Chapter 2 Maxwell's Equations

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Chapter 2 Maxwell’s Equations

The form of transformation which preserve Maxwell’s equations is not straightforward from their standard (non-covariant) from. The transformation of field equations must be consistent with transformation rule of electric and magnetic field. The existence of such transformation rule is expected. For instance, electrically charged particles at rest in the laboratory reference frame S are identified with sources of the electric field (electrostatic in S). On the other hand, these particles remain in motion in S0 and thus they form electric current which is the source of magnetic field in S0.

2.1 Electromagnetic potentials

The set of Maxwell’s equations read

E = 4pr,(2.1) r· 1 4p B ∂ E = J,(2.2) r⇥ c t c B = 0, (2.3) r· 1 E + ∂ B = 0. (2.4) r⇥ c t It is known fact that for E and B given in terms of electromagnetic potentials the sourceless Maxwell’s equations (2.3) and (2.4) have the form of identities. The vector potential A is a vector ﬁeld such that magnetic ﬁeld is obtained as B := A. The equation (2.3) is an identity,1 1 Independently on particular form of A. r⇥ ( A) 0, because it contains antisymmetric combinations of r· r⇥ ⌘ symmetric second-order partial derivatives. The Faraday’s law (2.4) can be cast in the form 1 E + ∂ A = 0. r⇥ c t ✓ ◆ 1 This equation reduces to identity when expression E + c ∂tA is proportional to the gradient of certain function j(t, x). In order to establish the correspondence with electrostatic potential we choose the minus sign in 54 lecture notes on classical electrodynamics

E + 1 ∂ A := j. The scalar potential j depends on variables t and c t r xi. In some special cases it can be a function of only xi (electrostatic potential). In terms of electromagnetic potentials one gets the electric ﬁeld strength and the magnetic ﬁeld in the form 1 E = ∂ A j,(2.5) c t r B = A.(2.6) r⇥ The electromagnetic potentials possess the gauge freedom i.e. they can be substituted by new potentials 1 j0 = j ∂ c, A0 = A + c (2.7) c t r

leaving the ﬁelds unchanged i.e. E0 = E oraz B0 = B. The gauge transformation (2.7) is the electromagnetic ﬁeld internal symmetry.

Four-current Figure 2.1: Spherical electric charge density r0 = qn0 at its rest frame. The density of electric charge r and three components of the electric current density Ji combined together Jµ (cr, Ji). (2.8) ! transform as components of contravariant four-vector. This statement follows from the fact that electric charge is invariant under the Lorentz transformations2. 2 According to experiments hydrogen Let us consider static configuration of electric charges in certain atoms and deuteruim atoms are electrically neutral. Protons and neutrons in 3 inertial reference frame S0. The electric charge density of this configu- deuterium atoms interact via strong inter- ration is given by r0 = qn0 where n0 is the concentration of electrically action what significantly increases theirs kinetic energy. If the motion of the pro- charged particles in their rest frame. For simplicity, we shall assume ton would have any affect on its elec- n0 = const. According to the Lorentz-FitzGerald contraction the electric tric charge then it would not be possible the existence of neutral deuterium atoms. charge density is higher by the factor g i.e. r = qn0g in the reference Neutrality of the deuterium means that frame S in which the configuration moves with a velocity v. motion of electrically charged particle has Moreover, the moving free charges contribute to the electric cur- no influence on their electric charge µ rent density J = qn0gv in this reference frame. It follows that J is proportional to a four-velocity uµ (gc, gv). It allows to conclude ! that Jµ transform exactly as contravariant components of some four- vector. If the electric charge configuration has some velocity in S then its four-current has the form (2.8). µ µ n Note, that the four-current transformation law J0 = L n J can be i also deduced from the continuity equation ∂0(cr)+∂i J = 0. This equation describes the conservation of electric charge and so it must have the same form in all inertial reference frames. The l.h.s. of the Figure 2.2: Electric charge density r = qn0g at the laboratory frame. The charge equation of continuity can be written in two inertial reference frames configuration has velocity v in this refer- µ n ∂µ0 J0 = ∂n J . Since partial derivatives transform as components of ence frame. 3 covariant four-vector, then components of four-current density must For instance, it can be represented by a charged spherical conductor transform as components of contravariant four-vector. maxwell’s equations 55

2.2 Electromagnetic ﬁeld tensor

We shall consider components with respect to Cartesian basis e { i}i=1,2,3 of electric and magnetic ﬁeld and components of the vector potential

E Ei, B Bi, A Ai. ! ! ! Partial derivatives with respect to the coordinates xi, i = 1, 2, 3, and with respect to time t are denoted by ∂i and ∂t. The gradient operator has Cartesian components ∂ . r! i For further convenience we deﬁne the symbol

1 ∂ ∂ := 0 c ∂t 0 The set of derivatives of any function f (x , x) i.e. (∂0 f , ∂1 f , ∂2 f , ∂3 f ) transform as covariant components of a four-vector i.e.

n ∂µ0 f = Lµ ∂n f .

Taking into account that function f is arbitrary we can write the last n formula in the form ∂µ0 = Lµ ∂n. We shall assume that the sequence of four elements j and A1, A2, A3 transform as contravariant components of a four-vector. Thus, we denote Aµ (A0, Ai) where A0 j. ! ⌘ In order to justify this assumption we note that Maxwell’s equations imply that the electromagnetic potentials must obey equations 4p 4p (∂2 2)j = (cr), (∂2 2)Ai = Ji (2.9) 0 r c 0 r c

4 i 5 4 where the Lorenz gauge condition ∂t j + ∂i A = 0 has been imposed. Ludvig Valentin Lorenz 1829 - 1891. In Since the d’Alembert operator ∂2 2 is Lorentz invariant and the literature frequently confused with 0 r Hendrik Antoon Lorentz. i µ 5 i the components (cr, J ) J transform as contravariant components If the original potentials j0 and A0 give ! i i i of the four-current then the sequence (j, A ) must also transform as ∂t j0 + ∂i A0 = f (t, x ) then the function c(t, xi) which is a solution of equation contravariant components of a four vector. Note, that the Lorenz (∂2 2)c(t, xi)= f (t, xi) allows to ob- µ 0 r condition is invariant under Lorentz transformations when A0 = tain new set of potentials j and Ai that al- µ n ready satisfy the Lorenz condition. Note L n A . that Lorenz condition do not ﬁxes the po- The covariant components A (A0, Ai) can be obtained from µ ! tentials completely. New potentials has contravariant ones by contraction with components of the metric tensor still gauge freedom given by functions c = n i which are solutions of the wave equation Aµ hµn A . Note, that expression A appears in two different contexts. (∂2 2)c(t, xi)=0. 0 r 1. Spatial components µ = 1, 2, 3 of the four-potential that differ from covariant components A = h An = d Aj = Ai. In this case i in ij the involved metric tensor is hµn because the four vector lives in the Minkowski spacetime. 56 lecture notes on classical electrodynamics

