Rotational Mechanics & Special Relativity

Rotational Mechanics Rotational Mechanics & Special Relativity& Special Relativity

The Department of Physics

Part IA Natural Sciences Tripos 2013/14

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Lecture 1

Course materialsCourse materialsCourse materialsCourse materials

Part IA Physics Lent Term, 2014

Rotational Mechanics & Special Relativity

Prof. Steve Gull

Lecture notes 2014

2

2

dm

dtr

F

Newton

Einstein

2mcE

Part IA Physics Lent Term, 2014

Rotational Mechanics & Special Relativity

Prof. Steve Gull

Examples book 2014

2

2

dm

dtr

F

Newton

Einstein

2mcE

(d) The time interval is the same in any frame. Thus tAB= (tB tA) = tAB= (tB tA). In fact we have a strong notion that time and space are absolute quantities. We think that we can define a point in ‘absolute’ space and ‘absolute’ time, and that space and time are the same for everyone, no matter how they are moving with respect to each other. These ideas obviously work very well in everyday life, but need closer examination.

(e) We can express the transformation between coordinates seen in one inertial Cartesian

frame and those in another using the ‘Galilean’transformation. Suppose that frame S' is moving along the positive x axis of frame S at constant speed v, and their origins coincide at t = t = 0.

An event, A, occurs in S at (xA,yA,zA,tA), and in S at (xA,yA,zA,tA ).

y

x

v

vtAAxA

xAy

xS S

MO 173TM 1322

7

This space is for your own notes

Text-book references:MO: Mansfield &

O’SullivanTM: Tiplar & Mosca

You can also find these slides, in colour, on the Physics Teaching website:


A question about calculating the position of the centre of mass etc. by integration. 1. (a) A uniform solid cone has a height b and a base radius a. It stands on

a horizontal table.

(i) Draw a diagram showing the cone divided into thin horizontal discs, each of thickness h. Find an expression for the volume of the disc at height h above the base. Integrate over all

the discs to show that the total volume, V, is given by 2

3V a b

.

(ii) The height, b

0, of the centre of mass is defined by the relation

0 i ii

b m h M ,

where M is the total mass, m

i is the mass of the disc at height h

i

and the sum is taken over all the discs. Treat the sum as an integral, and hence show that 0 4b b .

(b) A uniform solid cylinder of radius r and length l is cut into two equal parts along its cylindrical axis. Find the position of the centre of mass of either part.

{4 3r } The rotational equivalent of the mass is the moment of inertia, and is the subject of the following example. 2. State the parallel and perpendicular axis theorems.

(a) Calculate the moment of inertia of a uniform square plate of side a and mass m about an axis through its centre and parallel to a side.

(b) Use the perpendicular axis theorem to find the moment of inertia through the centre and perpendicular to its plane.

(c) More generally, show that the moment of inertia of a square plate about any axis in its plane through its centre is the same.

(d) Use the theorems of parallel and perpendicular axes to find the moment of inertia of a hollow cubical box of side a and total mass M about an axis passing through the centres of two opposite faces.

{(a) ma2/12; (b) ma2/6; (d) 5Ma2/18}

How to solve Physics problems Here is some general advice which you should apply to every Physics problem you tackle. Now is the time to develop the techniques which make the problems easier to understand and which can provide you with a framework within which to think about Physics.

1. Draw a diagram

A diagram always helps to clarify your thoughts. Use it to define the symbols you need to use (see 3 below). Make it big enough and be tidy.

2. Think about the Physics Ask yourself what is going on, and write it down in words in just one or perhaps two sentences. Try to understand the problem qualitatively before writing down any equations. Do not just write down equations!

3. Stay in symbols until the end At school you may have been taught to make calculations numerically rather than algebraically. However, you usually give yourself a big advantage if you delay substitution of numerical values until the last line as it enables you to check dimensions at every stage, and quantities often cancel before the last line. An exception to this rule arises where some terms are dimensionless factors which are simple fractions.

4. Check the dimensions Think about the dimensions of every quantity even as you write it down. You will find this a discipline which helps enormously to avoid errors and helps understanding. Make sure that the dimensions of your final equation match on each side before you make a numerical substitution. Write down the units of your answer at the end e.g. 4.97 J kg–1.

5. Does the answer make sense? You will probably have an idea of what looks about right, and what is clearly wrong. Many mistakes are simple arithmetic errors involving powers of ten. If in doubt, check your substitutions.

5 6

Mechanics in Mechanics in Rotational MotionRotational Motion

Mechanics in Mechanics in Rotational MotionRotational Motion

The Department of PhysicsThe Department of Physics

The centre of massThe centre of massThe centre of massThe centre of mass

Suppose that we have a distribution of masses, all within one rigid body. Where should we put the fulcrum such that there is no rotation when suspended in a uniform gravitational field? This point is the ‘centre of weight’, but it also has special properties when there is no gravitational field and so is usually called the centre of mass. For the simple case of a uniform light rigid rod of length l connecting two point masses m1 and m2, it is obvious where the centre of mass must be:

MO 128TM 149

y

x

m1 m2

x1 x2

l2l1

x0

Centre of mass

We require that the total turning moment about C is zero, so we define its position to be such that m1l1 = m2l2. . Since Since ll11 = = xx00 – – xx11, and , and ll22 = = xx22 – – xx00, we can write this , we can write this asas

where M = m1+m2.

C

, so

,

1 0 1 2 2 0

1 1 2 20

m x x m x x

mx m xx

M

We can generalise this to a one-dimensional rigid body of N masses as follows:

For a three-dimensional rigid body, this expression must be satisfied in each dimension simultaneously, and we may write:

Now we can express this in vector form, writing the position vector of the centre of mass as R = (x0, y0, z0), and of each contributing mass as ri = (xi, yi, zi):

1 1 2 2 3 30

1

1.

N

i ii

mx m x m xx mx

M M

; ; 0 0 01 1 1

1 1 1.

N N N

i i i i i ii i i

x mx y my z mzM M M

0 0 01 1

1 1, , , ,

N N

ii i i i ii i

x y z m x y z mM M

R rback

We are now in a position to find the centre of mass of a continuous body using integration rather than summation. Write the summation for R using δmi = mi:

1 1 1.i i i i i i

i i i

m m δmM M M

R r r r

Hence, for continuous bodies, we can replace the sum over i elements with an integral:

body

1.dm

M R r

In 1 dimension, dm is l dl, where l is the mass per unit length and l is the length variable.

In 2 dimensions, dm is a da, where a is the mass per unit area and a is the area variable.

In 3 dimensions, dm is v dv, where v is the mass per unit volume and v is the volume variable.

We can also use the principle of superposition to work out what happens if the rigid body is ‘lumpy’. The centre of mass of a composite system of several lumps of mass can be calculated by first finding the centre of mass of each lump separately, and then finding the centre of mass of the whole considering each lump as a point mass at its individual respective centre of mass.

m1

m2

m3

m1m2

m3

+ +=

1i i

i

mM

R r

body

1dm

M R rFor continuous bodies:

For discrete masses:

y

x

Example 1: find the centre of mass of a lamina shaped like an isosceles triangle

δx

x

b

h

2

b h xy

h

Divide the lamina into thin strips, each having a width of δx. Take moments about the origin. The moment of the element shown is xδm, where δm is the mass of the element. The area of the element is 2yδx, and the area of the lamina is bh/2.

If the mass of the lamina is M, then

2

222

M h xyδxδm M δx

bh h

x0

y

x

Example 1: find the centre of mass of a lamina shaped like an isosceles triangle

δx

x

b

h

2

b h xy

h

Divide the lamina into thin strips, each having a width of δx. Take moments about the origin. The moment of the element shown is xδm, where δm is the mass of the element. The area of the element is 2yδx, and the area of the lamina is bh/2.

If the mass of the lamina is M, then

2

222

M h xyδxδm M δx

bh h

Let δx tend to zero, and integrate to find the total moment. Then

0 20

20 2

0

3 3

0 2

21 1

2

22 3 3

h

h

M h xx xdm x dx

M M h

x xh x dxh

h h hx

h

x0

The lever balanceThe lever balanceThe lever balanceThe lever balance

We find experimentally, that the lever is balanced only when m1l1 = m2l2

y

x

m1m2

m1gm2g

(m1+m2)g

x1 x2

l2

l1

x0

This is the case even though the sum of all the forces is zero wherever we put the fulcrum, i.e.

We conclude that a second condition is necessary for equilibrium, that is that “the sum of all the turning moments must be equal to zero”.

This is expressed in the law of the lever:

where li is the perpendicular distance and fi is the force.

ext . ii

F F 0

ext ,0i i ii i

G G fl

. 0i i i

i i

G FlGext

, and ii

F F 0ext

The two conditions which must both be satisfied simultaneously for a body in static equilibrium are:

(1) The vector sum of all the external forces acting on the body is exactly zero.

(2) The sum of all the turning moments about an axis through any point is exactly zero.

MO 149TM 397

Circular motionCircular motionCircular motionCircular motionWhen a particle changes its position from P to P1

relative to a fixed point, O, we can describe the change in position using position vectors. The incremental vector δr is added to the position vector, r, describing the position P, to find the new position P1:

This constitutes a complete description of the change in position. Note, however, that the

O r

r + δrδrP

P1

change has two parts: a change in the radial distance from O, and a change in the angle about an axis OQ perpendicular to the plane of the triangle OPP1:

Here, we shall be concerned with the rotational aspects of the motion, i.e. things to do with

O r

r + δrδrP

P1

θ

Q

changes in the angle, θ. Note that this always requires you to define an axis about which the rotation takes place, in this case the axis OQ. The rotation can be in two senses, clockwise and counter-clockwise (or anti-clockwise). We adopt the right-handed convention to describe the positive direction of the axis of rotation (as shown by the arrow from O to Q). When looking along the axis in the direction of the arrow, the positive rotation is clockwise. If you look along the axis in the other direction, such that the arrow is pointing directly at you, rotation is anticlockwise.

The angle is best measured in radians. If it is small, i.e. << 1 radian, we can describe the rotation by a vector, θ , through O and along OQ, in the direction of OQ and of magnitude equal to θ.

We can see that small rotations can be represented by vectors as follows:

Oθ

Q

θ

For very small angles, the segment of the circle AB (on the surface of the sphere of radius a centred on O) becomes approximately straight, and can therefore be represented by vector s of length a. Three small rotations can be performed such that the segments

form a closed vector triangle. The rotation axes are perpendicular to each rotation, so straight lines proportional to the angles also form a closed triangle, i.e. they add up as vectors.

O

Q

A

Bs

θa

a 1

a 2a

3

therefore describe the angular velocity by the vector

Using a similar argument, the angular acceleration may also be represented by a vector which points along the axis about which the angular acceleration is happening:

.ddt

θ

θ ω

2

2.

d ddt dt

ω θ

θ ω

The moment of a force as a vectorThe moment of a force as a vectorThe moment of a force as a vectorThe moment of a force as a vectorWe have already met the concept of the “moment of

a force”, or the “turning moment”. We defined this as the force times the perpendicular distance of its point of action from the axis of rotation. We can represent the moment as a vector which, just as in the case of the angular velocity etc., points in a clockwise sense along the axis of rotation, and has a magnitude equal to the magnitude of the turning moment.

We consider a force, F, acting at point P which is described by position vector r from origin O:

The moment of the force about O is G = Fr sin( ) about the axis through O perpendicular to both r and F, clockwise looking in the direction of the arrow.

We define the moment vector GG == rr FF

Qθ

r sin(θ)

GForce F acts at point P,which is at position vector r from point O.

r

F

O P

Qθ

r sin(θ)

G

GG = = rr FF

Example 2: find a vector expression for the moment of a couple.

