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The Department of Physics
Part IA Natural Sciences Tripos 2014/15
Rotational Mechanics
& Special Relativity
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Course materials
Part IA Physics
Lent Term, 2014
Part IA Physics
Lent Term, 2014
Rotational Mechanics &
Special Relativity
Rotational Mechanics &
Special Relativity
Prof. Steve Gull
Prof. Steve Gull
Lecture notes 2014
Examples book 2014
d2 r
F m 2
dt
d2 r
F m 2
dt
Newton
Newton
E   mc2
Einstein
E   mc2
Einstein
(d) The time interval is the same in any frame.
Thus tAB = (tB tA) = tAB = (tB tA). 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.
y
y
vtA
S
S
v
MO 173
TM 1322
xA
xA
A
x
You can also find these slides, in
colour, on the Physics Teaching
website:
wwwteach.phy.cam.ac.uk/teaching/h
andouts.php
Text-book references:
MO: Mansfield & O’Sullivan
TM: Tiplar & Mosca
x
An event, A, occurs in S at (xA,yA,zA,tA), and in
S at (xA,yA,zA,tA ).
7
A question about calculating the position of the centre of mass etc. by integration.
How to solve Physics problems
1.
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.
(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 V 

3
a 2b .
1. Draw a diagram
(ii) The height, b0, of the centre of mass is defined by the relation
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.
b0   mi hi M ,
i
2. Think about the Physics
where M is the total mass, mi is the mass of the disc at height hi
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!
and the sum is taken over all the discs.
integral, and hence show that b0  b 4 .
(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.
{ 4r 3 }
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
Treat the sum as an
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}
6
The Department of Physics
Mechanics in
Rotational Motion
The centre of mass
MO 128
TM 149
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:
y
l2
l1
m1
C
m2
Centre of mass
x1
x0
x2
x
We require that the total turning moment about C is zero,
so we define its position to be such that m1l1 = m2l2. Since
l1 = x0 – x1, and l2 = x2 – x0, we can write this as
m1  x0  x1   m2  x2  x0  , so
m1 x1  m2 x2
x0 
, where M = m1+m2.
M
We can generalise this to a one-dimensional rigid body of
N masses as follows:
m1x1  m2 x2  m3 x3
1 N
x0 

mi xi .

M
M i 1
For a three-dimensional rigid body, this expression must
be satisfied in each dimension simultaneously, and we
may write:
1 N
1 N
1 N
x0 
mi xi ; y0 
mi yi ; z0 
mi zi .



M i 1
M i 1
M i 1
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 N
1 N
R   x0 , y0 , z0  
mi  xi , yi , zi  
mi r i


M i 1
M i 1
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
R
mi r i 
r i mi 
r i δmi .



M i
M i
M i
Hence, for continuous bodies, we can replace the sum
over i elements with an integral:
1
R
r dm.

M body
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.
m2
m1
=
+
m3
m1
m2
+
m3
For discrete masses:
1
R
mi r i

M i
For continuous bodies:
1
R
r dm

M body
Example 1: find the centre of mass of a lamina shaped like an isosceles triangle
y
y
δx
b
x0
b  h  x
If the mass of the lamina is M, then
2h
x
x
h
2 yδx 2 M  h  x 
δm  M

δx
2
bh 2
h
Let δx tend to zero, and integrate
to find the total moment. Then
h
2M  h  x
1
1
x0 
xdm 
x
dx


2
M
M0
h
Divide the lamina into thin strips,
each having a width of δx. Take
h
moments about the origin. The
2
x0  2  xh  x 2 dx
moment of the element shown is
h 0
xδm, where δm is the mass of
3
3

 h
2
h
h
the element. The area of the
x0  2    
h 2 3 3
element is 2yδx, and the area of
the lamina is bh/2.


The lever balance
y
(m1+m2)g
l1
l2
m2
m1
m 2g
m1g
x1
x0
x2
x
We find experimentally, that the lever is balanced
only when m1l1 = m2l2
This is the case even though the sum of all the
forces is zero wherever we put the fulcrum, i.e.
F ext   F i  0.
i
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:
Gext   Gi   fi li  0,
i
i
where li is the perpendicular distance and fi is
the force.
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.
F ext   F i  0,
i
and
Gext   Gi   Fi li  0.
i
i
MO 149
TM 397
Circular motion
When 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:
r + δr
O
r
P1
δr
P
This constitutes a complete description of the
change in position. Note, however, that the
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:
Q
P1
r + δr
δr
θ
P
O
r
Here, we shall be concerned with the rotational
aspects of the motion, i.e. things to do with
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.
Q
θ
θ
O
We can see that small rotations can be represented
by vectors as follows:
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
Q

