Chapter 12
Seeing Is Not Measuring
Doppler colors, the starbow that isn’t there, and why a fast cube looks rotated.
What this chapter does
Chapters 8 and 9 measured moving clocks and rulers with a lattice: a clock at every place, so no light has to travel anywhere. This chapter asks what you see with one eye, one telescope or one camera, where the light’s travel time is always mixed in.
- Two surprises. A camera photographs a fast cube, and a lamp flashes while it races toward Lin. You predict, then test.
- One number, \(k\). Radar finds it. It explains why an approaching clock looks fast but is measured slow, and why colors shift. Add one rule for directions, and you also get why the stars crowd ahead and why the cube looks turned.
- Nature’s checks, then practice and code.
The thread through all of it: seeing is measuring plus the light’s trip. Take the trip out, and Chapter 8’s \(\g\) is still there.
Warm-up
Three quick questions from earlier chapters. Each one is a tool this chapter leans on.
From Chapter 8: Rae coasts past Lin at \(v = 4/5\). Her wristwatch ticks once every 3 m of time. How long does one of those ticks last by Lin’s lattice?
That divides by \(\g\). Rae’s wristwatch is the one clock present at both ticks. Does it read the shorter time or the longer one?
Hint
At \(v = 0.8\) the stretch factor is \(\g = 5/3\).
Solution
Rae’s watch reads the wristwatch time \(\tau = 3\). Lin’s lattice measures \(\g\tau = \tfrac{5}{3} \times 3 = 5\) m of time.
From Chapter 3: a firecracker bursts 300 m from Lin. The lattice clock beside it reads \(t = 1000\) m at the burst. What does Lin’s own clock read when Lin sees the flash?
That’s when it happened, by the lattice. The light still has 300 m to travel. How long does that take?
That would have the light arrive before it left. Which way does the travel time push the reading?
Hint
Light covers 1 m of distance in 1 m of time.
Solution
The flash leaves at \(t = 1000\) and needs 300 m of time to reach Lin, so Lin sees it at \(1300\). Seeing is the event plus the light’s trip.
From Chapter 4: Lin sends a radar pulse at \(t = 2\) and receives its echo at \(t = 10\). When and where did the pulse bounce, by Lin’s lattice?
That’s when the echo came back. The bounce happened partway through the round trip. Which part?
That’s the whole round trip, out and back. The pulse only went out once. How far did it get?
Hint
The bounce is halfway through the round trip in time, and the distance is half the round trip.
Solution
\(t = (2 + 10)/2 = 6\) and \(x = (10 - 2)/2 = 4\). We’ll use exactly this radar rule to find a new number.
The puzzle
A cube, 2 m on a side, streaks past a camera at \(v = 0.9\). Chapter 9 told us what Lin’s lattice reports. Every clock along the track notes where its piece of the cube is at one instant, and along the motion the cube is only \(2/\g = 0.87\) m long. Squashed.
But a camera is not a lattice. A camera is one eye. Its shutter opens for an instant and records whatever light happens to arrive together, from wherever and whenever that light set off.
The shutter clicks just as the cube’s light reaches the lens head-on, so the cube appears straight in front of the camera. What does the photograph show?
Predict
The photograph shows the cube…
Experiment
The lab draws two pictures of the same instant. The Lattice panel shows where every part of the cube is at \(t = 0\), the moment the shutter opens. The Camera panel shows the photo. For each pixel we follow the light that reaches the lens at \(t = 0\) backward along its path, to the place and the time it left the cube.
At v = 0.9 the lattice shows the side face, the one facing the camera, squashed to 0.87 m. The photo shows that narrow side face plus the tail beside it: a cube turned by about 64°, not a squashed one.
How to read it. The small triangle in each panel marks where the lattice says the object really is at the shutter instant. The photo lags behind that mark, because it shows each part where its light left from. While the object is still far to the left, the photo can be empty although the lattice shows the object: its light hasn’t reached the camera yet.
Describe this lab
Two panels show the same instant, the moment the shutter opens. The camera is 4 m from a straight track. The Lattice panel draws the object where Lin’s lattice says it is at that instant: contracted along the motion by the stretch factor. The Camera panel draws what the shutter records: for each pixel the light ray is followed back to the object, so every part of the object appears where it was when its light left. The cube’s faces are lettered S (the side facing the camera), T (the tail, the trailing face) and N (the nose, the leading face). Controls: Cube, Rod or Sphere; the speed v from 0 to 0.99; the lattice position of the object’s center; Center the photo, which puts the object where its light reaches the camera head-on; and Fly past, which sweeps the object from 9 m left to 9 m right, approaching and then going away. Readouts give the length at rest, the lattice length and, for the cube and the sphere, the angle by which a centered photo looks turned, which is the angle whose sine is v. A small triangle in each panel shows where the lattice says the center is at the shutter instant: labeled “is here now” in the Lattice panel and “lattice position now” in the Camera panel. A small arrow at the bottom left of each panel shows which way the object moves. When the photo of the object lies outside the Camera panel, a note says which side it is on. Keys: Space plays or pauses, C centers the photo, 1, 2 and 3 pick the object.
Try it
Set \(v = 0\) and press Center the photo. Which lettered faces does the camera show? Now set \(v = 0.9\) and center again. Which new face appears in the camera panel? And in which panels is the side face S narrower than it was at rest?
The camera sits only 4 m from the track. That close, light from different parts of the box reaches the lens from different directions, so at high speed the photo also bends the box a little. The bending is real, not a drawing error, but it isn’t the point: look for the faces. The small arrow in each panel shows which way the object moves.
Try it
Pick the Sphere and slide the speed from 0 up to 0.99. In the camera panel, is the outline ever narrower than it is tall? Now watch the black dot, which faced the camera when the sphere was at rest.
The camera mixes light that left at different times. Your eyes do the same to everything they look at, including clocks and lamps. Here is a lamp on Rae’s rocket. It flashes once per unit of Rae’s own time, and it is either coming toward Lin or going away. Before you turn it on, one more guess.
Predict
Rae’s lamp flashes once a year by her own wristwatch, and she races toward Lin at 0.6. Chapter 8 says her clock runs slow. Through Lin’s telescope, Lin sees the lamp flash…
Rae’s lamp flashes once per unit of her time. Approaching at v = 0.6, its flashes reach Lin every 0.5 units of her time, and its yellow 589 nm light arrives as ultraviolet. Receding at 0.6, the flashes arrive every 2 units, and the light shifts past the red into the infrared (1178 nm): invisible too.