2. On the other hand, they appear as Cartesian contravariant components Ai of the vector potential A. Such components are mapped on the covariant components by the metric tensor of a Euclidean space

which has the form gij = dij in the Cartesian coordinates. It follows that 6 6 In order to avoid confusion one could j i define the four-potential as µ and pre- A = d A = A . A i ij serve the letter Ai exclusively for three Components of electric and magnetic field can be cast in the form components of the vector potential. In such a case = i where i Ai = Ai A A ⌘ Ai. Ei = ∂ Ai ∂ A0 = ∂ A ∂ A F ,(2.10) 0 i 0 i i 0 ⌘ 0i 1 1 Bi = e ∂ Ak = e (∂ A ∂ A ) e F (2.11) ijk j 2 ijk j k k j ⌘2 ijk jk where, by construction, F0i and Fjk are antisymmetric expressions. They constitute components of the electromagnetic field tensor (called also electromagnetic strength tensor, Faraday tensor7 ) 7 See Gravitation Charles W. Misner, Kip S. Thorne, and John Archibald Wheeler. F = F eµ en µn ⌦ where

F := ∂ A ∂ A . (2.12) µn µ n n µ

Components Fµn transform as covariant components of a second-rank tensor because Aµ and ∂n transform as covariant components of a four-vector. In terms of electric and magnetic ﬁeld they read

F = Ei, F = e Bk.(2.13) 0i ij ijk In order get the second relation (2.13) we contract both sides of (2.11) with the Levi-Civita symbol. It gives 1 1 1 e Bi = e e F = (d d d d )F = (F F )=F . abi 2 abi ijk jk 2 aj bk ak bj jk 2 ab ba ab Covariant components of the electromagnetic ﬁeld tensor can be identi- ﬁed with the following matrix

0 E1 E2 E3 E1 0 B3 B2 Fµn 0 1 , (2.14) ! E2 B3 0 B1 B C B E3 B2 B1 0 C B C @ A whereas the matrix

0 E1 E2 E3 E1 0 B3 B2 Fµn 0 1 . (2.15) ! E2 B3 0 B1 B C B E3 B2 B1 0 C B C @ A µn µa nb contains its contravariant components F = h h Fab. maxwell’s equations 57

2.3 Covariant form of Maxwell’s equations

With help of the electromagnetic ﬁeld tensor we can put Maxwell’s equations in their covariant form. • Gauss’s law for the electric ﬁeld can be cast in the following form

i i0 4p 0 µ0 4p 0 ∂ E = 4pr ∂ F = J ∂µF = J . i ! i c ! c • Ampere-Maxwell law can be also written in terms of the electromagnetic ﬁeld tensor 4p 4p # ∂ Bk ∂ Ei = Ji ∂ Fij ∂ Fi0 = Ji, ijk j 0 c !j 0 c k ∂j(#ijkB ) which| gives{z } µi 4p i ∂µF = J . c Thanks to tensor approach Gauss’s law and Ampere-Maxwell’s can be wrapped up together in a single, frame-independent law First pair of Maxwell’s equations

µn 4p n ∂µF = J . (2.16) c

• Gauss’s law for magnetic ﬁeld reads ∂ Bi = 0 ∂ F + ∂ F + ∂ F = 0. i ! 1 23 2 31 3 12 • Faraday’s law is of the form k i #ijk∂jE + ∂0B = 0 and it is equivalent to the following set of equations

∂ E3 ∂ E2 + ∂ B1 = 0 ∂ F + ∂ F + ∂ F = 0 2 3 0 2 03 3 20 0 32 ∂ E1 ∂ E3 + ∂ B2 = 0 ∂ F + ∂ F + ∂ F = 0 . (2.17) 8 3 1 0 ! 8 3 01 1 30 0 13 > ∂ E2 ∂ E1 + ∂ B3 = 0 > ∂ F + ∂ F + ∂ F = 0 < 1 2 0 < 1 02 2 10 0 21 Gauss’:> law for magnetic ﬁeld and:> Faraday’s law form the second pair of Maxwell’s equations. They can be wrapped up together in a single law Second pair of Maxwell’s equations ∂aFbg + ∂bFga + ∂gFab = 0. (2.18)

or equivalently as

∂[aFbg] = 0 where [...] stands for anti-symmetrization of any group of indices. In the case of three indices it reads 1 ∂ F := ∂ F + ∂ F + ∂ F ∂ F ∂ F ∂ F . [a bg] 3! a bg b ga g ab b ag a gb g ba The expression(2.18) is known as Bianchi identities in electromagnetism. Indeed, for Fµn given in terms of four-potential Aµ the expression (2.18) vanishes identically. 58 lecture notes on classical electrodynamics

The Levi-Civita symbol

The second pair of Maxwell’s equations can be put in similar form as the first pair. It can be done in terms of dual electromagnetic field tensor8. 8 This object naturally appears in the ap- First, we define the antisymmetric symbol in four dimensions proach based on differential forms Permutation symbol +1 for even permutation of 0123, e eµnab := 1 for odd permutation of 0123, (2.19) µnab ⌘ 8 <> 0 for repetition of indices known also as the Levi-Civita:> permutation symbol. In order to establish its transformation law we consider some general transformation of coordinates Diffeomorphism

µ µ x x0 (2.20) { }!{ } which, in contradistinction to Lorentz transformations, is not necessarily globally constant. We assume that transformation (2.20) is a diffeomorphism (invertible and has at least ﬁrst partial derivatives). The coordinates without prime are called “old” whereas those with prime are called “new”. The Jacobian matrix Jˆ of the transformation contains ﬁrst partial derivatives and it has the form9 9 We use convention in which the Jaco- bian matrix contains derivatives of old µ µ coordinates with respect to the new ones. ∂x 1 ∂x0 Jˆ := , Jˆ = , In such approach the Jacobian determi- ∂x n ∂xn  0  nant appears in the volume element. For instance, when changing of coordinates where the existence of the inverse transformation is assured by our from Cartesian xi to polar x i ones { } { 0 } assumption about the form of the transformation. The upper index then ∂(x, y) numbers the lines of the matrix and the lower one numbers its columns. J detJˆ = r The Jacobian determinant must not be zero ⌘ ⌘ ∂(r, j) and so dxdy = rdrdj. µ n a b ˆ ∂x ∂x ∂x ∂x J := det(J)=eµnab 0 1 2 3 = 0(2.21) ∂x0 ∂x0 ∂x0 ∂x0 6 and 1 1 1 J := det(Jˆ )= . J The Levi-Civita symbol is not a tensor because it has not tensor law of transformation. Its transformation law can be obtained from an alternative expression for the Jacobian determinant

∂x0 ∂x1 ∂x2 ∂x3 = µnab J e0 µ n a b (2.22) ∂x0 ∂x0 ∂x0 ∂x0 µnab where there appears a symbol e0 with indices that label new coordinates on the rhs of (2.22). Any permutation of two rows changes the sign of the determinant (2.22). Thus we can write

∂xr ∂xs ∂xg ∂xd rsgd = µnab Je e0 µ n a b .(2.23) ∂x0 ∂x0 ∂x0 ∂x0 maxwell’s equations 59

k l w c ∂x0 ∂x0 ∂x0 ∂x0 Contracting the equation (2.23) with r one gets Transformation formula of the ∂x ∂xs ∂xg ∂xd Levi-Civita symbol eµnab k l w c rsgd ∂x0 ∂x0 ∂x0 ∂x0 klwc Je = e0 .(2.24) ∂xr ∂xs ∂xg ∂xd The presence of Jacobian determinant spoils the tensor law of transformation.10 Similarly, taking arbitrary permutation of columns in (2.21) 10 Such objects are called relative tensors of and dividing by J we get weight W =+1. Transformation formula of the 1 ∂xµ ∂xn ∂xa ∂xb = ersgd0 eµnab r s g d .(2.25) Levi-Civita symbol eµnab J ∂x0 ∂x0 ∂x0 ∂x0 In order to deﬁne the Levi-Civita tensor we observe that Jacobian determinant of any rank-two tensor cannot be a scalar. In particular we are interested here in the metric tensor. Its covariant components transform under (2.20) according to

∂xa ∂xb gµ0 n(x0)= µ n gab(x) (2.26) ∂x0 ∂x0 where x0 on the lhs is a function of x i.e. x0 = x0(x). The formula (2.26) can be cast in the matrix form

T gˆ0(x0)=Jˆ gˆ(x)Jˆ.(2.27)

Taking determinant of both sides of (2.27) we get

2 g0(x0)=J g(x).(2.28) where g(x) := det(gˆ(x)), g0(x0) := det(gˆ0(x0)).

