LocusBlog

Time dilation, derived from a light clock

Time dilation is the result that a clock moving past you ticks more slowly than one at rest beside you, by a factor that depends only on its speed:

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

$\Delta t_0$ is the proper time, the interval measured by a clock present at both events. $\Delta t$ is the interval measured in a frame moving at speed $v$ relative to that clock. Since $\gamma \ge 1$ always, $\Delta t$ is never smaller than $\Delta t_0$, and the whole difficulty of using this formula is deciding which of the two times in front of you is the proper one.

The light clock

Build a clock out of two mirrors facing each other a distance $L$ apart with a light pulse bouncing between them. One tick is one round trip. In the frame where the clock sits still, the pulse covers $2L$ at speed $c$:

$$\Delta t_0 = \frac{2L}{c}$$

Now watch the same clock slide past at speed $v$, with the mirrors perpendicular to the motion. The pulse still leaves the bottom mirror and returns to it, but in this frame the clock has moved $v\,\Delta t$ during the round trip, so the light traces a shallow V rather than a straight line up and down. Each leg of the V is the hypotenuse of a right triangle with legs $L$ and $v\,\Delta t/2$, so the total path is:

$$2\sqrt{L^2 + \left(\frac{v\,\Delta t}{2}\right)^2}$$

Here is the only physics in the derivation: light travels at $c$ in this frame too. Not $c$ plus the clock’s speed, not $c$ along the diagonal only. That is the postulate, and everything else is algebra. Set the path equal to $c\,\Delta t$ and square both sides:

$$c^2\,\Delta t^2 = 4L^2 + v^2\,\Delta t^2 \quad\Longrightarrow\quad \Delta t^2\left(c^2 - v^2\right) = 4L^2$$

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

The mirror separation $L$ drops out, which matters: it means the result is about time and not about this particular gadget. A light clock and a wristwatch riding next to it must agree, or you could tell which of them was moving by comparing them, and no experiment inside a uniformly moving lab is allowed to reveal that. $L$ being unchanged is also load-bearing, and it is safe only because the mirrors are separated perpendicular to the motion; lengths along the direction of motion are the ones that contract.

Which clock reads the proper time

Proper time belongs to a pair of events, not to a person’s opinion. It is the interval measured by a single clock that is present at both of them, which means both events happen at the same place in that clock’s frame. Every other frame sees the two events at different places, needs two separated clocks to time them, and gets a longer answer.

Send a ship to a star 4.0 light years away, as measured from Earth, at $v = 0.8c$. Then:

$$\gamma = \frac{1}{\sqrt{1 - 0.8^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3}$$

which is exact, one of the handful of speeds where $\gamma$ is a clean fraction. Earth-based clocks give the trip a duration of $\Delta t = 4.0/0.8 = 5.0$ years. Departure and arrival happen 4.0 light years apart in that frame, so 5.0 years is not a proper time. The ship is at both events, so the ship’s clock is the one measuring proper time:

$$\Delta t_0 = \frac{\Delta t}{\gamma} = \frac{5.0}{5/3} = 3.0\ \text{years}$$

Check that from inside the ship, where nothing is moving and no clock is running slow. In the ship’s frame the Earth-star distance is contracted by the same $\gamma$, to $4.0/(5/3) = 2.4$ light years, and the star arrives at 0.8c, so the crossing takes $2.4/0.8 = 3.0$ years. Same number by a different route, using length contraction instead of time dilation. The two effects are not independent claims; they are one geometry described from two sides, and a calculation that gets different answers from the two routes has a mistake in it.

The muons that should not reach the ground

Cosmic rays striking the upper atmosphere produce muons, which decay with a mean lifetime of about 2.2 microseconds measured at rest. Take one moving at $0.98c$:

$$\gamma = \frac{1}{\sqrt{1 - 0.98^2}} = \frac{1}{\sqrt{0.0396}} = 5.03$$

Without dilation, a mean lifetime buys a distance of $0.98 \times 3.00\times10^{8} \times 2.2\times10^{-6} = 647$ m. The muons are created kilometres up, so on that arithmetic nearly all of them should decay long before sea level, and the flux at the ground should be a tiny fraction of the flux at altitude. With dilation the mean lifetime in the ground frame is $5.03 \times 2.2 = 11.1$ microseconds, and the distance becomes $0.98 \times 3.00\times10^{8} \times 11.1\times10^{-6} = 3.26$ km, five times further. The measured sea-level muon flux is far too large for the non-relativistic estimate and consistent with the dilated one.

Read it from the muon’s frame and nothing is dilated at all: the muon lives its ordinary 2.2 microseconds, but the atmosphere is rushing past and contracted by the same factor of 5.03, so the distance it has to survive is five times shorter. Different story, identical prediction, which is the same consistency check as the ship and the star.

Multiplying the wrong number

The formula only ever makes a time longer, so the failure mode is applying $\gamma$ to the interval that was already the dilated one. Take the 5.0 years Earth measures for the trip, call it the proper time because it came from the frame you are standing in, and multiply: $5.0 \times 5/3 = 8.33$ years for the crew. That is wrong by more than a factor of two against the correct 3.0 years, and it is wrong in the worse direction, since it has the moving clock running fast.

The test that catches it costs one sentence. Ask which single clock was present at both events. Here it is the ship’s, so the ship’s reading is the small one, and any answer that gives the crew more elapsed time than Earth is already dead before the arithmetic is checked. Problems on the time dilation skill hand you the interval from whichever frame is least convenient, which is the only way to find out whether you are identifying proper time or pattern-matching on the order of the numbers.

This is also where the twin paradox comes from and where it dissolves. If motion is relative, each twin should see the other’s clock running slow, and the situation looks symmetric. It is not. The travelling twin turns around, and during the turnaround they are accelerating, which means they never occupy one inertial frame for the whole journey; the light-clock derivation assumes an inertial clock and simply does not apply to them end to end. The stay-at-home twin does stay in one frame, so their accounting holds throughout. Run the trip above out and back and Earth logs 10.0 years while the ship logs 6.0. The twins are in the same place at the start and at the end, so there is nothing to reconcile: they compare two clocks side by side and the traveller’s shows less.

The simulations browser has relativity problems where you can push $v$ toward $c$ and watch $\gamma$ climb, which is the fastest way to see how little happens below about $0.1c$ and how violently it happens above $0.9c$.

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