A couple is a combination of two equal and opposite forces which are not in line with each other.

F

F

ar1

r2 O

The forces act at position vectors r1 and r2 with respect to an arbitrary origin, O.

Example 2: find a vector expression for the moment of a couple.

A couple is a combination of two equal and opposite forces which are not in line with each other.

F

F

ar1

r2 O

The forces act at position vectors r1 and r2 with respect to an arbitrary origin, O.

1 2

1 2

( )

G r ×F r × F

r r ×F

a×F

The moment of the couple is independent of the origin O, and depends only on the vector forces and the position vector of the point of action of one of them with respect to the other.

Angular acceleration of a rigid bodyAngular acceleration of a rigid bodyAngular acceleration of a rigid bodyAngular acceleration of a rigid body

Newton’s second law gives us the relationship between the linear force applied to a body of mass m and its acceleration. We have seen that the rotational equivalent of the force, F, is the moment of the force (or torque), G, and the rotational equivalent of the linear acceleration, , is the angular acceleration, . Can we find an equivalent to Newton’s second law for rotational mechanics, and if so, what is the equivalent of the mass, m ?

rθ

To answer this question, consider a particle of mass m rotating about an axis OQ:

Its linear speed is v = rω, but let us suppose that its speed is increasing, i.e. it has an acceleration in the direction of v. N2 tells us that there must therefore be a force acting on the particle in this

O

Q

r m

v = rω

ω

direction of magnitude F where

The magnitude of the moment of this force about OQ is directed along OQ. For this particle we may therefore write

Now we can consider a rigid body rotating about OQ as made up of the sum of N such elementary particles. Let the ith such particle be a distance ri from OQ and have a mass of mi. Then summing over the whole body, we have:

( )

d rωd dv dωF mv m m mr

dt dt dt dt

2 ,G Fr mr ω

2

1

,N

ext i ii

mr

G ω

.2mr G r×F ω

where Gext is the total external vector moment acting on the body about OQ.

If G is the rotational equivalent of F, and is the rotational equivalent of , then we must conclude that the rotational equivalent of the mass of the body is:

We call this quantity the moment of inertia, I, and then the rotational equivalent of N2 is

θ

r

2

1

N

i ii

mr

ext IG ω

Moment of inertiaMoment of inertiaMoment of inertiaMoment of inertia

In linear mechanics, the mass is measure of a body’s reluctance to change its state of linear motion. The larger the mass, the slower the rate of change of velocity for a given applied force. In rotational mechanics, the moment of inertia takes the place of the mass, and it is a measure of a body’s reluctance to change its state of angular motion. The larger the moment of inertia, the slower the rate of change of angular velocity for a given applied moment of a force. Like the mass, the moment of inertia is a scalar quantity, usually

MO 154TM 293

given the symbol I. (Do not confuse this with the vector impulse, .) Thus, for a rigid body which can be thought of as being composed from N particles:

In terms of I, we can write the rotational equivalent of N2 as

2

1

.N

i ii

I mr

.IG ω

pI

Example 3: an electric motor is attached to the axis of a massive flywheel of moment of inertia 70 kg m2. When an electric current is switched on, the motor applies a constant torque of 150 N m. What is the rotation rate of the flywheel after 30 s?

The torque causes an angular acceleration which increases the angular velocity of the flywheel.

Gωmotor

flywheel

Example 3: an electric motor is attached to the axis of a massive flywheel of moment of inertia 70 kg m2. When an electric current is switched on, the motor applies a constant torque of 150 N m. What is the rotation rate of the flywheel after 30 s?

The torque causes an angular acceleration which increases the angular velocity of the flywheel.

Gωmotor

flywheel, so .

After time the angular

speed is .

rad s

rev s

1

1

15030

7064.3

10.2

GI ω

It

Gω ωt t

I

ω

G ω

Moments of inertia of continuous bodiesMoments of inertia of continuous bodiesMoments of inertia of continuous bodiesMoments of inertia of continuous bodiesThe expression we have obtained for the moment of

inertia, I, of a rigid body involves a summation over N elemental mass contributions:

For continuous bodies, it is more convenient to use an integral form. We imagine that the body is divided into a very large number of very small masses, δm, all joined together to make up the whole. A particular one is at distance r from the axis, so it contributes an amount δI to the total

2

1

N

i ii

I mr

where δI = r2δm. Now go to the limit as δm tends to zero, and integrate over all the contributions to get the total moment of inertia, thus:

The trick in applying this expression is often to express dm in terms of the distance variable (r) using the density. You should also take advantage of the spatial symmetry of the problem to simplify the expression that you must integrate. Some examples follow.

2I r dm

Let the rod’s density be per unit length. Consider an element of the rod, length δx, at distance x from one end. The mass of the element is δm = δx. The contribution to the total moment of inertia from this element is δI, given by:

Example 4: find an expression for the moment of inertia of a rod of length l about an axis through one end perpendicular to the rod.

x

δx

l

rod

But , so

2 2

2 3

0

2

13

13

l

x

δI x δm x ρδx

I ρ x dx ρ l

M ρ l I Ml

x

δx

l

Let the rod’s density be per unit length. Consider an element of the rod, length δx, at distance x from one end. The mass of the element is δm = δx. The contribution to the total moment of inertia from this element is δI, given by:

Example 4: find an expression for the moment of inertia of a rod of length l about an axis through one end perpendicular to the rod.

rod

δr

ra

Let the disc have a surface density of per unit area. Consider a radial element of the disc at radius r, thickness δr. This has mass δm which is equal to 2rδr. It makes a contribution, δI, to the total moment of inertia given by:

Example 5: find an expression for the moment of inertia of a solid disc of mass M and radius a about an axis through its centre perpendicular to the plane of the disc.

δr

ra

Let the disc have a surface density of per unit area. Consider a radial element of the disc at radius r, thickness δr. This has mass δm which is equal to 2rδr. It makes a contribution, δI, to the total moment of inertia given by:

But , so

2 3

3 4

0

2 2

2

12

2

12

a

r

δI r δm πσr δr

I πσr dr πσa

M πσa I Ma

Note that this formula also applies to a cylinder.

Example 5: find an expression for the moment of inertia of a solid disc of mass M and radius a about an axis through its centre perpendicular to the plane of the disc.

O

P

A

centre of massrigid body

a

B

ri miRi

Parallel axes theoremParallel axes theoremParallel axes theoremParallel axes theoremThis theorem allows us to obtain the moment of inertia, I, of a rigid body about any axis, AB, parallel to an axis, OP, through the centre of mass about which the moment of inertia of the body is I0.Consider the contribution from the element mi:

MO 155TM 297

Then

2

2 2

2 2

20

( ) ( )

( 2 )

2

i ii i ii i

i iii

ii ii

ii i i ii i i

I mr m

m

m a R

ma mR m

I Ma I

r r

a R a R

R a

a R

Let I be the moment of inertia about axis AB.

Let I0 be the moment of inertia about an axis, OP, parallel to AB and through the centre of mass.

zero for

centre of mass

Then

2

2 2

2 2

20

( ) ( )

( 2 )

2

i ii i ii i

i iii

ii ii

ii i i ii i i

I mr m

m

m a R

ma mR m

I Ma I

r r

a R a R

R a

a R

Let I be the moment of inertia about axis AB.

Let I0 be the moment of inertia about an axis, OP, parallel to AB and through the centre of mass.

Let the moment of inertia about the axis through the centre be I0. We have already calculated the moment of inertia, I, about a parallel axis through one end of a rod to be Ml2/3. Thus:

A

B

O

Pa = l/2

Example 6: find an expression for the moment of inertia of a rod of length l and mass M about an axis through its centre perpendicular to the rod.

Let the moment of inertia about the axis through the centre be I0. We have already calculated the moment of inertia, I, about a parallel axis through one end of a rod to be Ml2/3. Thus:

A

B

O

Pa = l/2

22

0

20

13 4

112

lMl M I

I Ml

Example 6: find an expression for the moment of inertia of a rod of length l and mass M about an axis through its centre perpendicular to the rod.

z

x

y

Perpendicular axes theoremPerpendicular axes theoremPerpendicular axes theoremPerpendicular axes theoremThis is another useful theorem which relates the

moments of inertia about three perpendicular axes through any point in a lamina, one of which is perpendicular to the plane of the lamina.

Consider the contribution by an elemental mass mi:

mi

ri

2 2 2

2 2

( )z i i i i ii i

i i i ii i

z y x

I mr m x y

mx my

I I I

Let the lamina be in the xy plane. Then the moment of inertia about the z axis is

We have already found the moment of inertia of the disc about the z axis to be Ma2/2. Thus

Example 7: find an expression for the moment of inertia of a disc of radius a and mass M about an axis through its centre in the plane of the disc.

a

z

y

x

2

2

2

1;

21

22

14

z x y D

x y D

D

I Ma I I I

I I I Ma

I Ma

We have already found the moment of inertia of the disc about the z axis to be Ma2/2. Thus

Example 7: find an expression for the moment of inertia of a disc of radius a and mass M about an axis through its centre in the plane of the disc.

a

z

y

x

Sounds of Pulsars Sounds of Pulsars Sounds of Pulsars Sounds of Pulsars

• Single pulses are usually very variable !• Single pulses are usually very variable !

0329+54 P=0.71452s

0833-45 P=0.089s

0531+21 P=0.033s

0437-4715 P=0.00575s

1937+21 P=0.00167s

0329+54 P=0.71452s

0833-45 P=0.089s

0531+21 P=0.033s

0437-4715 P=0.00575s

1937+21 P=0.00167s






Lecture 3

Angular momentumAngular momentumAngular momentumAngular momentum

We define the angular momentum, L, to be the moment of the momentum about a point.

Suppose that a particle, A, of mass m and at position vector r relative to point B, has momentum P = mv, where v is its velocity:

r

P, v

B Aθ

r sin(θ)

m

The moment of the momentum (by direct analogy with the moment of a force) is given by

L = r P = r mv = mr v

r

P, v

B Aθ

r sin(θ)

L

m

MO 103TM 331

Conservation of angular momentumConservation of angular momentumConservation of angular momentumConservation of angular momentum

The angular momentum, like the linear momentum, is conserved in an isolated system. To show this, consider a system of N interacting particles. The angular momentum of the ith particle is Li = miri vi. The total angular momentum of the system is therefore given by

This is a vector addition of the angular momenta of all the particles about a given point. We can differentiate with respect to time to find the rate of change of L:

.1

ii

N

iim vrL

Now the value of Gint is zero for the following reason:

1 1

1

1

1 1 1

( ) ( )

( )

( )

N N

i i i ii ii i

Ni i

i iii

N

i i i iii

N N N

ii i i iii i i

tot ext

d d dm m

dt dt dtd d

mdt dt

m

m

Lr v r v

r vv r

v v r a

r a r F G

G G Gint

The internal interaction on the ith particle by the jth particle is in line and oppositely directed to the interaction on the jth particle by the ith particle by N3. The moment about O is ri Fij + rj Fji

= (ri – rj) Fij = b Fij . This is zero since b is in line with Fij. All the internal interactions are in

O

j

irj

ri

b

Fji

Fij

similar pairs, each of which comes to zero. Hence Gint must be zero. Therefore

and we conclude that the rate of change of the total angular momentum of a system is just the vector sum of the external moments applied to that system.

When the system is isolated, Gext = 0, so we conclude that the total angular momentum of an isolated system is constant.

,ext

ddt

L

G

Notes:

(a) This statement is always true, no matter what dissipative forces there might be internally.

(b) This statement is always true about any axis, not just one through the centre of mass.

