Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: special relativity (203-SN3-RE)

Q: How do I know who measures the proper time?

The proper time is measured by the clock for which the two events happen at the same place. The practical test comes down to one question: is a single clock enough to measure this duration, or are two synchronized clocks at different places needed? If one is enough, it is the proper time, and it is always the smallest of the measured durations.

Q: Can time dilation and length contraction be used at the same time?

No, never. They are two descriptions of the same disagreement, seen from two different frames, not two effects that add up. You choose one frame, stay in it, and use only one of the two. Combining both divides the result by gamma twice, and it is the most frequent mistake of the whole chapter.

Q: Does mass increase with speed?

No. Mass is a relativistic invariant: it has the same value in all frames, like electric charge. What increases without limit are the energy and the momentum. The notion of relativistic mass was abandoned by modern physics because it sustains exactly this confusion.

Q: Why can an object not exceed the speed of light?

Because its total energy equals gamma times its rest energy, and gamma diverges as the speed approaches that of light: an infinite energy would have to be supplied. The velocity addition law confirms it, never producing more than the speed of light from two speeds that do not exceed it.

Q: Do two rockets moving apart at 0.9 times the speed of light each violate relativity?

No. In Earth's frame, the distance between them does grow at the rate of 1.8 times the speed of light, and that is allowed because it is the speed of no object and nothing is carried at that rate. The speed of one measured from the other is 0.9945 times the speed of light.

Q: From what speed do I need to use relativity?

Below about 0.14 times the speed of light, that is 42,000 kilometres per second, Newtonian mechanics is off by less than one percent and is enough. Beyond that, relativity is needed. In dynamics, the best criterion is the ratio of kinetic energy to rest energy: very small compared with one, the classical calculation is fine.

This sheet does not redo the special relativity course, it lists what loses points on test day. Almost every mistake in this chapter comes from the same missing step: a formula is applied before deciding WHO measures WHAT. One factor $\gamma$ placed on the wrong side of the equation, and the result is off by a factor $\gamma^{2}$ with nothing to warn you.

It is written for CEGEP students in the Natural Science program taking Waves and Modern Physics, 203-SN3-RE (formerly 203-NYC), and for first-year university students, where the chapter is taught almost identically with the same canonical examples of the muon and the train.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Time and space are not a backdrop, they are MEASUREMENTS: before writing any formula, ask who holds the stopwatch, and whether the two events happen at the same place for that observer.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Postulates of special relativity
Lorentz factor
Proper time
Time dilation
Proper length
Length contraction
Relativity of simultaneity
Lorentz transformations
Relativistic velocity addition
Rest energy
Energy-momentum relation
Spacetime invariant

The essentials

Proper time and proper length: the two questions to ask

• $\Delta t_{0}$ is the proper time: it is the one measured by the clock for which BOTH events happen at the same place. All other observers measure more: $\Delta t=\gamma\,\Delta t_{0}$ , with $\gamma\ge 1$ .
• $L_{0}$ is the proper length: it is the one measured in the frame where the object is AT REST. All others measure less: $L=\dfrac{L_{0}}{\gamma}$ . Contraction acts only along the direction of motion.
• Lorentz factor: $\gamma=\dfrac{1}{\sqrt{1-\beta^{2}}}$ with $\beta=\dfrac{v}{c}$ . A useful inverse in dynamics: $\beta=\sqrt{1-\dfrac{1}{\gamma^{2}}}$ .
• The test that settles it in five seconds: is a single clock enough to measure this duration? If so, it is the proper time. Is a single ruler enough, the object being at rest? If so, it is the proper length.

