Speed Is the Wrong Thing to Add

Warm-up

Three quick ones from earlier chapters. Each is a tool this chapter leans on.

The puzzle

Rae’s rocket passes Lin’s lab at 3/5 of light speed. Rae points a launcher straight ahead and fires a probe. By Rae’s own rods and clocks, the probe leaves her at 3/5 of light speed.

Galileo would add: \(3/5 + 3/5 = 6/5\). That’s faster than light. Chapter 2 promised that a flashlight on a rocket still sends its light out at exactly 1, for Lin too, and Chapter 7 showed that anything faster than light would let replies arrive before questions. Something in that sum has to give.

So: how fast does Lin’s lattice measure the probe?

Experiment

Here is a number line of speeds, as Lin’s lab reckons them. Lin stands at 0. Light sits at the gold lines, 1 and −1. Everything beyond is hatched: nothing that carries a clock goes there.

Two markers show the probe. Galileo’s ghost adds the two speeds. Einstein’s probe follows Nature. Move the sliders and watch them part company.

One probe is a warm-up. The probe could carry its own launcher, and fire a second probe at 0.6 relative to itself. That one could fire a third. Galileo’s sums grow without limit: 0.6, 1.2, 1.8, … Let’s see what Nature does with a whole stack.

The lab tracks three numbers for the top stage: its speed in Lin’s lab, its Doppler factor \(k\) from Chapter 12, and a third number, \(\th\) (“theta”). We’ll name \(\th\) properly a few sections below. For now, just watch how it grows.

Then play. Try stages of 0.1, or 0.9. The speeds change; the habits of the last two columns don’t.

What just happened?

Sagredo is right to ask. The answer was in the stack. Look at its \(k\) column: 2, then 4, then 8. Nature isn’t adding speeds. It’s multiplying something else.

Multiply, don’t add

Doppler factors multiply

Start with something we trust: radar from Chapter 12. Rae moves away from Lin at 0.6. If Lin sends a flash every 1, Rae receives one every 2 of her own time. That 2 is the Doppler factor \(k\) for 0.6.

Now the probe. It moves away from Rae at 0.6, by Rae’s reckoning. So if Rae sent a flash every 2, the probe would get one every \(2 \times 2 = 4\).

Here’s the trick: Rae doesn’t need to send anything. Lin’s flashes pass Rae and keep going. They reach Rae every 2, so for the probe they are exactly “flashes sent from Rae every 2”. Light doesn’t care who lit it. The probe receives Lin’s flashes every 4. Plate 13.1 shows one flash doing all of this.

So the Doppler factor from Lin to the probe is 4. Doppler factors multiply:

\[ \begin{gathered} k_\text{probe} = k_\text{Rae} \times k_\text{probe, from Rae} \\ = 2 \times 2 = 4. \end{gathered} \]

From k back to a speed

A Doppler factor of 4 belongs to some speed. Which one? Start from Chapter 12’s formula and run it backward. Need a refresher?

\[ \begin{gathered} k^2 = \frac{1+v}{1-v} \\ k^2 - k^2 v = 1 + v \\ k^2 - 1 = v\,(k^2 + 1) \\ v = \frac{k^2 - 1}{k^2 + 1}. \end{gathered} \]

Check it on a speed we know. For \(k = 2\): \((4 - 1)/(4 + 1) = 3/5\). Good. Now \(k = 4\): \((16 - 1)/(16 + 1) = 15/17\). There’s Sagredo’s 17.

1. Combine \(0.6 \oplus 0.8\). First, the Doppler factor of 0.6: \(\sqrt{1.6/0.4} = \sqrt{4} = 2\). Doppler factors are the numbers that combine cleanly, so we translate into them first.
2. The Doppler factor of 0.8: \(\sqrt{1.8/0.2} = \sqrt{9} = 3\). Same recipe. Friendly speeds have whole-number factors; the formula card lists them.
3. Multiply: \(K = 2 \times 3 = 6\). One boost after another multiplies the Doppler factors, as the relay showed.
4. Translate back: \(w = (36 - 1)/(36 + 1) = 35/37 \approx 0.946\). This is the speed whose Doppler factor is 6. Galileo would have said 1.4.

