Week 1, solution to exercise 2

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1 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 t E=j (inhomogeneous), (1a) B=0, E+ t B=0 (homogeneous). (1b) Homogeneous equations. The first one, B=0, is the Gauss law for magnetism, and can be recast in the integral form as B dn=0(thanks to the Stokes theorem), which means that the net flux of the magnetic field through any closed surface Σ Σ is always zero; in other words, there are no magnetic monopoles. The second one, E+ t B=0, is the Faraday s law of induction, in the integral form, E dl= Σ Σ tb dn, whereσis 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 R 3 (or onr 3 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 A µ = (φ, A). Its gauge freedom is thus A µ A µ + 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 A ν ν A µ. (3) This is obviously a gauge-invariant antisymmetric four-tensor of rank 2; it is enough to compute its time-space components, using (2), while its space-space components, where the last equality follows from (2) and the identity, since, F 0i = 0 A i i A 0 = t A i + i A 0 = ( E i i φ ) + i φ= E i, F i j = i A j j A i = i A j + j A i = ǫ i jk B k, ǫ i jk ǫ lmk =δ il δ jm δ im δ jl, (4) ǫ i jk B k = ǫ i jk ǫ klm l A m = ( δ il δ jm δ im δ jl ) l A m = i A j + j A i. Finally thus, in matrix form, 0 E 1 E 2 E 3 F µν E = 1 0 B 3 B 2 E 2 B 3 0 B 1. (5) E 3 B 2 B 1 0 Inhomogeneous equations in covariant form. To write the inhomogeneous Maxwell s equations in a manifestly Lorentzcovariant form, first combine the electric charge densityρand electric current density j into the four-current, j µ = (ρ, j). Then, both these equations can be cast as µ F µν = j ν. (6)

2 2 Indeed, its time component, ρ= j 0 = i F i0 = i E i = E, which is the Gauss law for electricity (integral form: E dn= Q, i.e. the net flux of the electric field through a closed surface Σ Σ is equal to the total charge enclosed withinσ). Its space component, j i = µ F µi = 0 F 0i + j F ji = t E i +ǫ i jk j B k = ( t E+ B) i, which is the Ampère s circuital law (integral form: Σ B dl= Σ (j+ te) 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 j ν = 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, S [A]= d 4 x ( c 1 F µν F µν + c 2 A µ j µ), (8) where c 1,2 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 d 4 x( µ Γ) j µ = d 4 x( µ j µ )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 A µ A µ (A µ + µ Γ)(A µ + µ Γ)=A µ A µ + 2( µ Γ)A µ + ( 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, A µ = ν ( ν A µ ). 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, A µ = c 2 j µ. The r.h.s. comes from the first term; denote for short, f µν µ A ν ; then, = c 1 ( f αβ f βα )( f αβ f βα )=2c 1 ( f αβ f αβ f αβ f βα )=..., f νµ f νµ f νµ 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),...=2c 1 ( f µν f µν f µν f νµ + f νµ f νµ f νµ f µν )=2c 1 (0 f µν + 2 f νµ f µν )=4c 1 F νµ. f νµ Consequently, the Euler-Lagrange s equations become 4c 1 ν F νµ = c 2 j µ,

3 which is equivalent to the inhomogeneous Maxwell s equation provided that c 1 = c 2 /4, i.e., ( ) 1 S [A]=c 2 d 4 x 4 F µνf µν + A µ j µ. (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, where the identity, F µν F µν = F 0i F 0i + F i0 F i0 + F i j F i j = 2E 2 +ǫ i jk ǫ i jl B k B l = 2(E 2 B 2 ), (10) ǫ i jk ǫ i jl = 2δ kl, (11) 3 has been used. Therefore, S [A]=c 2 ( d 4 x 1 ) 2 (E2 B 2 )+φρ A j. (12) II. THE HAMILTONIAN FOR PURE CLASSICAL ELECTRODYNAMICS Conjugate momenta. Consider now a pure electromagnetic field, i.e. without any sources of electric charge or current, j µ = 0. In order to pass to the Hamiltonian formalism, calculate first the momenta conjugate to the dynamical variables A µ, π µ = ( 0 A µ ) = = c 2 F 0µ = c 2 (0, E), (13) f 0µ 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 A 0 =φ is zero! This is a straightforward consequence of the fact that the Lagrangian density does not contain a time derivative of A 0 (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, Now, H=π µ 0 A µ + c 2 2 (E2 B 2 ). π µ 0 A µ = π t A=c 2 E ( φ E)= c 2 (E φ+e 2 ), 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= c 2 2 (E2 + B 2 ) c 2 (Eφ). 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, H= d 3 xh= d 3 x c 2 2 (E2 + B 2 ). (14) Thus in order to recover the known expression for the energy density of the electromagnetic field, one has to set c 2 = 1.

