Download Solution to problem 2

Week 1, solution to exercise 2
I. THE ACTION FOR CLASSICAL ELECTRODYNAMICS
A. Maxwell’s equations in relativistic form
Maxwell’s equations in vacuum and in natural units (c = 1) are,
∇ · E = ρ,
∇ · B = 0,
∇ × B − ∂t E = j
∇ × E + ∂t B = 0
(inhomogeneous),
(homogeneous).
(1a)
(1b)
Homogeneous
equations. The first one, ∇ · B = 0, is the Gauss’ law for magnetism, and can be recast in the integral form
as Σ B · dn = 0 (thanks to the Stokes’ theorem), which means that the net flux of the magnetic field through any closed surface
Σ is always zero; in other words,
H there are no!magnetic monopoles. The second one, ∇ × E + ∂t B = 0, is the Faraday’s law of
induction, in the integral form, ∂Σ E · dl = − Σ ∂t B · dn, where Σ is a surface bounded by the closed contour ∂Σ; it means that
a time-varying magnetic flux through a surface spanned over a closed circuit induces the electromotive force in this circuit.
Scalar and vector potentials. Both these laws can be rewritten using the concept of scalar and vector potentials, φ and A,
using the Helmholtz decomposition theorem, which states that any twice continuously differentiable vector field F on a bounded
domain V ⊂ R3 (or on R3 itself, and F decays faster than 1/r at infinity) can be decomposed into a a curl-free component and a
divergence-free component, F = −∇φ + ∇ × A. Therefore,
B = ∇ × A,
E = −∇φ − ∂t A.
(2)
Notice that the potentials are not uniquely determined; one may always perform a gauge transformation, φ → φ + ∂t Γ and
A → A − ∇Γ, and find the same electric and magnetic fields.
Four-potential. To write it in the special relativistic language, one introduces the four-potential Aµ = (φ, A). Its gauge
freedom is thus Aµ → Aµ + ∂µ Γ, where recall ∂µ = (∂t , ∇) and ∂µ = (∂t , −∇) (our convention for the metric tensor is + − −−).
One essentially postulates that this is a Lorentz four-vector, postulates an action for it (see below), and derives from it the
Maxwell’s equations, thus a posteriori identifying its components with the scalar and vector potentials.
Define now the field strength four-tensor,
F µν ≡ ∂µ Aν − ∂ν Aµ .
(3)
This is obviously a gauge-invariant antisymmetric four-tensor of rank 2; it is enough to compute its time-space components,
using (2),
F 0i = ∂0 Ai − ∂i A0 = ∂t Ai + ∂i A0 = −E i − ∂i φ + ∂i φ = −E i ,
while its space-space components,
F i j = ∂i A j − ∂ j Ai = −∂i A j + ∂ j Ai = −ǫ i jk Bk ,
where the last equality follows from (2) and the identity,
ǫ i jk ǫ lmk = δil δ jm − δim δ jl ,
(4)
since,
Finally thus, in matrix form,
−ǫ i jk Bk = −ǫ i jk ǫ klm ∂l Am = − δil δ jm − δim δ jl ∂l Am = −∂i A j + ∂ j Ai .
F µν


 0 −E 1 −E 2 −E 3 
 E 1 0 −B3 B2 
 .
=  2
0 −B1 
 E B3
E 3 −B2 B1
0
(5)
Inhomogeneous equations in covariant form. To write the inhomogeneous Maxwell’s equations in a manifestly Lorentzcovariant form, first combine the electric charge density ρ and electric current density j into the four-current, jµ = (ρ, j). Then,
both these equations can be cast as
∂µ F µν = jν .
(6)
2
Indeed, its time component,
ρ = j0 = ∂i F i0 = ∂i E i = ∇ · E,
which is the Gauss’ law for electricity (integral form: Σ E · dn = Q, i.e. the net flux of the electric field through a closed surface
Σ is equal to the total charge enclosed within Σ). Its space component,
ji = ∂µ F µi = ∂0 F 0i + ∂ j F ji = −∂t E i + ǫ i jk ∂ j Bk = (−∂t E + ∇ × B)i ,
H
!
which is the Ampère’s circuital law (integral form: ∂Σ B · dl = Σ (j + ∂t E) · dn, which may be interpreted as the fact that the
current, including the displacement current ∂t E, generates a magnetic field).
