DYNAMICS AND RELATIVITY(PART II)

Lecturer: Dr. D.J. MillerRoom 535, Kelvin Buildingd.miller@physics.gla.ac.uk

Location: 257, Kelvin Building

No. of Lectures: 9

Recommended Text: Young & Freedman, 11th Edition

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Outline

Special Theory of RelativityInvariance of the physical laws

Relative nature of simultaneity

Time dilationLength contraction

The Lorentz transformation

9.

Angular Momentum and EnergyAngular momentum and torqueRotational energy

Analogy to linear quantities

8.

Linear Momentum and CollisionsMomentum and impulse

Conservation of momentum

The centre-of-mass

Elastic and inelastic collisions

7.

Conservation of EnergyPotential Energy

Conservative and non-conservative forces

6.

Work & Kinetic EnergyWork

Work done by a varying force

Kinetic energy

Power

5.

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A B C D

Question: A moving pendulum reaches the bottom of its swing. Which diagram correctly represents the forces (in red) on the pendulum at this point?

The correct answer was B.

The acceleration is towards the pivot, therefore the force must be towards the pivot.

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5. Work and kinetic energy

5.1 The definition of “work” in physics

“Work” is done whenever a force is applied to move an object from one place to another.

If the force is F, and the displacement of the object is s, then the work done W is given by

Y&F: Ch 6.1

So we also have

Note that both F and s are vectors, while W is a scalar. Recall the scalar product of two vectors

θ

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Units:Force, F : Newtons [N]Displacement, s : metres [m]Work, W : Joules [J]

James Prescott Joule1818-1889

1 Joule = 1 Newton metre

Example: Consider a force of 5 N pushing an object 2 m in the same direction as the direction of the force.

1 J = 1 N m

The work done is

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If the motion is in a different direction from the force, we have to take the direction of the vectors into account.

Example: What is the work done by gravity when a 5kg object moves 2m down a frictionless 30o slope?

30o

θ Displacement down the slope is

g ≈ 9.8 ms-2

Angle between force and displacement is θ = 60o

1 N = 1 kg m s-2

Force downwards is Newton’s second law

Work done is

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Work can also be negative

work is positive

work is negative

work is zero

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Question: A train is heading towards a bridge over a river. At the last minute the train driver notices the bridge is out! He slams on the breaks and manages to stop the train just before it goes over the edge…… If he is 100m away from the bridge when he puts on the breaks and the brakes apply a force of 10,000N how much work does the frictional force of the brakes do?

What is the work done by the frictional force of th e brakes?

A: None C: 50,000JB: 1,000,000J D: -1,000,000J

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The correct answer was D: -1,000,000 J

The work done is

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5.2 Work done by a varying force

So far, the force applied has been constant, but this need not be the case. Let’s consider a force which changes throughout the motion.

Y&F: Ch 6.3

We can divide the displacement into lots of little (infinitesimal) displacements such that

Let the force on the object for a displacement be

Then the work done on segment i is

The total work done is

Taking the limit this turns into an integral:

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Imagine a simpler case, where the motion is along only the x direction, starting at x1

and ending at x2.

Now and

unit vector in the x direction

component of Fin the x direction

⇒

The two shaded areas above are equal when ∆x → 0

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Notice that when Fx is independent of the position, then we return to our old formula:

Alternatively

Example: An example of a force which varies with displacement is the force exerted by a spring when you pull it.

k is known as the spring constant

The force applied by the spring is given by Hooke’s Law

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Robert Hooke1635-1703

Hooke’s Law tells us that the force required to stretch a spring increases linearly with the distance that you pull it.

The work done in stretching the spring by an amount X, is the area under the curve to the left.

Alternatively:

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Example: Work done pushing a swing

Let’s say we push a child in a swing. The child has a weight w, and the length of the chain is R. Assuming that we push him very slowly, what is the work done by each of the forces acting on the child?

“slowly” ⇒

If T is the tension on the rope and F is the force with which we push him, then

Y&F: Ex 6.9

First we need to know how the forces vary with θ.

x-direction:

y-direction:

all the forces are (approximately) in equilibrium (i.e. sum to zero)

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For the force F :

Now we can calculate the work done by each force moving from A (θ=0) to B (θ = θ0).

For the force w (gravity):

R

A

B

For the force T (tension): the angle between ds and T is always 90o, so and the work done by T is zero.

Notice that since the total net force is zero, so can do no work!

Remember that it is only the force in the direction ds which counts!

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5.3 Kinetic energy Y&F: Ch 6.2

Consider an object of mass m, under a constant force F as it moves from x1 to x2.

Since F = m a, the object is accelerated. Its velocity changes from v1 to v2.

Displacement:

Velocities:

But Newton’s second Law tells us that

The work done by the force is

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Definition: The Kinetic Energy of an object of mass m moving with velocity v is given by

It’s units are kg (ms-1)2 = N m = J

Work-energy theorem: The work done on an object by the total net force is given by the change of the object’s kinetic energy.