Describe this lab
On the left, a spacetime diagram with time up, numbered once per unit of Lin’s time, and Lin’s worldline at x = 0. Rae’s lamp moves toward Lin (arriving beside her at the top) or away from her (leaving from beside her at the bottom), and flashes once per unit of her own time. Gold lines at 45 degrees carry each flash to Lin, and a brace on Lin’s worldline shows how often they arrive. On the right, a disc shows the lamp’s color as Lin receives it, and a spectrum bar with a logarithmic wavelength scale from 200 to 1600 nm shows the hydrogen lines H-alpha, H-beta and H-gamma at their rest positions (dashed) and as received (solid). Controls: Approaching or Receding, the speed v from 0 to 0.95, and the lamp’s color as made, from 400 to 700 nm. Readouts give how many times as often the flashes reach Lin (the frequency factor: twice as often approaching at 0.6, half as often receding), Rae’s flash spacing (1, by her clock), Lin’s received spacing, and the lamp’s wavelength as made and as received. The Hear it button plays 440 hertz, then the tone as Lin would receive it, for under two seconds. Keys: A switches direction, H plays the tone.
Try it
Pick Approaching and set \(v = 0.6\). Count on the diagram: how many flashes reach Lin per unit of Lin’s time? Now pick Receding. Last, approaching again: at about what speed does the yellow lamp slip into the ultraviolet?
Play freely. Make the lamp red and drive it at Lin. Send hydrogen’s red line off the infrared edge. Then come back and we’ll make sense of the cube.
What just happened?
The photograph shows the cube turned, by about 64° at \(v = 0.9\). Its side face S, the one facing the camera, is narrowed to 0.87 m, just as the lattice says. But beside it sits the tail T, the trailing face, which no squashed box could ever show you. Seen from far away, no photograph shows the squashed box of the lattice. It shows the cube turned.
Why would a camera see the cube’s tail? Light from the far corner of the tail had farther to travel to the lens than light from the side face, so to arrive at the same instant it had to leave earlier. Earlier, the cube was further back, and that light’s path was clear. By the time the side face’s light left, the tail’s light was already streaming past behind the cube. A photo stitches together pieces of the object from different moments.
And the lamp? Approaching at 0.6, Rae’s yearly flashes reached Lin twice a year. Through Lin’s telescope, Rae’s clock looks fast. Chapter 8 is still right: Lin’s lattice measures each of Rae’s years as 1.25 of Lin’s, so her clock runs slow. Both are true, because the telescope also counts the light’s trip, and that trip gets 0.75 years shorter with every flash as Rae comes closer: \(1.25 - 0.75 = 0.5\). The full story, with the numbers, is in Seen fast, measured slow.
This course calls this pair of words reckoning and seeing. We met it in Chapter 3 with firecrackers. This chapter is about what seeing does at high speed. We’ll come back to the cube with a pencil and the 3-4-5 triangle. First, the same mixing of times happens to clocks and colors, and there it comes with a single number attached.
What the light carries
Radar gives k
Back to the lamp’s numbers. When Rae approached at \(v = 0.6\), her flashes reached Lin twice as often as she sent them. When she receded, they arrived half as often. That 2 is the star of this chapter. We call it \(k\), the Doppler factor. Each speed has one \(k\), and it is never less than 1. At 0.6, \(k = 2\).
\(k\) at a glance
| What Lin receives | Coming closer | Going away |
|---|---|---|
| Time between arrivals | ÷ \(k\): every 0.5 | × \(k\): every 2 |
| How often they arrive (frequency) | × \(k\): twice as often | ÷ \(k\): half as often |
The two rows always go opposite ways. A wavelength is a spacing (you’ll see why under Colors), so it follows the top row.
Quick check, using the table. At \(v = 0.8\), \(k = 3\). (You’ll see where that comes from in a moment.) Rae flies toward Lin at that speed and flashes once a day by her clock. How many days apart do the flashes reach Lin?
That’s the spacing while she goes away. Coming closer, do the flashes bunch up or spread out?
That’s \(\g\) at 0.8: how long each of her days lasts by Lin’s lattice. The question asks what Lin sees, with the shrinking light trip still inside.
Hint
Coming closer, the time between arrivals is divided by \(k\).
Solution
Every \(1/3\) of a day, so three flashes a day. Coming closer divides the spacing by \(k\) and multiplies the frequency by \(k\), as the table above shows.
Here is where the 2 comes from, using nothing but the radar of Chapter 4. Rae leaves Lin at the origin, receding at \(v = 0.6\). Lin sends a flash at \(t = 1\). It catches her where her worldline \(x = 0.6\,t\) meets the light line \(x = t - 1\): at \(t = 2.5\), \(x = 1.5\). How much has Rae aged by then? Her wristwatch time from the origin is \(\sqrt{2.5^2 - 1.5^2} = \sqrt{4} = 2\).
Lin sent at 1; Rae received at 2. A flash sent at \(t = 2\) gives the same picture, twice as big, so Rae receives it at 4. Flashes sent every 1 arrive every 2: going away, the spacing is stretched by \(k = 2\).
Lin, at x = 0, sends flashes at t = 1, 2 and 3. Rae recedes at v = 0.6 and receives them when her clock reads 2, 4 and 6. Her reflections return to Lin at t = 4, 8 and 12.
Describe this lab
A spacetime diagram with time up and x to the right. Lin’s worldline is the vertical line x = 0. Rae’s worldline starts at the origin and recedes at speed v, with tick marks every unit of her wristwatch time. Lin sends flashes at t = 1, 2 and 3. Gold light lines carry each flash out to Rae and back. At v = 0.6 she receives them at her times 2, 4 and 6, and the echoes return to Lin at 4, 8 and 12. A speed slider changes v, from 0.2 to 0.8. A line under the diagram gives k: Rae’s clock reading at R1, divided by 1, the time Lin sent that flash. A table under it lists, for each of the three flashes, when Lin sends it (by Lin’s clock), what Rae’s wristwatch reads when it reaches her, and when the echo gets home (by Lin’s clock). For any v, Rae receives a flash sent at T at her time kT, and the echo returns at k squared times T.
Try it
Read Rae’s wristwatch at R1, R2 and R3: the numbers on her worldline, or the middle column of the table under the diagram. Lin sent those flashes at 1, 2 and 3. Now drag the speed to 0.8. What does Rae’s clock read at R1? That reading is \(k\) at 0.8. The line under the diagram does the division, so you can check.
Now the rule that makes \(k\) powerful. The principle of relativity says Rae’s situation and Lin’s are mirror images: each measures the other receding at 0.6. So if Lin’s flashes, sent every 1, reach Rae every \(k\), then Rae’s flashes, sent every 1, reach Lin every \(k\) too. That lets us follow one flash out and back. Lin sends at \(T\). Rae receives it when her clock reads \(kT\) and reflects it at once, which is the same as sending a fresh flash at \(kT\). That reaches Lin at \(k \cdot kT = k^2 T\). With \(k = 2\): out at 1, echo at 4.
