Turning Points Tutorial 6 - Einstein’s Theory of Special Relativity

 

This is quite a challenging tutorial.  Take it slowly and carefully.

 

Spoiler Alert:  There is some abysmal science fiction used in some of the examples!

 

Contents

Special Relativity

Time Dilation

Length Contraction
Mass Increase

Speed of Light

Mass and Energy

Bertozzi's Experiment

A Fly in the Ointment?

Forces on Charges

 

Special Relativity

The Theory of Special Relativity was first published as a paper in 1905.  A more general theory of relativity was brought out in 1916, which is beyond our scope.  The theory of special relativity is based on two statements or postulates:

  1. The laws of physics have the same form in all inertial frames of reference.  This does not just apply to Newton’s Laws, but all laws.  No experiment could show absolute motion or absolute state of rest.

  2. The speed of light in free space is the same in all inertial frames of reference.  It does not matter whether the light is coming from a moving or stationary source, or whether the observer is moving or stationary.

The speed of light is not affected by the relativity equations we are studying in this tutorial.  It shows physical invariance.

 

In special relativity, you may see the terms proper length, proper time, and rest mass.  These are length, time, and rest mass as observed by a stationary observer.

 

 

Think about this:

 

 

The speed of sound in air is 330 m s-1 relative to the air.

 

Question 1

What is the speed of sound relative to:                                                         

 

A: 

 

B: 

 

C: 

Answer

 

The observer at B and at C will hear a change of frequency of the siren as the police car goes past them.  This is called the Doppler effect.

 

Now consider this:

 

 

The speed of light does not depend on the observer.  All observers, whether at A, B, or C observe that light travels at 3 × 108 m s-1.  There is NO Doppler effect.

 

These postulates have far reaching implications for Physics, and all have been confirmed by experiment.  These are:

 

 

Maths Note

The equations of relativity use the Lorentz Transformation.  It was devised by a Dutch mathematician, Hendrik Lorentz (1853 - 1928).  Initially it was little more than a mathematical curiosity - a solution waiting for a problem.  It soon turned out to be the key to solving why nothing can travel faster than the speed of light.  The equation is written as:

[g - Lorentz factor; v - speed (m s-1); c - speed of light (m s-1).]

The term g is "gamma", a Greek lower case letter 'g'.

 

It can also be written as:

 

Another way of writing it is:

where:

In some books the authors use a, the reciprocal of g, to give:

This is used with length contraction.  Alpha is always less than 1.

 

 

If we plot gamma against speed, the graph looks like this:

 

At 3.0 × 108 m s-1 the line would be infinity.  This is not shown on this graph.  From this graph we can see that the Lorentz factor is always greater than 1.  When an object is stationary, g = 1.  When the object is travelling at less than 5.0 × 107 m s-1, g is very close to 1.  The increase to anything significantly above 1 takes place at about 1 × 108 m s-1.

 

The equation looks horrendous, but it is not difficult to work with provided you follow a strategy:


1. Work out the term
v2/c2.

2.  Take the number you work out away from one.  You will get a fraction.

3. Find out the square root of the answer to step 2.

4. Take the reciprocal of your answer to Step 3.  This is gamma (g).

5. Now multiply whatever relativistic quantity you are working with by gamma.

6.  Except for length, which you multiply by alpha.

 

It does not matter how you use the equation, as long as you show each step, and get the answer right!

 

Note that speeds of often given in terms of a fraction of c, the speed of light.  So 2.40 × 108 m s-1 is often given as 0.80 c.  In the equation, the c terms cancel out.


Time Dilation

If the speed of light is invariant, it can be shown that an observer will observe a moving clock as running slower than a stationary clock.  This has been shown by starting two very accurate clocks at exactly the same time.  This is synchronisation of the clocks.  One of the clocks stays in the lab, while the other is taken on an aeroplane which is flown for a very long time.  It is found that the moving clock is slightly behind the other clock.  The difference is not big (about 1 ns).  Just in case one clock was wrong, the experiment is repeated with the clocks swapped over.

 

Suppose the time on the stationary clock is t0, and the time on the moving clock is t, (or t' ) the two are related by:

 

 

The term t0 is the proper time, the time between two events as measured, as if the moving object were travelling slowly.  We could call it the "rest time".  The next part of the equation is called the Lorentz factor, and is often represented by g, "gamma", a Greek letter 'g'.  At normal speeds, g = 1, but increases rapidly as we approach the speed of light.

