Download F33OT2 Symmetry and Action and Principles in Physics Contents

F33OT2 Symmetry and Action and Principles in Physics
notes by Juan P. Garrahan and Paul Saffin
Contents
1 Lecture one: Revision of Classical Mechanics
1.1 Motivation for this module . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Revision of Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2 Lecture two: Continuous symmetries in mechanical
2.1 Coordinate transformations . . . . . . . . . . . . . .
2.2 Symmetry transformations . . . . . . . . . . . . . . .
2.2.1 Translational invariance . . . . . . . . . . . .
2.2.2 Rotational invariance . . . . . . . . . . . . . .
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3 Lecture three: Canonical transformations
3.1 Phase space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Transformations in phase space . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Generating functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 Lecture four: Poisson brackets
4.1 Definition . . . . . . . . . . . . . . . . . . . . .
4.2 Properties . . . . . . . . . . . . . . . . . . . . .
4.3 Time evolution . . . . . . . . . . . . . . . . . .
4.4 Infinitesimal transformations . . . . . . . . . . .
4.5 Poisson brackets and canonical transformations
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5 Lecture five: relativistic electromagnetism
5.1 A recap of special relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 question set 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6 Lecture six: particle dynamics
6.1 A free-particle action . . . . .
6.2 relativistic forces . . . . . . .
6.3 action for a charged particle .
6.4 question set six . . . . . . . .
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7 Lecture seven: symmetry and conservation laws - Lagrangian
7.1 conservation of momentum and angular momentum . . . . . . . .
7.2 Nöther’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2.1 time shift symmetry . . . . . . . . . . . . . . . . . . . . .
7.2.2 position shift symmetry . . . . . . . . . . . . . . . . . . .
7.2.3 Galilean boost . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 question set 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8 Lecture eight: symmetry and conservation laws - Hamiltonian
8.1 Type 2 generating function . . . . . . . . . . . . . . . . . . . . . .
8.2 transformation of phase space functions . . . . . . . . . . . . . . .
8.3 transformation of the Hamiltonian . . . . . . . . . . . . . . . . . .
8.4 question set eight . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9 Lecture nine: a little group theory
9.1 question set 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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10 Lecture ten: Symmetry breaking and phase transitions
10.1 Recap of statistical mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Ising model and magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2.1 Mean-field solution of the Ising model . . . . . . . . . . . . . . . . . . . .
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List of Figures
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Two-body central force problem. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Passive and active coordinate transformations. . . . . . . . . . . . . . . . . . . .
Rotation in 2d. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Phase space orbits of a harmonic oscillator. . . . . . . . . . . . . . . . . . . . . .
Figure showing the passive transformation at work. . . . . . . . . . . . . . . . .
Figure showing the active transformation at work. . . . . . . . . . . . . . . . . .
Figure showing the phase-plane. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Figure showing the phase-plane. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Figure showing the phase-plane. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Figure showing the phase-plane. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Ising model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Graphical solution of Eq. (10.290). There is only one solution, m = 0, for T > Tc .
There are two solutions m± 6= 0, with m− = −m+ , for T < Tc . The critical
temperature is Tc = 2Jz/kB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Magnetisation squared m2 as a function of temperature. The magnetisation is
zero in the paramagnetic phase and nonzero in the ferromagnetic phase. . . . . .
List of Tables
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1
Lecture one: Revision of Classical Mechanics
1.1
Motivation for this module
This module is the second one of the Theoretical Physics strand. It follows the introductory
module on Classical Mechanics, F33OT1 Principles of Dynamics, which is a pre-requisite for
the current module. The aim of the module is to highlight two central concepts in theoretical
physics: (i) symmetry and (ii) variational, or action, principles; and to consider how these are
important both in the development of theories of physical phenomena and in the solution of
physical problems.
Symmetry plays a central role in modern Physics. You can read more about this in the
following two articles by two of the foremost theorists of the 20th century, who made fundamental
discoveries in particle physics intimately related to the concept of symmetry:
• The role of symmetry in fundamental physics,
by D.J. Gross (Nobel Prize in Physics 2004)
Proc. Natl. Acad. Sci. USA 93, 14256 (1996).
• Symmetry and Physics,
C.N. Yang (Nobel Prize in Physics 1957)
Proc. Am. Phil. Soc. 140, 267 (1996).
The idea of the least-action principle, which you studied in F33OT1 in the context of Classical Mechanics, originates with Fermat in the 1660s. Fermat’s principle states that the path
followed by a ray of light between two points is the one that takes the least time. This variational principle allows to derive the laws of Geometrical Optics such as those of reflection and
Snell’s law of refraction. Mapertuis, Euler and Lagrange in the 1750s reformulated Newtonian
mechanics by means of the calculus of variation. As modern Physics emerged it was found
that physical theories could be formulated starting from an action principle. Such is the case
of Maxwell’s equations for Electromagnetism (1860s) which can be obtained from an action
written in terms of the electric and magnetic fields, and Einstein’s General Relativity (1915),
whose equations can be derived from the so-called Einstein-Hilbert action. The action principle
even plays a role in Quantum Mechanics through Feynman’s path integral formulation (1940s).
Symmetry and the action principle are thus two of the pillars on which the whole edifice of
modern Theoretical Physics stands. The aim of this module is to explore this two concepts
building on your knowledge of Classical Mechanics.
1.2
Revision of Classical Mechanics
F33OT1 discussed the elegant formalism of Lagrange, Hamilton and others, by which Newtonian mechanics is reformulated starting from a least-action (or Hamilton) principle: Newton’s
equations are obtained by extremising the action,
Z tf
Z tf
S≡
dt L(q, q̇) =
dt [pq̇ − H(p, q)] .
(1.1)
ti
ti
3
The action is a functional (i.e. a function of a function) of the paths of the system between
times ti and tf . The first equality corresponds to Lagrange’s definition of the action, where
the Lagrangian L is a function of the generalised coordinates and velocities, which we denote
collectively by q, q̇. The second equality corresponds to Hamilton’s definition of the action,
where the Hamiltonian is now a function of the generalised coordinates and momenta, q, p.
The Lagrangian and Hamiltonian formulations are equivalent, and L(q, q̇) and H(p, q) are related
by a Legendre transform. In the Lagrangian formulation, the minimisation of the action, δS = 0,
gives the Euler-Lagrange equations:
d ∂L ∂L
−
= 0.
dt ∂ q̇
∂q
(1.2)
These are Newton’s equations of motion. The derivatives of the Lagrangian w.r.t. the velocities
define the momenta:
∂L
p≡
,
(1.3)
∂ q̇
and the Hamiltonian is obtained from the Lagrangian via:
H(p, q) = p q̇(p, q) − L[q̇(p, q), q] ,
(1.4)
where the velocities are expressed as functions of the momenta and coordinates, q̇ = q̇(p, q), by
solving Eqs. (1.3). In the Hamiltonian formulation, minimisation of the action, δS = 0, gives
Hamilton’s equations of motion:
q̇ =
∂H
∂H
, ṗ = −
.
∂p
∂q
(1.5)
These equations are equivalent to the Euler-Lagrange equations (1.2): the Euler-Lagrange equations are N second order differential equations (N =number of degrees of freedom), while Hamiton’s equations are 2 × N first order differential equations.
rA
mA
k
r
R
y
P
rB
mB

