Preprint typeset in JHEP style - HYPER VERSION
Classical Field Theory
Gleb Arutyunovaa
Institute for Theoretical Physics and Spinoza Institute, Utrecht
University, 3508 TD Utrecht, The Netherlands
Abstract: The aim of the course is to introduce the basic
methods of classical eld theory and to apply them in a variety of
physical models ranging from classical electrodynamics to
macroscopic theory of ferromagnetism. In particular, the course
will cover the Lorentz-covariant formulation of Maxwells
electromagnetic theory, advanced radiation problems, the
Ginzburg-Landau theory of superconductivity, hydrodynamics of ideal
liquids, the Navier-Stokes equation and elements of soliton theory.
The students will get acquainted with the Lagrangian and
Hamiltonian description of innite-dimensional dynamical systems,
the concept of global and local symmetries, conservation laws. A
special attention will be paid to mastering the basic computation
tools which include the Green function method, residue theory,
Laplace transform, orthogonal polynomials and special
functions.
Last Update 09.06.10
Email: [email protected] Correspondent fellow at Steklov
Mathematical Institute, Moscow.
Contents1. Classical Fields: General Principles 1.1 Lagrangian
and Hamiltonian formalisms 1.2 Noethers theorem in classical
mechanics 1.3 Lagrangian for continuous systems 1.4 Noethers
theorem in eld theory 1.5 Hamiltonian formalism in eld theory 2.
Electrostatics 2.1 Laws of electrostatics 2.2 Laplace and Poisson
equations 2.3 The Green theorems 2.4 Method of Greens functions 2.5
Electrostatic problems with spherical symmetry 2.6 Electric dipole
moment 3. Magnetostatics 3.1 Laws of magnetostatics 3.2 Magnetic
(dipole) moment 3.3 Gyromagnetic ratio. Magnetic moment of
electron. 4. Relativistic Mechanics 4.1 Einsteins relativity
principle 4.2 Lorentz transformations 4.3 Rotation and Lorentz
groups in more detail 5. Classical Electrodynamics 5.1 Relativistic
particle 5.2 Relativistic particle in electromagnetic eld 5.3
Maxwells equations and gauge invariance 5.4 Fields produced by
moving charges 5.5 Electromagnetic waves 5.6 Hamiltonian
formulation of electrodynamics 5.7 Solving Maxwells equations with
sources 5.8 Causality principle 3 3 9 11 15 19 20 20 25 26 28 30 35
36 36 37 39 41 41 41 44 48 48 50 52 54 57 61 63 68
1
6. Radiation 6.1 Linard-Wiechert Potentials e 6.2 Dipole
Radiation 6.3 Applicability of Classical Electrodynamics 6.4
Darvins Lagrangian 7. Advanced magnetic phenomena 7.1 Exchange
interactions 7.2 One-dimensional Heisenberg model of ferromagnetism
7.3 Landau-Lifshitz equation 8. The Ginzburg-Landau Theory 9.
Elements of Fluid Mechanics 9.1 Eulers equation 9.2 Bernoullis
equation 9.3 The Navier-Stokes equation 10. Non-linear phenomena in
media 10.1 Solitons 11. Appendices 11.1 Appendix 1: Trigonometric
formulae 11.2 Appendix 2: Tensors 11.3 Appendix 3: Functional
derivative 12. Problem Set 12.1 Problems to section 1 12.2 Problems
to section 2 12.3 Problems to section 7
69 70 73 83 83 87 88 90 102 102 102 102 102 102 102 103 107 107
107 109 110 110 115 123
2
1. Classical Fields: General PrinciplesClassical eld theory is a
very vast subject which traditionally includes the Maxwell theory
of electromagnetism including electromagnetic properties of matter
and the Einstein theory of General Relativity. The main scope of
classical eld theory is to construct the mathematical description
of dynamical systems with an innite number of degrees of freedom.
As such, this disciple also naturally incorporates the classics
aspects of uid dynamics. The basic mathematical tools involved are
partial dierential equations with given initial and boundary
conditions, theory of special functions, elements of group theory.
1.1 Lagrangian and Hamiltonian formalisms We start with recalling
the two ways the physical systems are described in classical
mechanics. The rst description is known as the Lagrangian formalism
which is equivalent to the principle of least action1 (Maupertuiss
principle). Consider a point particle which moves in a
n-dimensional space with coordinates (q 1 , . . . , q n ) and in
the potential U (q). The Newtons equations describing the
corresponding motion (trajectory) are mi = q U . q i (1.1)
These equations can be obtained by extremizing the following
functionalt2 t2
S=t1
dt L(q, q, t) = t1
dt
mq 2 U (q) . 2
(1.2)
Here S is the functional on the space of particle trajectories:
to any trajectory i which satises given initial q i (t1 ) = qin and
nal q i (t2 ) = qfi conditions it puts in correspondence a number.
This functional is called the action. The specic function L
depending on particle coordinates and momenta is called Lagrangian.
According to the principle of stationary action, the actual
trajectories of a dynamical system (particle) are the ones which
deliver the extremum of S. Compute the variation of the
actiont2
S = t1
dt
U d (mq i ) + i q i + total derivative , dt q
where we have integrated by parts. The total derivative term
vanishes provided the end points of a trajectory are kept xed under
the variation. The quantity S vanishes for any q i provided
eq.(1.1) is satised. Note that in our particular example, the
Lagrangian coincides with the dierence of the kinetic and the
potential energy L = T U and it does not explicitly depend on
time.1
More accurately, the principle of stationary action.
3
In general, we simply regard L as an arbitrary function of q, q
and time. The equations of motion are obtained by extremizing the
corresponding action S d L L = i =0 i i q dt q q and they are
called the Euler-Lagrange equations. We assume that L does not
depend ... on higher derivatives q , q and so on, which reects the
fact that the corresponding dynamical system is fully determined by
specifying coordinates and velocities. Indeed, for a system with n
degrees of freedom there are n Euler-Lagrange equations of the
second order. Thus, an arbitrary solution will depend on 2n
integration constants, which are determined by specifying, e.g. the
initial coordinates and velocities. Suppose L does not explicitly
depend2 on t, then dL L L = i qi + i qi . dt q q Substituting hereL
q i
from the Euler-Lagrange equations, we get dL L d L i d L i = i
qi + q = q . i dt q dt q dt q i
Therefore, we nd that d L i q L =0 dt q i as the consequence of
the equations of motion. Thus, the quantity H= L i q L, qi (1.4)
(1.3)
is conserved under the time evolution of our dynamical system.
For our particular example, mq 2 2 H = mq L = + U (q) = T + U E . 2
Thus, H is nothing else but the energy of our system; energy is
conserved due to equations of motion. Dynamical quantities which
are conserved during the time evolution of a dynamical system are
called conservation laws or integrals of motion. Energy is our rst
non-trivial example of a conservation law. Introduce a quantity
called the (canonical) momentum pi =2
L , qi
p = (p1 , . . . , pn ) .
This is homogenuity of time.
4
For a point particle pi = mq i . Suppose that U = 0. Then pi = d
L dt q i =0
by the Euler-Lagrange equations. Thus, in the absence of the
external potential, the momentum p is an integral of motion. This
is our second example of a conservation law. Now we remind the
second description of dynamical systems which exploits the notion
of the Hamiltonian. The conserved energy of a system expressed via
canonical coordinates and momenta is called the Hamiltonian H H(p,
q) = 1 2 p + U (q) . 2m
Let us again verify by direct calculation that it does not
depend on time, dH 1 U 1 U = pi pi + q i i = m2 qi qi + q i i = 0
dt m q m q due to the Newton equations of motion. Having the
Hamiltonian, the Newton equations can be rewritten in the form qj =
H , pj pj = H . q j
These are the fundamental Hamiltonian equations of motion. Their
importance lies in the fact that they are valid for arbitrary
dependence of H H(p, q) on the dynamical variables p and q.In the
general setting the Hamiltonian equations are obtained as follows.
We take the full dierential of the Lagrangian L i L i dL = dq + dq
= pi dq i + pi dq i = pi dq i + d(pi q i ) q i dpi , q i qi where
we have used the denition of the canonical momentum and the
Euler-Lagrange equations. From here we nd d(pi q i L) = q i dpi pi
dq i . | {z }H
From the dierential equality the Hamiltonian equations follow.
Transformation H(p, q) = pi q i L(q, q)|qi pi is the Legendre
transform.
The last two equations can be rewritten in terms of the single
equation. Introduce two 2n-dimensional vectors x= p q , H=H pj H q
j
5
and 2n 2n matrix J: J= 0 1 1 0 .
Then the Hamiltonian equations can be written in the form x=J H,
or J x= H.
In this form the Hamiltonian equations were written for the rst
time by Lagrange in 1808. A point x = (x1 , . . . , x2n ) denes a
state of a system in classical mechanics. The set of all these
points form a phase space P = {x} of the system which in the
present case is just the 2n-dimensional Euclidean space with the
metric (x, y) = 2n xi y i . i=1 To get more familiar with the
concept of a phase space, consider a one-dimensional 2 example: the
harmonic oscillator. The potential is U (q) = q2 . The Hamiltonian
2 2 H = p2 + q2 , where we choose m = 1. The Hamiltonian equations
of motion are given by ordinary dierential equations: q = p, p = q
= q = q .
Solving these equations with given initial conditions (p0 , q0 )
representing a point in the phase space3 , we obtain a phase space
curve p p(t; p0 , q0 ) , q q(t; p0 , q0 ) .
Through every phase space point there is one and only one phase
space curve (uniqueness theorem for ordinary dierential equations).
The tangent vector to the phase space curve is called the phase
velocity vector or the Hamiltonian vector eld. By construction, it
is determined by the Hamiltonian equations. The phase curve can
consist of only one point. Such a point is called an equilibrium
position. The Hamiltonian vector eld at an equilibrium position
vanishes. The law of conservation of energy allows one to nd the
phase curves easily. On each phase curve the value of the total
energy E = H is constant. Therefore, each phase curve lies entirely
in one energy level set H(p, q) = h. For harmonic oscillator p2 + q
2 = 2h and the phase space curves are concentric circles and the
origin. The matrix J serves to dene the so-called Poisson brackets
on the space F(P) of dierentiable functions on P:n
{F, G}(x) = ( F, J G) = J i F j G =j=13
ij
F G F G j . pj q j q pj
The two-dimensional plane in the present case.
6
The Poisson bracket satises the following conditions {F, G} =
{G, F } , {F, {G, H}} + {G, {H, F }} + {H, {F, G}} = 0 for
arbitrary functions F, G, H. Thus, the Poisson bracket introduces
on F(P) the structure of an innitedimensional Lie algebra. The
bracket also satises the Leibnitz rule {F, GH} = {F, G}H + G{F, H}
and, therefore, it is completely determined by its values on the
basis elements xi : {xj , xk } = J jk which can be written as
follows {q i , q j } = 0 , {pi , pj } = 0 ,j {pi , q j } = i .
The Hamiltonian equations can be now rephrased in the form xj =
{H, xj } x = {H, x} = XH .
