Events, clocks, and spacetime
Muons are made when cosmic rays strike the upper atmosphere, around 15 km up. A muon at rest lives on average only 2.2 millionths of a second, and even at the speed of light it could cover only 660 m in that time. Yet muons arrive at the ground in great numbers. How do they make a journey more than twenty lifetimes long?
This is the first of four optional essays. The core of the book treated particles moving slowly compared with light. Here we drop that restriction, which needs Einstein's special relativity. This essay builds its central ideas from two principles and one thought experiment: clocks in motion run slow, lengths along the motion shrink, and "at the same time" depends on who is asking. The next essay turns these into the relation between energy, momentum and mass that relativistic quantum mechanics needs.
Two principles that seem to clash
The principle of relativity, going back to Galileo, says that the laws of physics are the same in every frame moving at constant velocity. No experiment done inside a smoothly moving train with the blinds down can tell you the train is moving. The second principle is an experimental fact about light: its speed in empty space is always $c = 3.00\times10^{8}$ m/s, whatever the motion of its source, and the most careful attempts, most famously Michelson and Morley's in 1887, found no dependence on the motion of the observer either.
Put together, the two seem to contradict each other. A light ray passes a train moving at $v$ in the same direction. By the rule that speeds subtract, someone on the train should measure it moving at $c - v$. By the principle of relativity, the train is as good a frame as the ground, so the law "light moves at $c$" must hold there too. In 1905 Einstein kept both principles and gave up the assumption that made them clash: that time runs the same for everyone.
A clock made of light
The cleanest way to see what happens to time is a clock that ticks by bouncing a pulse of light between two mirrors a distance $d$ apart. At rest, one tick, up and down, takes $2d/c$. Now put the clock on a train moving sideways at speed $v$. Seen from the ground, the pulse no longer goes straight up and down: the mirrors move while it travels, so it follows a longer diagonal path. It still travels at $c$, the second principle insists, so each tick takes longer. For half a tick of ground time $t$, the diagonal has length $ct$, the sideways distance is $vt$, and the height is still $d$, so by Pythagoras
$$(ct)^2 = d^2 + (vt)^2,\qquad t = \frac{d/c}{\sqrt{1 - v^2/c^2}} .$$The moving clock's tick, measured by ground clocks, is longer than its tick at rest by the factor
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} .$$By the principle of relativity every kind of clock, including a muon's internal decay, must keep time with a light clock moving alongside it, or the two could be compared to detect the motion. So a moving clock of any kind runs slow by $\gamma$. This is time dilation. At everyday speeds $\gamma$ differs from 1 by a fraction $\tfrac12 v^2/c^2$: about $3\times10^{-13}$ for an airliner, which is why nobody noticed.
Worked example: the muon's journey
Take a muon made 15 km up and moving at $v = 0.999c$, and suppose it keeps that speed all the way down. (Real muons lose some energy in the air; this example neglects that.) Then $\gamma = 1/\sqrt{1 - 0.998} = 22.4$.
From the ground. The trip takes $15\ \text{km}/(0.999c) = 50.1$ μs, which is 22.8 of the muon's rest lifetimes of 2.197 μs. Without time dilation, the fraction surviving would be $e^{-22.8} \approx 10^{-10}$. But the muon's own clock runs slow by 22.4, so for the muon only $50.1/22.4 = 2.24$ μs pass, about one lifetime, and a fraction $e^{-1.02} = 0.36$ survives.
From the muon. The muon is entitled to regard itself as at rest, with its clock running normally. Then it is the atmosphere that rushes past at $0.999c$, and the only way the two accounts can agree on the same outcome is if the atmosphere's thickness, measured in the muon's frame, is shorter by the same factor: $15\ \text{km}/22.4 = 670$ m, which it crosses in 2.24 μs. Lengths along the direction of motion contract by $\gamma$. This is length contraction, the other face of time dilation.
Check. Both frames predict the same thing that can be counted at the ground, about 36 percent of the muons, and disagree only about how to describe the journey. Experiments comparing muon counts at a mountain top and at sea level confirm that far more survive than a calculation without time dilation allows.
"At the same time" depends on the frame
Einstein's own route to all this, in his 1916 popular book, goes through simultaneity. Lightning strikes a railway embankment at two points, A and B. An observer standing on the embankment exactly midway sees the two flashes at the same moment and, since the light from each travelled the same distance at the same speed, calls the strikes simultaneous. An observer on a train, passing the midpoint just as the lightning strikes, is moving towards B and away from A. She meets the light from B first. The light from both strikes moves at $c$ in her frame too, and she is at rest midway between the two points on the train where the lightning struck, so she concludes that B struck first.
Neither is wrong. Two events at different places that are simultaneous in one frame are, in general, not simultaneous in another moving relative to it. There is no absolute "now" stretching across space. This is the deepest of the changes, and time dilation and length contraction both follow from it.