The signature11 of the quadratic form cannot be changed by transforma- 11 Numbers of positive and negative ele- tion of coordinates hence the sign of the determinant must be consistent ments when the form has its canonical form. with the sign of determinant of metrics in any frame. In particular, we know that in Cartesian basis det(hˆ)= 1 so g(x) < 0 and g (x ) < 0. 0 0 Multiplying (2.28) by 1 and taking the square root of its both sides we get g (x )=sgn(J) J g(x).(2.29) 0 0 q q Next, dividing the lhs (rhs) of equation (2.24) by the lhs (rhs) of (2.29) one gets expression

k l w c 1 klwc 1 rsgd ∂x0 ∂x0 ∂x0 ∂x0 e0 = sign(J) e g p g ∂xr ∂xs ∂xg ∂xd 0 which is thep pseudo-tensor transformation law. When restrict considerations to transformations with sign(J)=1 one gets that expression 1 eµnab transforms as a tensor. The Levi-Civita pseudotensor is p g(x) deﬁned as12 12 The minus sign is optional. Our choice agrees with Landau Lifshitz convention adopted in The Classical Theory of Fields. 60 lecture notes on classical electrodynamics

1 #µnab(x) := eµnab. (2.30) g(x) Since (2.30) is a tensor we can deﬁnep another tensor contracting (2.30) with components of the metric tensor (lowering its indices)

µnab #rsgd(x) :=grµgsnggagdb# (x)

µnab 1 = grµgsnggagdbe g(x) 1 p = det(gˆ(x)) ersgd. g(x) Thus p # (x) := g(x)e . (2.31) rsgd rsgd q Using (2.25) and (2.29) we get that (2.31) transforms as

#0 (x0)= g (x )e0 rsgd 0 0 rsgd q 1 ∂xµ ∂xn ∂xa ∂xb = ( ) ( ) sgn J J g x eµnab r s g d J ∂x0 ∂x ∂x ∂x q 0 0 0 ∂xµ ∂xn ∂xa ∂xb = ( ) ( ) sgn J #µnab x r s g d .(2.32) ∂x0 ∂x0 ∂x0 ∂x0 It shows that (2.31) is a pseudotensor. Note that contraction of pseudotensors (2.30) and (2.31) is a scalar

#µnab(x)# (x)= 4! µnab

Dual electromagnetic ﬁeld tensor

Now we are ready to express the second pair of Maxwell’s equations in different but equivalent form. We are going back to Cartesian coordinates where p g = 1 and thus #µnab = eµnab and # = µnab +eµnab. The electromagnetic dual tensor is deﬁned in the following way Dual electromagnetic tensor

µn 1 µnab ⇤F := # F (2.33) 2 ab

and it has the form Fµn = 1 eµnabF in Cartesian coordinates. ⇤ 2 ab It gives

01 1 0123 1 0132 0123 1 ⇤F = e F e F = e F = F = B , 2 23 2 32 23 23 02 1 0231 1 0213 0231 2 ⇤F = e F e F = e F = F = B , 2 31 2 13 31 31 03 1 0312 1 0321 0312 3 ⇤F = e F e F = e F = F = B , 2 12 2 21 12 12 maxwell’s equations 61

and similarly

12 1 1230 1 1203 0123 3 ⇤F = e F e F = e F = F = E , 2 30 2 03 03 03 23 1 2310 1 2301 0123 1 ⇤F = e F e F = e F = F = E , 2 10 2 01 01 01 31 1 3120 1 3102 0123 2 ⇤F = e F e F = e F = F = E . 2 20 2 02 02 02 The contravariant components of the dual tensor can be arranged in the form of matrix

0 B1 B2 B3 0 1 3 2 1 µn B 0 E E ⇤F . (2.34) ! B 2 3 1 C B B E 0 E C B C B C B B3 E2 E1 0 C B C @ A Comparing (2.34) with (2.15) we conclude that dual transformation maps electric field into negative of magnetic field and magnetic field into electric field

⇤E = B and ⇤B = E.(2.35) Double dual transformation changes the sign of the ﬁeld. Contraction of Bianchi identities (2.18) with constant expression 1 eanbg 3! gives 1 1 1 ∂ eanbgF + ∂ eanbgF + ∂ eanbgF = 0. a 6 bg b 6 ga g 6 ab    Changing labels of indices in the above equation one gets 1 1 1 ∂ eµnbgF +∂ eanµgF + ∂ eanbµF = 0. µ 6 bg µ 6 ga µ 6 ab    1 ⇤ µn 1 ⇤ µn 1 ⇤ µn 3 F 3 F 3 F Finally,| the second{z pair} of| Maxwell’s{z equation} | reads{z } Alternative form of second pair

µn of Maxwell’s equations ∂µ ⇤F = 0 (2.36) where n = 0 corresponds with magnetic Gauss’s law and n = 1, 2, 3 gives three components of Faraday’s law.

Covariance of Maxwell’s equations

The expression Fµn = ∂µ An ∂n Aµ represent components the rank-two 2 (0) tensor so it transforms according to µn µ n ab F0 = L a L bF . 62 lecture notes on classical electrodynamics

µ n 1 where A and ∂ transform as components of rank-one (0) tensor (four-vector). First pair of Maxwell’s equation in the inertial reference frame S0 which moves with the velocity V with respect to the reference frame S have the form ∂ F µn 4p J n = 0. This equations are equivalent to the µ0 0 c 0 following ones

1 r µ n ab 4p n b (L ) L L ∂rF L J = 0 µ a b c b r da | {z } which can be written in the form

n ab 4p b L ∂aF J = 0. (2.37) b c 

Each equation in (2.37) in S0 (with given value of n) is linear combination of Maxwell’s equations in S where coefficients of the combination are elements of the Lorentz matrix. First pair of Maxwell’s equations (2.37) takes the form Transformation of first pair of Maxwell’s equations n a0 4p 0 n ai 4p i L ∂aF J +L ∂aF J = 0. (2.38) 0 c i c   Gauss0s law in S Ampere0s law in S | {z } | {z } Similarly, components of the dual electromagnetic tensor in S0 are given by components of this tensor in S13 13 The sign 1 = sgnJ, where J 1 = ± det(Lˆ ) is a Jacobian of a transforma- µn µ n ab tion, appears due to transformation rule ⇤F0 = L L ⇤F . µnab ± a b of the Levi-Civita pseudotensor # (x) given by (2.32). In further part we shall Hence the second pair of Maxwell’s equations reads choose +1 because we are interested in boost transformations which are proper Lorentz transformations. 1 r µ n ab (L ) µ L a L b∂r⇤F = 0 r da | {z } and it can be cast in the form Transformation of second pair of Maxwell’s equations n a0 n ai L 0 ∂a⇤F +L i ∂a⇤F = 0. (2.39) h i h i Gauss0 law in S Faraday0s law in S | {z } | {z } The important fact about transformation of Maxwell’s equation is that left hand sides of these equations in a given reference frame are linear combinations of left hand sides of the equations in another frame. The coefficients of linear combinations are elements of the Lorentz matrix. The Lorentz transformations do not mix equations belonging to different pairs. maxwell’s equations 63

2.4 Maxwell’s equations and external differential forms

The formalism of external differential forms provides natural frame- work for representation of Maxwell’s equations. Among others, especial useful are applications of Stokes theorem. Using this theorem one gets integral form of Maxwell’s equations which is explicitly Lorentz invariant.