It follows from Jacobi identity that the Poisson bracket of two
integrals of motion is again an integral of motion. The Leibnitz
rule implies that a product of two integrals of motion is also an
integral of motion. The algebra of integrals of motion represents
an important characteristic of a Hamiltonian system and it is
closely related to the existence of a symmetry group. In the case
under consideration the matrix J is non-degenerate so that there
exists the inverse J 1 = J which denes a skew-symmetric bilinear
form on phase space (x, y) = (x, J 1 y) . In the coordinates we
consider it can be written in the form =j
dpj dq j .
This form is closed, i.e. d = 0. A non-degenerate closed
two-form is called symplectic and a manifold endowed with such a
form is called a symplectic manifold. Thus, the phase space we
consider is the symplectic manifold.
7
Imagine we make a change of variables y j = f j (xk ). Then yj =
y j k x = Aj J km k xkAj k x mH
= Aj J km k
y p xm
y pH
or in the matrix form y = AJAt yH
.
The new equations for y are Hamiltonian with the new Hamiltonian
is H(y) = H(f 1 (y)) = H(x) if and only if AJAt = J . Hence, this
construction motivates the following denition. Transformations of
the phase space which satisfy the condition AJAt = J are called
canonical4 . Canonical transformations do not change the symplectic
form : (Ax, Ay) = (Ax, JAy) = (x, At JAy) = (x, Jy) = (x, y) . In
the case we considered the phase space was Euclidean: P = R2n .
This is not always so. The generic situation is that the phase
space is a manifold. Consideration of systems with general phase
spaces is very important for understanding the structure of the
Hamiltonian dynamics. Short summary A Hamiltonian system is
characterized by a triple (P, {, }, H): a phase space P, a Poisson
structure {, } and by a Hamiltonian function H. The vector eld XH
is called the Hamiltonian vector eld corresponding to the
Hamiltonian H. For any function F = F (p, q) on phase space, the
evolution equations take the form dF = {H, F } = XH F . dt Again we
conclude from here that the Hamiltonian H is a time-conserved
quantity dH = {H, H} = 0 . dt Thus, the motion of the system takes
place on the subvariety of phase space dened by H = E constant.In
the case when A does not depend on x, the set of all such matrices
form a Lie group known as the real symplectic group Sp(2n, R) . The
term symplectic group was introduced by Herman Weyl. The geometry
of the phase space which is invariant under the action of the
symplectic group is called symplectic geometry.4
8
1.2 Noethers theorem in classical mechanics Noethers theorem is
one of the most fundamental and general statements concerning the
behavior of dynamical systems. It relates symmetries of a theory
with its conservation laws. It is clear that equations of motion
are unchanged if we add to a Lagrangian a total time derivative of
a function which depends on the coordinates and time only: d L L +
dt G(q, t). Indeed, the change of the action under the variation
will bet2
S S = S +t1
dt
d G G(q, t) = S + i q i |t=t2 . t=t1 dt q
Since in deriving the equations of motion the variation is
assumed to vanish at the initial and nal moments of time, we see
that S = S and the equations of motion are unchanged. Let now an
innitezimal transformation q q + q be such that the variation of
the Lagrangian takes the form (without usage of equations of
motion!)5 of a total time derivative of some function F : L = dF .
dt
Transformation q is called a symmetry of the action. Now we are
ready to discuss Noethers theorem. Suppose that q = q + q is a
symmetry of the action. Then L = L i L i L L d dF q + i q = i q i +
i q i = . i q q q q dt dt d L L d dF q i + i q i = . i dt q q dt dt
L =
By the Euler-Lagrange equations, we get L = This gives
d L i dF q = . i dt q dt As the result, we nd the quantity which
is conserved in time dJ d L i q F dt dt q i This quantity J= =
0.
L i q F = pi q i F i q is called Noethers current. Now we
consider some important applications.As we have already seen, a
variation of the Lagrangian computed on the equations of motion is
always a total derivative!5
9
Momentum conservation. Momentum conservation is related to the
freedom of arbitrary choosing the origin of the coordinate system.
Consider a Lagrangian L= Consider a displacement q i = q i + ai q i
= ai , q i = qi qi = 0 . Obviously, under this transformation the
Lagrangian remains invariant and we can take F = 0 or F = any
constant. Thus, J = pi q i = pi ai , Since ai arbitrary, all the
components pi are conserved. Angular momentum conservation.
Consider again L= and make a transformation q i = qi + Then, L = mq
i ij j ij j
m 2 q . 2 i
m 2 q 2 i
q
q i =
ij j
q .
q .
Thus, if ij is anti-symmetric, the variation of the Lagrangian
vanishes. Again, we can take F = 0 or F = any constant and obtain J
= pi q i = pi Since nentsij ij j
q ,
is arbitrary, we nd the conservation of angular momentum
compoJij = pi q j pj q i .
Particle in a constant gravitational eld . The Lagrangian L= m 2
z mgz . 2d (mgat). dt
Shift z z + a, i.e. z = a. We get L = mga = quantity J = mzz F =
mza + mgat
Thus, the
is conserved. This is a conservation law of the initial velocity
z + gt = const.
10
Conservation of energy. Energy conservation is related to the
freedom of arbitrary choosing the origin of time (you can perform
you experiment today or after a several years but the result will
be the same provided you use the same initial conditions). We
derive now the conservation law of energy in the framework of
Noethers theorem. Suppose we make an innitesimal time displacement
t = . The Lagrangian response on it is dL L = . dt On the other
hand, L = L i L i L d L L q + i q + t = q i + i q i , q i q t dt q
i q
where we have used the Euler-Lagrange equations and assumed that
L does not explicitly depends on time. Obviously, q i = q i and q i
= q i , so that L = d L i L dL q + i qi = . i dt q q dt
Cancelling , we recover the conservation law for the energy dH =
0, dt H = pi q i L .
Finally, it remains to note that in all the symmetry
transformations we have considered so far the integration measure
dt in the action did not transform (even for in the last example dt
d(t + ) = dt ). 1.3 Lagrangian for continuous systems So far our
discussion concerned a dynamical system with a nite number of
degrees of freedom. To describe continuous systems, such as
vibrating solid, a transition to an innite number of degrees of
freedom is necessary. Indeed, one has to specify the position
coordinates of all the points which are innite in number. The
continuum case can be reached by taking the appropriate limit of a
system with a nite number of discrete coordinates. Our rst example
is an elastic rod of xed length which undergoes small longitudinal
vibrations. We approximate the rod by a system of equal mass m
particles spaced a distance a apart and connected by uniform
massless springs having the force constant k. The total length of
the system is = (n + 1)a. We describe the displacement of the ith
particle from its equilibrium position by the coordinate i . Then
the kinetic energy of the particles isn
T =i=1
m 2 . 2 i
11
The potential energy is stored into springs and it is given by
the sum 1 U= k 2n
(i+1 i )2 .i=0
Here we associate 0 = 0 = n+1 with the end points of the
interval which do not U move. The force acting on ith particle is
Fi = i : Fi = k(i+1 + i1 2i ) . This formula shows that the force
exerted by the spring on the right of the ith particle equals to
k(i+1 i ), while the force exerted from the left is k(i i1 ). The
Lagrangian isn
L=T U =i=1
m 2 1 k 2 i 2
n
(i+1 i )2 .i=0
At this stage we can take a continuum limit by sending n and a 0
so that = (n + 1)a is kept xed. Increasing the number of particles
we will be increasing the total mass of a system. To keep the total
mass nite, we assume that the ratio m/a , where is a nite mass
density. To keep the force between the particles nite, we assume
that in the large particle limit ka Y , where Y is a nite quantity.
Thus, we have 1 L=T U = 2n
i=1
m 2 1 a a i 2
n
a(k a)i=0
i+1 i a
2
.
Taking the limit, we replace the discrete index i by a continuum
variable x. As a result, i (x). Also i+1 i (x + a) (x) x (x) . a a
Thus, taking the limit we nd L= 1 2 dx 2 Y (x )2 .0
Also equations of motion can be obtained by the limiting
procedure. Starting from i+1 + i1 2i m i ka = 0, a a2 and using 2
i+1 + i1 2i = xx a0 a2 x2 lim
12
we obtain the equation of motion Y xx = 0 . Jus as there is a
generalized coordinate i for each i, there is a generalized
coordinate (x) for each x. Thus, the nite number of coordinates i
has been replaced by a function of x. Since depends also on time,
we are dealing with the function of two variables (x, t) which is
called the displacement eld. The Lagrangian is an integral over x
of the Lagrangian density 1 1 L = 2 Y (x )2 . 2 2 The action is a
functional of (x, t):t2
S[] =t1
dt0
dx L ((x, t), (x, t), x (x, t)) .
It is possible to obtain the equations of motion for the eld (x,
t) directly from the continuum Lagrangian. One has to understand
how the action changes under an innitesimal change of the eld (x,
t) (x, t) + (x, t) . The derivatives change accordingly, (x, t) (x,
t) + (x, t) , t t t (x, t) (x, t) + (x, t) . x x x This givest2
(1.5)
(1.6) (1.7)
S[] = S[ + ] S[] =t1
dt0
dx
L L L + x . t + (x )
Integrating by parts, we ndt2
S[] =t1
dt0
dx
L L L x t (x ) t2
+0
dx
L |t=t2 + (t ) t=t1
dtt1
L |x= . (x ) x=0
(1.8)
The action principle requires that the action principle be
stationary w.r.t. innitezimal variations of the elds that leave the
eld values at the initial and nite time unaected, i.e. (x, t1 ) =
(x, t2 ) = 0 .
13
On the other hand, since the rod is clamped, the displacement at
the end points must be zero, i.e. (0, t) = ( , t) = 0 . Under these
circumstances we derive the Euler-Lagrange equations for our
continuum system L L L + = 0. t (t ) x (x ) Let us now discuss the
solution of the eld equation c2 xx = 0 , c= Y ,
where c is the propagation velocity of vibrations through the
rod. This equation is linear and, for this reason, its solutions
satisfy the superposition principle. Take an ansatz (x, t) = eikx
ak (t) + eikx bk (t) . If we impose (0, t) = 0, then bk (t) = ak
(t) and we can rene the ansatz as (x, t) = ak (t) sin kx .
Requiring that ( , t) = 0 we get sin k = 0, i.e. k kn = n .
Coecients ak (t) then obey ak + c2 k 2 ak (t) = 0 ak (t) = eik t ak
, where k = ck is the dispersion relation. Thus, the general
solution is (x, t) =n
sin kn x An cos n t + Bn sin n t ,
n = ckn ,
and the constants An , Bn are xed by the initial conditions.