(a)
(b)
L
(c)
Figure 1: Two-body central force problem.
Lets consider the mechanical problem illustrated in Fig. 1(a). Two particles, A and B, of
equal mass, mA = mB = m, move on a flat, frictionless, two dimensional surface. The two
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bodies interact through a linear spring of force constant k and zero rest distance. If r A and r B
indicate the positions of the two particles measured from a lab reference frame, Fig. 1(b), then
the Lagrangian reads:
1
1
(1.6)
L = m ṙr 2A + ṙr 2B − k |rr A − r B |2 .
2
2
The first term is the kinetic energy T and the second term minus the potential energy −V . A
more convenient set of coordinates is given by the center of mass and relative coordinates,
Fig. 1(b),
1
mA r A + mB r B
= (rr A + r B ) , r ≡ r A − r B .
(1.7)
R≡
mA + mB
2
The Lagrangian then reads,
2
1
1
1
R + µ ṙr 2 − k |rr |2 ,
L = M Ṙ
2
2
2
(1.8)
where the total and relative masses are defined by M = 2m and µ = m/2. It is also convenient
to represent the relative coordinate and velocity in polar variables (note in particular the form
of the velocity),
r = r r̂r ⇒ ṙr = ṙ r̂r + r θ̇ θ̂θ .
(1.9)
Replacing in the Lagrangian we obtain,
2
1
1
1
1
R + µ ṙ2 + I(r) θ̇2 − kr2 ,
L = M Ṙ
2
2
2
2
(1.10)
where I(r) = µr2 is an instantaneous moment of inertia. We can now understand the motion
of the system from the Euler-Lagrange equations. The centre of mass and relative momenta are
given by:
∂L
∂L
∂L
R , pr ≡
= M Ṙ
= µ ṙ , L ≡
= I(r) θ̇.
(1.11)
P ≡
∂ ṙ
R
∂ θ̇
∂Ṙ
The Lagrangian does not depends on R or on θ, so the EL equations for these variables read:
d ∂L
= 0 ⇒ P = const. ,
dt ∂Ṙ
R
d ∂L
= 0 ⇒ L = const. ,
dt ∂ θ̇
(1.12)
that is, conservation of linear momentum P , and conservation of angular momentum
L. This is consistent as there are no external forces or torques, so these quantities should indeed
be conserved. The Lagrangian does depend on the relative distance r, and the corresponding
EL equation reads,
L2
µ r̈ = −k r + 3 .
(1.13)
µr
This is the equation of motion of a harmonic oscillator in presence of a centrifugal force. Motion
of the system is that of uniform translation of the centre of mass, rotation around it of constant
angular momentum—but of varying angular velocity θ̇ due to the varying moment of inertia
I(r)—while at the same time internally oscillating according to Eq. (1.13), Fig. 1(c). [Ref:
Goldstein §3.1, §3.2, §3.3.]
5
To obtain the Hamiltonian from Eq. (1.10) via a Legendre transform (1.4) we first need to
invert Eqs. (1.11) to express the velocities in terms of the momenta (and coordinates):
R=
Ṙ
P
pr
L
, ṙ =
, θ̇ =
.
M
µ
I(r)
(1.14)
The Legendre transform (1.4) is then:
R + pr ṙ + L θ̇ − L
H = P Ṙ
2
2
P
p2r
L2
P
p2r
L2
1 2
=
+
+
−
+
+
− kr
M
µ
I(r)
2M
2µ 2I(r) 2
P2
p2
L2
1
⇒H =
+ r +
+ kr2
2M
2µ 2I(r) 2
(1.15)
Note that the Hamiltonian is independent of R and θ. Hamilton’s equations (1.5) read for this
case:
L2
p
P = 0 , L̇ = 0 , ṙ = r , ṗr = −kr + 3 .
(1.16)
Ṗ
µ
µr
The last two equations combine to form the equation of motion (1.13). The first two equations
state the conservation of total linear momentum and angular momentum. This conserved quantities arise as a consequence of the lack of restoring forces in R and θ. The relation between
conservation laws, symmetries, and the independence of the Hamiltonian on certain generalised
coordinates (or momenta) will be the subject of forthcoming lectures.
6
2
Lecture two: Continuous symmetries in mechanical
systems
Our first foray into symmetry will be the exploration of basic symmetries in mechanical systems.
The central ones are translational invariance and rotational invariance, which lead to conservation of linear and angular momentum, respectively, and invariance under time translation,
which relates to conservation of energy.
2.1
Coordinate transformations
Lets assume we have the Lagrangian for a mechanical problem, L(q, q̇), where q, q̇ collectively
denote all the coordinates and velocities of the system (for example, for a system of particles
A, B, . . . we would have q = {rr A , r B , . . .} and q̇ = {ṙr A , ṙr B , . . .}). Consider now the change
of coordinates, q → q 0 , that is, the old coordinates can be written in terms of the new ones,
q = q(q 0 ), and vice-versa. We can then write the Lagrangian for the new coordinates:
L(q, q̇) = L[q(q 0 ), q̇(q 0 , q̇ 0 )] = L0 (q 0 , q̇ 0 ) ,
(2.17)
where in general the new Lagrangian L0 is a different function of its arguments than the old
∂q
one. [In the equation above we have used the fact that q̇ = dtd q = q̇ 0 ∂q
0 , so q̇ is a function both of
0
0
q and q̇ .] One of the powers of the Lagrangian formalism is that we can describe the dynamics
of the system with whichever coordinate system we wish by writing the corresponding EulerLagrange equations (which in general will be different between the unprimed and the primed
representation).
Example:
Consider a two dimensional particle of mass m subject to a central force. The position of the
particle is r = (x, y) in Cartesian coordinates. The Lagrangian reads:
p
1 2
2
2
2
ẋ + ẏ − V
x +y ,
(2.18)
L(x, y, ẋ, ẏ) =
2
since the potential only depends on the distance to the origin, V (|rr |). We now transform to
polar coordinates, (x, y) → (r, θ), that is, r = (x, y) = (r cos θ, r sin θ). Replacing in (2.18) we
get the new Lagrangian,
i
p
1h
˙ θ)2 + (r sin
˙ θ)2 − V
(r cos
r2 cos2 θ + r2 sin2 θ
L(x, y, ẋ, ẏ) =
2
1 2
⇒ L0 (r, θ, ṙ, θ̇) =
ṙ + r2 θ̇2 − V (r) ,
(2.19)
2
which is a different function of its variables than L.
2.2
Symmetry transformations
When the coordinate transformation q → q 0 is such that L0 is equal to L we have a symmetry
transformation, i.e. the Lagrangian is invariant under q → q 0 ,
L(q, q̇) = L(q 0 , q̇ 0 ) .
7
(2.20)
This is one of the strengths of the Lagrangian (or Hamiltonian) formulation of classical mechanics: in Newton’s formulation we need to find the symmetries by exploring all the equations of
motion, while in Lagrange’s (of Hamilton’s) formulation we only need to establish the symmetry
transformations of a single function, the Lagrangian (or Hamiltonian).
2.2.1
Translational invariance
Consider a coordinate transformation corresponding to a uniform translation. For example if the
positions of the particles in a mechanical system are given by the vectors r A with A = 1, . . . , N ,
with N the total number of particles, then the coordinate transformation amounts to:
r A → r 0A = Ta (rr A ) = r A + a for all A = 1, . . . , N ,
(2.21)
i.e., we translate the coordinates of all bodies by the same amount. Here we have defined Ta to
be the operation that translates a vector by a .
Coordinate transformations like the one of (2.21) have two equivalent interpretations, illustrated in Fig. 2. The first one is termed passive: the transformation q → q 0 corresponds to
moving the reference frame while keeping the physical system in place, and q 0 are the coordinates as measured from this new frame. The second interpretation is termed active: here
we think of moving the whole physical system, but keeping the reference framed fixed; q 0 are
the coordinates of the new positions of all particles of the system with respect to the original
(unmoved) reference frame. In practice it does not make a real difference which interpretation
one takes.
y
r
Passive

y
r!
y!


Active
y
r

r!
r


!
Figure 2: Passive and active coordinate transformations.
A translationally invariant Lagrangian does not change under the transformation (2.21) for
any value of the translation vector a . Such a system has translational symmetry.
8
Example:
Consider the two body problem of Lecture 1, Eq. (1.6). If we perform a generic translation, Eq.
(2.21), we have that:
1
1
m ṙr 2A + ṙr 2B − k |rr A − r B |2
2
2
1
1 h 0
2
2
2
=
m (ṙr A − ȧa) + (ṙr 0B − ȧa) − k |rr 0A − a − r 0B + a |
2
2
1
1
2
r 02
r 0A − r 0B |
m ṙr 02
=
A + ṙ
B − k |r
2
2
= L(rr 0A , r 0B , ṙr 0A , ṙr 0B )
L(rr A , r B , ṙr A , ṙr B ) =
(2.22)
where we have used that ȧa = 0 as the vector a is constant. Since L is the same as L0 for any
choice of a this problem has translational invariance. We will see in forthcoming lectures that
conservation of linear momentum is a consequence of this invariance.
The translational invariance of the system of the example is a continuous symmetry: the
Lagrangian is unchanged under translations of the coordinates for any value of a . Mechanical
systems can also have discrete symmetries, i.e. symmetries under a finite set of transformations.
An example is inversion: r → r 0 = −rr . This is a discrete symmetry of the problem above.
2.2.2
Rotational invariance
Lets consider now rotations in two-dimensions (the three-dimensional case will be considered in
the Problems Class). If we rotate all position vectors by an angle α, see Fig. 4, we have:
r A → r 0A = Rα (rr A ) for all A = 1, . . . , N ,
(2.23)
where Rα indicates rotation of a vector on the plane by an angle α. In terms of Cartesian
component, r A = (xA , yA ) we have:
0 xA cos α − yA sin α
xA
xA
,
(2.24)
=
= R(α) ·
yA
yA0
xA sin α + yA cos α
i.e. we can represent a rotation in terms of a rotation matrix.
cos α − sin α
R(α) ≡
.
sin α cos α
y
r!
α
r