That out-and-back trick finds \(k\) at any speed. Numbers first, with \(k = 2\): Lin sends at 1, and the echo returns at 4. The radar rule puts the bounce at \(t = (1 + 4)/2 = 2.5\) and \(x = (4 - 1)/2 = 1.5\), so Rae’s speed is \(1.5/2.5 = 3/5\). The steps below do the same with letters. Steps 3 and 4 are algebra. On a first read you can take their results and come back later.
- Lin sends at \(T\); the echo returns at \(k^2T\). By the radar rule, the reflection happens at \(t = \tfrac{T + k^2T}{2}\) and \(x = \tfrac{k^2T - T}{2}\). Halfway through the round trip in time; half the round trip in distance.
- Rae’s speed is \(x/t\), so \(v = \dfrac{k^2 - 1}{k^2 + 1}\). Check with \(k = 2\): \(3/5\). The \(T\)’s cancel, as they must: her speed can’t depend on when Lin starts sending.
- Solve for \(k\): \(v k^2 + v = k^2 - 1\), so \(k^2 (1 - v) = 1 + v\), and \(k = \sqrt{\dfrac{1+v}{1-v}}\). Need a refresher? Multiply both sides by \(k^2 + 1\), then gather the \(k^2\) terms on one side.
- At the reflection Rae’s clock reads \(kT\), while Lin’s lattice says \(t = \tfrac{(k^2 + 1)T}{2}\). Their ratio is the stretch factor: \(\g = \dfrac{k + 1/k}{2}\). With \(k = 2\): \(\g = (2 + 0.5)/2 = 1.25\). So \(k\) and \(\g\) are two views of one speed: know one, and you can find the other. Rae’s clock is the one clock present at the origin and at the reflection, so it shows the wristwatch time; the lattice time is \(\g\) times longer.
One formula for both directions. Look at the bottom row of the table: the frequency is multiplied by \(k\) coming closer and by \(1/k\) going away. One formula does both. Give \(\sqrt{(1+v)/(1-v)}\) a negative \(v\) for a source going away, and it turns upside down: at \(v = -0.6\), \(\sqrt{0.4/1.6} = 1/2\), which is \(1/k\). We call that value \(D\), the frequency factor: \(D = k\) coming closer, \(D = 1/k\) going away. \(k\) itself is never less than 1. The function you’ll build is called dopplerFactor(v) for short, but it returns \(D\), the frequency factor.
\[ \begin{gathered} \lit{k} = \sqrt{\frac{1+v}{1-v}} \ \ (\text{speed } v \ge 0), \\ \g = \frac{\lit{k} + 1/\lit{k}}{2}, \\ f_\text{received} = D\, f_\text{emitted}, \\ D = \lit{k} \ \ \text{(approaching)}, \\ D = 1/\lit{k} \ \ \text{(receding)}. \end{gathered} \]
Which symbol is which?
- \(k\), the Doppler factor: how much the spacing of flashes stretches while two observers move apart. Always 1 or more.
- \(1/k\): the spacing squeeze while they come closer. Always 1 or less.
- \(D\), the frequency factor: \(k\) coming closer, \(1/k\) going away. So it can be less than 1. Code:
dopplerFactor(v). - \(\g\), the stretch factor of Chapter 8: what a lattice measures, with no light trip inside.
Quick check. At \(v = 12/13\), \(k = 5\). Rae recedes from Lin at that speed, and Lin sends her a flash once a day. How many days apart do the flashes reach her, by her clock?
That’s the frequency factor, \(1/k\): flashes per day. The question asks for the spacing between them.
Hint
Going away, the spacing is stretched by \(k\).
Solution
Every \(5 \times 1 = 5\) days of her time. Equivalently, the rate is multiplied by \(1/k = 1/5\): one flash per 5 days.
Why, really?
Why must Lin see Rae with the same \(k\) that Rae sees Lin? Because the flashes can’t know who is “really” moving. A train of flashes is fixed by two worldlines and the 45° light lines between them. Swap the names Lin and Rae and you get the same drawing, mirrored. If \(k\) came out different for the two of them, either one could count flashes and learn which of them was moving. Galileo’s ship says nobody can.
Seen fast, measured slow
Back to the lamp you predicted. Approaching at 0.6, Rae’s yearly flashes reach Lin twice a year: through the telescope her clock looks fast. Yet Chapter 8 measured moving clocks running slow, and it was right. Both are true. Here is how, step by step.
Now point the radar logic at a clock. Rae flies toward Lin at 0.6 with a clock that flashes once a year by her wristwatch. Lin’s telescope receives a flash every half year. Through the eyepiece Rae’s clock looks twice as fast as Lin’s own. That’s the right panel of Plate 12.1.
Meanwhile Lin’s lattice notes when each flash leaves, using the lattice clock right where Rae is. It measures one year of Rae’s clock as \(\g = 1.25\) of Lin’s years. Slow.
The difference is the trip the light takes. Between two flashes, 1.25 years by the lattice, Rae comes \(0.6 \times 1.25 = 0.75\) light-years closer. So the second flash has 0.75 years less to travel than the first. Seen spacing: \(1.25 - 0.75 = 0.5\). When she goes away, each flash has 0.75 years more to travel: \(1.25 + 0.75 = 2\).
Try it
Back in the Doppler lamp, set \(v = 0.6\). Read Lin’s received spacing approaching, then receding, and average the two before you read on. Which number from Chapter 8 comes out?
Here is the test of what’s real. Average the two views: \((0.5 + 2)/2 = 1.25\). The light delays cancel, and what’s left over is the stretch factor. Time stretching is whatever remains after every delay has been accounted for. In symbols, it’s the last line of the worked example: \(\g = (k + 1/k)/2\).
Myth
“Moving clocks look slow.” An approaching clock looks fast, by the factor \(k\); a receding one looks slow, by \(1/k\). Measured by a lattice, which takes the light’s travel time out, both run slow by the same \(\g\). “Looks” and “is measured” are different verbs, and they give different numbers.
Colors
A light wave is a clock too: its crests are ticks. In meters of time, a wave’s length and its period are the same number. Yellow light of 589 nm, the color of a sodium street lamp, is a clock that ticks every 589 nm of time. So everything we said about flashes applies to crests. Approaching, crests arrive \(k\) times as often: the frequency is multiplied by \(k\), and the wavelength divided by \(k\). Receding, the other way round.