N.B. Notice that this is the work done by the total net force . In our swing example, Wtot = 0 so the change in kinetic energy was zero. We could not have used this to calculate the work done by the individual forces.

This should be familiar from the last slide

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The work energy theorem is still true even if the force changes during the motion.

If we go back to our small steps ∆xi, then for step i the work done is

The total work done is the sum over all the steps:

change in kinetic energy for step i

So the total work done is still equal to the change in kinetic energy.

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Question: You throw a stone into a pile of nice soft mud. The mud exerts a constant resistive force on the stone, stopping it in 3cm. How much faster do you have to throw the stone to make it penetrate 12cm?

A: Twice as fast

B: Three times as fast

C: Four times as fast

D: Eight times as fast

E: Sixteen times as fast

The correct answer is…. AThe work done is force × distance, so we need 4 times as much energy.

But the energy is so we only need twice as much speed to make 4 times the energy.

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Example: work done by a pile-driver

a) the speed of the pile-driver, v, when it reaches point 2b) the average force, F, the pile-driver exerts on the beam

A pile-driver of mass m = 200kg drops a distance s12 = 3m

before hitting a beam. It drives the beam a distance s23 = 7.4cm into the ground. If the guide-rails exert a constant frictional force on the pile-driver of f = 60N, what is

a) Between points 1 and 2, the forces on the pile-driver are friction and gravity:

Total net force

Total work done

Change in kinetic energy =

Work-energy theorem ⇒

Y&F: Ex 6.5

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b) Between points 2 and 3, there is an extra force on the pile-driver, from the beam pushing it up. This is the force, F, that we need to work out.

Total net force

Total work done

Change in kinetic energy =

Work-energy theorem

from a)

So

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5.4 Power

It is often useful to know how the work done by a force changes over time. We define Power as the rate of change of work done with time:

Unit: Watt [W] 1 W = 1 Js-1

James Watt 1736-1819(Try not to confuse work W with Watts W!)

The work done between time 0 and time T is given by

If the power is constant then total work done is

Y&F: Ch 6.5

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Recall that the work done for a constant force is

So the power is

Power = Force × velocity

We can relate Power to Force and Velocity , just as we related Work to Force and Displacement :

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Example: resistive forces on a car

R F

A car has a maximum speed of 150km/h and produces a maximum power of 25kW. What is the resistive force it feels at its maximum speed?

At its maximum speed, v, the car is not accelerating, so must be under zero net force.

i.e.

So, the rate at which work is done against the resistance is the maximum power output of the car.

Work done by the engine is

number of seconds in an hour

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If the resistive force is constant and the car weighs 1000kg, what is its maximum speed up a 30o slope?

For no acceleration, there should be no force

Power output of the engine isR

F

w = mg

N

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Question: Bob’s drives off in his new car and applying a (leisurely) constant acceleration he does 0 to 30 mph in 10s. Later in the day, when trying to overtake, he needs to accelerate from 30 to 60 mph in 5s. The power required for the overtake is:

A: the same asB: twiceC: four timesD: six timesE: eight times

… the power required for the acceleration from 0 to 30mph.

For the overtake,

For the first acceleration, with v = 30mph

So we need 3 times the work done in half the time.

The correct answer is…. D

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6. Conservation of Energy

6.1 Potential Energy

Imagine an object of mass m, falling freely under gravity. Let’s say it falls from a height y1 to a height y2.

The work done by the force of gravity is

and this causes an increase in the kinetic energy of the object.

Where does this energy come from? The higher the object is at the start, the more potential it has to gain energy when it falls.

We say that an object of mass m, at a height y, has gravitational potential energy , given by

Notice that it doesn’t really matter where the point y = 0 is, since only changes in potential energy matter.

Y&F: Ch 7.1 & 7.2

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In the example just shown, the object loses potential energy, but objects can also gain potential energy by moving against the gravitational field.

If we apply an upwards force F to the object of mass m, moving it from y1 to y2, we will be doing work

If we choose then the object moves with zero acceleration. Its velocity doesn’t change, so it has the same kinetic energy as before, but it gains potential energy

The work done is stored as potential energy.

If F is larger than mg, then there will be a net upwards force and the object will accelerate upwards. The extra work done by the force will contribute to the kinetic energy of the object.

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Let’s return to our free-falling object (i.e. )

Change in potential energy

Work done by gravitational force

Work-energy theorem

We define the total energy to be the sum of the kinetic and potential energies:

The total energy is conserved . It does not change.

When a force conserves the total energy, we say it is a conservative force .

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Example: The maximum height of a projectile Y&F: Ex 7.4

A projectile is fired from point 1 with an initial speed v0 at an initial angle α0. What is the maximum height it reaches?

At point 1:

At point 2:

Conservation of energy

Therefore the maximum height is

Also, since there is no force in the x-direction, we know that the velocity in the x-direction cannot change.

and we have

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Example: Looping the loop!

