IB DP Physics · Theme A · Space, time and motion

Galilean and special relativity.

A.5 · Reference frames, the Lorentz transformations, time dilation, length contraction and space–time diagrams
Higher level only: there is no standard level content in A.5

Additional HL · 8 hours A.1 Kinematics ↗

The journey

Four parts.

Part 01

Galilean relativity

Reference frames, and the common-sense rules for changing between them

Part 02

Einstein’s postulates

The Lorentz transformations and velocity addition

Part 03

Time and length

Simultaneity, time dilation, length contraction and muons

Part 04

Space–time diagrams

World lines, and two frames on one diagram

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Part 01 · HL only

Galilean relativity.

How Newton’s physics compares the views of two observers

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Where and when

Reference frames and events.

An event: something that happens at one place and one time
A reference frame: axes and clocks used to give it coordinates, \((x, t)\)
An inertial frame is one that is not accelerating

TWO FRAMES, ONE EVENT

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

What a frame is

Imagine a grid of meter rules filling space, with a clock at every point, all synchronized with each other. An observer “in” that frame reads the position and time of any event from the rule and the clock that are right beside it when it happens. That is what the coordinates \((x, t)\) mean. A second observer, moving relative to the first, carries a grid and clocks of their own and writes \((x', t')\) for the same event.

Why inertial

In a frame that is not accelerating, a body with no resultant force on it moves in a straight line at constant speed: Newton’s first law holds. In an accelerating car, a ball on the floor rolls backwards with nothing pushing it. The whole of this topic deals with inertial frames only, moving at constant velocity relative to each other.

The convention

Frame S′ moves at velocity \(v\) along the \(x\) axis of frame S, and the two origins coincide at \(t = t' = 0\).

Before any calculation, decide which frame is S, which is S′, and what the sign of \(v\) is. Most errors start there.

Newton’s laws in every inertial frame

Galilean relativity.

Newton’s laws of motion are the same in all inertial frames
No experiment in mechanics can tell you whether you are “really” moving

Only relative motion has any meaning.

The Galilean transformations \(x' = x - vt\) \(t' = t\) Time is the same for everyone

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Galileo’s ship

Shut yourself in a cabin below decks on a ship sailing smoothly at constant velocity. Drop a ball: it falls straight down at your feet. Throw it to a friend: it needs the same effort in every direction. Nothing you can do inside the cabin reveals whether the ship is moving or tied up in harbor. Galileo used this, in 1632, to answer those who said the Earth could not be moving because we would feel it.

The transformations

In a time \(t\) the origin of S′ has moved a distance \(vt\) along the \(x\) axis. So an event at position \(x\) in S is \(vt\) closer to the origin of S′: \(x' = x - vt\). As for time, Newton took it for granted that a second is a second for every observer: \(t' = t\). It seems too obvious to write down. It is the assumption that special relativity overturns.

Accelerations come out the same in both frames, because the frames differ only by a constant velocity. With the same forces and the same masses, \(F = ma\) holds in both.

Velocities in two frames

Galilean velocity addition.

\(u' = u - v\)

\(u\): velocity of an object, measured in S
\(u'\): velocity of the same object, measured in S′
\(v\): velocity of S′ relative to S

A train passes a platform at 30 m s⁻¹ A ball rolls forward along the carriage at 5.0 m s⁻¹ \(u' = u - v\) so \(u = u' + v\) \(u = 5.0 + 30 = \class{res}{35 \,\mathrm{m} \,\mathrm{s}^{-1}}\) relative to the platform

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Common sense, written as an equation

Walk forward along a moving train and your speed over the ground is the train’s speed plus your walking speed. Walk towards the back and it is the difference. Divide the transformation \(x' = x - vt\) by the time and that is what comes out, provided the time is the same in both frames.

Signs do the work

Take one direction as positive for everything. A ball rolling towards the back of the carriage has \(u' = -5.0\) m s⁻¹ and \(u = 25\) m s⁻¹. Two cars approaching each other at 20 m s⁻¹ each have velocities \(+20\) and \(-20\); from the first, the second has \(u' = -20 - 20 = -40\) m s⁻¹.

What it predicts for light

Shine a torch forward from the train. By this rule, the light should pass the platform at \(c + 30\) m s⁻¹. That prediction is the subject of the next two slides.

Mixing up which velocity is which. \(v\) belongs to the frame; \(u\) and \(u'\) belong to the object.

Where Galilean relativity fails

The problem with light.