At \(v = 0.6\), approaching, 589 nm arrives as \(589/2 \approx 295\) nm, which is ultraviolet. The yellow lamp vanishes from human sight while it rushes toward you.
Quick check. How fast must a sodium lamp (589 nm) approach you for its light to arrive at 380 nm, the violet edge of human sight?
That’s \(k\), the factor the wavelength is divided by. The question asks for the speed. Step 2 of the worked example turns \(k\) into \(v\).
That’s \(1 - 380/589\), the everyday Doppler guess. Relativity’s \(k\) is \(\sqrt{(1+v)/(1-v)}\), not \(1/(1 - v)\).
Hint
First find \(k = 589/380\). Then use \(v = (k^2 - 1)/(k^2 + 1)\).
Solution
\(k = 589/380 = 1.55\), so \(k^2 = 2.403\) and \(v = 1.403/3.403 \approx 0.41\). Above about 0.41 the approaching yellow lamp has slipped out of sight into the ultraviolet, as the lamp lab showed.
Atoms are the most dependable lamps in the universe. Hydrogen glows at the same visible wavelengths (we track three of them) in a lab on Earth and in a galaxy far away. So when an astronomer finds the familiar pattern shifted, the shift gives the speed. Every wavelength changes by the same factor, which is why the lamp lab draws its spectrum on a log scale: there, all three lines slide by the same distance.
Real numbers
Hα light has a wavelength of 656.3 nm, so it is a clock with a period of 656.3 nm of time. In seconds, divide by light’s speed, \(299\,792\,458\un{m/s}\): \(656.3 \times 10^{-9}\un{m} \div 299\,792\,458\un{m/s} = 2.19 \times 10^{-15}\un{s}\).
A police radar gun is radar with \(k\). Aim one, working at about 24 GHz, at a car doing 30 m/s. In our units the car’s speed is \(v = 30/299\,792\,458 = 1.0 \times 10^{-7}\). The echo comes back shifted twice, by \(k^2 \approx 1 + 2v\) Need a refresher?, a change of about \(2v \times 24\un{GHz} \approx 4.8\un{kHz}\). The gun counts that beat and shows you the ticket.
The headlight
Light can’t change its speed. It can change its direction, depending on who measures it. Go back to the clock made of light (Chapter 5). In Rae’s frame its light goes straight up, at 90° to her motion. In Lin’s frame the same light runs along the slant of a 3-4-5 triangle: 0.6 forward for every 0.8 up, and 1 along the slant. Its angle from the forward direction has \(\cos\theta = 0.6/1\). That’s 53°, not 90°.
So light that Rae sends sideways, Lin measures leaning forward. Everything Rae shines into her forward half, all the angles up to 90°, lands for Lin inside a cone of 53°. A fast lamp throws its light ahead of itself, like a headlight. Physicists call the change of direction aberration.
Optional algebra, for any angle. If you skip the next paragraph, keep the 3-4-5 picture and its result: light sent sideways lands at \(\cos\theta_\text{lab} = v\). For a ray at any angle \(\theta'\) in Rae’s frame, take the event one unit of her time along the ray: \(t' = 1\), \(x' = \cos\theta'\), \(y' = \sin\theta'\). The Lorentz machine of Chapter 10, run backward, gives Lin’s \(t = \g(1 + v\cos\theta')\) and \(x = \g(\cos\theta' + v)\). Light travels 1 per unit of time, so \(\cos\theta_\text{lab} = x/t\):
\[ \cos\theta_\text{lab} = \frac{\cos\theta' + v}{1 + v\cos\theta'}. \]
Check it. Sideways light, \(\cos\theta' = 0\), gives \(\cos\theta_\text{lab} = v\): 53.13° at 0.6, 36.87° at 0.8. Straight ahead stays straight ahead (\(1 \to 1\)), and straight back stays straight back.
Quick check. At \(v = 0.6\), Rae’s lamp sends a ray 120° from her forward direction, so it leans backward (\(\cos\theta' = -0.5\)). At what angle from forward does Lin measure it, in degrees?
That’s the angle in Rae’s frame. Put \(\cos\theta' = -0.5\) and \(v = 0.6\) into the rule: which way does the ray swing?
That has the sign of the top line flipped. The top is \(\cos\theta' + v = -0.5 + 0.6\).
Hint
The top is \(-0.5 + 0.6 = 0.1\); the bottom is \(1 + 0.6 \times (-0.5) = 0.7\).
Solution
\(\cos\theta_\text{lab} = 0.1/0.7 = 1/7\), so \(\theta_\text{lab} = 81.79°\). A ray Rae aims backward of sideways, Lin measures leaning forward: that is the headlight.
Now turn it around and ride with Rae, looking out at the stars, which sit at rest around Lin. From Rae’s seat the stars are the moving lamps, all streaming backward past her at \(v\), so the same rule applies with the roles swapped. It moves every star’s image toward the direction she’s heading. The frequency factor \(D\) colors them: straight ahead \(D = k\), shifted toward the blue; straight behind \(D = 1/k\), toward the red.
The sky from Rae’s rocket: at rest the stars fill the sky evenly, and half of them lie within 90° of straight ahead. At v = 0.6 that half crowds into a cone of 53°, and at 0.9 into 25.8°. Stars ahead turn blue-white and brighten; stars behind turn red and fade.
Describe this lab
A round map of the whole sky around Rae’s rocket. The center is straight ahead; the distance from the center is the angle from the direction of motion, with rings at 30, 60, 90 (abeam), 120 and 150 degrees, and the rim is straight behind. About 400 stars sit at rest, spread evenly. As the speed v rises, aberration moves every star toward the center, and a gold dashed circle shows the cone that holds half of all the stars: 90 degrees at rest, 53.1 at 0.6, 25.8 at 0.9. With Doppler colors on, each star is treated as a glowing body whose temperature is multiplied by its shift: stars ahead turn blue-white and brighter, stars behind turn red and dim. Options: show trails from each star’s rest position, and mark 16 beacon stars that sit exactly abeam at rest. The Accelerate button raises the speed steadily to 0.99 in about four seconds. Readouts give the half-sky angle, the share of stars ahead of abeam, and the shift ahead and behind. Keys: Space accelerates or pauses, C toggles colors, T toggles trails, B toggles beacons.
Try it
Read the half-sky angle at \(v = 0.6\) and at 0.9. Then find the speed that squeezes half of all the stars into a cone of 30° around straight ahead. (It’s one of the friendly speeds, \(\sqrt{3}/2 \approx 0.866\). The slider gets as close as 0.87.) The marked stars, the beacons, sat exactly sideways (“abeam”) at rest. Where are they at that speed?