On the left the flash returns to the SAME place, one clock is enough: proper time,

10.0

ns. On the right the path is slanted, hence longer:

16.7

ns at

0.800c

Lorentz, simultaneity and the light cone

• Lorentz transformations, from $S$ to $S'$ moving at $v$ along $x$ : $x'=\gamma(x-vt)$ and $t'=\gamma\left(t-\dfrac{vx}{c^{2}}\right)$ , with $y'=y$ and $z'=z$ . The inverse is obtained by changing $v$ to $-v$ .
• Two events that are simultaneous in $S$ and a distance $\Delta x$ apart are separated in $S'$ by $\Delta t'=-\gamma\dfrac{v\,\Delta x}{c^{2}}$ : the event located at the FRONT, in the direction of motion of $S'$ , happens first there.
• Inside the light cone, $s^{2}>0$ : a signal slower than light can connect the two events, and their order is the same for everyone. Outside the cone, $s^{2}<0$ , and the order depends on the observer.

The front flash, simultaneous on the platform with the rear flash placed at O, at

240

m, lies outside the cone: their order depends on the observer. C, inside the cone, follows O for everyone.

Dynamics: what changes and what does not

• What CHANGES with speed: $E=\gamma mc^{2}$ , $p=\gamma mv$ , $K=(\gamma-1)mc^{2}$ . What does NOT change: the mass $m$ , an invariant, a property of the particle just like its charge.
• Energy-momentum relation: $E^{2}=(pc)^{2}+(mc^{2})^{2}$ . The combination $E^{2}-(pc)^{2}$ equals $(mc^{2})^{2}$ in ALL frames: it is the invariant that checks any collision calculation.
• The ratio that decides everything: $\dfrac{K}{mc^{2}}$ . Very small compared with 1, classical mechanics is enough; of the order of 1 or more, it is disqualified. Rest energies: electron $0.511$ MeV, proton $938.3$ MeV.
• Spacetime invariant: $s^{2}=(c\Delta t)^{2}-(\Delta x)^{2}$ , identical in all frames. Positive, timelike, the order of the events is absolute; negative, spacelike, the order depends on the frame.

The triangle of

E^{2}=(pc)^{2}+(mc^{2})^{2}

for the

2.00

MeV electron: the side

mc^{2}

never changes,

pc

and

E

grow with speed.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. A spaceship travels toward a star $6.00$ light-years away at $0.600c$ . How long does the trip last for the crew?

the whole question, 4 to 6 points

What not to write

“The distance is contracted to $4.80$ ly, so the trip lasts $8.00$ years, and since time is also dilated by a factor $1.25$ , it actually lasts only $6.40$ years on board.”

What to write

“I place myself in the spaceship's frame. The distance there is contracted to $\dfrac{6.00}{1.25}=4.80$ ly, so the trip lasts $\dfrac{4.80}{0.600}=8.00$ years on board. Full stop.”

From Earth,

10.0

years; on board,

8.00

years, by contraction alone or by dilation alone (green). Combining both divides by

\gamma

twice:

6.40

years (red).

Why: Dilation and contraction are TWO DESCRIPTIONS of the same disagreement, seen from two different frames, never two effects that add up. You choose one frame, stay in it, and use only one of the two. The check that catches it: redo the calculation from Earth, $\dfrac{6.00}{0.600}=10.0$ years, then divide by $\gamma$ , and recover $8.00$ years.

2. A rocket moves away from Earth at $0.800c$ and fires a projectile forward at $0.700c$ in its own frame.

2 points, and a physically impossible answer that a marker penalizes heavily

What not to write

“The speed seen from Earth is $0.700c+0.800c=1.50c$ .”

What to write

“ $u=\dfrac{u'+v}{1+\dfrac{u'v}{c^{2}}}=\dfrac{1.500}{1+0.560}=0.9615c$ , less than $c$ as it should be.”

The red line, the Galilean prediction, crosses

u=c

; the exact curve, in green, hugs the asymptote without crossing it, and gives

0.9615c

instead of

1.50c

Why: The denominator is the entire relativistic content of the law: it is almost 1 as long as one of the speeds is small, which explains three centuries of Galilean mechanics, and it brakes sharply as soon as both are comparable to $c$ . Survival reflex: any speed greater than $c$ at the end of a calculation signals an error, no exception.

3. A question asks why a massive object cannot reach the speed of light.

1 to 2 points, and a confusion that is paid for afterward on the whole dynamics block

What not to write

“Because its mass increases with speed and becomes infinite as $v\to c$ , so it becomes infinitely heavy.”

