**Alan C. Pickwick**

**"EAAE Summerschools" Working Group**

**Manchester Grammar School (UK)**

# Abstract

This workshop aims to introduce the topic of time dilation and cosmic rays by way of demonstrations, experiments and calculations that teachers may use in their classrooms.

In the first half of the workshop we will explore how time is affected when clocks move past us at high speed. In our daily lives we perceive time as a steady flow and the same in all places. However, if clocks move relative to us at high speed, time does not behave so simply. The theory of special relativity is needed to explain what is observed.

In the second half of the workshop we will use equipment to count cosmic ray muons. The fact that these muons can be detected at ground level is perhaps a surprise. We will explore the calculations required to explain the observations and show that the muons are expected to reach the ground if we use the appropriate theory.

# Introduction

'Just let me get this right, you say moving clocks run slow? I don't believe it!!'

The first part of the workshop introduces the fact that moving clocks run more slowly than their stationary equivalents.

# Part 1 - How do we measure time?

**Investigate a Stationary Clock.**

We will make and observe a super-bouncy ball clock.

To do this we use a shallow wooden box, one metre square.

We roll a ball from one side to the other and let it bounce back to the point from which it started.

This is the basic unit of time for our experiment.

Key Point: You stand next to the box. This means that you measure the 'Proper Time' for this event. Put another way, you do not move relative to the box so you measure the 'Proper Time' for the event.

**Investigate a moving clock.**

Using the super-bouncy ball clock, we pull the 'clock' across the floor at a steady speed whilst repeating the experiment.

The ball has to be sent on a diagonal path across the board. It will take longer to complete its journey.

This demonstrates that our moving clock runs slower than our stationary one.

More precisely, a clock moving relative to you runs more slowly.

Caution: This model clock should use flashes of light not super-bouncy balls. The speed of light does not depend on the motion of the observer and is the same in all directions. However, the speed of the ball does depend on the motion of the observer. Be aware of this serious limitation of this model.

**An alternative demonstration - The 'People Clock'.**

Two experimenters stand about 2 metres apart. They throw a ball between each other.

We measure the time for four complete exchanges of the ball.

The two experimenters then walk along parallel lines about 2 metres apart. They throw a ball from one to the other.

Again we measure the time for four complete exchanges of the ball.

In the second experiment the ball takes longer to go from one experimenter to the other and back again because the path is longer.

Caution: This model clock should use flashes of light not super-bouncy balls. The speed of light does not depend on the motion of the observer and is the same in all directions. However, the speed of the ball does depend on the motion of the observer. Be aware of this serious limitation of this model.

**An Apparent Contradiction - The Railway Experiment.**

Take two identical clocks. Fasten one on the platform wall of a railway station. Fasten the other on the inside of an express train.

Stand on the station platform and wait for the train to pass through the station at high speed.

Observe the clock on the train. The clock on the train will appear to tick more slowly than the one in the station.

Now reverse the experiment.

Travel on the train and compare the clock on the train with the one on the wall of the station.

The clock on the wall of the station will appear to tick more slowly than the one in the train.

When you first read about this experiment you probably think either:

- That you have misread the description or
- That the description is wrong.

I can assure you that the description is accurate. However the effect is very small. At a speed of 50 m/s the difference is about one part in 72 million million (7.2 x 10^{13}). You could not measure this difference with any clock!!

**To calculate this difference in time.**

From the theory of Special Relativity, the time dilation formula links the proper time (tp), an improper time (ti), the speed of the clock (v) and the speed of light (c):

tp = ti ( 1 - ( v^{2} / c^{2} ) )^{1/2} (Equation 1)

Substituting tp = ti ( 1 - ( 50^{2} / ( 3 x 10^{8} )^{2} ) )^{0.5} =

Try evaluating this equation using your calculator. You will probably find that your calculator cannot evaluate such a small change.

To obtain a result we need to use the Binomial Approximation. You might remember it from your school or university days!!