Maxwell’s theory, 1860s

Light travels at one speed, \(c\)

Fixed by the constants of electricity and magnetism. No frame is mentioned

Galilean velocity addition

Its speed should depend on the observer

\(c - v\) if you chase it, \(c + v\) if you run to meet it

Both cannot be right. Either there is one special frame in which light has speed \(c\), or the rule for adding velocities is wrong.

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The nineteenth-century answer

Every wave then known traveled in a medium: sound in air, ripples on water. Physicists assumed that light did too, and called its medium the ether. The speed \(c\) would be the speed of light relative to the ether, just as 340 m s⁻¹ is the speed of sound relative to the air. An observer moving through the ether would measure a different speed, exactly as the Galilean rule says.

A testable prediction

The Earth moves round the Sun at 30 km s⁻¹, so it must be moving through the ether. Light sent along the direction of that motion and light sent across it should take slightly different times to cover the same distance and return. The difference would be tiny, one part in a hundred million, but an instrument using the interference of light (C.3) could detect it.

If the ether existed, there would be one frame, at rest in it, that was different from all others, and Galilean relativity would not apply to light.

The story · Cleveland, Ohio, 1887

Background story · not assessed

The most famous failed experiment.

Black and white photograph of a large square stone slab carrying optical parts, resting on a round trough in a brick-walled basement — The **interferometer** of 1887 on its stone slab, which floated on mercury

Albert Michelson and Edward Morley sent two beams of light at right angles, back and forth across a sandstone slab, and made them interfere
Turning the slab should have shifted the fringes by about 0.4 of a fringe as the Earth moved through the ether
They saw less than 0.02. The speed of light was the same in every direction

Photograph: Case Western Reserve University, about 1887, public domain.

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Quick check

Galilean relativity.

01What is an inertial reference frame?

One that is not accelerating

02What does Galilean relativity say about time?

\(t' = t\): the same in every frame

03S′ moves at 4.0 m s⁻¹. An event is at \(x\) = 50 m, \(t\) = 5.0 s. Find \(x'\).

\(50 - 4.0 \times 5.0 = 30 \,\mathrm{m}\)

04A car at 25 m s⁻¹ is overtaken by one at 35 m s⁻¹. Velocity of the second, seen from the first?

10 m s⁻¹

05What did the Galilean rule predict for the speed of light?

That it depends on the observer’s motion

06What did Michelson and Morley find?

No difference: the same speed in all directions

PRESS → TO REVEAL EACH ANSWER

Part 02 · HL only

Einstein’s postulates.

Keep the principle of relativity, keep the speed of light, and change everything else

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The story · Bern, 1905

Background story · not assessed

A patent clerk’s paper.

Photograph of the young Albert Einstein standing beside a desk in a suit, with dark curly hair and a mustache — **Albert Einstein** at the patent office, about 1905

Einstein was working as a patent examiner in Bern, with no university post
In one year he published four papers that changed physics: among them the photon (E.2) and \(E = mc^{2}\) (E.3)
“On the Electrodynamics of Moving Bodies” was received on 30 June and printed on 26 September 1905
It needed no ether, and started from two short statements

He kept the speed of light and gave up absolute time.

Photograph: Lucien Chavan, 1904 or 1905, public domain.

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The whole theory rests on these

The two postulates of special relativity.

First postulate

The laws of physics are the same in all inertial frames

Not just mechanics: electricity, magnetism and light as well

Second postulate

The speed of light in a vacuum is the same in all inertial frames

Whatever the motion of the source or of the observer

Two observers moving relative to each other measure the same speed for the same beam of light.

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The first is Galileo’s idea, widened

Galilean relativity said that no mechanical experiment can detect uniform motion. Einstein extended it to every law of physics. If Maxwell’s equations are among those laws, and they give a definite speed for light, then that speed must come out the same in every inertial frame. The second postulate follows naturally from the first.

How strange the second is

Chase a beam of light at 99% of its speed and measure how fast it is leaving you. It is still leaving at \(3.00 \times 10^{8}\) m s⁻¹, not at 1% of that. Nothing else behaves this way. A speed is a distance divided by a time. If two observers are to agree on the speed of the same light, they must disagree about the distances and the times. The rest of this deck is the working out of that sentence.

Learn both postulates as statements. “State the postulates of special relativity” is a standard question.

One number that appears everywhere

The Lorentz factor.