R

h Fg

Fc

1

2

3

If we don’t switch on the engine(!), how high (in terms of R) does the car have to start in order to loop-the-loop successfully?

The crucial thing for survival is that the upward force should be bigger than the downward force at point 3. We have two competing forces – the gravitational force Fg, pulling down, and the centrifugal force Fc, pushing up.

Gravitational force at point 3 is:

Centrifugal force at point 3 is:we need

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So what condition on h gives ? Let’s use energy conservation again.

At point 1:

At point 2:

At point 3:

Conservation of energy

This tells us that

The requirement then becomes

So we need a starting height

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Question: A bottle is dropped from a balcony and hits the pavement with a speed v. To double the speed of impact it would have to be dropped from a balcony

A: twice as highB: three times as highC: four times as highD: five times as highE: six times as high

Conservation of energy,

So to double the speed we need 4 times the height

The correct answer is…. C

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There are many other forms of potential energy too.

For example, a spring can contain elastic potential energy .

Remember Hooke’s Law ?

So when we extend the spring from x1 to x2 we do work

The elastic potential energy stored in a spring is therefore given by

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What happens if we take away our hand and let the spring pull the mass back?

Work done by the spring is

Work-energy theorem tells us that the work done is equal to the change in kinetic energy

So

Again, total energy (potential + kinetic) is conserved.

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Potential energy is the area under the curve of

Compare this with gravityNotice that the position of O is no longer arbitrary

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We can work in the opposite direction, and derive the force from the potential

In 1 dimension,

For our two examples:

More generally, in 3 dimensions,

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The gravitational force exerted on a mass m by another mass M at a distance r, is

G is Newton’s constant

Isaac Newton 1643-1727

The potential energy is

constant

Choose at

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If with y small

with

The potential we used before was an approximation which is valid when the change in height is small

radius of the Earth

height above the Earth’s surface

Then

“small” means

Radius of the Earth = 6,380,000 mMass of Earth = 5.97 × 1024 kg

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Example: Escape Velocity of a satellite

A satellite of mass m is launched from the Earth’s surface with velocity v. What v do we need, in order that the satellite escapes the Earth’s pull?

At launch and

“Escape” means that the satellite reaches with

Conservation of energy

So

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Question: I want to mountain-bike a distance of 1km up a hill. I can choose to go straight up the hill (1km) or I can choose to zigzag my way up, taking twice the distance (2km). The average force which I must exert on the zigzag route is

A: 1/4B: 1/3C: 1/2D: equal to

the average force I exert going straight up.

The correct answer is…. C

The work I need to do is the same, to provide the gravitational potential energy (independent of the path). But W=Fs so making s twice as long means that I can make F half as big.

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6.2 Conservative & non-conservative forces Y&F: Ch 7.3

We have already seen two examples of a conservative force : gravity and a spring.

If the force is conservative, then we can convert potential energy into kinetic energy and vice versa, with no energy lost.

potentialenergy

kineticenergy

potential energy

kinetic energy

potential energy

kinetic energy

Total mechanical energy (K +U) is conserved.

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AB

path 2

path 1

Imagine a conservative force F, which depends on position, acting on an object. What is the work done by the force when an object moves from point A to point B?

Work done taking path 1 is

Work done taking path 2 is

For a conservative force we know the potential energy at both A and B, so the work done must simply be

The work done depends only on the endpoints, not on the path taken.

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This is really the definition of a conservative force. It is this definition which allows us to define potential energy in the first place.

AB

Imagine we now regard A → B → A as the complete path

Work done by F isand is independent of the path.

This means that the potential energy at point A (or B) is well defined and does not change depending on the past history of the object.

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A non-conservative force is a force where the work done by a force moving an object from point A to point B depends on the path taken.

This is sometimes called a dissipative force .

A good example of a non-conservative force is friction .

The frictional force always opposes the motion so always, so the work

AB

done moving A → B → A is

The work done depends on the path taken

⇒ we cannot define potential energy

The work done is usually dissipated as heat.

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Y&F: Ex 7.12Example: “Futon” fun

You are rearranging your furniture and want to move your 40kg “futon” 2.5m across the room. The straight line path is blocked by a coffee table, so you need to move the “futon” around the table as shown. How much work do you do moving the “futon”around the table compared to the straight line path? (The coefficient of friction is 0.2)

Frictional force is opposing the motion,so

Work done moving the “futon” is

For the straight path:

For the path around the table:

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Question: Two identical balls are rolled (from rest) down the slopes shown. Which ball will be travelling faster when it reaches the other end?

1 2

A: Ball 1 will be faster.B: Ball 2 will be faster.C: The two balls will have the same speed.

The correct answer is…. CThe total energy is conserved, so

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7. Linear momentum and Collisions

7.2 Momentum and impulse Y&F: Ch 8.1

Definition: The momentum of an object is its mass × its velocity.