( 1 + x )^{n} = 1 + nx + ( very small terms ) if x << 1.

Applying this approximation to Equation 1:

We can completely ignore the small terms because v^{2} / c^{2} is very very small.

tp = ti ( 1 - ( 0.5 v^{2} / c^{2} ) ) = ti - ( ti x 0.5 v^{2} / c^{2} )

Rearranging. Try evaluating this on your calculator:

( tp - ti ) / ti = - 0.5 v^{2} / c^{2} = - 0.5 x 50^{2} / ( 3 x 10^{8} )^{2} =

(Answer: - 1.389 x 10^{-14})

This is the fractional difference in the times.

Take the reciprocal of 1.389 x 10^{-14} and confirm that it is one part in 7.2 x 10^{13}.

**To derive the Time Dilation formula using Pythagoras's Theorem.**

Referring to Figure 3, consider the time for a flash of light to travel from one side of the 'clock' to the other. When the clock is stationary, this is the 'Proper Time' (tp). The flash of light travels at speed (c), so the perpendicular distance across the 'clock' is ctp.

When the clock is in motion at speed (v), it tells the 'Improper Time' (ti). These two times are linked in the right-angled triangle in Figure 3 by Pythagoras's Theorem. I have written out the algebra in the following paragraph but I suggest that you do it yourself to ensure that you have understood the underlying principles.

c^{2} t_{i}^{2} = c^{2} t_{p}^{2} + v^{2} t_{i}^{2}

Rearranging: c^{2} t_{i}^{2} - v^{2} t_{i}^{2} = c^{2} t_{p}^{2}

Rearranging: t_{i}^{2} ( c^{2} - v^{2} ) = c^{2} t_{p}^{2}

Rearranging: t_{i}^{2} = c^{2} t_{p}^{2} / ( c^{2} - v^{2} ) = t_{p}^{2} / 1 - v^{2} / c^{2}

Taking square roots of both sides: t_{i} = t_{p} / ( 1 - v^{2} / c^{2} )^{1/2}

When there is relative motion, the denominator is less than unity (1) and so the Improper Time (t_{i}) is always greater than the Proper Time (t_{p}).

**Summary:**

Moving clocks run slower than stationary ones.

**More Precise Summary:**

A clock measuring the proper time interval between two events (i.e. it is present at both events) measures a smaller interval than a clock measuring an improper interval between these same two events.

# Part 2 - Counting Cosmic Ray Muons

**Introduction**

Cosmic rays create fast-moving particles that reach the surface of the Earth all the time. We can investigate the muons in this 'cosmic-rain' to help us understand how moving clocks behave. The second part of the workshop will show an experiment to confirm that moving clocks do actually run more slowly than their stationary equivalents.

**What are muons?**

Muons are elementary particles from the same family as the electron.

- They have the same charge as that of an electron.
- They have a mass about 200 times that of an electron.
- They are unstable and decay with a half-life of 2.2 microseconds.

The muons that we are going to investigate are created 60 km above the surface of the Earth as follows:

- Cosmic rays (very fast protons and light nuclei) arrive from the Sun and from the Galaxy (probably from supernova explosions).
- These cosmic rays collide with nitrogen or oxygen nuclei in the top of the atmosphere.
- The collisions produce many sub-nuclear fragments that travel down towards the surface.
- The fragments travel at just less than the speed of light.

The muons in these showers of particles should decay very rapidly. None should reach the surface of the Earth.

**To verify this statement we will perform a calculation.**

Consider a muon travelling at just less than the speed of light.

Calculate the time it takes to reach the surface of the Earth, 60 km below.

Time = distance / speed =

(Speed of Light = 3.0 x 10^{5} km/s)

(Answer: Time = 200 microseconds)

How many half-lives of a muon take place in 200 microseconds?

Number of half-lives = time / half-life = 200 / 2.2 =

(Answer: 91 half-lives)

During one half-life, the number of muons halves. Therefore in 91 half-lives, the fraction remaining will be 0.5 * 0.5 * 0.5 ………… repeated 91 times.