(c) The rotational equivalent of N2 is and we have just shown that . Therefore we can write

,IG ωd dtG L

d dI

dt dtI

L ωG

L ω

Angular impulseAngular impulseAngular impulseAngular impulse

We defined the impulse of a force as the integral of the force with respect to time, i.e.

The action of the impulse was to change the linear momentum, thus:

Now the moment of the force is r F, and the angular momentum is r P. Taking the cross product with r in the above equation we get:

.p dtI F

2 2

2 1

1 1

Δt t

p

t t

ddt dt

dt

PI F P P P

Thus we see that when a moment of a force acts for a finite time, it causes the total angular momentum of the system to change, the change being equal to the integral of the angular moment with respect to time.

2 2 2

1 1 1

2

2 1

1

t t t

ppt t t

t

t

dt dt dt

ddt

dt

J r I r F r F G

LL L L

Rotational kinetic energyRotational kinetic energyRotational kinetic energyRotational kinetic energy

The translational kinetic energy for a particle of mass m moving in a straight line at speed v is mv2/2. We can obtain the equivalent rotational quantity by considering the ith particle of a rigid object in rotation about an axis at at angular speed ω. Let the particle have mass mi and be at distance ri from the axis. Then:

22 2 21 1 12 2 2i i i i i i iKE mv m rω mr ω

Summing over the entire body (N particles), we get

Note that, for an individual point particle, we can consider its kinetic energy either as the translational quantity mv2/2, or as the rotational quantity mr22/2. However, in general the motion of a solid object must be analysed in terms of the linear motion of its centre of mass, plus the rotation around the centre of mass, so the total KE is the sum of the translational and rotational components.

2 2 2

1

1 12 2

N

tot i ii

KE mr ω Iω

Rotational oscillations: the physical pendulumRotational oscillations: the physical pendulumRotational oscillations: the physical pendulumRotational oscillations: the physical pendulum

A

Cl

mg

A

C

l

Consider a rigid body of arbitrary shape that is suspended from, and free to rotate about, a horizontal frictionless axis (A). It is displaced slightly from equilibrium. What is the period of oscillation?

l y

y

ω

Point C, the centre of mass, is raised relative to its equilibrium position by an amount y. Conserving energy we have:

Now Substituting and differentiating with respect to time gives us

Hence the period of small oscillations is

21.

2E Iω mgy const

, so cos (1 cos ).l y l θ y l θ

so

, if is small.2

2

1(2 ) sin 0,

2

0

I ωω mgl θω

mgldθθ θ

dt I

2 .I

T πmgl

Example 8: the pendulum of a grandfather clock is made from a brass disc of diameter 2a and mass M with its centre attached to the end of a thin metal rod having a mass of m so that the centre of the disk is L below the point of suspension. What is the period of the pendulum?

The moment of inertia of the disc about an axis through its centre and perpendicular to its face is Ma2/2. Using the parallel axes theorem we have:

The moment of inertia of the rod is mL2/3, so the total moment of inertia is

Now the centre of mass is at distance l below

A 2 21

2I ML Ma

A

2 2 21 12 3

I ML Ma mL

L

2aM

ml

C

A

the pivot such that

if (i.e. a light rod). If

as expected for a simple pendulum.

22 2

2

( )2

22

22

32 2( ) (2 )

2 0 0,2

2

LML m M m l

M m Ll

M m

mLML MaI

T π πM m gl gL M m

L aT π m a

g gL

LT π

g

L

2aM

ml

C

A






Lecture 4

The general motion of a rigid bodyThe general motion of a rigid bodyThe general motion of a rigid bodyThe general motion of a rigid bodyWe have seen already that the linear motion of a

rigid body can be analysed by considering only the linear motion of its centre of mass which moves as if it carries all the mass and is acted upon by the sum of all the external forces. This is a required condition, but is not sufficient to describe the entire motion of the body because is does not take into account the rotational motion of the body about the centre of mass. We need a second statement, added to the first, to define the general motion of a body, linear plus rotational.

The additional statement is that there is rotation about an axis through the centre of mass of a body which is the result of the sum of the external moments of all the forces acting on the body about that axis, and the moment of inertia of the body about that axis. This second statement, added to the first, is sufficient to define the general motion of a body.

We will clarify this with the example of a cylinder rolling down an inclined plane:

The cylinder is acted on by the body force, mg, by the normal reaction force, N, and by the frictional force, F, as shown in the diagrams. The net result is (a) the centre of mass accelerates down the plane, and (b) the cylinder rolls down the plane.

Example 9: what is the acceleration of a uniform solid cylinder rolling, without slipping, down a plane inclined at angle α to the horizontal?

α

v

ω

mg

mg

Fa

ω

v

N

α

a

(b) The rotation about the cylindrical axis is as if acted upon by the sum of the moments of the external forces:

where I is the moment of inertia about the cylindrical axis. We also have v = ωa. Thus

(a) The linear motion of the centre of mass is as if it is a point mass equal to the total mass of the cylinder, acted upon by the sum of the external forces. Resolving down the slope:

sin( )dv

mg α F mdt

,dω

Fa Idt

mg

Fa

ω

v

N

α

2.

I dω I dvF

a dt dta

Substituting for F gives us

Now I = ma2/2, so we deduce that

so 2 2

sin( )sin( ) , .

mg αI dv dv dvmg α m

dt dt dta m I a

Substituting for F gives us

Now I = ma2/2, so we deduce that

so 2 2

sin( )sin( ) , .

mg αI dv dv dvmg α m

dt dt dta m I a

sin( ) 2sin( ).

2 3mg αdv

g αdt m m

(b) The linear KE of the centre of mass is

(c) Thus when the centre of mass has dropped through a distance h we have

In terms of energy:

(a) Rotational KE of the cylinder is2 2 2 21 1 1

2 4 4rotKE Iω ma ω mv

212linKE mv mg

Fa

ω

v

N

α

αh

s

(b) The linear KE of the centre of mass is

(c) Thus when the centre of mass has dropped through a distance h we have

In terms of energy:

(a) Rotational KE of the cylinder is2 2 2 21 1 1

2 4 4rotKE Iω ma ω mv

212linKE mv mg

Fa

ω

v

N

α

, so

We conclude that as before.

2

22

34

23sin 2 sin

4 3

2sin

3

rot linmgh KE KE mv

gvh s α v α s

g

gdvα

dt

αh

s

Rotating frames of referenceRotating frames of referenceRotating frames of referenceRotating frames of reference

Suppose that we have a particle, P, which is rotating at constant angular speed in a circle about O:

PO

ω

ω

r

y

x

y

xP

rO ω t

ω

ω

Oblique view Plan view

We can represent the position of P with respect to O by the position vector r = (x,y), where x is r

cos(ω t) and y is r sin(ω t). We differentiate to find the velocity, and again to find the acceleration, thus:

22 2

2

22

2

(cos( ),sin( ))

( sin( ), cos( ))

( cos( ), sin( ))

r ωt ωt

dr ω ωt ω ωt

dtd

r ω ωt ω ωtdtd

ωdt

r

r

r

rr

This tells us that the point P has a constant acceleration (since ω and r are constant) which is of magnitude ω

2 r and is directed along r in the

negative direction, i.e. towards O. The actual speed of the particle is constant and equal to ω r in a tangential direction, but the acceleration arises from the fact that P is constantly changing direction towards the centre of the circle, so the vector velocity is constantly changing.

Newton’s second law of motion tells us that there must be a force associated with the acceleration.

Centripetal forceCentripetal forceCentripetal forceCentripetal forceThe centripetal force is the force on a particle which

is directed towards the axis of rotation and which is required to maintain the rotational motion of the particle. From N2:

where is the unit vector along r. Note that the centripetal force does no work as the velocity and force are orthogonal to each other.

Since , we can also write

22

2

2 2

2

.

,̂

dm mω

dtv rω

d mvm

rdt

rF r

rF r

F

m

r

r̂

Linear and rotational equivalentsLinear and rotational equivalentsLinear and rotational equivalentsLinear and rotational equivalents

Linear Rotational

mass msmall displacement drvelocity v = dr/dtacceleration a = dv/dt = d2r /dt2

linear momentum p = mvforce F = dp/dt = malinear kinetic energy mv2/2work done dW = Fdrlinear impulse dp = Fdt

moment of inertia Ismall angular displacement dθangular velocity ω = dθ/dtangular acceleration dω /dt = d2θ /dt2

angular momentum L = Iωmoment G = dL/dt = Id2θ/dt2

angular kinetic energy Iω2/2work done dW = Gdθangular impulse dL = Gdt

We can identify quantities in linear mechanics and rotational mechanics which behave in equivalent fashions. If you are not sure what to do in a rotational problem, think what you would do in the equivalent linear problem, and then use the table below.

a) b)B4 After

v0

ucentreof mass

b

a) Linear momentum before = linear momentum after

Example 10: (Tripos 2000). A rod of mass M and length L lies on a smooth horizontal table. A small particle of mass m travels at speed v0 on the table at 90° to the rod. It collides with the end of the rod and sticks to it. Calculate the speed of the centre of mass of the combined rod and particle after the collision, and find the new position of the centre of mass …

Example 10: (Tripos 2000). A rod of mass M and length L lies on a smooth horizontal table. A small particle of mass m travels at speed v0 on the table at 90° to the rod. It collides with the end of the rod and sticks to it. Calculate the speed of the centre of mass of the combined rod and particle after the collision, and find the new position of the centre of mass, …

B4 After

v0

ucentreof mass

b

a) Linear momentum before = linear momentum after

som

mv m M u u vm M0 0

,

b) New position of the centre of mass: moments about it sum to zero

u

centreof mass

b

b) New position of the centre of mass: moments about it sum to zero

02 2

02 2

L b L b b bM M bm

L L

LM LMbM bm b

M m

, so

u

centreof mass

b

Example 10: (Tripos 2000) … and the angular speed of rotation about the centre of mass.

The principle of the conservation of angular momentum may be applied about the centre of mass:

B4 After

v0

ucentreof mass

b ω

22 2

0

00

32 3

1 13 3

63( 4 )3

L bbv mb Iω mb Mb M L b ω

L L

mvLω v b

M M L M mb L b L bm m






Lecture 5

The GyroscopeThe GyroscopeThe GyroscopeThe Gyroscope

A gyroscope is heavy flywheel which is rapidly rotating about an axis. It is usually mounted so that it can be turned about either of the two other orthogonal axes:

L = I

z

y

x

MO 164TM 339

Whilst the flywheel is stationary, there is nothing unexpected about the gyroscope, so that a moment applied about the y or z axes produces rotation about the y or z axes.

When the flywheel is rapidly rotating about the x axis, a moment about y produces slow rotation (called precession) about z and vice-versa. This is apparently counter to expectations, but is readily understood, either in terms of angular momentum, or in terms of the forces acting on particles in the flywheel.

A precessing gyroscopeA precessing gyroscopeA precessing gyroscopeA precessing gyroscopeThe gyroscope is precessing about the

z axis. Look down from on top along –z :

Ω

x, L

tδt t+δtt

z, Ω

y

ω

A

C

B

The blue particle moves from A to B to C. At B, it is moving in a curved path. Therefore it must feel a force to the left

Consider the motion of a particle (blue square) which is just coming up towards you at A, moving over the top at B, and disappearing at C

z

y

x

z

y

x

Looking from the side along –y :

yx, L

z, F

F

The forces form a couple which acts along y

z

y

x

z

y

x

Precession using vectorsPrecession using vectorsPrecession using vectorsPrecession using vectorsApply a couple of moment G to a rotating flywheel:

Ω

tG

L

In time δt the moment, G, of the couple exerts a change in angular momentum of Gδt, so that δL = Gδt. Note that this change is directed along G. If G is perpendicular to L (as here) then the change in angular momentum is also perpendicular to the angular momentum.

