This lesson is beyond the Class 12 syllabus. It follows from the two postulates of lesson 3 with nothing more than Pythagoras' theorem.
A clock made of light
The simplest clock to analyse is a light clock. It is a flash of light bouncing between two mirrors a distance \(d\) apart. One crossing from one mirror to the other is one tick. For a clock at rest, the flash moves straight across at speed \(c\), so one tick lasts \(\Delta t_0 = d/c\). With the mirrors \(3\,\text{m}\) apart, that is \(10\,\text{ns}\).
Now let an identical clock move sideways at speed \(v\). Seen from the ground, the mirrors move while the flash crosses. So the flash runs along a slanting line, longer than \(d\). By the second postulate the flash still moves at \(c\) on the slant. A longer path at the same speed takes longer. So each tick of the moving clock takes longer than a tick of the clock at rest.
One light clock is at rest; an identical one moves past at \(0.6c\). By the time the clock at rest has ticked 5 times, how many times has the moving clock ticked?
Press play in the figure above and count. In \(50\,\text{ns}\) the clock at rest ticks 5 times. At \(0.6c\) the moving clock ticks 4 times, and at \(0.8c\) only 3 times. The zigzag of the moving flash is the extra path that costs the time.
\[ \Delta t_0 = \dfrac{d}{c} \]
Why. in its own frame the flash crosses the gap \(d\) straight, at speed \(c\).
\[ (c\,\Delta t)^2 = d^2 + (v\,\Delta t)^2 \]
Why. seen from the ground, during one tick \(\Delta t\) the mirrors move \(v\Delta t\) sideways, and the flash covers the slant \(c\Delta t\). The three form a right triangle.
\[ \Delta t^2\,(c^2 - v^2) = d^2 \]
Why. collect the terms in \(\Delta t^2\).
\[ \Delta t = \dfrac{d/c}{\sqrt{1 - v^2/c^2}} = \dfrac{\Delta t_0}{\sqrt{1 - v^2/c^2}} \]
Read it. take the square root, divide top and bottom by \(c\), and use step 1.
Proper time
Two ticks of a clock happen at the same place in the clock's own frame. The time between two events at the same place, read on a clock at that place, is the proper time \(\Delta t_0\). Every other inertial frame sees the two events at different places, and measures a longer time. This is time dilation.
A clock moving at speed \(v\) runs slow. The time between two of its ticks, measured in a frame where it moves, is \(\gamma\) times its proper time, where \(\gamma = 1/\sqrt{1 - v^2/c^2}\).
A clock on a spaceship moving at \(0.6c\) shows that \(1\) hour has passed. How much time has passed for observers on the Earth?
The ship's clock measures the proper time, because its ticks happen at one place on the ship: \(\Delta t_0 = 1\,\text{h}\).
At \(0.6c\), \(\gamma = 1/\sqrt{1 - 0.36} = 1/0.8 = 1.25\).
So \(\Delta t = \gamma\Delta t_0 = 1.25\,\text{h} = 75\) minutes. The moving clock runs slow.
Values used: \(c = 3 \times 10^8\,\text{m/s}\), written \(300\,\text{m/μs}\) where that is easier.
The size of \(\gamma\)
The Lorentz factor \(\gamma\) measures how strong the effects of relativity are. At \(0.6c\) it is \(1.25\), at \(0.8c\) it is \(1.67\), at \(0.96c\) it is \(3.57\), and at \(0.995c\) it is \(10.0\). Turning the formula round gives the speed for a given \(\gamma\): \(v/c = \sqrt{1 - 1/\gamma^2}\). So \(\gamma = 2\) needs \(v = 0.87c\).
Fig 4.1 has two lessons in it. At the left the curve is flat: even at \(0.2c\), sixty thousand kilometres per second, \(\gamma\) is only \(1.02\). At \(300\,\text{m/s}\), the speed of an aircraft, \(\gamma - 1\) is about \(5 \times 10^{-13}\). That is why Newton's mechanics serves every everyday motion. At the right, as \(v\) nears \(c\), \(1 - v^2/c^2\) nears zero and \(\gamma\) grows without limit. The line \(v = c\) is never reached.