\(\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}\)

Always 1 or more, with no unit
Almost exactly 1 at everyday speeds
Grows without limit as \(v \rightarrow c\)

GAMMA AGAINST SPEED

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Why we never notice

For an aircraft at 250 m s⁻¹, \(\frac{v^{2}}{c^{2}}\) is about \(7 \times 10^{-13}\) and \(\gamma\) differs from 1 only in the thirteenth decimal place. Even for the Earth in its orbit, at 30 km s⁻¹, the difference is five parts in a thousand million. Our intuitions about space and time were formed at speeds where relativity makes no measurable difference.

Work with \(\frac{v}{c}\)

Speeds in this topic are nearly always given as fractions of \(c\). For \(v = 0.60c\): \(\frac{v^{2}}{c^{2}} = 0.36\), so \(\gamma = \frac{1}{\sqrt{0.64}} = \frac{1}{0.80} = 1.25\). For \(0.80c\): \(\gamma = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} = 1.67\). These two turn up constantly because they give exact answers.

Why nothing reaches \(c\)

At \(v = c\) the bottom of the fraction is zero. For \(v > c\) it is the square root of a negative number.

Forgetting to square: it is \(\frac{v^{2}}{c^{2}}\) under the root, not \(\frac{v}{c}\).

Replacing the Galilean ones

The Lorentz transformations.

\(x' = \gamma (x - vt)\)

\(t' = \gamma \left(t - \dfrac{vx}{c^{2}}\right)\)

The coordinates of the same event in two frames
Time is no longer the same: \(t'\) depends on where the event is as well as when
At low speed, \(\gamma \approx 1\) and \(\dfrac{vx}{c^{2}} \approx 0\): back to \(x' = x - vt\), \(t' = t\)

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

They are forced on us

These equations are not a guess. They are the only way of relating two frames that keeps the speed of light the same in both and still treats the two frames even-handedly. You do not need to derive them.

The new feature is in the time equation

In the Galilean version every observer shared one time. Here the time of an event in S′ involves its position in S. Two events that happen at the same time \(t\) but at different places \(x\) get different values of \(t'\). That one term, \(\frac{vx}{c^{2}}\), is the source of the relativity of simultaneity, of time dilation and of length contraction.

Newton is still inside

For everyday speeds the Lorentz transformations give the Galilean ones to very high accuracy. Relativity does not overthrow Newtonian mechanics; it contains it as the low-speed case.

For intervals between two events the same equations hold with \(\Delta x\) and \(\Delta t\) in place of \(x\) and \(t\). That is how they are most often used.

Worked example

One event, two frames.

S′ moves at \(v = 0.60c\) · \(\gamma = 1.25\) In S: \(x = 1.2 \times 10^{9} \,\mathrm{m}\), \(t = 5.0 \,\mathrm{s}\) \(x' = \gamma (x - vt)\) \(x' = 1.25 \times (1.2 \times 10^{9} - 0.60 \times 3.00 \times 10^{8} \times 5.0)\) \(x' = \class{res}{3.8 \times 10^{8} \,\mathrm{m}}\)

\(t' = \gamma \left(t - \dfrac{vx}{c^{2}}\right)\) \(\dfrac{vx}{c^{2}} = \dfrac{0.60 \times 1.2 \times 10^{9}}{3.00 \times 10^{8}} = 2.4 \,\mathrm{s}\) \(t' = 1.25 \times (5.0 - 2.4)\) \(t' = \class{res}{3.3 \,\mathrm{s}}\)

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

A short cut for \(\frac{vx}{c^{2}}\)

Write \(v\) as \(0.60c\) and one \(c\) cancels: \(\frac{vx}{c^{2}} = 0.60 \times \frac{x}{c}\). The quantity \(\frac{x}{c}\) is the time light takes to cover the distance \(x\): here \(\frac{1.2 \times 10^{9}}{3.00 \times 10^{8}} = 4.0\) s. Many questions give distances in light seconds or light years for exactly this reason.

Compare with Galileo

The Galilean answers would have been \(x' = 3.0 \times 10^{8}\) m and \(t' = 5.0\) s. The position differs by the factor \(\gamma\). The time differs completely: the two observers do not agree on when the event happened, though their clocks were set together at the origin.

Keep these numbers. A later slide uses both sets of coordinates to check that the space–time interval is the same in the two frames.

Velocities that never exceed c

Relativistic velocity addition.