The formula, as shorthand

Let’s do that recipe once with letters, for any \(u\) and \(v\). The combined Doppler factor squared is a product, \(K^2 = \frac{(1+u)(1+v)}{(1-u)(1-v)}\). Put it into \(w = (K^2 - 1)/(K^2 + 1)\), then multiply the top and the bottom by \((1-u)(1-v)\):

\[ \begin{gathered} w = \frac{(1+u)(1+v) - (1-u)(1-v)}{(1+u)(1+v) + (1-u)(1-v)} \\ = \frac{2u + 2v}{2 + 2uv} \\[4pt] \boxed{\;w = \frac{u + v}{1 + uv}\;} \end{gathered} \]

That’s the velocity addition formula. It’s the shorthand for “multiply the Doppler factors”. Three checks:

• Everyday speeds. A jet at 300 m/s is \(u = 10^{-6}\). Two of them make \(uv = 10^{-12}\), so the bottom is 1.000000000001. Galileo was right about every speed he could ever measure. That also answers the number line’s third Try it: the two rules agree whenever \(uv\) is close to 0 (one speed is small), or when \(u + v\) is close to 0 (the two speeds nearly cancel, so there is almost nothing to shrink).
• Light. Put \(u = 1\): \(w = (1 + v)/(1 + v) = 1\), for any \(v\). A flash fired from a rocket moves at 1 for Lin too. The formula passes the warm-up’s test.
• Never past light. A line of algebra shows \(1 - w = \frac{(1-u)(1-v)}{1 + uv}\). If \(u\) and \(v\) are both between −1 and 1, then \(1 - u\), \(1 - v\) and \(1 + uv\) are all positive, so \(w\) stays below 1. Always. (The same argument with \(1 + w\) keeps it above −1.)

Rapidity: the number that really adds

Doppler factors multiply. Multiplying is awkward; adding is easy. Is there a way to turn one into the other? Yes: logarithms. \(\ln(2 \times 2) = \ln 2 + \ln 2\). Need a refresher?

So give every speed the number \(\th = \ln k\). Physicists call it the rapidity. For 0.6 it is \(\ln 2 = 0.693\). Two boosts of 0.6 make \(0.693 + 0.693 = 1.386 = \ln 4\). Rapidities add:

\[ \th_w = \th_u + \th_v, \qquad k = e^{\th}. \]

Speed is squeezed below 1. Rapidity is not. Ten stages at 0.6 give a rapidity of 6.93, a perfectly ordinary number, while the speed is stuck at 0.999998. Plate 13.2 shows how the endless rapidity line gets squeezed onto the short speed ruler.

cosh, sinh and tanh, built from \(e^\th\)

Numbers first. Take 0.6 again, where \(k = 2\) and \(\g = 1.25\). Average \(k\) and \(1/k\), then take half their difference:

\[ \begin{gathered} \frac{2 + \tfrac12}{2} = 1.25 = \g, \\ \frac{2 - \tfrac12}{2} = 0.75 = \g v. \end{gathered} \]

(That last one is \(1.25 \times 0.6\).) Try 0.8, where \(k = 3\) and \(\g = 5/3\): \((3 + \tfrac13)/2 = 5/3\), which is \(\g\), and \((3 - \tfrac13)/2 = 4/3\), which is \(\g v = \tfrac53 \times 0.8\). It works for every speed: half the sum of \(k\) and \(1/k\) is \(\g\), and half their difference is \(\g v\). (Chapter 12 found the first one with radar.)

Now put \(k = e^\th\), so \(1/k = e^{-\th}\). These two combinations of \(e^\th\) and \(e^{-\th}\) come up so often that they have names, “hyperbolic cosine” and “hyperbolic sine”: Need a refresher?

\[ \begin{gathered} \g = \cosh\th = \frac{e^{\th} + e^{-\th}}{2}, \\ \g v = \sinh\th = \frac{e^{\th} - e^{-\th}}{2}, \\ v = \tanh\th = \frac{\sinh\th}{\cosh\th}. \end{gathered} \]

At \(\th = \ln 2\) these are the numbers we started with: \(\cosh\th = 1.25\), \(\sinh\th = 0.75\) and \(\tanh\th = 0.75/1.25 = 0.6\). Every friendly speed works the same way:

So the whole chapter fits on one line. To combine two boosts, add their rapidities and take the tanh: \(w = \tanh(\th_u + \th_v)\).

Turning versus boosting

A circle \(x^2 + y^2 = 1\) has a cousin with a minus sign, \(t^2 - x^2 = 1\). You met it in Chapter 10 as a calibration hyperbola: every frame agrees on it. On the circle, a point is \((\sin\varphi, \cos\varphi)\), and turning adds angles. (We measure \(\varphi\) from the vertical axis, the way rapidity is measured from the time axis, so sine and cosine trade their usual places.) On the hyperbola, a point is \((\sinh\th, \cosh\th)\), and boosting adds rapidities. The line from the origin to that point is a worldline. Its speed is \(\tanh\th\). One difference: turn far enough and the circle brings you back where you started; boost as far as you like and the hyperbola never comes back.

The shaded sectors carry one more fact, which we state without proof: a sector’s area is always half its angle (on the circle) and half its rapidity (on the hyperbola). That’s why physicists call \(\th\) an angle. It measures area the way \(\varphi\) does, on a curve with a minus sign.

Faster than light, legally

Two ships fly toward each other, at \(+0.9\) and \(-0.9\) in Lin’s lab. In one unit of Lin’s time the gap between them shrinks by \(0.9 + 0.9 = 1.8\). Is that illegal?