 

This can be rewritten as:

 

 

Or as:

 

 

 

 

The term t is the time between the two events as measured by an observer in a frame which moves with a constant velocity v relative to the other observer.

 

The equation looks horrendous, but it is not difficult to work with provided you follow a strategy:


1. Work out the term
v2/c2.

2.  Take the number you work out away from one.  You will get a fraction.

3. Find out the square root of the answer to step 2.

4. Divide the term t0 by the answer to step 3 to get t.

 

The time t is always longer than t0.  If it isn’t, something has gone wrong.  This effect is called time dilation.

 

Worked example

A spaceship passes the Earth at a speed of 0.8 c (c = 3.0 × 108 m s-1) and flashes a signal lamp for 2.0 ms.  What is the duration of the signal on Earth?

Answer

Don’t bother to convert 0.8 c into m s-1!  It will “come out in the wash”.

 

Work out v2/c2 = (0.8 c)2 ÷ (1 c)2 = 0.64

 

Take away 0.64 from 1: 1 – 0.64 = 0.36

 

Take the square root of 0.36:  (0.36)0.5 = 0.60

 

Now work out g g = 0.60-1 = 1.667

 

Since:

 

time t = 2.0 ms × 1.667 = 3.33 ms

(Author's note: I would like to thank my student Joseph Reffitt who got me thinking hard about the wording of this example.)

Not that hard is it?

 

Time dilation has been shown even in relatively slow moving objects like aeroplanes.  A clock was taken up in an aeroplane and flown about for several hours, while a second identical clock was left running on the ground.  There was a tiny but measurable difference between the two.  In our sedate life styles, the difference is so tiny as to be negligible.

 

Muon decay gives us more tangible evidence for time dilation.  Muons (see Particle Physics) are unstable subatomic particles that are formed in the upper atmosphere by cosmic rays.

 

Question 2

What is a muon?

Answer

 

 

Muons can be detected at an observatory at the top of a mountain, at A, and at the bottom of the mountain at B.  Here are some data about muons:

 

Question 3

Use the data above to work out:

(a)    The time taken for a muon to travel 2000 m;

(b)   The number of rest half lives elapsed in this time;

(c)    The expected intensity at B compared with that at A;

(d)   The half life of moving muons relative to a stationary observer;

(e)    The number of moving half lives elapsed in the time (a)

(f)     The intensity at B assuming time dilation. 

Answer


Length Contraction

Another consequence of the invariance of the speed of light is that an observer measuring a rod moving parallel to its length will find that it is shorter relative to its stationary length.  In effect if you measure a moving car, you will find that it is shorter than the stationary car.  This effect is called length contraction.

 

As with time dilation, the change is so tiny as to be negligible.  But this is not the case for objects moving close to the speed of light.

 

The relationship is a little easier than the one for time:

(See the Maths Note above for what is meant by alpha)

This can be written as:

 

 

This is more easily written as:

 

 

The term l0 is the proper length as measured in the frame that is at rest relative to the object.

The term l is the length as measured by an observer in a frame of reference that moves at a constant relative velocity of v.

 

Again this looks a fairly horrendous equation, but use the problem solving strategy below and it’s not that difficult:

1. Work out the term v2/c2.

2.  Take the number you work out away from one.  You will get a fraction.

3. Find out the square root of the answer to step 2.

4. Multiply the term l0 by the answer to step 3 to get l.

 

Your moving length will be less than the stationary length.  It it’s more, you’ve made a mistake.

 

Question 4

A spaceship of length 60.0 m passes the Earth at a speed of 0.98 c.  What is the length of the spaceship as seen by an observer on Earth? 

Answer

Question 5

An observer measures the length of a metre rule to be 80 cm.  What is the speed relative to the observer? (c = 3.0 × 108 m s-1)

Answer

 

Let’s go back to our mountain:

 

Muon decay can be explained by applying the contraction equation.  We consider the problem from the moving frame of reference of the muon instead of the fixed frame of reference of the Earth.  In this case the muon sees the Earth and the mountain as a giant pancake with a pimple on it.  Let’s remind ourselves of the data:

 

Question 6

Use the data to calculate:

(a)    the height of the mountain as it passes a muon;

(b)   the time taken for the mountain to pass the muon;

(c)    the number of rest half lives elapsed in the time in part (b)

(d)   the percentage of muons remaining when point B is passed compared to point A

Answer

 

Does this mean that my 30 km train journey to work in the morning is shorter than it really is?  In theory yes, by less than 1 micrometre.


