The Lorentz Machine

Warm-up

Three quick ones from earlier chapters. Today we build a machine out of exactly these parts.

The puzzle

Look at your toolbox. Chapter 5 gave you \(\g\). Chapter 6, the interval. Chapter 7, the tilted “now”. Chapter 8, stretched time. Chapter 9, shrunken rulers. Five tools, and each one came with a ritual: which clock is present at both events? Whose “now” are we using?

Here is a bolder idea. One machine. You feed it any event, written in Lin’s lab coordinates, plus Rae’s speed. Out comes the same event in Rae’s coordinates. No rituals. If such a machine exists, every answer from Chapters 6 to 9 should fall out of it, with Chapter 5’s \(\g\) as one of its parts.

Let’s test it on an old friend: the firecracker at \(\lab{t = 5}\), \(\lab{x = 3}\) from Chapter 6. In this chapter it is event A. Rae’s rocket passes the origin at \(v = 0.6\).

Experiment

Here is Lin’s spacetime diagram with Rae’s axes drawn on top. Slide Rae’s speed and her axes close like scissors. Switch the drawing to Rae’s frame and every event glides to its new coordinates. The table keeps score in both frames. The teal curves are hyperbolas, each one drawn as a double line: that is this course’s mark for something every frame agrees on.

Now the same machine with every event at once. The second lab feeds a whole grid of Lin’s events through the machine and plots where they land on Rae’s graph paper. Drag sideways on the picture, or use the slider. The readout \(k\) tells you how much the lattice is squashed along the up-right light line \(t = x\), and stretched along the up-left one, \(t = -x\). Let’s call it the light-line factor. The gold box shows the squeeze. Turn on Hyperbola trails (T) to see the path each event took: each one slides along its own hyperbola.

Then play. Pick events off the light lines and watch the ledger: one number goes down, one goes up, and their product stays put.

What just happened?

What’s new is that the machine doesn’t care whether Rae is there. Event B at \(\lab{(3, 5)}\) happens nowhere near her, and the machine still answers: \(\rkt{(0, 4)}\). By her own lattice’s reckoning, B is 4 meters from Rae, and the lattice clock standing there reads 0 when B goes off. That is Chapter 7’s tilted “now” in one line.

You also watched Rae’s axes in Lin’s picture. They lean toward the light line from both sides, and the faster the rocket, the tighter they close.

Why scissors and not a rigid turn? Because the light line must stay a light line for Rae too: neither blade may cross it. Next, let’s find out why the machine has to work this way.

Building the machine

Two lattices, one event

Lin has a lattice of synchronized clocks at rest in the lab. Rae has her own lattice riding with the rocket. When a firecracker goes off, the nearest clock of each lattice writes down a time and a place: \(\lab{(t, x)}\) for Lin, \(\rkt{(t', x')}\) for Rae. The machine we want, the Lorentz transformation, turns one record into the other. Changing frames like this is called a boost.

Friendly numbers first: the light-line ledger

Before any algebra, take the lab’s answers at \(v = 0.6\) and, for each event, add the coordinates and subtract them.

Every sum got cut in half, and every difference doubled. That’s the whole machine at \(v = 0.6\). The 2 is the light-line factor from the squeeze machine: at \(v = 0.6\), \(k = 2\).

Why would anyone add and subtract coordinates? Because \(t + x\) and \(t - x\) are the light lines’ own measuring sticks. Along a light ray moving right, \(t - x\) stays fixed. Along one moving left, \(t + x\) does. And their product is an old friend: \((t + x)(t - x) = t^2 - x^2\), the squared interval. Halve one factor, double the other, and the product can’t change.

Why it has to work this way

We ask the machine for four things, and each one buys one fact. Follow along with a pencil. Refresher: rearranging (Toolkit, opens in a new tab)