What to write

“Because its energy $E=\gamma mc^{2}$ diverges as $\beta\to 1$ : an infinite energy would have to be supplied. The mass $m$ does not change, it is an invariant.”

Why: The notion of relativistic mass $\gamma m$ was abandoned precisely because it produces this confusion: it gives the impression that mass is an observer-dependent quantity, when it is the only one that is not. Practical corollary: in $E^{2}=(pc)^{2}+(mc^{2})^{2}$ , the $m$ is always the rest mass, never anything else.

4. An electron is accelerated through $2.00$ MV. Calculate its speed.

the whole question, plus the sub-questions that depend on it

What not to write

“ $\dfrac{1}{2}mv^{2}=2.00$ MeV gives $\dfrac{v}{c}=\sqrt{\dfrac{2K}{mc^{2}}}=\sqrt{7.83}=2.80$ , that is $2.80c$ .”

What to write

“I first form $\dfrac{K}{mc^{2}}=\dfrac{2.00}{0.511}=3.91$ , much greater than 1: the regime is relativistic. So $\gamma=1+3.91=4.91$ and then $\beta=\sqrt{1-\dfrac{1}{\gamma^{2}}}=0.979$ .”

For

K/mc^{2}=3.91

, the classical parabola (red) gives

2.80c

, beyond the wall

\beta=1

; the relativistic curve (green) gives

0.979c

Why: The correct path is energy, then $\gamma$ , then $\beta$ : never the other way around. The reflex that saves you is to form the ratio $\dfrac{K}{mc^{2}}$ BEFORE any calculation. The same $2.00$ MeV given to a proton gives $\dfrac{2.00}{938.3}=0.0021$ , and there the classical formula is excellent to within $0.2$ %.

5. Two lightning flashes that are simultaneous for the station master strike the front and the rear of a moving train.

3 points, and a misunderstanding of the rest of the chapter

What not to write

“They are simultaneous for everyone: two events that are simultaneous are so in all frames.”

What to write

“In the train, they are separated by $|\Delta t'|=\dfrac{vL_{0}}{c^{2}}$ , and it is the FRONT flash, in the direction of motion, that happens first.”

Seen from the platform, the passenger moves toward the front: its light reaches them first, the rear's long after. In the train, seated in the middle, they conclude that the front was struck first.

Why: Simultaneity is the first casualty of the two postulates, and it is the CAUSE of the two other effects: length contraction is nothing but the disagreement about what it means to measure both ends “at the same time”. Check by the invariant: on the platform $s^{2}=0-240^{2}$ , in the train $s^{2}=180^{2}-300^{2}$ , and both equal $-5.76\times 10^{4}$ m $^{2}$ .

6. Two rockets move away from each other, each at $0.900c$ relative to Earth, in opposite directions.

2 points, often on a reasoning question at the end of the test

What not to write

“That is impossible, they would separate at $1.800c$ and relativity forbids it.”

What to write

“In Earth's frame, the separation rate is indeed $1.800c$ and nothing forbids it: it is the speed of no object. The proper relative speed is $\dfrac{1.800}{1.810}c=0.9945c$ .”

Seen from Earth, the gap grows at

1.800c

, the speed of no object; measured from rocket A, the speed of B is

0.9945c

, below

c

Why: What relativity forbids is for an OBJECT to exceed $c$ in a given frame, not for a distance to grow quickly. No information travels at $1.800c$ , nothing is carried at that rate, and each of the two rockets stays well below $c$ where it is measured.

Which method to choose

Which formula to apply, depending on what the problem gives

Read the problem and first spot what varies: a duration, a length, a combined speed, or an energy.

\gamma

stays glued to 1 over the first two thirds of the axis then takes off: it is not a gradual correction, it is a wall. Points:

1.25

1.667

7.089

If The problem gives a DURATION and asks for the other one → Identify which one is the proper time: the one measured where the two events happen at the same place. Then $\Delta t=\gamma\,\Delta t_{0}$ , the proper time always being the SMALLEST.

