Special Relativity: Kinematics

Muons are made about 15 km up, when a cosmic ray smashes into the top of the atmosphere. A muon at rest survives about 2.2 microseconds before it decays. Even moving at almost the speed of light, 2.2 microseconds buys it about 660 m. It should never get anywhere near the ground. Detectors at sea level catch them by the thousand.

Something in that accounting is wrong, and it is not the muon. The last unit ended with light behaving in ways that only made sense if you took the wave picture seriously. This one starts with light behaving in a way that only makes sense if you give up something you have never questioned: that a second is a second and a metre is a metre for everybody.

The two things you are asked to believe

Einstein's whole argument rests on two statements, and everything strange that follows is forced by them.

The first is the principle of relativity: the laws of physics are identical in every inertial frame. An inertial frame is one where a free object moves in a straight line at constant speed: a coasting spaceship, a train on smooth track, not a car going round a bend. There is no experiment you can do inside a sealed box that tells you how fast the box is moving, only whether it is accelerating.

The second is harder to swallow: light travels at $c = 3.00 \times 10^8$ m/s in every inertial frame, no matter how the source moves and no matter how the measurer moves. Chase a light pulse at $0.9c$ and it still retreats from you at $c$, not at $0.1c$.

Hold both at once and something has to give. Speed is distance over time. If everyone gets the same speed for light while moving at different speeds through the same room, then they cannot be agreeing about distances and times. The disagreement is the physics.

A clock made of light

Here is the cleanest way to see what the second postulate costs you. Build a clock out of two parallel mirrors a distance $d$ apart, with a pulse of light bouncing between them. One tick is one round trip. In the frame where the clock sits still, the light goes straight up and straight back, so a tick takes

$\Delta t_0 = \frac{2d}{c}.$

Now watch the same clock from a frame it is moving through, sideways at speed $v$, perpendicular to the mirror separation. The mirrors drift while the light is in flight, so the pulse has to travel along a slanted path to catch the top mirror where it will be, not where it was. The slant is longer than $d$. But the second postulate says that light still covers this longer path at $c$, not faster. A longer path at the same speed takes more time. The moving clock ticks slower.

Put numbers on it. Let the tick take $\Delta t$ in this frame. In that time the clock moves sideways by $v\Delta t$, and the light covers $c\Delta t$ along the slant, in two equal legs. Each leg is the hypotenuse of a right triangle with vertical side $d$ and horizontal side $v\Delta t/2$:

$\left(\frac{c\Delta t}{2}\right)^2 = d^2 + \left(\frac{v\Delta t}{2}\right)^2.$

Multiply out and gather the $\Delta t$ terms: $c^2\Delta t^2 - v^2\Delta t^2 = 4d^2$, so $\Delta t^2 (1 - v^2/c^2) = 4d^2/c^2$. Take the square root and recognise $2d/c$ as $\Delta t_0$:

$\Delta t = \frac{\Delta t_0}{\sqrt{1 - v^2/c^2}} = \gamma \, \Delta t_0 .$

That factor gets its own name. The Lorentz factor is

$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1-\beta^2}}, \qquad \beta \equiv \frac{v}{c}.$

It is 1 when $v = 0$, and it grows without limit as $v \to c$. At $0.6c$ it is 1.25; at $0.8c$ it is 1.667; at $0.995c$ it is 10.0. Nothing about the argument used the fact that the clock was made of mirrors. Any clock carried alongside it (a pendulum, a decaying muon, a heartbeat) must agree with it, or the crew could tell they were moving by watching their two clocks drift apart, and the first postulate forbids that.

Example 1. A muon's mean lifetime at rest is $2.20\ \mu$s. It moves through the lab at $0.980c$. What lifetime does the lab measure?

Birth and decay happen at the same place in the muon's frame, so $2.20\ \mu$s is proper time. First $\gamma$: $\beta^2 = 0.980^2 = 0.9604$, so $1 - \beta^2 = 0.0396$, and $\sqrt{0.0396} = 0.1990$. Then $\gamma = 1/0.1990 = 5.03$. So $\Delta t = 5.03 \times 2.20\ \mu\text{s} = 11.1\ \mu$s. The lab sees the muon live five times longer than it does.