Scalar and Vector Fields The generalization to continuous systems
in more space dimensions is now straightforward. In two-dimensions
one can start with two-dimensional lattice of springs. The
displacement of a particle at the site (i, j) is measured by the
quantity ij , which is a two-dimensional vector. In the limit when
we go to a continuum, this becomes a displacement eld (x, y, t) of
a membrane subjected to small vibrations in the (x, y)-plane. In
three dimensions we get a vector ijk . The continuous limit yields
a three-dimensional displacement eld (x, y, z, t) of a continuous
solid vibrating in the x, y, z directions with eoms of a partial
dierential equation type: c1 xx c2 yy c3 zz c4 xy c5 yz c6 xz = 0
,
14
the coecients ci encode the properties of the solid. In general
elds depending on the space-time variables are tensors, i.e. they
transforms under general coordinate transformations in a denite
way. Namely, a i ...i tensor eld j1 ...jp of rank (p, q) under
general coordinate transformations of the q 1 coordinates xi : xi x
i (xj ) transforms as follows1 l1 ...lq p (x ) =
k ...k
x k1 x kp xj1 xjq i1 ...ip (x) . xi1 xip x l1 x lq j1 ...jqi
Here tensor indices are acted with the matrices x j which form a
group GL(d, R). x This is a group of all invertible real d d
matrices. A simplest example is a scalar eld that does not carry
any indices. Its transformation law under coordinate
transformations is (x ) = (x). 1.4 Noethers theorem in eld theory
In order to fully describe a dynamical system, it is not enough to
only know the equations of motion. It is also important to be able
to express the basic physical characteristics, in particular, the
dynamical invariants, of the systems via solutions of these
equations. Noethers theorem: To any nite-parametric, i.e. dependent
on s constant parameters, continuous transformation of the elds and
the space-time coordinates which leaves the action invariant
corresponds s dynamical invariants, i.e. the conserved functions of
the elds and their derivatives. To prove the theorem, consider an
innitezimal transformation xi x i = xi + xi , i = 1, . . . , d, I
(x) I (x ) = I (x) + I (x) . As in the nite-dimensional case, the
variations xi and I are expressed via innitezimal linearly
independent parameters n : xi =1ns i Xn n ,
I (x) =1ns
I,n n .
(1.9)
Here all n are independent of the coordinates x. Such
transformations are called k global. Particular cases arise, when
Xn = 0 or I,n = 0. In the rst case the coordinates xi do not change
under symmetry transformations at all, while the elds are
transformed according to I (x) I (x) = I (x) + I (x) . In the
second case the symmetry acts on the space-time coordinates only,
but the elds (being dependent on the space-time coordinates)
undergo the induced transformationk I (x ) = (x + x) = I (x) + k I
(x)xk = I (x) + k I (x)Xn n = I ,
15
where the last equality is due to I = 0. In the general case
symmetry transformations act on both the space-time coordinates and
the elds, cf. eq.(1.9). Considerk I (x ) = I (x + x) = I (x) + k I
(x)xk + . . . = I (x) + k I (x) Xn n + . . .
It is important to realize that the operations and /x do not
commute. This is because is the variation of the elds due to both
the change of their form and their arguments xi . We therefore
introduce the notion of the variation of the form of the eld
function k I (x) = I (x) I (x) = (I,n k I Xn )n . Variation of the
form does commute with the derivative /x. For the variation of the
Lagrangian we, therefore, have L (x ) = L (x) + The change of the
action is6 S = dx L (x ) dx L (x) = dL dx [L (x) + L (x) + k xk ]
dx dx L (x) . dL k dL x = L (x) + L (x) L (x) + k xk . dxk dx L
(x)
Transformation of the integration measure is x 1 d 1 x 1 1 + x1
x1 x x . . dx = det . . . dx = J dx det . . . .x 1 xd
xd x1
dx .
. . .
x d xd
x1 xd
1 +
xd xd
Jacobian
Thus, at leading order in n we have dx = dx(1 + k xk + . . .).
Plugging this into the variation of the action, we nd S = dL dx L
(x) + k xk + k xk L = dx d dx L (x) + k (L xk ) . dx
We further note that L L L L I + k I = k I + k I = L (x) = I (k
I ) (k I ) (k I ) L = k I , (k I )We consider a eld theory in
d-dimensions, so that the integration measure dx must be understood
as dx = dx1 dx2 . . . dxd dd x.6
16
where we have used the Euler-Lagrange equations. Thus, we arrive
at the following formula for the variation of the action S = dx d L
I +L xk = k ( ) dx k I dx L d m k (I,n m I Xn )+L Xn n . k ( ) dx k
I
Since the integration volume is arbitrary we conclude thatk dJn
=0 dxk
divJn = 0 ,
wherek Jn =
L k m (I,n m I Xn ) L Xn (k I )
and n = 1, . . . s. Thus, we have shown that the invariance of
the action under the sparametric symmetry transformations implies
the existence of s conserved currents. k An important remark is in
order. The quantities Jn are not uniquely dened. One can add k k Jn
Jn + m km , n where km = mk . Adding such anti-symmetric functions
does not inuence the n n k conservation law k Jn = 0. Now we are
ready to investigate concrete examples of symmetry transformations
and derive the corresponding conserved currents. Energy-momentum
tensor. Consider the innitezimal space-time translationsk x k = xk
+ xk = xk + n n
=
k k Xn = n
k and I,n = 0. Thus, the conserved current Jn becomes in this
case a second k rank tensor Tn L k k Tn = n I n L . (k I ) k Here,
as usual, the sum over the index I is assumed. The quantity Tn is
the so-called stress-energy or energy-momentum tensor. If all the
elds vanish at spacial innity then the integral7
Pn =
0 dn1 xTn
is a conserved quantity. Here 0 signies the time direction and
the integral is taken over the whole (n 1)-dimensional space.
Indeed, dPn = dt7
dx
0 dTn = dt
dn1 x
i dTn = dxi
d (Tn n) ,
Here we explicitly distinguished a time direction t and write
the integration measure in the action as dx = dt dn1 x.
17
where is a (n 2)-dimensional sphere which surrounds a n
1-dimensional volume; its radius tends to innity. The vector n is a
unit vector orthogonal to . Angular momentum. Consider innitezimal
rotations x n xn + xm nm , where nm = mn . Because of
anti-symmetry, we can choose nm = nm with n < m as linearly
independent transformation parameters. We ndk xk = Xj j = n 0 or
< 0 implying that in the other x2 x2 i i direction the second
derivative must have an opposite sign. 2.6 Electric dipole moment
On large distances eld of a neutral system is well approximated by
the so-called electric dipole moment given byN
p=i=1
e i xi ,
(2.28)
where ei is the magnitude of a charge at some distance Ri taken
from an arbitrary point, in this case chosen to be the origin.
Neutrality means thatN
ei = 0 .i=1
35
Note that for such a system, the dipole moment does not depend
on the choice of the origin of a reference frame, i.e. shifting all
Ri Ri a givesN N N N
pa =i=1
ei (xi a) =i=1
e i xi ai=1
ei =i=1
ei xi = p .
We get (x x )2 = x2 2x x + x 2 x2 1 2 xx x2 |x| 1 xx |x|2 = |x|
xx . |x|
Thus, for the potential we nd =i
ei 1 = |x xi | |x|
ei +i i
ei
(x xi ) (x p) + ... = + ... , 3 |x| |x|3
where we have used neutrality of the system of charges. Thus,
the electric eld is E= (x p) 3n(n p) p = . 3 |x| |x|3
Thus, for a neutral system the electric eld at large distances
from this system behaves itself as 1/r3 !
3. Magnetostatics3.1 Laws of magnetostatics In the case when
electric eld is static, i.e. it does not depend on time, the second
pair of the Maxwell equations take the form div H = 0 , The rst
equation allows one to write H = rot A . Substituting this in the
second equation, we obtain grad div A A = 4 j. c rot H = 4 j. c
Because of gauge invariance the vector potential is not uniquely
dened, therefore, we can subject it to one additional constraint,
which will chose to be div A = 0 .
36
Then, the equation dening the vector potential of
time-independent magnetic eld takes the form 4 A = j . c Obviously,
this is the Poisson equation, very similar to the equation for the
electrostatic potential. Therefore, the solution reads as A(x) = 1
c d3 x j(x ) . |x x |
Now we can determine the corresponding magnetic eld H = rot A =
1 c d3 x 1 1 , j(x ) = |x x | c d3 x [j(x ), R] , R3
where the bracket means the vector product12 . This is the
Biot-Savart law. It describes the magnetic eld produced by
time-independent currents. e The integral form of Maxwells equation
rot H = 4 j is called Amp`res law. To c derive it, consider a
surface S enclosed by a contour C. The ux of both sides of the last
equation through S is (rot H n)dS =S
4 c
(j n)dS .S
Application of the Stocks theorem gives H d =C
4 c
(j n)dS =S
4 I, c
where I = law.
S
(j n)dS is the full current through the surface S. This is the
Amp`re e
3.2 Magnetic (dipole) moment Free magnetic charges do not exist.
The really existing object which plays the basic role13 in study of
magnetic phenomena is the so-called magnetic dipole. A small
magnetic dipole is a magnetic arrow (like the compass arrow) which
aligns along the direction of an external magnetic eld. Consider
the magnetic eld created by a system of stationary moving charges
on distances large in comparison with the size of this system. We
choose a center of a reference frame somewhere inside the system of
moving charges. Then x 0), d) the sign of energy UM is determined
by the general formula UM =(M1 M2 )3(M1 n12 )(M2 n12 ) 3 R12
,
n12 =
R12 R12
.
We, therefore, nd A(x) = M x . |x|3
This is the leading term in the expansion of the vector
potential for a bounded stationary current distribution. As a
result, the magnetic eld of a magnetic dipole is 3n(n M ) M H = rot
A = , |x|3 where n is the unit vector in the direction of x. This
expression for the magnetic eld coincides with the formula for the
electric eld of an electric dipole. 3.3 Gyromagnetic ratio.
Magnetic moment of electron. Suppose that the current I ows over a
closed at loop C on an arbitrary shape. For the magnetic moment we
have M= d3 x M(x ) = 1 2c d3 x x j(x ) = 1 2c dS d x j(x ) ,
where dS is an area dierential corresponding the transverse
section of the (thin) loop C. Since the current I is dened as
I=S
(j n)dS ,
we have
1 dS x (j(x ) n)d 2c so that the magnetic moment can be written
in the form M= M= I 2c xd .C
Since x d = 2 dS, where dS is the area of an elementary triangle
formed by the radii drawn from the origin of the coordinate system
to the end points of the element
39
d , the integral above is equal to the total area S enclosed by
the current loop C. Therefore, IS |M | = c independently of the
shape of the contour. Here |M | is a magnitude of the magnetic
dipole moment of a current loop. If the current is formed by
particles of masses mi with charges ei moving with velocities vi t
We shall see that the above only represents one of the possible
Greens functions, since a dierent treatment of the poles produces
dierent Greens functions - an advanced or a retarded one: Retarded
Green function states G = 0 if t < t Advanced Green function
states G = 0 if t > t Notice that the dierence of the two Gadv
Gret , called the Pauli Greens function GP auli , satises the
homogenous equation . Consider the retarded Greens function. For t
> t , it should give a wave propagating from a point-like
source. Let us dene = t t , R = x x and R = R . Then we have ei(tt
) e , since > 0. Thus we need to require that < 0 in order to
have a decaying function at large , hence we have to integrate over
the lower complex plane. In opposite, for t < t , the contour
over which we integrate in the upper half of the complex plane
should give zero contribution due to the aforementioned physical
reasons. As a result, one could innitesimally shift the poles into
the lower half plane when performing the analytic continuation.