Example: Mean lifetime of a muon, $2.20$ μs at rest, $\gamma\tau_{0}=22.0$ μs seen from the ground at $0.9950c$ .
If The problem gives a LENGTH or a distance and asks for the other one → Identify the proper length: the one measured where the object is at rest. Then $L=\dfrac{L_{0}}{\gamma}$ , the proper length always being the LARGEST.

Example: Earth-star distance of $6.00$ ly for Earth, $4.80$ ly for the spaceship at $0.600c$ .
If The problem chains TWO speeds, one measured in a moving frame → Addition law, $u=\dfrac{u'+v}{1+\frac{u'v}{c^{2}}}$ . Then check that the result is indeed less than $c$ .

Example: Projectile at $0.700c$ fired by a rocket at $0.800c$ : $0.9615c$ seen from Earth.
If The problem gives an ENERGY, an accelerating voltage or a momentum → Form $\dfrac{K}{mc^{2}}$ , then $\gamma=1+\dfrac{K}{mc^{2}}$ , then $\beta$ . For $p$ , go through $E^{2}=(pc)^{2}+(mc^{2})^{2}$ .

Example: Electron through $2.00$ MV: $\gamma=4.91$ , $\beta=0.979$ , $pc=2.46$ MeV.
If The problem asks whether TWO EVENTS can be linked by a cause → Calculate the invariant $s^{2}=(c\Delta t)^{2}-(\Delta x)^{2}$ . Positive, timelike, causality possible and order absolute. Negative, spacelike, no causality and relative order.

Example: The two flashes of the train, $s^{2}=-5.76\times 10^{4}$ m $^{2}$ : spacelike, the order depends on the frame.

The first two branches NEVER combine. A problem that speaks of both a duration and a distance is handled by choosing one frame, never by applying both formulas.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Handling a dilation or contraction problem

When to use it: As soon as a problem involves two observers in relative motion and asks for a duration or a length.

1 Name the two frames and say which one moves relative to the other, with the sign of $v$ .
2 Explicitly identify the proper time or the proper length, and JUSTIFY in one sentence why it is that one.
3 Calculate $\gamma$ and write it down, with at least three significant figures.
4 Apply the single formula corresponding to the chosen frame, never combining two.
5 Check the direction of the result: the duration measured by the observer in relative motion is LARGER, the length SMALLER.

Concluding sentence

“Both events happen at the same place in the spaceship's frame, so it is the one measuring the proper time $\Delta t_{0}$ ; for Earth, $\Delta t=\gamma\,\Delta t_{0}$ .”

The trap: Writing $\Delta t_{0}=\gamma\,\Delta t$ instead of $\Delta t=\gamma\,\Delta t_{0}$ : the result is off by a factor $\gamma^{2}$ and nothing in the calculation signals it.

Marking: 1 mark for the justified identification of the proper time, 1 mark for $\gamma$ , 2 marks for the application, 1 mark for the check of the direction.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

No speed exceeds $c$

Reread all the speeds written on the paper. A single value greater than $c$ is enough to condemn the calculation.

$2.80c$ for an electron, or $1.50c$ for a projectile: in both cases the classical formula was applied outside its domain.
The proper time is the smallest, the proper length the largest

Compare the two durations obtained, then the two lengths. If the moving observer measures a shorter duration, the $\gamma$ is on the wrong side.

$10.0$ ns in the car against $16.7$ ns from the platform: consistent. The opposite would have signalled an inverted formula.
$\gamma$ is consistent with $\beta$

Check the order of magnitude in your head: $\gamma=1.25$ at $0.600c$ , $\gamma=1.667$ at $0.800c$ , $\gamma\approx 7$ at $0.990c$ . These three are worth knowing by heart.

A $\gamma$ of $2$ announced for $0.500c$ is wrong: the true value is $1.155$ .
The invariant is the same in both frames

After any Lorentz transformation, recompute $s^{2}=(c\Delta t)^{2}-(\Delta x)^{2}$ on both sides. It is the only reliable numerical check of the chapter.

