Download Mechanics and Electromagnetism

Mechanics and Electromagnetism
D. Atkinson and O. Scholten
∗
(August 1991)
∗
These lecture notes were originally written by D.Atkinson for the courses of 1988/1989 and
1989/1990 and have been modified by O. Scholten for the course 1991/1992.
1
Chapter 1
Introduction
The purpose of this course is to explore mechanics and electromagnetism in the
formalisms of Lagrange and of Hamilton. The unification that this gives to the
subject is very satisfying; and the pivotal rôle that Special Relativity has in the
treatment of electromagnetic phenomena is especially clear.
It is assumed that the student has some knowledge of classical mechanics in the
formulation of Newton: from this springboard the method of Lagrange is developed
in Chapter 2, and that of Hamilton in Chapter 3. Two classical problems are handled
with the new tools in Chapter 4: the Kepler problem of planetary motion, that was
first cracked by Newton, and the Rutherford scattering of α-particles by an atomic
nucleus.
The reader probably knows the Lorentz transformation of Special Relativity
already: an interesting derivation is given in Chapter 5, and special importance
is attached to contravariant and covariant Lorentz vectors. The starting point of
Chapter 6 is the Maxwell system of electromagnetic equations, and these are cast
into manifestly relativistically covariant form. Lagrangian and Hamiltonian methods
are used consistently. Chapter 7 is devoted to the theorem of Emmy Noether, and
the conservation of electromagnetic energy, momentum, and angular momentum is
shown to follow from invariance of the lagrangian under respectively translations in
time and space, and under Lorentz transformations.
The effect of a field on a point charged particle, and the inverse effect of such
a charged particle on the field, are considered in Chapters 8 and 9. Some of the
technical details of calculations are relegated to the decent obscurity of a half dozen
appendices.
The main reference book used with the lectures is:
L.D. Landau and E.M. Lifshits, Mechanics and Electrodynamics, Pergamon Press.
In the notes at some places this book will be referred to. It is strongly recommended
to try to solve the problems given in the book. Other books that provide a more
extensive treatment of the material at approximately the same level are:
K.R. Symon, Mechanics, Addison-Wesley,
J.B. Marion, Classical Dynamics, Academic Press,
J.R. Reitz, F.J. Milford and R.W. Christy, Foundations of Electromagnetic Theory,
Addison-Wesey,
R.K. Wangsness, Electromagnetic fields, Wiley,
A much more indepth treatment of the material is given in the following two classics,
2
H. Goldstein, Classical Mechanics, Addison-Wesley,
J.D. Jackson, Classical Electrodynamics, Wiley.
3
Chapter 2
Euler-Lagrange Equations
2.1
Point Particle
AIM: derive equations of motions without the explicit introduction of forces.
REASON: Forces are often clumsy to deal with since it are vectors.
Consider a point-particle of mass,m, in a conservative force-field: that is, suppose
that there exists a scalar potential energy, V (~r ), that depends only on the position,
~r , of the particle, such that the force is given by
~
F~ = −∇V.
(2.1)
Generally, it is convenient to express vectors in terms of Cartesian components,
~r = (x1 , x2 , x3 ) , thus:
Fi = −
∂V
.
∂xi
(2.2)
Newton’s second law can then be written
∂V
d
[mẋi ] = −
,
dt
∂xi
(2.3)
where dots mean time-derivatives.
In the alternative approach the non-relativistic kinetic energy of the particle is
introduced,
T = 12 mẋi ẋi ,
(2.4)
the convention being made that a repeated index (here i) is to be summed from 1
to 3. Accordingly,
∂V
d ∂T
[
]=−
.
dt ∂ ẋi
∂xi
(2.5)
The same equation (2.5) holds for N particles in a conservative force-field. We
write the kinetic energy in the form
T = 21 mi ẋi ẋi ,
(2.6)
where we sum over the Cartesian coordinates of all N particles, with the convention
that m3i−2 = m3i−1 = m3i is the mass of the i’th particle, the x, y, z coordinates
4
of which are x3i−2 ,x3i−1 and x3i respectively. The potential energy may include
particle-particle interactions, so long as these are conservative:
V =
3N
X
Vi +
i
2.2
1
2
3N
X
Vij .
(2.7)
i6=j
Canonical Transformation
AIM: rewriting equation (2.5) to obtain an equation of motion that is not only
valid for a cartesian coordinate system but for any set of generalized coordinates
describing the system.
We shall now perform a (possibly explicitly time-dependent) transformation of
the 3N Cartesian coordinates, xi , to a set of 3N independent coordinates, qn . A simple example would be the polar coordinates of the N particles, but much more exotic
possibilities exist, as we will see. The purpose is eventually to obtain coordinateindependent equations of motion. By differentiating Eq. (2.6) with respect to qn
and to q̇n , we find
∂ ẋi
∂T
= mi ẋi
,
∂qn
∂qn
(2.8)
∂T
∂ ẋi
= mi ẋi
.
∂ q̇n
∂ q̇n
(2.9)
dT
∂T
6= dq
and that care should be taken with the
Please note that in general ∂q
n
n
difference between straight and partial derivatives.
Since the qn are functions of the xi and t, but not explicitly of the q̇n , it follows
that xi may be written as a function only of the qn and t, so that
∂xi
∂xi
dxi
≡ ẋi =
q̇k +
.
dt
∂qk
∂t
(2.10)
Now the only dependence on the variables, q̇k , here is in the explicit occurrence of
q̇k on the right-hand side, so that
∂ ẋi
∂xi
=
.
∂ q̇n
∂qn
(2.11)
Hence, combining eqs.(2.9,2.11),
∂T
∂xi
= mi ẋi
,
∂ q̇n
∂qn
(2.12)
so that
d ∂T
d
∂xi
∂ ẋi
[
] = [mi ẋi ]
+ mi ẋi
.
dt ∂ q̇n
dt
∂qn
∂qn
(2.13)
By using Eq. (2.3), we can write the first term on the right-hand side as
−
∂V ∂xi
∂V
=−
,
∂xi ∂qn
∂qn
(2.14)
5
since V does not depend on the ẋi explicitly. The second term on the right of
Eq. (2.13) is equal to ∂T /∂qn , as can be seen from Eq. (2.8). Therefore
∂V
∂T
∂L
d ∂T
[
]=−
+
=
,
(2.15)
dt ∂ q̇n
∂qn ∂qn
∂qn
where the Lagrangian is defined by L = T − V . This is quite different from the
total energy, E = T + V . Since V does not depend explicitly on ẋi , neither does it
depend explicitly on q̇n , which means that the partial q̇n -derivative of T is equal to
that of L. Hence Eq. (2.15) can be rewritten as follows:
∂L
d ∂L
[
]−
= 0,
(2.16)
dt ∂ q̇n
∂qn
which is the famous Euler-Lagrange equation, the derivation of which was the goal
of this chapter. The Lagrangian L should be regarded as a function of q, q̇, and t;
L(q, q̇, t).
2.3
Hamilton’s Variational Principle
AIM: use a variational principle to arrive at the Euler-Lagrange equation and
introduce the principle of ”least action”.
Finally, we introduce the action, as the time-integral of the Lagrangian:
S(t2 , t1 ) =
Z
t2
dtL(q, q̇, t).
(2.17)
t1
Hamilton’s variational principle states that, if we vary the functions q(t) and q̇(t),
for t1 < t < t2 , but in such a way that q(t1 ) and q(t2 ) remain fixed at their physical
values, then, of all the possible values that q(t) and q̇(t) can have, those values that
make S extremal satisfy the Euler-Lagrange equations. We shall now prove this
statement.
δS(t2 , t1 ) =
Z
t2
t1
Z t2
dt[L(q + δq, q̇ + δ q̇, t) − L(q, q̇, t)]
∂L
∂L
δqn +
δ q̇n ].
(2.18)
∂qn
∂ q̇n
t1
By an integration by parts, we find
Z t2
∂L
d ∂L
δS(t2 , t1 ) =
dt[
− (
)]δqn ,
(2.19)
∂qn dt ∂ q̇n
t1
where the integrated terms vanish, since δq is constrained to be zero at t1 and t2 .
Since δq is arbitrary, it follows that for δS to be zero and thus for S to be extremal,
the expression between square brackets must vanish.
There now appear to be two different axioms from which the same Euler-Lagrange
equation can be derived;
=
dt[
• Newton: F = ma.
• Hamilton: The action S is minimal.
Since these give rise to the same equation of motion, which determines the physical
observables, they are equivalent. For many generalizations used later, Hamiltons
principle is more versitile and it will therefore be used frequently.
6
Chapter 3
Hamilton’s Equations
3.1
Canonical Momentum
AIM: introduce momenta associated with the generalized coordinates, the generalization of pi = mẋi to non-cartesian coordinates.
In terms of Cartesian components, the Lagrangian of our system of N particles,
interacting with one another, and with an external, conservative field, can be written
L(x, ẋ, t) = 12 mi ẋi ẋi − V (x, t),
(3.1)
where V may depend explicitly on t (for example, a time-dependent external potential), but not on ẋ. Then clearly
∂L
= m(i) ẋ(i) ,
∂ ẋi
(3.2)
where there is no summation over the bracketed index, (i). We recognize the righthand side of Eq. (3.2) as a component of the linear momentum, and we now define
the generalized, or canonical momentum, by
pn =
∂L
.
∂ q̇n
(3.3)
It is important to realize that the canonical momenta are not necessarily momenta
in the Cartesian sense: for example, in terms of spherical polar coordinates, the
“momentum” conjugate to the angle, θ, is the angular momentum.
The rate of change of the Lagrangian can be written
dL
∂L
∂L
∂L
=
q̇n +
q̈n +
.
dt
∂qn
∂ q̇n
∂t
(3.4)
Now from the Euler-Lagrange equation, we know that
d ∂L
∂L
= [
],
∂qn
dt ∂ q̇n
(3.5)
so that Eq. (3.4) can be rewritten
dL
d ∂L
∂L
= [
q̇n ] +
.
dt
dt ∂ q̇n
∂t
(3.6)
7
If the potential and thus L has no explicit time dependence,
Eq. (3.6) we obtain
d ∂L
[
q̇n − L] = 0.
dt ∂ q̇n
∂L
∂t
= 0, then from
(3.7)
Since the time derivative of q̇n pn −L is zero, it is associated with a conserved quantity
which is the total energy (see Eq. (3.10)) of the system.
3.2
The Hamiltonian
AIM: derive first order equations of motion for the generalized coordinates and
the conjugate (canonical,generalized) momenta.
The Hamiltonian is defined by
H = pn q̇n − L,
(3.8)
and we use Eq. (3.3) to rewrite Eq. (3.6) in the form
dH ∂L
+
= 0.
(3.9)
dt
∂t
If L does not depend explicitly on the time, the second term above is absent, and
hence the Hamiltonian is constant in time. If, moreover, the potential energy is
conservative, Eq. (3.2) holds,so that
H = mn ẋn ẋn − L = 2T − [T − V ] = T + V,
(3.10)
which means that the Hamiltonian is equal to the total energy, which is timeindependent. Note that two conditions are necessary for these conclusions to be
true: (1) L must not explicitly depend on the time, and (2) V must not explicitly
depend on the q̇.
From the definition, Eq. (3.8), we deduce
dH = q̇n dpn + pn dq̇n −
∂L
∂L
∂L
dqn −
dq̇n −
dt.
∂qn
∂ q̇n
∂t
(3.11)
By using the definition (3.3), we see that the second and the fourth terms above
cancel. Moreover, the same definition, combined with the Euler-Lagrange equation,
implies that
∂L
= ṗn ,
∂qn
(3.12)
so that
∂L
dt.
(3.13)
∂t
The form of this increment suggests that we consider H to be a function of qn , pn
and t (and not independently of q̇n ). Then the following partial derivatives can be
read off from Eq. (3.13):
dH = q̇n dpn − ṗn dqn −
q̇n =
∂H
∂pn
(3.14)
8
ṗn = −
∂H
∂qn
(3.15)
∂L
∂H
dH
=−
=−
.
∂t
∂t
dt
(3.16)
These are the Hamilton equations, which constitute an alternative to the EulerLagrange system.
One of the advantages of Hamilton’s approach is that one can readily define more
general transformations of the variables. The set, qn ,pn , may be replaced by another
set, Qn ,Pn , in which the Q’s and the P ’s can both be functions of all the q’s and the
p’s, and possibly of t. If the Hamilton equations, Eq. (3.14)–Eq. (3.16), remain valid
in terms of the new variables (i.e. the transformation is such that these equations,
with qn and pn replaced respectively by Qn and Pn ,are true), then we speak of a
canonical transformation. The new variables are just as acceptable as the old ones.
Pn is called the momentum canonically conjugate to Qn .
3.3
Conservation Laws
AIM: relate the existence of conservation laws to presence of symmetries (invariants) in the system. Book: chapter 2
We have seen already that when L has no explicit time dependence i.e. L is
invariant under time translations, there is a conserved quantity, E, the total energy
= 0 (see also book, §6 ).
of the system, dE
dt
It can also be shown (see book, §7) that if the system is invariant under translations, the operation ~ri0 = ~ri +~ε with ~ε being a constant (space and time independent)
vector, there is a conserved quantity, the momentum of the system. If the system
is invariant under rotations (see book §9) there is a conserved quantity, the angular
momentum.
In general it is the case that if the system has a symmetry, its properties are
invariant under a class of operations (space translations, rotations etc.) there will be
an associated conserved quantity. Finding the conserved quantities greatly facilitates
solving the equations of motion. for this reason it is important to find the symmetries
of the system, the Lagrangian.
3.4
Poisson Brackets
AIM: derive the -so called- Poisson brackets which can be regarded as the classical
equivalent of quantum mechanical commutation relations.
Lastly, we introduce the Poisson bracket between two arbitrary functions of the
canonical variables ((q, p, t) and not q̇), say F and G:
{F, G} =
∂F ∂G
∂F ∂G
−
∂qn ∂pn ∂pn ∂qn
(3.17)
The time-derivative of the arbitrary function, F , can be written
dF
dt
=
∂F
∂F
∂F
q̇n +
ṗn +
∂qn
∂pn
∂t
9
∂F ∂H ∂F
∂F ∂H
−
+
∂qn ∂pn ∂pn ∂qn
∂t
∂F
= {F, H} +
∂t
=
(3.18)
In particular, the total time-derivative of a function that does not depend explicitly on the time is given by the Poisson bracket of that function with the Hamiltonian:
we say that the Hamiltonian generates time-translations. In particular,
q̇n = {qn , H},
(3.19)
ṗn = {pn , H}.
(3.20)
These equations are impressively symmetrical between the q’s and the p’s, and they
replace the first two of the Hamilton equations, (3.14) and (3.15). It is easy to see
that
{qk , pl } = δkl ,
(3.21)
and Dirac’s recipe for the intuitive leap from classical to quantum mechanics is to
represent dynamical variables by linear operators on a Hilbert space, and Poisson
brackets by commutators (multiplied by 2π/(ih)). With this interpretation, the
general equation (3.18), as well the “quantization condition”, Eq. (3.21), have the
same forms in classical and in quantum mechanics. Both Poisson brackets and
commutators satisfy the same algebra: in particular the Jacobi identity is satisfied
by both, namely
{E, {F, G}} + {F, {G, E}} + {G, {E, F }} = 0.
10
(3.22)
Chapter 4
Central field and Two-Body
Problems
4.1
One Dimensional Problem
AIM: show application of E-L equation of motion.
As a first problem we will consider the case of a potential depending on a single
coordinate only, U (x),
L = 12 mẋ2 − U (x)
(4.1)
Since there is no explicit time dependence the total energy E = 12 mẋ2 + U (x) is a
constant of motion. Using this we arrive at
q
dx q
= E − U (x)/ 12 m
dt
which is straightforward to integrate yielding
(4.2)
ẋ =
t=
q
1
2m
dx
Z
q
E − U (x)
+ constant.
As a specific example the harmonic oscillator potential U (x) =
taken, giving
t=
q
1
2m
dx
Z
q
E − 12 kx2
+ constant.
(4.3)
2
1
2 kx
will be
(4.4)
q
With the standard substitution x = 2E/k sin φ the integration can be performed.
The full solution of the problem is thus
t =
q
x =
q
m/k(φ + φ0 ),
2E/k sin φ,
(4.5)
or more familiarly,
x=
q
2E/k sin
q
k/mt + φ0 ,
(4.6)
where φ0 is determined by the initial conditions.
11
4.2
Two-Body Problem
AIM: show equivalence 2-body and central-field problems (see book §11).
In this chapter, we shall consider a system of just two particles, at positions ~ra
and ~rb , with an interaction potential energy, V , that depends only on the relative
positions of the particles. The Lagrangian can be written
L = 12 ma~r˙ a 2 + 12 mb~r˙ b 2 − V (~ra − ~rb ).
(4.7)
We reduce this to an effective one-particle system by introducing the relative and
the centre-of-mass vectors:
~r = ~ra − ~rb ,
(4.8)
~ = ma~ra + mb~rb ,
R
M
(4.9)
and
where
M = ma + mb .
(4.10)
In terms of these variables, the Lagrangian becomes
~˙ 2 + 1 m~r˙ 2 − V (~r),
L = 21 M R
2
(4.11)
where m is defined by
1
1
1
≡
+
.
m
ma mb
(4.12)
The motion thus separates into a trivial one for the center of mass and one that is
more interesting for the relative coordinate. The latter is equivalent to that of a
particle of mass m moving in a potential V (~r).
4.3
Central Field Problems
AIM: show the method of solution for a very general class of problems (see book
§12).
A very common class of problems is that in which the force between two particles
depends on their relative distance only. This problem is equivalent to that of a
particle moving in a central field, a potential V (r) depending only on r = |~r| and
not on its orientation, not on θ and φ in polar coordinates. In polar coordinates the
kinetic energy of a particle with mass m is
T = 21 m(ṙ2 + r2 θ̇2 + r2 sin2 θ φ̇2 ).
(4.13)
The Lagrangian of the problem is
L = T − V (r) = 21 m(ṙ2 + r2 θ̇2 + r2 sin2 θ φ̇2 ) − V (r).
12
(4.14)
The first observation to make is that this L has no explicit φ dependence (is
cyclic in φ) and thus there should be an associated conserved quantity. From the
Euler-Lagrange equation in the coordinate φ one deduces that
∂L
d ∂L
=
= 0.
dt ∂ φ̇
∂φ
(4.15)
The conserved quantity, a constant of motion, is therefore
∂L
= 21 mr2 sin2 θ 2φ̇,
∂ φ̇
Mz ≡
(4.16)
The z-component of the angular momentum.
By choosing a coordinate system in which initially φ = 0 and φ̇ = 0 one obtains
Lz = 0 at t = 0 and at all later times since it it is a constant of motion. It is now
simple to solve Eq. (4.16), giving
φ̇(t) = 0
(4.17)
and thus φ(t) = 0.
Substituting Eq. (4.17) in the expression for the Lagrangian one observes that
(for this choice of coordinates) the lagrangian is cyclic in θ. From the Euler-Lagrange
equation one obtains the associated constant of motion,
M=
∂L
= mr2 θ̇
∂ θ̇
(4.18)
which is the angular momentum. From the conservation of angular momentum one
can easily derive Keplers second law (book §12). The value of M is known once the
initial conditions, θ̇(t = 0) and r(t = 0) are specified.
Next the radial motion is solved, realizing that the problem is now reduced to
one dimension. The energy of the system equals
E = 12 m(ṙ2 + r2 θ̇2 ) + V (r) = 12 mṙ2 + 21 M 2 /mr2 + V (r)
(4.19)
Using the approach discussed in the first section one arrives at
t=
dr
Z
q
2
(E
m
− V (r)) −
M2
m2 r2
+ constant
(4.20)
Alternatively the angular motion can be expressed as a function of r by starting
from
M = mr2
dθ
dθ dr
= mr2
,
dt
dr dt
(4.21)
giving
dθ
1
= M/mr2 .
dr
ṙ
(4.22)
Integrating this gives the equation for the orbit of the particle,
θ(r) =
M dr/r2
Z
q
2m(E − V (r)) − M 2 /r2
+ constant.
13
(4.23)
The integration constant is determined by the boundary condition at t = 0.
At the turning points the radial velocity ṙ = 0. In general the kinetic energy
of the particle does not vanish since the angular velocity remains finite because of
angular momentum conservation. The position of the turning points can be obtained
from solving
E = M 2 /2mr2 + V (r).
(4.24)
Please note that the direction of the angular motion (sign of θ̇) remains fixed.
4.3.1
Kepler Problem
A common problem (gravitation and coulomb) is one in which the potential is
proportional to 1/r, a problem investigated extensively by Kepler in the study of
planetary motion. The equation for the orbit of the particle, Eq. (4.23) can be solved
analytically using V (r) = −α/r ,

