wrog

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).

3 verticies: bottom left, bottom right, top rightforming a right trianglebottom (horizontal) side is labeledvₘ = v, vₛ = 0, vₛ−vₘ = −vhypoteneuse (sloped ~35°) is labeledvₘ = 0, vₛ = v, vₛ−vₘ = +v(vertical) right side is a blue upwards arrowAdd a slanted downwards arrow from the top vertexto a bit past the middle of the bottom sideAdd hints of possible lines through bottom-left vertex ranging from horizontal to hypoteneuse slope.Emphasize two of them:(*) one intersects upwards/vertical arrow half way up, labeled vₛ = v/2(*) one intersects slanted arrow half way up, labeled vₘ = v/2;Third line, in-between, colored red, labeled vₛ−vₘ = 0

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