B
a
O
a 2
s
a 1
a 3
A
θ
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.
therefore describe the angular velocity by the
vector
dθ
θω
.
dt
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:
2
dω d θ
θω
 2.
dt
dt
The moment of a force as a vector
We 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:
Force F acts at point P,
which is at position
vector r from point O.
Q
G=rF
O
G
F
r
θ
P
r sin(θ)
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 G = r  F
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.
The forces act at position
vectors r1 and r2 with
respect to an arbitrary
origin, O.
F
a
F
G  r 1 × F  r 2 × ( F )
r1
r2
O
  r1  r 2  × 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 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, r ,
is the angular acceleration, θ . Can we find an
equivalent to Newton’s second law for rotational
mechanics, and if so, what is the equivalent of
the mass, m ?
To answer this question, consider a particle of mass
m rotating about an axis OQ:
Q
v = rω
O
r
m
ω
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
direction of magnitude F where
d  rω 
d
dv
dω
F  ( mv)  m  m
 mr
dt
dt
dt
dt
The magnitude of the moment of this force about
OQ is G  Fr  mr 2 ω, directed along OQ. For this
2
particle we may therefore write G  r × F  mr ω.
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:
N


Gext   m r ω,
i 1
2
i i
where Gext is the total external vector moment
acting on the body about OQ.
If G is the rotational equivalent of F, and θ is the
rotational equivalent of r , then we must conclude
that the rotational equivalent of the mass of the
body is:
N
2
m
r
 ii
i 1


We call this quantity the moment of inertia, I, and
then the rotational equivalent of N2 is
Gext  I ω
Moment of inertia
MO 154
TM 293
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
given the symbol I. (Do not confuse this with the
vector impulse, I p .) Thus, for a rigid body which
can be thought of as being composed from N
N
particles:
I   mi ri2 .
i 1
In terms of I, we can write the rotational equivalent
of N2 as G  I ω.
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?
flywheel
motor
ω
G
The torque causes an angular
acceleration which increases
the angular velocity of the
flywheel.
G
G  Iω, so ω  .
I
After time t the angular
G
speed is ω  ωt  t.
I
150
 ω
 30
70
 64.3 rad s 1
 10.2 rev s 1
Moments of inertia of continuous bodies
The expression we have obtained for the moment of
inertia, I, of a rigid body involves a summation
over N elemental mass contributions:
N
I   mi ri
2
i 1
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
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:
I   r dm
2
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.
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
δx
x
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:
δI  x 2 δm  x 2ρδx

l
1 3
I   ρ x dx  ρ l
3
x 0
2
1 2
But M  ρ l, so I  Ml
3
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.
Let the disc have a surface density of 
per unit area. Consider a radial element
of the disc at radius r, thickness δr. This
r
has mass δm which is equal to 2rδr. It
a
makes a contribution, δI, to the total
moment of inertia given by:
δr


δI  r 2 δm  2πσr 3δr
a
1
I   2πσr dr  πσa 4
2
r 0
3
But M  πσa 2 , so I 
1
Ma 2
2
Note that this formula also applies to a cylinder.
Parallel axes theorem
This 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:
B
centre of mass P
ri
a
Ri
O
rigid body
A
mi
MO 155
TM 297
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.
Then I   mi ri   mi r i r i
2
i
i
zero for
centre of mass
  mi (a  R i ) (a  R i )
i
  mi ( a 2  2 R i a  Ri2 )
i
  mi a   mi R  2a   mi R i
2
i

2
i
i
I  Ma 2  I 0
i
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.
B
P
a = l/2
A
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:
O

1 2
l2
Ml  M  I 0
3
4
1
I0 
Ml 2
12
Perpendicular axes theorem
This 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:
z
y
ri
mi
x
Let the lamina be in the xy plane. Then the
moment of inertia about the z axis is
I z   mi ri   mi ( x  y )
2
2
i
i
i
  mi x   mi y
2
i
i

i
Iz  Iy  Ix
2
i
2
i
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.
We have already found the moment of
inertia of the disc about the z axis to
be Ma2/2. Thus
z
a
y
x
1
I z  Ma2 ; I x  I y  I D
2
1

I x  I y  2 I D  Ma 2
2
1

I D  Ma2
4
Sounds of Pulsars
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
• Single pulses are usually very variable !
MO 103
TM 331
Angular 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:
L
r
B
r sin(θ)
The moment of the
P, v
momentum (by direct
analogy
with
the
moment
θ
m
of a force) is given by
A
L = r  P = r  mv = mr  v
Conservation 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
N
L   mi r i  v i .
i 1
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:
N
dL d N
d
 (  mi r i  v i )   mi ( r i  v i )
dt dt i 1
dt
i 1
N
dr i
dv i
  mi (
 vi  r i 
)
dt
dt
i 1
N
  mi (v i  v i  r i  ai )
i 1
N
N
N
i 1
i 1
i 1
  r i  mi ai   r i  F i   Gi
 Gtot  Gext  Gint
Now the value of Gint is zero for the following reason:
Fji
j
rj
O
b
i
ri
Fij
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
similar pairs, each of which comes to zero. Hence
Gint must be zero. Therefore
dL

