Your Spacetime Simulator

Warm-up

Three quick ones. Each is a part your simulator runs on today.

The puzzle

Fifteen chapters ago you wrote one small function: the hypotenuse of a triangle. Since then you’ve written a stretch factor, an interval, a Lorentz machine, an odometer for wristwatch time, a Doppler factor, a rule for adding speeds and a recipe for energy. Each one is a law of Nature in your own hand, checked against Nature’s numbers.

Today we bolt them together into a small universe: ships, flashes of light, firecrackers, and clocks that age the way real clocks do. Build your own universe. Then we take it to court. Six famous paradoxes are waiting, each a place where common sense says relativity contradicts itself. Your simulator is the witness.

“What I cannot create, I do not understand.”

Written on Richard Feynman’s blackboard at the time of his death, 1988

Experiment

Below is your simulator, loaded with the trip from Chapter 11. Lin stays home. Rae flies 3 light-years out at \(v = 3/5\) and comes back. Each ship is a worldline with a tick for every year on its own wristwatch. The dashed line is the now-line: all the events that the frame you’re viewing from calls “now”. The clocks read each ship’s wristwatch time where its worldline crosses that line.

Other kinds of line turn up as you explore. Each time a new one appears, the key under the diagram says what it is.

• Drag the time slider, or press Space to play.
• Click a ship’s clock (or its worldline) to jump into its frame. The diagram redraws itself in that ship’s coordinates, and the now-line becomes the ship’s now.
• In “What they see”, pick whose eyes to use. It starts with Lin’s. The strip shows where each ship appears to be, which is not where it is.

Three things to try, each about a minute. Read one, do it in the simulator just below, then come back for the next.

Then make it yours. The scene list holds the six court cases and four challenges. “Edit the scene” adds ships, turning points, events and flashes. Export your universe as a file and send it to a friend; import theirs. A scene file is plain data, numbers and names, and never code.

How your simulator thinks

Nothing in the simulator is new physics. Every number it shows comes from one of ten small functions, each built in an earlier chapter. The simulator holds two versions of each. One is ours: the reference library. The other is yours, if you wrote it and it passed its tests. Your code runs in a sandbox, a sealed corner of the page where it can compute numbers but can’t touch anything else.

The table below is the parts list. Each part is in one of three states:

• Yours. You wrote it and it passes its tests, so the simulator runs your code. A Use reference button appears beside it in the table. Press it and the simulator runs ours instead. If yours passes, nothing on screen moves: that is what passing means.
• Failing. You wrote it, but it doesn’t pass yet. The simulator uses ours until it does.
• Not written yet. The simulator uses ours. The button appears once you build the part in its chapter (linked in the table) and it passes. Until then, the swap test under the table shows the same thing with a sample solution.

One more part, trip(d, v), isn’t in the simulator. You build it at the end of this chapter, and it runs the trip planner there.

That is what passing the tests meant. Ours and the sample part company only around the twelfth or thirteenth decimal place (the Difference line shows it), where a computer rounds its arithmetic.

To see the parts cooperate, let’s do by hand what the simulator did when you jumped into Rae’s frame. The question is the one that worried Sagredo: what does Lin’s clock read on Rae’s now-line, just before and just after the turn?

1. The turn is the event T = (t 5, x 3). Rae’s outbound frame moves at \(v = 0.6\), so \(\g = 1.25\). In that frame, T happens at \(\rkt{t'} = 1.25\,(5 - 0.6 \times 3) = 1.25 \times 3.2 = 4\). This is your lorentz(). The outbound frame’s clocks were set to 0 at O, where Rae’s watch also read 0, and she rides with that lattice. So its clock beside her agrees with her watch: 4.
2. Lin sits at \(x = 0\), so any event on her worldline, (t, 0), has \(\rkt{t'} = 1.25\,t\). Put \(x = 0\) into the same formula.
3. “Now”, for Rae, means the same \(t'\) as T. So \(1.25\,t = 4\), and \(t = 3.2\): Lin’s clock reads 3.2. Lin is at rest in the lab, so her wristwatch reads lab time. wristwatchTime() of her straight path from (0, 0) to (3.2, 0) is 3.2.
4. After the turn, Rae’s frame moves at \(v = -0.6\). Now T has \(\rkt{t'} = 1.25\,(5 + 0.6 \times 3) = 8.5\), and the event on Lin’s worldline with \(t' = 8.5\) has \(t = 6.8\). Same machine, new frame: only the sign of \(v\) changed. The inbound lattice was also set to \(t' = 0\) at O, so its clock beside T reads 8.5, even though Rae’s watch there reads 4. A frame’s time is its lattice’s, not the traveler’s; only the slope of the now-line matters for the question.
5. So Rae’s “now at Lin” leaps from 3.2 to 6.8 as she turns. Lin’s clock never leapt. Rae changed which of Lin’s events she calls “now”. The 3.6 years between were always there, ticked off one by one on Lin’s worldline. Rae’s outbound now-lines sweep up to 3.2, her inbound ones start from 6.8.