θ(r) = cos−1  q
M/r − mα/M
2mE + m2 α2 /M 2

 + constant
(4.25)
In the book, §13, an extensive discussion of this problem and its special cases is
given and will not be repeated here.
14
4.3.2
Rutherford Scattering
book: §§14,15,16
In the Kepler problem the interest was focussed on closed orbits or boundstates.
In Rutherford scattering one considers a charged probe coming in from a large
distance and scattering off a test charge (target). Since the size of each of the
particles is taken small compared to the typical size in the problem (the distance of
closest approach), the particles are taken to be point charces.
In a typical experiment one observes the scattered particles, in particular their
deviation in angle from a straight line trajectory. This deviation is the scattering
angle χ. In an imaginary experiment the scattering angle for each particle can be
determined with infinite precision. In practice one aims a homogeneous beam of
particles with n particles per unit area, and measures the number of particles dN
scattered between the angles χ and χ0 = χ + dχ. The partial cross-section for
scattering in this angle bin is defined as
dσ = dN/n.
(4.26)
Please note that dσ carries the units of area.
To calculate this we use Eq. (4.23) to obtain θ0 , the angle at which the projectile
reaches the distance of closest approach rmin ,
θ0 = θ(rmin ) =
Z
M dr/r2
∞
rmin
q
2m(E − V (r)) − M 2 /r2
+ constant.
(4.27)
where the constant in Eq. (4.23) is choosen such that the particle comes in at θ = 0.
Instead of using the energy and angular momentum as constants of motion it proves
to be easier to introduce instead the initial velocity of the particle,
2
,
E = 21 mv∞
(4.28)
and the impact parameter ρ,
M = mρv∞ .
(4.29)
The scattering angle can be expressed in terms of θ0 as
χ = |π − 2θ0 |,
(4.30)
where time reversal symmetry has been used (incoming part of the orbit is same as
out going part). Since θ0 depends on E and M or equivalently on ρ and v∞ , also
the scattering angle can be regarded as a function of these two, χ(ρ, v∞ ). At this
point we can return to the problem of calculating the scattering cross-section.
Assume that a particle with impact parameter ρ (ρ0 = ρ + dρ) scatters at an
angle χ (χ0 = χ + dχ), then dN equals the number of particles that were impinging
at the scatterer with impact-parameters between ρ and ρ0 ,
dN = n2πρ|dρ|.
(4.31)
The partial cross-section is
dρ dσ = 2πρ dχ.
dχ (4.32)
15
As a last step the solid angle, dω = 2π sin χdχ is introduced to obtain the differential
cross-section,
ρ dρ dσ
=
dω
sin χ dχ (4.33)
which is a function of the scattering angle.
For the case of the coulomb potential, V (r) = α/r the analytic solution can
be used for the orbits given in the previous section. After some manipulation the
details of which are given in §16 of the book, the famous Rutherford equation for
the differential cross-section can be derived,
dσ
=
dω
α
2
2mv∞
!2
1
.
sin4 21 χ
(4.34)
to describe the scattering of α particles off a heavy nucleus.
The expression for the cross-section is the same for repulsive and attractive forces,
even though the orbits of the particles are very different.
The expression for the cross-section diverges for χ = 0, even the total crosssection diverges. The reason is that even at large distances an 1/r potential is
non-negligible. The large impact parameters give rise to small angle scattering.
4.4
Small Oscillations
book §§17,19,23
Many physical systems are in a quasi-stationary state in the sense that they
are close to their lowest energy state but have not reached it yet. The system can
be described in terms of small oscillations around this equilibrium state. Also for
investigating the stability of certain solutions one often considers the resonse of the
system under small oscillations.
Assume that a system has a equilibrium state for the coordimate q = q0 . The
potential has a minimum at q0 which will be normalized to zero, V (q0 ) = 0. Introducing the coordinate x = q − q0 , for small values of x the potential can be written
as
V (x) = 12 kx2 + O(x3 ).
(4.35)
In this expantion the linear term in x vanishes since the potential is required to have
a minimum at x = 0. Similarly the kinetic energy can be expressed as
T =
1
2
a(q) q̇ 2 =
1
2
[a(q0 ) + O(x)] ẋ2 ≈ 21 mẋ2 .
(4.36)
To leading order the Lagrangian is
L = 12 mẋ2 − 12 kx2 .
(4.37)
The equation of motion is
mẍ + kx = 0,
(4.38)
16
which has the well known solutions
x(t) = a cos(ωt + α),
q
ω =
k/m
(4.39)
or equivalently
x(t) = <(Ae−iωt ),
A = ae−iα .
(4.40)
From the Lagrangian the momentum can be constructed,
p=
∂L
= mẋ.
∂ ẋ
(4.41)
As a next step the Hamiltonian is obtained,
H = pẋ − L = 12 mẋ2 + 21 kx2 = 12 p2 /m + 12 kx2 ,
(4.42)
where the r.h.s. should be used for obtaining the Hamilton equations of motion. For
this system the energy, E = 12 mω 2 a2 is a conserved quantity, a constant of motion.
For a system with N degrees of freedom the approach is analogous, see book
§19. The eigen frequencies (in general many, but not more than N ) of the system
are obtained from solving the characteristic equation,
|kik − ω 2 mik | = 0
(4.43)
with solutions
ωα ; α = 1, · · · , s ; s ≤ N.
(4.44)
For each frequency there is at least one eigen-mode or normal-mode Qα in which the
system can oscillate with frequency ω = ωα . The general solution of the problem
can be expressed in terms of these normal-modes,
xk =
X
Akα Qα .
(4.45)
α
In problems like these it often will be advantageous to use the normal-modes as the
generalized coordinates of the system.
perturbations As mentioned in the begining of this section, the harmonic oscillator Lagrangean represents only a lowest order (in the deviation from equilibrium)
approximation to the full problem. As an example of the use of perturbation theory
the following Lagrangean is considered,
L = 21 mẋ2 + 12 m0 xẋ2 − 12 kx2
(4.46)
with a small (m0 m) anharmonic term in the kinetic energy. The equation of
motion for this problem is
mẍ + kx + m0 ( 21 ẋ2 + xẍ) = 0.
(4.47)
17
Since m0 is small the solution is expanded in its powers, x = x0 + m0 x1 + m02 x2 + · · ·.
To zeroth order in m0 the equation of motion is:
mẍ0 + kx0 = 0,
(4.48)
with the solution
x0 (t) = A0 e−iωt ; ω 2 = k/m.
(4.49)
The part of the equation of motion proportial to m0 reads
mẍ1 + kx1 + 12 x˙0 2 + xx¨0 = 0.
(4.50)
Since the solution of the homogeneous equation is the solution for x0 , only the special
solution needs to be obtained,
x1 (t) = − 21 A20 /m[1 + e−2iωt ] .
(4.51)
CHECK THIS EQN.
where the explicit form of the solution for x0 (t), Eq. (4.49), has been used. This
procedure can be extended to higher order terms, but beware of some pittfalls, see
book §23.
18
Chapter 5
Special Relativity
5.1
Lorentz Transformation
The transformation between one coordinate system, (t, x, y, z), and another,
(t , x0 , y 0 , z 0 ), such that the space axes of the two systems are coincident at t0 = 0 = t,
and parallel thereafter, and such that the primed system has velocity v, along the
x-direction, with respect to the unprimed system, can be written
0
x0 = γ(x − vt)
y0 = y
z0 = z .
(5.1)
The classical assumption (Galileo, Newton) was that γ = 1, but if we only require
that the origin of the primed system have velocity v with respect to the unprimed
system, the above more general transformation is possible. What is γ? From the
uniformity of space, we see that γ must not depend on the coordinates (in a nonuniform gravitational field, near a star, for example, the properties of space are not
uniform, i.e. translation-invariant, but that is the domain of general, not of special
relativity). In the absence of gravity, γ is independent of t, x, y, z, but it can depend
on v. Space is also invariant under rotations: a rotation by 180 degrees about the
y-axis should have no measurable effect; but it changes the sign of the x and the
z coordinates, and also of the relative velocity of the primed system. Rotational
invariance therefore means that γ must be independent of the sign of v, i.e. it is a
function only of v 2 .
The unprimed system has a velocity −v, with respect to the primed system,
along the x-direction, so the inverse relation between the two systems is
x = γ(x0 + vt0 )
y = y0
z = z0 ,
(5.2)
where γ is the same as in Eq. (5.1), since it is unaffected by the change in the sign
of v.
Note that we have not assumed that the coordinate, t, in the unprimed system, is
the same as t0 , the coordinate in the primed system. Such an assumption was made,
quite explicitly, by Newton, in his Principia: “Absolute, true, and mathematical
time, of itself, and from its own nature, flows equably without relation to anything
external”. If we set t0 = t, with Newton, and do not enquire too closely what, if
anything, ‘equable flow of time’ means, then Eq. (5.1) and Eq. (5.2) together imply
γ = 1. Now if light is emitted at the origin of space- and travels along the x-axis with
19
velocity c, then at t, the light will have reached the point x = ct. From Eq. (5.1),
with γ = 1, we see that x0 = (c − v)t, which means that the speed of light, as
measured in the primed system, should be c − v.
The experimental fact that the measured speed of light is however independent
of the motion of the source, and of the observer (Michelson-Morley experiment and
the tests of ‘æther-drag’ in the solar-system), forces the conclusions t0 6= t and γ 6= 1.
Since the speed of light in the primed system is actually c, and not c − v, it follows
that x = ct in the unprimed system must correspond to x0 = ct0 in the primed
system. Filling in these values into Eq. (5.1) and Eq. (5.2), we find
ct0 = γ(c − v)t
ct = γ(c + v)t0 ;
(5.3)
and by multiplying these two equations together, we obtain
s
γ = 1/ 1 −
v2
.
c2
(5.4)
By eliminating x0 between Eq. (5.1) and Eq. (5.2), we see that
t0 = γ(t −
v
x) .
c2
(5.5)
From now on, we choose units of distance and time in such a way that c = 1, so
that we may rewrite the Lorentz transformation, as it is called, in the form
t0 = γ(t − vx)
x0 = γ(x − vt) .
(5.6)
Further, by introducing the parametrization v = tanh u, so that γ = cosh u and
γv = sinh u, we can rewrite Eq. (5.6) in the form
t0 = t cosh u − x sinh u
x0 = −t sinh u + x cosh u .
(5.7)
This form looks very much like a rotation: if the space axes are rotated by an angle
θ about the z-axis, we have
y 0 = y cos θ − x sin θ
x0 = y sin θ + x cos θ .
(5.8)
There are two important differences between Eq. (5.7) and Eq. (5.8), namely that,
in the former, hyperbolic functions replace circular ones, and the signs of the two
hyperbolic sines are the same, whereas those of the two circular sines are different.
One can regard the Lorentz transformation as a ‘hyperbolic rotation’ between the
t-axis and the x-axis.
20
5.2
Contravariant and Covariant Vectors
AIM: introduce a compact notation without which formulation of Electromagnetism will be very clumsy. The notation makes space-time equivalence manifest.
With c = 1, the relativistically invariant interval, ds, between two infinitesimally
separated points is given by
ds2 = dt2 − dx2 − dy 2 − dz 2
(5.9)
In the following, we write x0 = t, x1 = x, x2 = y, x3 = z. In terms of the metric
tensor, gµν , which is defined to be equal to be +1 if µ = 0 = ν, to be −1 if µ = i = ν,
for i = 1, 2, or 3, and to be 0 if µ 6= ν, we can rewrite Eq. (5.9) compactly as follows:
ds2 = gµν xµ xν .
(5.10)
Here the Einstein summation convention has been used, that is, the repeated indices,
µ and ν, are implicitly summed from 0 to 3 . If a new set of coordinates, x0µ ,are
linearly related to the old ones,
x0µ = Λµν xν ,
(5.11)
where the matrix, Λµν , is constant (i.e. independent of x), and is such that
µ
ν
gµν x0 x0 = gµν xµ xν ,
(5.12)
then we speak of a Lorentz transformation. In Eq. (5.11) and Eq. (5.12), there is an
implicit summation over the repeated indices. This will be a general rule: if a Greek
index occurs above (a contravariant index), and below (a covariant index), in the
same term, then a summation over the values 0, 1, 2, 3 is implied. Any quadruple of
numbers, aµ , together with the transformation law,
a0µ = Λµν aν ,
(5.13)
is said to be a contravariant Lorentz four-vector.
Since the matrix, Λµν , is independent of x, it follows, by differentiation of
Eq. (5.11), that
∂x0 µ
,
=
∂xν
so that the transformation law, Eq. (5.13), can also be written
Λµν
(5.14)
∂x0 µ ν
a .
(5.15)
∂xν
Suppose now that Φ is a Lorentz invariant function of x. Then, by the usual
chain rule for partial derivatives,
a0µ =
∂Φ
∂x0 µ
Hence the
law as the
vector, bµ ,
∂xν ∂Φ
.
(5.16)
∂x0 µ ∂xν
partial differentiation operator does not have the same transformation
contravariant vector Eq. (5.15). Rather, it is an example of a covariant
which transforms as follows:
=
∂xν
bµ =
bν .
∂x0 µ
0
(5.17)
21
The product of any contravariant vector, aµ , and any covariant vector, bµ , which we
often write simply as ab, is Lorentz invariant:
µ
a0 b 0 = a0 b 0 µ =
∂x0 µ ∂xσ ρ
a bσ = aρ bρ = ab.
∂xρ ∂x0 µ
(5.18)
If aν is a contravariant vector, and we define
aµ = gµν aν ,
(5.19)
then clearly
a0µ = gµν a0ν = gµν
∂x0ν ρσ
g aσ ,
∂xρ
(5.20)
where g ρσ , the contravariant metric tensor, is equal to gρσ , the covariant tensor that
we introduced above. However, by differentiating Eq. (5.12) with respect to xρ , we
obtain
gµν x0µ
∂x0ν
= xµ gµρ .
∂xρ
(5.21)
Next, differentiate both sides with respect to x0λ , and recall that the matrix Eq. (5.14)
is independent of x:
gµν
∂x0ν
∂xτ
=
g
ρτ
∂xρ
∂x0µ
(5.22)
Multiply both sides by g ρσ , and use the obvious fact that
gρτ g ρσ = δτσ ,
(5.23)
where the Kronecker δ is equal to 1 if both indices are equal, and to 0 if they are
not. Hence
gµν
∂xσ
∂x0ν ρσ
g
=
.
∂xρ
∂x0µ
(5.24)
By using Eq. (5.20), we see that aµ transforms as a covariant vector. In a similar
way, if bµ is a covariant vector, then
bµ = g µν bν ,
(5.25)
is a contravariant vector.
22
CONTRA- & CO-variant Summarry
Take a point at time t and at space coordinates (x, y, z). In space-time coordinates this point is specified as
Contravariant : xµ = (t, x, y, z) , by definition.
Covariant : xµ = (t, −x, −y, −z) , µ = 0, 1, 2, 3.
(5.26)
(5.27)
The relation between co- and contra- variant vectors is thus given by
x0 = x0 xi = −xi ; i = 1, 2, 3.
(5.28)
The invariant quantity is s2 = t2 − x2 − y 2 − z 2 . The metric tensor is thus given by
gµν = g
µν