According to this prescription, the Greens function is specied as
follows G(x, t; x , t ) = 1 4 3 d3 k d eikRi . k 2 c12 ( + i)2
65
We can conveniently rewrite the previous statement, by making
use of partial fractions G (x, t; x , t ) = 1 = 3 d3 k 4 (5.31)
deikR
c 1 1 ei . 2k ck i ck i
In the limit 0, using Cauchys theorem25 , we nd G (x, t; x , t )
= = 1 4 3 0
d3 keikR 2i d3 k
c ick e eick 2k
(5.32)
c 2 2 2c = R 1 = R =
eikR sin ck k (5.33) sin ((ck) ) (5.34) (5.35)
dk sin(kR) sin(ck )
d (ck) sin
(ck) R c
R R 1 dx eix c eix c eix eix 4R R R 1 = dx eix( c ) eix( + c )
2R 1 R 1 R = + R c R c 1 R = R c
(5.36) (5.37)
Note that in the meantime we have used: partial fractions
(5.31), the Cauchy theorem in (5.31-5.32), switched to spherical
coordinates and integrated over the angles(5.33), substituted ck =
x (5.34), expanded the trigonometric functions in terms of their
complex exponentials (5.35), and identied Fourier transforms of
delta functions (5.36). On the last step we have rejected + R ,
because for , R, c > 0, the c result will always be zero.
Substituting back our original variables, we get Gret (x, t; x , t
) = t +|xx | c
t
|x x |
.
The result can be understood as the signal propagating at the
speed of light, which was emitted at t and will travel for |xx |
and will be observed at time t. Thus, c25
Cauchy integral formula reads f (a) = 1 2i f (z) dz, za
C
where a function f (z) is holomorphic inside the region
surrounded by a contour C and integration is performed in
counter-clockwize direction.
66
this Green function reects a natural causal sequence of events.
The time t is then expressed in terms of the retarded time t t=t +
|x x | . c
Substituting this solution and integrating over t , we obtain
the retarded potentials (x, t) = = t +|xx | c
t
|x x | x ,t |xx | c
(x , t ) d3 x dt + 0 d3 x + 0 , (5.38)
|x x | t +|xx | c
1 A (x, t) = c 1 = c
t
|x x | j x ,t |xx | c
j (x , t ) d3 x dt + A0 d3 x + A0 , (5.39)
|x x |
where 0 and A0 are the solutions of the homogeneous dAlambert
equations (those corresponding to the free electromagnetic eld).
Note that for in the case of time-independent and j we have = (x )
3 dx . |x x |
This is just the electrostatic formula for the scalar potential.
Moreover, if the current j is time-independent, we obtain A(x) = 1
c j(x ) 3 dx . |x x |
This potential denes the following magnetic eld H = rotx A = 1 c
rotx j(x ) + |x x |x
1 j(x ) d3 x . |x x |
(5.40)
Note the use above of the following identity rot(a) = rot a +
a.
The rst term in (5.40) vanishes, because curl is taken with
respect to coordinates x, while the current j depends on x . This
leaves H= 1 c R j(x ) 3 1 dx = 3 R c j(x ), x x |x x |3 d3 x .
67
This is the famous law of Biot-Savart, which relates magnetic
elds to their source currents. Let us now show that Gret is Lorentz
invariant. We write Gret (x, t; x , t ) = (t t ) t +|xx | c
t
|x x |
.
Here the extra term (t t ) ensures that Gret (x, t; x , t ) = 0
for t < t , because (t t ) = When we use (f (x)) =i
0, t < t 1, t t (x xi ) . |f (xi )|
In the last formula the derivative is evaluated at the set of
points xi , such that f (xi ) = 0. Realizing that for a wave
propagating at the speed of light ds2 = 0 and using some algebraic
trickery26 , we get Gret (x, t; x , t ) = 2c (t t ) (|x x | c (t t
)) 2 |x x | (|x x | c (t t )) = 2c (t t ) |x x | + c (t t )2 2
= 2c (t t ) |x x | c2 (t t )
,
where the argument of the delta function is the 4-interval
between two events (x, t) and (x , t ), which is a Lorentz
invariant object. From this we can conclude that the Greens
function is invariant under proper orthochronical (ones that
maintain causality) Lorentz transformations. 5.8 Causality
principle A quick word on intervals. A spacetime interval we have
already dened as ds2 = c2 dt2 dx2 i We refer to them dierently
depending on the sign of ds2 :26
(5.41)
Introduce u = |x x | c (t t ). Then |x x | c2 (t t )2 2
= u(u + 2c(t t )) = u2 + 2uc(t t )) .
Thus, we introduce f (u) = u2 + 2uc(t t ) with f (u) = 2u + 2c(t
t ). Equation f (u) = 0 has two solutions: u = 0 and u = 2c(t t ).
The second one will not contribute into the formula describing the
change of variables in the delta-function because of (t t ). Thus,
|x x | c2 (t t )2 2
=
(|x x | c (t t )) (|x x | c (t t )) = . (2u + 2c(t t ))|u=0 2c(t
t )
68
t t absolute t future t t t tq X space-like t t pastt t qX t t
time-like t
light-like
Figure 10: At every point in time every observer has his past
light cone, which is a set of all events that could have inuenced
his presence, and a future light cone, the set of events which the
observer can inuence. The boundaries of the light cones also dene
the split between dierent kinds of space-time intervals. On the
light cone itself the intervals are all light-like, time-like on
the inside and space-like on the outside.
time-like intervals if ds2 > 0 space-like intervals if ds2
< 0 light-like intervals (also called null intervals) if ds2 = 0
Consider Figure 10 representing the light-cone built over a point
X. Signals in X can come only from points X , which are in the past
light-cone of X. We say X > X (X is later than X ). The inuence
of a current j in X on potential A at X is a signal from X to X.
Thus, the causality principle is reected in the fact that A(X) can
depend on 4-currents j(X ) only for those X for which X > X .
Thus, A(X) G(X X ) = 0 j(X ) (5.42)
for X < X or points X that are space-like to X. Hence, the
causality principle for the Green function is G(X X) = 0 , (5.43)
in terms of the conditions described above. The retarded Greens
function is the only relativistic Greens function which has this
property.
6. RadiationThe last part of these lectures will treat two
classical radiation problems: Linarde Wiechert potentials and the
dipole radiation.
69
6.1 Linard-Wiechert Potentials e The charge distribution in
space and time of a single point-like charge is given by (x, t) = e
(x r (t)) , j (x, t) = ev (x r (t)) . Here x is the position of the
observer, r (t) is the trajectory of the charge and v = r (t), its
velocity. The potential then reads (x, t) = t +|xx | c
t
|x x |
e (x r (t )) d3 x dt
(6.1)
Let us take x = r (t ), because only then the integrand is
non-zero. Then eq.(6.1) can be integrated over x and we get (x, t)
= e Take f (t ) = t +|xr(t )| c
t +
|xr(t )| c
t
|x r (t )|
dt .
(6.2) f (x) is evaluated at =0
t and use (f (x)) =
the point were f (x) = 0, i.e. at t which solves t
(x) , where |f (x)| |xr(t )| + t c
df (t ) 1 (x r(t )) r(t ) 1Rv =1 =1 . dt c |x r (t )| c R In the
last equation we have used the fact that R = x r (t ) and v = r
(t). The potential then becomes (x, t) = e e 1 = . R 1 1 Rv R Rv c
R c (6.3)
We can use the same line of reasoning to show A (x, t) = e v . c
(R Rv )c
(6.4)
The formulae (6.3) and (6.4) are the Linard-Wiechert potentials.
Let us compute e the corresponding electric and magnetic elds.We
have 1 A c t H = rotA . E = ;
Moreover, R(t ) is given by the dierence in the times t and t
with an overall factor of c ` ` R t =c tt .
70
Therefore,
R (t ) R v t R (t ) t t = = =c 1 . t t t R t t t 1 . = t 1 Rv
Rc
(6.5)
From this relation, it follows that
Analogously, one can also start from the expressions R(t ) = c(t
t ) and t = t (t, x), such that ` R t = c t t = 1 c ` 1 R t = c ` x
r t (x, t) ! 1 R R = + t , c R tx
where one can again identify
R t
with the previous result from (6.5) and nally obtain R t = c R
Rv c R RRv c
and
R=
.
Now we have all the necessary ingredients, which we can use to
nd E and H, i.e. to obtain the Linard-Wiechert e elds. First lets
calculate the quantity , = e (R Rv 2 ) c
(R
Rv ). c
The rst term is R = c t and we can rewrite the second term by
using of the vector identities (R v) = (R )v + (v )R + R ( v) + v (
R).
Now we have to calculate these quantities one at a time. A
dicult quantity is (v )R = (v )x (v )r(t ).
Switching to index notation hugely simplies this vm m Ri = vm m
xi vm m ri = vm mi vm vi m t = vi vi vm m t vi . Here I have used
that m ri =dri dxm
=
dri dt dt dxm
= vi m t . Going back to vector notation )R = v (v t )v.
(v Similarly
(R Now we calculate (
)v = (R
t )v.
v)i = =
ijk j vk ijk j t
vk
= (( t ) v)i , and similarly R= x r = ( t ) v.
71
Now use an identity A (B C) = B(A C) C(A B), and we nally get (R
v) = v + Substituting all the quantities nally gives = e c2 (R Rv 3
) c dA dt
t (R v v 2 ).
! + cv(R R v ) . R(c v + R v) c2 2
A similar (but a little bit easier) exercise for e dA = dt c(R
Rv )3 c Putting these together we obtain E =
gives ! .
vR 2 Rv (R )(vR cv) + (c v 2 + R v) c c
vR v2 1 (R )(1 2 ) + 2 (R(R v) R2 v) c c c (R R 3 (v(R v) v(R
v)) . c eRv 3 ) c
By using R2 = R R and again the relation A (B C) = B(A C) C(A B)
we now nd E = For the magnetic eld we use H= Substituting the
quantities gives H = vR R e v2 1 ( )(1 2 ) + 2 (R2 v) Rv 3 R c c c
(R c ) R 3 (v(R v) v(R v)) . c A= 1 c (v) = 1 (( c v) + ( ) v) . e
(R Rv 3 ) c
(R
v2 vR 1 Rv )(1 2 ) + 2 (R ((R ) v)) . c c c c
We see that we almost have the electric eld (from the equation
just above the nal result for E), but we are missing the quantities
R(1 v2 ) and c1 R(R v). However, the cross product with these
quantities will vanish, since R R = 0, 2 c and therefore we can
simply add these quantities. We nally have H= R E. R2
To summarize, the Linard-Wiechert elds are given by the
following expressions e H = 1 R, E , R 2 1 v2 e R, R v R, v R vR c
c c + E=e 3 3 Rv R c c2 R Rv c
.
Notice that in the last equation the rst term only depends on
the velocity of the 1 moving particle and is proportional to R2
(short distance), whereas the second term
72
P ( x, y , z )
r R0
r R
V
0 l
r x'
(x ', y ', z ')
Figure 11: A diagrammatic representation of a dipole1 depends on
acceleration and is proportional to R providing, therefore, the
longdistance dominating contribution, the so-called wave-zone. Note
also that ux is 1 proportional to E 2 hence is also proportional to
R2 . Therefore,
E 2 dV =
1 2 R d = 4 , R2
which is a constant ux of E at large distances. It is worth
stressing that there is no energy (radiation) coming from a charge
moving at a constant velocity, because we can always choose a frame
where it is stationary, hence H = 0 E H = 0, consequently it cannot
emit energy. 6.2 Dipole Radiation Field of a neutral system is
expressed with the help of the so-called electric moment given in
its discretized form asN
d=i=1
ei Ri ,
(6.6)
where ei is the magnitude of a charge at some distance Ri taken
from an arbitrary
73
point, in this case chosen to be the origin. For a neutral
system we require thatN
ei = 0 .i=1
Note that for such a system, electric moment does not depend on
the choice of the origin of the reference frame, i.e. shifting all
Ri Ri a givesN N N N
da =i=1
ei Ri a =i=1
ei Ri ai=1
ei =i=1
e i Ri = d .