L then keeps constant magnitude, but is constantly changing direction in the plane of G and L, so precession Ω takes place.

t+δtL+δL

L δL

L+δL

Hence, there is precession about an axis perpendicular to both L and G. The rate of precession, Ω rad s1, is easily calculated. In a small time, δt, it precesses through small angle Ωδt.

In this triangle, we can see that, as δt is small, we can write

G δt = Ωδt L G = Ω L

Ωδt

L δL=Gδt

L+δL

In fact, we can write this in vector form as follows:

I G Ω×L Ω×ω

Hence, there is precession about an axis perpendicular to both L and G. The rate of precession, Ω rad s1, is easily calculated. In a small time, δt, it precesses through small angle Ωδt.

In this triangle, we can see that, as δt is small, we can write

G δt = Ωδt L G = Ω L

Ωδt

L δL=Gδt

L+δL

In fact, we can write this in vector form as follows:

To see this, consider a gyroscope at an angle to the horizontal plane:

I G Ω×L Ω×ω

In the case where the gyroscope is at an angle, first resolve the angular momentum into vertical and horizontal components:

θ

L

Ω

G

LV = L cos(θ ) is constant.LH = L sin(θ) is affected by the couple and is therefore changing direction but not magnitude. G = Ω L sin(θ ) G is perpendicular to the plane containing Ω and L, so we may write

G = Ω L = I Ω ω

Gyroscope examplesGyroscope examplesGyroscope examplesGyroscope examples

(i) Luni-solar precession

Moon, Sun

LN

S

G

The effect of the unequal ‘pull’ of gravity from the Moon and Sun on the non-spherical Earth applies a moment which causes the N-S axis to precess with a period of about 26,000 y. We see this a slow change in the positions of the stars with time.

(ii) Atomic precession

BL, m

An atom has an angular momentum and a magnetic moment. The magnetic moment subjects the atom to a couple when a magnetic field is applied which results in small changes in the energies of its electronic states. This results in the splitting of spectral lines – the Zeeman effect.

Einstein’s theory ofEinstein’s theory ofSpecial RelativitySpecial Relativity

Einstein’s theory ofEinstein’s theory ofSpecial RelativitySpecial Relativity

MO: 193-227TM: 1319-1356

The Department of PhysicsThe Department of Physics

Frames of referenceFrames of referenceFrames of referenceFrames of reference

A frame of reference is just a set of axes which we can use to define points (or ‘events’). We are all familiar with Cartesian frames (x,y,z), but there are others commonly used.

We need a frame of reference in which to define positions, velocities, and accelerations. For example, a position vector r might have the coordinates (2,4,7) in one Cartesian frame. In another, the same vector might be (6,5,11). A reference frame helps us to be specific about our measurements.

We use this concept of a ‘frame of reference’ widely in Physics. The ‘laboratory frame’, for example, is often the one in which you, the observer, are situated. You must imagine a set of axes fixed to the floor, and you are standing stationary at the origin. It is often helpful, however, to change your viewpoint to another frame, say one which is moving at a steady speed through the laboratory frame parallel to the x axis. For example, you can imagine standing on the platform of a station (the ‘laboratory frame’) watching someone run past you at 10 mph. Coming into the station is a train,

also moving at 10 mph parallel to the runner. If you were to transform your point of view into the train’s frame (the ‘moving frame’) you would see the runner apparently running on the spot, not making any progress at all relative to the train. We make wide use of the concept of a frame of reference in the theory of special relativity

Example 1: a particle, P, is situated at position vector (5,3) in a two-dimensional coordinate frame S. What are its coordinates measured in frame S’ which has the same origin as S but is rotated clockwise by 30 degrees relative to S?

S

y

x

P

5

3

y’

x’

S’

xP ’

yP ’

30°

The rotation matrix for clockwise rotation about the origin by angle θ is given by

cos sin.

sin cos

θ θ

θ θ

R

Example 1: a particle, P, is situated at position vector (5,3) in a two-dimensional coordinate frame S. What are its coordinates measured in frame S’ which has the same origin as S but is rotated clockwise by 30 degrees relative to S?

S

y

x

P

5

3

y’

x’

S’

xP ’

yP ’

30°

The rotation matrix for clockwise rotation about the origin by angle θ is given by

cos sin.

sin cos

θ θ

θ θ

R

The new coordinates are therefore

0.87 0.5 5 2.83.

0.5 0.87 3 5.10P P

PP

x x

yy

R

Before coming on the the development of Einstein’s Special Theory of Relativity, it will be useful to examine in general terms some assumptions we have been making about the frames of reference in which we have been thinking about physics. Thus:

(a) A frame of reference is just a set of calibrated axes or coordinate system against which we can measure positions, velocities, accelerations, and times. Typically, we might

S1 2 3 4 5

metres

say that an ‘event’ occurs at a particular instant in space and time defined in one coordinate system, S, by (x,y,z,t). The same event, seen in another coordinate system, S, is defined by the coordinates (x,y,z,t ).

(b) The geometry of the space in which we are working obeys the axioms derived from the postulates of Euclid: in this space, two parallel lines meet at infinity, the sum of the angles in a triangle is 180 degrees, etc. We refer to this as ‘Euclidean space’.

(c) The distance between two events in space seen in one frame is the same when viewed in any other frame.

S

S

S‘event’

(x,y,z,t)(x,y,z,t )(x,y,z,t )

x

y

x

yx

y

Thus if events A and B occur in Cartesian frame S at positions (xA,yA,zA,tA) and (xB,yB,zB,tB), then the space interval, sAB, between the two events viewed in S is given by Pythagoras’ Theorem.

y

x

A

B

xy

S

yB

yA

xBxA

sAB

The symbol indicates the interval. Thus sAB is the interval in s between points A and B

Note: do not confuse frame names, S, S’ etc, with the variable s

Thus s2AB

= (xB xA)2 + (yB yA)2 + (zB zA)2.

Now view the same two events from the point of view of another frame, S.

y

x

A

B

x

y

SsAB

y

xS

The two events occur in S at coordinates of (xA,yA,zA,tA) and (xB,yB,zB,tB). Again the space interval, s AB

, between the two events, viewed in S, is given by Pythagoras’ Theorem, i.e.

s 2AB = (xB xA)2 + (yB yA)2 + (zB zA)2.

Our assumption is that s2AB

= s 2AB .

Euclidean space

y

x

y

x

A

B

x

x

y

y

SS

Our assumption is that s2AB

= s 2AB .

(d) The time interval is the same in any frame. Thus tAB

= (tB tA) = tAB = (tB tA). In fact

we have a strong notion that time and space are absolute quantities. We think that we can define a point in ‘absolute’ space and ‘absolute’ time, and that space and time are the same for everyone, no matter how they are moving with respect to each other. These ideas obviously work very well in everyday life, but need closer examination.

(e) We can express the transformation between coordinates seen in one inertial Cartesian

frame and those in another using the ‘Galilean’ transformation. Suppose that frame S' is moving along the positive x axis of frame S at constant speed v, and their origins coincide at t = t = 0.

An event, A, occurs in S at (xA,yA,zA,tA), and in S at (xA,yA,zA,tA ).

y

x

v

vtA

AxA

xAy

xS S

MO 173TM 1322

We see xA = xA vtA, which is the only coordinate affected, so the Galilean transformation is:

x = x vty = yz = zt = t

This is the transformation which applies to all the Newtonian Physics you’ve done so far. As we shall see, it only works for transformations between frames in which v << c, the speed of light.

Note that space

and timeare separate

(f) Note that the quantities x, x, etc. are really intervals between the two events: (i) the origins coinciding with each other, and (ii) event A. Even here, we are expressing the transformation between space and time intervals, not between absolute space and absolute time positions. We could therefore equally well write: x = x vt and equally x = x’ + vt’ y = y y = y’ z = z z = z’ t = t t = t’where denotes the interval between events.

Example 2: an observer in a high-speed train, travelling at 575 km h‒1 (currently the speed record held by the TGV) measures the time between his passing two signals as precisely 3 s. What is the distance between the two signals measured by a second observer on the track using a tape measure?

S (track frame)

A BΔxAB

S’ (train frame)

A,B Δx’AB = 0v

Δt’AB = 3

The observer in the train is present at both events, so he measures a space interval of zero.The Galilean transformation gives:

Example 2: an observer in a high-speed train, travelling at 575 km h‒1 (currently the speed record held by the TGV) measures the time between his passing two signals as precisely 3 s. What is the distance between the two signals measured by a second observer on the track using a tape measure?

S (track frame)

A BΔxAB

S’ (train frame)

A,B Δx’AB = 0v

Δt’AB = 3

The observer in the train is present at both events, so he measures a space interval of zero.The Galilean transformation gives:

m

AB AB ABΔ Δ Δ

5750000 3

3600479

x x v t






Lecture 6

Problems with classical physicsProblems with classical physicsProblems with classical physicsProblems with classical physicsNineteenth-century physicists sought mechanical explanations for everything. Even Maxwell’s electromagnetic theory was based on mechanical interactions. In particular, it was thought that light waves, like all other known waves, must travel in a medium. Thus water waves travel on the surface of water; sound waves travel through the air; waves on my mother’s washing line travelled down the line (waves on a string). What about light, radio etc.? The work of Huygens, Young,


2 2

2 2 2

10

ψ ψx c t

1D wave equation for waves on a string, or water waves, or sound waves:

represents the amplitude of the wave, that is the displacement of a point in the medium, and c is the speed of the wave. For example, for waves on a string:T

cρ

ψ

2 2

0 02 20

E Eε μ

x t

James Clerk Maxwell showed that, for electromagnetic waves (light, radio, x-ray etc.) the equation was

E represents the electric field of the wave


and Fresnel seemed to account for the properties of light in terms of waves. Therefore, (it was thought) there must be a corresponding medium for them? They made the hypothesis that there was indeed such a medium, and they called it the luminiferous aether. Thus, light travelled at 3108 m s1 through this medium which was all-pervading. This hypothesis could be tested by looking for effects caused by motion through the aether.One such piece of evidence was supplied by Bradley in 1725 who observed stellar aberration.

E was thought to be the displacement of a point in the aether

Stellar aberrationStellar aberrationStellar aberrationStellar aberration

Imagine you are in vertically-falling rain. Now get on your bike – the rain appears to come down at you from an angle to the vertical.

v

c v/c 104 radians

This was easily measured by Bradley, and appeared to show evidence for a stationary aether.

aether wind

Earth’s orbital velocity through the stationary

aether

The Michelson-Morley experimentThe Michelson-Morley experimentThe Michelson-Morley experimentThe Michelson-Morley experiment

One of the most-famous attempts to measure motion through the aether was the Michelson-Morley experiment. The Earth moves at about 30 km s1 in its orbit around the Sun, which is an appreciable fraction of the speed of light, 300,000 km s1. M & M set up an optical interferometer which would have been easily sensitive enough to detect this motion. The principle was that a coherent light beam was divided into two parts, and each part sent along perpendicular paths as follows:

S A

B

C

T

O

aether wind

v

d

dThis is a simplified diagram of a MM interferometer. You will meet this again in more detail next year.

A beam of light from a coherent light source, S, is split by the half-silvered mirror, A, into two beams, one travelling towards B, and the other towards C. The beams are reflected by the fully-silvered mirrors B and C. That from B passes through A to T, and that from C is reflected at A towards T. The two beams combine in the telescope, T, and interfere to produce an interference pattern, which is measured by the observer O.