Lengths shrink along the motion

Now go back to the muon, and insist that both frames get the same answer for the one thing they cannot disagree about: whether the muon reaches the ground.

In the ground frame, the muon covers 15 km because its clock is slowed by $\gamma$. In the muon's own frame its clock is perfectly normal and it really does only live 2.2 microseconds. The only way it still arrives is if the distance from creation point to ground is shorter in its frame. It has to be shorter by exactly the factor that saves the accounting, which is $\gamma$:

$L = \frac{L_0}{\gamma}.$

$L_0$ is the proper length: the length measured in the frame where the object is at rest. Every frame that sees the object move measures something shorter, and only along the direction of motion. Transverse dimensions are untouched: the light clock's mirror separation was perpendicular to $v$ and never entered the contraction.

Note the asymmetry in where the zero subscript goes. Proper time is the shortest; proper length is the longest. If you remember one of them by "multiply by gamma", you will get the other one backwards. Better to think about which frame is the simple one each time.

Example 2. A ship of proper length 100 m flies at $0.800c$ past a platform of proper length 50.0 m that is at rest in the lab. In the lab frame, how long between the nose reaching the near end of the platform and the tail clearing the far end?

Work entirely in the lab frame, and be careful about which lengths are proper there. The platform is at rest in the lab, so its lab length is the full 50.0 m. The ship is moving, so its lab length is contracted: $\gamma = 1.667$ at $0.800c$, giving $L = 100/1.667 = 60.0$ m.

Track the nose. At the start it is at the near end. At the finish the tail is at the far end, so the nose is 60.0 m past the far end. The nose has therefore travelled $50.0 + 60.0 = 110$ m at $v = 0.800 \times 3.00\times10^8 = 2.40\times10^8$ m/s:

$\Delta t = \frac{110}{2.40\times10^8} = 4.58\times10^{-7}\ \text{s} = 458\ \text{ns}.$

Two frames cannot agree on "at the same time"

The strangest consequence is the one that dissolves most of the paradoxes. Put a lamp at the exact middle of a train and flash it once. In the train's own frame the two ends are equally far away, the light goes both ways at $c$, and it hits both ends at the same instant.

Watch from the ground. The light still goes both ways at $c$: that is the postulate, and it does not care that the source is moving. But now the rear of the train is rushing toward the leftgoing pulse while the front is running away from the rightgoing one. The rear gets hit first. Two events that are simultaneous in one frame are not simultaneous in the other.

How big is the effect? Take a train of proper length $L_0$ whose clocks are synchronised in the train frame. Seen from the ground, those clocks are not synchronised: the rear clock reads ahead of the front clock by

$\Delta t = \frac{v L_0}{c^2} = \frac{\beta L_0}{c}.$

Memorise the shape of that, and memorise that $L_0$ is the proper length of the moving object, with no $\gamma$ attached. The rule of thumb for the sign: the rear clock is ahead, where "rear" means the trailing end. For a 1.00 km train at $0.600c$, the offset is $0.600 \times 1000 / (3.00\times10^8) = 2.00\ \mu$s.

The transformation, and adding velocities

The three results above are all consequences of one pair of equations. If frame $S'$ moves at speed $v$ along the $+x$ axis of $S$, with origins coinciding at $t = t' = 0$, then an event at $(x, t)$ in $S$ sits at

$x' = \gamma(x - vt), \qquad t' = \gamma\left(t - \frac{vx}{c^2}\right).$

The $y$ and $z$ coordinates are unchanged. Set $\Delta x = 0$ and the second equation gives time dilation. Set $\Delta t' = 0$ and the pair gives length contraction. Set $\Delta t = 0$ and the $vx/c^2$ term gives the simultaneity offset. That term is the whole of relativity's weirdness in four symbols.

Differentiate the pair and you get the velocity addition rule. If an object moves at $u$ in $S$ along $x$, its speed in $S'$ is

$u' = \frac{u - v}{1 - uv/c^2}.$

Run it the other way (a missile fired at $u'$ from a ship moving at $v$) and the sign flips: $u = (u' + v)/(1 + u'v/c^2)$. A ship at $0.600c$ firing forward at $0.500c$ gives $(1.100)/(1 + 0.300) = 0.846c$, not $1.1c$. The denominator is what keeps you under $c$ no matter how many boosts you stack. Do three boosts of $0.500c$: first $0.500 \oplus 0.500 = 1.000/1.250 = 0.800$, then $0.800 \oplus 0.500 = 1.300/1.400 = 0.929$.