\(u' = \dfrac{u - v}{1 - \dfrac{uv}{c^{2}}}\)

The top is Galileo’s rule
The bottom is the correction

Two ships approach: \(u = 0.80c\), \(v = -0.60c\) in S \(u' = \dfrac{0.80c - (-0.60c)}{1 - (0.80)(-0.60)}\) \(u' = \dfrac{1.40c}{1.48} = \class{res}{0.95c}\) Light instead: \(u = c\) \(u' = \dfrac{c - v}{1 - \dfrac{v}{c}} = \class{res}{c}\)

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The two ships

Seen from the Earth, one ship moves at \(+0.80c\) and the other at \(-0.60c\). Take the second ship as the frame S′, so \(v = -0.60c\). Galileo would give \(u' = 1.40c\) for the speed at which the first ship approaches. The correct answer is \(0.95c\). However close to \(c\) two velocities are, they never combine to more than \(c\).

The second postulate, recovered

Put \(u = c\) into the equation: \(u' = \frac{c - v}{1 - \frac{v}{c}} = \frac{c (c - v)}{c - v} = c\), whatever \(v\) is. The equation has the constancy of the speed of light built into it.

At low speeds the product \(\frac{uv}{c^{2}}\) is negligible, the bottom is 1 and the rule becomes \(u' = u - v\) again.

Losing a minus sign. Decide the positive direction first, give every velocity its sign, and substitute with brackets.

Quick check

Einstein’s postulates.

01State the second postulate.

The speed of light in a vacuum is the same in all inertial frames

02\(\gamma\) for \(v = 0.60c\)?

1.25

03\(\gamma\) for \(v = 0.80c\)?

1.67

04What do the Lorentz transformations become at low speed?

The Galilean transformations

05A ship at \(0.50c\) fires a probe forward at \(0.50c\) relative to itself. Probe’s speed from the Earth?

\(\dfrac{0.50c + 0.50c}{1 + 0.25} = 0.80c\)

06The ship shines a laser forward. Speed of the light from the Earth?

\(c\)

PRESS → TO REVEAL EACH ANSWER

Part 03 · HL only

Time and length.

What two observers no longer agree about

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

At the same time, for whom?

The relativity of simultaneity.

A lamp at the center of a carriage flashes
Light travels at \(c\) in both frames
On the platform, the back wall moves to meet its flash

Simultaneous in one frame, not in another.

ONE PAIR OF EVENTS, TWO FRAMES

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Nobody is being fooled

This is not about how long light takes to reach someone’s eyes, and neither observer is mistaken. Each has corrected for the travel time of the light. In the carriage frame the two arrivals happen at the same time. In the platform frame they happen at different times. Both descriptions are correct, each in its own frame. There is no third, “true” answer.

Why the second postulate forces this

In Galilean physics the platform observer would say that the forward flash travels faster than \(c\) and the backward one slower, which would exactly make up for the motion of the walls, and everyone would agree. With light at \(c\) in both directions in both frames, that escape is closed.

From the equations

Two events with the same \(t\) but a separation \(\Delta x\) have \(\Delta t' = -\gamma \frac{v \Delta x}{c^{2}}\), which is not zero. Events at the same place, \(\Delta x = 0\), stay simultaneous in every frame.

Simultaneity is relative only for events at different places along the direction of motion.

Moving clocks run slow

Time dilation.

\(\Delta t = \gamma \Delta t_{0}\)

\(\Delta t_{0}\): the proper time, in the frame where both events happen at the same place
Every other frame measures a longer time

A LIGHT CLOCK

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The light clock

A pulse of light bounces between two mirrors; each return trip is one tick. For someone traveling with the clock the light goes straight up and down. For someone watching it fly past, the mirrors move while the light is in flight, so the light follows a longer, slanting path. The light’s speed is \(c\) for both observers, so a longer path means a longer time. The moving clock ticks more slowly.

Not just light clocks

If a light clock on a spaceship ran slow compared with a wristwatch beside it, the crew could tell they were moving, which the first postulate forbids. So every process on the ship runs slow by the same factor as seen from outside: watches, heartbeats, the decay of unstable particles.

Finding the proper time

Ask: in which frame do the two events happen at the same place? The two ticks of a clock happen at the same place for whoever is carrying the clock. That observer’s time is \(\Delta t_{0}\), the shortest that anyone measures.

Thinking that the crew feel time passing slowly. To them everything is normal, and it is the clocks left behind that run slow.

Moving objects are shorter

Length contraction.