Continuity equation. Notice that the inhomogeneous Maxwell’s equation implies the continuity equation,
∂ν jν = ∂ν ∂µ F µν = 0,
(7)
since the part with partial derivatives is symmetric in the indices, while the field strength is antisymmetric. This expresses the
electric charge conservation.
B.
The action
Action in covariant form. Postulate the following action, treating as the dynamical variables the components of the fourpotential,
Z
(8)
S [A] =
d4 x c1 Fµν F µν + c2 Aµ jµ ,
where c1,2 are some constants. It is Lorentz-invariant in an obvious way. It is also gauge-invariant: the first term in an obvious
way (as the field strength is gauge-invariant), the second term also, since
Z
Z
d4 x(∂µ Γ) jµ = − d4 x(∂µ jµ )Γ = 0,
by virtue of integration by parts and the continuity equation (7). As an example of a more complicated term, which is however
not gauge-invariant, take
Aµ Aµ → (Aµ + ∂µ Γ)(Aµ + ∂µ Γ) = Aµ Aµ + 2(∂µ Γ)Aµ + (∂µ Γ)(∂µ Γ).
Note: This is the mass term in the Proca action describing a massive spin-1 field.
Euler-Lagrange’s equations for this action,
∂L
∂L
= ∂ν
.
∂Aµ
∂(∂ν Aµ )
Now, the l.h.s. comes only from the second term in the Lagrangian density, since the first term depends only on the field strength,
which in turn depends only on the derivative of the four-potential,
∂L
= c2 jµ .
∂Aµ
The r.h.s. comes from the first term; denote for short, fµν ≡ ∂µ Aν ; then,
∂
∂
∂L
= c1
( fαβ − fβα )( f αβ − f βα ) = 2c1
( fαβ f αβ − fαβ f βα ) = . . . ,
∂ fνµ
∂ fνµ
∂ fνµ
and since the non-zero contributions can come only from the terms with (α, β) = (µ, ν) or (α, β) = (ν, µ), so (the summation
convention suspended below),
. . . = 2c1
∂
( fµν f µν − fµν f νµ + fνµ f νµ − fνµ f µν ) = 2c1 (0 − f µν + 2 f νµ − f µν ) = 4c1 F νµ .
∂ fνµ
Consequently, the Euler-Lagrange’s equations become
4c1 ∂ν F νµ = c2 jµ ,
3
which is equivalent to the inhomogeneous Maxwell’s equation provided that c1 = c2 /4, i.e.,
!
Z
1
S [A] = c2 d4 x Fµν F µν + Aµ jµ .
4
(9)
Action in three-dimensional form. It is instructive to rewrite the Lagrangian density of (8) in the language of electric and
magnetic fields. Consider first the Lorentz scalar,
Fµν F µν = F0i F 0i + Fi0 F i0 + Fi j F i j = −2E2 + ǫ i jk ǫ i jl Bk Bl = −2(E2 − B2 ),
(10)
ǫi jk ǫi jl = 2δkl ,
(11)
where the identity,
has been used. Therefore,
S [A] = c2
II.
Z
!
1 2
2
d x − (E − B ) + φρ − A · j .
2
4
(12)
THE HAMILTONIAN FOR PURE CLASSICAL ELECTRODYNAMICS
Conjugate momenta. Consider now a pure electromagnetic field, i.e. without any sources of electric charge or current,
jµ = 0. In order to pass to the Hamiltonian formalism, calculate first the momenta conjugate to the dynamical variables Aµ ,
πµ =
∂L
∂L
= c2 F 0µ = −c2 (0, E),
=
∂(∂0 Aµ ) ∂ f0µ
(13)
where we have used an intermediate formula in the derivation of the Euler-Lagrange’s equations above, as well as the matrix
expression for the field strength.
Note: The momentum conjugate to A0 = φ is zero! This is a straightforward consequence of the fact that the Lagrangian
density does not contain a time derivative of A0 (its velocity), so this is not a dynamical field.