4 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 fieldφ i (x), with some Lagrangian density depending only on the field and its first derivative, L=L[φ i (x), µ φ i (x)]. Consider an infinitesimal joint transformation of the space-time and the field, x µ = x µ +δx µ, φ i (x )=φ i (x)+δφ i (x). (15a) (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 δ which is concerned merely with the change in the field s value, Since there is to first order, we find, i.e., δφ i (x) φ i (x) φ i (x). (16) φ i (x )=φ i (x+δx)=φ i (x)+( µ φ i )δx µ =φ i (x)+( µ φ i )δx µ, δφ i (x) δφ i ( µ φ i )δx µ. (17) Second, the relation between position derivatives before and after the transformation reads, µx ν = µ(x ν δx ν )=δ ν µ µδx ν =δ ν µ µ δx ν, Third, notice that the derivative µ and the variationδdo not commute, Now to first order, so that finally, µ= ( µx ν ) ν = µ ( µ δx ν ) ν. (18) [ µ,δ]φ i (x)= µ (δφ i (x)) { µφ i (x ) µ φ i (x) }. µ φ i (x )= { µ ( µ δx ν ) ν }{ φ i (x)+δφ i (x) } = µ φ i (x) ( µ δx ν )( ν φ i (x))+ µ (δφ i (x)), [ µ,δ]φ i (x)=( µ δx ν )( ν φ i (x)). (19) On the other hand, if the variation δ is used instead, the commutator vanishes. Indeed, from (19) and (17), [ µ,δ]φ i (x)= µ { δφ i (x)+( ν φ i (x))δx ν} { δ( µ φ i (x))+( ν µ φ i (x))δx ν} = [ µ, δ]φ i (x)+( µ δx ν )( ν φ i (x)), which proves the assertion that [ µ, δ]φ i (x)=0. (20) Fourth, the variationδof the Lagrangian density consists of varying the position and the field, δl=( µ L)δx µ + δl. (21) The latter reads δl= φ i δφ i + ( µ φ i ) δ( µ φ i )=...

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 { }...= µ ( µ φ i ) δφ i. Finally, using (17), { ( δl= µ δφ i ( µ φ i ( ν φ i )δx ν)}. (22) ) Fifth, vary the Lagrangian density multiplied by the volume element (i.e., the integrand of the action), δ(dωl)=δ(dω)l+dωδl. Now the volume element transforms by the Jacobian, which to first order gives i.e., Finally therefore, using (21) and (22), dω = det [ ν x µ] dω=det [ δ µ ν+ ν δx µ] dω= { 1+ µ δx µ} dω, δ(dω)=( µ δx µ )dω. (23) δ(dωl)=dω { ( µ δx µ )L+δL } = dω { µ (Lδx µ )+ δl } { = dω µ Lδx µ + ( δφ i ( µ φ i ( ν φ i )δx ν)}. (24) ) Before we summarize, decompose the infinitesimal changes in the position and the field [(15a), (15b)] in s small parameters δω a, a=1,..., s, δx µ = X µ aδω a, δφ i =Φ i aδω a. (25) Assume also that the transformation in question is a symmetry of the action, i.e.δ(dωl)=0. Then (24) implies that there exist s Noether s currents, which are conserved, j µ a LX a µ ( ) Φ i ( µ φ i ) a ( ν φ i )Xa ν, (26) µ j µ a= 0. (27) 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 j µ a j µ a+ ν K µν a, K µν a = K νµ a, (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 w.r.t. space-time translations, X µ ν=δ µ ν andφ i ν= 0. There are thus 4 conserved currents, forming the so-called canonical energymomentum tensor, and its conservation means that µ T µν = 0. T µν = Lη µν + ( µ φ i ) ( ν φ i ), (29) 5

6 6 B. Electromagnetic energy-momentum tensor Canonical. Let now the field be the electromagnetic four-potential A µ, with the corresponding action without sources, S [A]= 1 d 4 x F αβ F αβ. (30) 4 The canonical energy-momentum tensor thus reads from (29), T µν = 1 4 ηµν F αβ F αβ F µσ ( ν A σ ). (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, Then, so that the symmetric tensor, K µσν F µσ A ν. (32) F µσ ( ν A σ )+ σ (F µσ A ν )= F µσ ( ν A σ )+( σ F µσ ) A } {{ } ν + F µσ ( σ A ν )=F µσ Fσ, ν =0 T µν = 1 4 ηµν F αβ F αβ + F µσ F ν σ. (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, T 00 = 1 4 F αβf αβ + F 0σ F 0 σ = 1 2 (E2 B 2 )+ Choosing on the other handν=igives, 3 (F 0i ) 2 = 1 2 (E2 B 2 )+E 2 = 1 2 (E2 + B 2 ). i=1 T 0i = F 0 j F i j = F 0 j F ji = E j ǫ i jk B k = (E B) i.