Let us now consider a neutral system of moving charges. From
diagram 11 using Pythagorean theorem and assuming that l R0 , l
being the characteristic size, we 27 get R= R0 R2 R0 2
=
2 R0 2R0 R + R 2
12
R0 R2 R0
R0
1
R0 R2 R0
= R0
R0 R . R0
By using (5.38), we then nd the retarded scalar potential = x ,t
R 3 c dx = R x , t R0 R0 R x , t c = d3 x R0 R0 R0 R0 R0 1 R0 = d3
x R x , t , R0 R0 R0 c
R0 c
+ =
where the rst term vanishes because it is proportional the
complete charge of the system, which we have set to zero, by dening
the system to be neutral. In the remaining term we will write the
integral as d t R0 , the electric moment at time c t R0 , which is
just a continuous version of (6.6) c d t Therefore28 , R d t = R R
R27 28
R0 c
=
d3 x R x , t
R0 c
.
(6.7)
R c
.
Here R (x , y , z ). To simplify our further treatment, the have
changed the notation R0 R.
74
Further, we nd d t div R 1 1 dR 1 + div d = 3 + div d , R R R R
di di R R d div d = = = , i i x R x R R = dR c
so that d t div R On the other hand, = Thus, d t R c = div . R
Here divergence is taken over coordinates of the point P (x, y, z)
where the observer is located. Using expression (5.39), the vector
potential becomes A= = 1 c 1 c j x ,t R 3 c dx = R j x , t R0 R0 R
j x , t c d3 x R0 R0 R0 R0 dR R d 2 . R3 R RR c
=
dR R d + 2 . 3 R R R
R0 c
+ .
First integral can also be expressed via electric moment, which
can be achieved by using the continuity equation R0 x ,t t c = div
j x , t R0 c .
Multiplying both sides of this equation by time independent R ,
integrating over entire space and using the denition (6.7), we can
then state that R0 p t t c = d3 x R div j x , t R0 c .
To proceed, let us sidetrack and consider an arbitrary unit
vector a, i.e. |a| = 1. Then aR divj = div j aR = div j aR j j a,
aR
75
where the last step follows from a being a constant and can
write a R0 d t t c = d3 x div j aR +a
R = 1. Based on that we R0 c
d3 x j x , t
.
Since currents do not leave the volume V , we nd that d3 x div j
aR = (aR ) jn dS = 0
as the normal component jn of the current vanishes (all currents
never leave the integration volume V ). This gives a R0 d t t c =a
d3 x j x , t R0 c .
Since the last relation is valid for any unit vector a, we
obtain that R0 d t t c Therefore, we arrive at29 A= R 1 d t cR t c
. = d3 x j x , t R0 c .
We see that both the scalar and the vector potential of any
arbitrary neutral system on large distances are dened via the
electric moment of this system. The simplest system of this type is
a dipole, i.e. two opposite electric charges separated by a certain
distance from each other. A dipole whose moment d changes in time
is called an oscillator (or a vibrator ). Radiation of an
oscillator plays an important role in the electromagnetic theory
(radiotelegraphic antennae, radiating bodies, proton-electron
systems, etc.). To advance our investigation of a dipole, let us
introduce the Hertz vector P (t, R) = It is interesting to see that
P (t, R) =2
d t R
R c
.
(6.8)
P (t, R) =
1 2P . c2 t2
This can be derived as follows. First, we notice that 1 R 1 d R
x x d P = 2 d = 3d 2 , x R x cR t x R cR t29
Here we again changed the notation R0 R.
76
since
R x
=
x . R
Dierentiating once again, we get
1 x2 3 x2 d 1 d 1 x2 2 d 2 P = 3d + 3 5d + 2 + 2 3 2, x2 R R c
R4 t cR t c R t so that3
i=1
1 2d 2 P = 2 , x2 c R t2 i
which represents the spherically symmetric solution of the wave
equation. Consider the retarded potentials (R, t) = divP (t, R) ,
A(R, t) = 1 P (t, R) ; c t
The potentials are spherically symmetric, i.e. they depend on
the distance R only. For the electromagnetic elds we have H = rotA
(t) = 1 rotP (t, R) ; c t 1 A (t) 1 2 P (t, R) E= = 2 divP (t, R) c
t c t2 1 2 P (t, R) = 2 + 2 P (t, R) + rot rot P (t, R) . c t2
On the last line the sum of the rst two terms is equal to zero
by virtue of the wave equation. This results in E = rot rot P (t,
R) . (6.9) Assume that the electric moment changes only its
magnitude, but not its direction, i.e. d (t) = d0 f (t) . This is
not a restriction because moment d of an arbitrary oscillator can
be decomposed into three mutually orthogonal directions and a eld
in each direction can be studied separately. Based on this we have
f t R c , P (t, R) = d0 R f f rot P = rot d0 + , d0 = R R R
f t R 1 = R R
R c
R , d0 = R f t RR c
R, d0
77
as rot d0 = 0. In the spherical coordinate system we compute the
corresponding components R, d0 R, d0 R, d0 and get30 rot P rot PR
R
= Rd0 sin , = R, d0
= 0,
= Rd0 sin .
= rot P
= 0, R f t RR c
= d0 sin
= sin
P (t, R) . R
Since the magnetic eld components are the components of the curl
of the vector potential, the latter is written in terms of the
Hertz vector (6.8), where we nd HR = H = 0 1 2 P (t, R) H = sin . c
t R The components of curl of any vector eld a in spherical
coordinates are given by (rot a)R = 1 a (sin a ) ; R sin R aR 1 (R
sin a ) (rot a) = R sin R 1 aR (rot a) = (Ra ) . R R
;
Using these formulae together with equation (6.9), we also nd
the components of the electric eld ER = = E = = E =30
1 sin ( sin ) P (t, R) R sin R 1 P 2 cos P ; sin2 = R sin R R R
1 sin P (t, R) = R ( sin ) R sin R R sin P R ; R R R 0.
Note that P here is the numerical value of the Herz vector P
.
78
From the above expressions we can see that electric and magnetic
elds are always perpendicular; magnetic lines coincide with circles
parallel to the equator, while electric eld lines are in the
meridian planes. Now let us further assume that f (t) = cos t or in
a complex form d t Then P = R R = and R R P R = R 1+ iR c P = 1 i 2
R + R c c P. d0 ei(t c ) RR
d t
R c
= d0 cos t
R c
R c
= d0 ei(t c ) .R
(6.10)
=
R R 1 i 1 d ei(t c ) d0 ei(t c ) = 2 0 R c R
1 i + R c
P (R, t) ,
Thus, for this particular case we get the following result 1 i +
R c 1 i ER = 2 cos + 2 R cR 1 i E = sin + 2 R cR H = i sin c P (R,
t) ; P (R, t) ; 2 c2 P (R, t) .
These are the exact expressions for electromagnetic elds of a
harmonic oscillator. They are complicated and we will look more
closely only on what happens close and far away from the
oscillator. To do that we will aid ourselves with the concept of a
characteristic scale, which is determined by the competition
between 1 R and 2 2 = = , c Tc
where T and are the period and the wavelength of the
electromagnetic wave, respectively. Close to the oscillator By
close to the oscillator we mean: R 2 or 1 R 2 = , c
79
i.e. distances from oscillator are smaller than the wavelength.
Thus we can simplify t so that R c = t R 2R = t t , c
d t R d (t) c . R R Using the close to oscillator condition,
elds are determined by the electric moment d (t) and its derivative
d without retarding t P (t, R) = H 1 sin d (t) i P i d (t) sin sin
2 = , c R c R c R2 t
because id (t) = d(t) , which follows from the particular choice
of the time depent dence of the oscillator that we have made in
(6.10). Similarly in this limit the electric eld components become
ER = 2 cos 2 cos P = d (t) ; 2 R R3 sin sin E = P = 3 d (t) . R2
R
At any given moment t, this is a eld of a static dipole. For the
magnetic eld we nd 1 d (t) J H= ,R = ,R . 3 cR t cR3 Given that
this introduced current J obeys J = d(t) , this expression gives
the t magnetic eld of a current element of length . This is known
as the Biot-Savart law31 . Far away from the oscillator Let us now
consider the region far away from the oscillator, i.e. the region
where R 2 or 1 R 2 = . c
Distances greater than the wavelength are called wave-zone. In
this particular limit our eld components become d t 2 2 sin P = 2
sin 2 c c R ER = 0 ; d t R 2 c = H . E = 2 sin c R H = 31
R c
;
Note that E
1 R3
and H
1 R2 .
80
Thus summarizing we get E R = E = HR = H = 0 , and E = H = or
sin 2 d t E = H = 2 cR t2R c
2 sin R d0 cos t 2R c c
,
.
This last result is valid for any arbitrary d (t), not
necessarily d0 f (t), because we can always perform a harmonic
Fourier decomposition of any function. Thus in the 1 wave zone the
electric and magnetic elds are equal to each other and vanish as R
. Additionally, vectors E, H, and R are perpendicular32 . Note that
the phase of E and H, i.e. t R moves with the speed of light. c
Thus, in the wave zone of the oscillator an electromagnetic wave is
propagating! = cT = 2c .
This wave propagates in the radial direction, i.e. its phase
depends on the distance to the center. Let us now look at the
Poynting vector c S= 4 E, H 1 sin2 c EH = = 4 4 c3 R2 2d t t2R c
2
,
where on the rst step we have used the fact that the electric
and the magnetic elds are perpendicular. Additionally note that the
second derivative with respect to time inside the square is an
acceleration. Energy ux through the sphere of radius R is2
=0 0 2
SR2 sin dd = 1 sin2 4 c3 R2 2d t t2R c 2
=032
2 2d t R sin dd = 3 3c t22
R c
2
=
2 2 d . 3c3
0
Note that E, H and R have completely mismatching components i.e.
if one vector has a particular non-zero component, for the other
two this component is zero.
81
For d t
R c
= d0 cos t T
R c
the ux for one period isT
2 dt = 3 d2 4 3c 00
cos2 t
R c
dt = 2 3
0
=
d2 4 T 0 3c3
=
2d2 3 2d2 0 0 = 3c3 3
.
The averaged radiation in a unit time is then 1 = T0 T
cd2 dt = 0 3
2
4
.
(6.11)
Thus, the oscillator continuously radiates energy into
surrounding space with average 1 rate d2 4 . In particular this
explains that when transmitting radio signals by 0 telegraphing one
should use waves of relatively short wavelengths33 (or equivalently
high frequencies ). On the other hand, radiation of low frequency
currents is highly suppressed, which explains the eect of the sky
appearing in blue, which is to the high frequency end of the
visible light34 spectrum. Lastly, let us nally focus on the concept
of resistance to radiation, which is given by R such that = R J 2 .