External differential forms

Differential forms can be defined in spaces with any finite number of dimensions. The case N = 4 is of special importance in application to electromagnetism. Let ( , (4)) be 4 dimensional affine space where is an infinite A V A collection of points (more precisely – manifold) and (4) is a vector V space. 1 One-form w is a linear function which maps vectors v T(p) on real One-form definition 2 numbers 1 w : v R 7! such that Linearity in argument of the

1 1 1 form w(a1 v1 + a2 v2)=a1 w(v1)+a2 w(v2). (2.40)

A set of all such linear forms (covectors) at any point p form a vector space provided that Linear space of covectors

1 1 1 1 (w1 + w2)(v) := w1(v)+w2(v), (2.41) 1 1 (a w)(v) := a w(v). (2.42)

This space, denoted by T⇤(p), is dual to T(p) and it is called tangent Cotangent space space of covectors or cotangent space. Components of one-forms are denoted by wa and they are deﬁned as sequences of numbers representing values of one-form on basis vectors e at the point p i.e., Components of one-form { a}a=0,...,3 1 wa := w(ea).

In the space of one-forms one can deﬁne the dual basis which consists on one-forms eb that act on the basis vectors e T(p) giving { }b=0,...,3 a 2 b b e (ea) := da . (2.43)

1 Elements of dual basis ea and components of one-forms w { }a=0,...3 transform in the following way

∂x a ∂xb a = 0 ( ) b = ( ) e0 b p e , wa0 a p wb. (2.44) ∂x ∂x0 64 lecture notes on classical electrodynamics

Geometric interpretation of covectors

Let be certain region of spacetime and f a differentiable function O f : R. The differential of this function df is a liner operation O! which associates a number ∂ f df = (p)dxa ∂xa with the function f at any ﬁxed point p . This number depends 2O on some (arbitrary) sequence of numbers dx0, , dx3 . It can be { ··· } ∂ f interpreted as linear form with N = 4 variables. Derivatives ∂xa (p) form a sequence of real numbers which depends on f . Let v T(p) be a vector which has components v0, , v3 in a 2 { ···0 } 3 given basis ea a=0, ,3. The mapping between sequences dx , , dx { } ··· { ··· } and v0, , v3 allows to associate a linear form with the differential Differential of a function at the { ··· } df at p (tangent covector). It acts on vectors v T(p) according to ﬁxed point as a model of one- 2 form ∂ f df(v) := (p)va.(2.45) ∂xa The formula (2.45) allows us to associate with the differential df at p the linear differential form

1 ∂ f a w := (p)e T⇤(p).(2.46) ∂xa 2 Such association does not depend on the reference frame. Moreover, it is invertible – with any sequence of numbers it can be associated a Invertibility sequence of derivatives ∂ f . Since p is ﬁxed the sequence of deriva- a |p tives is a numerical sequence. The simplest function which allows to associate a sequence of numbers with a sequence of partial derivatives at p has the form f (x0, , x3)=w xa ··· a where wa is a sequence of numbers. It gives

b ∂ f = w d = w . a |p b a a It means that, there is one-to-one correspondence (isomorphism) between the space T⇤(p) and the differential of function f at p. This correspondence preserves the structure of both spaces. Basis 1 a Any covector w can be decomposed in basis of covectors e a=0,...,3 a a { a} dual to vector basis i.e. e (eb) := db. In particular, covectors e can be cast in the form (2.46) what gives

∂ f a ea = (p)eb (2.47) ∂xb where f a is a sequence of functions. Each function f a deﬁnes { }a=0,...,3 one covector ea. Components b of this covector in the basis that the maxwell’s equations 65

covector belongs to read a a ∂ f a (e )b = (p)=d . ∂xb b The simplest set of functions satisfying this requirement is given by

f a(x0, , x3) := xa. ··· Differentials of such functions, associated with covectors, read

ea = dxa.

We stress that dxa in this expression have the meaning of linear forms Differentials dxa as linear forms that act on elements of the space T(p) according to

a a b b a b a a dx (v)=dx (v eb)=v dx (eb)=v db = v . Hence 1 a a w(v)=wadx (v)=wav .

Linear differential forms

In previous section we managed to establish one-to-one correspondence between differentials of functions at fixed point and linear forms with a constant components wa in the basis dx . In this section we shall extend the idea of differential forms on expressions with non constant components wa. Linear one-form in a region of We consider certain region of Minkowski spacetime14 and linear spacetime O forms at each point p . We say that field of covectors (or field of linear 14 Here we concentrate on definition of 2O differential forms) is given in . linear forms in Minkowski spacetime, O however, it must be stressed that such 1 a Linear one-form w(p)=wa(p)e is defined only by its coordinates considerations do not depend of this particular fact. w (x0, , x3) after setting the coordinate reference frame whole region a ··· . In such fixed coordinate frame an arbitrary linear differential one-form Linear differential one-form O is given by expression

1 0 3 a w = wa(x ,...,x )dx . (2.48)

Note, that wa in (2.48) form a sequence of functions and not a sequence of numbers. This fact has some immediate consequences. Namely, Each differential of a function is for a given sequence of numbers we can always ﬁnd such a function differential form but not each dif- f (x0,...,x3) that the sequence of its partial derivatives at p corre- ferential form is a differential of sponds with wa. It is not possible for generic sequence of functions a function 0 3 wa(x ,...,x ). It follows that each differential of function is a differential 1 form whereas the inverse is not necessarily true. The differential form w acts on vector ﬁeld v T(p). It maps sequence of differentials dxa 2 { } onto sequence of functions

dx0,...,dx3 v0,...,v3 (2.49) { }7!{ } 66 lecture notes on classical electrodynamics

where va are components of certain vector v in vector basis e . { } { a}a=0,...,3 This mapping deﬁnes the function wv(p) : T(p) R which is given 7! by Function p wv(p) R ! 2 1 a wv(p) := w(v)=wav . (2.50)

Linear two-form is a function which associates a real number with Linear two-form a pair of vectors from T(p)

2 w : (v , v ) R. 1 2 ! This function has the following properties

• it is linear in its ﬁrst argument Linearity

2 2 2 w(a1 v1 + a2 v2, v3)=a1 w(v1, v3)+a2 w(v2, v3), (2.51)

where a1 and a2 are some real numbers,

• it is anti-symmetric Skew-symmetry

2 2 w(v , v )= w(v , v ). (2.52) 1 2 2 1 4 The set of all such forms on T(p) is a (2) = 6 dimensional real vector space provided that Linear space of two-forms

2 2 2 2 (w1 + w2)(v1, v2) := w1(v1, v2)+w2(v1, v2) (2.53) 2 2 (a w)(v1, v2) := a w(v1, v2).(2.54)