 Gext ,
dt
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.
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 G  I ω, and
we have just shown that G  dL dt . Therefore
we can write
dL
dω
G I
dt
dt
 L  Iω
Angular impulse
We defined the impulse of a force as the integral of
the force with respect to time, i.e.
I p   Fdt.
The action of the impulse was to change the
linear momentum, thus:
t2
t2
dP
I p   Fdt  
dt  P 2  P 1  ΔP
dt
t1
t1
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:
t2
t2
t2
t1
t1
t1
J p  r  I p  r   Fdt    r  F  dt   G dt
t2
dL

dt  L2  L1  L
dt
t1
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.
Rotational 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:
2
1
1
1
2
2 2
KEi  mi vi  mi  ri ω  mi ri ω
2
2
2
Summing over the entire body (N particles), we get
KEtot
1 N
1 2
2 
2
   mi ri  ω  Iω
2  i 1
2

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 mr22/2. 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.
Rotational oscillations: the physical pendulum
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?
A
A

l y
l

l
C
y
mg
ω
C
Point C, the centre of mass, is raised relative to its
equilibrium position by an amount y. Conserving energy we
have:
1 2
E  Iω  mgy  const.
2
Now l  y  l cos θ, so y  l(1  cos θ). Substituting and
differentiating with respect to time gives us
1
I (2ωω)  mgl sin θω  0, so
2
d 2θ mgl

θ  0, if θ is small.
2
dt
I
I
.
Hence the period of small oscillations is T  2π
mgl
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:
1
2
2
A
l
m
L
C
I A  ML  Ma
2
The moment of inertia of the rod is mL2/3, so
the total moment of inertia is
1
1 2
2
I A  ML  Ma  mL
2
3
2
M
2a
Now the centre of mass is at distance l below
the pivot such that
A
l
m
2
2
mL
2 ML2  Ma 2 
I
L
3
 T  2π
 2π
( M  m) gl
gL(2 M  m)
C
M
L
ML  m  ( M  m)l
2
2M  m L
 l

Mm 2
L a2
 T  2π

if m  0 (i.e. a light rod). If a  0,
g 2 gL
2a
L
T  2π
as expected for a simple pendulum.
g
The general motion of a rigid body
We 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:
Example 9: what is the acceleration of a uniform solid cylinder rolling,
without slipping, down a plane inclined at angle α to the horizontal?
a
N
ω
F
v
a
ω
v
mg
α
mg
α
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.
(a) The linear motion of the centre of
N
mass is as if it is a point mass equal
a
to the total mass of the cylinder,
ω
F
acted upon by the sum of the
external forces. Resolving down the
v
slope:
dv
mg sin(α)  F  m
mg
α
dt
(b) The rotation about the cylindrical axis is as if acted upon
by the sum of the moments of the external forces:
dω
Fa  I
,
dt
where I is the moment of inertia about the cylindrical
axis. We also have v = ωa. Thus
I dω I dv
F
 2
.
a dt a dt
Substituting for F gives us
I dv
dv
dv mg sin(α)
mg sin(α)  2
 m , so

.
2
dt
dt
a dt
m I a
Now I = ma2/2, so we deduce that
dv mg sin(α) 2

 g sin(α).
dt
mm 2 3
In terms of energy:
(a) Rotational KE of the cylinder is
1 2 1 2 2 1 2
KErot  Iω  ma ω  mv
2
4
4
(b) The linear KE of the centre of mass is
N
F
1
KElin  mv2
2
a
ω
v
mg
α
(c) Thus when the centre of mass has dropped through a
distance h we have
3
mgh  KErot  KElin  mv 2
h
4
 2g

3v 2
2
h 
 s sin α, so v  2   sin α   s
4g
 3

dv 2 g
We conclude that

sin α as before.
dt
3
s
α
Rotating frames of reference
Suppose that we have a particle, P, which is
rotating at constant angular speed in a circle
about O:
y
ω
O r
y
P
x
O
ω
r
P
ωt
ω
ω
Oblique view
Plan view
x
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:
r  r (cos(ω t ),sin(ω t ))
dr

 r ( ω sin(ω t ), ω cos(ω t ))
dt
2
d r
2
2

 r ( ω cos(ω t ), ω sin(ω t ))
2
dt
d2 r
2



ω
r
2
dt
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 force
The 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:

d2 r
2
m
F  m 2   mω r .
dt
r
Since v  rω, we can also write
F
d2 r
mv 2
F m 2 
rˆ ,
r
dt
where r̂ is the unit vector along r. Note that the
centripetal force does no work as the velocity and
force are orthogonal to each other.
Linear and rotational equivalents
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.
Linear
mass m
small displacement dr
velocity v = dr/dt
acceleration a = dv/dt = d2r /dt2
linear momentum p = mv
force F = dp/dt = ma
linear kinetic energy mv2/2
work done dW = Fdr
linear impulse dp = Fdt
Rotational
moment of inertia I
small angular displacement dθ
angular velocity ω = dθ/dt
angular acceleration dω /dt = d2θ /dt2
angular momentum L = Iω
moment G = dL/dt = Id2θ/dt2
angular kinetic energy Iω2/2
work done dW = Gdθ
angular impulse dL = Gdt
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, …
After
B4
v0
centre
of mass
u
b
a) Linear momentum before = linear momentum after
m
 mv0   m  M  u, so u 
v0
m M
b) New position of the centre of mass: moments about it sum to zero
 L  b  L  b   b  b