 1
µ=ν=0
= −1 µ = ν = 1, 2, or 3


0
µ 6= ν
(5.29)
or




g=
1 0
0
0
0 −1 0
0
0 0 −1 0
0 0
0 −1



 .

(5.30)
The metric tensor is obviously symmetric, gµν = gνµ . The definition of a covariant
vector is
xµ = gµν xν
(5.31)
with an implicit sum over repeated indices. Please check that using Eq. (5.26) this
leads to Eq. (5.27). By definition of the metric tensor the invariant length can be
written as
s2 = t2 − x2 − y 2 − z 2
= gµν xµ xν = xµ gµν xν = xν gρν xρ
= xnu xnu = xmu xmu
This last line is the reason for introducing the co- contra-variant notation. The
permutations and relabelling are allowed since we are not dealing with quantummechanical operators but only with summations over real numbers, which commute.
In other words, a summation over individual matrix elements is implied, not the
multiplication of matrices. Always make sure that with a substitution of repeated
indices the correct number of implicit summations is implied;
(s2 )2 = xµ xµ xν xν
(5.32)
with the implicit summation convention µ = 0, · · · , 3 and ν = 0, · · · , 3, the summation runs over 4 × 4 = 16 terms, is thus not equal to
xµ xµ xµ xµ
(5.33)
since the summation convention is only µ = 0, · · · , 3 and the implicit sum contains
a total of only 4 terms.
23
The contra/co variant notation has been introdeced because of the following rule:
RULE: If a quantity contains an equal number of upper (contravariant) and lower
(covariant) indices, and if each implicit summation runs over one upper and one
lower index and all indices are summed over in this manner, the quantity is an
invariant.
Please note that the introduction of upper and lower indices is only a matter of
bookeeping. It does not contain any magic, the above rule only ensures that the
correct metric tensors are included.
Lorentz transformations
Contravariant vectors transform as:
x0µ = Λµν xν
(5.34)
where Λµν are the matrixelements of a 4 × 4 matrix, or, in sloppy language, Λµν is
a 4 × 4 matrix. The matrix elements can be expressed as
Λµν =
∂x0µ
∂xν
(5.35)
Covariant vectors transform according to
x0µ = Λµν xν
(5.36)
where
Λµν = gµρ gντ Λρ τ 6= Λµν
(5.37)
24
5.3
Mechanics of a Free Particle
In order to guess the correct Lagrangian for a free particle in relativity theory,
and hence to infer the form of relativistic mechanics, we consider again Eq. (2.17),
which we can rewrite in the rest-frame of the particle, in which the -variable is τ ,
the proper of the particle, which is also the invariant interval:
(dτ )2 = (dx0 )2 − (dx1 )2 − (dx2 )2 − (dx3 )2 .
(5.38)
In the rest-system, we write Eq. (2.17) in the form
S=
Z
b
dτ L(xi , ẋi , τ ).
a
(5.39)
The integration is taken from a space- point point, a, to another one, b. The problem
is that we do not know what L should be. It depends asymmetrically on space and
, and certainly is not a relativistic invariant. However, the action only refers to the
two space- points, a and b. This is true in all Lorentz frames, although of course
the and space coordinates of a and b do depend on the frame that is chosen. The
Hamilton variational principle simply says that the physical trajectory between a
and b is the one for which δS = 0. This is a Lorentz invariant specification of the
dynamics, and it is consistent with the principle of relativity to require that the
action be a Lorentz invariant. However, the only invariants available, out of which
the action of a free particle could be made, are the mass, m, and the invariant
interval, τ . The action must be translationally invariant, so we cannot include τ
itself, but only the invariant measure, dτ . Hence Eq. (5.39) takes the form
S=κ
Z
b
dτ,
(5.40)
a
where κ is a function of the mass of the particle, m, only. Suppose that we now
transform from the rest-system to any other Lorentz frame. From Eq. (5.38), we see
that
q
√
dτ = dt 1 − ẋ21 − ẋ22 − ẋ23 = dt 1 − v 2 ,
(5.41)
where t ≡ x0 , and v is the velocity of the new frame, with respect to the restsystem. By comparing this equation with Eq. (2.17), we can read off the form of
the Lagrangian in a general Lorentz frame:
√
L = κ 1 − v2.
(5.42)
The constant, κ, can be identified by expanding the above expression to first order
in v 2 :
L = κ − 21 κv 2 + O(v 4 ),
(5.43)
from which it follows that κ = −m, so that L reduces in the low velocity limit to
the correct non-relativistic kinetic energy,
T = 12 mv 2 ,
(5.44)
25
aside from the constant, −m, which drops out of the Euler-Lagrange equation. The
Lagrangian is thus
√
L = −m 1 − v 2 .
(5.45)
We next calculate the canonical momenta:
pi =
∂ √
mẋi
∂L
j ẋj = √
=
−m
1
−
ẋ
.
∂ ẋi
∂ ẋi
1 − v2
(5.46)
The Hamiltonian is derived from the standard formula, Eq. (3.8), yielding
H = ẋi √
√
mẋi
m
1 − v2 = √
+
m
.
2
1−v
1 − v2
(5.47)
According to the general argument of Chapter 3, H will be a constant in time, and
we now assume that it is equal to the total energy, E, as in the non-relativistic case.
This assumption has far-reaching consequences, for we see that the energy of a free
particle in its rest-frame (v = 0), is not zero, but is equal rather to the rest mass
(in units for which c 6= 1, the rest-energy is mc2 ).
The form of Eq. (5.47) is not yet in canonical form, i.e. it is not expressed in
terms of the coordinates and momenta. To remedy this, we observe that
ẋi ẋi
v2 − 1 + 1
pi pi
=
=
= A2 − 1 ,
2
2
2
m
1−v
1−v
√
where A = 1/ 1 − v 2 . Hence
s
A=
1+
pi pi
;
m2
(5.48)
(5.49)
and by using the fact that H = mA, we obtain the Hamiltonian in canonical form:
H=
q
m2 + pi pi .
(5.50)
Note that, for p~ 2 = pi pi << m2 ,
H =m+
p~ 2
p~ 4
p~ 6
−
+
O(
).
2m 8m3
m5
(5.51)
We recognise the second term as the nonrelativistic kinetic energy. The first term,
the rest-mass of the free particle, is the equivalent energy that is locked up in
a particle at rest, and which can be liberated on the annihilation of matter and
antimatter.
26
5.4
Four-Momentum
Next, we introduce the 4-velocity,
uµ =
dxµ
,
dτ
(5.52)
which is clearly a contravariant 4-vector. From Eq. (5.41), we see that
uµ = √
1
dxµ
1 − v 2 dx0
(5.53)
so that
m
= E,
1 − v2
(5.54)
m
dxi
= pi .
1 − v 2 dx0
(5.55)
mu0 = √
and
mui = √
Hence
pµ ≡ muµ = (E, p~ ),
(5.56)
is a contravariant 4-vector. The invariant,
pµ pµ = E 2 − p~ 2 ,
(5.57)
has the same value in any Lorentz frame. In the rest-frame, it is clearly equal to
m2 , so in general
E 2 = p~ 2 + m2 .
(5.58)
As a simple application of this last formula, consider the decay at rest of a
π + meson, of mass mπ , into a µ+ lepton, of mass mµ , and a neutrino, which has
mass zero. Since the 3-momentum is conserved, and it is zero before the decay, the
momenta of the µ+ and of the neutrino must be equal and opposite. Suppose that
the magnitude of the momentum of the µ+ , which can be measured, is p. The zeroth
component of the 4-momentum, the relativistic energy, is also conserved. Before the
decay, the energy of the π + is just the pion mass (Eq. (5.58) with p~ = 0); and after
the decay, it is the sum of the the neutrino energy, which is equal to p (Eq. (5.58)
with m = 0), and the muon energy. That is,
mπ = p +
q
p2 + mµ 2 .
(5.59)
This equation can be solved to give
p=
mπ 2 − mµ 2
.
2mπ
(5.60)
27
Chapter 6
Maxwell’s Equations
AIM: Arrive at a lagrangian formulation of Electro-magnetism. This implies writing
the Maxwell equations as the equations of motion for the electromagnetic fields.
6.1
Electromagnetic Fields
AIM: write the Maxwell equations in a contravariant form.
The Maxwell equations, in the presence of a charge-density, ρ(x), and a currentdensity, ~ (x), are
~ E
~ = ρ
∇.
~ B
~ = 0
∇.
~
~ ∧B
~ − ∂ E = ~
∇
∂t
~
~ ∧E
~ + ∂B = 0
∇
∂t
,
(6.1)
,
(6.2)
,
(6.3)
.
(6.4)
Note that no polarization or magnetization has been included: these are the equations in vacuo, except in so far that charge distributions are taken into account. In
~ and E,
~ and
a polarizable and magnetizable medium, one distinguishes between D
~ and H.
~ When, however, one adopts the more fundamental view that all
between B
the charges should be explicitly taken into account, this distinction need no longer
be made. As in Chapter 5, we choose units such that c = 1 ; moreover the ugly
4π that often disfigures the right-hand sides of Eq. (6.2) and Eq. (6.4) has been
removed by suitably redefining the unit of electric charge.
~ is divergence-free (since there do not seem to be any magnetic monopoles),
Since B
it follows that there is a vector field, A, whose curl it is, i.e.
~ =∇
~ ∧ A.
~
B
(6.5)
A proof of this statement can be found in the Appendix A, from which it will be
~ does not determine A
~ uniquely. Define C
~ = −E
~ − ∂ A/∂t,
~
seen that B
so that
~ ∧C
~ = 0,
∇
(6.6)
28
and from Appendix A again, we know that this implies the existence of a scalar
field, Φ, such that
~ −
−E
~
∂A
~ = ∇Φ
~ .
=C
∂t
(6.7)
Substituting Eq. (6.5) and Eq. (6.7) into Eq. (6.2) and Eq. (6.4), we find
∂2Φ
∂ ∂Φ ~ ~
+ ∇.A ] = ρ ,
− ∇2 Φ − [
2
∂t
∂t ∂t
~
∂2A
~ A
~ ] = ~ .
~ + ∇[
~ ∂Φ + ∇.
(6.8)
− ∇2 A
2
∂t
∂t
The above equations can be cast into a more compact form by writing the operator
∂ 2 /∂t2 − ∇2 = ∂ µ ∂µ = ∂ 2 , and combining the scalar and the vector potentials into
one 4-potential:
~).
Aµ = (Φ, A
(6.9)
We shall show in a moment that this 4-potential has the transformation properties
of a contravariant Lorentz vector. We may write
∂ 2 A0 − ∂0 [∂µ Aµ ] = ρ ,
~ + ∇[∂
~ µ Aµ ] = ~ .
∂2A
(6.10)
These equations cry out to be combined into one covariant 4-dimensional equation, do they not? However, there is an awkward sign difference in front of the
second term on the left. If we suppose that electric charge is a relativistic invariant,
so that the charge, e, of an electron is the same in any inertial system, then charge
density is not invariant; rather, the product, e = ρ(x)d3 x, is Lorentz invariant.
Electric current is caused by the flow of electrons: it is given by the sum of the
electric charges, multiplied by their velocities. The current density is accordingly
the product of the charge density and the velocity:
~ = ρ~v ,
(6.11)
so that if we define the 4-current density by
dxµ
≡ (ρ, ~ ) ,
dt
(6.12)
edxµ = ρ dxµ d3 x = j µ d4 x .
(6.13)
jµ = ρ
then
Now since e and d4 x are Lorentz invariants, and dxµ is a contravariant 4-vector, it
follows that j µ must also be a contravariant vector.
We can rewrite Eq. (6.11) in component form as follows:
∂ 2 A0 − ∂0 [∂ν Aν ] = j 0
∂ 2 Ak + ∂k [∂ν Aν ] = j k .
(6.14)
Now we can understand the apparently awkward sign difference, for the derivative
operator is covariant, and if we cast it into the unnatural, contravariant form, ∂ µ =
29
−∂µ for µ = 1, 2, 3 , we pick up a minus sign! Hence Eq. (6.14) can be written in
the beautiful form
∂ 2 Aµ − ∂ µ [∂ν Aν ] = j µ
(6.15)
Since j µ is a contravariant vector, it follows that Aµ must also be a contravariant
vector. In fact, Eq. (6.15), which is merely (!) a rewriting of Eq. (6.2)-Eq. (6.4),
is in a relativistically covariant form. The equations knew more than their creator,
Maxwell, did, when he invented them! To do Maxwell and Lorentz justice, they
were worried that the electromagnetic equations are not consistent with Galilean
covariance; and they did their best to understand this fact.
6.2
Electromagnetic Field Tensor
The contravariant 4-potential changes under a Lorentz transformation as follows:
Aµ0 (x0 ) =
∂x0 µ ρ
A (x) ;
∂xρ
(6.16)
that is, the transformed field, at the transformed point, is equal to the old field, at
the old point, multiplied by the Lorentz-transformation matrix. A covariant version
of the 4-potential can be defined:
Aµ (x) = gµν Aν (x) ;
(6.17)
and of course the transformation law for this is
Aµ 0 (x0 ) =
∂xρ
Aρ (x) .
∂x0 µ
(6.18)
It is convenient to introduce the second-order tensor
Fµν = ∂µ Aν − ∂ν Aµ ,
(6.19)
which transforms as follows:
Fµν 0 (x0 ) =
∂xρ ∂xσ
Fρσ (x) .
∂x0 µ ∂x0 ν
(6.20)
After these book-keeping preliminaries, we can write the Maxwell equations
Eq. (6.15) in the still compacter form
∂µ F µν = j ν .
(6.21)
The field tensor, which is manifestly antisymmetric, can be expressed directly in
terms of the electric field and the magnetic induction, for if i,j and k are restricted
to the values 1, 2, 3, then
F0k = ∂0 Ak − ∂k A0 = −∂0 Ak − ∂k A0 = Ek ,
(6.22)
Fjk = ∂j Ak − ∂k Aj = −jkl Bl .
(6.23)
and
30
~ and B,
~ and vice-versa.
Thus the field tensor can be expressed wholly in terms of E
For future reference it is helpful to give the matrix elements Fµν in matrix form,