It is a vector quantity (it has magnitude and direction) usually denoted

It is measured in units of kg m s-1

Remember Newton’s second law . Do you see a similarity?

If the mass is constant,

⇒ Force is the rate of change of momentum

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Definition: The impulse applied to an object is the Force × the time the force is applied

This is also a vector quantity and is usually denoted

e.g. A force of 3N acting on an object for 4s gives an impulse of 3N × 4s = 12Ns in the same direction as the force.

It is measured in units of Ns

If the force is constant, then

so

The impulse given to an object is equal to its chan ge in momentum.

- this is the impulse-momentum theorem

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This impulse-momentum theorem also holds when the force is not constant, but we must slightly modify our definition of impulse.

Definition for a varying force:

Then

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m m

Y&F: Ch 8.27.3 Conservation of momentum

When two objects collide, the total momentum of the system is conserved .

This is a consequence of Newton’s third law: “For every action, there is an equal and opposite reaction.”

If is the force on the object with mass m, and is the force on m then N3 tells us

But so,

constant

momentum is conserved

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Example: Recoil of a rifle Y&F: Ex 8.4

A stationary rifle of mass mR = 3kg fires a bullet of mass mB = 5g at a velocity vB = 300 ms-1 relative to the ground. If the rifle is allowed to recoil freely, what is its recoil velocity? What are the kinetic energy and momenta of the bullet and rifle just after firing?

momentum before = 0 (everything at rest)

momentum after =

momentum before = momentum after ⇒

So, the velocity of the rifle in the x-direction is:

⇒

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Kinetic energy before

momentum before = momentum after = 0

Kinetic energy after

2 things to notice:

The kinetic energy of the bullet is much larger (velocity is more important to energy than mass since it is squared)

The kinetic energy is not conserved in this case, but momentum is.(the energy comes from the explosive reaction which launched the bullet)

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Conservation of momentum is true for more complicated systems too.

If there are no external forces on the system, then momentum is conserved.

constantthe sum here is over all the bodies in the system

7.3 The centre-of-mass Y&F: Ch 8.5

Definition: The centre-of-mass of a collection of n objects of mass at positions is

x

y The centre-of mass will move with a velocity

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Remember momentum , so

So as long as the masses don’t change,

constant

momentum conservation

constant

The velocity of the centre-of-mass is the same afte r the collision as it was before.

The centre-of-mass is also useful if there is a net force on the system and momentum is not conserved.

The acceleration of the centre-of-mass is

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The centre-of-mass moves like an object of mass under a force

e.g. Consider an object breaking up in flight.

The motion of the centre-of-mass after the collision is the same is if the object had stayed in one piece.

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7.4 Elastic and inelastic collisions Y&F: Ch 8.3 & 8.4

For the moment, lets assume there are no external forces.

In a collision total momentum is conserved , but total mechanical energy may not be .

Definition : An elastic collision is a collision in which total mechanical energy is conserved .

Example: A ball of mass 1 kg, travelling at 2 ms-1 collides elastically with another ball of mass 2 kg travelling in the opposite direction at 1 ms-1. What are their velocities after the collision?

Before:

After:

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I have two equations, and two unknowns ( and ), so I can solve the system.

Solve for : after some algebra (check it!)…. remember

The + sign gives (and ) which is the trivial case of no collision

The - sign gives

and

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Notice that I went about this more generally than I needed. In this case the momentum (in the x-direction) is

We are in the centre-of-mass frame (the frame where the centre-of mass is at rest), so

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Conservation of energy

So and . As before, the + sign gives the trivial solution.

This is a general result for an elastic 2 → 2 collision in the centre-of-mass frame.

Even away from the centre-of-mass frame this result is useful.

The centre of mass velocity is

And the velocity in a general frame is

velocity in centre-of-mass frame

(similarly for )

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This collision was only in one dimension, but it could in principle be in more dimensions.

For example, imagine the collision of snooker/pool/billiard balls:

Momentum is a vector quantity, so it is conserved in both directions

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Kinetic energy conservation ⇒now this is the magnitude

of the whole vector

rearrange

The velocity vectors form a right handed triangle, so the two balls fly off at 90o

Before the collision:

After the collision:

Momentum conservation

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Question: Two balls moving in a three dimensional space, collide are rebound in an elastic collision. The balls are lit from above (from a light source very far away) and cast shadows on a two-dimensional plane. Pretend the shadows have a mass equal to their ‘parent’ ball. Does the collision of the shadows conserve

A: kinetic energy onlyB: momentum onlyC: both kinetic energy and momentumD: neither kinetic energy nor momentum

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The correct answer is…. B momentum only

momentum conservation: I could define the plane (with the shadows) as being z=0, then momentum conservation for the shadow collision is the same as the x and y components of momentum conservation for the ball collision.

mom. conservation for balls

mom. conservation for shadows⇒

Kinetic energy:

For the shadows, the z-components are lost, so we can’t have kinetic energy conservation.

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Definition : An inelastic collision is a collision in which total mechanical energy is notconserved .