M 
   M   bm  0
 L
 2   L  2
LM
LM

 bM  bm  0, so b 
2
2  M  m
centre
of mass
u
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:
After
B4
v0
centre
of mass
u
b
ω
 2 1b
2 
1  L  b
2
v0 mb  Iω   mb 
Mb 
M  L  b  ω


3
L
3
L





  6mv0 
3L
 ω  v0 b 


3
M
M
L
(
M

4
m
)
 3b2 L  b3   L  b   



m
m


MO 164
TM 339
The 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
z
other orthogonal axes:
L = I
x

y
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.
z
A precessing gyroscope

The gyroscope is precessing about the x
z axis. Look down from on top along –z :
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
x,
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
y
Ω
C
ω
z, Ω
B
L
A
tδt
t
y
t+δt
Looking from the side along –y :
z, 
F
x, L
The forces form a
couple which acts
along y
y
z
x

y
F
Precession using vectors
Apply a couple of moment G to a rotating flywheel:
Ω
G
t
L+δL
L
t+δt
L+δL
L
δL
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.
Hence, there is precession about an axis perpendicular to both L
and G. The rate of precession, Ω rad s1, is easily calculated. In a
small time, δt, it precesses through small angle Ωδt.
Ωδt
L+δL
L
δL=Gδt
In this triangle, we can see that,
as δt is small, we can write
G δt = Ωδt L

G = ΩL
In fact, we can write this in vector form as follows:
G  Ω × L  IΩ × ω
To see this, consider a gyroscope at an angle to the horizontal plane:
In the case where the gyroscope is at an angle, first resolve the
angular momentum into vertical and horizontal components:
Ω
L
θ
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=IΩω
Gyroscope examples
(i) Luni-solar precession
L
G
N
Moon, Sun
S
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
B
L, 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.
The Department of Physics
Einstein’s theory of
Special Relativity
MO: 193-227
TM: 1319-1356
Frames 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 twodimensional 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?
The rotation matrix for clockwise rotation
about the origin by angle θ is given by
y
3
P
S
30°
5
 cos θ  sin θ 
R
.
 sin θ cos θ 
The new coordinates are therefore
x 
 xP   0.87 0.5  5   2.83 
P
  R   
   
.
x 
y 
yP   0.5 0.87  3   5.10 

 P 
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.
1
2
3
4
S
5
metres
‘event’
(x,y,z,t)
(x,y,z,t )
(x,y,z,t )
y
S
x
Typically, we might 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.
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
yB
yA
S
sAB
B
y A
x
x
xA
xB
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
B
A
sAB
S
x
The two events occur in S at coordinates of
(xA,yA,zA,tA) and (xB,yB,zB,tB). Again the space
interval, s AB , between the two events, viewed in
S, is given by Pythagoras’ Theorem, i.e.
s 2AB = (xB xA)2 + (yB yA)2 + (zB zA)2.
Our assumption is that s2AB = s 2AB .
Euclidean space
y
B
y
A
x
x
S
Our assumption is that s2AB = s 2AB .
(d) The time interval is the same in any frame.
Thus tAB = (tB tA) = tAB = (tB tA). 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.
y
y
vtA
S
S
v
MO 173
TM 1322
xA
xA
A
x
x
An event, A, occurs in S at (xA,yA,zA,tA), and in
S at (xA,yA,zA,tA ).
We see xA = xA vtA, which is the only coordinate
affected, so the Galilean transformation is:
x = x  vt
y = y
Note that space
and time
z = z
are separate
t = 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.
(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  vt and equally
x = x’ + vt’
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?
The observer in the train is
ΔxAB B
present at both events, so he
A
measures a space interval of
zero.
S (track frame)
The Galilean transformation
gives:
v
A,B
S’ (train frame)
Δx’AB = 0
Δt’AB = 3