Mass Increase

Two key points

 

Therefore a stationary observer will measure the mass of a moving object as being greater than that object when it is stationary relative to the observer.  This is described in the equation:

 

Or:

 

The term m0 is called the proper or rest mass.  It is the mass as measured by an observer in a frame of reference which is at rest relative to the observer.

 

The term m is the relativistic mass which is the mass as measured by an observer in a frame that is moving at a constant velocity v.

 

The equation looks horrendous, but you should be getting used to these equation by now.  It is not difficult to work with provided you follow a strategy:
1. Work out the term
v2/
c2.

2.  Take the number you work out away from one.  You will get a fraction.

3. Find out the square root of the answer to step 2.

4. Take the reciprocal to get g.

5. Multiply the term m0 by g to step 4 to get m.

 

The time m is always  bigger than m0.  If it isn’t, something has gone wrong.  

 Question 7 

The rest mass of an electron is 9.11 × 10-31 kg at rest.  What is its mass when it is travelling at 0.998 c?

Answer


Impossibility of Speeds Greater than the Speed of Light

As v approaches c, the mass m gets bigger and bigger.   The closer it gets, the more it tends to infinity.  Further acceleration requires a force approaching infinity.  So it is impossible for the speed of light to be reached, let alone exceeded by an object of non zero rest mass. We can even plot a graph:

 

 

Now think about this:

 

 

Although the immediate reaction is to say that the speed of C relative to A is 4 ´ 108 m s-1, or 4c/3, it can be shown using equations similar to the ones we have seen above that the relative velocities of the two rockets are 12c/13.  (See Einstein Velocity Additions at the end of this tutorial.) Therefore the radio waves from A can catch up with C.  This requires more complex relationships.

 

Question 8 

A powerful laser is set up on Earth which can shine a spot of light on the moon.  Suppose this laser is set up so that it sweeps through 180o a second, i.e. 30 rpm.  What is the speed at which the beam sweeps across the surface? 

Distance from Earth to the Moon 4 ´ 108 m.

Answer

 

This is above the speed of light, but no mass is being transferred so it is possible.

 

Particles with zero rest mass, photons and neutrinos, always travel at the speed of light.  It has been suggested that there may be a group of particles called tachyons for which v > c at the instant they are created.  They always travel faster than light and speed up as they lose energy.  They have never been detected.

 

 

Mass and Energy

In classical physics, we know that kinetic energy is given by:

If we accelerate an electron with a voltage V, we know that all the energy is kinetic, so we can write:

 

If we plot a graph of speed against accelerating voltage, we see:

 

 

We see from this graph that at about 2.6 × 105 V, the electron speed is 3.0 × 108 m s-1.

 

Question 9

Show that, when an electron is travelling at 3.00 × 108 m s-1, the accelerating voltage is about 2.6 × 105 V.

 

Mass of electron = 9.11 × 10-31 kg;

Magnitude of the charge of an electron = 1.60 × 10-19 C.

Answer

 

Since nothing can travel at more than the speed of light, clearly the classical model has broken down, so a more sophisticated approach is called for.  Einstein's Theory of Relativity provided the answer in that if energy is supplied to an object, its mass increases.  Conversely, if energy is transferred away from the object, the mass decreases.  Therefore, as an object speeds up, its kinetic energy increases and so does its mass.  According to the theory of special relativity, the increase in energy is proportional to the increase in mass:

 

DE µ Dm

DE = Dm c2

 

In this case c2 (= 9.00 ´ 1016 m2 s-2) is the constant of proportionality.  This can be extended to give us the total energy of a particle:

 

Ek = eV

E = m0c2 + eV

 Therefore:


So we can write an equation for the total energy in terms of the rest energy and the kinetic energy:

 

 

This again looks horrendous, but apply the problem solving strategy:

1. Work out the term v2/c2.

2.  Take the number you work out away from one.  You will get a fraction.

3. Find out the square root of the answer to step 2.

4. Divide the term m0 by the answer to step 3 to get m.

 

Worked example

Calculate the speed of an electron which has been accelerated from rest through a p.d of 2.0 × 106 V

What NOT to do!