Fµν :



µ↓ 
−→ ν
0
E1
E2
E3
−E1
0
−B3 B2
−E2 B3
0
−B1
−E3 −B2 B1
0

.




(6.24)
Please note that F µν is similar with some important sign changes,

F µν :
µ↓
0
 E
 1

 E2
E3
−→ ν

−E1 −E2 −E3
0
−B3 B2 


B3
0
−B1 
−B2 B1
0
.
(6.25)
~ and B
~ are not four vectors but three vectors, E1 = E 1 = Ex .
Also, since E
Despite the fact that the 4-potential, Aµ , is not uniquely determined by the field
tensor, it is an extremely useful quantity. If it is subjected to a gauge transformation,
i.e.
Aµ −→ A0µ = Aµ + ∂ µ G ,
(6.26)
where G is any Lorentz scalar field, then clearly the field tensor is unchanged. Such
a gauge-transformation has no physical consequences; and so any interactions with
the electromagnetic 4-potential must respect this gauge invariance. The restriction
turns out to be very important, with ramifications far outside the field of electromagnetism.
A particular restriction that is often made on the 4-potential is the so-called
Lorentz condition, viz.
∂µ Aµ = 0 .
(6.27)
By means of a gauge transformation, it is always possible to achieve the Lorentz condition, without changing the physics. For under the gauge transformation Eq. (6.26),
∂µ A0µ = ∂µ Aµ + ∂ 2 G .
(6.28)
and the right side can be made to vanish by choosing G such that ∂ 2 G = −∂µ Aµ .
The solution of this partial differential equation can be readily performed by Fourier
transformation, as in Appendix B.
When the Lorentz condition, Eq. (6.27), is satisfied, the Maxwell equations,
Eq. (6.21), become even simpler:
∂ 2 Aν = j ν .
(6.29)
31
6.3
Lagrangian Density
Let us first examine the free electromagnetic field. We shall see how the Maxwell
equation, Eq. (6.21), can be derived from a variational principle. In order to do this,
we regard the field, Aν (t, ~r ), as a continuous infinity of canonical variables. For a
given time, t, the canonical variables are labelled by ν, and the continuous variable,
~r. Since the expression for the Lagrangian will inevitably involve a summation over
all space, it is convenient to introduce a Lagrangian density:
L=
Z
d3 xL(x).
(6.30)
The action can accordingly be written
S=
Z
tb
dtL =
Z
tb
dt
Z
3
d xL(x) =
ta
ta
Z
b
d4 xL(x).
(6.31)
a
Since the action, S, is a Lorentz invariant, and d4 x is a Lorentz-invariant measure,
it follows that the Lagrangian density, L , is Lorentz-invariant.
Consider now a variation in the fields, Aµ , such that the values stay fixed at the
space-time points a and b. The resultant change in the action is
b
∂L
∂L
δAν +
δ(∂µ Aν )]
∂Aν
∂(∂µ Aν )
a
Z b
∂L
∂L
d4 x[
=
− ∂µ
]δAν .
∂Aν
∂(∂µ Aν )
a
δS =
Z
d4 x[
(6.32)
Since δAν is arbitrary between the end-points, the Hamilton variation principle,
δS = 0, implies
∂µ
∂L
∂L
−
=0
∂(∂µ Aν ) ∂Aν
(6.33)
This covariant expression is the continuum version of the Euler-Lagrange equation.
We shall now show that the Lagrangian density,
L = − 14 F µν Fµν − Aµ jµ = − 12 F µν ∂µ Aν − Aµ jµ ,
(6.34)
when inserted into Eq. (6.33), yields the Maxwell equation, Eq. (6.21). To do this,
we regard L as a function of the 4 variables Aν , and the 16 variables ∂ µ Aν . We find
∂L
= −j ν ,
∂Aν
(6.35)
and
∂Fρσ
∂L
= − 12 {F ρσ
}
∂(∂µ Aν )
∂(∂µ Aν )
= − 12 F ρσ [δρµ δσν − δσµ δρν ]
= −F µν .
(6.36)
It is clear now that the Euler-Lagrange field equations yield indeed the Maxwell
equation, when the above Lagrangian density is used. However, the Lagrangian
32
density is not uniquely determined by the requirement that the equation of motion
be correct.
In the above account, the current density, j ν , is simply treated as an externally
prescribed source of the electromagnetic field. In a more thoroughgoing theory, this
source is itself expressed in terms of canonical fields, for example those pertaining
to the electron. This leads to quantum electrodynamics, which is a very beautiful
and successful theory, which is, however, beyond the present scope.
6.4
Hamiltonian
The canonical momentum densities that correspond to the field variables, Aµ ,
are defined by
πν =
∂L
= −F 0ν = −∂ 0 Aν + ∂ ν A0 .
∂(∂0 Aν )
(6.37)
From this, it is clear that π 0 vanishes identically. Moreover, from Eq. (6.7), we see
that
Ei = −∂ 0 Ai + ∂ i A0 = −F 0i = π i .
(6.38)
Remember that ∂ i = −∂i , so that the signs above are correct. We have written the
~ = (E1 , E2 , E3 ), and of course E
~
components of the electric field as subscripts: E
and thus also ~π is emphatically not a Lorentz vector.
The Hamiltonian density is now defined by
H = π µ Ȧµ − L = π k Ȧk − L
= −F 0k ∂0 Ak + 41 F µν Fµν
= −π k [∂ k A0 − π k ] + 41 F j0 Fj0 + 14 F 0k F0k + 14 F jk Fjk
= 12 πk πk − πk ∂k A0 + 14 Fjk Fjk ,
(6.39)
in current free space. The last form is canonical, that is, the hamiltonian density
is expressed in terms of the fields, and their space derivatives, but not their time
derivatives, and the momentum densities.
In terms of the electric field and the magnetic induction, the hamiltonian may
be written in the form
H=
Z
3
d x H(x) =
Z
~ E
~ + 1 B.
~ B]
~ ,
~ ∇A
~ 0 + 1 E.
d3 x [E.
2
2
which is not a lorentz scalar.
33
(6.40)
Chapter 7
Conservation Laws
7.1
Noether Theorem
THEOREM:
For every continuous symmetry of the Lagrangian density, there is a conserved
physical quantity.
We shall illustrate this theorem by considering the invariance of the free electromagnetic Lagrangian density, Eq. (6.34) without the current density term, under
time translations, space translations, and Lorentz transformations. These invariances lead respectively to the conservation of energy, momentum, and angular momentum.
Suppose, in general, that L is unchanged under a transformation of the spacetime points x → x0 , and of the fields, Aµ (x) → A0 µ (x0 ). That is
µ
ν
L(∂ 0 A0 (x0 )) = L(∂ µ Aν (x)).
(7.1)
Define
µ
δAµ (x) = A0 (x) − Aµ (x),
(7.2)
and
ν
δL(x) = L(∂ µ A0 (x)) − L(∂ µ Aν (x)).
(7.3)
Note that A0 µ (x) and ∂ µ A0 ν (x) occur in these definitions, and not A0 µ (x0 ) and
∂ 0 µ A0 ν (x0 ) . We find therefore that
δL =
∂L
∂L
δ(Aµ ) +
δ(∂ µ Aν ),
µ
∂A
∂(∂ µ Aν )
(7.4)
where the space-time argument, x, has been suppressed. Now
∂µ[
∂L
∂L
∂L
δAν ] = ∂ µ [
]δAν +
∂ µ (δAν ),
µ
ν
µ
ν
∂(∂ A )
∂(∂ A )
∂(∂ µ Aν )
∂L
∂L
=
δAµ +
δ(∂ µ Aν ),
µ
µ
ν
∂A
∂(∂ A )
34
(7.5)
where the Euler-Lagrange equation, Eq. (6.33), has been used to obtain the last line.
On comparing this with Eq. (7.4), we see that
δL = ∂ µ [
∂L
δAν ],
∂(∂ µ Aν )
(7.6)
which is the general form of the Noether equation.
To evaluate δL(x) in detail, we replace x0 in Eq. (7.1) by x, which means that
we must replace x by x, where the transformation x → x is the inverse of x → x0 .
So in place of Eq. (7.1), we have
µ
ν
L(∂ µ A0 (x)) = L(∂ Aν (x)).
(7.7)
Thus Eq. (7.3) can be written
µ
L(∂ Aν (x)) − L(∂ µ Aν (x))
L(x) − L(x)
(x − x)ρ ∂ρ L(x) + O((x − x)2 )
−δxρ ∂ρ L(x) .
δL(x) =
=
=
=
(7.8)
In the last line use has been made of the fact that the transformation is infinetesimal
and continuous and thus (x − x)ρ = (x − x0 )ρ .
7.2
Energy Momentum Tensor
Under space-time translations, xµ → x0 µ = xµ +aµ , we know that the transformed
field at the transformed point is just the original field at the original point, i.e.
A0 µ (x0 ) = Aµ (x), so that A0 µ (x) = Aµ (x), where xµ = xµ − aµ . Thus
µ
A0 (x) = Aµ (x − a) = Aµ (x) − aν ∂ν Aµ (x) + O(a2 ).
(7.9)
Hence, for infinitesimal aµ ,
δAµ (x) = −aλ ∂λ Aµ (x),
δL(x) = −aν ∂ν L(x).
(7.10)
(7.11)
The Noether equation, Eq. (7.6), takes the form
"
ν
λ
a ∂ν L = a ∂µ
#
∂L
∂λ Aρ .
∂(∂µ Aρ )
(7.12)
Since λ only denotes an index for an implicit summation it may also be relabelled
by ν since this index has not been used on the right hand side. Bringing the terms
to one side and flipping positions of ν indices one obtains
"
0 = aν ∂µ
#
∂L
∂ ν Aρ − g µν L .
∂(∂µ Aρ )
(7.13)
Since this equation must hold for arbitrary aν , it follows that
∂µ T µν = 0,
(7.14)
35
where the energy-momentum tensor is defined by
T µν =
∂L
∂ ν Aρ − g µν L.
∂(∂µ Aρ )
(7.15)
By inserting the explicit form of the Lagrangian density for current free space,
Eq. (6.34), we obtain
T µν = −F µρ ∂ ν Aρ + 41 g µν Fρσ F ρσ .
(7.16)
The four-momentum (check that it is a contravariant vector indeed!) of the electromagnetic field is defined by
Pν =
Z
d3 xT 0ν ,
(7.17)
so that, by using Eq. (7.14), we find
Ṗ ν =
Z
d3 x∂0 T 0ν = −
Z
d3 x∂i T iν = 0,
(7.18)
on condition that the fields vanish at spatial infinity. The four quantities, P ν , are
the conserved quantities of the Noether theorem.
The zeroth component is just the Hamiltonian, since
P0 =
Z
d3 xT 00 =
Z
d3 x[−F 0ρ ∂ 0 Aρ + 41 Fρσ F ρσ ] =
Z
d3 xH = H,
(7.19)
where we have used Eq. (6.39).
The space components of the field four-momentum are
i
P =
Z
3
d xT
0i
= −
= −
Z
d3 xF 0ρ ∂ i Aρ
Z
d3 x[∂ 0 Aj − ∂ j A0 ]∂ i Aj .
(7.20)
However, from Appendix A, we see that
~ ∧ B]
~ i = [E
~ ∧ (∇
~ ∧ A)]
~ i = Ej [∂i Aj − ∂j Ai ]
[E
= [∂0 Aj − ∂j A0 ][∂i Aj − ∂ j Ai ]
= −[∂ 0 Aj − ∂ j A0 ]∂ i Aj − ∂ j {[∂0 Aj − ∂j A0 ]Ai }.
(7.21)
where we have used the free Maxwell equation to obtain the last line. Hence
P~ =
Z
~ ∧ B,
~
d3 xE
(7.22)
where P~ = (P 1 , P 2 , P 3 ). The integrand of the above equation, the field momentum
density, is called the Poynting vector.
36
7.3
Angular Momentum Tensor
The Lagrangian density is invariant under under a Lorentz transformation; but
the four potential is not, since it is a four vector. If we write
µ
(7.23)
µ
(7.24)
x0 = Λµν xν ,
then
A0 (x0 ) = Λµν Aν (x) .
In order to obtain δA, we need to calculate A0 µ (x) rather than A0 µ (x0 ). This is done
by replacing x0 by x, which means that x must be replaced by x = Λ−1 x :
µ
A0 (x) = Λµν Aν (x).
(7.25)
We consider an infinitesimal Lorentz transformation, and its inverse,
(Λ−1 )ρσ = δσρ − ερσ .
(7.26)
A0 (x) = [δνµ + εµν ][1 − ερσ xσ ∂ρ ]Aν (x),
(7.27)
Λµν = δνµ + εµν ,
Hence
µ
so that
δAµ = εµν Aν − ερσ xσ ∂ρ Aµ .
(7.28)
Since the Lagrangian density is Lorentz invariant, we find from Eq. (7.8) that
δL(x) = −ερσ xσ ∂ρ L(x) .
(7.29)
By inserting Eq. (7.28) and Eq. (7.29) into Eq. (7.6), we obtain the following:
ερσ xσ ∂ ρ L = ερσ ∂µ [
∂L
∂L
Aρ +
xσ ∂ ρ Aν ].
∂(∂µ Aσ )
∂(∂µ Aν )
(7.30)
Since εσρ is antisymmetric (see appendix C), but otherwise arbitrary, it follows that
the odd part of its coefficient in the above equation must vanish. Thus
∂µ Lµρσ = 0,
(7.31)
where
∂L
∂L
Aρ +
xσ ∂ ρ Aν − g µρ xσ L − [ρ ↔ σ]
∂(∂µ Aσ )
∂(∂µ Aν )
= J µρσ + S µρσ ,
Lµρσ =
(7.32)
where the extrinsic (or orbital) angular momentum density is
∂L
∂ ρ Aν − g µρ L] − [ρ ↔ σ]
∂(∂µ Aν )
= T µρ xσ − T µσ xρ ,
J µρσ = xσ [
37
(7.33)
and the intrinsic (or spin) angular momentum density is
∂L
Aρ − (ρ ↔ σ)
∂(∂µ Aσ )
= F µρ Aσ − F µσ Aρ .
S µρσ =
(7.34)
We define the angular momentum tensor by
Lρσ =
Z
d3 xL0ρσ .
(7.35)
In view of Eq. (7.31), we see that
L̇ρσ =
Z
d3 x∂0 L0ρσ = −
Z
d3 x∂i Liρσ = 0,
(7.36)
so that all the components of the angular momentum tensor are time independent
(they are the Noether currents that correspond to the invariance of the Lagrangian
density under Lorentz transformation). Consider
J
21
=
Z
3
d xJ
021
=
Z
d3 x[x1 T 0 2 − x2 T 0 1 ] .
(7.37)
Since T 0 i are the momentum densities, we see that J 2 1 is the third component of
the orbital angular momentum, which we often write J 3 . Then S 3 ≡ S 2 1 is the
third component of the intrinsic, or spin angular momentum. We write L2 1 ≡
L3 = J 3 + S 3 . In a similar way, by permuting 1, 2, 3 cyclically, we define the other
components: L1 = J 1 + S 1 and L2 = J 2 + S 2 .
7.4
The Photon
We will now use the tensors, that we have introduced via the Noether theorem,
to calculate the energy, momentum and angular momentum in a particular field
configuration, namely that specified by the following 4-potential:
A1 = ∆ cos Ω,
A2 = ∆ sin Ω,
A 0 = 0 = A3 ,
(7.38)
where
Ω = k µ xµ = k(x0 − x3 ).
(7.39)
This corresponds to wave propagation along the x3 axis, with unit velocity (i.e. the
speed of light). In terms of the electric and magnetic fields, and the field tensor,we
find
E1 = B2 = F 10 = F 13 = k∆ sin Ω
(7.40)
E2 = −B1 = F 20 = F 23 = −k∆ cos Ω
(7.41)
E3 = B3 = F 03 = F 12 = 0.
(7.42)
It is easy to see that ∂µ F µν = 0, so that the fields correspond to regions of space-time
in which there is no charge nor current density. Moreover, the electric and magnetic
38
fields are perpendicular to the direction of propagation, and to each other. Both
vectors have constant magnitude, but they rotate around the x3 axis. We say that
the radiation is circularly polarized.
Consider first the energy momentum tensor. We find
T 00 = 12 [(F 01 )2 + (F 02 )2 + (F 23 )2 + (F 13 )2 ] = k 2 ∆2 ,
(7.43)
so that the energy density is constant in space and time. Further,
T 0i = −∂ 0 Aj ∂ i Aj ,
so that T
01
=0=T
02
(7.44)
and
T 03 = −∂0 A1 ∂3 A1 − ∂0 A2 ∂3 A2 = k 2 ∆2 ,
(7.45)
which is the same as the energy density.
Now suppose that a target is placed in this electromagnetic radiation, with crosssectional area α orthogonal to x3 , and suppose that it absorbs all the radiation that
falls on it, during a time T . The energy that is transferred to the target is obtained
by integrating T 00 over all the radiation that will fall on to the target, during the
specified period. This is, however, a volume of cross-sectional area α and length T
(since the speed of light is unity). This absorbed energy is αT k 2 ∆2 . Similarly, we
can calculate the momentum that is transferred to the target, in the same time, by
integrating T 0i over the same volume. Clearly the momentum is purely in the x3
direction, and it is also equal to αT k 2 ∆2 .
In the 17th century, Newton postulated that light consists of a stream of particles.
This idea fell into disrepute, largely because of the success of the wave-hypothesis
of Huygens. The Maxwell equations are wave equations par excellence. At the
beginning of the 20th century, however, the particle theory of light was reintroduced
by Planck, in order to deal with the ultra-violet catastrophe of black-body radiation,
and by Einstein, in connection with the photo-electric effect, in which it was clear
that the particles of light, photons, each had an energy that was proportional to the
frequency.
Suppose that there are N photons, each of mass m, in the volume of the radiation
that falls on our target in the time, T . We calculated that both the energy and the
momentum of these N photons is αT k 2 ∆2 . Hence the energy, E, and the momentum,
p, per photon are equal to one another. However, from Eq. (5.58), we know that
m2 = E 2 − p2 , from which it follows that the mass of a photon is zero.
We will look lastly at the angular momentum transfer to the target. Since
T 01 = 0 = T 02 , it follows that J 3 vanishes. However,
S 021 = A1 F 02 − A2 F 01 = k∆2 .
(7.46)
The total angular momentum that is contained in the volume of radiation, that falls
on the target in the time T , is therefore wholly intrinsic, and it is equal to αT k∆2 .
The ratio of the energy to the angular momentum for the N photons, and thus for
each photon separately, is k = 2πν, where ν is the frequency of the radiation. Since
we know, by studying the energy of photo-electrons, that the energy of a photon is
proportional to the frequency, it follows that the intrinsic angular momentum, or
spin, of a photon is independent of the frequency. This spin does not in fact depend
on any of the accidental features of the field configuration. It is a fundamental
unit of angular momentum, usually written as h/(2π), where h is called Planck’s
constant.
39
Chapter 8
Point Charge
AIM: solve the dynamics of Electromagnetic fields interacting with massive particles.
8.1
Lagrangian
~ is eE.
~ The force
The force on a particle of charge e, at rest in an electric field E,
on a moving charge can be obtained by applying a Lorentz transformation to this
static situation, but some care is needed, since a three-dimensional force is not a
relativistic 4-vector. Consider the infinitesimal quantity defined by
dpµ ≡ eF µν dxν ,
(8.1)
where e is the charge, and x the coordinate of an elementary charge, say an electron,
in an electromagnetic field, F µν . This is manifestly a Lorentz 4-vector. If we now
divide by dt, which is not a Lorentz scalar, but rather the zeroth component of a
4-vector, we obtain the non-covariant relation,
dxν
dpµ
= eF µν
,
dt
dt
(8.2)
the spacelike components of which (i.e. µ = j = 1, 2, 3) are
dpj
= eF j0 + eF jk vk ,
dt
(8.3)
where ~v is the velocity of the electron. By means of Eq. (6.22) and Eq. (6.23), we
can write this, in 3-vector notation, as
~ + e~v ∧ B
~ .
p~˙ = eE
(8.4)
If the electron is instantaneously at rest, ~v = 0, then the right-hand side reduces to
~ the static force, which implies that p~ is the 3-momentum of the electron. We
eE,
know however that p~ constitutes the space components of the 4-momentum, which
is a contravariant 4-vector. Hence Eq. (8.1) is in fact the covariant relation between
an increment of the 4-momentum of an electron and the electromagnetic field that
interacts with it.
To summarize the sequence of the reasoning, Eq. (8.1) is the definition of a
contravariant vector, which we identify with an increment of the 4-momentum of
40
the electron, for in the rest-frame of the electron the space components coincide
with the known static force exerted on the electron by the electric field. Since
we know that the 4-momentum transforms under a Lorentz transformation as a
contravariant vector, we can assert that Eq. (8.1) identifies the incremental change
of the 4-momentum in an arbitrary inertial frame. In a frame in which the electron
has velocity ~v , we see from Eq. (8.4) that the Lorentz force on a moving point charge
in an electromagnetic field must be
~ + e~v ∧ B.
~
F~ = eE
(8.5)
The zeroth component of Eq. (8.1), is also interesting, since it gives the incremental
change in the energy of the electron, E. Since F 00 = 0, the zeroth component of
Eq. (8.2) can be written
dxk
d
E = eF 0k
dt
dt
~ .
= e~v .E
(8.6)
The rate of change of the energy of the electron is equal to the rate at which the
Lorentz force does work, which is F~ .~v , and so we see that Eq. (8.5) and Eq. (8.6)
are consistent with one another.
We shall now show that the Lorentz force, Eq. (8.5), can be derived from the
following action:
S = −m
Z
b
a
q
dxµ dxµ − e
Z
b
a
dxµ Aµ .
This can be rewritten
Z tb √
Z
i
i
S = −m
dt 1 − ẋ ẋ − e
ta
tb
dt[A0 − ẋi Ai ],
(8.7)
(8.8)
ta
where the indices of x and A are contravariant ones, and where the repeated Latin
index, i, is to be summed from 1 to 3.
From Eq. (8.8),we can read off the Lagrangian as
√
L = −m 1 − ẋi ẋi − eA0 + eẋi Ai ;
(8.9)
and so the canonical momentum is
p̃i ≡
∂L
mẋi
√
+ eAi .
=
∂ ẋi
1 − v2
(8.10)
Note that this canonical momentum, p̃i , is not the same as the ordinary momentum,
pi , the time derivative of which occurs in Eq. (8.4). Indeed p̃i = pi + eAi . L depends
explicitly on xi via the potentials A0 and Ai . Thus we calculate
j
∂L
∂A0
j ∂A
=
−e
+
e
ẋ
.
∂xi
∂xi
∂xi
(8.11)
The Euler Lagrange equation (2.16) reads then
j
d
mẋi
∂A0
i
j ∂A
[√
+
eA
]
+
e
−
e
ẋ
= 0.
dt 1 − v 2
∂xi
∂xi
41
(8.12)
Now
dAi
∂Ai ∂Ai j
=
+ j ẋ ,
dt
∂t
∂x
(8.13)
so it follows from Eq. (8.12) that
mẋi
∂A0
∂Ai j
∂Aj j
d
∂Ai
[√
−
e
−
e
ẋ
+
e
ẋ .
]
=
−e
dt 1 − v 2
∂t
∂xi
∂xj
∂xi
(8.14)
Since
Ei = −
∂Ai ∂A0
−
,
∂t
∂xi
(8.15)
and
i
j
~ i = (~v ∧ (∇
~ ∧ A))
~ i = ∂A ẋj − ∂A ẋj ,
(~v ∧ B)
∂xi
∂xj
(8.16)
we see that the right-hand side of Eq. (8.14) is indeed the i’th component of the
right-hand side of Eq. (8.5). This concludes the demonstration that the Lagrangian
(8.9) does indeed yield the Lorentz force.
8.2
Hamilton’s Equations
It is convenient to make the transition from the Lagrangian to the Hamiltonian
formalism. The Hamiltonian is
√
mẋi ẋi
i i
+
eA
ẋ
+
m
1 − v 2 + eA0 − eẋi Ai
H = p̃i ẋi − L = √
2
1−v
m
+ eA0 .
= √
2
1−v
(8.17)
This is not yet in the canonical form, in which the Hamiltonian must be expressed as
a function of the coordinates, xi , and the canonical momenta, p̃i . From Eq. (8.10),
we have
m2 + (p̃i − eAi )(p̃i − eAi ) = m2 +