1. Light stays light. A ray moving right through O satisfies \(t - x = 0\). It must be a light ray for Rae too, so \(t' - x' = 0\) exactly when \(t - x = 0\). The machine only multiplies and adds (the box above), so \(t' - x'\) is some amount of \((t - x)\) plus some amount of \((t + x)\). On the ray \(t - x\) is zero but \(t + x\) isn’t, so the second amount must be zero. A pure multiplier is left: \(\rkt{t' - x'} = k\,\lab{(t - x)}\). If light had another speed for Rae, she could detect her motion. Galileo’s promise, from Chapter 2.
2. The same for the other light line. A ray moving left keeps \(t + x = 0\), so \(\rkt{t' + x'} = m\,\lab{(t + x)}\) for some other number \(m\). We don’t assume \(m = k\). We let the next requirement decide.
3. The interval is the same for everyone. Multiply the two lines: \(\rkt{t'^2 - x'^2} = m k\,\lab{(t^2 - x^2)}\). For this to equal \(\lab{t^2 - x^2}\) for every event, \(m k = 1\), so \(m = 1/k\). Chapter 6: every free-float frame agrees on \(\inv{\tau^2}\).
4. Rae rides on her own axis. Rae’s worldline \(x = vt\) must land on \(x' = 0\), where \(t' + x' = t' - x'\). Put \(x = vt\) into both lines: \(\dfrac{(1 + v)\,t}{k} = k\,(1 - v)\,t\), so \[k^2 = \frac{1 + v}{1 - v},\] and taking the square root, \[k = \sqrt{\frac{1 + v}{1 - v}}.\] The speed v is the only thing that tells the machine which rocket we mean.
5. Check it on friendly numbers. At \(v = 0.6\): \(k^2 = 1.6/0.4 = 4\), so \(k = 2\), as the ledger said. At \(v = 0.8\): \(k^2 = 1.8/0.2 = 9\), so \(k = 3\). Never trust a formula you haven’t fed a number you already know.

From the squeeze to the formulas

The ledger shows what the machine does. But it hands you \(t' + x'\) and \(t' - x'\), and usually you want \(t'\) and \(x'\) on their own. Two moves get them back. Add the two results and halve: the \(x'\)s cancel and \(t'\) is left. Subtract and halve: the \(t'\)s cancel and \(x'\) is left. Watch them work on event A at \(v = 0.6\), with numbers only.

1. Add and subtract. A is at \(\lab{(5, 3)}\), so \(\lab{t + x} = 8\) and \(\lab{t - x} = 2\). These are A’s first two columns in the ledger.
2. Squeeze. At \(v = 0.6\), \(k = 2\). Halve the sum: \(8 \div 2 = 4\), so \(\rkt{t' + x'} = 4\). Double the difference: \(2 \times 2 = 4\), so \(\rkt{t' - x'} = 4\). Sums are divided by \(k\). Differences are multiplied by \(k\).
3. Add and halve. \(\rkt{t'} = (4 + 4)/2 = 4\). \((t' + x') + (t' - x')\) is \(2t'\). The \(x'\)s cancel.
4. Subtract and halve. \(\rkt{x'} = (4 - 4)/2 = 0\). \((t' + x') - (t' - x')\) is \(2x'\). This time the \(t'\)s cancel.
5. Check. \(\rkt{(4, 0)}\), the same answer the boost lab gave. And \(4^2 - 0^2 = 16 = 5^2 - 3^2\). The interval check costs nothing, so always do it.

Those moves never cared that the event was \((5, 3)\). So run them once more with the letters \(\lab{t}\) and \(\lab{x}\) in place of 5 and 3. Keep \(k = 2\) for now: one new thing at a time.

1. Squeeze. \(\rkt{t' + x'} = \tfrac12\,\lab{(t + x)}\) and \(\rkt{t' - x'} = 2\,\lab{(t - x)}\). Step 2 above, with letters.
2. Add and halve. The sum is \(\tfrac12 t + \tfrac12 x + 2t - 2x\), which is \(2.5\,t - 1.5\,x\). Halve it: \[ \rkt{t'} = 1.25\,\lab{t} - 0.75\,\lab{x}. \] Gather the \(t\)s: \(\tfrac12 + 2 = 2.5\). Gather the \(x\)s: \(\tfrac12 - 2 = -1.5\).
3. Subtract and halve. The difference is \(\tfrac12 t + \tfrac12 x - 2t + 2x\), which is \(2.5\,x - 1.5\,t\). Halve it: \[ \rkt{x'} = 1.25\,\lab{x} - 0.75\,\lab{t}. \] The same two numbers, with \(t\) and \(x\) trading places.
4. Test it on A. \(1.25 \times 5 - 0.75 \times 3\) \(= 6.25 - 2.25 = 4\), and \(1.25 \times 3 - 0.75 \times 5\) \(= 3.75 - 3.75 = 0\). The same \((4, 0)\) as the numbers-only route. Two routes, one answer.
5. Meet the two numbers. 1.25 is \(\g\) at 0.6, from the first warm-up. And \(0.75 = 0.6 \times 1.25\), which is \(\g v\). So at 0.6 the machine is \(\rkt{t'} = \g(\lab{t} - v\,\lab{x})\) and \(\rkt{x'} = \g(\lab{x} - v\,\lab{t})\). \(\g t - \g v\,x\) is \(\g(t - vx)\): take out the shared \(\g\).