Figure 3: Rotation in 2d.
9
(2.25)
Example:
Lets go back to the two body problem of Lecture 1, Eq. (1.6) and consider the change in the
Lagrangian under the rotation above. We can express q(q 0 ) by inverting the rotation:
0
0 xA
xA cos α + yA0 sin α
xA
=
= R(−α) ·
.
(2.26)
yA
−x0A sin α + yA0 cos α
yA0
Note that the inverse rotation is a rotation with angle −α. For the kinetic energy we need the
velocities,
0 ẋA
ẋA
= R(−α) ·
,
(2.27)
ẏA
ẏA0
since α is constant. In the kinetic energy we have the square of the modulus of the velocity,
which reads:
0 ẋA
ẋ0A
ẋA
2
T
02
0
0
0
0
ṙA = ẋA ẏA ·
= ẋA ẏA ·R(−α) ·R(−α)·
= ẋA ẏA ·
.
= ṙA
ẏA
ẏA0
ẏA0
Here we have used the fact that R is an orthogonal matrix so that the transpose is the inverse,
RT = R−1 . A similar calculation leads to,
|rr A − r B |2 = |rr 0A − r 0B |2 .
Since neither the terms with velocities nor that with the coordinates change under (2.26) the
Lagrangian (1.6) is rotationally invariant.
10
3
Lecture three: Canonical transformations
In Lecture 2 we discussed coordinate transformations and symmetries in the Lagrangian formulation of classical mechanics. In this Lecture and the next we will discuss and extend these ideas
in the Hamiltonian formulation. While the Lagrangian formulation is often the more practical
one for solving actual problems, the Hamiltonian one is more important conceptually (and also
is the starting point when going from classical to quantum mechanics).
3.1
Phase space
The Hamiltonian is a function of the generalised coordinates and momenta, H(p, q), where as
before p, q collectively denote all the momenta and coordinates of the system in question. The
momenta and coordinates are independent variables (this is in contrast to the velocities in the
Lagrangian formulation which are time derivatives of the coordinates). The variables p, q span
a 2N -dimensional space (where N is the number of degrees of freedom) known as phase space.
(In contrast, the N -dimensional space span by the coordinates q is called configuration space.)
A point (p, q) in phase space uniquely specifies the state of the system: given a point (p, q) and
the Hamiltonian all future (and past) evolution of the system can be uniquely determined.
Example:
Consider the one-dimensional harmonic oscillator of mass m and elastic constant k,
H=
1 2 1 2
p + kq .
2m
2
(3.28)
Energy is conserved, so for a given total energy E we get a relation between p and q,
p2 + mkq 2 = 2mE .
(3.29)
This is the equation for an ellipse. That is, for a given value of the total energy, the motion of
the harmonic oscillator traces an ellipse in phase space. Such a phase space curve is called an
orbit. Orbits never cross (due to the fact that every point in phase space uniquely determines
a state of the system).
p
q
Figure 4: Phase space orbits of a harmonic oscillator.
11
3.2
Transformations in phase space
We can now consider general variable changes in phase space, that is,
q → q 0 (p, q)
p → p0 (p, q)
(3.30)
The Hamiltonian will then change,
H(p, q) = H[p(q 0 , p0 ), q(p0 , q 0 )] = K(p0 , q 0 ) .
(3.31)
In general the function K(p0 , q 0 ) is not a Hamiltonian, i.e., the equations of motion expressed in
terms of p0 , q 0 are not Hamilton’s equations from K. [This contrasts with coordinate transformations q → q 0 which, as we saw in Lecture 2, take the Lagrangian L(q, q̇) to another Lagrangian
L0 (q 0 , q̇ 0 ).]
Example:
Lets go back to the harmonic oscillator of Eq. (3.28). Consider now the following two phase
space transformations (i.e. one would have the upper plus sign, the other the lower minus sign):
0
q
q = ±p
→
(3.32)
p
p0 = q
The Hamiltonian changes into
1 02 1 02
q + kp ,
(3.33)
2m
2
which is the same for the two transformations. In terms of p, q the system evolves according to
Hamilton’s equations:
p
∂H
∂H
=
, ṗ = −
= −kq
(3.34)
q̇ =
∂p
m
∂q
Lets now see what happens to these equations under the transformation:
0
q̇ = p/m
ṗ = q 0 /m
→
(3.35)
ṗ = −kq
q̇ 0 = ∓ kp0
H(p, q) → K(p0 , q 0 ) =
The transformed equations are obtained from −∂K/∂q 0 and ∂K/∂p0 only for the lower sign.
That is, only under the transformation:
0
q = −p
(3.36)
p0 = q
K(p0 , q 0 ) is a Hamiltonian. A phase space transformation that takes one Hamiltonian to another
is called a canonical transformation.
3.3
Generating functions
Canonical transformations are obtained from generating functions, i.e., one can express the
transformation equations in terms of derivatives of a single function of a combination of old and
new coordinates and momenta.
12
Example:
Consider the canonical transformation (3.36). A generating function for it is:
F1 (q, q 0 ) = −q q 0 ,
(3.37)
with the transformation obtained as:
p=
∂F1
= −q 0 ,
∂q
p0 = −
∂F1
=q
∂q 0
(3.38)
In general F1 is not the only generating function. Other combinations of old and new phase
space variables is possible. That is:
∂F1
,
∂q
∂F2
F2 (q, p0 ) → p =
,
∂q
∂F3
F3 (p, q 0 ) → q = −
∂p
∂F4
F4 (p, p0 ) → q = −
∂p
F1 (q, q 0 ) → p =
∂F1
∂q 0
∂F2
q0 =
∂p0
∂F3
, p0 = − 0
∂q
∂F4
, q0 =
∂p0
p0 = −
(3.39)
(3.40)
(3.41)
(3.42)
When they exist different generating functions are equivalent.
Example:
For the previous example an alternative generating function is:
F4 (p, p0 ) = −p p0 ,
(3.43)
From Eq. (3.42) we get:
q=−
∂F4
= p0 ,
∂p
q0 =
which is the same canonical transformation (3.36).
13
∂F4
= −p
∂p0
(3.44)
4
Lecture four: Poisson brackets
In this lecture we continue with our exploration of the formal properties of classical mechanics
in the Hamiltonian formulation. Here the analogies with quantum mechanics will be even closer
as we will discuss the concept of Poisson brackets.
4.1
Definition
The Poisson bracket (PB) between two functions of the coordinates and momenta, f (p, q)
and g(p, q), is defined as:
N X
∂f ∂g
∂f ∂g
−
,
{f, g} ≡
∂qα ∂pα ∂pα ∂qα
α=1
(4.45)
where the index α runs over the N degrees of freedom of the system.
Example:
Consider the functions f (p, q) = qβ and g(p, q) = pγ . Their PB is then:
X
N
N X
∂qβ ∂pγ
∂qβ ∂pγ
−
=
δβ,α δγ,α = δβ,γ ,
{qβ , pγ } =
∂qα ∂pα ∂pα ∂qα
α=1
α=1
(4.46)
where δα,β is the Kronecker delta (δα,β = 1 if α = β, and 0 otherwise). Notice the similarity
to the commutator of the position and momentum operators in quantum mechanics, [x̂, p̂] =
x̂p̂ − p̂x̂ = i~.
We also have,
{qα , qβ } = {pα , pβ } = 0 ∀α, β .
(4.47)
Equations (4.46,4.47) are called fundamental PBs [since all PBs can be calculated from them
by using properties (i–iv)].
4.2
Properties
PBs have the following algebraic properties:
(i) PB is antisymmetric: {f, g} = −{g, f }.
An obvious consequence is that the PB of any function with itself vanishes, {f, f } = 0.
(ii) PB is linear: {af + bg, h} = a{f, h} + b{g, h}, where a, b are constants.
(iii) PB obeys Leibnitz rule: {f g, h} = f {g, h} + {f, h}g.
A similar rule is obeyed by the derivative operator [the product rule, d(f g) = f dg + gdf ].
(iv) PB obeys the Jacobi identity: {f, {g, h}} + {g, {h, f }} + {h, {f, g}} = 0.
(These properties are also obeyed by the commutator, [fˆ, ĝ] ≡ fˆĝ − ĝ fˆ, in quantum mechanics.)
14
4.3
Time evolution
Consider an arbitrary function of the coordinates and momenta, f (p, q). Its equation of motion
reads:
N X
∂f
∂f
df
=
q̇α +
ṗα ,
(4.48)
dt α=1 ∂qα
∂pα
since f carries an implicit time dependence through that of q and p. But from Hamilton’s
equations we have that q̇ = ∂H/∂p and ṗ = −∂H/∂q, so the above equation becomes:
N
X
df
=
dt α=1
∂f ∂H
∂f ∂H
−
∂qα ∂pα ∂pα ∂qα
= {f, H} .
(4.49)
So the time evolution of any observable (any function of q, p) is given by its PB with the
Hamiltonian. A consequence of Eq. (4.49) is that any function f (p, q) that has {f, H} = 0 is a
conserved quantity. In particular, if the Hamiltonian is time independent then it is conserved,
as {H, H} = 0 due to property (i) above. [The equation of motion of an explicit function of
time, f (p, q, t), is: f˙ = ∂t f + {f, H}.]
Example:
1
p2x + p2y + p2z . Any component of the
Consider a free particle in three dimensions, H = 2m
momentum vector has vanishing PB with the Hamiltonian,
{px , H} =
3 X
∂px ∂H
α=1
∂px ∂H
−
∂qα ∂pα ∂pα ∂qα
=
3 X
α=1
0×
pα
− δα,1 × 0 = 0 ,
m
(4.50)
and similarly for py and pz . This means that {pp, H} = 0 and the linear momentum vector p is
a conserved quantity (as it should in this case).
4.4
Infinitesimal transformations
Using Eq. (4.49) we can write an infinitesimal time translation, t → t + δt, as
f → f + δt
df
= f + δt{f, H} .
dt
(4.51)
H is the generator of time translations. Note that H is both the operator that generates
translation in time, and the quantity that is conserved if the Hamiltonian is time-translationally
invariant.
Just like H is the operator that generates infinitesimal translations in time, other operators
generate other infinitesimal transformations via the PB. Consider the case of infinitesimal spatial
translations, r → r + δaa, where δaa is an infinitesimally small translation vector. A function of
position is transformed,
f (rr ) → f (rr ) + δaa · ∇f (rr ) = f (rr ) + δax ∂x f (rr ) + δay ∂x f (rr ) + δaz ∂x f (rr )
15
(4.52)
The momentum operator, p = (px , py , pz ), is the corresponding generator:

{f, px } = ∂x f 
{f, py } = ∂y f
⇒ {f, p } = ∇f

{f, pz } = ∂z f
(4.53)
so that,
f (rr ) → f (rr ) + δaa · ∇f (rr ) = f (rr ) + δaa · {f, p } .
(4.54)
Clearly p is invariant under translations, {pp, p } = 0. The Hamiltonian is translationally invariant
if {H, p } = 0. This means both that r is a cyclic (or ignorable) coordinate (since ∇H = 0), and
that momentum p is conserved. Note again the dual role of p as the generator of translations
and as the conserved quantity if the Hamiltonian is translationally invariant.
4.5
Poisson brackets and canonical transformations
Canonical transformations preserve the Poisson bracket relations. That is:

 {qα , pβ } = {qα0 , p0β } = δα,β
{qα , qβ } = {qα0 , qβ0 } = 0
only if (p, q) → (p0 , q 0 ) is a C.T.

{pα , pβ } = {p0α , p0β } = 0
(4.55)
Example:
The example of section 3.2 showed that only one of the two phase space transformations
0
q
q = ±p
→
(4.56)
p
p0 = q
was canonical. Lets prove this again using PBs:
1 = {q, p} → {p0 , ±q 0 } = ∓{q 0 , p0 } = 1 only for lower sign
16
(4.57)
5
Lecture five: relativistic electromagnetism
THEY HAVE SEEN THE LEVI-CIVITA ALTERNATING SYMBOL, AND THE SUMMATION CONVENTION
Aim: To illustrate the Lorentz symmetry in Maxwell’s equations
Learning outcomes: At the end of this lecture should
• know how to express Maxwell’s equations in a manifestly Lorentz covariant fashion
5.1
A recap of special relativity
We start by defining the Minkowski co-ordinates that we shall be using, in four-vector notation
dxµ = (cdt, dx, dy, dz)
(5.58)
i.e. the index µ take four possible values, with dx0 = cdt, dx1 = dx, dx2 = dy, dx3 = dz. Now
we introduce the matrix ηµν , called the metric,
ηµν = diag(−1, 1, 1, 1)
(5.59)
and write the invariant length as
ds2 = −c2 dt2 + dx2 + dy 2 + dz 2 =
X
ηµν dxµ dxν := ηµν dxµ dxν ,
(5.60)
µν
where we have used Einstein summation, i.e. repeated indices come with an implied summation
symbol. Now define a Lorentz transformation, Λµν , to be those transformations
dx0µ = Λµν dxν
(5.61)
that leave the invariant length the same
ηµν dxµ dxν = ds2 = ds02 = ησρ dx0σ dx0ρ = ησρ Λσµ Λρν .dxµ dxν
(5.62)
this gives the requirement on the transformation marix Λµν to be
Λσµ Λρν ησρ = ηµν
(5.63)
and defines what we mean by a Lorentz transformation.
e.g. Let’s have a look at the standard transformation we use in special relativity for a boost
in the x-direction