eV = ˝ mv2

v2 = (2 ´ 1.6 ´ 10-19 C ´ 2.0 ´ 106 V) ÷ 9.11 ´ 10-31 kg  = 7.02 ´ 1017 m2 s-2    

                    v = 8.38 ´ 10m s-1  

Answer

Total energy = rest energy + kinetic energy 

Kinetic energy = charge ´ voltage

Kinetic energy = eV = 1.6 ´ 10-19 C ´ 2.0 ´ 106 V = 3.2 ´ 10-13 J

Rest energy = m0c2 = 9.11 × 10-31 kg × 9.0 × 1016 m2 s-2 = 8.2 × 10-14 J

Total energy = 3.2 × 10-13 J + 8.2 × 10-14 J = 4.02 ×10-13 J

Now use:

Substituting gives us:

Rearranging gives us:

Square this to get rid of square root:

Now rearrange

Rearranging and  the square root gives us:

v = 0.979 c = 2.9 ´ 108 m s-1

 

There are a lot of steps in a calculation like this, but they are quite simple, so don’t panic when you see an example like this.

 

Now you have a go:

Question 10 

What p.d. is needed to accelerate a proton, rest mass 1.67 ´ 10-27 kg from rest to a speed of 0.95 c? 

Answer

 

Bertozzi's Experiment

In 1964, an American physicist, William Bertozzi, carried out an experiment to determine the relationship with speed and kinetic energy of accelerated electrons.  He used apparatus like this simplified arrangement:

 

 

Electrons from an electron gun are fired in bursts and they travel as a group in an evacuated tube.  They are accelerated using a linear accelerator.  They pass through a pair of plates at A and are detected as a pulse on the CRO.  They then strike an aluminium target at B, and a second pulse is shown on the CRO.  The main point of this experiment is that the actual kinetic energy of the electrons is measured by measuring the temperature rise of the aluminium target at B.  Previously the kinetic energy had been inferred from:

 

Ek = eV

 

Instead the electrons give up their kinetic energy to the aluminium target of mass m, and specific heat capacity c.  We then work out the energy transferred to the target by measuring its temperature change.  If we know how many electrons there are in each group, we can work out the kinetic energy for each electron.  The energy going into the target is:

 

If there are n electrons, we can say that the kinetic energy for each electron is:

 

Question 11

1.2 × 1015 electrons are accelerated by an accelerating voltage of 2.50 × 106 V.  They strike an aluminium plate of mass 100 g. 

(a)  Calculate the charge in coulomb (C)

(b)  Calculate the temperature change.

 

Magnitude of electronic charge = 1.60 × 10-19 C;

Specific heat capacity of aluminium = 900 J kg-1 K-1.

Answer

 

We can generate a graph of kinetic energy against speed:

Kinetic energy = Total energy - rest energy

In physics code:

 

The graph shows the kinetic energy in eV against the speed of the electron:

Bertozzi's results were within 10 % of the idealised data that were used to generate the graph above.  The conclusion was that as the kinetic energy of the electrons increased, the speed of the electrons approached a limiting speed.  Therefore nothing can exceed the speed of light.

 

Note that the traditional formula for kinetic energy:

is a solution for:

 

when the speed is very much less than the speed of light.  You are not expected to know this for the exam.

 

 

A Fly in the Ointment?

In Summer 2011 observations suggested that neutrinos could travel faster than light.  The neutrinos had their source at the CERN experiment, and travelled to be detected at a laboratory in Italy, 730 km away.  The little brutes were seen to arrive 60 nanoseconds before the expected time.  Many physicists remained skeptical, but the results were repeated.  If it were to be concluded that neutrinos could travel faster than light, then a fundamental tenet of physics would have been been undermined. 

 

Later it was discovered that there was a faulty connection between a GPS device and a computer, so the data were wrong all the time.  Neutrinos don't travel faster than light.  That's the little brutes put back in their place.  At least one physicist ended up "on the carpet".

 

 

Forces on a Charge from an Electric Current (IB students)

We know that when there is an electric current flowing through a wire, there is always a magnetic field.  It is a consequence of special relativity.  Remember that time and length are not absolute.  They are observed differently depending on whether the observer is in a stationary or moving frame of reference:

 

The argument goes like this.  Consider this piece of copper wire.  It is long and straight.  It is neutral as there are equal numbers of positive and negative charges.  (The model is a lattice of ions in a sea of free electrons.)  We can show them arranged like this for simplicity:

 

 

If we bring a positive charge to the wire, there is no electrical force:

 

Now suppose we connect the wire to a battery.  The electrons move to the right as shown.

