8.1 Conservation of Charge and Energy. ρ + J =0, (8.1) t can be derived by considering the flow of charges from a given volume, Q t =

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1 Chapter 8 Conservation Laws 8. Conservation of Charge and Energy As was already shown, the continuity equation ρ + J =, (8.) can be derived by considering the flow of charges from a given volume, Q = ρ(r)d 3 r = J(r) da = J(r)d 3 r (8.2) The continuity of charges expresses the fact that the total charge is conserved. Another conservation law can be derived from the Faraday s and Ampere s laws E + B B E c 2 = (8.3) = µ J. (8.4) By differentiating the energy density (7.9) withrespecttotimewegetthat u = 2 (ϵ E 2 + ( B 2 )=ϵ E B ) µ J + B ( E) µ ϵ µ ϵ µ µ (8.5) or 2 (ϵ E 2 + B 2 )= (E B) J E. (8.6) µ µ This is the Poynting s theorem which expresses the fact that energy is conserved. 8

2 CHAPTER 8. CONSERVATION LAWS 8 Indeed the energy required to assemble a static distribution ofcharges is given by (7.89) andthetheenergyrequiredtoassembleadistributionof currents is given by (7.9). Then from the Poynting s theorem (8.6), the rate of change of energy u ( = 2 (ϵ E 2 + ) B 2 ) d 3 r µ = (E B) da J Ed 3 r. (8.7) µ Note that the second term is the rate at which the total external work is done on all charges W = ρ (E + v B) vdt d 3 r = ρ (E + v B) v d 3 r = E (ρv) d 3 r = E J d 3 r. (8.8) If the system is closed (i.e. there are no external work), then thepoynting s theorem (8.6) canbewrittenas where u = S (8.9) u = 2 (ϵ E 2 + µ B 2 ) (8.) is the energy density and S = µ E B (8.) is the Poynting vector. What it describes is the energy flux similarly to how the current density describes the flux of charges. For example, consider a current I flowing down the wire which heats the environment with a power that we are to calculate. The electric field in the wire is E = V L ẑ (8.2) and the magnetic field at the surface is B = µ I 2πa ˆφ (8.3)

3 CHAPTER 8. CONSERVATION LAWS 82 and therefore the Poynting vector is pointing towards the center of the wire S = µ E B. (8.4) Then we can use the Poynting s theorem to find the power u d3 r = Sd 3 r = S da = VI 2πaL = VI (8.5) 2πaL in agreement with (7.85). 8.2 Conservation of Momentum When one considers motion of charged particles in electromagnetic field very often one encounters situations when the linear or angular momentum of particles is not conserved. The problem is that in addition to momentum carried by particle the momentum can also be carried by fields such as electric and magnetic fields. Consider the total electromagnetic force in some volume F = ρ (E + v B) d 3 r V = (ρe + J B) d 3 r (8.6) V Using two Maxwell s equations (i.e. ϵ E = ρ and µ B ϵ B = J) the force density can be rewritten as f = ρe + J B = ϵ ( E) E + ( ) B ϵ µ E B. (8.7) and the third Maxwell equation (i.e. E + B the time derivative of Poynting vector as =)canusedtorewrite µ S = (E B) = E B + E B = E B E ( E) (8.8)

4 CHAPTER 8. CONSERVATION LAWS 83 or E ϵ B = µ S ϵ + ϵ E ( E). (8.9) By substituting (8.9) into(8.7) andusingthefourthmaxwellequation(i.e. B =)weobtain f = ϵ ( E) E µ B ( B) ϵ E B = ϵ [( E) E E ( E)] + [( B) B B B] µ µ (8.2) S ϵ. This can be modified further using E 2 = 2(E ) E +2E ( E) (8.2) B 2 = 2(B ) B +2B ( B) (8.22) to obtain [ f = ϵ ( E) E +(E ) E ] 2 E2 + [ ( B) B +(B ) B ] S µ 2 B2 µ ϵ. (8.23) Now if we define the Maxwell stress tensor (spatial components oftheso called energy-momentum or stress-energy tensors) ( T ij ϵ E i E j ) 2 δ ije 2 + ( B i B j ) µ 2 δ ijb 2 2 (E2 E2 2 E2 3 ) E E 2 E E 3 = ϵ E 2 E E 2 + E2 2 E3 2 E 2 E 3 + E 3 E E 3 E 2 E 2 E2 2 + E2 3 2 (B2 B2 2 B3) 2 B B 2 B B 3 B 2 B B 2 µ + B2 2 B2 3 B 2 B 3 (8.24) B 3 B B 3 B 2 B 2 B2 2 + B3 2 then the force density can be expressed as f i = j=,2,3 S i T ji µ ϵ x j (8.25) or using the vector notation f = T µ ϵ S (8.26)