Hamiltonian. The Hamiltonian density is obtained from the Lagrangian density (written for convenience in the threedimensional form) by using the Legendre transformation,
H = π µ ∂ 0 Aµ +
c2 2
(E − B2 ).
2
Now,
πµ ∂0 Aµ = −π · ∂t A = c2 E · (−∇φ − E) = −c2 (E · ∇φ + E2 ),
and moreover, consider a quantity
∇ · (Eφ) = (∇ · E)φ + E · ∇φ = E · ∇φ,
as the first term vanishes by virtue of the electric Gauss’ law with zero charge density. Therefore,
H =−
c2 2
(E + B2 ) − c2 ∇ · (Eφ).
2
The Hamiltonian is the integral over three-space of the Hamiltonian density. By virtue of the Stokes’ theorem, the second
term turns into a surface integral of Eφ at infinity, which vanishes (under reasonable conditions for the electric field). Hence,
only the first term contributes,
Z
Z
−c2 2
(E + B2 ).
(14)
H=
d3 x H =
d3 x
2
Thus in order to recover the known expression for the energy density of the electromagnetic field, one has to set c2 = −1.
4
III. THE ENERGY-MOMENTUM TENSOR FOR PURE CLASSICAL ELECTRODYNAMICS
A. Noether’s theorem
General derivation. Let us recall how to derive the Noether’s theorem in a fairly general setting of a classical field
theory of a multi-component field φi (x), with some Lagrangian density depending only on the field and its first derivative,
L = L[φi (x), ∂µ φi (x)].
Consider an infinitesimal joint transformation of the space-time and the field,
x′µ = xµ + δxµ ,
′i
′
i
(15a)
i
φ (x ) = φ (x) + δφ (x).
(15b)
First, notice that the variation δ of the field accommodates both the change in the position and the field value. Define additionally a variation δ̄ which is concerned merely with the change in the field’s value,
δ̄φi (x) ≡ φ′i (x) − φi (x).
(16)
Since there is to first order,
φ′i (x′ ) = φ′i (x + δx) = φ′i (x) + (∂µ φ′i )δxµ = φ′i (x) + (∂µ φi )δxµ ,
we find,
δ̄φi (x) ≡ δφi − (∂µ φi )δxµ .
(17)
Second, the relation between position derivatives before and after the transformation reads,
∂′µ xν = ∂′µ (x′ν − δxν ) = δνµ − ∂′µ δxν = δνµ − ∂µ δxν ,
i.e.,
∂′µ = (∂′µ xν )∂ν = ∂µ − (∂µ δxν )∂ν .
(18)
Third, notice that the derivative ∂µ and the variation δ do not commute,
n
o
[∂µ , δ]φi (x) = ∂µ (δφi (x)) − ∂′µ φ′i (x′ ) − ∂µ φi (x) .
Now to first order,
on
o
n
∂′µ φ′i (x′ ) = ∂µ − (∂µ δxν )∂ν φi (x) + δφi (x) = ∂µ φi (x) − (∂µ δxν )(∂ν φi (x)) + ∂µ (δφi (x)),
so that finally,
[∂µ , δ]φi(x) = (∂µ δxν )(∂ν φi (x)).
(19)
On the other hand, if the variation δ̄ is used instead, the commutator vanishes. Indeed, from (19) and (17),
o
o n
n
[∂µ , δ]φi (x) = ∂µ δ̄φi (x) + (∂ν φi (x))δxν − δ̄(∂µ φi (x)) + (∂ν ∂µ φi (x))δxν = [∂µ , δ̄]φi (x) + (∂µ δxν )(∂ν φi (x)),
which proves the assertion that
[∂µ , δ̄]φi (x) = 0.
(20)
Fourth, the variation δ of the Lagrangian density consists of varying the position and the field,
δL = (∂µ L)δxµ + δ̄L.
The latter reads
δ̄L =
∂L
∂L i
δ̄φ +
δ̄(∂µ φi ) = . . .