 

 

If we bring a positive charge to the wire, there is no electrical force:

 

 

Although the electrons are moving, the charge density remains the same, so there is no electrical force acting on the charge.

 

Now suppose we move the electric charge at the same velocity as the electron flow:

 

 

For the stationary observer, there is no change in the charge density, and the wire is still neutral, so there is no electrical force.

 

Now let's move to the frame of reference of the charge.  We are observing from the point of view of the charge.

 

 

As far as the charge sees it, the positive charge are moving from right to left.  Since they are moving, the distance between the positive charges is reduced as a result of their motion.  The motion is not at all fast, but the principle of length contraction still applies.  Also, since the electrons are stationary, they are spread out.  So the situation according to the charge is this:

 

 

Therefore the positive charges have a higher charge density than the than the electrons, so to the charge the wire is not neutral, and gives out a repulsive electrical force:

 

 

Now the stationary observer will see the effect of the force on the charge.  And being a good classical physicist, the observer will conclude that the force is moving due to the action of a magnetic field. 

 

The electron drift speed in copper is less than 1 mm s-1.  The length contraction would be very small, and normally we would ignore it.  However small it is, it is still definite.  But we see a measurable effect because:

From special relativity, we can conclude that magnetic fields observed from a stationary frame of reference are actually electric fields from a moving frame of reference.

 

Einstein Velocity Additions (IB students)

As we saw above, the maximum relative velocity is c, 3.0 × 108 m s-1.  If an object is travelling at 0.6 c, and another object is approaching at 0.8 c, the relative approach speed is NOT 1.4 c, but 1.0 cWe can obtain expressions for relative velocities for moving objects.  These are called Einstein Velocity Additions

 

Consider this situation (Spoiler alert - some very bad science fiction):

 

A stationary observer is on a far-distant planet.  She sees a spacecraft travelling at a velocity of v.  It fires a probe in the direction it is travelling.  To the stationary observer the velocity is u.  In the spacecraft there is a moving observer.  He sees that the probe has a velocity of .

 

To the stationary observer the velocity of the probe u can be worked out as:

 

 

To the moving observer, the velocity of the probe is , which is worked out as:

 

 

Worked example

The spacecraft in the example above has a velocity 0.40 c as measured by the stationary observer on the planet.  The probe has a velocity of 0.50 c as measured by the stationary observer.  What is the velocity of the probe as measured by the moving observer in the spacecraft?

Answer

Work out the term uv/c2 = (0.40 c × 0.50 c) ÷ c2 = 0.20

The value downstairs = 1 - 0.20 = 0.80

Therefore:

= (0.50 c - 0.40 c) ÷ 0.80 = 0.125 c

 

Question 12

Use the value of u' in the example above to verify that the stationary observer's measurement of the velocity of the probe is correct at 0.5 c.

Answer

 

 

Spacecraft Moving in Opposite Directions

In the last example, the directions were all positive, i.e. from left to right.  Now let's consider two spacecraft approaching each other in opposite direction, as in the picture.

 

 

Our initial reaction is to say that the approach speed is given by:

 

But we can't use that in a relativity context.   The straight brackets show that we are talking about the values of the velocities (i.e., speeds). 

 

Suppose we have an observer on spacecraft A.  The relativity equation is as before:

 

The uʹ is because we are considering the moving frame of reference.

 

The stationary observer will see the velocity of spacecraft B as vB, therefore u = vB.  She will also see the velocity of spacecraft A as vA, therefore v = vA .

 

The moving observer in spacecraft A will see the velocity of spacecraft B as uʹ.

 

So we can change the equation to:

 

Let's put some numbers in to see what happens:

 

Worked Example

The spacecraft in the example above has a velocity 0.60 c as measured by the stationary observer on the planet.  The probe has a velocity of 0.50 c as measured by the stationary observer.  What is the velocity of the probe as measured by the moving observer in the spacecraft?

Answer

Use:

Work out the term uv/c2 = (0.60 c × -0.50 c) ÷ c2 = -0.30

 

The value downstairs = 1 - -0.30 = 1.30

 

Therefore:

= (0.60 c - -0.50 c) ÷ 1.30 = 1.10 c ÷ 1.30 = 0.85 c

 

 

If the velocities of the spacecraft are very much less than the speed of light, the "common sense" relationship:

 

is true.

 

 

Further material on Special Relativity can be found in Physics 6 Tutorial 1 and Physics 6 Tutorial 18.