For a velocity with a transverse part, the perpendicular component transforms too, because the time interval changed even though $y$ did not:

$u'_y = \frac{u_y}{\gamma\left(1 - u_x v / c^2\right)}.$

One more result worth carrying: a source receding at $\beta$ stretches the wavelength of its light by the relativistic Doppler factor $\lambda_{obs} = \lambda_0\sqrt{(1+\beta)/(1-\beta)}$, with the sign inside flipped for approach.

The twins, and where the missing years go

A traveller flies to a star 8.00 light-years away (Earth frame) at $0.800c$, turns around instantly, and comes back. Earth clocks say the round trip takes $16.0/0.800 = 20.0$ yr. The traveller's own clock reads $20.0/1.667 = 12.0$ yr. They meet again and the traveller is 8.0 years younger. That much is not controversial: only one of them is present at both the departure and the reunion, so only their clock reads proper time.

The confusion comes from the traveller's point of view. On each leg the traveller is in an inertial frame and judges the Earth clock to be the moving one, ticking slow. Each leg lasts 6.0 yr of traveller time, during which they judge Earth to age $6.0/1.667 = 3.6$ yr, so 7.2 yr over both legs. But Earth actually aged 20.0 yr. Where did the other 12.8 years go?

Into the turnaround. At the moment the traveller reverses, they switch from one inertial frame to another, and the two frames disagree about what "now" means 8.00 light-years away. That is the simultaneity offset, applied over the distance to Earth, twice: once for each direction:

$\Delta t_{jump} = \frac{2 v L_0}{c^2} = 2 \times 0.800 \times 8.00 = 12.8\ \text{yr}.$

Check: $7.2 + 12.8 = 20.0$ yr, which is what Earth read. The accounting closes. The asymmetry between the twins is real and it lives entirely in the fact that one of them changed frames.

The muon numbers follow the same pattern. At $0.995c$, $\gamma = 10.0$, so the lab lifetime is $22.0\ \mu$s and the mean travel distance is $0.995 \times 3.00\times10^8 \times 22.0\times10^{-6} = 6570$ m. Surviving from 15 km down to the ground means surviving $15000/6570 = 2.28$ mean distances, so the fraction is $e^{-2.28} = 0.102$. Roughly one in ten arrives, against $e^{-22.8} \approx 10^{-10}$ without dilation.

Recognising the technique

A problem is asking for relativistic kinematics when a speed is quoted as a multiple of $c$, or when a lifetime, a length, or a trip time is given in one frame and wanted in another. Once you are sure, the work is almost always the same four steps.

Compute $\gamma$ first, from $\beta$, before anything else. Then name your two frames and write down, for each quantity in the problem, which frame it was measured in. Most wrong answers in this topic are a $\gamma$ that should have been a $1/\gamma$, and naming the frames is what prevents that.

Then ask what kind of question it is. If two events happen at the same place in some frame, that frame reads the proper time and every other frame reads more: lifetimes, trip times, and light-clock ticks are all this. If an object has a rest length and you want its length in a frame it moves through, divide by $\gamma$. If the problem says "at the same time" in one frame and asks about another, you need $vL_0/c^2$, and the pole-and-barn, lightning-on-the-train and twin-turnaround problems are all that one formula wearing different hats. If two speeds are being combined, never add them: use the denominator.

And if a problem seems to contain a contradiction (a 20 m pole inside a 10 m barn, a twin younger than their twin) the resolution is almost always simultaneity rather than dilation. Ask which events the two frames are being asked to call simultaneous, and the paradox usually evaporates before you have done any arithmetic.

All of this was about where and when. None of it touched mass, force or energy, and those need repairs of their own: momentum is no longer $mv$ and kinetic energy is no longer $\tfrac{1}{2}mv^2$. That is relativistic energy and momentum, where $\gamma$ shows up again and finally earns the most famous equation in physics.