Why, really?
Run through rain that falls straight down and it hits your face: from where you are, the rain seems to come from ahead. Starlight does the same thing to a moving telescope. The rain picture is only a cartoon, because rain adds speeds the old way and light refuses to. But the forward lean is real, and for light the exact amount is the formula above.
Why the cube turns
Now the cube, at \(v = 0.6\) so the numbers come out friendly. Put the camera far away, looking straight at the track, so all the light it collects travels the same direction: straight toward the lens. Light from the two corners of the side face S, the face toward the camera, leaves at the same moment and travels the same distance. So the photo shows that face exactly as wide as the lattice measures it: \(L/\g = 0.8\,L\).
The tail T, the trailing face, is the surprise. Light from its far corner starts one cube-depth \(L\) farther from the camera, so to arrive with the rest it must leave \(L\) earlier. Back then, the cube was \(0.6\,L\) further back. That light slips past behind the cube, which has since moved out of its way, and lands in the photo \(0.6\,L\) to the left of the side face.
Side by side: 0.8 L of side face and 0.6 L of tail. That is exactly what an ordinary cube at rest looks like when it’s turned by the angle whose cosine is 0.8 and whose sine is 0.6. The 3-4-5 triangle again. At any speed, a head-on photo shows the cube turned by the angle with \(\sin\alpha = v\): 37° at 0.6, 64° at 0.9.
The same delays stretch or squeeze a moving rod’s photo. While it approaches, light from its far end left earlier, when that end was further back, so the photo is longer than the lattice length. While it goes away, the photo is shorter.
Myth
“Length contraction is something you can see, as a squashing.” A photograph of a fast sphere still shows a circle, and a fast cube photographed head-on looks turned, not flattened. Contraction is a fact about which events the lattice counts as “now”. A photograph is a fact about which light arrives together.
Try it
Go back to the camera lab, pick the cube, set \(v = 0.6\) and press Center the photo. Does the readout say 37°? Is the tail T in the camera panel narrower than the side face S, as the 0.6 and 0.8 above say it should be?
Deep dive: whose 90°? The transverse Doppler shift
The headlight rule hides a second gift: the color of every ray. Run the flash argument from “Seen fast, measured slow” along a ray that leaves at the angle \(\theta_\text{lab}\) in Lin’s frame. By Lin’s lattice, Rae’s lamp sends a crest every \(\g\) of its own periods (time stretching), and between crests it moves \(\g v\) along its motion. Only the part of that step along the ray, \(\g v\cos\theta_\text{lab}\), shortens the next crest’s trip to a far-away eye. So crests arrive every \(\g(1 - v\cos\theta_\text{lab})\) periods:
\[ \begin{aligned} f_\text{lab} &= \frac{f_\text{emitted}}{\g(1 - v\cos\theta_\text{lab})} \\ &= \g(1 + v\cos\theta')\, f_\text{emitted}. \end{aligned} \]
The second form comes from putting the aberration rule into the first. Check both ends: straight ahead, \(1/(\g(1 - v)) = \g(1 + v) = k\); straight behind, \(1/k\). As they must.
Now aim sideways. If “sideways” means 90° in Rae’s frame, \(\cos\theta' = 0\) and the second form gives \(\g\): at 0.6, the light arrives bluer by 1.25. If it means 90° in Lin’s frame, \(\cos\theta_\text{lab} = 0\) and the first form gives \(1/\g\): redder, by 0.8. Both are called “the transverse Doppler shift”, and they are different rays. The red one is pure time stretching: a lamp passing at right angles has no light delay to speed it up or slow it down, so you see its clock running slow by \(\g\).
Looking exactly sideways is hard, because a tiny error in the angle lets the much bigger first-order shift leak in. So the classic Ives–Stilwell method avoids it. It looks along the beam both ways and averages, as you’ll see in a moment.
That’s the toolkit: \(k\) for clocks and colors, the aberration rule for directions, and both together for pictures. Has anyone actually seen any of this?
Ask Nature
Nature has been running these experiments for us, in laboratories and across the sky.
Ask Nature
Atoms as moving clocks. In 1938 Herbert Ives and G. R. Stilwell sent a fast beam of hydrogen down a tube (molecular ions that broke up into glowing hydrogen atoms). Each atom is a tiny lamp, and its blue-green Hβ light, 486.1 nm at rest, is its clock. With a mirror they looked along the beam both ways at once. Light an atom sends forward is chased by the atom, like Rae’s lamp coming toward Lin. Light it sends backward is left behind, like Rae’s lamp going away.
Why look both ways? Go back to Rae’s lamp at 0.6. By Lin’s lattice each tick lasts 1.25. Coming closer, each flash’s trip is 0.75 shorter than the last, so Lin sees a tick every \(1.25 - 0.75 = 0.5\). Going away, each trip is 0.75 longer: every \(1.25 + 0.75 = 2\). The light-trip part is the same size both ways, with opposite signs. So averaging the two views cancels it: \((0.5 + 2)/2 = 1.25\). What is left is the clock’s own slowing, \(\g\), as Lin’s lattice measures it.
Now suppose moving clocks didn’t slow. By Lin’s lattice each tick would last 1, and the trip would change by 0.6 per tick: seen every 0.4 and every 1.6. Average: exactly 1, the rest value. The two shifted lines would sit evenly on either side of the line at rest. Time stretching multiplies both by 1.25, and the midpoint moves to 1.25. For light, a crest’s spacing is its wavelength: forward, the wavelength is divided by \(k\); backward, it is multiplied by \(k\).
See the trick with friendly numbers. A lamp of 500 nm is watched along its beam at \(v = 0.6\), forward and backward at once. Where is the midpoint of the two shifted lines, in nm?
That’s the midpoint if moving clocks didn’t slow: \(500 \times 0.4 = 200\) and \(500 \times 1.6 = 800\) average to 500. With time stretching the lines sit at \(500/k\) and \(500\,k\). Where are they?
That’s the two lines added together. The midpoint is halfway between them.
Hint
At 0.6, \(k = 2\): the forward line arrives at 250 nm, the backward one at 1000 nm.
Solution
\((250 + 1000)/2 = 625\) nm, which is \(\g = 1.25\) times 500 nm. Without time stretching the lines would sit at 200 and 800, with the midpoint at 500. Each real line is that value times 1.25 (\(200 \times 1.25 = 250\), \(800 \times 1.25 = 1000\)), so the midpoint is too. Averaging cancels the light trip; what is left is the stretch.