1 1 With each pair of one-forms w1, w2 there can be associated a two- External product of one-forms as 1 1 form w w which acts on vectors v , v according to a prototype of two-form 1 ^ 2 1 2 1 1 1 1 (w w )(v , v ) := e w (v )w (v ) 1 ^ 2 1 2 ij 1 i 2 j 1 1 w (v ) w (v ) det 1 1 2 1 .(2.55) ⌘ 2 1 1 3 w1(v2) w2(v2) 4 5 1 1 1 The form w1 w2 is obtained as external product of covectors w1 and 1 ^ w2. Each one-form maps vectors from T(p) on real numbers and these numbers are arranged in anti-symmetric expression. It follows from determinant properties that

1 1 1 1 w w = w w (2.56) 1 ^ 2 2 ^ 1 1 1 1 1 1 1 1 (w + w ) w = w w + w w . (2.57) 1 2 ^ 3 1 ^ 3 2 ^ 3 The external product can be used to deﬁne differential two-forms and more generally k–forms. maxwell’s equations 67

Linear differential two-form is given by formula Linear differential two-form

2 1 w = w dxa dxb. (2.58) 2 ab ^

It associates a real number with each pair of tangent vectors (v1, v2), according to

2 1 w(v , v )= w (dxa dxb)(v , v ) 1 2 2 ab ^ 1 2 1 = w e dxa(v )dxb(v ) 2 ab ij i j a b 1 dx (v1) dx (v1) wab det a b ⌘ 2 " dx (v2) dx (v2) # b 1 va v = w det 1 1 (2.59) 2 ab a b " v2 v2 #

a a µ µ a a where dx (vi)=dx (vi eµ)=vi dµ = vi . The coefﬁcients of this form are values which it takes on each pair of vectors from the set e , { a}a=0,...,3 b 2 1 da d w := w(e , e )= w det µ µ . (2.60) µn µ n ab a b 2 " dn dn #

k Linear differential k-form w, where 0 k N 4, is a covariant Linear differential k-form 0   ⌘ tensor of rank (k) given by

k 1 a a w = wa ...a dx 1 ... dx k (2.61) k! 1 k ^ ^ where aa = 0, 1, 2, 3. The antisymmetric expressions

dxa1 ... dxak ^ ^ are elements of basis in the space of differential k–forms. The form (2.61) associates a real number with each sequence of tangent vectors v1,...,vk according to the formula

a1 ak v1 ... v1 k 1 . . w(v ,...,v )= wa ...a det 2 . . 3 . (2.62) 1 k k! 1 k ..... a a 6 v 1 ... v k 7 6 k k 7 4 5 k Components of w are given by values which this form takes on each k-th element sequence of vectors e i.e. { a} k wa1...ak = w(ea1 ,...,eak ). (2.63) 68 lecture notes on classical electrodynamics

Dual forms (adjoint forms)

k With each given k–form w in N dimensions there can be associated N k the (N k)–form15 s . This operation is deﬁned in terms of Hodge 15 Here we are mostly interested in the operator “ ”, which depends on the scalar product (and so it requires case N=4. However, for generality we ⇤ shall keep explicitly N for denoting num- the metric tensor) and space (spacetime) orientation. ber of dimensions. The Hodge operator requires Levi-Civita pseudotensor which is proportional to Levi-Civita permutation symbol ea1...aN . The permutation symbol in N dimensions is introduced exactly in the same way as in the case N = 4 discussed before. Let

p(0, 1, . . . , N 1)=a1, a2,...,aN 1 be a permutation of numbers 0, 1, . . . , N 1 and n(p) stands for number of permutations i.e. number of mappings of the sequence 0, 1, . . . , N 1 onto a1, a2,...,aN 1. It is deﬁned as Levi -Civita permutation symbol ( 1)n(p), a1...aN ea1...aN e := . ⌘ ( 0 for repetition of indices

Levi-Civita pseudotensor is an expression whose components has pseu- dotensorial transformation law. The covariant components of Levi- Civita pseudotensor are deﬁned as follows Covariant components of Levi - Civita pseudotensor # := g e . (2.64) a1...aN | | a1...aN q Its contravariant components #a1...aN are given by expression Contravariant components of the Levi -Civita pseudotensor #a1...aN := ga1 b1 ...gaN bN g e | | b1...bN 1 q = g ea1...aN g | | q sgn(g) = ea1...aN g | | Hodge star operator is a mappingp between linear forms which associates (N k)–form with the form (2.61). This k-form is given by Hodge operator

k 1 1 a1...ak b1 bN k ⇤w = g w ea1...ak b1...bN k dx ... dx (N k)! k! | | ^ ^  q where

1 a1...ak ⇤wb1...bN k := g w ea1...ak b1...bN k (2.65) k! | | q are components of dual form (adjoint form). maxwell’s equations 69

Physical examples

The four-current one-form is a differential form with components Jµ i.e. Electric current one-form

µ J := Jµdx . (2.66)

Its dual is a 3 form J given by expression Dual there-form of the electric ⇤ current density 1 a b g ⇤J = ⇤ J dx dx dx (2.67) 3! abg ^ ^ µ where ⇤ Jabg = J #µabg. We shall restrict our considerations to Cartesian coordinates, so g = p g = 1. The electric current dual form has Cartesian coordinates | | components p 0 0 ⇤ J123 = J #0123 = J , 1 1 ⇤ J = J # = J , 023 1023 2 2 ⇤ J = J # = J , 031 2031 3 3 ⇤ J = J # = J 012 3012 and so

0 1 2 3 1 0 2 3 ⇤J = J dx dx dx J dx dx dx ^ ^ ^ ^ J2dx0 dx3 dx1 J3dx0 dx1 dx2.(2.68) ^ ^ ^ ^ Another important example are differential forms which involve components of electromagnetic tensor. For instance, Faraday’s two- form is defined as follows Two-form of the electromagnetic 1 field F := F dxµ dxn. (2.69) 2 µn ^ The explicit form of the Faraday’s two-form in terms of components of electric and magnetic fields reads

F = E1dx0 dx1 + E2dx0 dx2 + E3dx0 dx3 ^ ^ ^ B1dx2 dx3 B2dx3 dx1 B3dx1 dx2.(2.70) ^ ^ ^ Similarly, Maxwell’s form is a two-form dual to Faraday’s form. It Dual two-form of the electromag- reads netic ﬁeld

1 a b ⇤F := ⇤F dx dx , (2.71) 2 ab ^ where

1 µn ⇤F = F # . (2.72) ab 2 µnab They have their explicit form

0 B1 B2 B3 B1 0 E3 E2 ⇤Fab 0 1 . (2.73) ! B2 E3 0 E1 B C B B3 E2 E1 0 C B C @ A 70 lecture notes on classical electrodynamics

Thus, Maxwell’s two-form is given by expression

1 0 1 2 0 2 3 0 3 ⇤F = B dx dx B dx dx B dx dx ^ ^ ^ E1dx2 dx3 E2dx3 dx1 E3dx1 dx2.(2.74) ^ ^ ^ The adjoint operation interchanges electric and magnetic ﬁeld in such a way that E = B and B = E. Note, that in the alternative convention ⇤ ⇤ where # = 1 the duality relations read E = B and B = E. 0123 ⇤ ⇤