\(L = \dfrac{L_{0}}{\gamma}\)

\(L_{0}\): the proper length, in the frame where the object is at rest
Every other frame measures a shorter length
Only along the direction of motion

A ship of proper length 120 m passes at \(0.80c\) · \(\gamma = 1.67\) \(L = \dfrac{L_{0}}{\gamma} = \dfrac{120}{1.67}\) \(L = \class{res}{72 \,\mathrm{m}}\) Its width and height are unchanged

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

It follows from time dilation

A ship travels from the Earth to a star at speed \(v\). Earth observers say the distance is \(L_{0}\) and the trip takes \(\Delta t = \frac{L_{0}}{v}\). The crew’s clocks record less time, \(\frac{\Delta t}{\gamma}\). But the crew agree that the star approached them at speed \(v\). The only way both can be true is if, for the crew, the distance to the star was shorter: \(\frac{L_{0}}{\gamma}\).

What a length is

To measure a moving object you must mark where its two ends are at the same time. But observers in different frames disagree about what “the same time” means. Length contraction is a consequence of the relativity of simultaneity, not a squashing of the object by any force.

Each sees the other contracted

The crew find the Earth-to-star distance shortened; Earth observers find the ship shortened. There is no contradiction, because the two are measuring different things with different ideas of simultaneity.

Proper time is the shortest time; proper length is the longest length. Check every answer against that.

Worked example

A journey to a star.

Star 4.0 light years away · ship at \(0.80c\) · \(\gamma = 1.67\) Earth frame \(\Delta t = \dfrac{L_{0}}{v} = \dfrac{4.0 \,\mathrm{ly}}{0.80c} = \class{res}{5.0 \,\text{years}}\) Ship’s clock: the proper time \(\Delta t_{0} = \dfrac{\Delta t}{\gamma} = \dfrac{5.0}{1.67} = \class{res}{3.0 \,\text{years}}\)

Ship frame: the distance is contracted \(L = \dfrac{L_{0}}{\gamma} = \dfrac{4.0}{1.67} = 2.4 \,\mathrm{ly}\) \(\Delta t_{0} = \dfrac{L}{v} = \dfrac{2.4 \,\mathrm{ly}}{0.80c} = \class{res}{3.0 \,\text{years}}\) Both frames agree on the ship’s clock

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

One fact, two explanations

Everyone agrees on what the ship’s clock reads on arrival: three years. They disagree about why. On Earth they say the trip took five years and the ship’s clocks ran slow. On the ship they say their clocks were fine and the trip was short because the distance was only 2.4 light years. Time dilation in one frame is length contraction in the other.

Which is the proper time?

Departure and arrival both happen at the ship, as far as the crew are concerned: at the same place in the ship’s frame. So the ship’s three years is the proper time. The Earth frame needs two clocks in different places, one on Earth and one at the star, to time the trip.

Light years make it easy

A distance in light years divided by a speed in units of \(c\) gives a time in years directly.

Applying \(\gamma\) twice: shortening the distance and also slowing the clock in the same frame. Use one effect per frame.

Experimental evidence

Muon decay.

Made high in the atmosphere, moving at \(0.995c\)
Mean lifetime at rest: 2.2 μs
In one lifetime: about 660 m

Without relativity almost none should reach sea level. In fact most do.

THE LAST 1920 m · \(\gamma = 10\)

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The puzzle

Muons are unstable particles made when cosmic rays strike the top of the atmosphere. At rest they last 2.2 μs on average. Even at almost the speed of light, a muon would cover only \(0.995 \times 3.00 \times 10^{8} \times 2.2 \times 10^{-6} = 660\) m in one lifetime. A descent of 1920 m takes 6.4 μs, nearly three lifetimes, and after three lifetimes only about one in twenty would be left.

The measurement

In 1963 David Frisch and James Smith counted muons of this speed on Mount Washington, 1917 m up, and again at sea level. On the mountain: 563 an hour. At sea level, without relativity, they should have found about 27. They found 412.

Two explanations, one result

In the Earth frame the muons’ clocks run slow by \(\gamma = 10\): they live 22 μs, and 6.4 μs is less than a third of a lifetime. In the muons’ frame they live 2.2 μs as usual, but the mountain is only 192 m high and rushes past in 0.64 μs: again less than a third of a lifetime. The same fraction survives either way.

Muon decay is the evidence named in the guide for both time dilation and length contraction. Be able to tell it from either frame.

Worked example

Muons in two frames.