Platform: $0-240^{2}=-5.76\times 10^{4}$ . Train: $180^{2}-300^{2}=-5.76\times 10^{4}$ . Identical, the transformation is right.

The typical problem, taken apart

The atmospheric muon, worked in both frames

A muon is created at an altitude of $15.0$ km and descends at $v=0.9950c$ . Its mean lifetime at rest is $\tau_{0}=2.20$ μs, and a population decays as $N=N_{0}e^{-t/\tau}$ , where $\tau$ is the mean lifetime expressed in the frame in which $t$ is counted. What fraction reaches the ground?

The situation: a muon created at an altitude of

15.0

km descends at

0.9950c

toward a detector on the ground; its mean lifetime at rest is

2.20

μs.

Step 1

$\gamma=\dfrac{1}{\sqrt{1-0.9950^{2}}}=\dfrac{1}{\sqrt{0.009975}}=10.01$

Why

We calculate $\gamma$ first, always, because it is what then decides everything. Note that four decimals must be kept on $\beta^{2}$ : rounding $0.9950^{2}$ to $0.99$ would change $\gamma$ from $10.0$ to $7.1$ .

Step 2

Without relativity: $d_{0}=\beta c\,\tau_{0}=657$ m, that is $\dfrac{15{,}000}{657}=22.8$ mean lifetimes and $e^{-22.8}=1.2\times 10^{-10}$

Why

We first establish the CLASSICAL result, because it is the one the experiment contradicts. One muon in ten billion: the detector at sea level should be empty, and it is not.

Step 3

Ground frame: $\tau=\gamma\tau_{0}=22.0$ μs, flight distance $\beta c\gamma\tau_{0}=6.58$ km, so $\dfrac{15{,}000}{6{,}575}=2.28$ mean lifetimes and $e^{-2.28}=0.102$

Why

Here the muon's clock is moving, so it runs slowly: we use DILATION, and only it. The thickness of the atmosphere stays $15.0$ km, it is measured in the frame where it is at rest.

Step 4

Muon frame: $L'=\dfrac{15{,}000}{10.01}=1.50$ km, proper time $\dfrac{1{,}498}{0.9950\times 3.00\times 10^{8}}=5.02$ μs, that is $\dfrac{5.02}{2.20}=2.28$ mean lifetimes

Why

Here the muon's clock ticks normally, it is its own: we use CONTRACTION, and only it. We recover exactly $2.28$ , and that is the check of the whole exercise.

Step 5

Surviving fraction: $e^{-2.28}=0.102$ , that is about $10$ %

Why

The number of mean lifetimes elapsed is a COUNTABLE quantity, the number of turns of the muon's internal clock: everyone must agree on it, which is why both frames give the same $2.28$ .

The conclusion, written out

About $10$ % of the muons reach the ground, against $1.2\times 10^{-10}$ in classical physics. The two observers tell incompatible stories, one a slowed clock, the other a flattened atmosphere, and agree on the only measurable thing: the detector's count.

The classic mistake on this problem: Combining both effects, by contracting the atmosphere to $1.50$ km AND dilating the mean lifetime to $22.0$ μs, gives $0.228$ mean lifetimes and $80$ % survivors, which would contradict the experiment. One frame, one effect.

Learn by heart

• $\gamma=\dfrac{1}{\sqrt{1-\beta^{2}}}$ and $\beta=\sqrt{1-\dfrac{1}{\gamma^{2}}}$ . To know by heart: $\gamma(0.600c)=1.25$ , $\gamma(0.800c)=1.667$ , $\gamma(0.990c)=7.09$ .
• $\Delta t=\gamma\,\Delta t_{0}$ and $L=\dfrac{L_{0}}{\gamma}$ . The proper quantity is the short duration and the long length.
• $u=\dfrac{u'+v}{1+\frac{u'v}{c^{2}}}$ , and $u'=c$ always gives back $u=c$ whatever $v$ is.
• $E=\gamma mc^{2}$ , $K=(\gamma-1)mc^{2}$ , $p=\gamma mv$ , and the invariant $E^{2}=(pc)^{2}+(mc^{2})^{2}$ .
• Rest energies: electron $0.511$ MeV, proton $938.3$ MeV, muon $105.7$ MeV, charged pion $139.6$ MeV.
• Simultaneity: two events that are simultaneous in one frame and a distance $L_{0}$ apart are separated by $\dfrac{vL_{0}}{c^{2}}$ in the other, the FRONT first.
• Below $0.14c$ , Newtonian mechanics is off by less than 1%: that is the real content of “at ordinary speeds”.