Exterior derivative

We have seen that with a differential of certain function f : R there O! can be associated differential one-form in the region of Minkowski O spacetime. Since the function f can be interpreted as zero-form then differential of function associates one-form with each zero-form. The exterior derivative (exterior differential) is a generalization of this Exterior differential operation for an arbitrary k form. This operation associates (k + 1)– form of class r 1 with k–form of class r: C C k 1 a a dw = d wa ...a dx 1 ... dx k k! 1 k ^ ^ ✓ ◆ 1 a a = dwa ...a dx 1 ... dx k k! 1 k ^ ^ ^ 1 a a a = ∂a wa ...a dx 0 dx 1 ... dx k .(2.75) k! 0 1 k ^ ^ ^ The exterior derivative has the following properties Properties of exterior differentiation k k k k d(w1 + w2)=dw1 + dw2, (2.76) k l k l k l d(w w)=dw w +( 1)kw dw, (2.77) ^ ^ ^ k ddw = 0 for any form, (2.78) 0 a dw = ∂aw dx . (2.79)

k A differential form w is called closed if Closed form

k dw = 0. (2.80)

k A differential form w is called exact if Exact form

k k 1 w = d s . (2.81)

k 1 Each exact form is closed because dd s 0. The inverse statement ⌘ holds only in regions contractible to a point. k Poincaré Lemma. Let be a region contractible to a point and w, Poincaré Lemma O⇢A where k = 1, 2, , be an arbitrary closed differential k form deﬁned in k ··· k 1 k k 1 O i.e. dw = 0. There exists a form s such that w = d s . maxwell’s equations 71

Examples of exterior differentiation

Let us consider three examples of differential forms of degree 0, 1, 2 which depend only on spatial coordinates x1, x2 and x3:

0 w = w,(2.82)

1 1 2 3 w = w1 dx + w2 dx + w3 dx ,(2.83) 2 w = w dx1 dx2 + w dx2 dx3 + w dx3 dx1.(2.84) 12 ^ 23 ^ 31 ^ The exterior derivatives of these forms read Exterior derivatives of time- independent zero-, one- and two- 0 1 2 3 dw = ∂1w dx + ∂2w dx + ∂3w dx ,(2.85) form 1 dw =(∂ w ∂ w ) dx1 dx2 +(∂ w ∂ w ) dx2 dx3 1 2 2 1 ^ 2 3 3 2 ^ +(∂ w ∂ w ) dx3 dx1,(2.86) 3 1 1 3 ^ 2 dw =(∂ w + ∂ w + ∂ w ) dx1 dx2 dx3.(2.87) 1 23 2 31 3 12 ^ ^ k One can associate components of forms w, k = 0, 1, 2 with components Relation with gradient, curl and of some vector ﬁelds. In such a case the differential forms obtained divergence by exterior derivatives of given forms have components which can be identiﬁed with components of well-known operations: gradient, curl and divergence.

Physical examples of application of the Poincaré Lemma

The Poincaré Lemma applied to closed forms has some implications on associated vector fields – with components equal to components of differential forms. 1 1 • If the form w is closed i.e. dw = 0 (their components correspond Existence of scalar potential with a components of a certain curl-free vector field or irrotational vector 0 field), then there exist an exact form s (scalar potential) in the region 1 0 contractible to a point, such that w = ds. The vector field associated 1 0 with w is the gradient of this scalar potential j s. ⌘ 2 2 • If the form w is closed i.e. dw = 0 (their components correspond Existence of vector potential with components of solenoidal vector field), then there exist an exact 1 form s (vector potential) in the region contractible to a point, such 2 1 2 that w = ds (vector field associated with w is the curl of the vector field).

Let us consider the field Electrostatic field of linear charge density of infinitely long line rˆ x y E = 2l = 2l xˆ + yˆ 0 r 0 x2 + y2 x2 + y2  72 lecture notes on classical electrodynamics

This field correspond with the electric field of infinitely long uniformly charged lines where l0 stands for linear density of electric charge. The second field has the form Magnetic field of infinitely long conducting wire 2I fˆ 2I y x H = 0 = 0 xˆ + yˆ c r c x2 + y2 x2 + y2  and it corresponds with magnetic field of infinitely long straight conductor with current intensity I0 = const. We define two differential one-forms with components equal to components of the above vector fields

1 x y e = 2l dx + dy ,(2.88) 0 x2 + y2 x2 + y2  1 2I y x h = 0 dx + dy .(2.89) c x2 + y2 x2 + y2 

1 1 1 1 It follows that both forms e and h are closed Both e and h are closed

1 de = 0 E = 0, (2.90) ,r⇥ 1 dh = 0 H = 0 in E2 0, 0 .(2.91) ,r⇥ \{ } A closed form is not necessarily exact. This can be seen as follows. We shall integrate each form along the circle belonging to the z = const plane. Their center is located at the z–axis. We parametrize the circle in the following way C(t) (cos t, sin t) where t [0, 2p]. The integrals ! 2 read

1 2p e = 2l0 0 dt = 0, (2.92) IC(t) Z0 1 2p 2I0 4p h = 1 dt = I0.(2.93) IC(t) c Z0 c The second integral does not vanish. It means that the integral between two points depends on the integration path that connects these points. 1 Hence, the differential form h is not exact (it cannot be represented by 1 the exterior derivative of some zero-form). The form h is not definite at r = 0 (vector fˆ is not definite at the origin) what requires exclusion of this point from domain of the form. Consequently, the form is definite in the region which is not contractible to a point i.e. there is no satisfied 1 the assumption of Poincaré Lemma. On the other hand, the form e is exact and it reads

1 2 2 e = d 2l0 ln x + y + const .  q maxwell’s equations 73

Volume form

The form 4 w = w dx0 dx1 dx2 dx3 0123 ^ ^ ^ 4 is a 4-form in Minkowski spacetime. It is closed, dw = 0, because its degree is equal to number of spacetime dimmensions. The example of 4 4-form is volume form w which is given by Volume form

1 vol = 1 = g e dxµ dxn dxa dxb. (2.94) ⇤ 4! µnab ^ ^ ^ p

Maxwell’s equations in formalism of differential forms

Exterior derivative of Faraday’s form F reads Exterior derivative of Faraday’s form 1 1 dF = dF dxµ dxn = ∂ F dxl dxµ dxn 2 µn ^ ^ 2 l µn ^ ^ 1 = ∂ F dxl dxµ dxn + ∂ F dxµ dxn dxl 3! l µn ^ ^ µ nl ^ ^ h + ∂ F dxn dxl dxµ n lµ ^ ^ 1 i = [∂ F + ∂ F + ∂ F ] dxl dxµ dxn.(2.95) 3! l µn µ nl n lµ ^ ^ 0

The coefﬁcients| of the resulting{z differential} form are equal to electromagnetic Bianchi identities. It allows to represent second pair of Maxwell’s equations as exterior derivative of Faraday’s form, namely Second pair of Maxwell’s equations dF = 0. (2.96)

On the other hand, ﬁrst pair of Maxwell’s equations can be expressed using Maxwell’s form ⇤F. Its exterior derivative has the form Exterior derivative of Maxwell’s form 1 a b d⇤F = d⇤F dx dx 2 ab ^ ^ 1 g a b = ∂ ⇤F + ∂ ⇤F + ∂ ⇤F dx dx dx .(2.97) 3! g ab a bg b ga ^ ^ ⇥ ⇤ In spite of formal similarity between expressions (2.95) and (2.97) their physical content is quite different. The expression (2.97) reads

