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Maxwell's Stress Tensor and Electromagnetic Momentum

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Maxwell's Stress Tensor and Electromagnetic Momentum Maxwell's Stress Tensor and Electromagnetic Momentum Dr. Priti Mishra Momentum conservation is rescued in electrodynamics by the realization that the fields themselves carry momentum. This is not so surprising when you consider that we have already attributed energy to the fields. In the case of the two point charges in Fig. 1, whatever momentum is lost to the particles is gained by the fields. Only when the field momentum is added to the mechanical momentum of the charges is momentum conservation restored. Let us see how this works out quantitatively in the following sections. Figure 1: A point charge q1 moving along x-axis encounters an identical one, q2 proceeding in at the same speed along the y-axis. 1 Maxwell's Stress Tensor Let us calculate the total electromagnetic force on the charges in volume V: As we know the force acting on a point charge q moving with velocity v in an electric field E and magnetic field B is given by Lorentz force F = q(E + v × B) (1) 1 Similarly, the force per unit volume acting on a charge distribution ρ in a volume V : f = ρ(E + v × B) (2) As we know ρv = J where J is the current density. ) f = ρE + J × B (3) Let us write the above equation in terms of fields alone, eliminating ρ and J by using Maxwell's equations (i) and (iv): ρ r:E = 0 ) ρ = 0r:E (4) @E r × B = µ J + µ 0 0 0 @t 1 @E ) J = r × B − 0 (5) µ0 @t Substituting ρ and J from equations (4) and (5) into equation (3) we get 1 @E f = 0(r:E)E + r × B − 0 × B (6) µ0 @t Now @ @E @B (E × B) = × B + E × (7) @t @t @t or @E @ @B × B = (E × B) − E × (8) @t @t @t and from Maxwell's third equation (Faraday's law) @B = −r × E (9) @t Substituting above in equation (8) we get @E @ × B = (E × B) + E × (r × E) (10) @t @t Substituting equation (10) in equation (6) we obtain 1 @ f = 0(r:E)E+ (r × B) × B −0 (E × B) + E × (r × E) (11) µ0 @t or 1 @ f = 0 [(r:E)E − E × (r × E)] + (r × B) × B − 0 (E × B) µ0 @t (12) 2 Just to make things look more symmetrical, let's add a term (r:B)B; since r:B = 0, this costs us nothing. 1 @ f = 0 [(r:E)E − E × (r × E)]+ [(r:B)B − B × r × B]−0 (E × B) µ0 @t (13) Meanwhile, from the property of gradient we know that r(A:B) = A × r × B + B × r × A + (A:r)B + (B:r)A (14) ) r(E2) = r(E:E) = E × r × E + E × r × E + (E:r)E + (E:r)E = 2E × r × E + 2(E:r)E (15) So 1 E × r × E = r(E2) − (E:r)E (16) 2 Similarly 1 B × r × B = r(B2) − (B:r)B (17) 2 Substituting equations (16) and (17) in equation (13) 1 2 1 1 2 f = 0 (r:E)E − r(E ) + (E:r)E + (r:B)B − r(B ) + (B:r)B 2 µ0 2 @ − (E × B) (18) 0 @t Or 1 f = 0 [(r:E)E + (E:r)E] + [(r:B)B + (B:r)B] µ0 1 2 1 2 @ − 0r(E ) + r(B ) − 0 (E × B) (19) 2 2µ0 @t Above equation can be simplified by introducing the Maxwell stress ten- sor, 1 2 1 1 2 Tij = 0 EiEj − δijE + BiBj − δijB (20) 2 µ0 2 The indices i and j refer to the coordinates x; y, and z, so the stress ten- sor has a total of nine components (Txx;Txy;Txz;Tyx;Tyy;Tyz;Tzx;Tzy;Tzz). 3 The Kronecker delta, δij, is 1 if the indices are the same (δxx = δyy = δzz = 1) and zero otherwise (δxy = δyz = δzx = 0). Thus 1 2 2 2 1 2 2 2 Txx = 0 Ex − Ey − Ez + Bx − By − Bz (21) 2 2µ0 1 2 2 2 1 2 2 2 Tyy = 0 Ey − Ex − Ez + By − Bx − Bz (22) 2 2µ0 1 2 2 2 1 2 2 2 Tzz = 0 Ez − Ex − Ey + Bz − By − Bx (23) 2 2µ0 and 1 Txy = Tyx = 0ExEy + BxBy (24) µ0 1 Tyz = Tzy = 0EyEz + ByBz (25) µ0 1 Tzx = Txz = 0EzEx + BzBx: (26) µ0 Because it carries two indices, where a vector has only one, Tij is sometimes ! ! written with a double arrow: T . T is a rank-2 tensor, it is represented by a 2-dimensional, 3 × 3 matrix: 2T T T 3 ! xx xy xz T = 4Tyx Tyy Tyz5 Tzx Tzy Tzz where Txx;Txy;Txz;Tyy;Tzz;Tyz etc are given in equations (21)-(26). ! One can form the dot product of T with a vector a: ! (a: T )j = Σx;y;z(aiTij) (27) the resulting object, which has one remaining index, is itself a vector. In ! particular, the divergence of T has as its jth component ! 