5 CHAPTER 8. CONSERVATION LAWS 84 where the divergence is taken with respect to either first or second index in T ij would not make a difference due to symmetry of the stress tensor T ij = T ji. (8.27) This equation represent the flow of momentum that can be exchanged between charged particles and electric and magnetic field without violating the momentum conservation. We can now go back to the volume integral to express the external force as F = f d 3 r = T da µ ϵ Sd 3 r. (8.28) V and in the static case as V S F = S T da. (8.29) Thus, T is nothing but a force per unit area, or stress tensor whose diagonal terms describe pressure and off-diagonal terms describe shear. For example, consider a hemisphere of a uniformly charged sphere of radius R and charge Q. Since this is a static case the stress tensor (a local quantity) calculated over surface should be sufficient to calculate the net force on the hemisphere, F = T da + T da = disk 2π dφ R bowl rdr T ẑ + 2π dφ In Cartesian coordinates the radial unit vector is π/2 sin θdθr 2 T ˆr (8.3) ˆr = sin θ cos φˆx + sin θ sin φŷ +cosθẑ (8.3) and the relevant components of the stress tensor on the bowl aregivenby where T zx = ϵ E z E x = ϵ (ρr) 2 sin θ cos θ cos φ (8.32) T zy = ϵ E z E y = ϵ (ρr) 2 sin θ cos θ sin φ (8.33) T zz = ϵ ( E 2 2 z Ex 2 ) ϵ ( E2 y = 2 (ρr)2 cos 2 θ sin 2 θ ) (8.34) ρ = Q 4πϵ R 3 (8.35)

6 CHAPTER 8. CONSERVATION LAWS 85 or (T ˆr ) z = T zx da x + T zy da y + T zz da z = ϵ (ρr) 2 (sin 2 θ cos φ 2 cos θ + sin 2 θ sin 2 φ cos θ + 2 cos3 θ 2 sin2 θ cos θ) = ϵ 2 (ρr)2 cos θ. (8.36) Similarly on the disk and thus, (T ẑ ) z = T zz = ϵ ( E 2 2 z Ex 2 ) ϵ E2 y = 2 (ρr)2 (8.37) F z = 2π dφ R R rdr ϵ 2π 2 (ρr)2 + dφ π/2 r 3 dr + ϵ ρ 2 πr 4 π/2 sin θdθr 2 ϵ 2 (ρr)2 cos θ = ϵ ρ 2 π sin θ cos θdθ ( = ϵ ρ 2 πr π ) sin(2θ)d(2θ) 4 = 3 ( ) 2 Q 4 ϵ πr 4 4πϵ R 3 3Q 2 = 4πϵ 6R. (8.38) 2 Note that the stress tensor can be generalized to the stress-energy (or energy momentum) tensor u µ ϵ S µ ϵ S 2 µ ϵ S 3 T µν = µ ϵ S T T 2 T 3 µ ϵ S 2 T 2 T 22 T 23 (8.39) µ ϵ S 3 T 3 T 32 T 33 and then the conservations of momentum and energy can be combined in a single covariant equation f µ + µ T µν = (8.4) where f µ is the external four-force of the change of external (let say mechanical) momentum per unit time. The momentum stored in electromagnetic field or in the momentum density g = µ ϵ S = ϵ (E B). (8.4)

7 CHAPTER 8. CONSERVATION LAWS 86 In the absence of the external momentum (i.e. f =)theconservationof momentum implies T da = µ ϵ Sd 3 r (8.42) S V or in a differential form g = T. (8.43) Moreover, one can also define the angular momentum density l = r g = ϵ [r (E B)] (8.44) that must also be conserved. For example consider a long solenoid with current I, radiusr and n turns per unit length and two coaxial cylindrical conducting shells of length l and radii a < Rand b > Rwith charges Q and Q respectively. When the current is switched off the cylindrical shells start to rotate due to flow of angular momentum that was initially stored in electromagnetic fields. Initially the electric field between shellswas and the magnetic field inside solenoid was This corresponds to momentum density and angular momentum density E = Q 2πϵ l sŝ (8.45) B = µ niẑ. (8.46) g = ϵ (E B) = µ niq 2πls ˆφ (8.47) l = µ niq 2πls s ˆφ = µ niq ẑ (8.48) 2πl in the region a<s<ror the total angular momentum L = µ niq ẑ 2πl R a sds 2π dφ l dzẑ = 2 µ niq ( R 2 a 2) ẑ. (8.49) This angular momentum went into the angular momentum of charged cylinders due to induced electric field (Faraday s law), { 2 E = µ n di R 2 ˆφ for s>r dt s µ 2 n di sˆφ (8.5) for s<r. dt

8 CHAPTER 8. CONSERVATION LAWS 87 Then the total torque on the cylinders is N b = r ( QE) = 2 µ nq di R 2 dt b b ˆφ = 2 µ nqr 2 di dt ẑ (8.5) N a = r (QE) = 2 µ nq di dt a a ˆφ = 2 µ nqa 2 di dt ẑ (8.52) and the total angular momentum L = I 2 µ nq ( R 2 a 2) di dt ẑ = 2 µ niq ( R 2 a 2) ẑ. (8.53) The angular momentum that was initially stored in the electric and magnetic field (8.49) is now in the angular momentum of the charges on the cylinders (8.53).