m2
m2 v 2
=
,
1 − v2
1 − v2
(8.18)
so that
0
H = eA +
q
m2 + (p̃k − eAk )(p̃k − eAk ) ,
(8.19)
which is indeed canonical.
The Hamilton equations (3.8) take the form
ẋi =
∂H
p̃i − eAi
=q
,
∂ p̃i
m2 + (p̃k − eAk )(p̃k − eAk )
∂H
∂A0
e(p̃j − eAj )
∂Aj
p̃˙ i = − i = −e i + q
.
i
∂x
∂x
m2 + (p̃k − eAk )(p̃k − eAk ) ∂x
42
(8.20)
(8.21)
If the Lagrangian (8.9) does not depend explicitly on the time, the Hamiltonian
is independent of time: it is one of the constants of the motion. In such a case,
it is advantageous to use Eq. (8.19) to simplify the above two equations. From
Eq. (8.20), we have immediately
[H − eA0 ]ẋi = p̃i − eAi ,
(8.22)
j
0
∂A
∂A
p̃˙ i + e i = eẋj i .
∂x
∂x
8.3
(8.23)
Constant Magnetic Field
In this section, we consider a charged point mass in a constant magnetic field.
This is of relevance to the motion of elementary particles in a bubble chamber, or
in an accelerator. Suppose that A0 = A2 = A3 = 0 but A1 = −Bx2 , where B is
~ and B
~ vanish, except B3 = −∂2 A1 = B.
constant. Then all the components of E
The Hamilton equations, Eq. (8.22) and Eq. (8.23), are
H ẋi = p̃i + eBx2 δi1 ,
(8.24)
p̃˙ i = −eB ẋ1 δi2 .
(8.25)
These equations reduce to
H ẍ1 =
eB ẋ2
H ẍ2 = −eB ẋ1
H ẍ3 = 0.
(8.26)
The velocity in the x3 direction is constant, and to solve the first two equations, we
set ζ = x1 + ix2 , so that
H ζ̈ = eB(ẋ2 − iẋ1 ) = −ieB ζ̇.
(8.27)
The solution is
ζ = α + βe−iωt ,
(8.28)
√
where ω = eB/H. However, H = m/ 1 − v 2 , so that
√
eB 1 − v 2
ω=
.
m
(8.29)
From Eq. (8.28), we see that
ζ̇ = −iωβe−iωt ;
(8.30)
and hence
2
v⊥
≡ (ẋ1 )2 + (ẋ2 )2 = |ζ̇|2 = ω 2 |β|2 .
(8.31)
Thus v⊥ is constant, and since
2
v⊥
|ζ − α| = |β| = 2 ,
ω
2
2
(8.32)
43
it follows that the motion in the (1, 2) plane is a circle, of radius R = v⊥ /ω. In view
of the constant velocity in the x3 direction, it follows then that the trajectory of the
particle is a spiral, or helix, around the direction of the constant magnetic field.
~ so that v⊥ = v,
In the case that ẋ3 = 0, i.e. the motion is perpendicular to B,
the particle travels with constant speed in a circle of radius
vm
√
.
(8.33)
R=
eB 1 − v 2
In a synchrotron, in order to keep the radius of the trajectory constant, one has
to increase the bending magnetic field, as the speed of the particle gets closer and
closer to that of light.
8.4
Constant Electric Field
Consider next a point charge in a constant electric field, E. For convenience, we
choose the x3 axis to be parallel to the field, and the origin to be such that xi = 0
when t = 0, i = 1, 2, 3. Let the initial value of the particle velocity be (v1 , v2 , v3 ),
~ = 0, and
and choose the x2 axis such that v2 = 0. The fields can be specified by A
Φ = A0 = −Ex3 . The Hamilton equations of motion are
(H + eEx3 )ẋi = p̃i ,
(8.34)
p̃˙ i = eEδi3 .
(8.35)
We can integrate Eq. (8.35) immediately and then combine it with Eq. (8.34):
(H + eEx3 )ẋi = Hvi + eEtδi3 ,
(8.36)
and the Hamiltonian,
m
− eEx3 ,
H=√
2
1−v
(8.37)
is a constant.
For i = 2 this implies that x2 ≡ 0, i.e. that the motion is wholly in the (x1 − x3 )
plane. For i = 3, we rewrite Eq. (8.36) as follows:
d
d
[(H + eEx3 )2 ] = [(Hv3 + eEt)2 ],
dt
dt
(8.38)
from which we deduce
(H + eEx3 )2 − (Hv3 + eEt)2 = H 2 [1 − v32 ].
(8.39)
By dividing the i = 1 and the i = 3 components of Eq. (8.34) , we obtain
Hv3 + eEt
dx3
=
.
1
dx
Hv1
(8.40)
On using Eq. (8.39) to eliminate t in favour of x3 , we find the solution
1+
eEx3 q
eEx1
= 1 − v32 cosh[
],
H
Hv1
(8.41)
44
which is a catenary, the same figure as that assumed by a uniform cord, hanging in
a uniform gravitational field.
In the above work, we have relied heavily on the fact that Ḣ = 0. The actual
value of the Hamiltonian can be found by computing Eq. (8.17) at t = 0:
m
,
(8.42)
H=q
1 − v12 − v32
since at t = 0, x3 = 0, so that then also A0 = 0. It is important to realise that
the velocities do change, and that although H is indeed time-independent, the two
terms in the last line of Eq. (8.17) separately do depend on the time.
At the risk of being tedious, but in the hope of avoiding future errors, let us
~ =0
now treat the configuration A0 = A1 = A2 = 0 and A3 = −Et. Since again B
and Ei = Eδi3 , this is the same physical problem, although at first sight it looks
different, since now
m
H=√
,
(8.43)
1 − v2
but here H is not time-independent, since L depends on t explicitly, through A3 , so
Ḣ = −∂L/∂t 6= 0.
The Hamilton equations of motion are
√
mẋi
= p̃i + eEtδi3 ,
1 − v2
(8.44)
together with p̃˙ i = 0, so now all the p̃i are time-independent, but v is not. Since
m
√
=
1 − v2
s
=
q
m2 +
m2 ẋi ẋi
1 − v2
m2 + p̃21 + p̃22 + (p̃3 + eEt)2 ,
(8.45)
we can cast Eq. (8.44), i = 3, into the form
ẋ3 = q
p̃3 + eEt
m2 + p̃21 + p̃22 + (p̃3 + eEt)2
.
(8.46)
This equation can be integrated to give
(C + eEx3 )2 − (p̃3 + eEt)2 = C 2 − p̃23 ,
(8.47)
where C is an integration constant. This equation has the same algebraic form as
Eq. (8.39), the only difference being that the constants have other names. We invite
the reader to complete the solution, and to show that the catenary, Eq. (8.41), can
be recovered.
It is instructive to observe that the two configurations are related by a gauge
~ = 0, and A0 = −Ex3 (our first example),
transformation of the form Eq. (6.26). If A
then it is sufficient to take G = Ex3 t, in Eq. (6.26), to see that A00 = 0 and
A03 = ∂ 3 G = −∂3 G = −Et (our second example).
Although the above problem is relatively straightforward, two lessons can be
learned:
45
[1]
The Hamiltonian method is easier if ∂L/∂t = 0 ,
for then the Hamiltonian is a constant of the motion.
[2]
One can sometimes use the gauge freedom to ensure that Ḣ = 0.
46
Chapter 9
Radiation from a Moving Charge
9.1
Solution of the Wave Equation
In a region of space-time in which there are no sources (i.e. where j ν = 0), the
Maxwell equations Eq. (6.29) in Lorentz gauge Eq. (6.27) reduce to
∂ 2 Aµ = 0 .
(9.1)
Solutions of this equation can be found by Fourier transformation:
Aµ (x) =
Z
d3 keikx
3
X
a(~k, λ)εµ (~k, λ),
(9.2)
λ=0
where kx = kν xν = k 0 t − ~k.~x. The four polarization vectors, ε(~k, λ), λ = 0, 1, 2, 3,
correspond to the four Lorentz components of Aµ . The spacelike components are
as in Appendix B, that is, they are mutually orthogonal unit spacelike vectors,
such that ε(~k, 3) is parallel to ~k. ε(~k, 0) is a unit vector in the timelike direction,
(1, 0, 0, 0).
From Eq. (9.1) we have
2
µ
∂ A =−
Z
3
ikx 2
d ke
k
3
X
a(~k, λ)εµ (~k, λ) = 0 .
(9.3)
λ=0
Since the Fourier components are independent, k 2 = (k 0 )2 − ~k.~k = 0, or equivalently,
k 0 = |~k| ≡ k .
(9.4)
This constraint is called the mass-shell condition; it corresponds to the fact that
photons are massless.
The Lorentz condition Eq. (6.27) implies
∂µ Aµ =
Z
h
i
d3 keikx k a(~k, 0) − a(~k, 3) = 0 ;
(9.5)
and this means simply that a(~k, 0) = a(~k, 3). Under this constraint, any solution of
the Maxwell equations can be written in the form Eq. (9.2).
47
9.2
Retarded Potentials
The scalar potential engendered by a static charge, e, at a distance r is e/(4πr).
If we have a static charge distribution, ρ(~x 0 ), then the scalar potential at the point
x is given by the integral
Φ(~x ) =
1 Z 3 0 ρ(~x 0 )
dx 0
.
4π
|~x − ~x |
(9.6)
If however the charge distribution is time-dependent, we expect the above integral
to be modified, since the potential should not be determined by the instantaneous
charge-distribution, but rather by the retarded charge-distribution, i.e. the distribution as it was when a light-signal was emitted from ~x 0 , that has just arrived at ~x .
Thus
Φ(t, ~x ) =
1 Z 3 0 ρ(t − |~x 0 − ~x |, ~x 0 )
dx
.
4π
|~x 0 − ~x |
(9.7)
In Appendix E it is shown that Eq. (9.6) is indeed a solution of the differential
equation
∇2 Φ(~x ) = −ρ(~x ) ;
(9.8)
while Eq. (9.7) is a solution of the time-dependent wave-equation
∂ 2 Φ(t, ~x ) = ρ(t, ~x ) ,
(9.9)
where ∂ 2 = ∂ µ ∂µ . The rather intuitive derivations given above are thus put on a
firm foundation. Moreover, since the Maxwell equation for the 4-potential has the
form
∂ 2 Aµ = j µ ,
(9.10)
when the Lorentz gauge condition Eq. (6.27) is satisfied, it follows that we can use
the analysis of Appendix E to write the retarded solution to Eq. (9.10) in the form
Aµ (x) =
1 Z 3 0 j µ (t − |~x 0 − ~x |, ~x 0 )
dx
.
4π
|~x 0 − ~x |
(9.11)
This is not the most general solution of the Maxwell equation, Eq. (9.10), since any
solution of the homogeneous equation, Eq. (9.1), may be added to Eq. (9.11). The
retarded solution is assumed, however, to give the physical, or causal solution of the
Maxwell equation.
9.3
Liénard-Wiechert Fields
Let us specialize the general expression Eq. (9.11) for the retarded potential to
the case that the source is a single point charge, for example an electron, which is
~ that depends on the time. The corresponding charge density is
at a position X,
~ ,
ρ(x0 ) = j 0 (x0 ) = eδ 3 (~x 0 − X)
(9.12)
48
where e is the electron’s charge and the δ-function is Dirac’s distribution, which
can be thought of as the limit of an infinitely high, infinitely sharply-peaked function, located at the point where its argument is zero. The corresponding current
distribution is
~ ,
~ (x0 ) = e~v δ 3 (~x 0 − X)
(9.13)
~˙ the velocity of the electron.
where ~v = X,
With these expressions for the charge and current densities, we can evaluate the
integral in Eq. (9.11). A difficulty, however, is that the time retardation means that
the zeroth, or time component of the argument of j µ depends on ~x 0 . The easiest
way to keep things straight is to introduce a t0 -integration, and a “time-retardation”
delta-distribution. For the scalar potential, this yields
1
e Z 4 0 3 0 ~ 0
d x δ (~x − X (t ))δ(t0 − t + |~x 0 − ~x |) 0
4π
|~x − ~x |
Z
e
1
~ 0 ) − ~x |)
=
dt0 δ(t0 − t + |X(t
.
~ 0 ) − ~x |
4π
|X(t
A0 (x) =
(9.14)
Before the last integration is performed, the integration variable is changed from
~ 0 ) − ~x |. However,
t to ť = t0 − t + |X(t
0
"
#
∂ ~ 0
) − ~x | dt0
dť = 1 + 0 |X(t
∂t