Real numbers
Ives and Stilwell’s atoms were far slower. Take an illustrative \(v = 0.005\). If moving clocks didn’t slow, Hβ would arrive at \(486.1 \times 0.995 \approx 483.67\) nm forward and \(486.1 \times 1.005 \approx 488.53\) nm backward, with the midpoint at exactly 486.1. Time stretching moves both lines about 0.006 nm toward the red, and the midpoint to 486.106 nm. That tiny extra shift is the wavelength times \(v^2/2\), because at small speeds \(\g \approx 1 + v^2/2\) (it’s on the formula card). Ives and Stilwell saw that tiny extra shift. Ives himself read it as support for an ether theory. Physicists today read it as the clocks inside the atoms running slow, measured in the lab: the same time stretching as Chapter 8’s muons.
H. E. Ives and G. R. Stilwell, J. Opt. Soc. Am. 28, 215 (1938)Why did Ives and Stilwell compare the forward and backward lines, instead of measuring one of them?
Hint
The lamp at 0.6 was seen every \(1.25 - 0.75\) and every \(1.25 + 0.75\). What happens to the 0.75 when you average?
Solution
The light-trip part, \(\pm 0.75\) at 0.6, is equal and opposite, so the average cancels it and leaves 1.25, the time stretching. One line alone would mix the two effects.
Ask Nature
The sky is full of shifted hydrogen. Astronomers read speeds this way every night. A star’s whole pattern of lines, shifted together, tells how fast it moves toward or away from us. Two stars orbiting each other swing back and forth between bluer and redder. Many planets around other stars were found from the tiny back-and-forth shift of their star’s lines. For very distant galaxies the redshift is mostly something else, the stretching of space itself, which belongs to general relativity.
Ask Nature
Faster than light, apparently. Some quasars shoot out blobs of glowing gas. Radio telescopes following them for years have measured apparent sideways speeds across the sky of several times light speed, and in a few jets more than thirty times. Nothing outruns light. It’s seeing again.
Take a blob moving at \(v = 0.8\), aimed 37° from our line of sight, so it comes 0.64 toward us and 0.48 sideways per year. The light from its second position has 0.64 light-years less to travel, so it arrives only \(1 - 0.64 = 0.36\) years after the first. Apparent sideways speed: \(0.48/0.36 = 1.33\). In general \(v_\text{app} = \dfrac{v\sin\varphi}{1 - v\cos\varphi}\), and its biggest value is \(\g v\). First seen in the quasar 3C 279: A. R. Whitney et al., Science 173, 225 (1971)
Practice
◇ back of the envelope, ◆ pencil and paper, ★ head-scratcher. Answers like 5/3 or sqrt(3)/2 are fine.
A lamp moves at \(v = 0.6\). By what factor is the frequency you receive multiplied, first when it approaches you, then when it recedes? (These are the values of \(D\) at \(v = 0.6\) and \(v = -0.6\).)
That’s \(\g\), the stretch factor. \(k\) includes the light delay too. What did the radar picture give?
That’s the receding value. Does an approaching source deliver its flashes more often or less often?
That’s \(k\) itself: in the radar picture, going away stretches the spacing by \(k = 2\). The frequency goes the other way. What is it multiplied by?
Hint
Approaching, the frequency is multiplied by \(k = \sqrt{(1 + v)/(1 - v)}\); receding, by \(1/k\). Same as putting \(v = -0.6\) into the formula.
Solution
Approaching: \(\sqrt{1.6/0.4} = \sqrt{4} = 2\). Receding, \(v = -0.6\): \(\sqrt{0.4/1.6} = 1/2\). The two are reciprocals.
A star emits light of 500 nm. You fly toward it at \(v = 0.6\). What wavelength do you receive?
That’s the receding answer. Approaching, do the crests arrive more often or less often?
That divides by \(\g = 1.25\). The light delay matters too. What is \(k\) at 0.6?
Hint
Approaching multiplies the frequency by \(k\), so it divides the wavelength by \(k\).
Solution
\(k = 2\), so \(500/2 = 250\) nm: ultraviolet, invisible to your eye. It doesn’t matter that you move and the star sits still; only the relative speed counts.
A lamp moving at \(v = 0.8\) sends light at 90° to its motion, in its own frame. At what angle from the direction of motion does Lin measure that light, in degrees?
That’s the angle at \(v = 0.6\). Which triangle goes with 0.8?
That’s the angle in the lamp’s own frame. In Lin’s frame the lamp moves forward while the light climbs. Does the light lean?
Hint
Sideways light has \(\cos\theta' = 0\), so \(\cos\theta_\text{lab} = v\).
Hint
It’s the 3-4-5 triangle with the 4 lying forward: 0.8 forward, 0.6 up, 1 along the slant.
Solution
\(\cos\theta_\text{lab} = (0 + 0.8)/(1 + 0) = 0.8\), so \(\theta_\text{lab} = 36.87°\). Half the lamp’s light now fits in a cone of 37°.
A rod moves along its length at \(v = 0.6\). Lin’s lattice measures it as 4 m long. Is its photographed length 4 m?
Hint
In the camera lab, pick the rod, set 0.6, and drag the lattice position from left to right (or press Fly past).
Solution
Not in general. Light from the two ends that arrives together left at different times. While the rod approaches, the far end’s light left earlier, when that end was further back, so the image is stretched. While it recedes, the image is squeezed. Only when its light reaches the camera head-on does the photo show the lattice length, 4 m, and even then the rod looks turned.
A Doppler factor of \(k = 3\) goes with what stretch factor \(\g\)?
\(k\) and \(\g\) are different numbers. Which formula joins them?
That’s \(k/2\). The formula averages \(k\) with its reciprocal. What’s \(1/k\)?
Hint
\(\g = (k + 1/k)/2\).
Solution
\(\g = (3 + 1/3)/2 = (10/3)/2 = 5/3\). That’s the friendly speed \(v = 0.8\): check with \(v = (k^2 - 1)/(k^2 + 1) = 8/10\).
Rae approaches Lin at \(v = 0.6\). Her clock ticks once per year of her own time. How many ticks per Lin-year does Lin see through her telescope?
That’s \(1/\g\): the rate Lin’s lattice measures, once the light delay is taken out. The question asks what she sees. Is the delay growing or shrinking?
That would be a receding clock. Rae is coming closer.
That’s \(\g\): how many of Lin’s years one of Rae’s ticks lasts, by Lin’s lattice. The question asks how many ticks per year Lin sees.
Hint
Seeing is counting arriving flashes. What does \(k\) count?
Solution
Seen rate \(= k \times 1 = 2\) ticks per year. Her clock looks twice as fast, even though Lin’s lattice measures it running slow, at \(1/1.25 = 0.8\) ticks per year.