1 2 3 d⇤F =[∂ ⇤F + ∂ ⇤F + ∂ ⇤F ] dx dx dx 1 23 2 31 3 12 ^ ^ 0 2 3 +[∂ ⇤F + ∂ ⇤F + ∂ ⇤F ] dx dx dx 0 23 2 30 3 02 ^ ^ 0 3 1 +[∂ ⇤F + ∂ ⇤F + ∂ ⇤F ] dx dx dx 0 31 3 10 1 03 ^ ^ 0 1 2 +[∂ ⇤F + ∂ ⇤F + ∂ ⇤F ] dx dx dx , 0 12 1 20 2 01 ^ ^ 74 lecture notes on classical electrodynamics

where dual components ⇤Fµn are given in terms of components Fab. It leads to expression containing left hand sides of Maxwell’s equations 1 2 3 1 2 3 d⇤F =[ ∂ E ∂ E ∂ E ] dx dx dx 1 2 3 ^ ^ E r· +[ ∂ E1 + ∂ B3 ∂ B2] dx0 dx2 dx3 | 0 {z2 3 } ^ ^ ∂ E1+( B)1 0 r⇥ +[| ∂ E2 + ∂{zB1 ∂ B3}] dx0 dx3 dx1 0 3 1 ^ ^ ∂ E2+( B)2 0 r⇥ +[| ∂ E3 + ∂{zB2 ∂ B1}] dx0 dx1 dx2.(2.98) 0 1 2 ^ ^ ∂ E3+( B)3 0 r⇥ Making use of Maxwell’s| equations{z we} substitute right hand sides of (2.98) by sources what gives

4p 0 1 2 3 1 0 2 3 d⇤F = [ J dx dx dx + J dx dx dx c ^ ^ ^ ^ + J2 dx0 dx3 dx1 + J3 dx0 dx1 dx2] ^ ^ ^ ^ 4p 1 2 3 0 2 3 = [⇤ J dx dx dx + ⇤ J dx dx dx c 123 ^ ^ 023 ^ ^ 0 3 1 0 1 2 + ⇤ J dx dx dx + ⇤ J dx dx dx ] 031 ^ ^ 012 ^ ^ 4p 1 g a b 4p = ⇤ J dx dx dx = ⇤J. c 3! gab ^ ^ c The ﬁrst pair of Maxwell’s equations takes the form First pair of Maxwell’s equations 4p d⇤F = ⇤J. (2.99) c Note, that the property dd = 0 implies the continuity equation Continuity equation

d⇤J = 0, (2.100) where 0 1 2 3 1 0 2 3 d⇤J = dJ dx dx dx dJ dx dx dx ^ ^ ^ ^ ^ ^ dJ2 dx0 dx3 dx1 dJ3 dx0 dx1 dx2 ^ ^ ^ ^ ^ ^ =(∂ Jµ) dx0 dx1 dx2 dx3.(2.101) µ ^ ^ ^

2.5 Integral form of Maxwell’s equations

Integral form of Maxwell’s equations which commonly appears in the literature is, in fact, not a fully integral form because it contains temporal derivatives. In order to get fully integral form one needs to integrate Ampere-Maxwell’s law and Faraday’s law over temporal coordinate. It leads to surface integral calculated on two-dimensional surfaces parametrizes by temporal coordinate. Differential forms allow us to obtain explicitely invariant formulation of fully integral Maxwell’s equations. maxwell’s equations 75

Integration of differential forms

The integral of differential form is given by integration of a function representing values that differential form takes on vectors v T(p). i 2 Let us consider k dimensional regular surface sector embedded in P some region of Minkowski spacetime (k N) Parametrization of the surface O⇢A  sector F Rk (t1,...,tk) x(t1,...,tk) RN (2.102) P D3 7! 2 such that for all values of parameters (t1,...,tk) the matrix 2D ∂xa qˆ := , a = 0, 1, 2, 3 j = 1, . . . , k (2.103) ∂ j  t has rank k. Rank k transformation matrix A regular surface sector is called oriented if a certain parametrization of the type (2.102) has been chosen, and admitted changes of parametrization Fixing orientation of the surface sector ti = ti(t1,...,tk), i = 1, . . . , k (2.104) are such that their Jacobian is positive

∂ti det > 0. ∂ j  t

Figure 2.3: Oriented surface sector embedded in . O

For the differential form of degree k

k 1 0 3 a a w = wa ...a (x ,...,x ) dx 1 ... dx k , (2.105) k! 1 k ^ ^ given in some region , the integral of this form over some regular O surface sector is deﬁned as follows Deﬁnition of the integral of P k form 76 lecture notes on classical electrodynamics

k k 1 k w := w(v1,...,vk)dt ...dt , (2.106) ZP ZD k where w(v1,...,vk) is the value of the form on vectors ∂xa vj := ea(p), a = 0, 1, 2, 3, j = 1, . . . , k. (2.107) ∂tj Since ∂xai ai ai b b ai b ai ai dx (vj)=dx (v eb)=v dx (eb)=v d = v = (2.108) j j j b j ∂tj are partial derivatives, then applying (2.62) one gets in (2.106) the determinant of partial derivatives. This determinant is just the Jacobian of transformation xa = xa(t1,...,tk) i.e.

∂xai ∂(xa1 ...xak ) det i, j = 1, 2, . . . , k. (2.109) ∂ j ⌘ ∂(t1 ...tk)  t Thus the integral (2.106) reads Integral of k form

a1 ak k 1 ∂(x ...x ) 1 k w := wa ...a dt ...dt . (2.110) k! 1 k ∂(t1 ...tk) ZP ZD

k When the integral of the form w splits into a ﬁnite or countable number Splitting the integral into inte- of integrals over orientable surfaces i which have no common points, grals over distinct orientable sur- k P then the integral of the form w over hypersurface is a sum of integrals faces P over Pi k k w := Â w. (2.111) ZP i ZPi Let us consider an oriented hypersurface ∂ that consists on regular Induced orientation at the border P sectors ∂ ,...,∂ . The hypersurface ∂ = F(∂ ) is the image of P1 PM P D faces S , i = 1, . . . , M of the polyhedron in the space of parameters i D t ,...t i.e. { 1 k} ∂ = S S ... S . D 1 [ 2 [ [ M The orientation introduced on allows to introduce invariantly the D orientation on ∂ . This is so-called induced orientation. One can choose P outward unit vector n of the polyhedron . It is perpendicular to S at D i any point p localised in its interior. The orientation on Si is determined by the basis f1,...,fk 1, and chosen in such a way that n, f1,...,fk 1 has orientation compatible with the orientation of Rk.

Integral of volume form

We shall consider the integral of volume form 1 Figure 2.4: Induced orientation. vol = 1 = g e dxµ dxn dxa dxb. ⇤ 4! µnab ^ ^ ^ p maxwell’s equations 77

over some four-dimensional region of Minkowski spacetime . P⇢A This integral is given by expression

1 ∂(xµxnxaxb) vol = # dt1dt2dt3dt4 = d4W (2.112) 4! µnab ∂(t1t2t3t4) ZP ZD ZD where

1 ∂(xµxnxaxb) d4W := # dt1dt2dt3dt4 (2.113) 4! µnab ∂(t1t2t3t4) is the four-dimensional volume element in Minkowski spacetime.