∂φi
∂(∂µ φi )
(21)
5
In the first term, use the Euler-Lagrange’s equations, while in the second term commute the derivative and variation according
to (20); this yields
(
)
∂L
i
. . . = ∂µ
δ̄φ .
∂(∂µ φi )
Finally, using (17),
δ̄L = ∂µ
(
)
∂L i
i
ν
δφ − (∂ν φ )δx .
∂(∂µ φi )
(22)
Fifth, vary the Lagrangian density multiplied by the volume element (i.e., the integrand of the action),
δ(dΩ L) = δ(dΩ)L + dΩ δL.
Now the volume element transforms by the Jacobian, which to first order gives
o
i
n
h
dΩ′ = det ∂ν x′µ dΩ = det δµν + ∂ν δxµ dΩ = 1 + ∂µ δxµ dΩ,
i.e.,
δ(dΩ) = (∂µ δxµ )dΩ.
(23)
Finally therefore, using (21) and (22),
(
n
o
n
o
δ(dΩ L) = dΩ (∂µ δxµ )L + δL = dΩ ∂µ (Lδxµ ) + δ̄L = dΩ ∂µ Lδxµ +
)
∂L i
i
ν
.
δφ
−
(∂
φ
)δx
ν
∂(∂µ φi )
(24)
Before we summarize, decompose the infinitesimal changes in the position and the field [(15a), (15b)] in s small parameters
δωa , a = 1, . . . , s,
µ
δxµ = Xa δωa ,
δφi = Φia δωa .
(25)
Assume also that the transformation in question is a symmetry of the action, i.e. δ(dΩ L) = 0. Then (24) implies that there exist
s Noether’s currents,
jµa ≡ −LXaµ −
∂L i
Φa − (∂ν φi )Xaν ,
i
∂(∂µ φ )
(26)
∂µ jµa = 0.
(27)
which are conserved,
This implies in particular (via Stokes’ theorem) that the time component of each current integrated over the whole three-space
is time-independent.
Note: The Noether’s currents are not uniquely determined. One may always modify them by
jµa → jµa + ∂ν Kaµν ,
Kaµν = −Kaνµ ,
(28)
as this does not influence the conservation laws (27).
Canonical energy-momentum tensor. The simplest application of Noether’s theorem is found when the field is symmetric
µ
µ
w.r.t. space-time translations, Xν = δν and Φiν = 0. There are thus 4 conserved currents, forming the so-called canonical energymomentum tensor,
T µν = −Lηµν +
and its conservation means that ∂µ T µν = 0.
∂L
(∂ν φi ),
∂(∂µ φi )
(29)
6
B. Electromagnetic energy-momentum tensor
Canonical. Let now the field be the electromagnetic four-potential Aµ , with the corresponding action without sources,
Z
1
S [A] = −
d4 x Fαβ F αβ .
4
(30)
The canonical energy-momentum tensor thus reads from (29),
T µν =
1 µν
η Fαβ F αβ − F µσ (∂ν Aσ ).
4
(31)
Symmetric. This is obviously not symmetric, but we may exploit the freedom (28) to attempt to find an object K µσν , antisymmetric in the first two indices, such that T µν + ∂σ K µσν is symmetric. Choose,
K µσν ≡ F µσ Aν .
(32)
Then,
−F µσ (∂ν Aσ ) + ∂σ (F µσ Aν ) = −F µσ (∂ν Aσ ) + (∂σ F µσ ) Aν + F µσ (∂σ Aν ) = F µσ Fσν ,
| {z }
=0
so that the symmetric tensor,
T̃ µν =
1 µν
η Fαβ F αβ + F µσ Fσν .
4
(33)
Conserved charges. The four conserved charges are the time components (µ = 0) of (33), integrated over the whole threespace. The integrand of the charge corresponding to ν = 0,
3
T̃ 00 =
X
1
1
1
1
Fαβ F αβ + F 0σ Fσ0 = − (E2 − B2 ) +
(F 0i )2 = − (E2 − B2 ) + E2 = (E2 + B2 ).
4
2
2
2
i=1
Choosing on the other hand ν = i gives,
T̃ 0i = F 0 j F j i = −F 0 j F ji = E j ǫ i jk Bk = (E × B)i .