1 (r: T )j = 0 [(r:E)Ej + (E:r)Ej] + [(r:B)Bj + (B:r)Bj] µ0 1 2 1 2 − 0rj(E ) + rj(B ) (28) 2 2µ0 Thus the force per unit volume in equation (19) can be written in the much simpler form ! @S f = (r: T ) − µ (29) 0 0 @t 4 where S = E × H = E × B is the Poynting vector. µ0 The total force on the charges in volume V is evidently Z Z ! Z @S F = fdV = (r: T )dV − 0µ0 dV (30) V V V @t Or, Z I ! Z @S F = fdV = T :da − 0µ0 dV (31) V S V @t (I used the divergence theorem to convert the first term to a surface inte- R gral.) In the static case (or, more generally, whenever V SdV is independent of time), the second term drops out, and the electromagnetic force on the charge configuration can be expressed entirely in terms of the stress tensor at the boundary. ! Physically, T is the force per unit area (or stress) acting on the surface. More precisely, Tij is the force (per unit area) in the ith direction acting on an element of surface oriented in the jth direction. The "diagonal" elements (Txx;Tyy;Tzz) represent pressures, and ”off-diagonal” elements (Txy;Tyz;Tzx , etc.) are shears. 5 2 Conservation of momentum According to Newton's second law, the force on an object is equal to the rate of change of its momentum: p F = mech (32) dt Equation (31) can therefore be written in the form Z I dpmech @S ! F = = −0µ0 dV + T :da (33) dt V @t S where pmech is the total (mechanical) momentum of the particles contained in the volume V. This expression is similar in structure to Poynting's theo- rem, and it invites an analogous interpretation: The first integral represents momentum stored in the electromagnetic fields themselves: Z pem = 0µ0 SdV (34) V while the second integral is the momentum per unit time flowing in through the surface. Equation (33) is the general statement of conservation of mo- mentum in electrodynamics: Any increase in the total momentum (mechanical plus electromagnetic) is equal to the momentum brought in by the fields.(If V is all of space, then no momentum flows in or out, and pmech + pem is constant.) As in the case of conservation of charge and conservation of energy, con- servation of momentum can be given a differential formulation. Let }mech be the density of mechanical momentum, and }em the density of momentum in the fields: }em = 0µ0S (35) Then Equation (33), in differential form, says @ ! (} + } ) = r: T (36) @t mech em 6 ! Evidently - T is the momentum flux density, playing the role of JJ (current density) in the continuity equation, or S (energy flux density) in Poynting's theorem. Specifically, -Tij is the momentum in the i direction crossing a surface oriented in the j direction, per unit area, per unit time. Notice that the Poynting vector has appeared in two quite different roles: S itself is the energy per unit area, per unit time, transported by the elec- tromagnetic fields, while µ S is the momentum per unit volume stored in !0 0 those fields. Similarly, T plays a dual role: It itself is the electromagnetic ! stress (force per unit area) acting on a surface, and T describes the flow of momentum (the momentum current density) transported by the fields. 7.
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