ΔxAB  ΔxAB  vΔtAB
575000
0
3
3600
 479 m
Problems with classical physics
Nineteenth-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,
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 3108
m s1 through this medium which was allpervading. 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.
Electromagnetic waves
1D wave equation for waves on a string, or water waves, or
sound waves:
2ψ 1 2ψ
 2 2 0
2
x c t
ψ 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
c
ρ
James Clerk Maxwell showed that, for electromagnetic waves
(light, radio, x-ray etc.) the equation was
E represents the
2 E
2 E
 ε0μ0 2  0
electric field of
2
x
t
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 3108
m s1 through this medium which was allpervading. 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 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.
c
 v/c  104
radians

aether wind
v
Earth’s orbital
velocity through
the stationary
aether
This was easily
measured by
Bradley, and
appeared to show
evidence for a
stationary aether.
The Michelson-Morley experiment
One of the most-famous attempts to measure
motion through the aether was the MichelsonMorley experiment. The Earth moves at about 30
km s1 in its orbit around the Sun, which is an
appreciable fraction of the speed of light, 300,000
km s1. 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:
aether wind
B
v
d
S
This is a simplified diagram of
a MM interferometer. You will
meet this again in more detail
next year.
C
A
d
T
O
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’
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/(cv).
tACA = d/(c+v) + d/(cv).
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.
v
aether wind
B
v
c v
2
c
2
B
d
S
A
C
A
d
T
O
The speed from B to A is the same as that from
A to B, so the total time of flight is
t ABA  2d / c  v .
2
2
 the time difference between ACA and ABA is
d
d
2d
t 

 2
cv c v
c  v2
d(c  v)  d(c  v)
2d

 2
2
2
c v
c  v2
2d
2d


1
2
2
2
2
c(1  v / c ) c(1  v / c ) 2

1
2d
2
2 1
2
2  2

(1  v / c )  (1  v / c )
c

Now, if x is much less than 1 we can use the first
two terms of the binomial expansion of
(1+x)n  1+nx, so that
2
2
2d
v
2d
v
t 
(1  2 ) 
(1  2 )
c
c
c
2c
2
2
2d v
dv

 2  3 .
c 2c
c
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.
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 electromagnetism
Newtonian 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
1
by the expression c 
, where ε0 and μ0
(ε0 μ0 )
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 postulates
MO 194
TM 1321
Einstein’s ideas about electromagnetism and the
nature of physical laws may be summarised in
his two postulates:
NB!
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
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?
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?
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.
Time dilation
MO 195
TM 1324
One 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
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
First view the situation in the upper
B
clock’s rest frame. Let events A, B, C
be a photon leaving the base mirror,
h
arriving at the upper mirror, and
arriving back at the base mirror
respectively. In the clock’s rest frame,
A
C
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
rest frame S
time interval be t AC . Then
2
2

c

t
2h
AC
tAC 
, or h 2 
c
4
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:
B
cΔtAC
2
A
h
D
vΔtAC
2
C
v
In the triangle ABD, by Pythagoras’ theorem, we
have
2
2
2
2
v ΔtAC c ΔtAC
2
h 

4
4
But in the moving clock’s rest-frame, S, we have
2
2

c
Δ
t
2
AC
h 
4
So substituting for h gives us
2
2
2 v 2 ΔtAC
c 2 ΔtAC
c 2 ΔtAC


4
4
4
Rearranging, we find
where

ΔtAC  γΔtAC
 
1
(1  v c )
2
2
.
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.
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 lightclocks (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 fourfifths 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)
MO 204
TM 1335
The 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:
A
E
v
C
B
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
ΔtAB between events A and B, and the time
interval ΔtBC, 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.
Simultaneity
MO 205
TM 1330
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.
blue beacon
first observer
A
C
B
red beacon
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).
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:
blue beacon
first observer
v
A
v
C
v
B
v
red beacon
Second observer’s frame
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 …
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.
MO 201
TM 1326
Length contraction
A 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.
A
v
L0
B
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 ΔtAB = L/v. Now ΔtAB = γ ΔtAB
(the captain measures a proper time
L0
interval).So we must conclude that L 
.

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.
Summary
Einstein’s postulates lead to the following
conclusions:
Time dilation: t   t 
Simultaneity: events simultaneous in one
frame are not necessarily so in another.
Length contraction parallel to motion:
L0
L

No length contraction perpendicular to motion.
Summary
Einstein’s postulates lead to the following
conclusions:
Time dilation: t   t 
Simultaneity: events simultaneous in one
frame are not necessarily so in another.
Length contraction parallel to motion:
L0
L

No length contraction perpendicular to motion.
Measuring the speed of light
Lorentz transformation
MO 198
TM 1322
We 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.
At time t (measured in S), the origin of the moving
frame is at position x = vt in S:
v measured in S
vt
A
r
x
S' (moving frame)
S (laboratory frame)
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.
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:
 vx 

vx 
t  γ  t  2  ;
t  γ  t  2 
c 
 c 

x  γ  x  vt  ;
x  γ  x  vt 
y  y ;
z  z ;
with γ 
y  y
z  z
1
1 v c
2
2
.
Back
Now x, t etc. are really intervals between the two
events ‘the origins coincide’ and event A. We
can therefore equally well write:

vΔx 
Δt  γ  Δt  2  ;
c 

Δx  γ  Δx  vΔt  ;

vΔx 
Δ t  γ  Δt   2 
c 

Δx  γ  Δx  vΔt 
Δy  Δy;
Δz  Δz;
Δy  Δy
Δz  Δz
with γ 
1
1  v2 c 2
.
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:
 ct    
 
 x     v
c
 y   
   0
 z  
   0
0 0  ct 
c
 

0 0  x 
 
0
1 0  y 
z

0
0 1  
 v
(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
 

 v
A
c
 0
 0

0 0 
c

0 0

0
1 0
0
0 1 
 v
We will return to this later.
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?
A
ΔxAB
B
S (track frame)
v
A,B Δx’AB = 0
Δt’AB = 1.198×10‒6
S’ (train frame)
The observer in the train is
present at both events, so he
measures a space interval of
zero.
The Lorentz transformation
gives:
 