~ 0 ) − ~x ).X(t
~˙ 0 )
(
X(t
 dt0 .
= 1 +
0
~
|X(t ) − ~x |
(9.15)
In other words,
dt0 = dť/κ ,
(9.16)
where
κ = 1 − ~n.~v ,
(9.17)
with
~n =
~ 0)
~x − X(t
.
~ 0 )|
|~x − X(t
(9.18)
Thus ~n is a unit vector pointing from the point charge, as it was at the retarded
time, to the field point, ~x . The integration over ť can now be performed trivially,
and we find
e
0
.
(9.19)
A (x) =
4πκR ret
Here
~ 0 )| ,
R = |~x − X(t
(9.20)
49
and the suffix ‘ret’ means that t0 , which occurs implicitly in the definition of κ and
of R, must be replaced by the retarded time, i.e. by the solution of the equation
t = t0 + R(t0 ) .
(9.21)
Similarly, for the vector potential,
"
e~v
~
A(x)
=
4πκR
#
.
(9.22)
ret
The expressions Eq. (9.19) and Eq. (9.22) are called the Liénard-Wiechert potentials.
~ of
They give the 4-potential at the point ~x , at time t, in terms of the position X
0
0
~ ) − ~x |.
the point charge, as it was at the time t = t − |X(t
From the 4-potential, the Liénard-Wiechert fields can be calculated. They are
~ =
E
e
4πκ3
(
~n ∧ [(~n − ~v ) ∧ ~v˙ ] (1 − v 2 )(~n − ~v )
+
R
R2
2
˙
˙
~ = − e ~n ∧ ~v + ~n ∧ (~v ∧ ~v ) + (1 − v )~v
B
4πκ3
R
R2
(
)
,
(9.23)
)
.
(9.24)
In these expressions, as in subsequent ones, the specification that the quantities on
the right are to be evaluated at the retarded time is implicit. The details of this
calculation are to be found in Appendix F, as is the proof that the above LiénardWiechert fields are related by
~ = ~n ∧ E
~ .
B
(9.25)
As can be seen from the above expressions, both the electric field and the magnetic induction have contributions that behave like R−2 , and others that behave like
R−1 . The former give the near-field, and are all that is left if the acceleration of the
charge, ~v˙ , is zero:
2
~ = e(1 − v ) (~n − ~v ) ;
E
4πκ3 R2
(9.26)
~ can always be obtained from E
~ by means of Eq. (9.25). These terms
while B
correspond to electrostatics and magnetostatics, i.e. to a charge moving at constant
speed and so giving rise to a constant current. The fields fall off according to the
famous inverse square law of distance. Indeed, with v << 1, and to lowest order
~ and O(v) for B),
~ and with ~v˙ = 0, we find
(O(1) for E
~ = e~n ,
E
4πR2
(9.27)
~ = − e~n ∧ ~v .
B
4πR2
(9.28)
These terms are not surprising: they arise essentially from the application of a
Lorentz transformation to the static situation.
50
The really amazing thing about the Liénard-Wiechert fields lies in the R−1 or
radiation terms. These are present only if the charge undergoes acceleration. For
large R, we can neglect the R−2 terms, leaving
~ =
E
e
~n ∧ [(~n − ~v ) ∧ ~v˙ ] ,
4πκ3 R
(9.29)
together with Eq. (9.25). In the radiation zone (far from the accelerated source), the
electric field is at right angles to the direction of the source (i.e. orthogonal to ~n),
and also to the magnetic induction, which is also at right angles to the direction of
the source. In a word, the radiation is transverse. That we have to do with radiation
is clear, since the energy and momentum densities are proportional to the squares
of the field strengths, so to R−2 , and this means that there is a continual streaming
of energy and momentum into space. We shall illustrate this in the nonrelativistic
limit.
For nonrelativistic motion of the electron, i.e. v << 1, the Liénard-Wiechert
fields reduce to
~ = e ~n ∧ (~n ∧ ~v˙ ) + e ~n ,
(9.30)
E
4πR
4πR2
~ = − e ~n ∧ ~v˙ .
B
4πR
(9.31)
For large R, we neglect the static R−2 term, and we find
˙
~ E
~ = e~n ∧ ~v
E.
4πR
"
#2
~ B
~ .
= B.
(9.32)
Now according to Eq. (6.40), the energy density of the field (the hamiltonian density)
is
~ E
~ + B.
~ B]
~ + E.
~ ∇A
~ 0.
H = 12 [E.
(9.33)
The last term here does not contribute in the radiation zone. Indeed, even without
the nonrelativistic approximation, we can calculate from Eq. (9.19) that
2
~ ∇A
~ 0 = e(1 − v ) (~n − ~v ).~v .
E.
16π 2 κ5 R4
(9.34)
Hence the hamiltonian density in the radiation zone is
H=
e2 v̇ 2 sin2 θ
,
16π 2 R2
(9.35)
where θ is the angle between ~n and ~v˙ , and this means that, in a shell about the
source of radius R and infinitesimal thickness δR, the instantaneous radiant energy
δE is
δE = 2π
Z
R+δR
2
R dR
R
Z
0
π
sin θdθ
e2 v̇ 2 sin2 θ
16π 2 R2
1 2 2
=
e v̇ δR .
6π
(9.36)
51
The energy in the shell is independent of R: it propagates outwards, without loss,
with the speed of light — indeed it is light!! (Or radio waves, or IR, UV, γ-radiation,
or whatever!) Note that the R−2 terms in the fields do not contribute to this energy
in the limit of large R: they fall off too rapidly.
~ n . We define the
Consider lastly a collection of point charges, en , located at X
dipole moment of this collection of charges by
~ =
D
X
~n .
en X
(9.37)
n
In the nonrelativistic limit, and in the radiation zone (i.e. v << 1 and R large compared with the dimensions of the source), we have, from Eq. (9.30) and Eq. (9.31),
~¨
~n ∧ (~n ∧ D)
~
,
E=
4πR
(9.38)
~¨
~n ∧ D
~
B=−
.
4πR
(9.39)
and
As we have seen, the electric field and the magnetic induction are transverse to the
propagation direction, are orthogonal to each other, and are equal in amplitude. At
~¨ we have
an angle θ from the direction of D,
E2 = B2 =
D̈2 sin2 θ
.
16π 2 R2
(9.40)
Thus the radiation is most intense at right angles to the second derivative of the
dipole moment, as anyone who has played with a radio transmission antenna knows!
52
Appendix A
Triple Vector Product
The vector product of two three-vectors can be written
(~a ∧ ~b )i = εijk aj bk ,
(A.1)
where the repeated indices, j and k, are summed over 1, 2, 3. This is a general rule:
if a Latin index is repeated in the same term, a summation is implied. The symbol,
εijk , is defined to be +1 if i, j, k is a cyclic permutation of 1, 2, 3, and −1 for an
anticyclic permutation. It is 0 if two or more of the indices are the same. From
Eq. (A.1), we see that, for example, (~a ∧ ~b )1 = a2 b3 − a3 b2 .
A triple vector product can be calculated as follows:
[~a ∧ (~b ∧ ~c )]i =
=
=
=
εijk aj εklm bl cm
εijk εlmk aj bl cm
(δil δjm − δim δjl )aj bl cm
aj b i c j − aj b j c i .
(A.2)
The summation over j in the last line corresponds to a scalar product of two vectors.
In vector notation, we have shown that
~a ∧ (~b ∧ ~c ) = ~b (~a .~c ) − ~c (~a .~b ).
(A.3)
From Eq. (A.2), we can immediately evaluate
~ ∧A
~ )]i = vj ∂i Aj − vj ∂j Ai ,
[~v ∧ (∇
(A.4)
Note that the order of the factors is important. Further,
~ ∧ (∇
~ ∧A
~ )]i = ∂i ∂j Aj − ∂j ∂j Ai .
[∇
(A.5)
In vector notation, this last equation reads
~ ∧ (∇
~ ∧A
~)=∇
~ (∇
~ .A
~ ) − ∇2 A
~.
∇
(A.6)
53
Appendix B
Potentials
B.1
Vector Potential
~
~ B
~ = 0, then there exists a vector field, A(x),
~
If a vector field, B(x),
satisfies ∇.
~ =∇
~ ∧ A.
~
such that B
Proof
~
Given B(x),
we write a three-dimensional Fourier transform as follows,
~
B(x)
=
Z
d3 keikx
3
X
b(~k, λ)~ε (~k, λ).
(B.1)
λ=1
Here ~ε (~k, λ), λ = 1, 2, 3, are three orthonormal vectors, ~ε (~k, 3) being parallel to ~k,
the others being therefore orthogonal to it. kx means ~k.~x, in this, and in subsequent
formulas. Clearly
~ B(x)
~
∇.
=i
Z
d3 keikx kb(~k, 3),
(B.2)
~ B
~ = 0 is accordingly
where k = |~k|. The necessary and sufficient condition that ∇.
~
that b(~k, 3) = 0, k 6= 0. Define now A(x)
as follows:
~
A(x)
=
Z
d3 keikx
3
X
a(~k, λ)~ε (~k, λ),
(B.3)
λ=1
where a(~k, 1) = −ib(~k, 2)/k, a(~k, 2) = ib(~k, 1)/k, for k 6= 0, while a(~k, 3) is arbitrary.
Then
~ ∧ A(x)
~
∇
= i
Z
d3 keikx
3
X
a(~k, λ)~k ∧ ~ε (~k, λ)
λ=1
= i
=
Z
Z
d3 keikx k[a(~k, 1)~ε (~k, 2) − a(~k, 2)~ε (~k, 1)]
d3 keikx [b(~k, 1)~ε (~k, 1) + b(~k, 2)~ε (~k, 2)].
~
The last expression can be seen to be equal to B(x),
since b(~k, 3) = 0, k 6= 0.
54
(B.4)
B.2
Scalar Potential
~
~ ∧C
~ = 0, then there exists a scalar field, Φ(x),
If a vector field, C(x),
satisfies ∇
~ = ∇Φ.
~
such that C
Proof
In terms of the Fourier transform,
~
C(x)
=
Z
d3 keikx
3
X
c(~k, λ)~ε (~k, λ),
(B.5)
λ=1
we have
~ ∧ C(x)
~
∇
=i
Z
d3 keikx k[c(~k, 1)~ε (~k, 2) − c(~k, 2)~ε (~k, 1)].
(B.6)
For this to vanish, it is necessary and sufficient that c(~k, 1) = 0 = c(~k, 2), for k 6= 0.
Define now the scalar field, Φ(x), by means of the Fourier transform
Φ(x) =
Z
d3 keikx φ(~k),
(B.7)
where φ(~k) = −ic(~k, 3)/k, for k 6= 0. Then it follows that
~
∇Φ(x)
=
Z
d3 keikx c(~k, 3)~ε (~k, 3).
(B.8)
~
We recognize the right-hand side of the above equation as C(x).
55
Appendix C
Invariant Measure
If x0 and x are related by a Lorentz transformation, and F (x) is a Lorentz scalar,
then
Z
4 0
0
d x F (x ) =
Z
d4 xF (x)
(C.1)
Proof
By the usual rules for changing variables in a multiple integral, we have
Z
d4 x0 F (x0 ) =
Z
d4 x
∂(x0 0 , x0 1 , x0 2 , x0 3 )
F (x) ,
∂(x0 , x1 , x2 , x3 )
(C.2)
where the Jacobian can be written
∂(x0 0 , x0 1 , x0 2 , x0 3 )
= det(Λµν ) ,
∂(x0 , x1 , x2 , x3 )
(C.3)
in which the matrix Λ is defined by
Λµν
∂x0 µ
=
,
∂xν
(C.4)
that is, it is the Lorentz transformation matrix itself. We shall show that the
determinant of this matrix is always unity, thus completing the proof of Eq. (C.1).
For a rotation about one of the space axes by an angle θ,
det(Λµν ) = cos2 θ + sin2 θ = 1 ;
(C.5)
while a Lorentz transformation along one of the space axes by a hyperbolic boost,
u, gives
det(Λµν ) = cosh2 u − sinh2 u = 1 .
(C.6)
Since a general Lorentz transformation can be built up by compounding rotations
about the space axes (Euler angles), and a pure Lorentz boost along one space axis,
and since the determinant of the product of a number of matrices is equal to the
product of the determinants of each matrix, each of which is equal to unity, we see
finally that the determinant of the most general Lorentz transformation matrix is
also equal to unity.
56
Consider next an infinitesimal Lorentz transformation along the x1 axis, described by a hyperbolic angle, u. The transformation matrix is
Λµν = δνµ + εµν ,
(C.7)
where ε01 = −u = ε10 , and the other elements of εµν are zero. For an infinitesimal
rotation by an angle, θ, about the x3 -axis, on the other hand, Eq. (C.7) holds, with
ε12 = −θ = −ε21 , and the other elements are zero.
Now consider the doubly contravariant form,