γ
−βγ 0 0
 −βγ
γ
0 0 

Λµν = 
(5.64)
 0
0
1 0 
0
0
0 1
this gives



cdt0
γ(cdt − βdx
 dx0 
 γ(dx − βcdt) 




 dy 0  = 

dy
0
dz
dz

17
(5.65)
and does indeed satisfy (5.63).
We also need the inverse metric, denoted η µν , this has components
η µν = diag(−1, 1, 1, 1),
(5.66)
η µν ηµσ = δσν .
(5.67)
and satisfies
Given (5.61) we see that one also gets
∂µ0 =
∂
−1 ν ∂
=
(Λ
) µ ν = (Λ−1 )νµ ∂ν ,
∂x0µ
∂x
(5.68)
where we have introduced the standard definition of a matrix inverse
(Λ−1 )µν Λνσ = δσµ
(Λ−1 )µν Λσµ = δνσ .
(5.69)
(5.70)
In fact, we may multiply (5.63) by (Λ−1 )µκ η νξ to find an explicit expression for the inverse of Λµν
(Λ−1 )ξκ = η νξ Λρν ηρκ .
(5.71)
In the example of (5.64) we find that
(Λ−1 )00
(Λ−1 )01
(Λ−1 )10
(Λ−1 )11
=
=
=
=
η ν0 Λρν ηρ0
η ν0 Λρν ηρ1
η ν1 Λρν ηρ0
η ν1 Λρν ηρ1
= η 00 Λ00 η00
= η 00 Λ10 η11
= η 11 Λ01 η00
= η 11 Λ11 η11
= Λ00
= −Λ10
= −Λ01
= Λ11
(5.72)
(5.73)
(5.74)
(5.75)
giving

(Λ−1 )µν

γ βγ 0 0
 βγ γ 0 0 

= 
 0
0 1 0 
0
0 0 1
(5.76)
which may be checked explicitly to be the inverse Lorentz transformation, just by doing the
matrix multiplication.
This result is useful because it shows that
η µν ∂ν
(5.77)
transforms as a four-vector. That is to say
(η µν ∂ν )0 = η µν (Λ−1 )σν ∂σ
= η µν [η σρ Λξρ ηξν ]∂σ
= Λµρ (η ρσ ∂σ )
18
(5.78)
(5.79)
(5.80)
meaning that contraction of a downstairs index with the inverse metric gives something that
transforms in a standard way, i.e. it makes relativistic sense.
In detail,
 1∂ 
− c ∂t
∂


∂
∂x

η µν ν = 
(5.81)
∂


∂x
∂y
∂
∂z
and if we use the chain-rule we may
 1 ∂
− c ∂t0
 ∂0
 ∂x∂

0
∂y
∂
∂z 0
confirm that
 
∂
∂
− β ∂x
γ − 1c ∂t
  γ ∂ + β1 ∂
∂x
c
∂x
=
∂
 
∂y
∂
∂z




(5.82)
which is just how a vector should transform.
5.2
electromagnetism
In electromagnetism we start with the vectors E and B, with the following relations
ρ
0
∇.B = 0
∂B
∇×E+
= 0
∂t
∂E
∇ × B − µ0 0
= µ0 j
∂t
∂ρ
+ ∇.j = 0
∂t
∇.E =
(5.83)
(5.84)
(5.85)
(5.86)
(5.87)
The first hint that we may be able to write this in a relativistically covariant way comes from
the last relation, which we see may be rewritten as
∂
(cρ) + ∂i j i = ∂µ j µ = 0
∂(ct)
(5.88)
if we introduce the four-vector j with components
j µ = (cρ, j).
(5.89)
We are then left with the two three-vectors E and B, there is no way to turn these into fourvectors, as there is nothing left that could be the zero-component. However, we do have another
vector in the problem, the vector-potential, A,
B = ∇×A
19
(5.90)
and a scalar, the electric potential φ,
∂
A − ∇φ
(5.91)
∂t
So we should in fact think of E and B as derived quantities, rather than fundamental, and we
now have the possibility of combining φ with A to make a four-vector. To see how this works
note that (5.90) and (5.91) may be written
E = −
B1
B2
B3
E 1 /c
E 2 /c
E 3 /c
=
=
=
=
=
=
η 22 ∂2 A3 − η 33 ∂3 A2
η 33 ∂3 A1 − η 11 ∂1 A3
η 11 ∂1 A2 − η 22 ∂2 A1
η 00 ∂0 A1 − η 11 ∂1 A0
η 00 ∂0 A2 − η 22 ∂2 A0
η 00 ∂0 A3 − η 33 ∂3 A0
(5.92)
(5.93)
(5.94)
(5.95)
(5.96)
(5.97)
and we have introduced
Aµ = (φ/c, A)
(5.98)
This compels us to introduce the field strength tensor F µν , defined by
F µν = η µσ ∂σ Aν − η νσ ∂σ Aµ
(5.99)
and now everything is written with spacetime indices, so is fully covariant. It is also worth
commenting that the field strength tensor has components


