8th graders (i.e., "Study your math, kids. Keys to the universe!" or maybe just, "It works, bitches").

But still a bit frustrating, because, years later I learned some Actual Physics, and a few questions remain. Not that I'm particularly surprised that this worked, but,... what's the gimmick? Some particular sweet spot where to put the basket so that it doesn't matter quite so much where you put the knife or where you let the ball drop from? Or maybe it's a sweet spot in the knife placement or the original drop point? Was the scenario chosen the one that was easiest to calculate? Or the one that had the basket the furthest forward? Or the one had the longest flight time for the ball?

I suppose I could just call John up and ask, "Hey, that thing your dad did with the knife and the swinging shot put; what was the deal with that, anyway?" But this would be cheating. So I'm going to analyze this a bit. . .

( bet you didn't see *this* coming... )

So I've arbitrarily decided that more people need to know about spherical trigonometry. (e.g., just in case the GPS gets destroyed and we're stuck having to do our own navigation again.)

It's really the same solving of triangles that you learned to do in high school geometry/trig, i.e., gimme side-angle-side or angle-side-angle to nail down what the triangle actually is, then use Law of Sines or Law of Cosines or some combination thereof to work out the previously unknown sides/angles that you care about.

It's just that some of the rules for spherical trig are a Little Bit Different.

So, jumping right into the deep end, here's an application inspired by recent events:

You, an observer stationed on planet Earth at some particular latitude, want to know where the sun is going to be in the sky at some particular time of day, some particular day of the year.

And, cutting to the chase, here's a triangle to solve:

If you know b, A, c (side, angle, side), you can solve for a (opposite side) using the Law of Cosines

cos a = cos b cos c + sin b sin c cos A

and then B (next angle) using the Law of Sines

sinB/sinb = sinA/sina = sinC/sinc

( ...math wanking continues... )

You may not have been aware of this, but III.F.1 of the WSDCC's 2008 Delegate Selection Plan reads as follows:

Alternates shall be listed and seated in the order in which they were elected,
and must be of the same presidential preference, and, to the extent possible,
of the same gender and from the same electing jurisdiction as the delegate being
replaced.

This rule applies at all levels of caucus starting with the LD caucus, where the (bottom-level) precinct delegates go to elect the (2nd-level) delegates to the CD caucus and state convention.

Now, the interesting thing about this is that, unlike at the higher levels, where male and female delegates/alternates are strictly segregated (i.e., you get so many delegates of each gender for a given jurisdiction and there are are likewise separate alternate lists for each gender), the delegation from the precinct caucus to the LD caucus is co-ed; and for each precinct for each candidate there's likewise just a single ordered list of alternates where the M-F pattern can be anything.

( ...they'll probably just cut me up and put me in a recycle bin under the Washington State Convention Center )

Current Mood: accomplished

So I guess that was too easy.

( if I turn into The Time Cube Guy, please shoot me... )

You have one (1) ordinary six-sided die (i.e., the traditionally marked, uniform-density, cubical kind that you could find in any Vegas casino within about 5 minutes).

You need to make a random choice between four (4) alternatives, same probability for each (25%, i.e., barring the occasional coin-landing-on-edge weirdness that supposedly can happen in real life, which we won't worry about).

You could, e.g., roll the die and if you don't get 1,2,3,or 4, just roll again and keep rolling until you do. That would be too easy, and also could conceivably take a while if you get unlucky with the 5s and 6s.

But the real problem is that, due to various contrivances entirely beyond your control, it just so happens that if you roll the die a second time, you will be plagued by frogs, and (trust me on this) you really do not want to be plagued by frogs.

So you absolutely have to do it in one (1) roll. Good luck.

( hint and potential spoilage... )

Tune her piano to Kirnberger III ( ** )

Of course, this being the 2nd time in my life that I've attempted to tune a piano, I imagine getting it to sound even remotely correct will be something of a triumph. (Last time around, people were saying "Gee, you ought to get that tuned" after I was done). On the other hand, last time around was 20 years ago and back then I didn't have clue #1 about temperament issues ( * ).

Suffice it to say, having a sense of absolute pitch --- and yes, I know people call it "perfect" pitch, but it just isn't, okay? --- is almost completely useless for this sort of job, just in case you were wondering.

On the other hand, given the state that her piano is in now (hasn't been tuned for maybe 20 years), almost anything will be an improvement, and since she's mostly tone-deaf anyway, I could probably tune it completely randomly and she wouldn't notice a thing (though

emmacrew and I almost certainly would...).

Also, I have this theory that Kirnberger III, with all of its perfect fifths, will be a lot easier to do than Equal Temperament. We'll see.

( more math/music wanking . . . )

Current Music: The Well-Tempered Clavier (what else?)

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wrog

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Space Exploration
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January 2026

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Relativity (3 of n): The Interval
Relativity (2 of n): Meet the Moving People
Relativity redux (1 of n): E pluribus immobiles
Kepler Orbit Cheat Sheet
Getting Trilorne to Hover
Arthur C. Clarke and the Projective Plane
Spherical Geometry 4 — Mommy, Where Does Trigonometry Come From?
Spherical Geometry 3 -- A bit of Circular Reasoning
Spherical Geometry 2 -- Deficits Matter
The Essence of Spherical Geometry, Part 1
towards a truly chromatic Horn
Public Service Math Lesson: Hyperbolic Trigonometry
Today's game theory intro
physics is cool
public service math lesson: spherical trigonometry
Something I probably won't present to the state Rules Committee in 2012
more on S4 (really A5, but who cares)
Today's Puzzle a.k.a. Stupid S4 Tricks
Mean tricks to play on your mother-in-law #27

Entries tagged with math

For any given pair of distinct events, the interval between them is an invariant

Why FTL Sucks, part 2

Velocities do not combine the way you'd think they do

About proper time

About proper distance

Why FTL Sucks, part 3

Why FTL Sucks, Stargate SG-1/Atlantis Edition

Meet the Moving People

Why FTL Sucks, part 1

The Stationary People are Not Special

The Moving People's trajectories / floor histories are all lines with the same slope v (velocity)

The Moving People's snapshots are all lines with slope 1/v

The Moving People's distance units in their snapshots are spaced the same as the time units in their histories

The Moving People and The Stationary People agree on what their relative velocity is.

Moving people also see light rays going at velocity 1

There is an intermediate trajectory that looks the same to both the Moving and Stationary People

The intermediate velocity is not actually half of the Moving People velocity.

Velocities do not combine the way you'd think they do

The Intermediate People see the Moving and Stationary People going the same speed in opposite directions

The speed of light as a constant

I am Stationary

Measuring Time

Measuring Distance

What "Stationary" means

1-Dimensional Universe

What Nice Units We Have

Meet the Moving People

(why I should not be allowed to write textbooks, part 342)

The 2 body problem

One body inverse-square force problem

Napier's Rules

How circles work

Let's talk about Area

The Geometry Axiom Everybody Hates

Brief review of the physics of wind instruments

So how do valves actually get used?

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

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The Moving People's trajectories / floor histories are all lines with the same slope $v$ (velocity)

The Moving People's snapshots are all lines with slope $1 / v$