The shorthand for steps 1 to 3 is the now-line of a frame moving at \(v\), written in lab coordinates:

\[ \lab{t} = v\,\lab{x} + \frac{\rkt{t'}}{\g}. \]

That is your simultaneityLine(v, tPrime) from Chapter 7: slope \(v\), intercept \(t'/\g\). The simulator draws that line, slides it up as time runs, and finds where each worldline crosses it. Each clock then reads wristwatchTime() of its path, cut off at the crossing. In code, the clock bar is two of your functions with a little glue between them:

// yours (Ch 7): which events are "now"
const line = simultaneityLine(v, tNow);
// a helper of ours: where this ship's worldline meets that line
const where = crossing(path, line);
// yours (Ch 11): the wristwatch time along the path, cut there
const clock = wristwatchTime(pathUpTo(where));

{ "version": 1, "units": "years",
  "ships":   [{ "id": "rae", "name": "Rae", "kind": "rkt",
                "path": [{ "t": 0, "x": 0 }, { "t": 5, "x": 3 }, { "t": 10, "x": 0 }] }],
  "flashes": [{ "t": 2, "x": 0, "dir": 1 }],
  "events":  [{ "id": "T", "t": 5, "x": 3, "label": "T" }] }

Now you know what the witness knows. Time to call it.

Practice

Welcome to the Paradox Court. Each case is a famous argument that relativity contradicts itself. For each one, load the scene, look at it from both frames (click a clock to jump), measure what you need, and give your verdict. Then write the judgment in your own words and compare it with the court’s.

The court has one rule, the rule of the whole course: frames may disagree about coordinates. An argument that needs them to agree is the argument that’s wrong.

Case 1 · The pole in the barn

The charge. A 10 m pole is carried at \(v = 4/5\) toward a barn 8 m long, open at both ends. The farmer reckons the pole contracted to 6 m, so at one instant she flicks both doors shut, and open again, with the whole pole inside (events A and B). The pole carrier reckons the barn contracted, to 4.8 m. A 10 m pole can’t fit in a 4.8 m barn. Somebody is lying.

In the scene, R is the pole’s rear passing the entrance and F its front reaching the exit. Put the ruler on the pole, then on the barn, in both frames. Then pick A and B in the measure panel and read their \(\Delta t\): first in the barn’s frame, then after pressing a pole clock.

Case 2 · The twins

The charge. Lin stays home; Rae flies to a star 3 light-years away at \(3/5\) and back. Rae argues: “From my rocket, Lin was the one moving the whole time. So Lin’s clock ran slow, so Lin should be younger.” The setup looks symmetric. Is it?

Scrub to the reunion R. Jump into Rae’s frame on each leg. Try Lin’s eyes and Rae’s eyes in the strip; the gold lines are Lin’s yearly birthday flashes.

Case 3 · The tachyon telephone

The charge. Suppose Lin had a tachyon telephone: its signal is instantaneous in the sender’s frame. At S she calls Rae, who is passing 3 light-years away at 0.6, and the call arrives at Q. Rae replies at once, instantaneous in her frame, and the reply reaches Lin at B. Harmless gadget, or broken universe?

In the measure panel, pick S and Q, then Q and B. Then compare S and B in the lab, and in Rae’s frame.

Case 4 · Bell’s spaceships

The charge. Two identical rockets wait 100 m apart, a thread tied between them. At A and B, simultaneous in the launch frame, both fire identical engines and reach 0.6 almost at once. In the launch frame they stay 100 m apart forever, and their engines did exactly the same thing at the same time. Surely the thread is fine?