0
E 1 /c E 2 /c E 3 /c
 −E 1 /c
0
B 3 −B 2 

F µν = 
(5.100)
 −E 2 /c −B 3
0
B1 
−E 3 /c B 2 −B 1
0
So, (5.99) gives us both of the source-free Maxwell relations (5.84), (5.85) in a manifestly
Lorentz-covariant language.
Now we have all the objects in relativistic form, we should consider the sourced Maxwell
relations, (5.83) and (5.86). As an example we note that (5.83) may now be written as
∇.E = ρ/0
1
1
⇒ ∂i E i /c =
ρ/0 = 2 j 0 /0
c
c
0i
0
⇒ ∂i F
= µ0 j .
µ0
⇒ −∂µ F
= µ0 j 0 .
(5.101)
(5.102)
(5.103)
(5.104)
Moreover, the x component of (5.86) may be written as
1 ∂E 1
c2 ∂t
12
31
⇒ ∂2 F − ∂3 F − ∂0 F 01
⇒ −∂2 F 21 − ∂3 F 31 − ∂0 F 01
⇒ −∂µ F µ1
∂2 B 3 − ∂3 B 2 −
20
= µ0 j 1
(5.105)
= µ0 j 1
= µ0 j 1
= µ0 j 1
(5.106)
(5.107)
(5.108)
(recall that c2 = 01µ0 ). In fact, we may look at the remaining components of (5.86) and conclude
that (5.83) and (5.86) may both be written in
−∂µ F µν = µ0 j ν .
(5.109)
In summary, all four of Maxwell’s relations may be encoded in the manifestly relativistically
covariant way given in (5.99) and (5.109).
5.3
question set 5
1. take the definition of the field strength tensor and calculate the following in terms of the
vector potential
F 00 , F 01 , F 10 , F 12
(5.110)
2. what is the value of F µν if the indices µ and ν are the same?
P
3. we know that a four-vector transforms as V 0µ = σ Λµσ V σ under Lorentz transformations,
so how does the product of four-vectors, V µ W ν , transform?
P
4. As F µν has two indices it transforms as F 0µν = σρ Λµσ Λνρ F σρ . Take the standard Lorentz
transformation matrix for boosts along the x-axis, and your knowledge of how F µν relates
to E and B to calculate the following
•
•
•
•
•
E 01 ,
E 02 ,
E 01 ,
E 02 ,
B 03 ,
if
if
if
if
if
the
the
the
the
the
only
only
only
only
only
non-zero
non-zero
non-zero
non-zero
non-zero
term
term
term
term
term
in
in
in
in
in
the
the
the
the
the
original
original
original
original
original
frame
frame
frame
frame
frame
is
is
is
is
is
E 1.
E 1.
E 2.
E 2.
E 2.
5. confirm that your results above are consistent with the general expressions
E 0k = E k , B 0k = B k
E 0⊥ = γ(E ⊥ + β × B), B 0⊥ = γ(B ⊥ − β × E),
(5.111)
(5.112)
6. in the first frame of reference we have a line of electric charges along the x-axis, producing
an electric field
y
z
3
E 1 = 0, E 2 = 2
,
E
=
(5.113)
y + z2
y2 + z2
now boost this in the x-direction and find the electric and magnetic fields in that frame,
i.e. the frame that measures a moving line of electric charges - a curent. Ten sketch
the magnetic field lines and confirm that they are in the direction expected for a flowing
current.
7. If we have two gauge potentials related by
A0µ = Aµ + η µν ∂ν λ
for some function λ, how are their field strengths, F 0µν , F µν , related?
21
(5.114)
6
Lecture six: particle dynamics
Aim: To illustrate the utility of the action in coupling charged particles to the electromagnetic
field
Learning outcomes: At the end of this lecture should
• know how to construct the action for an uncharged relativistic particle
• know how to couple a charged relativistic particle to electromagnetism
6.1
A free-particle action
If we have an uncharged particle, the only quality it has is its mass. So if we are to construct
an action with dimensions
[S] = kg.m2 .s−1
(6.115)
we must use c to take care of the length units as it is the only object available with length in
it. We are driven to something of the form
S ∼ mc2 T
(6.116)
where T is something with dimensions of time. The only thing that fits the bill is the particle’s
proper time, so we try
Z
2
S = −mc
dτ.
(6.117)
This also makes geometric sense, what we are saying is that the evolution of a particle is such
that its proper time is extremized. If we think in a four-dimensional sense of a particle having a
world-line, then what we want to do is extremize the world line, just like a rubber band wants
to extremize its length. Now we use the expression for the proper time to see if this quess makes
sense,
Z r
1
2
dt2 − 2 dx2
(6.118)
S = −mc
c
r
Z
1
2
= −mc
dt 1 − 2 v 2
(6.119)
c
Z
v2
2
∼ −mc
dt 1 − 2 + ...
(6.120)
2c
Z
m
(6.121)
∼
dt v 2 + const + O(v 4 /c2 )
2
This is just the action of a free particle in Newtonian theory, so we recover Newtonian dynamics
in the small-speed limit.
22
Another way to write this is
Z
Z
dτ
dτ = −mc
dτ
dτ
s 2
Z
2
dt
1 dx
2
= −mc
dτ
− 2
dτ
c
dτ
S = −mc
2
2
(6.122)
(6.123)
So we find a Lagrangian
2
s
L = −mc
dt
dτ
2
1
− 2
c
dx
dτ
2
and we may use the Euler-Lagrange equations
∂L
∂L
d µ
d
µ
=
0,
ẋ
:=
x
−
∂xµ dτ ∂ ẋµ
dτ
(6.124)
(6.125)
to find
d2 xµ
= 0
dτ 2
(6.126)
i.e.
d µ
U = 0,
dτ
Uµ =
dxµ
dτ
(6.127)
which is just what we knew from special relativity, so we do indeed have the correct action.
6.2
relativistic forces
In Newtonian dynamics we have
m
dv
= F
dt
(6.128)
In special relativity this is modified to
dU µ
dpµ
m
=
= Fµ
dτ
dτ
(6.129)
where F µ represents the components of some four-force. The first thing to notice is that
ηµν U µ U ν = −c2
(6.130)
and because both c and ηµν are constant
ηµν U µ
d ν
1 d
1 d
U =
(ηµν U µ U ν ) =
(−c2 ) = 0
dτ
2 dτ
2 dτ
23
(6.131)
So
ηµν U µ F ν = 0.
(6.132)
U 0 F 0 = δij U i F j ,
1
F0 =
v.F
c
(6.133)
which gives
(6.134)
which, given that work=force X distance, we interpret as the power input to the particle by the
force - up to the factor of 1c .
This is all very nice, but do we have an example of a relativistic force? We certainly know
that electromagnetism is relativistic, so we should start there, and consider the Lorentz force.
dp
= F = q(E + v × B).
dt
(6.135)
Rather than give a messy derivation involving all components at once, let’s look at the z
component
dp3
dt
dp3
dt
dp3
dt
dp3
dt
dp3
dt
dp3
dτ
= q(E 3 + v 1 B 2 − v 2 B 1 )
(6.136)
= q(cF 03 − v 1 F 13 − v 2 F 23 )
(6.137)
q 0 03
(U F − U 1 F 13 − U 2 F 23 − U 3 F 33 )
γ
q
=
(−η00 U 0 F 03 − η11 U 1 F 13 − η22 U 2 F 23 − η33 U 3 F 33 )
γ
q
= − ηµν U µ F ν3
γ
=
= −qηµν U µ F ν3 .
(6.138)
(6.139)
(6.140)
(6.141)
And if we were to repeat this for the other components we would get
dpi
= −qηµν U µ F νi .
dτ
(6.142)
If we now use the energy-change equation for a particle in an electric field
d
= qv.E,
dt
as well as remembering that the energy, is given by = cp0 , we find
give the final result
(6.143)
dp0
dτ
= −qηµν U µ F ν0 to
dpρ
= −qηµν U µ F νρ .
dτ
This equation is written entirely in terms of relativistic quantities, so is fully covariant.
24
(6.144)
6.3
action for a charged particle
We found the action for a free, uncharged particle in the first section of this lecture, now we try
to construct one for a charged particle, i.e. one that reproduces the relativistic force equation we
have just derived. We shoud certainly expect the same term involving the world-line as before,
but now the particle has an extra attribute, its charge. This allows it to talk to electromagnetism
which, as we have seen, is described in terms of a four-vector Aµ . What we need therefore is
for some attribute of the particle to couple to this four-vector and produce a scalar that may
be used in the action. The only vector that the particle has on offer is its four-velocity U µ , so
it is natural to consider the quantity ηµν U µ Aν . Therefore we write down the action
Z
Z
2
S = −mc
dτ + q dτ ηµν U µ Aν .
(6.145)
Noting that the dimensions of Aµ are [Aµ ] = m.Kg.C −1 .s−1 , which come from [0 ] = Kg −1 .m−3 .s2 .C 2 ,
[µ0 ] = Kg.m.C −2 , [jµ ] = C.m−2 .s−1 , [E] = Kg.m.C −1 .s−2 , [B] = Kg.C −1 .s−1 The Euler Lagrange equations are then found using
∂L
∂L
d
= 0
(6.146)
−
dτ ∂ ẋµ
∂xµ
d
(mηµν ẋν + qηµν Aν ) − qηνµ ẋν ∂µ Aν = 0
(6.147)
dτ
dpσ
= −qηµν U µ F νσ .
(6.148)
dτ
which is just what we needed.
6.4
question set six
1. when we took the small-speed limit if the action we wrote down only the first term,
complete the series up to order v 6 .
2. The canonical momenta are defined in the usual way
∂L
∂ ẋα
take the action for a charged particle and show its canonical momentum is
Pα =
Pµ = ηµν (pν + qAν )
(6.149)
(6.150)
3. express the interaction part of the action in terms of the electric potential, and the threevector potential, using dt instead of dτ as the integral measure. Use this, combined with
the small-speed version of the kinetic term, to write down the potential that a charged
particle moves in.
4. take the non-relativistic Lagrangian for a charged particle
L =
1 2
mv − qφ + qv.A
2
and evaluate the Hamiltonian
25
(6.151)
5. consider the gauge transformation
A0µ = Aµ + η µν ∂ν λ
(6.152)
for some function λ, and show that the integrand of the interaction term for the charged
particle changes by a total derivative
6. a charged particle is placed at the origin, x = 0, at t = 0 such that it is initially stationary;
we set the origin of time such that t(τ = 0) = 0. Now suppose that there is a constant
electric field in the x-direction of strength E. Write down the non-zero components of the
field-strength tensor. Calculate x(τ ), t(τ ) in terms of the charge, rest-mass, electric field
and c. Evaluate dx/dt, and note that although it is constantly accelarating, it nevertheless
has c as its speed limit.
7. A charged particle, with a velocity v in the x-direction, enters a region of homogeneous
magnetic field pointing in the z-direction with strength B; the particle enters at t(τ =
0) = 0. Write down the non-zero component of the field strength tensor. Calculate x(τ ),