Nothing in those steps needed \(k\) to be 2. The 1.25 came from \((2 + \tfrac12)/2\), and the 0.75 from \((2 - \tfrac12)/2\). For any speed, put \(k\) where the 2 was and \(1/k\) where the \(\tfrac12\) was:

\[ \begin{aligned} \rkt{t'} &= \frac{k + 1/k}{2}\,\lab{t} - \frac{k - 1/k}{2}\,\lab{x} \\[4pt] \rkt{x'} &= \frac{k + 1/k}{2}\,\lab{x} - \frac{k - 1/k}{2}\,\lab{t} \end{aligned} \]

At \(v = 0.8\), \(k = 3\). Then \((3 + \tfrac13)/2 = \tfrac53\) and \((3 - \tfrac13)/2 = \tfrac43\). Those are \(\g\) and \(\g v\) at 0.8, straight from the friendly table.

Why are the two numbers always \(\g\) and \(\g v\)? Give them names: call the first one \(a\) and the second one \(b\). At 0.6, \(a = 1.25\) and \(b = 0.75\). Five short steps pin them down, each checked on those numbers.

1. Add them. \(a + b = k\). In \(\tfrac{k + 1/k}{2} + \tfrac{k - 1/k}{2}\) the \(1/k\)s cancel. Check: \(1.25 + 0.75 = 2\).
2. Subtract them. \(a - b = 1/k\). This time the \(k\)s cancel. Check: \(1.25 - 0.75 = 0.5\).
3. Multiply the two results. \((a + b)(a - b)\) \({} = a^2 - b^2\), and that is \(k \times 1/k = 1\). So \(a^2 - b^2 = 1\). The ledger’s trick again. Check: \(1.5625 - 0.5625 = 1\).
4. Rae rides on her own axis. On her worldline \(x = vt\), the machine must give \(x' = a\,x - b\,t = 0\). So \(a\,vt = b\,t\), which means \(b = v\,a\). Step 4 of the derivation. Check: \(0.6 \times 1.25 = 0.75\).
5. Solve. Put \(b = v\,a\) into step 3: \(a^2 (1 - v^2) = 1\), so \(a = 1/\sqrt{1 - v^2} = \g\) and \(b = \g v\). Only one pair passes both tests: \(\g\) and \(\g v\).

So, at every speed, the machine is:

1. Event \(\lab{(10, 8)}\), rocket at \(v = 0.8\). From the friendly table: \(\g = 5/3\) and \(\g v = 4/3\). Look these up before computing. The 3-4-5 triangle hands them over for free.
2. \(\rkt{t'} = \tfrac{5}{3}(10) - \tfrac{4}{3}(8) = \tfrac{50 - 32}{3} = 6\). \(\g(t - vx)\) is the same as \(\g t - \g v\,x\).
3. \(\rkt{x'} = \tfrac{5}{3}(8) - \tfrac{4}{3}(10) = \tfrac{40 - 40}{3} = 0\). Zero: the event is on Rae’s worldline, since \(8 = 0.8 \times 10\).
4. Check the interval: \(6^2 - 0^2 = 36\) and \(10^2 - 8^2 = 36\). If these ever disagree, the arithmetic is wrong. This check is free, so always do it.
5. Check the ledger: \(t + x = 18\) shrinks by \(k = 3\) to \(6 = t' + x'\), and \(t - x = 2\) grows to \(6 = t' - x'\). Two independent routes to the same answer.

Running the machine backward

Rae has a question too: where do her events land in Lin’s lab? She doesn’t need a second machine. In Rae’s frame, Lin’s lab moves at \(-v\). The principle of relativity says her frame is as good as Lin’s, so the same machine must work for her with the speed flipped:

\[ \begin{aligned} \lab{t} &= \g\,(\rkt{t'} + v\,\rkt{x'}) \\ \lab{x} &= \g\,(\rkt{x'} + v\,\rkt{t'}) \end{aligned} \]

Try it on A, with the backward machine and \(v = 0.6\): \(\rkt{(4, 0)}\) gives \(t = 1.25 \times (4 + 0) = 5\) and \(x = 1.25 \times (0 + 0.6 \times 4) = 3\). Home again. In light-line form it’s even plainer: the forward machine divides \(t + x\) by \(k\), and flipping \(v\) turns \(k\) into \(1/k\), which multiplies it back.