The Lorentz transformation, and what stays the same
The precise rule connecting the two frames' coordinates, replacing Galileo's, is the Lorentz transformation. For a frame moving at $v$ along $x$,
$$\begin{aligned} x' &= \gamma\,(x - vt),\\[4pt] t' &= \gamma\left(t - \frac{v\,x}{c^2}\right). \end{aligned}$$The first line is Galileo's rule scaled by $\gamma$. The second is where everything new lives: the term $vx/c^2$ means that events at the same $t$ but different $x$ get different $t'$, which is the relativity of simultaneity. When $v \ll c$, $\gamma \approx 1$ and $vx/c^2$ is negligible, and Galileo's rules return.
Distances and times each change between frames, but one combination does not. For two events separated by $\Delta t$ and $\Delta x$, the interval
$$s^2 = (c\,\Delta t)^2 - (\Delta x)^2$$is the same in every frame. For an event 5 units of $ct$ and 3 units of $x$ from the origin, $s^2 = 25 - 9 = 16$. In a frame moving at $0.6c$ ($\gamma = 1.25$) the same event is at $ct' = 1.25(5 - 0.6\times3) = 4$ and $x' = 1.25(3 - 0.6\times5) = 0$, and $s^2 = 16 - 0 = 16$. That frame is the one moving with a clock that goes from the origin to the event, and $s/c$ is the time the clock itself records. Space and time, separately, depend on the frame; spacetime, with this interval, does not.
Optional: deriving the Lorentz transformation
Assume the transformation is linear (so that uniform motion in one frame is uniform in the other) and write $x' = \gamma(x - vt)$, with $\gamma$ to be found; by the principle of relativity the reverse must have the same form with $v$ reversed, $x = \gamma(x' + vt')$. A light flash sent from the origin at $t = t' = 0$ along $x$ satisfies $x = ct$ and, by the light postulate, $x' = ct'$. Substituting: $ct' = \gamma(c - v)t$ and $ct = \gamma(c + v)t'$. Multiplying the two equations and cancelling $tt'$ gives $c^2 = \gamma^2(c^2 - v^2)$, so $\gamma = 1/\sqrt{1 - v^2/c^2}$. Solving the pair $x' = \gamma(x - vt)$, $x = \gamma(x' + vt')$ for $t'$ then gives $t' = \gamma(t - vx/c^2)$.
Two questions
1. A slower muon. Repeat the worked example for a muon moving at $0.99c$ from the same height. What fraction survives to the ground, and how thick is the atmosphere in the muon's frame?
Worked answer
$\gamma = 1/\sqrt{1 - 0.9801} = 7.09$. The ground-frame trip takes $15\ \text{km}/(0.99c) = 50.5$ μs; the muon's own time is $50.5/7.09 = 7.1$ μs, 3.25 lifetimes, so $e^{-3.25} = 0.04$ survive: about 4 percent. In the muon's frame the atmosphere is $15/7.09 = 2.1$ km thick.
Reducing the speed from $0.999c$ to $0.99c$ looks like a tiny change but reduces $\gamma$ threefold, because $\gamma$ grows steeply as $v$ approaches $c$. The slip to avoid is treating "almost the speed of light" as a single regime.
2. Whose clock is really slow? The ground says the muon's clock runs slow. The muon, entitled to regard itself as at rest, says ground clocks run slow. Can both be right?
Worked answer
Yes, because they are comparing different things. To say a moving clock runs slow, the ground compares it with two different ground clocks as it passes them, and that comparison relies on the ground's synchronisation of those two clocks. The muon judges the ground clocks to be out of synchronisation, by the relativity of simultaneity, and does the reverse comparison with its own clocks. Each measurement is correct within its own frame's definition of "at the same time".
The tempting wrong turn is to look for which clock is "really" slow, as if there were an absolute time. The question has an absolute answer only when the same two clocks meet twice, which requires at least one of them to change its motion; if one stays in uniform motion throughout, the other has recorded less time.
Further reading
- Einstein, Relativity: The Special and the General Theory (1916, tr. Lawson), §VII "The Apparent Incompatibility of the Law of Propagation of Light with the Principle of Relativity", §IX "The Relativity of Simultaneity", §XI "The Lorentz Transformation" and §XII "The Behaviour of Measuring-Rods and Clocks in Motion". Written for a general reader, and in the public domain.
- Feynman, Lectures on Physics, Vol. I, Ch. 15 "The Special Theory of Relativity", §15-4 "Transformation of time" and §15-6 "Simultaneity", and Ch. 17 "Space-Time", §17-2 "Space-time intervals".
Where this leaves us
Muons survive the journey because their clocks, seen from the ground, run slow by $\gamma$, about 22 at $0.999c$; from the muon's point of view, the same fact appears as an atmosphere contracted to a few hundred metres. Both follow from keeping the speed of light the same in every frame, which forces simultaneity, time and length to depend on the frame, while the spacetime interval stays fixed.
If time and space change between frames, so must the quantities built from them: momentum and energy. Newton's formulas fail near the speed of light, and the replacements contain the most famous equation in physics and the relation between energy, momentum and mass that relativistic quantum mechanics is built on. That is the next essay.