t(τ ) in terms of the charge, rest-mass, electric field and c. Evaluate the zero component
of te relativistic force, and interpret the result.
26
7
Lecture seven: symmetry and conservation laws - Lagrangian
Aim: To show that a symmetry of the action leads to a conservation law in the equations of
motion
Learning outcomes: At the end of this lecture should
• know how to derive Nöther’s theorem
7.1
conservation of momentum and angular momentum
Suppose we have the following Lagrangian
L =
1
mẋ2
2
(7.153)
then we see that the Euler-Lagrange equations give
d
p = 0
dt
. This follows directly from L not having an explicit x dependence,
where p = ∂L
∂ ẋ
other words, the action S,
Z t2
S =
dt L
(7.154)
∂L
∂x
= 0. In
(7.155)
t1
is invariant if we replace x with x + a.
Now consider the Lagrangian relevant to bodies orbiting a star
L =
1 2 1 2 2
mṙ + mr θ̇ − V (r)
2
2
(7.156)
we now see that the Euler-Lagrange equations imply
d
J = 0
dt
(7.157)
where
J =
∂L
= mr2 θ̇
∂ θ̇
(7.158)
again, this conservation law follows because the Lagrangian has no explicit dependence on one
of the co-ordinates, this time θ, ∂L
= 0. An equivalent statement is that the action is invariant
∂θ
under the relacement θ → θ + α.
27
7.2
Nöther’s theorem
Suppose we make some shift in time, and some change in position according to
t̃ = t + ξ(t)
q̃(t̃) = q(t) + η(q, t)
(7.159)
(7.160)
where is a small parameter, and ξ(t), η(q, t) are arbitrary smooth functions. We shall say that
the action is invariant under such transformations if
Z t̃2
Z t2 d
d
dG
dt̃L q̃(t̃), q̃(t̃), t̃
dt L q(t), q(t), t + =
(7.161)
dt
dt
dt̃
t̃1
t1
Where we have added a total derivative to the second integral, because doing so does not alter
dt̃
to find
the equations of motion. Now change the measure of the first integral to dt̃ = dt dt
Z t2 d
d
dG
dt̃
dt L q̃(t̃), q̃(t̃), t̃
− L q(t), q(t), t − = 0 (7.162)
dt
dt
dt
dt̃
t1
Z t2 dG
˙
˙
dt L q + η, q̇ + η̇ − ξ q̇, t + ξ (1 + ξ) − L (q, q̇, t) − ⇒
= 0 (7.163)
dt
t1
where we have used
dt d
d
q̃ =
q̃ ' q̇ + η̇ − ξ˙q̇.
(7.164)
dt̃
dt̃ dt
Now use a Taylor expansion to find
Z t2 ∂L
∂L
dG
∂L
˙ − L (q, q̇, t) − + (η̇ − ξ˙q̇)
+ ξ
(1 + ξ)
= 0(7.165)
dt L + η
∂q
∂ q̇
∂t
dt
t1
Z t2
∂L
∂L
∂L dG
˙
˙
⇒
+η
+ (η̇ − ξ q̇)
−
dt Lξ + ξ
= 0(7.166)
∂t
∂q
∂ q̇
dt
t1
In fact, this integral expression must vanish for all choices of intergration region, giving that
the integrand itself vanishes.
Using the expression for the total time derivative of L
∂L ∂L
∂L
dL
=
+
q̇ +
q̈
(7.167)
dt
∂t
∂q
∂ q̇
we find that the integrand may be written
∂L
d ∂L
∂L
d
ξL + (η − ξ q̇)
− G − (η − ξ q̇)
−
= 0
(7.168)
dt
∂ q̇
dt ∂ q̇
∂q
The second term in this expresion vanishes due to the Euler-Lagrange equations so
d
∂L
ξL + (η − ξ q̇)
−G = 0
dt
∂ q̇
(7.169)
giving the following constant of motion
∂L
− G = constant
(7.170)
∂ q̇
This is Nöther’s theorem, i.e. if an action is invarant under (7.159), (7.160)then there is a
constant of motion given by (7.170).
ξL + (η − ξ q̇)
28
7.2.1
time shift symmetry
Suppose that we have a system that is invariant under a shift in time,
t̃ = t + t0 , x̃(t̃) = x(t),
d
d
x̃(t̃) = x(t)
dt
dt̃
(7.171)
then we see that
Z
t̃2
t̃1
Z t2 1 ˙2
1
2
dt̃ mx̃ − V (x̃) =
dt mẋ − V (x)
2
2
t1
(7.172)
i.e. invariant, with η = 0, ξ = const, leaving
∂L
q̇ − L = const
∂ q̇
(7.173)
but this is just the Hamiltonian, and so invariance under time shifts just corresponds to energy
conservation.
7.2.2
position shift symmetry
Now suppose that we have a system that is invariant under shifts of position
t̃ = t, x̃(t̃) = x(t) + X,
d
d
x̃(t̃) = x(t)
dt
dt̃
(7.174)
then we see that
Z
t̃2
t̃1
Z t2 1 ˙2
1
2
dt̃ mx̃ =
dt mẋ
2
2
t1
(7.175)
with ξ = 0, η = const leaving
∂L
= const
(7.176)
∂ ẋ
but this is just momentum conservation, so if the laws of physics do not depend on position,
then momentum is conserved.
7.2.3
Galilean boost
A more interesting example is the case of a Galilean boost, this is similiar to the Lorentz boost,
but is applied to Newtonian systems. Suppose we have two particles interacting such that their
potential only depends on their separation,
1
1
m1 ẋ21 + m2 ẋ22 − V (x1 − x2 )
2
2
Then we see that this changes by a total time derivative under
L =
t̃ = t
x̃1 = x1 + vt
x̃2 = x2 + vt
29
(7.177)
(7.178)
(7.179)
(7.180)
and the action changes according to
Z t2
Z t2
d
˙2 =
dt L(x1 , x2 , ẋ1 , ẋ2 ) + ([m1 x1 + m2 x2 ]v)
dt L(x̃1 , x̃2 , ˜˙x,
1 ˜x)
dt
t1
t1
(7.181)
Comparing this to the general formula we find
ξ = 0, η = vt, G = (m1 x1 + m2 x2 )v
(7.182)
yielding the following conserved quantity
(m1 ẋ1 + m2 ẋ2 )t − (m1 x1 + m2 x2 ) = const
⇒ M X = P t + const
(7.183)
(7.184)
where we have introduced the total momentum P , the centre of mass (m1 x1 + m2 x2 )/M , and
the total mass M = m1 + m2 . A further calculation shows that as the action is invariant under
x1 → x1 + a, x2 → x2 , then the total momentum P is conserved. So, the symmetry under
Galilean boosts implies that the centre of mass moves along at constant speed.
7.3
question set 7
1. take the Lagrangian
L =
1 2 1 2
ẋ + ẏ − x2 − y 2 − 2xy
2
2
(7.185)
and show that it is invariant under the shifts
x̃ = x + ,
ỹ = y − (7.186)
then calculate the corresponding Nöther constant.
2. consider the simple harmonic oscillator with Lagrangian
L =
1
1
mẋ2 − kx2
2
2
(7.187)
and consider the change
t̃ = t
x̃ = x + exp(−iωt)
(7.188)
(7.189)
where k = ω 2 m.
• Calculate the change in the Lagrangian, show it is a total derivative, and find the
corresponding conserved quantity.
• by finding the equations of motion for the system, use direct substitution to show
that
(ẋ + iωx) exp(−iωt)
is constant.
30
(7.190)
3. consider the Lagrangian
k
1 2
ẋ + 2
2
x
L =
(7.191)
• show that this system leaves the action invariant, in the sense given in the lectures,
under the scaling transformation
t̃ = λt
√
x̃(t̃) =
λx(t)
(7.192)
(7.193)
• take λ = 1 + to find the conserved quantity associated with this scaling symmetry.
• show that
1
−Et + xẋ
2
is constant, where E = 12 ẋ2 −
k
x2
(7.194)
is the total energy, which is constant.
4. consider the Lagrangian
L =
k
1 2 2
x ẋ + 4
2
x
(7.195)
and show that the action is invariant, in the sense given in the lectures, under the scaling
transformation
t̃ = λ4 t
x̃(t̃) = λx(t)
(7.196)
(7.197)
take λ = 1 + to find the conserved quantity associated with this scaling symmetry.
5. consider the Lagrangian
L = ẋ4 +
k
x4/3
(7.198)
and show that the action is invariant, in the sense given in the lectures, under the scaling
transformation
t̃ = λ4 t
x̃(t̃) = λ3 x(t)
(7.199)
(7.200)
take λ = 1 + to find the conserved quantity associated with this scaling symmetry.
31
8
Lecture eight: symmetry and conservation laws - Hamiltonian
Aim: To show that a symmetry of the Hamiltonian leads to a conservation law in the equations
of motion
Learning outcomes: At the end of this lecture should
• know how to use canonical transformations to construct conserved quantities from systems
with symmetries
8.1
Type 2 generating function
Recall that the type-2 generating function has
p =
∂F2
∂F2
∂F2
, Q=
, K=H+
, F2 = F2 (q, P )
∂q
∂P
∂t
(8.201)
In particular we have the identity canonical transformation
F2id = qP
(8.202)
which gives
p = P, Q = q
(8.203)
F2 = qP + G(q, P, t)
(8.204)
Now suppose we introduce
then we have to leading order in ∂G
= p + {p, G}
∂q
∂G
= q + {q, G}
Q = q + δq = q + ∂p
∂G
K = H +
∂t
P = p + δp = p − (8.205)
(8.206)
(8.207)
i.e. G generates the transformations
δq = {q, G}
δp = {p, G}
(8.208)
(8.209)
N.B. as q̇ = {q, H} we see that the Hamiltonian generates time evolution.
Now suppose we have some phase space function, U (q, p), how does this change under
such transformations? We define the change to be the difference between active and passive
transformations
δU = U (B) − U 0 (A0 )
where
32
(8.210)
• A0 is the location in the new chart (i.e. the chart with co-ords Q, P ) of the same point,
i.e. passive interpretation
• B is the location in the same chart (i.e. the chart with co-ords q, p) of the transformed
point, i.e. active interpretation
e.g. passive transformations.
These correspond to simply moving the co-ordinate axes around, and do not cause any
physical change in the state, only our description of the same state. Consider the function
f (x) = x2 , and define the shifted variable x̃ = x − a, f (x) = x2 = (x̃ + a)2 = f˜(x̃), i.e. leave
the function where it is, and move the axes (FIG. 5)
f(x)
f(x)
x
x
Figure 5: Figure showing the passive transformation at work.
e.g. active transformations
These transformations leavev the axes alone, but move the actual objects, and they correspond to physical changes. We have that f˜(x̃) = f (x), so we see that f˜(x) = f (x+a) = (x+a)2 ,
this defines a different function on the same set of axes. (FIG. 6).
f(x)
f(x)
x
Figure 6: Figure showing the active transformation at work.
33
e.g. phase space.
In terms of a phase space example consider the standard simple harmonic potential, this
has a phase space portrait given in FIG. 7, and we have given two curves, one representing
a pendulum with total energy E1 , and another with total energy E2 . We have also marked a
particular point, namely the point that is at the extreme of the pendulum’s swing, with energy
E1 .
p
E1
original
point
E2
q
Figure 7: Figure showing the phase-plane.
Now suppose we introduce
q̃ = q − α,
p̃ = p − β
(8.211)
