Topic 6 - Einstein’s Theory of Special Relativity

In the exam you are expected to:

 

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.

 

Think about this:

 

 

The speed of sound in air is 330 m/s 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.  There is NO Doppler effect.

 

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

·        Time dilation;

·        Length contraction;

·        Dependence of mass on velocity;

·        Equivalence of mass and energy;

·        The impossibility of acceleration beyond the speed of light.


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.  Suppose the time on the stationary clock is t0, and the time on the moving clock is t, the two are related by:

 

 

This can be rewritten as:

 

 

Or as:

 

 

The term t0 is the proper time, the time between two events as measured by an observer in the frame at which they occur at the same point.

 

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.

 

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) and flashes a signal lamp for 2.0 ms.  What is the duration of the signal on Earth?

Don’t bother to convert 0.8 c into m/s!  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:  Ö0.36 = 0.6

time t = 2.0 ms ¸ 0.6 = 3.33 ms

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 are unstable subatomic particles that are formed in the upper atmosphere by cosmic rays.

 

Question 2  What is a muon (from Physics AS Module 1)?  ANSWER

 

 

Muons can be detected at an observatory at the top of a mountain, at A, and at the bottom of the mountian at B.  Here is 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 sationary observer;

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

(f)     The instensity 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:

 

 

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 realative to the observer? (c = 3.0 ´ 108 m/s) 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 car 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:

 

 

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. Divide the term m0 by the answer to step 3 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, 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.  Therefore the radio waves from A can catch up with C.


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

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 ´ 1016 m2/s2) is the constant of proportionality.  This can be extended to give us the total energy of a particle:

 

 


Total energy = Rest energy + kinetic energy

[Learn this; it is useful]

 

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 = ½ mv2

v2 = 2 ´ 1.6 ´ 10-19 C ´ 2.0 ´ 106 V = 7.02 ´ 1017 m2/s2     O 

               9.11 ´ 10-31 kg

v = 8.38 ´ 108 m/s  

Total energy = rest energy + kinetic energy 

Kinetic energy = charge ´ voltage

Kinetic energy = e V = 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 m2s-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

 

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 9   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

 

Summary

 

Time dilation:

 

Length Contraction

 

 

Mass Increase

 

Mass Energy Equivalence

 

 

 

 

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