\(v = 0.995c\) · \(\gamma = 10\) · height 1920 m · \(\Delta t_{0} = 2.2 \,\mu\mathrm{s}\) Earth frame Time of flight \(= \dfrac{L_{0}}{v} = \dfrac{1920}{0.995 \times 3.00 \times 10^{8}} = 6.4 \,\mu\mathrm{s}\) Lifetime \(\Delta t = \gamma \Delta t_{0} = 10 \times 2.2 = \class{res}{22 \,\mu\mathrm{s}}\) \(\dfrac{6.4}{22} = 0.29\) of a lifetime

Muon frame \(L = \dfrac{L_{0}}{\gamma} = \dfrac{1920}{10} = \class{res}{192 \,\mathrm{m}}\) Time \(= \dfrac{L}{v} = \dfrac{192}{0.995 \times 3.00 \times 10^{8}} = 0.64 \,\mu\mathrm{s}\) \(\dfrac{0.64}{2.2} = 0.29\) of a lifetime

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The agreement is the point

Each frame uses one relativistic effect and gets 0.29 of a lifetime. With the exponential decay law of E.3, the fraction surviving is \(e^{-0.29} = 0.75\). Frisch and Smith measured \(\frac{412}{563} = 0.73\). Without relativity the flight would last \(\frac{6.4}{2.2} = 2.9\) lifetimes and the fraction would be \(e^{-2.9} = 0.055\), about 30 an hour. Frisch and Smith’s own estimate, from the speeds they actually selected, was 27.

Which quantities are proper

The muon’s lifetime in its own frame, 2.2 μs, is a proper time: its creation and its decay happen at the same place for the muon. The height of the mountain measured from the ground, 1920 m, is a proper length: the mountain is at rest in the Earth frame.

\(\gamma = \frac{1}{\sqrt{1 - 0.995^{2}}} = \frac{1}{\sqrt{0.00998}} = 10.0\).

Something everyone agrees on

The space–time interval.

\((\Delta s)^{2} = (c \Delta t)^{2} - (\Delta x)^{2}\)

For any two events
Invariant: the same in every inertial frame
Though \(\Delta t\) and \(\Delta x\) each differ from frame to frame

The event of Part 2, measured from the origin S: \(c \Delta t = 1.5 \times 10^{9}\) m, \(\Delta x = 1.2 \times 10^{9}\) m \((\Delta s)^{2} = (1.5^{2} - 1.2^{2}) \times 10^{18} = \class{res}{8.1 \times 10^{17} \,\mathrm{m}^{2}}\) S′: \(c \Delta t' = 9.75 \times 10^{8}\) m, \(\Delta x' = 3.75 \times 10^{8}\) m \((\Delta s)^{2} = (9.75^{2} - 3.75^{2}) \times 10^{16} = \class{res}{8.1 \times 10^{17} \,\mathrm{m}^{2}}\)

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Like a distance, with a minus sign

Two people using maps turned at different angles disagree about how far east and how far north a town is, but agree on its distance, because \(\text{east}^{2} + \text{north}^{2}\) does not depend on how the map is turned. Two observers in relative motion disagree about the time and the distance between two events, but agree on \((c \Delta t)^{2} - (\Delta x)^{2}\). Space and time are mixed together by motion, and the interval is what survives.

Its link to proper time

In the frame where the two events happen at the same place, \(\Delta x = 0\) and the interval is just \(c \Delta t_{0}\). So for any pair of events that one clock could be present at, \(\Delta s = c \times\) the proper time between them. In the example, \(\Delta s = \sqrt{8.1 \times 10^{17}} = 9.0 \times 10^{8}\) m, and the proper time is 3.0 s.

Time dilation in one line

\((c \Delta t)^{2} - (v \Delta t)^{2} = (c \Delta t_{0})^{2}\) gives \(\Delta t = \gamma \Delta t_{0}\) directly.

Use the interval when a question gives \(\Delta x\) and \(\Delta t\) in one frame and one of them in another. It avoids the Lorentz transformations altogether.

Quick check

Time and length.

01What is a proper time interval?

Measured where both events happen at the same place

02A clock moves at \(0.60c\). 8.0 s pass on it. Time in the lab?

\(1.25 \times 8.0 = 10 \,\mathrm{s}\)

03A 100 m rod moves along its length at \(0.60c\). Length in the lab?

\(\dfrac{100}{1.25} = 80 \,\mathrm{m}\)

04Which dimension of a moving object is contracted?

Only the one along its motion

05Which quantity is the same in all inertial frames?