but interpret this as a passive transformation, so just a different choice of axes, which we have
drawn in FIG 8. So, if we look at this system in terms of co-ordinates (q̃, p̃) we get the picture
FIG 9, and note that the point we marked as the extrema of the swing is still the extrema of
the pendulum’s swing, the location of this point is the A0 used above.
Now we take
q̃ = q − α,
p̃ = p − β
(8.212)
but interpret it as an active transformation, so we stay with the same co-ordinates, but move
points from (q, p) to (q − α, y − β) as shown in FIG 10. This has the effect of moving our point
from the energy E1 curve to the energy E2 curve, and it no longer corresponds to an extrema
of the swing - this is a physically distinct point on the phase portrait, and is the point B used
above.
8.2
transformation of phase space functions
We have seen how the position and momenta change, but what about functions of position and
moment? To see how functions change note that for a scalar function we have that
U (A) = U 0 (A0 )
34
(8.213)
p
p
E1
E2
β
q
q
original
point
α
Figure 8: Figure showing the phase-plane.
p
E1
E2
q
original
point
Figure 9: Figure showing the phase-plane.
35
p
new
point
E2
original
point
E1
q
β
α
Figure 10: Figure showing the phase-plane.
so the change in U is
δU = U (B) − U 0 (A0 ) = U (B) − U (A)
= U (q + δq, p + δp) − U (q, p)
∂U
∂U
δq +
δp
=
∂q
∂p
= {U, G}
(8.214)
(8.215)
(8.216)
(8.217)
and this is certainly consistent with what we had for the variation of q and p.
8.3
transformation of the Hamiltonian
The Hamiltonian is a bit more subtle, as the Hamiltonian is not just any old function, rather
it is the function that keeps the equations of motion in canonical form. If we make a canonical
transformation using F2 = qP + G, then we have that the Hamiltonian becomes
H 0 (A0 ) = H(A) + ∂G
∂t
(8.218)
so
δH = H(B) − H 0 (A0 ) = H(B) − H(A) − = {H, G} − = −
∂G
∂t
dG
dt
∂G
∂t
(8.219)
(8.220)
(8.221)
Then we see that if H is invariant (i.e. δH = 0), G is a constant of motion. i.e. constants of
motion generate the symmetries.
e.g. H generates time translation
e.g. p generates space translation
36
e.g. L generates rotation
e.g. Galilean boosts.
It may seem rather strange that the definition of δU involves this idea of active and passive
transformations, why not just evaluate H at each point and if it is the same, we have a symmetry.
Well, one place this will clearly not work is Galilean boosts. Here one takes a system of particles
and gives them all a boost in velocity of the same magnitude and direction. We would certainly
expect this to be a symmetry of Newtonian dynamics (for potentials that only depend on the
separation of the particles), but the boosted system has a higher energy, so simply subtracting
Hinitial from Hf inal would not give zero.
8.4
question set eight
1. In the expression for the change in phase-space co-ordinates q, p induced by the generator
G, take
• G = p, to find δq, δp.
• G = H, and write as δt, to find δq, δp.
• G = pt − mq, to find δq, δp, but write v in place of .
• for the Galilean boost (G = pt − mq) take the Hamiltonian H =
{H, G}. Then calculate {H, G} − ∂G/∂t
1 2
p,
2m
and calculate
2. for the simple harmonic oscillator with Hamiltonian
1 2 1 2
p + kq
2m
2
H =
(8.222)
calculate δq, δp, δH for the generator
G = ln(p + imωq) − iωt
(8.223)
noting that k = mω 2 . What does this symmetry correspond to?
3. Take the Hamiltonian
1 2 1 2
p + kq
2m
2
H =
(8.224)
and the generator
G = ωmq cos(ωt) = p sin(ωt), ω 2 = k/m
(8.225)
and evaluate δq, δp, δH.
4. Take the Hamiltonian
H =
1 2
p + λq
2m
(8.226)
and the generator
1
G = pt − qm + λt2
2
and evaluate δq, δp, δH.
37
(8.227)
5. Take the Hamiltonian
1 2
p + λq
2m
H =
(8.228)
and the generator
G = p + λt
(8.229)
and evaluate δq, δp, δH.
6. Take the Hamiltonian
1 2 1 2
p − kq
2m
2
H =
(8.230)
and the generator
G = ln(p + mλq) − λt, λ2 = k/m
(8.231)
and evaluate δq, δp, δH.
7. Take the Hamiltonian
H =
1 2 1 2 2 1 4
p + q p + q
4
2
4
(8.232)
and the generator
G = ln(p + iq) −
1
ln(p2 + q 2 ) − i(p2 + q 2 )t
2
(8.233)
1 2 1 4
p + kq
2m
2
(8.234)
and evaluate δq, δp, δH.
8. Take the Hamiltonian
H =
and the generator
G = ωmq cos(ωt) − p sin(ωt)
and evaluate δq, δp, δH.
38
(8.235)
9
Lecture nine: a little group theory
We saw from the lecture on the symmetries of the Hamiltonian that the generating function
G(q, p, t) induced the following infinitesimal change on a phase-space function
δU = {U, G}
(9.236)
for some small parameter , which if we write as = δα and take the limit δα → 0 then becomes
dU
dα
= {U, G}.
(9.237)
Now suppose we define
µ = {U, G} =
dU
.
dα
(9.238)
This is also a phase-space function so we have
dµ
= {µ, G}.
dα
(9.239)
which gives
d2 U
dα2
=
dµ
= {µ, G} = {{U, G}, G}
dα
(9.240)
This process continues in the obvious way.
d3 U
dα3
= {{{U, G}, G}, G}, etc
Now we consider the Taylor expansion of a function
1 2 d2 U dU + ...
+ α
U (α) = U0 + α
dα 0 2
dα2 0
1
= U0 + α{U, G}|0 + α2 {{U, G}, G}|0 + ...
2
= exp[αĜ]U |0
(9.241)
(9.242)
(9.243)
(9.244)
where we have introduced the operator
Ĝ = {., G}
(9.245)
ĜU = {U, G}
(9.246)
which acts according to
This sort of expression should be familiar from quantum mechanics, e.g. the evolution operator
is
i
|ψ(t)i = exp − (t − t0 )Ĥ ψ(t0 )i
(9.247)
~
39
for a system with Hamiltonian Ĥ.
e.g. Taking our generator to be angular momentum in z-direction, Lz = xpy − ypx we see
that
{x, Lz } = −y, {y, Lz } = x
(9.248)
so if we apply the generator Lz to some point (x, y)|0 = (x0 , y0 ) we find
1
x(ψ) = x|0 + ψ{x, Lz }|0 + ψ 2 {{x, Lz }, Lz }|0 + ...
2
1 2
= x0 − ψy0 − ψ {y, Lz }|0 ...
2
1 2
1
= x0 − ψy0 − ψ x + ψ 3 y + ...
2
3!
1 2
1
= x[1 − ψ ...] − y[ψ − ψ 3 ...]
2
3!
= x cos ψ − y sin ψ
(9.249)
(9.250)
(9.251)
(9.252)
(9.253)
which we recognise as a rotation by an angle ψ, a similar calcuation to find y(ψ) confirms this.
So, angular momentum just generates rotations. Similar calculations show that rotations by
angles θ, φ, ψ about the x, y, z axis are generated by
Rx (θ) = exp(θL̂x )
Ry (φ) = exp(φL̂y )
(9.254)
(9.255)
Rz (ψ) = exp(ψ L̂z )
(9.256)
(9.257)
One of the key points about this is that all the relevant information about finite rotations is
contained within generators, (Lx , Ly , Lz ).
e.g. Consider using the momentum as a generator, Ĝ = {., px }, then we see that a phase
space function becomes
U (a) = exp(ap̂x )U |0
(9.258)
1
= U0 + a{U, px }|0 + a2 {{U, px }, px }|0 + ...
2
∂U 1 ∂U
+ ...
= U0 + a
+
,
p
x
∂x 0 2 ∂x
0
∂U 1 2 ∂ 2 U = U0 + a
+ a
+ ...
∂x 0 2 ∂x2 0
= U (x0 + a).
(9.259)
(9.260)
(9.261)
(9.262)
In other words, the momentum generator just translates objects.
e.g. Now consider the generator of galilean boosts, Ĝ = B̂ = {., px t − mx}, then we see that
40
a phase-space function becomes
U (v) = exp(v B̂)U |0
= U0 + v{U, px t − mx}|0 + ...
∂U ∂U = U0 + vt
+ mv
+ ...
∂x ∂px 0
(9.263)
(9.264)
(9.265)
0
= U (x0 + vt, px0 + mv).
(9.266)
Which just corresponds to boosting the object by speed v.
Now suppose we want to perform two different finite transformations e.g. a rotation about
the x-axis followed by a rotation about the y-axis, can we still get away with only knowing
information about the generators? To see how this works let’s consider performing two transformations, one generated by Â, and the other by B̂.
1
1
exp(aÂ) exp(bB̂)U = (I + a + a2  + ...)(I + bB̂ + b2 B̂ B̂ + ...)U
2
2
1 2
1 2
= I + a + bB̂ + b B̂ B̂ + a  + abÂB̂
2
2
1 2
1
1
1
= I + a + bB̂ + b B̂ B̂ + a2  + abÂB̂ + abB̂ Â
2
2
2
2
1
1
+ abÂB̂ − abB̂ Â
2
2
1
1
= I + a + bB̂ + (a + bB̂)2 + ab[Â, B̂] + ...)U
2
2
1
= exp(a + bB̂ + ab[Â, B̂] + ...)U
2
(9.267)
(9.268)
(9.269)
(9.270)
(9.271)
(9.272)
So we need to find an expression for [Â, B̂] 1 . To do this, consider
[Â, B̂]U =
=
=
=
=
ÂB̂U − B̂ ÂU
Â{U, B} − B̂{U, A}
{{U, B}, LA } − {{U, A}, B}
{U, {A, B}
ĈU
(9.273)
(9.274)
(9.275)
(9.276)
(9.277)
C = {A, B}
(9.278)
where
and we have used the Jacobi identity. So, we can perform finite transformations using different
generators, and all we need to know is the Poisson-bracket commutation relation of the generators. Given that the symmetries of Newtonian mechanics are generated by momentum (P − i),
1
if you perform this to higher orders, one gets more commutators, the is the Baker-Campbell-Haussdorf
formula
41
angular momentum (Li = 12 ijk xj pk ), and boosts (bi = ii t − mxi ), all we need to know are thye
following Poisson-bracket commutation relations
{pi , pj }
{pi , Bj }
{pi , Lj }
{Li , Lj }
{Bi , Bj }
{Bi , Lj }
=
=
=
=
=
=
0
mδij
ijk pk
ijk Lk
0
ijk Bk .
(9.279)
(9.280)
(9.281)
(9.282)
(9.283)
(9.284)
The fact that the momentum generators commute means that if you perform two different
translations the order doesn’t matter, you get to the same place either way. However, if you
rotate about the x-axis then the y-axis, you get a different answer to rotating first about the
y-axis and then the x-axis.
9.1
question set 9
1. suppose we consider the transformation of complex number z by gθ = exp(iθ), z 0 = gθ z,
• what does this correspond to on an Argand diagram?
• show that gθ gφ = gθ+φ
• show that [gθ , gφ ] = 0.
• show that gθ gθ† = 1.
the objects gθ are known as group elements, and in this case they are elements of the
group U(1), the group consisting of 1×1 unitary matrices. Also, as the group elements
commute the group is called Abelian.
2. consider the matrix generators