Let us suppose that the apparatus is moving through a stationary aether from right to left, at speed v, so that there is an ‘aether wind’

S

T

A

B

C

O

d

d

blowing from left to right parallel to AC. The two light paths ABA and ACA are then not identical. We can see this as follows. First, the path ACA:

AC is with the flow, so tAC = d/(c+v);

CA is against the flow, so tCA = d/(cv).

tACA = d/(c+v) + d/(cv).

Now consider the path ABA:

Both AB and BA are across the wind. The light gets ‘blown’ to the right, so the path from A to B is slightly against the flow and takes longer.

S

T

A

B

C

O

d

d

The speed from B to A is the same as that from A to B, so the total time of flight is

aether wind

v

c

v

A

B

22 vc

./2 22 vcdt ABA

S

T

A

B

C

O

d

d

the time difference between ACA and ABA is

Now, if x is much less than 1 we can use the first two terms of the binomial expansion of

2 2

2 2 2 2

12 2 2 2 2

12 2 1 2 2 2

2

( ) ( ) 2

2 2(1 / ) (1 / )

2(1 / ) (1 / )

d d dt

c v c v c vd c v d c v d

c v c vd d

c v c c v c

dv c v c

c

(1+x)n 1+nx, so that

With the apparatus used by MM, this corresponded to a shift in the fringe pattern of about half a fringe, very easily seen if it existed.

.2

2

)2

1(2

)1(2

3

2

2

2

2

2

2

2

cdv

cv

cd

cv

cd

cv

cd

t

However, no shift was ever seen, despite the experiment being repeated with first AC then AB parallel to the Earth’s orbital velocity, and at six month intervals (just to check that the aether wasn’t coincidentally moving at the same velocity as the Earth when the experiment was first done).

Although this experiment is often cited as evidence that the aether does not exist, Einstein was probably not aware of it when he formulated his theory of special relativity.

Classical electromagnetismClassical electromagnetismClassical electromagnetismClassical electromagnetismNewtonian physics appears to operate in accord

with the Galilean transformation, i.e. we can transform the (x,y,z,t) coordinates of events seen in one inertial frame into those seen in another using this transformation.

However, Einstein was aware that Maxwell’s laws of classical electromagnetism (which you will come to next year) did not transform in the same way. In particular, the speed of radio or light waves was predicted to be given

by the expression , where ε0 and μ00 0

1

( )c

ε μ

not accelerated

are constants associated with the vacuum. This seems to say that the speed of light in a vacuum is independent of the motion of the source or the observer, since there is no meaning to motion relative to a vacuum. No need for an aether – it just doesn’t exist. He postulated that this was not some quirk of electromagnetism, but that its consequences applied to the whole of physics. He formulated the theory of relativity – special relativity (1905) applying to un-accelerated frames, and general relativity (1916) which is about gravity. Here we do the special theory.

Einstein’s postulatesEinstein’s postulatesEinstein’s postulatesEinstein’s postulates

Einstein’s ideas about electromagnetism and the nature of physical laws may be summarised in his two postulates:

1. All of the laws of physics are the same in every inertial (un-accelerated) frame.

2. The speed of light in a vacuum is the same for all observers.

The first of these postulates is quite consistent with the Newtonian mechanics that you have already met in school. You made the assumption that

NB!

MO 194TM 1321

Newton’s laws were true and obeyed in any inertial frame, and were consistent with the Galilean transformation. However, the new thing is that all the laws of Physics, including electromagnetism and anything else that you can mention, are the same in every inertial frame.

Turning that around, the first postulate states that, if you are in an inertial frame, there is no internal experiment which you can do which can distinguish between that frame and any other inertial frame – all are equivalent. For example, there is no such thing as an absolute rest frame.

When I am at rest in my inertial frame, that state of rest is the same as any other in any inertial frame no matter how fast it is moving relative to mine a revolutionary concept for people seeking an all-pervading aether.

The second postulate really follows on from the first. If there is no internal experiment I can do to tell which particular inertial frame I am in, then the speed of light in a vacuum must be the same for me as for anyone else. This has far-reaching implications for space and time.

What is time?What is time?What is time?What is time?Time is a notion or a concept. We know that it is not

a substance. We know that it always moves in one direction, i.e. ‘time passes’ or ‘time advances’. We assume that its rate of flow is uniform, and that it is universal.

We measure time in terms of intervals, that is we identify events and then we measure how many ‘ticks’ (assumed equally spaced) there are between the events using a machine – a clock – that has been designed to produce ticks at as uniform a rate as possible.

Time is the most-accurately measured of all physical quantities by many orders of magnitude.

How is time measured?How is time measured?How is time measured?How is time measured?

Time intervals are measured using processes which are assumed to be exactly periodic: the swinging of a pendulum; the oscillation of an escapement mechanism; the rotation of the Earth; the orbit of the Earth around the Sun; the vibration of an excited atom.

Since 1967, we have used atomic clocks to measure time. The SI second is defined to be exactly 9,192,631,770 cycles of vibration in an atomic clock controlled by one of the characteristic frequencies of caesium 133.

Notes:

(a) Atomic time is now independent of astronomy.

(b) We keep times consistent with astronomy by inserting leap seconds up to twice per year at midnight on June 30th or December 31st.

(c) The pulses received from highly regular ms pulsars may supersede atomic clocks in the future.

(d) The time scale disseminated by radio (UTC) – e.g. the ‘time pips’ on the (analogue) BBC – is an average over many atomic clocks in many countries.






Lecture 7

Time dilationTime dilationTime dilationTime dilationOne of the consequences of Einstein’s postulates

is that identical clocks appear to run at different rates depending on their relative motion. To see how this comes about, we will consider a special kind of clock. Its ‘pendulum’ is a photon reflected back and forth between mirrors. Since the speed of light is directly involved with the mechanism of this clock, we can see quite easily how Einstein’s postulates, especially the second one, affect the performance of the clock. His first postulate tells us that all clocks, whatever the mechanism, must be affected in

MO 195TM 1324

the same way (as otherwise we would be able to measure the ‘speed’ of our inertial frame relative to absolute rest by comparing this clock with one having a different mechanism):

mirrors

photon

clock face

C:\lazarus\units\relativity demo\windows_reldem.exe

First view the situation in the upper clock’s rest frame. Let events A, B, C be a photon leaving the base mirror, arriving at the upper mirror, and arriving back at the base mirror respectively. In the clock’s rest frame, S, the photon takes a time 2h/c to travel from A to B and back to C (where c is the speed of light). Let this time interval be . Then

A

B

C

h

rest frame S

4 or ,

2 222 AC

AC

tch

ch

t

ACt

back

Now let’s look at the same events as seen in the lower frame, S. The time interval between events A and C in this frame is tAC. Remember that the photon still travels at speed c, but has further to go, so takes longer:

A

B

C

v

h

ACΔ

2

v t

ACΔ

2

c t

D

In the triangle ABD, by Pythagoras’ theorem, we have

But in the moving clock’s rest-frame, S, we have

So substituting for h gives us

2 2 2 22 AC ACΔ Δ

4 4

v t c th

2 22 ACΔ

4

c th

2 2 2 2 2 2AC AC ACΔ Δ Δ

4 4 4

c t v t c t

Rearranging, we find

where .

Note that is greater than one for all speeds such that 0 v c, and is undefined for speeds of c or greater. This equation therefore shows us that the time interval between events A and C is shortest in the rest frame of the clock, and that the time interval measured between the same two events viewed in a frame in which that clock is moving is longer.

AC ACΔ Δt γ t

2 2

1

(1 )v c

Notes(a) We are forced to conclude that the rate of the

passage of time depends on relative motion.(b) The shortest time interval between two events

is measured by a clock which is present at both events.

(c) Such a time interval is called a proper time interval, and such a clock measures proper time.

(d) The time interval measured in another moving frame, using two clocks each of which is at only one of the events, is always longer.

(e) This is sometimes summarised in the statement ‘moving clocks run slow’. Be careful with this statement as it can cause confusion. Always ask: “Which clock was present at both events?” (It measures the shortest time interval.)

(f) This true for every kind of clock, not just light-clocks (remember Einstein P1).

(g) The effect is tiny in every day life: 70 mph for 6 years causes a 1 s shift

(h) The effect is larger for space travellers: at four-fifths of c, the shift is from 3 s to 5 s

Two quotations on the relativity of time:

Einstein 1905: “Thence we conclude that a clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.”

Rosalind 1599: “Time travels in divers paces with divers persons.”

(As you like it, Act 3 Scene 2, by William Shakespeare)

The twin paradoxThe twin paradoxThe twin paradoxThe twin paradox

We have seen that the time interval between two events measured by the captain of a space ship is shorter than the time interval between the same two events measured by observers at rest relative to the Earth.

Consider the following train of events:

v

E

A B

C

MO 204TM 1335

A rocket leaves Earth (event A) and travels to a distant point at very high speed. It turns around (event B) and travels back to Earth, arriving (event C) several years after leaving. We know from what we have done already that the time interval ΔtAB between events A and B, and the time interval ΔtBC, between events B and C, measured by the rocket captain, will be shorter than the corresponding intervals, ΔtAB and ΔtBC, measured on the Earth. (Remember that only the rocket captain’s clock is present at all the events.)

Thus we conclude (correctly) that the twin brother of the rocket captain left on the Earth will be older than his rocket-captain sister who has made the return journey.

But now consider the same events as seen by the rocket captain. She sees the Earth receding at high speed v behind her on the outward leg, and then approaching at high speed v on the return leg. According to the rocket captain, her brother left on Earth has been travelling relative to her, and so will be younger than her when she gets back to Earth. This is the twin paradox.

The twin paradox arises through sloppy thinking. Actually there is no paradox because the two situations are not symmetrical. Just ask yourself the question: “Who’s clock was present at all the events, A, B, and C ?” Also, the twin brother left on Earth remains the whole time in a single inertial frame of reference. The twin sister in the rocket changes from one inertial frame to another mid-course. On the way out, the frame travels at speed v; on the way back it travels at speed –v. We therefore need to analyse the situation carefully. We will return to this a bit later.

Another consequence of Einstein’s postulates is that if two events are simultaneous in one frame, they are not necessarily simultaneous when viewed in another moving frame. Consider the following example: a first observer is at rest relative to, and exactly half-way between, two flashing beacons, one blue and one red. He observes that a flash from the red beacon arrives at exactly the same instant as one from the blue beacon. He therefore deduces that the two flashes were emitted by the beacons at the same moment in his frame.

SimultaneitySimultaneitySimultaneitySimultaneity MO 205TM 1330

A BC

First observer’s frame

The beacons and first observer are stationary in this frame. The flashes arrive at the same instant (event C), and since the distances are the same, they must have left the two beacons at the same instant (events A and B are deduced to be simultaneous in this frame).

first observerblue beacon

red beacon

Second observer’s frame

v

v

vv

Now consider the same circumstances from the point of view of a second observer who is moving at speed v relative to the first observer:

A BC

first observerblue beacon

red beacon

In the second observer’s frame, the beacons and first observer are all moving to the left at speed v.

The second observer must agree that the two photons arrive at the same time at the first observer’s position (event C) since this is an event, and the nature of the event cannot be changed by relative motion. But the first observer is moving at speed v to the left relative to the photons in the two flashes, and so the red photons have to make up extra distance whilst the blue photons have less distance to travel (remember that both sets of photons travel at the same speed, c, in any frame). The second observer must conclude that the red photons left (event B) before the blue photons (event A) since they arrive together (event C) and the reds have

further to go than the blues. Thus, we have the first observer saying that A and B must have been simultaneous, whilst the second observer says that B happened before A. This is not a contradiction as we allow time to be relative, just as space is relative. We can think of space and time together as making up ‘spacetime’, and that two events, points in spacetime, can be considered as being joined by a vector called a ‘four vector’. All we are doing when we change frames is that we are viewing the four vector from a different point of view.