Stokes Theorem

k 1 Given (k 1)–form w of class 1 on ∂ the integral of this form over Stokes theorem for differential C P[ P ∂ is equal to the integral of its exterior derivative over forms P P k 1 k 1 w = d w . (2.114) ∂ Z P ZP This theorem is of extremal importance in physics. As example we shall consider application of the Stokes theorem for one-form and two-form that are defined in Euclidean space E3. 1 The one-form w and its exterior derivative read Stokes theorem for one-form and its relation with Stokes theorem 1 1 2 3 w = w1dx + w2dx + w3dx , for vector fields 1 dw =(∂ w ∂ w )dx2 dx3 +(∂ w ∂ w )dx3 dx1 2 3 3 2 ^ 3 1 1 3 ^ +(∂ w ∂ w )dx1 dx2.(2.115) 1 2 2 1 ^ If one associates Cartesian components of certain vector field with 1 components of the form w

H (w , w , w ) ! 1 2 3 (i.e. H w ), then Stokes theorem for differential forms takes the form i ⌘ i of familiar Stokes theorem for vector fields 1 1 w = dw H dx = H dS, Z∂S ZS , IC · ZS r⇥ · where components of the curl operator H are given directly by r⇥1 components of the exterior derivative dw. The physical example of such a vector field is magnetic field strength. Similarly, in the case of two-form one gets Stokes theorem for two-form and its relation with Gauss-Green the- 2 w = w dx2 dx3 + w dx3 dx1 + w dx1 dx2, 23 ^ 31 ^ 12 ^ orem for vector fields 2 dw =(∂ w + ∂ w + ∂ w )dx1 dx2 dx3. 1 23 2 31 3 12 ^ ^ 78 lecture notes on classical electrodynamics

If one associates components of a certain vector field D with compo- Components of strength vectors nents of this two-form, correspond with component of one-forms D (w23, w31, w12), ! 1 (E, H) w then the exterior derivative of this form corresponds with divergence of $ this vector field D. In such a case the Stokes theorem for differential whereas components of induc- r· forms results in the Gauss theorem for vector fields tions correspond with compo-

2 2 nents of two-forms w = dw D dS = D dV. ∂ , · r· 2 Z V ZV IS ZV (D, B) w $ The electric dislocation vector is an example of such a vector ﬁeld.

Integral form of Maxwell’s equations

Let be a three-dimensional orientable surface sector in Minkowski P P spacetime. A border of this surface sector is some piecewise regular two- dimensional closed surface ∂ . Integrating the ﬁrst pair of Maxwell’s P equations over the region ∂ and applying Stokes theorem one P[ P gets First pair of Maxwell’s equations

4p ⇤F = ⇤J. (2.116) ∂ c Z P ZP The second pair of Maxwell’s equations can be integrated in a similar way and it takes the form Second pair of Maxwell’s equations F = 0. (2.117) ∂ Z P Applying formula (2.110) to rhs of (2.116) one gets

a b g 1 1 1 ∂(x x x ) 1 2 3 ⇤J = ⇤ J dt dt dt (2.118) c c 3! abg ∂(t1t2t3) ZP ZD a b g 1 µ 1 ∂(x x x ) 1 2 3 = J #µa g dt dt dt c 3! b ∂(t1t2t3) ZD  1 µ 3 = J d Sµ (2.119) c ZP where three-volume element in Minkowski spacetime reads Three-volume element in the Minkowski spacetime a b g 3 1 ∂(x x x ) 1 2 3 d Sµ := #µa g dt dt dt . (2.120) 3! b ∂(t1t2t3)

If the four-vector Jµ has a single non-vanishing component J0 in the laboratory reference frame, than the integral over three-dimensional 3 region will contain only d S0 – three–volume element on hypersurface 0 1 0 3 x = const. Hence the expression c J d S0 has interpretation of total D R maxwell’s equations 79

electric charge contained in the region . It allows us to conclude that D the expression Invariant deﬁnition of electric charge 1 µ 3 Q := J d Sµ (2.121) c ZD represents total electric charge in the spacetime region . This expres- P sion is Lorentz invariant and thus it must remain unchanged under the change of inertial reference frame (i.e. under the Lorentz transformations). For generic situation all components Jµ are different from zero. Applying formula (2.110) to lhs of (2.116) one gets

a b 1 ∂(x x ) 1 2 ⇤F = ⇤Fab du du ∂ 2 ∂ ∂(u1u2) Z P Z D a b 1 µn 1 ∂(x x ) 1 2 = F #µnab du du 2 ∂ 2 ∂(u1u2) Z D  1 µn 2 = F d Sµn (2.122) 2 ∂ Z P where Two-area element at the surface in the Minkowski spacetime a b 2 1 ∂(x x ) 1 2 d Sµn := # du du (2.123) 2 µnab ∂(u1u2) is two-dimensional surface element and parameters u1, u2 are chosen in such a way, that orientation of faces Si (border of three dimensional region) induces the orientation which is consistent with orientation of R3 given by t1, t2, t3. Thus, lhs of the equation (2.117) takes the form

a b 1 ∂(x x ) 1 2 F = Fab du du ∂ 2 ∂ ∂(u1u2) Z P Z D a b 1 1 µ n µ n ∂(x x ) 1 2 = Fµn(da db db da) du du 2 ∂ 2 ∂(u1u2) Z D a b 1 1 µnls 1 ∂(x x ) 1 2 = Fµn# #abls du du 2 ∂ 2 2 ∂(u1u2) Z D   1 ls 2 = ⇤F d Sls (2.124) 2 ∂ Z P We define two generalized fluxes Generalized fluxes

1 µn 2 F(∂ ) := F d Sµn P 2 ∂ I P and

1 µn 2 F(∂ ) :=+ ⇤F d Sµn. P 2 ∂ I P e 80 lecture notes on classical electrodynamics

They are invariant under Lorentz transformations. Maxwell’s equations in its integral form are given as generalized fluxes eveluated on surfaces ∂ Maxwell’s equations as general- P ized fluxes F(∂ )=4pQ, F(∂ )=0. (2.125) P P This form of Maxwell’s equation is physicallye very sound. It shows that the content of Maxwell’s equations is the relation between generalized fluxes and the electric charges. Moreover, the explicit invariance of integral Maxwell’s equations follows from the fact that they are built of Lorentz invariants. For a particular choice of the surface ∂ (border of purely P spatial surface) these equations reduce to electric and magneticGauss’s law in its integral form. On the other hand, when the surface is the cylinder with height measured by temporal coordinate x0, then the equations give integral version of Ampere-Maxwell law and Faraday’s law.

Example

Let us consider purely spatial surface in the form of cube with side a P and the center located at the origin of reference frame. We choose the following parametrization

x0 = const, x1 = t1, x2 = t2, x3 = t3.

3 a The vector normal to the upper face x = 2 is given by n = e3. The inﬁnitesimal area element at this surface reads Electric ﬂux

1 2 ∂(x x ) 1 2 1 2 dS03 = # dt dt = dt dt . 0312 ∂(t1t2)

Let E (0, 0, E) be the electric field F03 = E (uniform in space). It ! 3 a follows that the flux of the electric field through the face x = 2 reads Magnetic flux

a/2 a/2 3 a 1 µn 2 1 2 2 F(x = 2 )= F d Sµn = dt dt E = a E. 2 a/2 a/2 Z Z Z If we substitute the electric ﬁeld by some uniform magnetic ﬁeld with components B (0, 0, B), where F03 = B, then ! ⇤ a/2 a/2 3 a 1 µn 2 1 2 2 F(x = 2 )=+ ⇤F d Sµn = dt dt B = a B. 2 a/2 a/2 Z Z Z e