Example: A ballistic pendulumA bullet of mass mB is fired with velocity at a wooden pendulum of mass mA, as shown. It sticks in the wood and the pendulum swings up a vertical distance y. What was the initial speed of the bullet?

momentum conservation velocity of block (with bullet) immediately

after impact

kinetic energy before collision is

kinetic energy after collision is

Kinetic energy is not conserved!

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Now we need to calculate v, which we can do using energy conservation after the collision.

kinetic energy after collision is

potential energy after collision is (this is just a definition of where ‘0’ height is)

kinetic energy after “upswing” is

potential energy after “upswing” is

(the pendulum comes to rest)

Energy conservation ⇒

⇒

⇒

Putting this together, we have the velocity of the bullet:

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Question: A 1kg lump of clay travelling at 1ms-1 smashes into another 1kg lump of clay which is not moving. They stick together and become one 2kg lump. What is the speed of the 2kg lump after the collision?

1kg 1kg

1kg 1kg

1 ms-1

v ms-1

Before:

After:

A: 0 ms-1

B: 0.25 ms-1

C: 0.5 ms-1

D: 1 ms-1

E: 2 ms-1

The correct answer is…. C

momentum is conserved (but not kinetic energy)

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8. Angular momentum and energy

Definition: The angular momentum of a point particle about O is given by

where is the vector from O to the object and is its momentum.

8.1 Angular momentum and torque Y&F: Ch 10.4

This is a vector product. A vector product of two vectors and is given by

is a vector perpendicular to the plane of and , and is of length

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O

The vector is perpendicular to both and (i.e. out of the slide)

For circular motion with angular velocity , is at right angles to so the magnitude of the angular momentum is

O

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Definition: If a point particle with position vector relative to O, feels a force , then the torque of the force (sometimes called the “moment of the force”) on the particle with respect to O is given by

O

The torque is a vector perpendicular to both andwith magnitude

Torque is the rate of change of angular momentum with time:

In the absence of an applied torque, angular moment um is conserved.

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Example: A plumber uses a long 80cm wench to unscrew a pipe fitting. If the angle between the wench and the horizontal is 19o and he applies his full weight of 900N to the wench, what is the torque on the pipe?

Torque is into the page (by right hand rule) and is of magnitude

Y&F: Ex 10.1

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Y&F: Ch 10.6

An interesting example of torque and angular momentum is a gyroscope .

This is a spinning object balanced on a pivot.

The object is set to spin about its axis, and instead of falling down as you would expect, it “precesses” about the pivot.

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Consider first a gyroscope which is not spinning

Gravity pulls the centre-of-mass down ( ), applying a torque

at right angles to both and .

The change in the angular momentum in a time dt is

which is also at right angles to both and .

With each dt, the changes in angular momentum add up, the angular momentum gets bigger and the gyroscope falls more quickly.

(This is exactly what we naively expect.)

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Now consider a gyroscope which is spinning

As before, Gravity pulls the centre-of-mass down ( ),

applying a torque

at right angles to both and .

Now, as before, but this time the torque only changes the

direction of the angular momentum, not its magnitude.

The angular momentum is always in the same direction as so the gyroscope doesn’t fall – it precesses .

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Downwards force on centre-of-mass

For a gyroscope of mass m:

Torque

Change of angular momentum

So

The gyroscope precesses with an angular speed

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8.2 Rotational energy Y&F: Ch 9.5

Recall, a point particle moving with velocity has kinetic energy .

If it is moving in a circle of fixed radius r, then so

Usually it is more useful to discuss the rotations of extended objects , for example, a rotating disk. Imagine that the object is made up of lots of particles with mass , circulating at a radius with velocity . Then the total energy is

This is the moment of inertia of the body

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Definition: The Moment of Inertia of a system of particles about O is given by

where the particle i has mass mi and its distance from O is ri.(Notice that this definition requires a specific axis of rotation – the moment of inertia of a body is dependent on the body’s situation.)

The kinetic energy of a body rotating about an axis O with angular speed ω and moment of inertia I is

It’s units are kg m2

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Usually we will be interested in bodies that are continuous, rather than a discrete number of point particles. Then the summation becomes an integral.

Example : The Moment of Inertia of a solid disk

O

First, lets think about a very thin annulus:

All segments of the annulus of equal mass have the same moment of inertia (since they are the same distance from O) so the annulus has moment of inertia

Total Volume is

Total Mass is

Density is

O

volume

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To get the moment of inertia of the disk, we must add up the moments of inertia for lots of annuli.

O

But recall so the moment of inertia of the whole disk about the axis O is

(Notice that l dropped out.)

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Question : What is the Moment of Inertia of a (non-thin) annulus about an axis through its centre, perpendicular to its plane?

O

A:

B:

C:

D:(Hint: Think about the limits)

The correct answer is…. B

If I must get (a disk), so A and C are both wrong.