An example in Euclidean space …An example in Euclidean space …An example in Euclidean space …An example in Euclidean space …

I am turning right

I am turning left

Who is correct? Both are. From the red car’s point of view, the road is on the left. From the green car’s point of view, the road is on the right. We are used to this and so don’t find it strange. So it is with events in space time. Whether one event happens before another depends on your point of view – i.e. which inertial frame you are in.

Length contractionLength contractionLength contractionLength contractionA third consequence of Einstein’s postulates is that

the length of a rigid bar measured in its rest frame is always greater than the length of the same bar measured in a frame moving parallel to its length.

To see this, consider the following. A spaceship travels from one interplanetary beacon to another.

vA B

L0

MO 201TM 1326

Event A is that of passing the first beacon, and event B is that of passing the second. The beacons are L0 apart in their rest frame (measured with a tape-measure). An observer at rest with respect to the beacons finds the time interval between A and B is ΔtAB = L0/v.

The captain of the spaceship measures the distance between the beacons as L, and the time interval as ΔtAB = L/v. Now ΔtAB = γ ΔtAB (the captain measures a proper time interval).So we must conclude that .

0LL

This means that the captain of the spaceship measures a shorter distance between the two interplanetary beacons than is measured by a tape-measure (or rigid bar) fixed between them. This result applies to all measurements of length, no matter how they are made. The length of an object appears to be contracted in the frame in which it is moving along its length, and rulers are longest when measured in their rest frames.

What about lengths perpendicular to the motion? The answer is ‘no change’. We can see this by considering a relativistic train running on a section

of straight track at a steady speed v. When viewed in the rest-frame of the track, the train runs smoothly by at speed +v. When viewed in the rest-frame of the train, the track runs smoothly by at speed v. In neither case is the train seen to come off the track. Therefore, we must conclude that lengths perpendicular to the motion do not change. If they did, an observer could test whether he is ‘fixed’ or ‘moving’ by observing whether the train stayed on the track or came off it, and that would violate Einstein’s first postulate.

Einstein’s postulates lead to the following conclusions:

Time dilation:

Simultaneity: events simultaneous in one frame are not necessarily so in another.

Length contraction parallel to motion:

No length contraction perpendicular to motion.

SummarySummarySummarySummary

t t

0LL

Einstein’s postulates lead to the following conclusions:

Time dilation:

Simultaneity: events simultaneous in one frame are not necessarily so in another.

Length contraction parallel to motion:

No length contraction perpendicular to motion.

SummarySummarySummarySummary

t t

0LL

Measuring the speed of light

C:\Documents and Settings\Peter Duffett-Smith\My Documents\teaching\Part IA Mechanics and Relativity\2008\Foucault experiment\7expts.xls






Lecture 8

Lorentz transformationLorentz transformationLorentz transformationLorentz transformationWe have already seen how the Galilean

transformation transforms the coordinates of events between different inertial frames which are moving with respect to each other on the basis of classical Physics, i.e. before Einstein proposed his revolutionary postulates. We now need to revise that transformation to incorporate the effects of Special Relativity. The new formulation is called the Lorentz Transformation, and it can be used to solve relativistic problems in a straightforward manner.

MO 198TM 1322

We consider an event, A, which occurs at position (x,y,z,t) as measured in the laboratory frame S.

The same event is also viewed in another frame, S’, moving at speed v along x, with its axes parallel to those of S, and such that x = x’ = 0 at t = t’ = 0.

v

S' (moving frame)

S (laboratory frame)

Ax

At time t (measured in S), the origin of the moving frame is at position x = vt in S:

Imagine that a rigid ruler is fixed to the y’ axis of S’ and is of just the right length so that its far end is instantaneously at event A. Its length in S is r.

r

v

S' (moving frame)

vt

S (laboratory frame)

Ax

measured in S

We see that x = r + vt, so r = x ‒ vt.Event A occurs at position (x,y,z,t ) measured

in the moving frame S. Now we know that a ruler is longest when measured in its rest frame, and is contracted by a factor γ when measured in a frame in which it is moving. In this case, the rigid ruler has a length of r in S and r0 in S’ (it is stationary in this frame) such that r0 = γr. But r0 is the x’-coordinate of event A measured in S’.

x = (x – vt) . (1)We could have done all this the other way around,

i.e. start in frame S’ and then move to S. We do

not need to start again, however, as all we need to do is to switch the dashes and replace v with v, so we get

x = (x + vt ) . (2)To find the way the times transform, we can

substitute in equation (2) for x from equation (1):

x = (x + vt ) = ( (x – vt) + vt )After a bit of algebraic manipulation, we get

t = (t – vx/c2). Replacing v with –v, and switching the dashes:

t = (t + vx /c2).

We have already noted that lengths perpendicular to the motion do not change. Collecting everything together, we have:

with

2 2

2 2

;

;

;

;

1.

1

vx vxt γ t t γ t

c c

x γ x vt x γ x vt

y y y y

z z z z

γv c

Back

Now x, t etc. are really intervals between the two events ‘the origins coincide’ and event A. We can therefore equally well write:

with

2 2

2 2

Δ ΔΔ Δ ; Δ Δ

Δ Δ Δ ; Δ Δ Δ

Δ Δ ; Δ Δ

Δ Δ ; Δ Δ

1.

1

v x v xt γ t t γ t

c c

x γ x v t x γ x v t

y y y y

z z z z

γv c

Notes:

(a) This set of equations is called the Lorentz Transformation. Use it - it will help you to solve relativity problems and get the correct answer.

(b) The sets (ct,x,y,z) and (ct,x,y,z ) are examples of 4-vectors. The LT can be written:

z

y

x

ct

cv

cv

z

y

x

tc

1000

0100

00

00

(c) There are many example of 4-vectors in special relativity, all of which transform using the same transformation matrix i.e.

b = A.b, where b and b are 4-vectors and

We will return to this later.

1000

0100

00

00

cv

cv

A

Example 3: advances in technology enable the high-speed train of Ex. 2 to travel at 0.8 c. An observer in the train measures the time between his passing two signals as 1.198 μs. What is the distance between the two signals measured by a second observer on the track using a tape measure?

S (track frame)

A BΔxAB

S’ (train frame)

A,BΔx’AB = 0v

Δt’AB = 1.198×10‒6

The observer in the train is present at both events, so he measures a space interval of zero.The Lorentz transformation gives:

Example 3: advances in technology enable the high-speed train of Ex. 2 to travel at 0.8 c. An observer in the train measures the time between his passing two signals as 1.198 μs. What is the distance between the two signals measured by a second observer on the track using a tape measure?

S (track frame)

A BΔxAB

S’ (train frame)

A,BΔx’AB = 0v

Δt’AB = 1.198×10‒6

The observer in the train is present at both events, so he measures a space interval of zero.The Lorentz transformation gives:

m

AB AB AB

8 6

2

AB

Δ Δ Δ

0 0.8 3 10 1.198 10

11.667

1 (0.8)

Δ 479

x γ x v t

γ

γ

x

Example 4: (part of Ex 11 in book). The coordinates of two events measured in frame S are A: xA = x0, tA = x0 /c; B: xB = 2x0, tB = x0 /(2c). (a) What is the speed and direction of travel of an inertial frame S in which both events occur at the same time? (b) When do the events occur as measured in S ?

(i) Events are already identified in the question.

(ii) Draw diagrams:

S

t

x

A

Bx0

2x0

x0/c

x0/2c

S

t

x

A

xA = ?tA = ?

vB

xB = ?

(iii) Write down the intervals (remember that the origins coinciding is event 0, so x, t etc are intervals already)

Event A: xA = x0, tA = x0 /c,

Event B: xB = 2x0, tB = x0 /(2c)

(iv) Use the Lorentz Transform:

(a)

(b)

, so

0 0 0 0A BA A B B2 2 2 2

0 0 0 0 0 0

2 2 2

2;

2

2 1;

2 2 2

x vx x v xvx vxt γ t γ t γ t γ

c c c c c c

x vx x v x x vx vγ γ

c c c c c c c

(iv) Use the Lorentz Transform:

(a)

(b)

, so

0 0 0 0A BA A B B2 2 2 2

0 0 0 0 0 0

2 2 2

2;

2

2 1;

2 2 2

x vx x v xvx vxt γ t γ t γ t γ

c c c c c c

x vx x v x x vx vγ γ

c c c c c c c

2 2

0 0 0AA B A 02 2

1 1 2

1 (1/ 4) 31 /

32 2 3( )

2 23 3

γv c

x cx xvxt t γ t x

c c c c c

Example 15: (NST – Physics Part IA 1999). Twins Alice and Bob go travelling in space. They each carry a clock to record how much they age during the trip. Alice leaves Earth and travels at a steady speed of 5c/13 to a space-station which is 1 light year away. Bob leaves Earth at the same time as Alice, but travels at a speed of 5c/13 in the opposite direction. When Alice reaches the space-station, she immediately turns around and travels towards Bob at a new speed of 12c/13, eventually catching up with him. Which twin is older, and by how much, when they meet in space?

Example 15: (NST – Physics Part IA 1999). Twins Alice and Bob go travelling in space. They each carry a clock to record how much they age during the trip. Alice leaves Earth and travels at a steady speed of 5c/13 to a space-station which is 1 light year away. Bob leaves Earth at the same time as Alice, but travels at a speed of 5c/13 in the opposite direction. When Alice reaches the space-station, she immediately turns around and travels towards Bob at a new speed of 12c/13, eventually catching up with him. Which twin is older, and by how much, when they meet in space?

Identify the events:

Event A

Alice leaves Earth Bob leaves Earth

Event B

turns around

Event C

meet in space

ES

AB

(B)

C

11 xAC

In S:

Alice S

Bob S

distance yr

speed

ly

yr

AB

AC AC AC

AC BC AC AB

AC AC AC

1 135 13 5

55 13 13

12 121

13 135 12 12 13 221

113 13 13 5 35

AC

ct

c

xt x t

c

c cx t t t

t t t

In Bob’s frame S:

yr

ACAC AC AC AC2

AC AC

0

221 12 20435 13 35

v xx t t t

c

t t

13 13 = ; =

12 5

In Alice’s frame S:

yr

yr

Alice has aged by yr

Bob is older by yr

AB AB

BC BC AC AB

13 12 125 13 5

107

12 10 1345 7 35

204 134 702

35 35 35

t t

t t t t

Experimental evidence for relativityExperimental evidence for relativityExperimental evidence for relativityExperimental evidence for relativity

(a) Time dilation in the decay of muons

Cosmic rays produce showers of muons at the top of the atmosphere. These have lifetimes of only about 2 s, so should travel only a few hundred metres before decaying. (Their speeds are close to c.) In practice, we measure most of them at ground level after travelling through many tens of km of atmosphere. This is because the muon clocks measure proper time intervals between events ‘enter atmosphere’ and ‘decay’, (which are about 2 s), whilst the Earth-based clocks measure much longer time intervals.

(b) Michelson-Morley experiment

This is evidence for the absence of the aether. Jaseja, Javan, Murray & Townes (1964) showed that any effect is less that 0.1% of that expected if there was an Aether.

(c) Magnetic effects

The magnetic force between two parallel current-carrying wires can be calculated instead from relativistic modifications of the electrostatic forces between the charges in the wires. This demonstrates the consistency between electromagnetism and mechanics brought about by Einstein’s postulates.