Rae’s approaching clock looks fast. Why do we still say it runs slow?
Model answer
What Lin sees includes the light’s travel time, and while Rae approaches that travel time shrinks with every tick, so the flashes bunch up and her clock looks fast by \(k\). Lin’s lattice takes the travel time out: it records when each tick happens, using the clock right where it happens. Then each of Rae’s ticks lasts \(\g\) of Lin’s. Seeing and measuring answer different questions. Averaging the approaching and receding views, \((k + 1/k)/2\), cancels the delays and leaves exactly \(\g\).
- Seeing includes a light-travel delay that shrinks as she approaches
- Correcting for the delay (or averaging k and 1/k) leaves the factor γ, slow
- Seeing and measuring answer different questions
Finish the worked example. Rae recedes from Lin at \(v = 0.8\). Lin sends a flash at \(t = 2\).
- At 0.8, \(k = \sqrt{1.8/0.2} = 3\). Rae is going away, so the spacing is stretched by \(k\): Lin’s flashes, sent every 1, reach her every 3 of her time.
- Rae receives Lin’s flash when her clock reads ___.
- She reflects it at once; the echo reaches Lin at \(t = \) ___.
- Check: the radar rule puts the reflection at \(t = (2 + \text{echo})/2\) and \(x = (\text{echo} - 2)/2\), and \(x/t\) should be 0.8.
That multiplies by \(1/3\). In the radar picture Rae is moving away: do flashes reach her more often or less often than Lin sends them?
10 is when the flash reaches her by Lin’s lattice. The echo still has to come back. Try \(k \times\) her reading.
Hint
Out: Rae’s clock reads \(3 \times 2\). Back: the reflection is a flash she sends at that reading.
Solution
Her clock reads \(3 \times 2 = 6\); the echo arrives at \(3 \times 6 = 18\), which is \(k^2 = 9\) times 2. Check: reflection at \(t = 10\), \(x = 8\), so \(v = 0.8\). And Rae’s clock reads 6 where the lattice reads 10: \(\g = 10/6 = 5/3\).
A black hole in our own galaxy shoots out a blob of hot gas, which flies straight away from us at \(v = 5/13\). Its hydrogen glows at Hα, 656.3 nm. At what wavelength does that light arrive?
That divides by \(k = 3/2\), which would be the approaching answer. Receding stretches the wavelength.
Hint
\(v = 5/13\) is on the friendly list: approaching, \(k = 3/2\).
Solution
Receding, the wavelength is multiplied by \(3/2\): \(656.3 \times 1.5 = 984.45\) nm. That’s infrared: the blob’s red hydrogen glow has slid out of sight.
Sort each statement: is it about what reaches one eye or camera, or about what the lattice reckons?
- Through a telescope, Rae’s approaching clock ticks twice a year.
- Each of Rae’s ticks lasts 1.25 years of Lin’s time.
- A photo shows a fast cube turned, with its tail showing.
- A cube moving at 0.9 is 0.87 m long along its motion.
- A quasar blob crosses the sky at 1.33 times light speed.
- The blob moves 0.8 light-years per year through space.
Hint
If the light’s travel time is still inside the number, it’s seeing.
Solution
Seeing: the telescope rate, the photo and the blob’s sideways speed across the sky. Measuring: the 1.25 years per tick, the 0.87 m and the 0.8 light-years per year.
Rae switches from receding at 0.6 to approaching at 0.6. Tick every quantity that changes.
Hint
Anything that includes the light delay can tell coming from going. Anything that removes it can’t.
Solution
The frequency factor (from \(1/k\) to \(k\)) and the seen color change; \(\g = 1.25\) and the measured tick (1.25 years) don’t.
A quasar blob moves at \(v = 0.8\) at the angle \(\varphi\) from our line of sight with \(\cos\varphi = 0.8\). What sideways speed across the sky do we see?
That’s the true sideways speed, \(v\sin\varphi\). The pictures of the blob arrive closer together than they were taken. By how much?
Hint
Per year it comes \(0.8 \times 0.8 = 0.64\) toward us and \(0.8 \times 0.6 = 0.48\) sideways.
Hint
The second picture’s light has 0.64 light-years less to travel.
Solution
The pictures arrive \(1 - 0.64 = 0.36\) years apart, so the blob seems to cross \(0.48\) light-years of sky in 0.36 years: \(0.48/0.36 = 4/3 \approx 1.33\). This angle, \(\cos\varphi = v\), gives the biggest apparent speed, \(\g v = 4/3\).
At what speed do half of all the stars crowd into a cone of 30° around Rae’s direction of travel?
Close, but the slider’s nearest value is 0.87. The exact speed is \(\cos 30°\). Can you write it with a square root?
That’s \(\sin 30°\). Which star marks the edge of the half-sky cone, and where does the formula put it?
Hint
The stars that sit abeam at rest (90°) mark the edge of the half. Where do they appear?
Solution
Stars at 90° appear at \(\cos\theta = v\). For a 30° cone, \(v = \cos 30° = \sqrt{3}/2 \approx 0.866\), the friendly speed with \(\g = 2\). The slider’s 0.87 counts too. Check it in the pilot’s window.
Keep the formula card nearby: \(k\), \(\g = (k + 1/k)/2\) and the aberration rule are all on it.
Build it
Two functions, and then the lamp and the pilot’s window run on your code. First \(D\), the frequency factor from Radar gives k.
Code you need
Nothing new, just one trick. Math.abs(x) is the size of x without its sign. Every comparison with NaN is false, so !(Math.abs(x) < 1) catches both “too big” and “not a number” at once:
if (!(Math.abs(x) < 100)) throw new RangeError('too big');
return x / 2;Build it: dopplerFactor(v)
Positive v means the source approaches; negative means it recedes. Your function returns \(D\): \(k = 2\) for dopplerFactor(0.6), and \(1/k = 0.5\) for dopplerFactor(-0.6). It must refuse v = 1: no lamp rides a light beam. A property test also checks that dopplerFactor(v) * dopplerFactor(-v) is 1 for 50 random speeds, because coming and going undo each other. Once it passes, the Doppler lamp and the pilot’s window run on your function.
The bonus moves the stars themselves.
Bonus: aberration(cosThetaPrime, v)
It takes the cosine of an angle in the source’s frame and returns the cosine of the angle in the lab: \(\dfrac{\cos\theta' + v}{1 + v\cos\theta'}\). Sideways light at 0.6 must come out at 0.6. Once both functions pass, every star in the pilot’s window is placed by your formula and colored by your \(D\).
Checkpoint
Five scored items plus one written answer. Feedback waits until you submit.