Find the Lorentz factor \(\gamma\) for a speed of \(0.8c\).
An astronaut's heart beats \(60\) times a minute, as the astronaut measures it. Her spaceship passes the Earth at \(0.8c\). How many beats per minute does an observer on the Earth count?
At what speed, as a fraction of \(c\), is the Lorentz factor equal to \(2\)?
Muons that should not reach the ground
A muon is a particle like a heavy electron, made when cosmic rays strike the top of the atmosphere. At rest it lives on average \(2.2\,\text{μs}\), then decays. Muons are made many kilometres up and travel down at about \(0.995c\). In \(2.2\,\text{μs}\) at that speed a muon covers only \(0.995 \times 300 \times 2.2 = 657\,\text{m}\). Yet large numbers of muons reach the ground. The two frames of Fig 4.2 give two views of how.
In the Earth frame, the muon's clock runs slow. At \(0.995c\), \(\gamma = 10.0\), so its mean life seen from the ground is \(22.0\,\text{μs}\). In that time it covers \(6.6\,\text{km}\). In the muon's own frame, its clock is normal and it lives \(2.2\,\text{μs}\). It is the atmosphere that moves, at \(0.995c\) upward. The ground must still arrive within \(2.2\,\text{μs}\), so the depth of air must be only \(657\,\text{m}\). Both frames agree on what happens; they explain it differently.
Moving lengths shrink
The length of a body measured in the frame where it is at rest is its proper length \(L_0\). The muon's trip shows what another frame measures. The Earth frame says the air is \(L_0\) deep and the trip takes \(\Delta t = L_0/v\). The muon's clock shows the proper time \(\Delta t_0 = \Delta t/\gamma\). In the muon's frame the ground covers the depth \(L\) at speed \(v\) in that time, so \(L = v\,\Delta t_0 = L_0/\gamma\). This is length contraction.
Only lengths along the motion contract. A spaceship \(100\,\text{m}\) long and \(20\,\text{m}\) wide passing the Earth at \(0.6c\) measures \(100/1.25 = 80\,\text{m}\) long from the Earth, and still \(20\,\text{m}\) wide. The effect is also mutual. A passenger measures a train car as \(60\,\text{m}\); from the ground at \(0.6c\) it measures \(48\,\text{m}\). A metre stick at rest on the ground measures only \(0.8\,\text{m}\) for the passenger.
A rod has a proper length of \(5\,\text{m}\). It moves along its length at \(0.8c\). What length does an observer at rest measure, in metres?
A muon is created and later decays. In which frame is the time between these two events the proper time?
A muon of mean life \(2.2\,\text{μs}\) moves at \(0.96c\), where \(\gamma = 25/7\). Take \(c = 3 \times 10^8\,\text{m/s}\). How far, in km, does it travel in one mean life as measured from the ground?
At Ooty in the Nilgiri hills, the GRAPES-3 experiment of the Tata Institute of Fundamental Research counts cosmic-ray muons with a large muon telescope. Most of these muons are made about \(15\,\text{km}\) up. Without time dilation, a muon at \(0.995c\) would need \(23\) mean lives to come down, and fewer than one in a billion would survive. With a dilated life of \(22\,\text{μs}\), about one in ten survives the same trip. The steady stream of muons counted at Ooty, day and night, is time dilation at work.
If one twin flies to a star at high speed and returns, the travelling twin comes home younger. This is not a contradiction. The traveller turns round, so the traveller's frame is not inertial for the whole trip, and the twins are not equivalent.
- Putting the dilated time in place of the proper time.Find the frame in which both events happen at one place. Its clock reads \(\Delta t_0\), the shortest time.
- Contracting a length across the motion.Only lengths along the direction of motion contract. Widths and heights stay the same.
- Thinking one frame is right and the other is fooled.Each frame is valid. They agree on every event, such as the muon reaching the ground.