Recall that we have previously dened J such that it obeys J =
denition, we get2 d(t R ) c . t
Using this
J
1 = T0
T
1 J dt = 2 T2 0 T
T
p t t R c
R c
2
dt = d2 2 d2 2 d2 2 0 0 = 2 2 = 0 2 . T 2 2
=
1 T 20
d2 2 sin2 t 0
dt =
Using the result (6.11), it is now easy to nd R cd2 R = 0 3 2
4
2c 2 2 = 2 2 2 d0 3
2
4
12 2 c
2 = 3c
2
2
.
Generally these range from tens of meters to tens of kilometers.
In this case charge polarized chemical bonds between the atoms in
the particles in the atmosphere act as little oscillators.34
33
82
6.3 Applicability of Classical Electrodynamics We conclude this
section by pointing out the range of applicability of classical
electrodynamics. The energy of the charge distribution in
electrodynamics is given by U= 1 2 dV (x)(x) .
Putting electron at rest, one can assume that the entire energy
of the electron coincides with its electromagnetic energy (electric
charge is assumed to be homogeneously distributed over a ball of
the radius re ) mc2 e2 , re
where m and e are the mass and the charge of electron. Thus, we
can dene the classical radius of electron re = e2 2.818 1015 m .
mc22
1 e In SI units it reads as re = 4 0 mc2 . At distances less
than re , the classical electrodynamics is not applicable.
In reality, due to quantum eects the classical electrodynamics
fails even at larger distances. The characteristic scale is the
Compton wavelength, which is the fundamental limitation on
measuring the position of a particle taking both quantum mechanics
and special relativity into account. Its theoretical value is given
by mc2
137 re 1013 m ,
1 where = 137 = e c is the ne structure constant for
electromagnetism. The most recent experimental measurement of
campton wavelenght (CODATA 2002) is one order of magnitude larger
and is approximately equal to 2.426 1012 m.
6.4 Darvins Lagrangian In classical mechanics a system of
interacting particles can be described by a proper Lagrangian which
depends on coordinates and velocities of all particles taken at the
one and the same moment. This is possible because in mechanics the
speed of propagation of signals is assumed to be innite. On the
other hand, in electrodynamics eld should be considered as an
independent entity having its own degrees of freedom. Therefore, if
one has a system of interacting charges (particles) for its
description one should consider a system comprising both these
particles and the eld. Thus, taking into account that the
83
propagation speed of interactions is nite, we arrive at the
conclusion that the rigorous description of a system of interacting
particles with the help of the Lagrangian depending on their
coordinates and velocities but do not containing degrees of freedom
related to the eld is impossible. However, if velocities v of all
the particles are small with respect to the speed of light, then
such a system can be approximately described by some Lagrangian. 2
The introduction of the Lagrangian function is possible up to the
terms of order v2 . c This is related to the fact that radiation of
electromagnetic waves by moving charges (that is an appearance of
independent eld) arises in the third order of v only. c At zero
approximation, i.e. by completely neglecting retarding of the
potentials, the Lagrangian for a system of charges has the form
L(0) =i 2 m i vi 2
i>j
ei ej . rij
The second term is the potential energy of non-moving charges.
In order to nd higher approximation, we rst write the Lagrangian
for a charge ei in an external electromagnetic eld (, A): Li = mc2
12 vi ei ei + (A vi ) . 2 c c
Picking up one of the charges, we determine electromagnetic
potentials created by all the other charges in a point where this
charge sits and express them via coordinates and velocities of the
corresponding charges (this can be done only approximately: 2 can
be determined up to the order v2 and A up to v ). Substituting the
found c c expressions for the potentials in the previous formula,
we will nd the Lagrangian for the whole system. Consider the
retarded potentials (x, t) = 1 A(x, t) = c d3 x dt3
t +
|xx | c
t
|x x | t +|xx | c
(x , t ) , j(x , t ) .
d x dt
t
|x x |
As before, integrating over t we get (x, t) = dx3
t
|xx | c
|x x |
,
1 A(x, t) = c
dx
3
j t
|xx | c
|x x |
.
If velocities of all the charges are small in comparison to the
speed of light, then the distribution of charges does not change
much for the time |xx | . Thus, the c
84
sources can be expanded in series in (x, t) = d3 x (t) 1 R c
t
|xx | . c
we have 1 2 2c2 t2 d3 x R(t) + . . .
d3 x (t) +
where R = |x x |. Since d3 x (t) is a constant charge of the
system, we have at leading and subleading orders the following
expression for the scalar potential (x, t) = d3 x (t) 1 2 + 2 2 R
2c t d3 x R(t) .
Analogous expansion takes place for the vector potential. Since
expression for the vector potential via the current already
contains 1/c and after the substitution in the Lagrangian is
multiplied by another power 1/c, it is enough to keep in the
expansion of A the leading term only, i.e. A= 1 c dx v . R
If the eld is created by a single charge, we have e e 2R = + 2 2
, R 2c t A= ev . cR
To simplify further treatment, we will make the gauge
transformation = where = This gives = e , R 1 , c t A =A+ e R . 2c
t ev e + cR 2c R . t ,
A =
R Here R = t x R and x R = R = n, where n is the unit vector
directed from the t charge to the observation point. Thus,
ev e A = + cR 2c t
R R
e ev + = cR 2c
R RR 2 R R
=
e ev + cR 2c
v RR 2 R R
.
Finally, since R2 = R2 , we nd RR = R R = R v. In this way we nd
= e , R A = e v + (v n) n . 2cR
If the eld is created by several charges then this expression
must be summed for all the charges.
85
Now substituting the potentials created by all the other charges
into the Lagrangian for a given charge ei we obtain Li =2 4 m i vi
1 m i vi + ei 2 8 c2
j=i
ei ej + 2 rij 2c
j=i
ej (vi vj ) + (vi nij )(vj nij ) . rij
Here we have also expanded the relativistic Lagrangian for the
point particle up to 2 the order v2 . From this expression we can
nd the total Lagrangian c L=i 2 mi vi + 2 4 m i vi 8c2
i
i>j
ei ej + rij
i>j
ei ej (vi vj ) + (vi nij )(vj nij ) . 2c2 rij
This Lagrangian was obtained by Darvin in 1922 and it expresses
an eect of electromagnetic interaction between charges up to the
second order in v . c It is interesting to nd out what happens if
we expand the potential further. For the scalar potential at third
order in 1/c and for the vector potential at second order in 1/c
one nds (3) = 1 3t 6c3 t3 d3 x R2 , A(2) = 1 c2 t d3 x j .
Performing a gauge transformation = with 1 , c t A =A+
1 2 = 2 2 6c t
d3 x R2 ,
we transform (3) into zero. The new vector potential will take
the form A (2) = 1 1 2 d3 x j 2 2 d3 x R2 c2 t 6c t 1 1 2 = 2 d3 x
j 2 2 d3 x R = c t 3c t 1 1 2 2 = 2 ev 2 2 d3 x (R0 r) = 2 c 3c t
3c
ev .
(6.12)
In the last formula we pass to the discrete distribution of
charges. This potential leads to a vanishing magnetic eld H = rot x
A (2) , as curl is taken with respect to the coordinates x of
observation point which A (2) does not depend on. For the electric
eld one nds E = A (2) /c, so that E= 2 ... d, 3c3
86
where d is the dipole moment of the system. Thus, additional
terms of the third order in the expansion of elds lead to the
appearance of additional forces which are not contained in Darvins
Lagrangian; these forces do depend on time derivatives of charge
accelerations. Compute the averaged work performed by elds for one
unit of time. Each charge experienced a force F = eE so that 2e ...
F = 3d. 3c The work produced is (F v) = 2e ... (d 3c3 ev) = 2 ... 2
d 2 ( d d) = 3 (d d) 3 d2 . 2 3c 3c dt 3c
Performing time average we arrive at (F v) = 2 2 d . 3c3
Now one can recognize that the expression of the right hand side
of the last formula is nothing else but the average radiation of
the system for one unit of time. Thus, the forces arising at third
order describe the backreaction which radiation causes on charges.
These forces are known as bracing by radiation or Lorentz friction
forces.
7. Advanced magnetic phenomenaMagnetic properties of all
substances admit a clear and logical systematization. At high
temperatures all of the substances are either diamagnetics or
paramagnetics. If some stu is put between the poles of a magnet,
the magnetic lines change in comparison to the situation when the
sta is absent. Under applying magnetic eld, all the substances get
magnetized. This means that every piece of volume behave itself as
a magnetic, while the magnetic moment of the whole body is a vector
sum of magnetic moments of all volume elements. A measure of
magnetization is given by M which is the magnetic moment density
(the magnetic dipole moment per unit volume). The product MV ,
where V is the volume, gives a total magnetic moment of a body M =
MV . A non-zero M appears only when external magnetic eld is
applied. When magnetic eld is not very strong, M changes linearly
with the magnetic eld H: M = H . Here is called magnetic
susceptibility (it is a dimensionless quantity). Then Paramagnetics
are the substances for which > 0
87
Diamagnetics are the substances for which < 0 Substances with
= 0 are absent in Nature Magnetic properties of substances are
often described not by but rather by the magnetic permeability: = 1
+ 4 . For paramagnetics > 1 and for diamagnetics < 1.
Introduce the magnetic induction B: B = H + 4 M . Then, B = H and =
1+4. Although vector B is called by a vector of magnetic induction
and H by a vector of magnetic eld, the actual sense of B is that it
is B (but not H!) is the average magnetic eld in media. For = 1/4
we have = 0. This is the situation of an ideal diamagnetic, in
which the average magnetic eld B = 0. Ideal magnetics do exists
they are superconductors. Absence of a magnetic eld inside a
superconductor is known as the Meissner-Ochsenfeld eect (1933). In
1895 Pierre Curie discovered that magnetic susceptibility is
inversely proportional to the temperature. The behavior of = (T )
is well described by the following Curie-Weiss law C (T ) = , T Tc
where C is a constant and Tc is known as the paramagnetic Curie
temperature. 7.1 Exchange interactions Identical particles behave
very dierently in classical and quantum mechanics. Classical
particles move each over its own trajectory. If positions of all
the particles were xed at the initial moment of time, solving
equations of motion one can always identify the positions of
particles at later times. In quantum mechanics the situation is
dierent, because the notion of trajectory is absent. If we x a
particle at a given moment of time, we have no possibility to
identify it among other particles at later moments of time. In
other words, in quantum mechanics identical particles are
absolutely indistinguishable. This principle implies that
permutation of two identical particles does not change a quantum
state of a system. Consider a wave-function of two particles (1,
2). Under permutation (1, 2) (2, 1) a state of a system should not
change. This means that (2, 1) = ei (1, 2) , where ei is a phase
factor. Applying permutation again, we get e2i = 1, i.e. ei = 1.