Use the ruler between the ships. Read it in the lab, then jump into the front ship’s frame once both are moving.

Case 5 · The muon’s side of the story

The charge. A muon is born 10 km up and falls at 0.995. In its own frame its clock ticks normally, as every clock does, and it lives about 2.2 μs, which is 659 m of time. So, argues the defense, in its own frame it can’t possibly cross 10 km of air. Yet the ground frame says about 22% reach the ground.

Read the muon’s clock when it lands. Then jump into the muon’s frame and put the ruler on the atmosphere.

Case 6 · The searchlight spot

The charge. A lighthouse sweeps its beam across a far wall. Sweep fast enough and the bright spot races across the wall faster than light. In the scene, the spot crosses 8 m in 4 m of time: twice light speed. Relativity is broken.

Compare spot events 1 and 5 in the measure panel. Then jump into the passing probe’s frame: which spot event comes first there?

Challenges

Four more scenes ship with the simulator as challenges. Work each one by hand first, then load it and check. In “Edit the scene”, every turning point has number boxes, so you can type your answer in exactly.

Stuck on a case? Every judgment above used the same three moves: ask which events are being compared, ask which frame each claim belongs to, and look for the invariant. The formula card has every equation on one page.

Build it

The last function is the shortest, because it stands on the others. trip(distanceLy, v) plans a round trip to a star and back at speed \(v\). It returns the years that pass on Earth and the years the traveler ages. Earth’s years are distance over speed, there and back. The traveler’s years are the same trip read by the one clock present at both departure and return, so divide by \(\g\). And \(\g\)? You wrote it in Chapter 5.

function square(n) { return n * n; }
function sumOfSquares(a, b) { return square(a) + square(b); }

Once trip() passes, the planner below runs on it. Pick a distance and a speed, then fly the trip in the simulator.

Checkpoint

Six items from across the course. Five are checked for you, and the sixth you compare with a model answer. Feedback waits until you submit.

Explain it to Sagredo

Epilogue: toward gravity

Everything in this course lived in free-float frames, and we kept adding fine print: free-float to a stated accuracy, over a stated region. Here is why the fine print matters. Drop two balls side by side in a small falling room, and they float together. Make the room wide, a few hundred kilometers, and both balls fall toward the center of the Earth, so they drift slowly together. No single free-float lattice covers both. That drift is a tide, and it is the signature of gravity: a spacetime that can’t be flattened into one lattice.

Your simulator already explains part of the clock budget of GPS satellites. Reckoned in a frame centered on Earth that doesn’t spin with it, the satellite clocks fall behind by about 7.2 μs a day because they move, as you found in Chapter 8. Yet compared with clocks on the ground they actually gain, about 38.5 μs a day net: an extra 45.7 μs a day comes from height. A clock higher up in Earth’s gravity runs faster than one lower down, as the two can confirm by exchanging signals.

Now put that next to the principle of Chapter 11: a free worldline between two events has the most wristwatch time. Let clocks run faster with height. A thrown stone that rises, spends a while where clocks run fast, and comes back down can age more than a stone that sits still. The path of greatest aging between “thrown” and “caught” turns out to be the arc a stone really follows. Falling is maximal aging, in a spacetime where height changes the pace of time. That is the doorway to general relativity, and you already know the hero waiting on the other side.

Where to read next

• Edwin F. Taylor and John Archibald Wheeler, Spacetime Physics (2nd ed., 1992), the book this course’s order of ideas follows. Free from the author’s site, eftaylor.com.
• Edwin F. Taylor, John Archibald Wheeler and Edmund Bertschinger, Exploring Black Holes: general relativity taught with the same maximal-aging idea. Also on eftaylor.com.
• The Feynman Lectures on Physics, Volume I, chapters 15 to 17. Free at feynmanlectures.caltech.edu.
• Hermann Bondi, Relativity and Common Sense: the Doppler-factor approach of Chapter 12, from one of its inventors.

Your Spacetime Surveyor card

One last thing to take with you: a card listing the parts you built and the checkpoints you passed. It’s made in your browser from your saved progress, and printing it sends nothing anywhere.

Going further

Two last problems that lean toward gravity: a head-scratcher and a pencil-and-paper one.

If you’ve come this far, you can compute, draw, explain and program special relativity. That was the promise on page one, and you kept your end of it.