(d) GPS Clocks

The rates of the clocks in the Global Satellite Positioning System (GPS) satellites need to be adjusted relative to those on the ground for both the time dilation of special relativity and the general relativistic effect of the difference in gravitational potential. Otherwise, the Earth-based clocks drift with respect to the satellite clocks, with corresponding growing errors in positions.






Lecture 9

The speed of light as the ultimate speedThe speed of light as the ultimate speedThe speed of light as the ultimate speedThe speed of light as the ultimate speed

The Lorentz transform shows that the speed of light places an upper limit to the speed which which two frames can move with respect to each other. As v c, so . Then lengths observed in a frame moving with respect to the observer shrink to zero, and time intervals between events (in the moving frame) extend to infinity. Note that this applies to moving objects, or more generally to moving energy. Virtual objects, such as the point of intersection of two

inclined rulers, can move at arbitrarily high speeds up to infinity.

fixed

move up at speed v

point of intersection

moves at speed > v

Addition of speeds formulaAddition of speeds formulaAddition of speeds formulaAddition of speeds formula

It is clear that the Galilean addition of speeds formula, , will not work if either or v is a large fraction of c. For example, if = c/2 and v = 3c/4, the result would be 5c/4, an impossible result since it must be less than c. Use the Lorentz transform to get the right formula.

To see how to do this, consider a rocket travelling at speed relative to a frame S between two markers, A & B. The markers are a distance apart fixed in this frame, and the rocket takes time

to travel between them.

x xu u v xu

xu

xu ABx

ABt

S

Markers fixed in S

xuA B

MO 208TM 1336

Clearly

From the point of view of an observer in frame S, relative to which frame S is moving parallel to at speed v, the corresponding intervals are xAB and tAB. The speed of the rocket in this frame is therefore given by

.AB

AB

Δ

Δx

xu

t

SS

v Markers fixed in S

xu

xu

A B

Dividing through by tAB and cancelling the gammas, we get

What about motion at an angle? This time, the rocket travels at velocity u in frame S, so we must first resolve u into x and y components:

2

x

γ x v txu

t v xγ t

c

AB ABAB

AB ABAB

21

xx

x

u vu

vu

c

as before.

2

( , ) ( , ); ( , ) ( , )

1

x y x y

xx

x

x y x yu u u u

t t t t

u vu

vu

c

AB AB AB AB

AB AB AB AB

u u

B

SS

v u

xu

yuA

ABx

ABy

2( )

y

y yu

v xtγ t

c

AB AB

ABABAB

Dividing through by gives

.

21

y

yx

uu

vuγ

c

Thus, the ‘vector’ velocity addition formula is

2 2

( , ) ( , ).1 1

yxx y

x x

uu vu u

vu vuγ

c c

u

ΔtAB

Example 6: further advances in technology enable the guard of the high-speed train of Ex. 2 to travel at 0.5 c from the back of the train towards the front. With what speed would an observer on the track see the guard moving when the train is travelling at a speed of 0.8 c?

0.5xu c

Guard

S (track frame)

S’ (train frame)

v

Guard

xu

The formula we have just derived for the addition of speeds gives

2

0.5 0.8 1.31 0.4 1.4

1

xx

x

u vu c c

u v

c

Aberration of lightAberration of lightAberration of lightAberration of lightWe have already seen that Bradley measured the

aberration of light, and took this to be evidence for the existence of the Aether. But aberration can also be understood in terms of relativity and the complete absence of the Aether. Here’s how. Consider a photon emitted from a light source L which is stationary in frame S’:

SS

vθ c.cos(θ )

c.sin(θ ) c L

What is the corresponding angle,θ, when the photon is viewed from frame S? We can use the addition of speeds formulae to find the components of the photon’s velocity as seen in S:

cos( ) sin( )( , ) ( , )

1 cos( ) 1 cos( )x y

c θ v c θc c

v vθ γ θc c

c

2 2

( , ) ( , ).1 1

yxx y

x x

uu vu u

vu vuγ

c c

u

In frame S we have cx = c cos(θ ), cy = c sin(θ).

, and

cos( ) sin( )( cos( ), sin( )) ( , )

1 cos( ) 1 cos( )

cos( )cos( )

1 cos( )

sin( )sin( )

1 cos( )

c θ v c θc θ c θ

v vθ γ θc c

vθ

cθv

θc

θθ

vγ θ

c

Note that you can get the corresponding formulae for by putting a negative sign in front of v and interchanging the dashes.

, and

cos( )cos( )

1 cos( )

sin( )sin( )

1 cos( )

vθ

cθv

θc

θθ

vγ θ

c

In frame S we have cx = c cos(θ ), cy = c sin(θ).

, and

cos( ) sin( )( cos( ), sin( )) ( , )

1 cos( ) 1 cos( )

cos( )cos( )

1 cos( )

sin( )sin( )

1 cos( )

c θ v c θc θ c θ

v vθ γ θc c

vθ

cθv

θc

θθ

vγ θ

c

, and

cos( )cos( )

1 cos( )

sin( )sin( )

1 cos( )

vθ

cθv

θc

θθ

vγ θ

c

Stellar aberrationStellar aberrationStellar aberrationStellar aberration

In Bradley’s case, θ = 90°, and the direction of the photon was reversed (i.e. incoming photon rather than outgoing). We conclude

v

c sin( ) = cos( )

v/c

as Bradley measured and used as evidence, incorrectly, of the existence of the Aether.

S S

Relativistic Doppler effectRelativistic Doppler effectRelativistic Doppler effectRelativistic Doppler effect

We are all familiar with the Doppler effect in relation to the changing pitch of wailing sirens on emergency vehicles as they pass by. Here we find a formula for the Doppler effect as applied to electromagnetic waves. We do this by considering a pulsing light source which is at rest in a frame S, and finding the time between two consecutive pulses measured in frame S, relative to which S’ is moving along the x-axis at a steady speed of v.

Let events A and B be the emission of consecutive pulses by the light source in S’.

MO 340TM 1328

pulses in frame S:

The observer in S sees the light source move between pulses, so the second pulse has further to go to reach the observer, O, than the first pulse.

S

A,BxAB = 0

tAB = T0

Slight source

v

S

v v

A B

xAB = ?

tAB = ?xAB

O

back

We use the Lorentz transform to find xAB and tAB:

Now the pulse from B has an extra distance, xAB, to travel before reaching the observer as compared with the pulse from A. Thus the time, T1, measured by the observer, O, between the arrival of the pulses is:

ABAB AB 02

AB AB AB 0

ΔΔ Δ

Δ Δ Δ

v xt γ t γT

c

x γ x v t γvT

We use the Lorentz transform to find xAB and tAB:

Now the pulse from B has an extra distance, xAB, to travel before reaching the observer as compared with the pulse from A. Thus the time, T1, measured by the observer, O, between the arrival of the pulses is:

ABAB AB 02

AB AB AB 0

ΔΔ Δ

Δ Δ Δ

v xt γ t γT

c

x γ x v t γvT

1 0 0 0

Δx γ vT γT γT T

c c AB

back

1 0 0 02

1 0

2

1 111

1

-

11

.c v

T

v vvc cv cT γT T T

c v vvc c

Tc v

c

If the frequency of the pulses, in the rest-frame of the light source, is 0, then the frequency, 1, measured by the observer in S is

Note that this is the change observed when the source is moving directly away from the observer. There is a (smaller) Doppler shift even for sources moving at right angles to the line of sight because of the time dilation factor.

1 0 .c v

ν νc v

Example 7: an observer on a space station sees two spacecraft passing each other, one coming directly towards him and the other moving directly away, both at a speed of c/2. Both are transmitting radio waves. What is the ratio of the frequencies received by the observer on the space station?

S – space station frame

v

v

The observer sees frequency ν1 from the spacecraft moving towards him, and ν2 from the one moving away. The frequency measured on board the spacecraft is ν0.

Example 17: an observer on a space station sees two spacecraft passing each other, one coming directly towards him and the other moving directly away, both at a speed of c/2. Both are transmitting radio waves. What is the ratio of the frequencies received by the observer on the space station?

S – space station frame

v

v

The observer sees frequency ν1 from the spacecraft moving towards him, and ν2 from the one moving away. The frequency measured on board the spacecraft is ν0.

1 0

2 0

1

2

/ 23

/ 2

c vν ν

c v

c vν ν

c vν c v c v

c v c vν

c v c cc v c c






Lecture 10

IntervalsIntervalsIntervalsIntervalsConsider two points (‘events’), P and Q, in

Euclidean space, observed in frames S and S’:

yx

P

Qy

xS x

yy

x

S

The interval, PQ, is the distance, Δs from P to Q, and is given by Pythagoras’ theorem:

Note that this is invariant in Euclidean space under rotation and translation – i.e. it has the same value whether viewed in frame S or S.

What is the equivalent interval in special relativity? Clearly, it is not the same since

x x whilst y = y. Furthermore, we can see that it must include time.

22222 yxyxs

The ‘equivalent’ of Pythagoras in SR is

You can show this quite easily by substituting for x and t using the Lorentz transform. The quantity s is invariant under the Lorentz transform – that is it has the same value when observed in any inertial frame. We can think of it as the ‘length’ (or norm) of the space-time four-vector. With all four dimensions it is

2222222 xtcxtcs

222222 zyxtcs

The space-time interval between the pulses in the rest-frame of the light source is given by

The corresponding interval measured in the S frame (the observer’s frame) is

2 2 2 2 2 20Δ Δ Δs c t x c T AB AB AB

2 2 2 2 2 2 2 2 2 20 0

2 22 2 2 2 2

0 02

2

Δ Δ Δ

Δ Δ

1

s c t x c γ T v γ T

c vs T c T s

v

c

AB AB AB

AB AB

Example 8: show that the space-time interval between consecutive pulses measured by observers in different frames, which are moving with respect to each other, is the same.

A

A

B

B

path of a photonpath of a photon

Space-time (Minkowski) diagramsSpace-time (Minkowski) diagramsSpace-time (Minkowski) diagramsSpace-time (Minkowski) diagramsWe can plot events and tracks. The geometry is

that of Minkowski space rather than Euclidean space:

ct

xS

QP

C

C

The photon meets observer A at event P, and observer B at event Q

If we chose the scales ct and x to be the same, then the path of a photon is a straight line at 45° to the axes. The vertical lines AA and BB represent the paths through space-time of stationary observers. The straight line CC is the path of a moving observer, with v < c.

All of these lines are called world lines.

This diagram is for the laboratory frame, S. How do we represent the view as seen in a moving frame, S, on the same diagram? The clue is given by CC – i.e. sloping lines for moving observers.

Let the origins coincide at t = t = 0:

ct

x

x

ct

path of a photon

in both frames

1

1

calibrate the axes using the x-t invariant,e.g:

x2 – c2t2 = 1

Pctp

xp

Pct

Px

Let the origins coincide at t = t = 0:

ct

x

x

ct

path of a photon

in both frames

1

1

calibrate the axes using the x-t invariant,e.g:

x2 – c2t2 = 1

Pctp

xp

Pct

Px

The twin paradoxThe twin paradoxThe twin paradoxThe twin paradox

Let us now revisit the twin paradox, and analyse it using space-time diagrams. We will take the specific case that the rocket travels outward for a distance of 4 light years (as measured on the Earth) before returning, travelling at a steady speed of 4c/5.

v = 4c/5

E

A B

C4 light years

The twin paradoxThe twin paradoxThe twin paradoxThe twin paradox The total journey time taken according to Earth clocks is 10 years: 5 out and 5 back. The twins agree to send radio messages to each other on the anniversaries of departure. According to the Earth-based twin, the rocket will not receive the first message until Earth year 5 after departure, and then it will receive eight other messages at regular intervals over the succeeding 5 years.