Thus, there are two types of particles:
88
1. (1, 2) = (2, 1) which corresponds to the Bose-Einstein
statistics 2. (1, 2) = (2, 1) which corresponds to the Fermi-Dirac
statistics Furthermore, an internal property which denes to which
class/statistics a particle belongs is the spin. Particles with
zero or integer spin obey the Bose-Einstein statistics, particles
with half-integer spin obeys the Fermi-Dirac statistics. Spin of
electron is 1/2, and, therefore, electrons are fermions. As such,
they obey the Pauli exclusion principle in each quantum state one
can nd only one electron. Consider a system consisting of two
electrons which interact only electrostatically. Neglecting
magnetic interaction between the electrons means neglecting the
existence of spins. Let (r1 , r2 ) be the orbital wave function.
Here r1 and r2 are coordinates of electrons. One cannot completely
forget about spins because the total wave function (1, 2) = S(1 , 2
)(r1 , r2 ) must be anti-symmetric. Here S(1 , 2 ) is the spin wave
function which describes a spin state of electrons. For two
electrons there are four states which lead to either anti-symmetric
wave function with the total spin S = 0: S = 0, or symmetric wave
function with S = 1: sz = 1 sz = 0 + sz = 1 Here sz is the
projection of spin on z-axis. For two electrons S = s1 + s2 and
taking square (quantum mechanically!) we obtain S(S + 1) = s1 (s1 +
1) + s2 (s2 + 1) + 2s1 s2 so that 1 s1 s2 = (S(S + 1) s1 (s1 + 1)
s2 (s2 + 1)) 2 From this formula we therefore nd that s1 s2 = 3 41
4
for S = 0 for S = 1
Returning back to the wave function we conclude that for S = 0
(r1 , r2 ) = s symmetric function for S = 1 (r1 , r2 ) = a
anti-symmetric function
89
Symmetric and anti-symmetric functions describe dierent orbital
motion of electrons and therefore they correspond to dierent values
of energies. Which energy is realized depends on a problem at hand.
For instance, for a molecule of H2 the minimal energy corresponds
to the symmetric wave function and, as a result, the electron spin
S is equal to zero.
Es S = 0 Ea S = 1 Spin Hamiltonian 1 Hs = (Es + 3Ea ) + (Ea Es
)s1 s2 4 1 Here the rst term 4 (Es + 3Ea ) E does not depend on
spin and represents the energy averaged over all spin states (three
states for S = 1 and one state for S = 0). The second term depends
on spins of electrons. Introducing A = Ea Es , we can write Hs = E
As1 s2 This allows to relate energetic preference of states with S
= 0 and S = 1 with the sign of A. For A < 0 the anti-parallel
conguration of spins is preferred, while for A > 0 parallel. The
parameter A is called an exchange integral. The Hamiltonian Hs
describes the so-called exchange interaction. 7.2 One-dimensional
Heisenberg model of ferromagnetism Here we will study in detail
so-called one-dimensional spin- 1 Heisenberg model of 2
ferromagnetism. We will solve it exactly by a special technique
known as coordinate Bethe ansatz. Consider a discrete circle which
is a collection of ordered points labelled by the index n with the
identication n n + L reecting periodic boundary conditions. Here L
is a positive integer which plays the role of the length (volume)
of the space. The numbers n = 1, . . . , L form a fundamental
domain. To each integer n along the chain we associate a
two-dimensional vector space V = C2 . In each vector space we pick
up the basis 1 0 | = , | = 0 1 We will call the rst element spin up
and the second one spin down. We introduce the spin algebra which
is generated by the spin variables Sn , where = 1, 2, 3, with
commutation relations [Sm , Sn ] = i Sn mn .
90
The spin operators have the following realization in terms of
the standard Pauli matrices: Sn = 2 and the form the Lie algebra
su(2). Spin variables are subject to the periodic boundary
condition Sn Sn+L .
Spin chain. A state of the spin chain can be represented as | =
|
The Hilbert space of the model has a dimension 2L and it isL
H =n=1
Vn = V1 VL
This space carries a representation of the global spin algebra
whose generators areL
S =n=1
I
Sn nth place
I.
The Hamiltonian of the model isL
H = Jn=1
Sn Sn+1 ,
where J is the coupling constant. More general Hamiltonian of
the formL
H=n=1
J Sn Sn+1 ,
where all three constants J are dierent denes the so-called XYZ
model. In what follows we consider only XXX model. The basic
problem we would like to solve is to nd the spectrum of the
Hamiltonian H. The rst interesting observation is that the
Hamiltonian H commutes with the spin operators. Indeed,L L
[H, S ] = J = i
[Sn Sn+1 , Sm ] n,m=1 L
= J
[Sn , Sm ]Sn+1 + Sn [Sn+1 , Sm ] n,m=1 Sn Sn+1 = 0 .
nmn,m=1
Sn Sn+1 n+1,m
91
In other words, the Hamiltonian is central w.r.t all su(2)
generators. Thus, the spectrum of the model will be degenerate all
states in each su(2) multiplet have the same energy. Sn
In what follows we choose = 1 and introduce the raising and
lowering operators 1 2 = Sn iSn . They are realized as S+ = 01 00 ,
S = 00 10 .
The action of these spin operators on the basis vectors are S +|
= 0 , S | = 0 , S +| = | , S | = | ,1 S 3| = 2 | , 1 S 3| = 2 |
.
This indices the action of the spin operators in the Hilbert
space+ Sk | k = 0 , Sk | k = 0 , + S k | k = | k , S k | k = | k ,
3 1 S k | k = 2 | k , 3 1 S k | k = 2 | k .
The Hamiltonian can be then written asL
H = Jn=1
+ 1 (Sn Sn+1 2
+ 3 3 + Sn Sn+1 ) + Sn Sn+1 ,
For L = 2 we have H = J S + S + S S + + 2S 3 S 30 0 1 0 2 1 0 =
J . 0 1 1 0 2 0 0 0 1 22
1
0
This matrix has three eigenvalues which are equal to 1 J and one
which is 2 Three states hw vs=1
3 J. 2
1 0 = , 0 0h.w.
0 1 , 1 0
0 0 0 1
corresponding to equal eigenvalues form a representation of
su(2) with spin s = 1 and the state hw vs=0
0 1 = 1 0h.w.
92
which corresponds to 3 J is a singlet of su(2). Indeed, the
generators of the global 2 su(2) are realized as0 0 S+ = 0 0 1 0 0
0 1 0 0 0 0 1 , 1 0 0 1 = 1 0 0 0 0 1 0 0 0 1 0 0 , 0 0 1 0 S3 = 0
0 0 0 0 0 0 0 0 0 . 0 0 0 1
S
hw hw The vectors vs=1 and vs=0 are the highest-weight vectors
of the s = 1 and s = 0 representations respectively, because they
are annihilated by S + and are eigenstates hw of S 3 . In fact,
vs=0 is also annihilated by S which shows that this state has zero
spin. Thus, we completely understood the structure of the Hilbert
space for L = 2.
In general, the Hamiltonian can be realized as 2L 2L symmetric
matrix which means that it has a complete orthogonal system of
eigenvectors. The Hilbert space split into sum of irreducible
representations of su(2). Thus, for L being nite the problem of
nding the eigenvalues of H reduces to the problem of diagonalizing
a symmetric 2L 2L matrix. This can be easily achieved by computer
provided L is sufciently small. However, for the physically
interesting regime L corresponding to the thermodynamic limit new
analytic methods are required. In what follows it is useful to
introduce the following operator: P = 1 II+ 2 = 2
1 II+ 4
S S
which acts on C2 C2 as the permutation: P (a b) = b a. It is
appropriate to call S 3 the operator of the total spin. On a state
| with M spins down we have S 3 | = 1 1 1 (L M ) M | = L M | . 2 2
2
Since [H, S 3 ] = 0 the Hamiltonian can be diagonalized within
each subspace of the full Hilbert space with a given total spin
(which is uniquely characterized by the number of spins down). Let
M < L be a number of overturned spins. If M = 0 we have a unique
state |F = | . This state is an eigenstate of the Hamiltonian with
the eigenvalue E0 = JL : 4L
H|F = Jn=1
3 3 Sn Sn+1 | =
JL | . 4
93
L! Let M be arbitrary. Since the M -th space has the dimension
(LM )!M ! one should nd the same number of eigenvectors of H in
this subspace. So let us write the eigenvectors of H in the
form
| =1n1 n1 n2 >n1 +2
a(n1 , n2 1)|n1 , n2 L8 2 n a(n1 , n2 )|n1 , n22 >n1 +1
a(n1 + 1, n2 )|n1 , n2 +n2 >n1
a(n1 , n2 + 1)|n1 , n2 +
a(n1 , n1 + 1) |n1 , n1 + 2 + |n1 1, n1 + 1 +
1n1 L
L4 |n1 , n1 + 1 . 2
Now we complete the sums in the rst bracket to run the range n2
> n1 . This is achieved by adding and subtracting the missing
terms. As the result we will getH| = J 2 1n1 L
a(n1 1, n2 ) + a(n1 , n2 1) + a(n1 + 1, n2 ) + a(n1 , n2 + 1)
+n2 >n1
L8 a(n1 , n2 ) |n1 , n2 2
a(n1 , n1 )|n1 , n1 + 1 + a(n1 + 1, n1 + 1)|n1 , n1 + 1 + + a(n1
, n1 + 1)|n1 , n1 + 2 + a(n1 , n1 + 2)|n1 , n1 + 2 + L8 a(n1 , n1 +
1)|n1 , n1 + 1 2 L4 a(n1 , n1 + 1) |n1 , n1 + 2 + |n1 1, n1 + 1 +
|n1 , n1 + 1 . 2
J 2
1n1 L
The underbraced terms cancel out and we nally getH| = J 2 a(n1
1, n2 ) + a(n1 , n2 1) + a(n1 + 1, n2 ) + a(n1 , n2 + 1) +n2
>n1
L8 a(n1 , n2 ) |n1 , n2 2 .
J + 2
a(n1 , n1 ) + a(n1 + 1, n1 + 1) 2a(n1 , n1 + 1) |n1 , n1 + 1
1n1 L
If we impose the requirement that a(n1 , n1 ) + a(n1 + 1, n1 +
1) 2a(n1 , n1 + 1) = 0 (7.1)
then the second bracket in the eigenvalue equation vanishes and
the eigenvalue problem reduces to the following equation 2(E E0
)a(n1 , n2 ) = J 4a(n1 , n2 ) =1
a(n1 + , n2 ) + a(n1 , n2 + ) . (7.2)
Substituting in eq.(7.1) the Bethe ansatz for a(n1 , n2 ) we get
Ae(p1 +p2 )n + Bei(p1 +p2 )n + Ae(p1 +p2 )(n+1) + Bei(p1 +p2 )(n+1)
2 Aei(p1 n+p2 (n+1)) + Bei(p2 n+p1 (n+1)) = 0 .
97
This allows one to determine the ratio B ei(p1 +p2 ) + 1 2eip2 =
i(p1 +p2 ) . A e + 1 2eip1Problem. Show that for real values of
momenta the ratioB A
is the pure phase:
B = ei(p2 ,p1 ) S(p2 , p1 ) . A
This phase is called the S-matrix. We further note that it obeys
the following relation S(p1 , p2 )S(p2 , p1 ) = 1 . Thus, the
two-magnon Bethe ansatz takes the form a(n1 , n2 ) = ei(p1 n1 +p2
n2 ) + S(p2 , p1 )ei(p2 n1 +p1 n2 ) , where we factored out the
unessential normalization coecient A. Let us now substitute the
Bethe ansatz in eq.(7.2). We get2(E E0 ) Aei(p1 n1 +p2 n2 ) +
Bei(p2 n1 +p1 n2 ) = J 4 Aei(p1 n1 +p2 n2 ) + Bei(p2 n1 +p1 n2 )
Aei(p1 n1 +p2 n2 ) eip1 + Bei(p2 n1 +p1 n2 ) eip2 Aei(p1 n1 +p2 n2
) eip1 + Bei(p2 n1 +p1 n2 ) eip2 Aei(p1 n1 +p2 n2 ) eip2 + Bei(p2
n1 +p1 n2 ) eip1 Aei(p1 n1 +p2 n2 ) eip2 + Bei(p2 n1 +p1 n2 ) eip1
.