The space–time interval (and \(c\))

06How do muons provide evidence for relativity?

Far more reach the ground than their lifetime at rest allows

PRESS → TO REVEAL EACH ANSWER

Part 04 · HL only

Space–time diagrams.

Relativity as a picture

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The story · Germany, 1908

Background story · not assessed

Einstein’s teacher draws the picture.

Portrait photograph of Hermann Minkowski, a man with a mustache and pince-nez in a dark suit — **Hermann Minkowski**

Minkowski had been one of Einstein’s mathematics teachers in Zurich
By 1908 he saw that special relativity is best understood as the geometry of four dimensions: three of space and one of time
His lecture “Space and Time”, on 21 September 1908, introduced the diagrams of this part
He died of appendicitis less than four months later, aged 44

“Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows.”

Photograph: unknown photographer, public domain.

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

A graph of where and when

Space–time diagrams and world lines.

\(x\) across, \(ct\) upwards: both in meters
A point is an event
A world line is the path of an object through space and time
Light: a line at 45°

FRAME S

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

A displacement–time graph on its side

In A.1 time ran across and position up. Here position runs across and time runs up, so the diagram reads like a film strip: each horizontal line is a snapshot of the \(x\) axis at one moment, and later moments are higher. An object at rest stays at the same \(x\) while time passes: a vertical line. A moving object leans over, more for a greater speed.

Why \(ct\)

Multiplying the time by \(c\) turns it into a distance, the distance light would travel in that time, so both axes have the same unit. Light then covers one unit of \(x\) for every unit of \(ct\): a line at 45°. Nothing material can have a world line that leans further than that.

Constant velocity only

A straight world line means a constant velocity. The course deals with straight world lines only.

Reading a steep line as fast. On these axes steeper means slower. A vertical line is at rest.

Speed from a diagram

The angle of a world line.

\(\tan \theta = \dfrac{v}{c}\)

\(\theta\): the angle between the world line and the \(ct\) axis
At rest: \(\theta = 0\). Light: \(\theta = 45^{\circ}\)

\(v = 0.60c\): \(\theta = \tan^{-1} 0.60 = \class{res}{31^{\circ}}\)

A PARTICLE MOVING AT CONSTANT VELOCITY

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Where it comes from

In a time \(t\) the particle moves a distance \(x = vt\). On the diagram that is a step of \(vt\) across for a step of \(ct\) up. The tangent of the angle from the vertical axis is across over up: \(\frac{vt}{ct} = \frac{v}{c}\).

Reading a speed

Pick any two points on a world line, read off \(\Delta x\) and \(\Delta (ct)\), and divide: \(\frac{v}{c} = \frac{\Delta x}{\Delta (ct)}\). A world line that goes 2 squares across for every 5 up belongs to something moving at \(0.40c\).

The scale is not linear in angle

\(0.60c\) is at 31°, \(0.90c\) at 42° and \(0.99c\) at 44.7°. The speeds close to \(c\) are crowded against the light line.

The angle is measured from the \(ct\) axis, not from the \(x\) axis.

Two frames on one diagram

The axes of a moving frame.

\(ct'\) axis: the world line of the origin of S′
\(x'\) axis: all the events at \(t' = 0\)
Each makes the same angle \(\theta\) with its unprimed axis
Light bisects both pairs of axes: the same speed \(c\)

S′ MOVING AT 0.6 c

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

The \(ct'\) axis

The time axis of any frame is the set of events at its origin, \(x' = 0\). Seen from S, the origin of S′ is an object moving at \(v\), so its world line is tilted at \(\theta\) from the \(ct\) axis, with \(\tan \theta = \frac{v}{c}\).

The \(x'\) axis

The space axis of a frame is the set of events that happen at its time zero, \(t' = 0\). From the Lorentz transformation, \(t' = 0\) means \(t = \frac{vx}{c^{2}}\), or \(ct = \frac{v}{c} x\): a line tilted up from the \(x\) axis by the same angle. That the \(x'\) axis is not horizontal is the relativity of simultaneity drawn on paper.

The light line bisects both

It lies half way between \(ct\) and \(x\), and half way between \(ct'\) and \(x'\). An observer in either frame finds that light covers equal amounts of their distance and their time.

To read an event’s coordinates in S′, draw lines through it parallel to the \(ct'\) and \(x'\) axes, not perpendicular to them.

Seeing it, not calculating it

Simultaneity on a diagram.