Rae recedes from Lin at \(v = 0.8\). By what factor is the rate of her flashes multiplied, as Lin receives them? (That’s \(D\) at \(v = -0.8\).)
That’s the approaching value. Receding flashes arrive less often.
That’s \(1/k\) at 0.6. This item uses \(v = 0.8\), where \(k = 3\). Receding, the rate is multiplied by \(1/k\).
That’s \(k\) at 0.6, coming closer. This item uses \(v = 0.8\), going away.
Solution
\(\sqrt{(1 - 0.8)/(1 + 0.8)} = \sqrt{1/9} = 1/3\), which is \(1/k\) for \(k = 3\).
Rae’s clock approaches Lin at \(v = 0.6\), ticking once a year by her wristwatch. Lin’s telescope shows it ticking twice a year. What does Lin’s lattice measure?
Solution
The lattice takes out the shrinking light delay and finds \(\g = 1.25\) years per tick: slow. The telescope’s 0.5 includes the delay.
Rae flies at 0.6, first away from Lin, then toward her. Which quantities stay the same, and which flip between \(k\) and \(1/k\)?
- The stretch factor \(\g\)
- How often Lin receives her flashes
- The frequency of her lamp’s light as Lin receives it
- How long Lin’s lattice measures each of her ticks
Solution
Anything that includes the light delay flips: the received flash rate and the received frequency. What the lattice measures, and \(\g\), depend only on the speed.
Rae leaves the origin, receding at \(v = 0.6\). Lin sends a flash at \(t = 1\). Place the event where the flash reaches Rae.
No pointer? Type t and x into the boxes under the diagram.
Solution
Her worldline \(x = 0.6\,t\) meets the light line \(x = t - 1\) at \(t = 2.5\), \(x = 1.5\). Her clock reads \(\sqrt{2.5^2 - 1.5^2} = 2 = k \times 1\) there.
A sphere flies past a camera at \(v = 0.9\). In the photograph, its outline is…
Solution
A circle. Contraction is a statement about the lattice; in a photo the light delays turn the sphere instead of squashing it, and a turned sphere still has a round outline.
In two sentences: why does a fast cube photographed head-on look turned, not squashed?
Model answer
The side face toward the camera shows its lattice width, \(L/\g\), because light from its corners leaves together. Light from the tail had farther to go, so it left earlier, when the cube was further back, and shows the tail beside the side face: together they look like a turned cube.
- Light from different parts leaves at different times
- The tail, the trailing face, becomes visible
Explain it to Sagredo
Sagredo says: “If a photograph of a fast cube shows it turned, then contraction is an illusion. Nothing really gets shorter.” Answer him.
Model answer
Contraction is a statement about the lattice: where both ends of the cube are at one instant in Lin’s frame. That measurement really gives \(L/\g\), and it has consequences you can count, like the muons’ shortened atmosphere. A photograph answers a different question: which light reaches one lens at one moment. That light left different parts of the cube at different times, which turns the picture. Neither is an illusion. They’re honest answers to different questions, and only the lattice’s is a length.
- Contraction is about simultaneous positions in one frame (the lattice)
- A photo is about light arriving together, which left at different times
- Both are real; they answer different questions
What’s invariant here?
- The count of flashes. Each flash is an event, and every frame agrees on which events happen: nobody disagrees about how many flashes are sent or received.
- The received spacing. While they move apart, flashes sent every 1 arrive every \(k\). Each number is read on one wristwatch: the sender’s, or the receiver’s. So every frame agrees on \(k\). The difference is sending versus receiving, not Lin versus Rae.
- The wristwatch time between two of Rae’s flashes: one year on her clock, \(\inv{\tau^2} = \lab{\dt^2} - \lab{\dx^2}\) for every observer.
- Light’s speed, in every direction. Its color (Doppler) and its direction (aberration) change with the observer; the 45° light lines never do.
- The same factor both ways. Lin receives Rae’s flashes exactly as Rae receives Lin’s: every \(1/k\) while they close in, every \(k\) while they separate. Only their relative speed and its direction matter: \(k = \sqrt{(1+v)/(1-v)}\), \(\g = (k + 1/k)/2\).
Still strange?
A photograph is an honest record of light, a lattice is an honest record of events, and they disagree about the shape of a cube. Nobody has to be wrong: relativity only promises to keep the events straight, and the light does whatever the events say.
Going further
Deep dive: is there really a starbow?
Science-fiction writers once imagined that a fast starship would see the stars gather into a rainbow ring: blue ahead, red behind, a band of every color between. It’s a beautiful picture, and it would be right if every star shone in a single color. Real stars glow at all colors at once, like any hot body. Shifting a hot body’s light by the frequency factor \(D\) for that direction (\(D = k\) straight ahead, \(1/k\) straight behind) gives the glow of a body \(D\) times hotter. So the forward sky turns blue-white and brighter, and the rear sky turns red, dims and finally drops out of sight. Careful calculations in the late 1970s found no rainbow ring. J. M. McKinley and P. Doherty, Am. J. Phys. 47, 309 (1979) The pilot’s window above draws the honest version.
★★ From the radar picture, show that \(\g v = \dfrac{k - 1/k}{2}\).
Model answer
The reflection happens at \(x = \tfrac{(k^2 - 1)T}{2}\) by Lin’s lattice, and Rae’s clock reads \(kT\) there. Distance per unit of her own time is \(\g v\), so \(\g v = \dfrac{(k^2 - 1)T}{2kT} = \dfrac{k - 1/k}{2}\). Check with \(k = 2\): \(0.75 = 1.25 \times 0.6\).
- Uses the radar position of the reflection
- Divides by Rae’s clock reading, kT
- Checks the friendly numbers
★★ Back to the twins of Chapter 11. Rae flies to a star 3 light-years away at 0.6 and back; Lin sends a birthday card every year. Rae’s outbound leg lasts 4 years of her time. How many of Lin’s cards reach her on the way out?
That’s the way home, when she flies toward the cards. Going away, do they arrive more often or less often?
That’s how long the outbound leg lasts on her clock. How many of Lin’s cards reach her per year of hers while she moves away?
Hint
Going away, the cards reach her at the rate \(1/k\) per year of her time.
Solution
\(4 \times \tfrac{1}{2} = 2\) cards. On the way home she meets them at \(k = 2\) per year: \(4 \times 2 = 8\). Total \(2 + 8 = 10\), one for each of Lin’s 10 years, as in Chapter 11.
If you liked the radar picture, Hermann Bondi’s Relativity and Common Sense builds all of special relativity from \(k\). Next, the factor \(k\) gets a superpower: when speeds combine, Doppler factors multiply.