We see that the dependence on A and B cancel out completely and
we get the following equation for the energy2
E E0 = J 2 cos p1 cos p2 = 2Jk=1
sin2
pk . 2
Quite remarkably, the energy appears to be additive, i.e. the
energy of a two-magnon state appears to be equal to the sum of
energies of one-magnon states! This shows that magnons essentially
behave themselves as free particles in the box. Finally, we have to
impose the periodicity condition a(n2 , n1 + L) = a(n1 , n2 ). This
results into ei(p1 n2 +p2 n1 ) eip2 L + which implies eip1 L = A =
S(p1 , p2 ) , B eip2 L = B = S(p2 , p1 ) . A B ip1 L i(p2 n2 +p1 n1
) B e e = ei(p1 n1 +p2 n2 ) + ei(p2 n1 +p1 n2 ) A A
98
The last equations are called Bethe equations. They are nothing
else but the quantization conditions for momenta pk .
Let us note the following useful representation for the
S-matrix. We haveS(p2 , p1 ) = = = eip2 eip1 1 + 1 eip2 eip2 e 2 p1
e 2 p1 e 2 p1 + e 2 p2 e 2 p2 e 2 p2 = i i i i i i eip1 eip2 1 + 1
eip1 eip1 e 2 p2 e 2 p2 e 2 p2 + e 2 p1 e 2 p1 e 2 p1 e 2 p2 sin p1
e 2 p1 sin p2 2 2 e 2 p1 sin p2 e 2 p2 sin p1 2 2i i i i i i i i i
i
=
cos p2 + i sin p2 sin p1 cos p1 i sin p1 sin p2 2 2 2 2 2 2 cos
p1 + i sin p1 sin p2 cos p2 i sin p2 sin p1 2 2 2 2 2 21 2 1 2
cos p2 sin p1 cos p1 sin p2 + 2i sin p1 sin p2 2 2 2 2 2 2 = cos
p1 sin p2 cos p2 sin p1 + 2i sin p1 sin p2 2 2 2 2 2 21 2 1 2
cot p2 2 cot p2 2
1 2 1 2
cot p1 + i 2 . cot p1 i 2
Thus, we obtained S(p1 , p2 ) =
cot p21 1 cot p22 + i 2 . cot p21 1 cot p22 i 2
It is therefore convenient to introduce the variable = 1 cot p
which is called 2 2 rapidity and get 1 2 + i S(1 , 2 ) = . 1 2 i
Hence, on the rapidity plane the S-matrix depends only on the
dierence of rapidities of scattering particles.
Taking the logarithm of the Bethe equations we obtain Lp1 = 2m1
+ (p1 , p2 ) , Lp2 = 2m2 + (p2 , p1 ) ,
where the integers mi {0, 1, . . . , L 1} are called Bethe
quantum numbers. The Bethe quantum numbers are useful to
distinguish eigenstates with dierent physical properties.
Furthermore, these equations imply that the total momentum is P =
p1 + p2 = Writing the equations in the form p1 = 2m1 1 + (p1 , p2 )
, L L p2 = 2m2 1 + (p2 , p1 ) , L L 2 (m1 + m2 ) . L
we see that the magnon interaction is reected in the phase shift
and in the deviation of the momenta p1 , p2 from the values of the
underbraced one-magnon wave numbers. What is very interesting, as
we will see, that magnons either scatter o each other or form the
bound states.
99
The rst problem is to nd all possible Bethe quantum numbers (m1
, m2 ) for which Bethe equations have solutions. The allowed pairs
(m1 , m2 ) are restricted to 0 m1 m2 L 1 . This is because
switching m1 and m2 simply interchanges p1 and p2 and produces 1
the same solution. There are 2 L(L + 1) pairs which meet this
restriction but only 1 L(L 1) of them yield a solution of the Bethe
equations. Some of these solutions 2 have real p1 and p2 , the
others yield the complex conjugate momenta p2 = p . 1 The simplest
solutions are the pairs for which one of the Bethe numbers is zero,
e.g. m1 = 0, m = m2 = 0, 1, . . . , L 1. For such a pair we have
Lp1 = (p1 , p2 ) , Lp2 = 2m + (p2 , p1 ) ,
which is solved by p1 = 0 and p2 = 2m . Indeed, for p1 = 0 the
phase shift vanishes: L (0, p2 ) = 0. These solutions have the
dispersion relation E E0 = 2J sin2 p , 2 p = p2
which is the same as the dispersion for the one-magnon states.
These solutions are nothing else but su(2)-descendants of the
solutions with M = 1. One can show that for M = 2 all solutions are
divided into three distinct classes Descendents ,L
Scattering States ,L(L5) +3 2
Bound StatesL3
so that
L(L 5) +3+L3= 2 gives a complete solution space of the
two-magnon L+
1 L(L 1) 2 problem.
Pseudovacuum
F
L onemagnon states
1 0
1 0
1 0
1 0
1 0 1 0
1 0 1 0
1 0 1 0
L(L1) twomagnon states 0 1 2
1 0
1 0 1 0
1 0 1 0
1 0 1 0
1 0
1 0
1 0 1 0
1 0 1 0
1 0 1 0
1 0 1 0
100
The su(2)-multiplet structure of the M = 0, 1, 2 subspaces.
The most non-trivial fact about the Bethe ansatz is that
many-body (multimagnon) problem reduces to the two-body one. It
means, in particular, that the multi-magnon S-matrix appears to be
expressed as the product of the two-body ones. Also the energy is
additive quantity. Such a particular situation is spoken about as
Factorized Scattering. In a sense, factorized scattering for the
quantum many-body system is the same as integrability because it
appears to be a consequence of existence of additional conservation
laws. For the M -magnon problem the Bethe equations readM
e
ipk L
=j=1 j=k
S(pk , pj ) .
The most simple description of the bound states is obtained in
the limit when L . If pk has a non-trivial positive imaginary part
then eipk L tends to and this means that the bound states
correspond in this limit to poles of the r.h.s. of the Bethe
equations. In particular, for the case M = 2 the bound states
correspond to poles in the two-body S-matrix. In particular, we nd
such a pole when1 2
cot p21 1 cot p22 = i . 2
This state has the total momentum p = p1 + p2 which must be
real. These conditions can be solved by taking p p p1 = + iv , p2 =
iv . 2 2 The substitution gives cos 1 2p 2
+ iv sin 1 2
p 2
iv cos 1 2
p 2
iv sin 1 2p 2
p 2
+ ivp 2
1 = 2i sin 2
+ iv sin 1 2
iv ,
which is cos The energy of such a state is p2 p1 + sin2 E = 2J
sin2 2 2 We therefore get E = 2J 1 cos
p = ev . 2 p p v v +i + sin2 i 4 2 4 2 .
= 2J sin2
p p cos2 p + 1 p 2 cosh v = 2J 1 cos = J sin2 . p 2 2 2 cos 2
2
Thus, the position of the pole uniquely xes the dispersion
relation of the bound state.
101
7.3 Landau-Lifshitz equation
8. The Ginzburg-Landau Theory 9. Elements of Fluid Mechanics9.1
Eulers equation 9.2 Bernoullis equation 9.3 The Navier-Stokes
equation
10. Non-linear phenomena in mediaRemarkably, there exist certain
dierential equations for functions depending on two variables (x,
t) which can be treated as integrable Hamiltonian systems with
innite number of degrees of freedom. This is an (incomplete) list
of such models The Korteweg-de-Vries equation u = 6uux uxxx . t The
non-linear Schrodinger equation i = xx + 2||2 , t
where = (x, t) is a complex-valued function. The Sine-Gordon
equation 2 2 m2 2+ sin = 0 t2 x The classical Heisenberg magnet S
2S =S , t x2 where S(x, t) lies on the unit sphere in R3 . The
complete specication of each model requires also boundary and
initial conditions. Among the important cases are 1. Rapidly
decreasing case. We impose the condition that (x, t) 0 when |x|
suciently fast, i.e., for instance, it belongs to the Schwarz
space L (R1 ), which means that is dierentiable function which
vanishes faster than any power of |x|1 when |x| .
102
2. Periodic boundary conditions. Here we require that is
dierentiable and satises the periodicity requirement (x + 2, t) =
(x, t) . The soliton was rst discovered by accident by the naval
architect, John Scott Russell, in August 1834 on the Glasgow to
Edinburg channel.37 The modern theory originates from the work of
Kruskal and Zabusky in 1965. They were the rst ones to call Russels
solitary wave a solition. 10.1 Solitons Here we discuss the
simplest cnoidal wave type (periodic) and also one-soliton
solutions of the KdV and SG equations For the discussion of the
cnoidal wave and the one-soliton solution of the non-linear
Schrodinger equation see the corresponding problem in the problem
set. Korteweg-de-Vries cnoidal wave and soliton By rescaling of t,
x and u one can bring the KdV equation to the canonical form ut +
6uux + uxxx = 0 . We will look for a solution of this equation in
the form of a single-phase periodic wave of a permanent shape u(x,
t) = u(x vt) , where v = const is the phase velocity. Plugging this
ansatz into the equation we obtain d vux + 6uux + uxxx = vu + 3u2 +
uxx = 0 . dx We thus get vu + 3u2 + uxx + e = 0 ,Russel described
his discovery as follows: I believe I shall best introduce this
phenomenon by describing the circumstances of my own rst
acquaintance with it. I was observing the motion of a boat which
was rapidly drawn along a narrow channel by a pair of horses, when
the boat suddenly stopped-not so the mass of the water in the
channel which it had put in motion; it accumulated round the prow
of the vessel in a state of violent agitation, then suddenly
leaving it behind, rolled forward with great velocity, assuming the
form of a large solitary elevation, a rounded, smooth and
well-dened heap of water, which continued its course along the
channel apparently without change of form or diminution of speed. I
followed it on horseback, and overtook it still rolling on at a
rate of some eight or nine miles an hour, preserving its original
gure
![ALGEBRAIC STRUCTURE OF CLASSICAL FIELD THEORY: FOR REAL SCALAR FIELDS · 2017-12-11 · arxiv:1209.2148v4 [math-ph] 7 dec 2017 algebraic structure of classical field theory: kinematics](https://img.dokumen.tips/doc/110x75/5b3c1ae97f8b9a5e1f8d2e70/algebraic-structure-of-classical-field-theory-for-real-scalar-fields-2017-12-11.jpg)