Old answers, one machine

Now the promise from the hook. Feed the machine the right questions, and the answers of Chapters 6 to 9 fall out, plus one from Chapter 2.

Time stretching. Rae’s clock sits at \(x' = 0\). Using the backward machine (you just met it), \(t = \g(t' + v \cdot 0) = \g t'\). Between O and the event where Rae’s wristwatch reads 8, Lin’s lattice reckons \(1.25 \times 8 = 10\) meters of time. That’s Chapter 8’s time stretching.

Contraction. Rae’s rod runs from \(x' = 0\) to \(x' = L_0\). Lin marks both ends at the same lab time, \(t = 0\). Then \(x' = \g x\), so the far end is at \(x = L_0/\g\). A 5 m rod at 0.6 measures 4 m: Lorentz contraction.

Simultaneity. Two lightning strikes at \(t = 0\), at \(x = 150\) and \(x = -150\), with the train at 0.6. The machine gives \(t' = 1.25(0 - 90) = -112.5\) and \(t' = +112.5\). The front strike comes first by 225 m of time, exactly Chapter 7’s answer.

Drawing Rae’s axes honestly

One last job: putting numbers on Rae’s scissored axes in Lin’s picture. Where is Rae’s tick \(t' = 1\)? It’s the event \((t', x') = (1, 0)\). Run the machine backward: \(t = \g\) and \(x = \g v\). At 0.6, that’s \((1.25, 0.75)\).

Measured with an ordinary ruler on the page, that point is \(\sqrt{1.25^2 + 0.75^2} = \sqrt{2.125} \approx 1.458\) from O. Rae’s unit tick is longer on paper than Lin’s. Refresher: squares and roots (Toolkit, opens in a new tab)

There’s a beautiful way to find every tick without arithmetic. Draw the calibration hyperbolas \(t^2 - x^2 = 1, 4, 9, \dots\). Each one crosses Lin’s time axis at 1, 2, 3 and Rae’s at her 1, 2, 3, because both frames agree on the interval from O. The hyperbolas are the rulers. The page lies; the hyperbolas don’t.

Ticks tell you where Rae’s axes read 1, 2, 3. To read an event that sits off her axes, use parallels. A line parallel to Rae’s \(x'\) axis is one of her lines of “now”: every event on it has the same \(t'\). So slide from the event along that direction until you hit her time axis, and read the tick there. That’s the event’s \(t'\). Slide parallel to her time axis until you hit her space axis, and that tick is its \(x'\). Plate 10.3 does this for event E with dotted lines. It’s the same trick you use on Lin’s square grid, where the parallels happen to be horizontal and vertical.

That’s the whole machine: two light lines, one light-line factor, and an interval that refuses to change. Next, let’s ask Nature whether she actually runs it.

Ask Nature

We built the machine from requirements. Each requirement is a claim about the world, and each one has been put on trial.

Today the machine runs every day at particle accelerators. Each collision is recorded in the lab, analyzed in the frame where the total momentum is zero, and translated back with the same machine, applied to energy and momentum instead of time and place (Chapter 14).

Practice

Friendly numbers first. Keep the ledger handy: \(t + x\) divides by \(k\), \(t - x\) multiplies by \(k\), and the product never changes. It’s the fastest check you own.

◇ Back of the envelope

◆ Pencil and paper

★ Head-scratchers

Want more? The formula card has the friendly-number table, and every exercise here can be checked in the boost lab above.

Build it

This is the keystone of your toolbox. Every chapter from here on leans on it, and so does the simulator you’ll assemble at the end. Once your lorentz passes, the boost lab at the top of this page will run your lorentz function instead of the built-in one, and so will the squeeze machine.

function area(box) {
  // same as: const w = box.w; const h = box.h;
  const { w, h } = box;
  return w * h;
}

A bonus for anyone who draws diagrams. You found the spacing of Rae’s ticks on paper by hand. Teach the computer.

Checkpoint

Five graded items and one written explanation. Feedback waits until you submit, so commit to your best answer for each.

Explain it to Sagredo

Going further

Two ★★ problems that peek at what’s coming.

Reading: Taylor and Wheeler’s Spacetime Physics derives the same formulas from the interval, with many more practice problems (free at eftaylor.com). Hermann Bondi’s Relativity and Common Sense builds all of special relativity from the factor \(k\).