The Earth-based twin knows about special relativity, and recognises that the clocks on the rocket measure proper time and therefore appear to run more slowly than those on Earth. He calculates that the value of is 5/3, so he expects the rocket-based twin to send just five messages, two on the outward leg, one at turn-around, and two on the return leg, since the rocket years are 5/3 Earth years long.

The Earth-based twin expects to find his twin sister to be four years younger than him on her return to Earth.S (Earth)

0

2

4

6

8

10

2 4 6

ct

x

CausalityCausalityCausalityCausality

x

ct

Future

Past

ElsewhereElsewhere

photon world lines x

ct

P

Q

R

If one event, P, causes another, Q, then Q must

lie in the future cone of P, as nothing can travel faster than light.The time interval, PQ, is ‘time-like’, i.e. ctPQ > xPQ. It is then possible to transform to another frame, S, in which P and Q both occur at the same place, separated only by a time interval.

If the interval is ‘space-like’, i.e. ct < x, as in PR, it is possible to find a frame in which P and R are simultaneous, and yet another in which R occurs before P. It is then not possible for R to have been caused by P. The event R lies in the ‘elsewhere’ of P.






Lecture 11

Relativistic mechanicsRelativistic mechanicsRelativistic mechanicsRelativistic mechanicsPerspective:

(a) We know that Einstein’s postulates have altered our views of space, time, and addition of speeds etc. Therefore, we must expect to have to modify our ideas of mechanics – momentum, energy etc.

(b) We have certain laws which hold true for slowly-moving frames, e.g. conservation of momentum, conservation of energy, conservation of mass etc. How can we modify them so that they hold good in all frames?

MO 210TM 1340

(c) The same modified laws have to hold good in the limit of small v and asymptotically approach the classical formulae.

(d) We can interface classical conditions with relativistic conditions by considering an elastic glancing collision between two particles of equal mass, m. We can apply the formula for the classical momentum in one frame in order to deduce what the relativistic equivalent must be in the other.

We assume that the principle of conservation of momentum applies. Then conserving momentum vertically implies that

2p sin(θ ) = 2mu p = mu /sin(θ)

Now transform into a frame travelling through the laboratory with speed v cos(θ ):

Relativistic momentumRelativistic momentumRelativistic momentumRelativistic momentum

θp, v relativistic

classicaluFrame S

The situation is exactly reversed. Apply the relativistic formula for addition of speeds to the lower mass:

θ

p, vrelativistic

classicalu

,

21

y

yx

uu

vuγ

c

Frame S

with uy = v sin(θ ), ux = v cos(θ), uy = u so

, with

, so

2 2 2 2

2 2

2 2

2

2 2 2 2

2 2

2 2

2

sin( ) 1

cos ( ) cos ( )1 1

cos ( )sin( ) 1

sin( )

cos ( ) cos ( )1 1

.sin( ) cos ( )

1

v θu γ

v θ v θγc c

v θv θ

v θcuv θ v θ

c cmu mv

pθ v θ

c

Now the interface between the purely classical (low-speed) motion and the relativistic (high-speed) motion is accurate in the limit as theta tends to zero. We must conclude that

This suggests that momentum is conserved when defined using the above formula. The only modification required to the classical definition is to multiply by .

2

2

.

1

mγm

vc

vp v

Relativistic energyRelativistic energyRelativistic energyRelativistic energyConsider the perfectly inelastic high-speed collision

between two equal masses:

S (ZMF)

v vv

S (laboratory)

Beforeu

at restm m

After

v

MS (laboratory)

at rest

S (ZMF)

v

The amalgamated mass M is stationary in the Zero Momentum Frame, but is travelling at speed v in the laboratory frame. We conclude that the ZMF is also travelling at speed v. Using the relativistic formula for the addition of speeds and cons. of momentum:

, and

2 2

2 2

2

2

2 2

2 2

2

1 1

1 2

1 1

u v

u

v

v v vu γ mu γ Mv

v vc c

vγ u cM m mγ v u v

c c

2 2

2 2

2 2 22 2

2 2 2 2

22

2

2

2

2 4 2

2 4 2

1 122

44 1 11

1

1 22 2

1 2 1v

v vc cM m mv v v vc c c cvc

vmcM m mγ

v v vc c c

Let’s expand this equation:

If v << c, then we can use the first two terms of the binomial expansion to get

12 2

22 2 1v

vM mγ m

c

)2

1(2 2

2

cv

mM

2 2 212 2

2Mc mc mv

these have the dimensions

of energythis is the classical

KE in the ZMF

2 2 212 2

2mc mv Mc

This equation tells us that we need to consider mass and energy together in order to conserve energy:

S (ZMF)

v vvat rest

S (ZMF)

v

Notes

(a) We deduce from this that mass and energy need to be considered together as different aspects of the same thing. Mass and energy are equivalent to each other.

(b) The conservation of energy is really the conservation of mass-energy.

(c) The mass-energy is calculated by E = γmc2.

(d) When the mass is at rest in a particular frame of reference, γ = 1. Therefore, the rest energy is given by E0 = mc2.

(e) The kinetic energy, K, is the extra energy that the mass has as a result of its motion. Therefore K = E – E0 = γmc2 mc2. We can expand this for v << c as follows:

12 2

2 22

22 2

2

1 1 1

11 1

22

vK γ mc mc

c

vK mc mv

c

(e) The kinetic energy, K, is the extra energy that the mass has as a result of its motion. Therefore K = E – E0 = γmc2 mc2. We can expand this for v << c as follows:

The classical result for the KE is just an approximation in the limit of zero speed.

12 2

2 22

22 2

2

1 1 1

11 1

22

vK γ mc mc

c

vK mc mv

c






Lecture 12

Relativistic mechanics - summaryRelativistic mechanics - summaryRelativistic mechanics - summaryRelativistic mechanics - summary

(a) The momentum of a particle is γmv.(b) The energy of a particle is γmc2.(c) The total momentum of a system is

(conserved)

(d) The total energy of a system is

(conserved)

(e) The kinetic energy of a particle is

(not conserved)

ii ii

γ mp v

2i i

i

E γ mc

2( 1)K γ mc

Before

u stationary

m, K 2m

S (lab)

A B

Afterv

M

S (lab)

Example 19: A particle of mass m and kinetic energy 2mc2 strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

Apply conservation of momentum and mass-energy:

M-E:

Mom:

2 2 22

5

5

v

v

u v

u

mc K mc γ Mc

M m γ

γ mu γ Mv

γ uv

Before

u stationary

m, K 2m

S (lab)

A B

Afterv

M

S (lab)

Example 9: A particle of mass m and kinetic energy 2mc2 strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

Apply conservation of momentum and mass-energy:

M-E:

Mom:

2 2 22

5

5

v

v

u v

u

mc K mc γ Mc

M m γ

γ mu γ Mv

γ uv

KE , so

, and , so

2 22 1 3

8 8 53 5 175

17

u u

v

v

mc γ mc γ

u c v c γ

mM m

γ

back

The E-p invariantThe E-p invariantThe E-p invariantThe E-p invariantWe saw earlier how we could think of x and ct as

two components of a space-time four-vector, and that the ‘length’ or norm of this four-vector was invariant under the Lorentz transformation, i.e.

In the same way, p and E/c are also components of another four vector called the energy-momentum four vector. This means that the components transform by the Lorentz transformation but it also means that there is an associated invariant quantity (i.e. its ‘length’ or ‘norm’) which remains

2222222 xtcxtcs

the same for a given system when viewed at any time in any inertial frame. This is very powerful and helps to simplify problems considerably. The invariant is

E2 p2c2 = E 2 p 2c2 = m2c4 ,where E is the total energy, p is the total momentum of the given system. The value of m is the total mass in a frame (if there is one) in which all the particles of the system are at rest. For a system of more than one particle, use:

is invariant.2 2

2i ii i

E c p

The energy-momentum 4-vectorThe energy-momentum 4-vectorThe energy-momentum 4-vectorThe energy-momentum 4-vector

We saw previously that the Lorentz transform for space and time may be written as

ct and x, y, z form the components of a 4-vector, and we can write the Lorentz transformation more generally as b = AA.b, where b and b are 4-vectors and A is the transformation matrix which

z

y

x

ct

cv

cv

z

y

x

tc

1000

0100

00

00

is given by

It can be shown that energy and momentum form a four vector, i.e. b = (E/c, px, py, pz):

1000

0100

00

00

cv

cv

A

z

y

x

z

y

x

p

p

p

cE

cv

cv

p

p

p

cE

1000

0100

00

00

Note

The ‘length’ or norm of a 4-vector is invariant under the Lorentz transformation. The square of the norm is bb which in Minkowski geometry is given by

bb = b12 b2

2 b32 b4

2

For the E-p 4-vector this is

(E2/c2) px2 py

2 pz2

and for the t-r 4-vector it is

(c2t2) x2 y2 z2

Before

u stationary

m, K 2m

S (lab)

A B

Afterv

M

S (lab)

Example 19 again: A particle of mass m and kinetic energy 2mc2 strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

Apply the left-hand side of the E-p invariant to ‘Before’, and the right-hand side to ‘After’:

Before

u stationary

m, K 2m

S (lab)

A B

Afterv

M

S (lab)

Example 18 again: A particle of mass m and kinetic energy 2mc2 strikes, and combines with, a stationary particle of mass 2m. Find the mass M of the composite particle.

Apply the left-hand side of the E-p invariant to ‘Before’, and the right-hand side to ‘After’:

Before:

After: RHS as before.

22 2 2 2 2 2 2

2 4 2 2 2 2 2 4 2 2 2 2 4

2 4

2 ( )

825 25 9 17

9. 17

u

u

E p c mc K mc muγ c

m c m u γ c m c m c c m c

M c M m

Nuclear binding energiesNuclear binding energiesNuclear binding energiesNuclear binding energies

The nucleus of an atom is held together by nuclear binding forces, and work must be done against these forces to split the nucleus into its constituents parts, or into fragments containing smaller numbers of nucleons. The binding energy appears as increased mass, so that the mass of the nucleus is smaller than the sum of the masses of its constituent parts. The binding energy per nucleon, however, varies with increasing atomic mass number as follows:

Binding energy per nucleonBinding energy per nucleonBinding energy per nucleonBinding energy per nucleon

The binding energy per nucleon increases sharply at first with atomic mass number, reaching a broad plateau with a peak at about Fe, and then falls again more slowly. We can extract some of the binding energy either in a nuclear fusion process or in a nuclear fission process.

The fusion of two nuclei with atomic mass numbers smaller than that of Fe generally releases energy, whilst the fusion of nuclei heavier than Fe generally absorbs energy. For example, nuclear fusion of H atoms to form He is the main process powering the Sun at the present time.

The fission into lighter fragments of a heavy nucleus with an atomic mass number larger than that of Fe generally releases energy, whilst the fission of nuclei lighter than Fe generally absorbs energy. Controlled nuclear fission is used in present-day atomic power stations to generate heat which is turned into steam to drive electricity generators. It may be possible in future to maintain the right conditions for nuclear fusion in a stable system so that electricity can be generated from water with the inert gas He as the by-product.

Splitting the atom:Splitting the atom:the Cockcroft-Walton experimentthe Cockcroft-Walton experiment

Splitting the atom:Splitting the atom:the Cockcroft-Walton experimentthe Cockcroft-Walton experiment

400 KV 200 KV 0 KV

alpha particles

protons injected

here

lithium target

fluorescent screen

C:\Program Files\Windows Media Player\wmplayer.exe

I wish you a pleasant Easter breakI wish you a pleasant Easter break

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