No. The gap isn’t a thing. Nothing rides along it, carrying a clock or a message at 1.8. That gives a test for every “faster than light” claim you will ever meet: is there an object, carrying a clock or a message, that moves at that speed? If not, the speed can be anything. If so, it has to be below 1.

So ask a real traveler instead: ship A, measuring ship B with A’s own lattice, gets \(0.9 \oplus 0.9 = 0.9945\). Below 1, as it must be.

Here is the searchlight in numbers. The Moon is about 384 000 km away, which is 1.28 light-seconds. Turn a laser once a second. Out at the Moon’s distance, the spot would go around a circle \(2\pi \times 384\,000\) km long in that second: about 2.4 million km per second, 8 times light speed. No law is broken. Every bit of light flies straight out from Earth at exactly 1. The spot at A and the spot at C are different light, sent at different moments, and nothing at A can cause anything at C.

The person at A can block the light, wave at it, paint the rock. None of it reaches C any faster than light from A to C could. Every change in the spot has to start back on Earth.

That is the test from the two ships, again: is there an object, carrying a clock or a message, that moves at that speed? Keep it handy.

Ask Nature

Velocity addition’s first-order effect was measured half a century before anyone wrote the formula down. In 1851 Hippolyte Fizeau sent light through water flowing in a tube, once with the current and once against it, and compared the two.

Now let velocity addition loose on it. In the water’s own frame the light moves at 0.75. In the lab, the formula gives

\[ \begin{gathered} w = \frac{0.75 + u}{1 + 0.75\,u} \\ \approx (0.75 + u)(1 - 0.75\,u) \\ \approx 0.75 + (1 - 0.75^2)\,u \\ = 0.75 + 0.44\,u. \end{gathered} \]

The first step uses the small-number shortcut \(1/(1 + x) \approx 1 - x\). Need a refresher? The second throws away the tiny \(0.75\,u^2\). So relativity predicts Fizeau’s fraction, \(1 - 1/n^2\), to first order in the water’s speed. Not exactly: the exact formula differs by terms in \(u^2\), far too small for any tube of water.

The cleanest test of \(u \oplus 1 = 1\) is one you met in Chapter 2, at CERN in 1964. A team timed gamma-ray flashes from neutral pions moving at 0.99975 of light speed. Galileo says the flashes fired forward should move at nearly 2. They moved at 1, within about one part in 10 000.

Light in glass, light in water, light from pions: every test so far says the same thing. Speeds combine by the formula, and light stays at 1.

Practice

Three tiers, from back-of-the-envelope (◇) through pencil-and-paper (◆) to one head-scratcher (★). Fractions and expressions such as 15/17 or ln(2) are welcome in the answer boxes.

One more idea before the head-scratcher: why does a steady push add rapidity steadily? Take a short stretch \(\Delta\tau\) of Rae’s wristwatch time, and judge it from the frame she was coasting in just before. In that frame she starts at rest, so her speed changes only by a small \(a\,\Delta\tau\), where Newton is fine. For small speeds, rapidity and speed are nearly equal. And rapidities add. So each short stretch adds \(a\,\Delta\tau\) of rapidity, in Lin’s reckoning too.

Keep the formula card nearby: the friendly-number table there turns most of these into mental arithmetic.

Build it

Three short functions and a bonus. Once addVelocities passes, the number line and the stack at the top of this page run on your code. If it fails, you can run the number line with it anyway and watch where your probe lands.

const grown = Math.exp(0.5);   // e to the power 0.5: 1.6487…
Math.log(grown);               // the natural log undoes it: 0.5
Math.tanh(100);                // 1, as close as a computer can tell
Math.atanh(0.5);               // 0.5493…, undoes Math.tanh

Checkpoint

Six items. Feedback waits until you submit. The first five are scored. The last one asks you to explain in your own words: it isn’t graded. You compare your answer with a model answer, and your text stays in the box.

Explain it to Sagredo

Going further

The head-scratcher at the end of Practice imagined a steady 1 g push as a long chain of tiny rapidity kicks. Here is that trip, drawn. Cut Rae’s time into \(n\) steps. Each dot is a kick; between kicks Rae coasts in a straight line, at the rapidity halfway through that step, so the trip starts and ends with a half kick. Add steps and the broken line settles onto a smooth curve, a hyperbola whose asymptote is a light ray. That ray is the dotted gold line: a flash Lin sends from home at \(t = 1/a\), about 0.97 years after Rae leaves at 1 g. (Equivalently, it is light that set off at the start from a point \(1/a\), about 0.97 light-years, behind Rae.) Rae’s curve creeps toward it forever and never meets it: that flash never catches her, and no message Lin sends after it ever reaches her. The line is Rae’s horizon. When your bonus rocketWorldline passes, the broken line is drawn from your points, in purple.

To read more, look for the “velocity parameter” in Taylor and Wheeler’s Spacetime Physics (a free PDF is at eftaylor.com), which is what they call rapidity.