If I increase I must increase I because I am moving mass outwards ⇒ D is wrong

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More formally,

O But now

so

Thus

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8.3 Analogy to linear quantities

You should have noticed parallels between this section and previous sections.

Momentum Angular momentum

ForceRate of change of momentum

TorqueRate of change of angular momentum

Mass m Moment of Inertia

Kinetic energy Kinetic energy

velocity angular velocity

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9. Special Theory of Relativity

9.1 Galilean relativity

“Relativity” in physics tells us how to compare measurements made in different frames of reference moving relative to one another.

Although there is a public perception that “relativity” belongs only to Einstein, classical physics also contains relativity, called Galilean Relativity .

Consider two reference frames moving at a velocity u in the x-direction relative to each other.

Y&F: Ch 37.1

u

O′

x′

y′

yx

utO x

y

x′ = y′

The coordinates are related by a Galilean Transformation :

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More generally, to find the coordinates in a new frame O′ moving with a velocity relative to the frame O:

i.e.

Everything else can be worked out from these. For example,

(We used this implicitly earlier in Section 7.)

Then the velocity is very large, this Galilean Transformation is wrong . We must replace it with the Lorentz transformation.

We will use Einstein’s Theory of Special Relativity to derive this Lorentz transformation.

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Question: A spaceship is traveling at 0.5c relative to the bad guys, when it fires its lasers. The laser light travels at speed c relative to the spaceship. What is the speed of the laser with respect to the bad guys?

A. 0.5cB. 1cC. 1.5cD. 2c

The correct answer is… B

When the velocity between frames is very large, the Galilean Transformation is wrong . We must replace it with the Lorentztransformation.

We will use Einstein’s Theory of Special Relativity to derive this Lorentztransformation, and we will see that the speed of light is c in all inertial frames .

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9.2 The Michelson-Morley Experiment

In 1864, Maxwell wrote down his theory of electromagnetism . This theory explains light in the context of electric and magnetic fields, and predicts the speed of light in a vacuum to be:

permittivity of free space

permeability of free space

So Maxwell showed that light travels at (roughly) 300,000kms-1. But relative to what?

By comparison, sound travels at roughly 330ms-1. Since sound is the vibration of molecules in the medium (e.g. air) then this speed is relative to the medium itself.

If you shout into a wind of speed 20ms-1, then the sound will move ‘downwind’ at 350ms-1

and ‘upwind’ at only 310ms-1.

So what is light moving in?

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Scientists suggested that light moved through the “luminiferous aether ”, some unknown background medium, and the speed of light in Maxwell’s equations must be relative to this.

Sun

Earth“Aetherwind”

In that case, then the speed of light should depend on the speed of the “aether wind” in exactly the same way as the speed of sound depends on the normal wind’s speed.

Even if the “aether wind” is constant with respect to the solar system we should find a different speed of light in different directions, and at different times of the day and/or year.

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Albert Michelson(1852-1931)

Edward Morely(1838-1923)

In 1887 Michelson & Morely set out to test this.

They split a coherent light beam (only one wavelength) into two using a semi-transparent mirror, and let the two beams travel in perpendicular directions before being bounced back by mirrors (11m away).

coherent light source

detector

semi-transparent

mirror

mirror

mirror

The beams were recombined, and if one has travelled faster than the other they would see interference fringes at the detector.

They found no fringes, indicating that the split beams travelled at the same speed.

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9.2 Einstein’s postulates of Special Relativity

E1:The laws of physics are the same in all inertial frames of reference.

E2:The speed of light in vacuum is the same in all inertial frames of reference and is independent of the source.

An inertial reference frame is a reference frame which is not undergoing any acceleration.

The Newtonian mechanics (and the associated Galilean transformation) that we have seen so far in this lecture course also obeys E1. For example, only holds if the frame is not accelerating, but holds for all non-accelerating (inertial) frames.

So postulate 1 is not so surprising…..

(Note that the Earth is not an inertial frame, since it is rotating. However, the acceleration from the rotation is so small that we can pretend it is an inertial frame in most cases.)

…. it is E2 which throws us the surprises!

Y&F: Ch 37.1

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Y&F: Ch 37.19.3 Simultaneity is relative

Einstein’s second postulates tells us that whether two events are simultaneous or not depends on the reference frame of the observer.

Example: Imagine a train travelling at a speed u along the track. Observer O is standing still on the platform while observer O′ is on the train, right in the middle of a carriage.

O

O′

uA B

A′ B′

Lightning strikes the carriage at points A′ and B′, when they are adjacent to points A and B on the ground.

Observer O observes the two lightning flashes at the same time. He marks the spot where he was standing, measures distances OA and OB (the lightning left scorch marks) and when he finds them equal concludes that the lightning stuck both ends of simultaneouly .

In O’s frame, if the distance OA=OB, then the light from both bolts must have travelled for time OA/c.

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Observer O′ must agree that the two light flashes reached observer O at the same time -after all, this is a real event which really happened.