Frequently asked questions

How do I know who measures the proper time?

The proper time is measured by the clock for which the two events happen at the same place. The practical test comes down to one question: is a single clock enough to measure this duration, or are two synchronized clocks at different places needed? If one is enough, it is the proper time, and it is always the smallest of the measured durations.

Can time dilation and length contraction be used at the same time?

No, never. They are two descriptions of the same disagreement, seen from two different frames, not two effects that add up. You choose one frame, stay in it, and use only one of the two. Combining both divides the result by gamma twice, and it is the most frequent mistake of the whole chapter.

Does mass increase with speed?

No. Mass is a relativistic invariant: it has the same value in all frames, like electric charge. What increases without limit are the energy and the momentum. The notion of relativistic mass was abandoned by modern physics because it sustains exactly this confusion.

Why can an object not exceed the speed of light?

Because its total energy equals gamma times its rest energy, and gamma diverges as the speed approaches that of light: an infinite energy would have to be supplied. The velocity addition law confirms it, never producing more than the speed of light from two speeds that do not exceed it.

Do two rockets moving apart at 0.9 times the speed of light each violate relativity?

No. In Earth's frame, the distance between them does grow at the rate of 1.8 times the speed of light, and that is allowed because it is the speed of no object and nothing is carried at that rate. The speed of one measured from the other is 0.9945 times the speed of light.

From what speed do I need to use relativity?

Below about 0.14 times the speed of light, that is 42,000 kilometres per second, Newtonian mechanics is off by less than one percent and is enough. Beyond that, relativity is needed. In dynamics, the best criterion is the ratio of kinetic energy to rest energy: very small compared with one, the classical calculation is fine.

Practise it

Corrected exercises: Special relativity, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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Revision sheet: special relativity (203-SN3-RE)

The essentials

Proper time and proper length: the two questions to ask

Lorentz, simultaneity and the light cone

Dynamics: what changes and what does not

The mistakes that cost marks

1. A spaceship travels toward a star 6.006.006.00 light-years away at 0.600c0.600c0.600c. How long does the trip last for the crew?

2. A rocket moves away from Earth at 0.800c0.800c0.800c and fires a projectile forward at 0.700c0.700c0.700c in its own frame.

3. A question asks why a massive object cannot reach the speed of light.

4. An electron is accelerated through 2.002.002.00 MV. Calculate its speed.

5. Two lightning flashes that are simultaneous for the station master strike the front and the rear of a moving train.

6. Two rockets move away from each other, each at 0.900c0.900c0.900c relative to Earth, in opposite directions.

Which method to choose

Which formula to apply, depending on what the problem gives

How the answer is expected to be written

Handling a dilation or contraction problem

Check before you hand in

The typical problem, taken apart

The atmospheric muon, worked in both frames

Learn by heart

Frequently asked questions

How do I know who measures the proper time?

Can time dilation and length contraction be used at the same time?

Does mass increase with speed?

Why can an object not exceed the speed of light?

Do two rockets moving apart at 0.9 times the speed of light each violate relativity?

From what speed do I need to use relativity?

Corrected exercises: Special relativity, 203-SN3-RE

See also

Looking for a 203-SN3-RE tutor in Montreal?

1. A spaceship travels toward a star $6.00$ light-years away at $0.600c$ . How long does the trip last for the crew?

2. A rocket moves away from Earth at $0.800c$ and fires a projectile forward at $0.700c$ in its own frame.

4. An electron is accelerated through $2.00$ MV. Calculate its speed.

6. Two rockets move away from each other, each at $0.900c$ relative to Earth, in opposite directions.