0 0 0
0 0 1
0 1 0
T 1 =  0 0 1  , T 2 =  0 0 0  , T 3 =  −1 0 0  ,
0 −1 0
−1 0 0
0 0 0
(9.285)
• calculate gx (θ) = exp(θT 1 ) = I + θT 1 + 12 θ2 (T 1 )2 + ...
• calculate gy (θ) = exp(φT 2 ) = I + φT 2 + 21 φ2 (T 2 )2 + ...
• calculate gz (θ) = exp(ψT 3 ) = I + ψT 3 + 21 ψ 2 (T 3 )2 + ...
• calculate det(gx (θ).


x
• show that gx (θ) rotates the vector X =  y  by an angle θ about the x-axis.
z
• calculate [T 1 , T 2 ] and express it in terms of T 2 , T 2 , T 3 .
42
The idea of this question is to show that the structure we saw coming from Poisson brackets
is in fact part of a much bigger picture. The group that is generated here, the group of
rotations in three-dimensions, is called SO(3), the group of 3×3 orthogonal matrices (i.e.
OOT = I) with determinant 1.
3. the elements we have been calling g, the group elements of group G, are reguired to satisfy
some properties if G is to be a group
• if g1 ∈ G, and g2 ∈ G, then g1 g2 ∈ G
• g1 (g2 g3 ) = (g1 g2 )g3
• there exists an element I such that Ig = gI = g
• for each g there exists an element called g −1 , where gg −1 = g −1 g = I
show that the U(1) is indeed a group.
4. The symmetry algebra of Newtonian mechanics includes the momentum, boost and angular momentum generators. Confirm the full set of Poisson-commutation relations amongst
these generators.
5. Using your knowledge of quantum mechanics operator commutator relations, postulate
a relation between the (Poisson bracket, classical generator) system, and the (quantum
commutator,quantum operator) system..
43
10
Lecture ten: Symmetry breaking and phase transitions
Throughout this lecture course we have discussed how to apply variational methods to problems beyond classical mechanics, including electromagnetism and relativity. In this lecture we
will extend this to applications in statistical physics. The second central topic of this lecture
course has been symmetry: how to identify symmetries, both continuous and discrete, in the
equations (or the actions) of physical problems, how to define symmetry transformations and
their mathematical properties via the study of groups, and how to exploit the consequences of
such symmetries for understanding the behaviour of physical systems.
In this lecture we will introduce the equally important concept of symmetry breaking,
that is, the (very frequent) situation where the equations of motion (or the action) of a physical
problem have a certain symmetry but the solutions to those equations do not. An example
in classical mechanics is that of Kepler’s orbits: the equations of motion for the gravitational
two-body problem (or the corresponding Lagrangian or Hamiltonian) are spherically symmetric;
the solutions to such equations, however, give elliptical orbits which do not have that symmetry.
The symmetry is said to be broken, in this case as a consequence of the initial conditions which
select the specific orbit. (Individual solutions are not symmetric which means that the symmetry
transformation that leaves the Lagrangian invariant transforms the orbit to a different one. The
transformed orbit is however also a solution—one with different initial conditions. This means
that while individual solutions are not symmetric, the whole set of solutions is. This is how the
symmetry properties of the equations manifest in the solutions.)
An even more dramatic instance of symmetry breaking is when the system, even if it can
explore all possible initial conditions, still chooses solutions which break the symmetry of the
Hamiltonian. This is termed spontaneous symmetry breaking, and occurs in systems with very
large number of degrees of freedom. It is the mechanism that underlies phase transitions, be it
those of everyday life such as the freezing of water into ice, or those that occurred at the very
early stages of the universe. (For a discussion of SSB in particle physics and cosmology see the
two essays by Gross and by Yang quoted in Lecture 1.) In this lecture we will focus on SSB in
statistical physics by studying a simple model for magnetic phase transitions.
10.1
Recap of statistical mechanics
Consider a system of N particles at volume V in contact with a thermal bath of temperature T .
The system has a mean energy which is a function of both volume and temperature, U (V, T ),
and an entropy S(V, T ). The key thermodynamic potential for this problem is the (Helmholtz)
free-energy, F (V, T ) ≡ U (V, T )−T S(V, T ). These quantities describe the macroscopic properties
of the system. Volume, temperature and number of particles are global variables which specify
the macrostate. If we have a microscopic description of the problem a microscopic state or
microstate is one where all the microscopic variables have a specific value (e.g. all the positions
and velocities of the N particles of the system). A macrostate is a collection or ensemble of
these microstates. The free-energy can be obtained from the microscopic description via the
partition sum:
X
Z(V, T ) ≡
e−βEα = e−βF (V,T ) .
(10.286)
α
44
Here α labels the microstates of the system, the sum is over all of these microstates, Eα is the
energy of microstate α, and β ≡ 1/kB T where kB is Boltzmann’s constant. This defines the
canonical ensemble, i.e., the probability distributions of microstates that gives the corresponding
macrostate at (V, T ). In the canonical ensemble the probability ρα of microstate α is:
ρα =
e−βEα
.
Z(V, T )
(10.287)
Example:
Consider a system with microstates α = 0, 1, 2, . . . , ∞, where the energy of microstate α is
Eα = αJ, where J is a constant. The partition function is then
Z=
∞
X
e
α=0
−βαJ
=
∞
X
e−βJ
α=0
α
=
1
,
1 − e−βJ
as this is a geometric sum. The free-energy is then:
F = −β −1 ln Z = β −1 ln 1 − e−βJ .
The probabilities in the canonical ensemble are given by ρα = e−βαJ 1 − e−βJ . From these we
can calculate the mean energy and entropy:
U =
∞
X
E α ρα =
α=0
S = −kB
Je−βJ
,
1 − e−βJ
∞
X
Je−βJ
ρα ln ρα = −kB ln 1 − e−βJ + T −1
,
1 − e−βJ
α=0
where we have used Gibbs entropy formula for the entropy given the probability distribution.
Note that F = U − T S gives the expression for the free-energy above, as it should.
Roughly speaking a thermal system with a fixed number of particles will try and minimise
the free-energy F = U − T S, which implies a compromise between reducing the mean energy
and increasing the entropy. In general terms reducing U tends to favour microscopic order, and
this is the term in the free-energy that dominates at low temperature. Conversely, in general
terms S tends to favour disorder, and −T S is the term that dominates the free-energy at high
temperatures. This implies that as temperature is decreased there is a changeover from entropy
dominance and disorder at high T to energy dominance and order at low T . This change can
be smooth or it can be sudden. In the latter case this leads to phase transitions which we
will explore now in the context of magnetism.
10.2
Ising model and magnetism
The Ising model is a simple model for a magnet. Consider a lattice of N sites. Lattice sites
are labelled by indices i, j, . . . = 1, 2, . . . , N . On each site sits a magnetic moment or spin,
represented as arrows in Fig. 11. Each spin can take one of two values,“up” or “down”. That
45
is, at site i the spin is σi , where σi = 1 or −1. A microstate, σ , is given by specifying all the
spins, σ = (σ1 , σ2 , . . . , σN ). The energy of a microstate is given by:
X
σ ) = −J
E(σ
σi σj ,
(10.288)
hi,ji
where J is a (positive) constant, and the sum is over all the pairs of nearest neighbouring spins
hi, ji. This energy function says that the energy of the system is decreased if neighbouring spins
are aligned (irrespective of their orientation). That means that the energy favours magnetic
order, i.e., spins pointing in the same direction.
Figure 11: Ising model.
The energy (10.288) has a fundamental up/down symmetry. If all the spins are reversed,
σ , the energy remains invariant:
σ → −σ
X
X
σ ) → E(−σ
σ ) = −J
σ ).
E(σ
(−σi )(−σj ) = −J
σi σj = E(σ
(10.289)
hi,ji
hi,ji
The lowest possible energy is Emin = −JN z. This is attained when all the spins are pointing
σ ) = (1, 1, . . . , 1) and σ (−) (σ
σ) =
in the same direction. There are two such microstates, σ (+) (σ
(−1, −1, . . . , −1). We can define a macrostate that has aQmean energy U = Emin . An example
is the probability distribution over microstates ρ(+) ≡ i δσi ,1 , which states that microstate
σ (+) has probability 1 and all other microstates have probability 0 (or similarly with σ (−) ).
This macrostate has the lowest possible mean energy,
but it also has the lowest entropy S = 0.
P
σ
Furthermore, the mean magnetisation, M ≡ h N
i=1 i i, is M = N for this macrostate, as all
spins are pointing up. [See Q2 in Problem Sheet 9.]
Conversely, the macrostate
with the largest entropy is that where all microstates are equally
Q
σ ) = i 12 (δσi ,1 + δσi ,−1 ). This macrostate has entropy S = kB N ln (2), mean
probable, ρ(rnd) (σ
energy U = 0, and mean magnetisation is M = 0. [See Q2 in Problem Sheet 9.]
σ) =
The macrostate ρ(rnd) has the same up/down symmetry of the energy function: ρ(rnd) (σ
(rnd)
σ ), i.e., the probability of any microstate is the same as that of its spin-reversed. That
ρ
(−σ
M = 0 for this macrostate is a manifestation of this symmetry. In contrast, in the macrostates
σ ) 6= ρ(±) (−σ
σ ), and therefore M 6= 0.
ρ(+) and ρ(−) the up/down symmetry is broken: ρ(±) (σ
Below we solve the Ising model in an approximate way via a variational method, and show that
at high temperatures the macrostate resembles ρ(rnd) and at very low temperatures it resembles
either ρ(+) or ρ(−) , indicating a phase-transition from a paramagnet, M = 0, to a ferromagnet,
M 6= 0.
46
10.2.1
Mean-field solution of the Ising model
[For details see Q3 of Problem Sheet 9.] Lets assume that the macrostate is one where all spins
are independent (this can only be an approximation to the exact macrostate):
Y 1 + m
1−m
σ) =
ρ(σ
δσi ,1 +
δσi ,−1 ,
2
2
i
that is, each spin has a probability 1+m
of being in the up state, and a probability 1−m
of
2
2
2
being in the down state. In this macrostate
the mean energy is U = −JN zm , the entropy is
1+m
1−m
1−m
ln
+
ln
, and the mean magnetisation is M = N m.
S = −kB N 1+m
2
2
2
2
The variable m is now only variable of the problem. That is, we will solve the Ising model
variationally by finding the value of m that minimises the free energy F = U − T S, where U
and S are the functions of m given above. By minimising the free energy w.r.t. the variational
parameter m we obtain a self-consistent equation:
m = tanh (2βJzm) .
(10.290)
This equation can be solved graphically, as shown in Fig. 12. For high temperatures, T > Tc ,
there is only one solution with zero magnetisation, m = 0. For low temperatures, T < Tc ,
there are two solutions of non-zero magnetisation, m+ and m− (the solution m = 0 is unstable
in this regime). These two low temperature solutions are related by symmetry: m− = −m+ .
The critical (or Curie) temperature is Tc = 2Jz/kB . It is the temperature at which we get a
phase transition from a paramagnet at high temperature (with zero spontaneous magnetisation,
M = 0), to a ferromagnet at low temperature (with non-zero spontaneous magnetisation, M 6=
0). See Fig. 13. The paramagnet has up/down symmetry (M = 0), while the ferromagnet
has not (M 6= 0 and therefore most spins point in one direction). The up/down symmetry is
spontaneously broken at the phase transition to the paramagnet.
m
m
tanh(2βJzm)
tanh(2βJzm)
T > Tc
T < Tc
Figure 12: Graphical solution of Eq. (10.290). There is only one solution, m = 0, for T > Tc .
There are two solutions m± 6= 0, with m− = −m+ , for T < Tc . The critical temperature is
Tc = 2Jz/kB .
47
2
m
ferromagnet
paramagnet
Tc
T
Figure 13: Magnetisation squared m2 as a function of temperature. The magnetisation is zero
in the paramagnetic phase and nonzero in the ferromagnetic phase.
48