It is much easier to use your calculator to work out 0.591 =

(Answer: 4.0 x 10^{-28} of the original)

Using this model, we would expect no muons to reach the surface. They would all have decayed on their journey.

**Demonstration of the detection of cosmic ray muons using a simple co-incidence detector.**

Two Geiger-Müller tubes are positioned one above the other.

The outputs from the tubes are connected to a coincidence detector (an AND gate) so that the counter only operates when both tubes trigger within about a microsecond of each other.

(You could do this at school with two independent counters and some of your pupils acting as coincidence detectors!!)

If the system is left to count for many minutes, a count rate of about 3 events per minute is observed.

If the tubes are now placed side-by-side, the count rate falls nearly to zero.

This suggests that the system is counting particles that are coming from above. It is most likely that these particles are muons.

However, this is a puzzle. All the muons should have decayed before they reached the surface of the Earth.

**Check Calculation:**

The flux of muons at sea level is about 170 per square metre per second. The area of the tubes we are using is ( Pi x 0.025^{2} m^{2}), so the expected count rate is predicted to be is ( 170 x Pi x 0.025^{2} m^{2}) which is 0.33 per second. This rate of 20 per minute is much larger than our observed value of 3 per minute, so our detectors are probably not very efficient!!

**Additional data about these muons:**

Their mean energy is about 2 GeV and they would penetrate about 2 metres of concrete!! At an energy of 2 GeV the speed of a muon is 2.9990 x 10^{8} m/s.

**Calculation of the Time Dilation Effect on Cosmic Ray Muons.**

For convenience of the calculations, assume the muons travel at 2.9984 x 10^{8} m/s.

Observed half-life = Proper half-life / ( 1 - ( 2.9984 x 10^{8} / c )^{2} )^{1/2}

= 2.2 microseconds / ( )

=

(Answer: 67.4 microseconds)

Now it takes 200 microseconds for the muons to travel down to the surface.

From our perspective on the ground, in these 200 microseconds we find that only about 3 half-lives have gone by.

This is why some muons reach the surface of the Earth.

On this basis, 1 in 8 muons are expected to reach the surface of the Earth because their 'moving half-life clocks' run slow.

# Questions - Test Yourself!

- You run past a radioactive source and you measure its half-life. Will it be:

a) Increased

b) Decreased

c) Unchanged.

(a) - A person runs past you carrying a radioactive source and you measure its half-life. Will it be:

a) Increased

b) Decreased

c) Unchanged.

(a) - You are in the park, listening to a military band. You start to walk away from the band. From your point of view, will the speed of the sound waves be:

a) Increased

b) Decreased

c) Unchanged.

(b) - You are in the park, listening to a military band. The band starts to march towards you. From your point of view, will the speed of the sound waves be:

a) Increased

b) Decreased

c) Unchanged.

(c) - You are in a car driving towards an illuminated traffic signal. From your point of view, will the speed of light from the signal be:

a) Increased

b) Decreased

c) Unchanged.

(c) - You are waiting to cross a road when you see a car approaching. Its lights are on. From your point of view, will the speed of light from the lights be:

a) Increased

b) Decreased

c) Unchanged.

(c) - Why is the super-bouncy ball model not a perfect one?

a) The apparent speed of the ball is affected by the motion of the observer.

b) The ball is affected by friction and so it slows down during the experiment.

c) The ball gains speed in the direction of the moving board because of friction.

(abc) - If you did not know about Special Relativity, why would you expect no cosmic ray muons to reach the surface of the Earth?

a) They travel too slowly.

b) They have a very short half-life.

c) They are absorbed by the air.

d) They have a negative charge.

(b) - Why do some cosmic ray muons reach the surface of the Earth?

a) They travel at just less than the speed of light.

b) A very large number are created at the top of the atmosphere.

c) Their half-lives appear to be lengthened by their speed.

d) The air is not very dense so it does not absorb them.

(c)