But the flashes do not reach observer O′ at the same time. He is travelling with speed uaway from observer O, and so O observes that the wavefront from bolt B reaches O′ first.

O

A B

O′

uA′ B′

utAgain, O′ must agree, but he disagrees on the interpretation .

O′ argues that since he is standing in the middle of the carriage, and the light from the two bolts both travel at the same speed c in O′’s reference frame (Einstein’s second postulate) then the lightning bolts could not have been simultaneou s.

Whether or not the lightning bolts were simultaneou s or not depends on the reference frame!

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9.4 Time Dilation Y&F: Ch 37.3

We would like to use Einstein’s postulates to derive a new transformation between coordinates in different inertial frames of reference.

Returning to our train, imagine that O′ observes a light from a source on the floor being projected vertically upwards. It bounces off a mirror on the ceiling and returns to the same position on the floor. He measures the time the light takes for the round trip

O

O′

uO′

ud

u∆t

Observer O measures the round trip as time ∆t. But since the source has moved, he reckons the light must travel further.

O′

source

mirrord

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O reckons the distance travelled by the light is

d

so in his frame, the light must take time

Time is not universal!

It is reference frame dependent!

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Definition: The proper time between two events is the time between the events in the frame where the events happen at the same point in space .

In our example, the two events (departure and arrival of light at the source) occur at the same point in space in O′’s reference frame, so the proper time between the events is ∆t0.

Generally, the time between two events in a frame moving with relative (constant) speed u with respect to the proper frame is

where

So clocks run slow if they are moving relative to you – this is known as time dilation .

�

� is ill-defined for

Note: � photons take no time to travel

� when ,

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The twin paradox

Imagine two twins, Bill and Ben. Ben decides to leave Earth in a spaceship which travels at a fixed speed of u relative to the Earth, and travels a distance d (again relative to the Earth) to a distant planet. Bill stays on the Earth during Ben’s journey.

In Bill’s reference frame the distance travelled is d, so Bill measures the time that Ben takes to get there as

How long does Ben measure that the journey has take n?

Bill

Ben

u

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In this example, we have two significant events:

• Ben’s departure from Earth• Ben’s arrival at the new planet

In Ben’s reference frame, both of these events happen at the same space point (by definition), so the time that Ben measures between these events is the proper time ∆t0 .

We know that the proper time interval is related to the (dilated) time interval in a reference frame moving with speed u by our time dilation formula:

Note that the sign of u doesn’t matter.

We know that the dilated time measured by Bill is so the proper time (as measured by Ben) between the events is:

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Bill

Ben

u

Now imagine that when Ben arrives at the planet, he immediately turns around and comes back at the same speed (u).

Exactly the same reasoning as before can be applied and the time to get back is the same as for the outward journey in both frames.

Therefore, the time elapsed for the total journey is:

In Bill’s frame: In Ben’s frame:

When he returns, Ben is younger than Bill!

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This seems paradoxical because what has distinguished Bill from Ben?

Relativity tells us that there is no ‘preferred’ reference frame – I could just have easily considered Bill (and the entire planet Earth) as moving away from Ben at a speed u, and when he returns, he would have been younger than Ben….or would he?

There is actually no paradox here, because there is something which tells us that it was Ben who travelled.

Ben accelerated when he turned around (changing his velocity from u to –u) – Ben was not in an inertial frame!

In fact, Ben knows this because he feels the g-force of the deceleration and acceleration while turning round.

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Example: Elementary particles called muons are created in the upper atmosphere by cosmic rays. If an average muon lives 1.56×10-6 s, will an average muon created at a height of 2km and at a (downward) speed of 0.98c reach the ground before it decays?

In the Earth’s rest frame the muon lives a time ∆t before decaying, where

Therefore an average muon travels a distance

Cosmic ray

muon

0.98c

2km

Ground

before decaying.

The average muon reaches the ground, travelling 5 times further than non-relativistic mechanics would otherwise predict.

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9.4 Length Contraction Y&F: Ch 37.4

E2 gives us a really good way to measure distances. Since the speed of light is c in every frame, we can time how long light takes to travel between two points and work out the distance.

O′

mirrorsource Back on the train, observer O′ decides to measure the

length of a box by using his source and mirror.

He measures a time ∆t0 for the light to reach the mirror and return and concludes the box is of length

∆t0 is the proper time between the light’s emission and re-arrival.

O

O′ O′

u

Observer O watches the experiment and measures a time ∆t1 for the light to reach the mirror and a time ∆t2 to return. These times are not the same because the train moves during the light’s journey.

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On the journey to the mirror , the light travels a distance

lu∆t1

length of the box in O’s frame

distance travelled by the train

But we know so

On the return journey , the light travels a distance

The total trip takes time

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So in frame O’s frame, , while in O′’s frame

But we know the relation between ∆t and ∆t0 from our earlier example.

The length of the box measured by O is contracted compared to the length measured by O′

Definition: The proper length between two points is the distance between the points in the frame where the points are at rest .