Λµν = Λµσ g σν = g µν + εµν .
(C.8)
In the case of the Lorentz transformation,
ε01 = ε0σ g σ1 = u
ε10 = ε1σ g σ0 = −u ,
(C.9)
so that εµν is antisymmetric. For the rotation,
ε12 = ε1σ g σ2 = θ
ε21 = ε2σ g σ1 = −θ ,
(C.10)
which means that here, too, εµν is antisymmetric.
Since any infinitesimal Lorentz transformation can be made by compounding
infinitesimal rotations around axes and a Lorentz transformation along an axis, it
follows that the general case can be written
Λµν = g µν + εµν ,
(C.11)
where the infinitesimal, ε, is antisymmetric, i.e.
εµν = −ενµ .
(C.12)
It is important to note that this antisymmetry applies to the doubly contravariant
form Λµν , or of course equally well for the doubly covariant form Λµν . It is not true
for the mixed form, Λµν , in terms of which the Lorentz transformation was originally
defined.
57
Appendix D
Laplacian
Consider a transformation of Cartesian coordinates, x1 , x2 , x3 , in three dimensions,
to another set of orthogonal coordinates, u1 , u2 , u3 , such that
(dx1 )2 + (dx2 )2 + (dx3 )2 = (h1 du1 )2 + (h2 du2 )2 + (h3 du3 )2 ,
(D.1)
where the h’s are functions of the u’s. Let τ be an infinitesimal volume. Then the
~ can be written
divergence of a vector, A,
1Z
1Z
1
2
3 ~ ~
~,
~
~
dx dx dx ∇.A =
dS ~n.A
∇.A =
τ τ
τ S
(D.2)
where S is the surface of the infinitesimal volume, and ~n is a unit vector normal to
this surface. In terms of the components in the u1 , u2 and u3 directions, this can be
written
1
τ
Z
u1 +du1
−
Z u1
h2 du2 h3 du3 A1 + permutations ,
where ‘permutations’ means the two terms obtained by replacing the indices 1, 2, 3
by respectively 2, 3, 1 and 3, 1, 2 . This is, however,
∂
1Z
du1 du2 du3 1 [h2 h3 A1 ] + permutations ;
1
τ u
∂u
and since du1 du2 du3 = (dx1 dx2 dx3 )/(h1 h2 h3 ), it follows that
~ A
~=
∇.
1
∂
[h2 h3 A1 ] + permutations .
h1 h2 h3 ∂u1
(D.3)
~ is the gradient of a scalar, A
~ = ∇Φ,
~
In the case that A
we find
1
∇ Φ=
h1 h2 h3
2
(
"
#
"
#
"
∂ h2 h3 ∂Φ
∂ h3 h1 ∂Φ
∂ h1 h2 ∂Φ
+ 2
+ 3
1
1
2
∂u
h1 ∂u
∂u
h2 ∂u
∂u
h3 ∂u3
#)
.(D.4)
The case of spherical polars is of particular interest. These are defined by
x1 = r sin θ cos φ
x2 = r sin θ sin φ
x3 = r cos θ ,
(D.5)
58
which yields
(dx1 )2 + (dx2 )2 + (dx3 )2 = (dr)2 + (rdθ)2 + (r sin θdφ)2 ,
(D.6)
so that
h1 = 1 ,
h2 = r ,
h3 = r sin θ .
(D.7)
Hence we can write
1
∂
∂Φ
1 ∂2Φ
1 ∂ 2 ∂Φ
+ 2
sin θ
+ 2 2
∇ Φ= 2 r
.
r ∂r ∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
2
(D.8)
A simple manipulation shows that the following operator identity holds:
∇2 =
1 ∂2
L2
r
+
,
r ∂r2
r2
(D.9)
where
L2 =
∂
1 ∂2
1 ∂
sin θ
+
.
sin θ ∂θ
∂θ sin2 θ ∂φ2
(D.10)
59
Appendix E
Charge Distributions
E.1
Static Potential
Suppose that ρ(~x 0 ) is a twice-differentiable function, such that
Z
∞
r0 dr0 |ρ(~x 0 )| < ∞,
(E.1)
0
where r0 = |~x 0 |. Define
Φ(~x ) =
1 Z 3 0 ρ(~x 0 )
dx 0
.
4π
|~x − ~x |
(E.2)
We shall show that
∇2 Φ(~x ) = −ρ(~x ).
E.1.1
(E.3)
Proof
~ = ~x − ~x 0 , so that
Change the integration variable from ~x 0 to R
1 Z d3 R
~ ),
Φ(~x ) =
ρ(~x − R
4π
R
(E.4)
~ |. Hence
where R = |R
∇2 Φ(~x ) =
1 Z d3 R 2
~ ),
∇ ρ(~x − R
4π
R
(E.5)
~ variable. In terms of polar
where ∇2 on the RHS can be shifted from the ~x to the R
coordinates,
∇2 =
1 ∂2
L2
R
+
,
R ∂R2
R2
(E.6)
1 ∂
∂
1 ∂2
sin θ
+
.
sin θ ∂θ
∂θ sin2 θ ∂φ2
(E.7)
where
L2 =
60
The integral in Eq. (E.5) can be separated into three parts:
∇2 Φ(~x ) = (Φ1 + Φ2 + Φ3 )/(4π),
(E.8)
where
Φ1
∂2
~ )]
=
dΩ
dR 2 [Rρ(~x − R
∂R
0
Z
∂
~ ) + ρ(~x − R
~ )]R=0
ρ(~x − R
= − dΩ[R
∂R
= −4πρ(~x ),
Φ2 =
Φ3 =
Z
∞
Z
0
∞
Z
0
Z
∞
(E.9)
dR Z 2π Z π
∂
∂
~ ) = 0,
dφ
dθ sin θ ρ(~x − R
R 0
∂θ
∂θ
0
(E.10)
dR Z π dθ Z 2π
∂2
~ ) = 0.
dφ 2 ρ(~x − R
R 0 sin θ 0
∂φ
(E.11)
The result Eq. (E.3) follows on combining the last four equations.
There is a subtlety here that ought not to be glossed over. Suppose that we were
to apply the Laplacian to Eq. (E.2), without the change of integration variable:
1 Z 3 0
1
.
d x ρ(~x 0 )∇2 0
∇ Φ(~x ) =
4π
|~x − ~x |
2
(E.12)
Apparently nothing prevents us from rewriting the Laplacian in terms of R, θ and
φ, as in Eq. (E.6), but without making the change in integration variable. However,
1 ∂2
1
1
∇ 0
=
R = 0,
2
|~x − ~x |
R ∂R R
2
(E.13)
so that one is sorely tempted to the erroneous conclusion that ∇2 Φ vanishes.
Let us see what has gone wrong by considering the case that ρ(~x 0 ) = 1 if |~x 0 | < ε
,and is zero otherwise. We can do an explicit calculation:
Φ(~x ) =
1
2
ε
Z
r dr
0
ε
02
0
Z
1
−1
02
√
r2
d cos θ
+ r02 − 2rr0 cos θ
r
max(r, r0 )
0
= 12 ε − 16 r2 ,
=
Z
dr0
(E.14)
on condition that r < ε. For r > ε, we have clearly
Φ(~x ) =
ε3
.
3r
(E.15)
It is now easy to check that ∇2 Φ(~x ) = −1 if r < ε. For r > ε, the Laplacian of Φ
obviously vanishes.
The illegal step is Eq. (E.12), because if one performs the x−differentiations
under the integral, the resulting integral is not absolutely convergent, and the formal vanishing of Eq. (E.13) does not guarantee the vanishing of the integral! The
objection clearly does not apply to Eq. (E.5).
61
We need to separate the integration domain into a sphere of radius ε around
|~x − ~x | = 0, and the rest. However, it is necessary to do this very smoothly. We
~ ), by the requirements that it be unity for |R
~ | < ε,
define the test-function, fε (R
~ | > 2ε. For ε < |R
~ | < 2ε, we wish fε (R
~ ) to go smoothly from 1 to 0.
and zero for |R
If we require it to be infinitely differentiable in (0 < R < ∞), then all its derivatives
at R = ε and at at R = 2ε must vanish.
Let us write
0
Φε (~x ) =
ρ(~x 0 )
1 Z 3 0
d x fε (|~x 0 − ~x |) 0
,
4π
|~x − ~x |
1 Z 3 0
ρ(~x 0 )
Φ⊥ (~x ) =
d x [1 − fε (|~x 0 − ~x |)] 0
.
4π
|~x − ~x |
(E.16)
(E.17)
Clearly, from Eq. (E.2),
Φ(~x ) = Φε (~x ) + Φ⊥ (~x ).
(E.18)
Moreover, Φ⊥ really is divergenceless, for Eq. (E.12) is unexceptionable, when a
sphere around the point ~x 0 = ~x is excised from the domain of the integral. Since
we have already demonstrated Eq. (E.3), by the legal proof Eq. (E.4) – Eq. (E.11),
it follows that
∇2 Φε (~x ) = −ρ(~x ),
(E.19)
and this holds, no matter how small ε is.
This result is summarized by the formula
∇2
1
= −4πδ 3 (~x 0 − ~x ),
0
|~x − ~x |
(E.20)
where δ 3 is the three-dimensional Dirac delta function (or more properly generalized
function, or distribution). This replaces Eq. (E.13). Within the more general framework of distribution theory, Eq. (E.12) is allowed, with the identification Eq. (E.20).
E.2
Retarded Potential
If the charge-distribution is time-dependent, we write ρ(x) ≡ ρ(t, ~x ), and we
define the retarded potential by
Φ(t, ~x ) =
1 Z 3 0 ρ(t − |~x 0 − ~x |, ~x 0 )
dx
.
4π
|~x 0 − ~x |
(E.21)
We shall show that
∂ 2 Φ(t, ~x ) = ρ(t, ~x ).
(E.22)
62
E.2.1
Proof
We first define Φε (t, ~x ) and Φ⊥ (t, ~x ) , analogously to Eq. (E.16) and Eq. (E.17),
i.e. in the former the integration domain of the integral in Eq. (E.21) is smoothly
restricted to a sphere of radius 2ε about the point ~x 0 = ~x , while Φ⊥ (t, ~x ) contains
the rest of the integral. In the latter,
ρ(t − R, ~x 0 )
1 Z 3 0 1 ∂2
dx
}
R{[1
−
f
(R)]
∇ Φ⊥ (t, ~x ) =
ε
4π
R ∂R2
R
1 Z 3 0
1 ∂2
=
d x [1 − fε (R)]
ρ(t − R, ~x 0 ) + O(ε)
4π
R ∂t2
∂ 2 Φ⊥ (t, ~x )
+ O(ε),
=
∂t2
2
(E.23)
i.e. ∂ 2 Φ⊥ (t, ~x ) → 0 as ε → 0. As to the contribution of the small sphere about
R = 0, it is easy to see that ∂ 2 Φε /∂t2 tends to zero as ε → 0. Moreover, just as in
the static case, we can apply the method of Eq. (E.8) to Eq. (E.11) to show that
~ )|R=0 = −ρ(t, ~x ).
∇2 Φε = −ρ(t − R, ~x − R
(E.24)
This concludes the demonstration of Eq. (E.22).
In a completely analogous way, an advanced potential can be defined by
Φ(t, ~x ) =
1 Z 3 0 ρ(t + |~x 0 − ~x |, ~x 0 )
dx
,
4π
|~x 0 − ~x |
(E.25)
and it is clear that this, too, is a solution of Eq. (E.22). That the retarded, and
not the advanced solution must be used, in order to describe the radiation from an
accelerating charge, is a prescription that is not contained in the Maxwell differential
equations: it is a deep question of boundary conditions.
63
Appendix F
Retarded Fields
We will sketch the rather tedious calculation of the Liénard-Wiechert fields, Eq. (9.23)
and Eq. (9.24), starting from the potentials, Eq. (9.19) and Eq. (9.22).
F.1
Lemmata
The following auxiliary theorems will first be proved:
dt0
dt
= κ1 ,
~ 0 = − ~n ,
∇t
κ
~
n∧(~
n∧~v )
∂~
n
=
,
0
∂t
R
~
[∇(κR)]t0 fixed = ~n − ~v ,
∂(κR)
= v 2 − ~n.~v − R~n.~v˙ .
∂t0
α:
β:
γ:
δ:
:
Here t = t0 + R(t0 ) ,
Further κ = 1 − ~n.~v ,
F.1.1
α:
β:
γ:
δ:
:
~ 0 )| ,
R(t0 ) = |~x − X(t
and ~v =
~ 0)
dX(t
dt0
~n =
~ 0)
~
x−X(t
R
.
~˙ .
≡X
Proofs
0)
~ 0) ˙
~ = −~n.~v .
= − ~x−X(t
.X
dt = [1 + Ṙ]dt0 , and Ṙ ≡ ∂R(t
∂t0
R
~ = ∇t
~ 0 + ~n.∇(~
~ x − X)
~ = ∇t
~ 0 + ~n − (~n.~v )∇t
~ 0
0 = ∇t
~
∂ ~
x− X
∂t0 R
~˙
= −X
− R~n Ṙ = (~n.~vR)~n−~v
R
~
~ − (~x − X).~
~ v ] = ~n − ~v
∇(κR)
= ∇[R
h
i
∂(κR)
∂~
n
˙
=
κ
Ṙ
−
.~
v
+
~
n
.
~
v
R = −R(~n.~v˙ ) − (~n − ~v ).~v .
∂t0
∂t0
64
F.2
Theorems
The main theorems are:
h
~
n.~v˙
R
1−v 2
R2
i
I:
~ 0 = − e3
∇A
κ
II:
e
~ ∧A
~ ]0
n ∧ ~v ,
[∇
t fixed = − κ2 R2 ~
III:
h
F.2.1
~
∂A
∂t0 ~
x fixed
i
=
+
e ˙
[~v
κ2 R
~n +
e
~v
κ2 R 2
,
+ ~n ∧ (~v ∧ ~v˙ )] +
e
[~n.~v
κ2 R 2
− v 2 ]~v .
Proofs
With the help of the Lemmata, and the following decompositions, the desired
formulae for the Liénard-Wiechert fields can be readily obtained:
n
I:
~
~ 0 = − 2e 2 [∇(κR)]
∇A
t0 fixed +
κ R
II:
e ~
~ ∧ A]
~ 0
[∇
v
t fixed = − κ2 R2 ∇(κR) ∧ ~
III:
h
~
∂A
∂t0 ~
x
i
fixed
=
e~v˙
κR
−
∂(κR) ~ 0
∇t
∂t0
o
e~v ∂(κR)
κ2 R2 ∂t0
~ from Theorem II, one must remember that
In calculating B
h i
~ ∧A
~ = [∇
~ ∧A
~ ]0
~ 0 ∧ ∂ A~0
*
∇
+
∇t
.
t fixed
∂t ~
x fixed
~ one uses
Similarly, in using Theorem III to calculate E,
*
F.3
~
∂A
∂t
=
h
~
∂A
∂t0 ~
x
i
dt0
fixed dt
.
Corollary
From Eq. (9.23) we have
2)
~ = 3e ~n ∧ {~n ∧ [(~n − ~v ) ∧ ~v˙ ]} + e(1−v
~n ∧ E
~n ∧ (~n − ~v )
3
2
κ R
κ R
However
~ = ~n{~n.C}
~ − C.
~
~n ∧ {~n ∧ C}
So with
~ = (~n − ~v ) ∧ ~v˙
C
we find
~n ∧ {~n ∧ [(~n − ~v ) ∧ ~v˙ ]} = −~n{~n.(~v ∧ ~v˙ )} − ~n ∧ ~v˙ + ~v ∧ ~v˙ = −~n ∧ {~v˙ + ~n ∧ (~v ∧ ~v˙ )}.
Hence
2)
~ = − 3e ~n ∧ {~v˙ + ~n ∧ (~v ∧ ~v˙ )} − e(1−v
~
~n ∧ E
~n ∧ ~v = B.
κ R
κ3 R 2
As we see from Eq. (9.24).
65