ΔxAB  γ  ΔxAB  vΔtAB

 γ 0  0.8  3  10 8  1.198  10 6
γ
1
1  (0.8)
 ΔxAB  479 m
2
 1.667

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:
t
t
v
A
x0/c
tA = ?
x0
B
xA = ?
x /2c
2x
0
0
A
B
xB = ?
x
x
S
S
(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)
 x0 vx0 
 x0 v2 x0 

vxA 

vxB 
tA  γ  tA  2   γ   2  ; tB  γ  t B  2   γ   2 
c 
c 
c 
c 


 c
 2c
 x0 vx0 
 x0 v2 x0  x0 vx0
v
1

γ  2   γ  2 ;
 2 , so  
c 
c  2c
c
c
2
 c
 2c
(b)
γ
1
1  v2 / c 2

1
1  (1 / 4)

2
3
cx0  2 3x0

vxA  2  x0
3


 tA  t B  γ  tA  2  

x0
  ( 2 )  
c 
2c 
c
3 c
3 2c

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.
space
Which twin is older, and by how much, when they meet in space?
Event C
Identify the events:
Event A
Event B
(B)
B
Bob S
A
C
Alice S
1
S
In S:
E
1
xAC
distance
1c
13
t AB 


yr
speed
5c 13
5
x AC
5
t AC 
 x AC 
t AC ly
5c 13
13
12c
12c
 x AC  1 
t BC 
 t AC  t AB 
13
13
5
12
12 13
221

t AC  1 
t AC 

 t AC 
yr
13
13
13 5
35
In Bob’s frame S:
=
13
13
;   =
12
5
 
vx AC

  0  t AC     t AC
 

x AC
   t AC
2
c


221 12 204
  t AC   
 t AC


yr
35 13
35
In Alice’s frame S:
13 12 12


t AB  t AB  


yr
5 13
5
10
  t BC     t AC  t AB    
t BC
yr
7
12 10 134
 Alice has aged by


yr
5
7
35
204 134 70
 Bob is older by


 2 yr
35
35
35
Experimental 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 Earthbased 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 currentcarrying 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.
The 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.
point of intersection
moves at speed > v
move up at speed v
fixed
Addition of speeds formula
MO 208
TM 1336
It is clear that the Galilean addition of speeds
fixed
in Sif either ux or v
v , will not
formula, ux  ux Markers
work

u
is a large fraction
of
c.
For
example,
if
B
x = c/2 and
A
u x an impossible
v = 3c/4, the result would be 5c/4,
resultS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 ux relative to a frame S between two

markers, A & B. The markers are a distance x AB
apart fixed in this frame, and the rocket takes time
 to travel between them.
t AB
Markers fixed in S
v
A
S
Clearly
S
ΔxAB
ux 
.

ΔtAB
B
u x
From the point of view of an observer in frame S,
relative to which frame S is moving parallel to ux
at speed v, the corresponding intervals are xAB
and tAB. The speed of the rocket in this frame is
therefore given by
 
xAB γ  xAB  vtAB
ux 

tAB

vxAB 
 
γ  tAB

2
c


Dividing through by tAB and cancelling the
gammas, we get
ux  v
ux 
vux
1 2
c
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:
B
u y

yAB
A
v
u x

x AB
S
S

xAB yAB
u  (ux , uy )  (
,
);


tAB
tAB
ux  v
ux 
as before.
vux
1 2
c
xAB yAB
u  (ux , uy )  (
,
)
tAB tAB
yAB
uy 

tAB

yAB

vxAB
 
γ( tAB
)
2
c
 gives
Dividing through by ΔtAB
uy
uy 
.

vux 
γ1 2 
c 

Thus, the ‘vector’ velocity addition formula is
uy
ux  v
u  (ux , uy )  (
,
).
vux

vux 
1  2 γ1  2 
c
c 

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?
v
Guard
ux  0.5c
S’ (train frame)
Guard
ux
S (track frame)
The formula we have just
derived for the addition of
speeds gives
ux  v 0.5  0.8
1.3
ux 

c
c
ux v
1  0.4
1.4
1 2
c
Aberration of light
We 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’:
c.sin(θ )
v
S
S
L
θ
c.cos(θ )
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:
uy
ux  v
u  (ux , uy )  (
,
vux