In our example, the box is at rest in frame O′, so its proper length is l0.

Generally, the distance between two points in a frame moving with relative (constant) speed u with respect to the proper frame is

where γ was defined earlier.

So lengths are contracted if they are moving relative to you – this is known as length contraction .

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The train in the tunnel paradox

A 25m long train approaches a 20m long tunnel at 0.6c. Some observers are standing on the ground beside the tunnel.

The proper length of the train is l0=25m, but an observer on the ground sees its length contracted to

The proper length of the tunnel is d0 = 20m, but the train driver see its length contracted to

So as the train passes through the tunnel, it the observers on the ground will see it momentarily entirely enclosed.

So the train driver never sees the train enclosed completely in the tunnel.

We seem to have a paradox: How can the train be in the tunnel but not in the tunnel?

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As before, this is not really a paradox if we define our expressions a little bit better.

What do we mean by “the train is in the tunnel”?

We mean that the front of the train is inside the tunnel and the back of the train is inside the tunnel simultaneously .

But we saw earlier how simultaneity is relative. Two events which are simultaneous in one frame may not be in another.

In our example, the two important events (front of train inside tunnel + back of train inside tunnel) are simultaneous in the rest frame of the tunnel, but they are not simultaneous in the rest frame of the train .

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9.4 The Lorentz transformation Y&F: Ch 37.5

We are now ready to write down our replacement for the Galilean transformation – the Lorentz transformation .

u

O′

x′

y′

yx

ut

O x

y

x′ = y′

x′ is a proper length in O′’s frame. In O’s frame this distance is length contracted to

So

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Alternatively, we could have compared the lengths in O’s frame:

So

from before

As an exercise you can show that , so the transformation becomes:

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Therefore the Lorentz transformation between the two frames, for a Lorentz “boost”along the x-axis is:

Notice:

• The y and z components remain unchanged (as for the Galilean transformation along the x-axis). We could find the general transformation by applying a rotation to the above.

• If u is small, and , so we return to the Galilean transformation

This tells us how to relate the coordinates of an event in one frame (t,x,y,z) to the coordinates of the same event in another frame (t′,x′,y′,z′)

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Also,

So the quantity doesn’t change with reference frame. We say that this is Lorentz invariant.

c.f. the length of a vector under a rotation

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We can differentiate the Lorentz transformation equations to see how velocities transform.

(chain rule)

Velocity transformations

But

and

So

Notice that we return to the Galilean velocity addition rule when we take u → 0

Also, if then as required by E2.

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Beware! Although the perpendicular (y and z) components don’t change under a boost (in the x-direction), the perpendicular velocities do.

This is because a velocity is measuring a change in position with respect to time , and our time coordinate changes with frame.

e.g.

Example: Two particles, A and B, approach each other with velocities vA=v and vB=–vwith respect to the ground. What is the velocity of B as measured in A’s rest frame?

A B

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1

Notice that the speed of B is always less than the speed of light in A’s frame.

If fact, our Lorentz transformation tells us that if an object has a speed in any frame , then it will have speed in all other frames (which travel with relative speed ______ compared to the first frame).

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9.4 Relativistic energy and momentum Y&F: Ch 37.7 & 37.8

Our classical definition of momentum was . This was useful because it was always conserved in collisions (elastic and inelastic).

Lets take a look at the previous example (particles A and B) imagine the particles have the same mass m.

Total momentum in the ground’s frame =

Total momentum in the A’s rest frame =

Now imagine they collide, and stick together (inelastic) to form a particle c of mass 2m.

Boost to A’s rest frame:

ButThis definition of momentum cannot be conserved in all frames!

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Definition: The relativistic momentum of a particle of mass m, moving with velocity is

This relativistic momentum is conserved in all frames (showing this from the velocity transformation is a bit tedious).

Sometimes people redefine mass , rather than momentum.

This would leave the definition of momentum as we had before, i.e. , but then mass becomes dependent on reference frame, which is very inconvenient.

Please do not do this!

A non-linear relation between momentum and velocity is much better!

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Since momentum is related to force and force is related to work and energy, we also need a new definition of energy .

Definition: The relativistic energy of a particle of mass m, moving with velocity (with no potential) is

Notice how similar this is to the new definition of momentum!

If we can expand the denominator as a series in

Does this look familiar?

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For

This is the rest energy of the object

This is the non-relativistic

kinetic energy of the object

Using our new definition of momentum to replace the velocity in

we find,

Notice that is a Lorentz invariant?

What does this remind you of?

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It is useful to think of time as another coordinate in the vector for the event.

Instead of having 3 components in a vector we have four

We collect these coordinates together as a four-vector

Position four-vector:

Momentum four-vector:

The “length” of this four vector is , which is a Lorentz invariant.

We can do the same thing for momentum & energy.

The “length” of this four vector is , which is also a Lorentz invariant.

This transforms under a Lorentz boost in the same way as the position vector.