Simultaneous in S: on a line parallel to the \(x\) axis
Simultaneous in S′: on a line parallel to the \(x'\) axis
A and B share the first kind of line but not the second

TWO EVENTS, TWO FRAMES

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Reading the diagram

A and B are at the same height, so they happen at the same \(ct\): simultaneous in S. Now find when each happens in S′. Slide from each event along a line parallel to the \(x'\) axis until you reach the \(ct'\) axis. B’s line lands lower on the \(ct'\) axis than A’s. In S′, B happened first.

The carriage again

This is the flash experiment of Part 3 in graphical form. The event that is further forward, in the direction S′ is moving, happens earlier in S′.

Order can change, cause cannot

The order of two events can differ between frames only if they are so far apart in space, and so close in time, that not even light could travel from one to the other. Then neither can have caused the other. If one event could have caused another, every observer agrees which came first.

Time dilation, length contraction and simultaneity can all be shown on these diagrams. Simultaneity is the one most often asked for.

Why the primed axes are stretched

The scales of the axes.

One unit on \(ct'\) is not the same length on the page as one unit on \(ct\)
The units are set by curves of constant space–time interval
On the tilted axes the marks are further out

WHERE “1” IS ON EACH AXIS

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Why the interval sets the scale

An event at one unit of time on a frame’s own time axis has \(x' = 0\) and \(ct' = 1\), so its interval from the origin is 1. Every observer agrees on that interval. In S the same event therefore satisfies \((ct)^{2} - x^{2} = 1\). All the “one unit of time” marks, for frames of every speed, lie on that one curve. It is a hyperbola, and it gets further from the origin as it nears the light line.

Time dilation on the diagram

Follow the \(ct'\) axis out to its mark “1′”: one tick of the moving clock. Now read that event’s height on the \(ct\) axis. It is above 1: it is at \(\gamma\). A clock moving with S′ takes \(\gamma\) units of S time to tick once.

Length contraction on the diagram

The curve \(x^{2} - (ct)^{2} = 1\) marks one unit of length on every \(x\) axis. A rod one unit long at rest in S′ has world lines along the \(ct'\) axis and parallel to it, through the mark “1′” on the \(x'\) axis. S measures its length from the two ends at the same time in S: along the \(x\) axis, where the far world line crosses at \(\frac{1}{\gamma}\), less than 1. A length is always measured between events that are simultaneous for the observer, which is why the two frames disagree.

Measuring along a primed axis with the unprimed scale. The two axes need their own tick marks.

Quick check

Space–time diagrams.

01What does a vertical world line show?

An object at rest in that frame

02At what angle is the world line of light?

45°

03A world line makes 20° with the \(ct\) axis. Speed?

\(v = c \tan 20^{\circ} = 0.36c\)

04What is the \(ct'\) axis, seen from S?

The world line of the origin of S′

05How do you find events simultaneous in S′?

On a line parallel to the \(x'\) axis

06What fixes the scales on the primed axes?

Curves of constant space–time interval

PRESS → TO REVEAL EACH ANSWER

Summary

Relativity on one slide.

Galilean

\(x' = x - vt\) · \(t' = t\)

\(u' = u - v\)
Fails for light

Postulates

Same laws · same \(c\)

In every inertial frame

Lorentz

\(x' = \gamma (x - vt)\)

\(t' = \gamma \left(t - \dfrac{vx}{c^{2}}\right)\)
Velocities never combine to more than \(c\)

Time and length

\(\Delta t = \gamma \Delta t_{0}\) · \(L = \dfrac{L_{0}}{\gamma}\)

Simultaneity is relative
Evidence: muon decay

Invariant

\((\Delta s)^{2} = (c \Delta t)^{2} - (\Delta x)^{2}\)

The same in every inertial frame

Space–time diagrams

\(\tan \theta = \dfrac{v}{c}\)

\(ct\) upwards, light at 45°

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY

Summary

Key words.

EventSomething that happens at one place and one time

Inertial frameA reference frame that is not accelerating

Lorentz factor\(\gamma\): 1 at low speed, larger as \(v\) nears \(c\)

Proper timeBetween events at the same place: the shortest

Proper lengthIn the object’s rest frame: the longest

SimultaneityRelative: it depends on the frame

Space–time intervalThe same for all inertial observers

World lineAn object’s path on a space–time diagram

MuonAn unstable particle whose survival shows time dilation

A.5 GALILEAN AND SPECIAL RELATIVITY · HL ONLY