vux
1  2 γ1  2
c
c




).
c cos(θ)  v
c sin(θ)
c  (cx , c y )  (
,
)
v

v

1  cos(θ) γ  1  cos(θ) 
c
c


In frame S we have cx = c cos(θ ), cy = c sin(θ).
c cos(θ)  v
c sin(θ)
 (c cos(θ), c sin(θ))  (
,
)
v

v

1  cos(θ) γ  1  cos(θ) 
c
c




v
v
cos(θ))
c
c
cos(θθ) 
and
cos(
,, and
v
1
 cos(θ))
c
sin(θ))
sin(θθ) 
sin(

v


γ 1
 cos(θ))
c


In frame S we have cx = c cos(θ ), cy = c sin(θ).
c cos(θ)  v
c sin(θ)
 (c cos(θ), c sin(θ))  (
,
)
v

v

1  cos(θ) γ  1  cos(θ) 
c
c



vv
cos(θ))
c
c

cos(
,,and
cos(θθ) 
and
v
1  cos(θ))
c
sin(θ))
sin(θθ) 
sin(

v

γ  1  cos(θ))
 c

Stellar aberration
In Bradley’s case, θ = 90°, and the direction of the
photon was reversed (i.e. incoming photon
rather than outgoing). We conclude
c
sin( ) = cos( )  

  v/c
v
S
S
as Bradley measured
and used as evidence,
incorrectly, of the
existence of the Aether.
Relativistic Doppler effect
MO 340
TM 1328
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’.
pulses in frame S:
S
v
A,B
xAB = 0
tAB = T0
light source
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.
v
v
xAB = ?
tAB = ?
O
B
A
xAB
S
back
We use the Lorentz transform to find xAB and tAB:

vΔxAB 
 
ΔtAB  γ  ΔtAB
  γT0
2
c


   γvT0
ΔxAB  γ  ΔxAB  vΔtAB
back
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:
ΔxAB
γv
T1  γT0 
 γT0 
T0
c
c

v
 T1  γT0  1    T0
c

v
1
c T
0
2
v
1 2
c

cv
.
c-v
T1  T0

v 
v
 1  c  1  c 




v 
v
 1  c  1  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
cv
ν1  ν0
.
cv
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.
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?
v
v
S – space station frame
ν1  ν0
cv
cv
ν 2  ν0
cv
cv
cv
The observer sees frequency ν1 from  ν1  c  v
cv
cv
the spacecraft moving towards him, ν2
and ν2 from the one moving away.
cv cc/2


3
The frequency measured on board
cv cc / 2
the spacecraft is ν0.
Intervals
Consider two points (‘events’), P and Q, in
Euclidean space, observed in frames S and S’:
y
Q
y
S
P
x
x
The interval, PQ, is the distance, Δs from P to Q,
and is given by Pythagoras’ theorem:
s  x  y  x   y
2
2
2
2
2
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.
The ‘equivalent’ of Pythagoras in SR is
2
2


s  c t  x  c t  x
2
2
2
2
2
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 fourvector. With all four dimensions it is
s  c t  x  y  z
2
2
2
2
2
2
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.
The space-time interval between the pulses in the rest-frame of the
light source is given by
 2  ΔxAB2  c 2T02
ΔsAB2  c 2 ΔtAB
The corresponding interval measured in the S frame (the observer’s
frame) is
ΔsAB2  c 2 ΔtAB2  ΔxAB2  c 2 γ2T02  v 2 γ2T02
 ΔsAB2 
c 2  v2
2
v
1 2
c
 2
T02  c 2T02  ΔsAB
Space-time (Minkowski) diagrams
We can plot events and tracks. The geometry is
that of Minkowski space rather than Euclidean
space:
A C B
ct
path
of
a
photon
Q
P
C
x
A
B
S
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
ctp
ct
path of a photon
in both frames
P
ct P
x
1
1
calibrate the axes using
the x-t invariant,e.g:
x 2 – c2 t 2 = 1
x P
xp
x
The 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
A
E
C
4 light years
B
The 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.
ct
10
8
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.
6
4
2
x
0
S (Earth)
2
4
6
The Earth-based twin expects to find his
twin sister to be four years younger than
him on her return to Earth.
Causality
ct
Future
photon world
lines
ct
Q
R
P
Elsewhere
x
x
Elsewhere
Past
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 ‘timelike’, i.e. ctPQ > 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. ct < 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.
Relativistic mechanics
MO 210
TM 1340
Perspective:
(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?
(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.
Relativistic momentum
p, v
Frame S
relativistic
θ
u
classical
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 ):
Frame S
u
θ
relativistic
p, v
The situation is exactly reversed. Apply the
relativistic formula for addition of speeds to the
lower mass:
uy
uy 
,

vux 
γ1 2 
c 

with uy = v sin(θ ), ux = v cos(θ), uy = u so
v sin(θ)
1
u
, with γ 
2
2
2
2
 v cos (θ) 
v
cos
(
θ
)
γ1
1

2
2
c
c


2
2
v cos (θ)
v sin(θ) 1 
2
v sin(θ)
c
 u

,
so
2
2
 v 2 cos 2 (θ) 
v cos (θ)
1

1


2
2
c
c


mu
p

sin(θ)
mv
v 2 cos 2 (θ)
1
c2
.
Now the interface between the purely classical
(low-speed) motion and the relativistic (highspeed) motion is accurate in the limit as theta
tends to zero. We must conclude that
p
mv
2
 γ mv .
v
1 2
c
This suggests that momentum is conserved when
defined using the above formula. The only
modification required to the classical definition is
to multiply by .
Relativistic energy
Consider the perfectly inelastic high-speed collision
between two equal masses:
Before
After
u
v
at rest
m
S (laboratory)
v
v
S (ZMF)
m
v
M
S (laboratory)
v
at rest
S (ZMF)
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:
vv
2v
u

, and γu mu  γv Mv
2
2
v
v
1 2 1 2
c
c
2 


v 
 1 2 

γu u
2
c 

 Mm
 m
2
2

γv v
u 
v 
 1 2  1 2 
c 
c 









2
v


1 2

c
M  m


v2
4 2


c
 1

2
2


v  
1 2  




c




 M  2m

 2

2
v
 1

2
c



  2m



v2
1 2
2m
c

 2mγv
2
4
2
v v
v
1 2 2  4
1 2
c
c
c
v2
1 2
c
2
2
2


v
4v
1 2   2
c 
c

Let’s expand this equation:
 v2 
M  2mγv  2m  1  2 
c 


1
2
If v << c, then we can use the first two terms of the
binomial expansion to get
2
v
M  2m (1  2 )
2c
1
2
2
2 
 Mc  2mc  2  mv 
2

these have the dimensions
of energy
this is the classical
KE in the ZMF
This equation tells us that we need to consider
mass and energy together in order to conserve
energy:
v
v
v
v
at rest
S (ZMF)
S (ZMF)
1
2
2
2mc  2  mv   Mc
2

2
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:
1



2
2
 v 


2
2
K   γ  1 mc    1  2   1  mc
c 





 2 1
v2
 K   1  2  1  mc  mv 2
2
2c


The classical result for the KE is just an
approximation in the limit of zero speed.
Relativistic 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
p   γi mi vi
(conserved)
i
(d) The total energy of a system is
2
E   γi mi c
(conserved)
i
(e) The kinetic energy of a particle is
(not conserved)
K  (γ  1)mc2
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.
Before
u
A
m, K
stationary
B
2m
After
v
M
S (lab)
S (lab)
Apply conservation of momentum
back
and mass-energy:
2
2
M-E: mc 2  K  2mc 2  γv Mc 2 KE  2mc   γu  1 mc , so γu  3


 M  5m γv
Mom: γu mu  γv Mv
γuu
v 
5
8
8
5
u 
c, and v 
c, so γv 
3
5
17
5m
M 
 17 m
γv
The E-p invariant
We 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.
s  c t  x  c t   x 
2
2
2
2
2
2
2
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
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:
2
2

 
 2
  Ei     p i  c is invariant.
 i   i 
The energy-momentum 4-vector
We saw previously that the Lorentz transform for
space and time may be written as
 ct    
 
 x     v
c
 y   
   0
 z  
   0
0 0  ct 
c
 

0 0  x 
 
0
1 0  y 
z

0
0 1  
 v
ct and x, y, z form the components of a 4-vector,
and we can write the Lorentz transformation
more generally as b = A.b, where b and b are 4vectors and A is the transformation matrix which
is given by
 

 v
A
c
 0
 0

0 0 
c

0 0

0
1 0
0
0 1 
 v
It can be shown that energy and momentum form a four
vector, i.e. b = (E/c, px, py, pz):
 E c  


 px     v
c
 p   
 y   0
 p  
 z   0
0 0  E c
c



0 0  px 


p
0
1 0  y 
p 

0
0 1  z 
 v
Note
The ‘length’ or norm of a 4-vector is invariant
under the Lorentz transformation. The square
of the norm is bb which in Minkowski
geometry is given by
bb = b12  b22  b32  b42
For the E-p 4-vector this is
(E2/c2)  px2  py2  pz2
and for the t-r 4-vector it is
(c2t2)  x2  y2  z2
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.
Before
u
After
v
stationary
A
m, K
B
2m
M
S (lab)
S (lab)
Apply the left-hand side of the E-p invariant to ‘Before’, and the
right-hand side to ‘After’:
Before:

 E   p c  mc  K  2mc
2
2 2
2
2

2
 ( muγu )2 c 2
8 2
 25m c  m u γ c  25m c  m c c 9  17 m2 c 4
9
After: RHS  M 2 c 4 .  M  17 m as before.
2 4
2
2
2 2
u
2 4
2 2
Nuclear 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 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:
the Cockcroft-Walton experiment
protons
injected here
400 KV
lithium target
200 KV
0 KV
alpha particles
fluorescent screen