Conservation Laws in E&M

Physics 460 Fall 2006 Susan M. Lea

1 Conservation laws in E&M

Last semester we noted that charge is conserved, and we were able to write both integral dQ d = ρdτ = ~j ^n dA dt dt ¡ ¢ Z Z and di¤erential ∂ρ + ~ ~j = 0 ∂t r ¢ equations to express charge conservation. Our next task is to do the same thing with energy, momentum and angular momentum.

1.1 Conservation of energy in E&M We have already found expressions for the energy stored in electric and magnetic ﬁelds.The total energy in a volume V is

2 1 2 1 B U = ε0E + dτ 2 2 µ ZV µ 0 ¶ The work done by electromagnetic forces on a charge q moving through a dis- placement d~l is

dW = q(E~ + ~v B~ ) d~l £ ¢ = q E~ + ~v B~ ~v dt £ ¢ = qE~³ ~v dt ´ ¢ As usual, the magnetic force does no work. Thus the rate at which work is done is dW = qE~ ~v dt ¢ Now consider a volume element with charge dq = ρdτ . The work done on this charge is dW = ρ~v E~ dτ = ~j E~ dτ dt ¢ ¢ Now we sum up over the whole volume to ﬁnd the total rate of doing work: dW = ~j E~ dτ dt ¢ ZV We can express this result in terms of the ﬁelds alone by using the Ampere- Maxwell law for ~j : 1 ∂E~ ~j = ~ B~ ε0 (1) µ0 r £ ¡ ∂t

1 Thus dW 1 ∂E~ = E~ ~ B~ ε dτ dt ¢ µ r £ ¡ 0 ∂t ZV Ã 0 ! Now from the front cover:

~ E~ B~ = B~ ~ E~ E~ ~ B~ r ¢ £ ¢ r £ ¡ ¢ r £ ³ ´ ³ ´ ³ ´ The last term is what we have in our integrand, and we can use Faraday’s law for ~ E~ to get: r £ ∂B~ E~ ~ B~ = B~ ~ E~ B~ ¢ r £ ¢ Ã¡ ∂t ! ¡ r ¢ £ ³ ´ ³ ´ Then ~ ~ dW 1 ~ ∂B 1 ~ ~ ~ ~ ∂E = B + E B + E ε0 dτ dt ¡ V "µ0 ¢ ∂t µ0 r ¢ £ ¢ ∂t # Z ³ ´ Using the divergence theorem on the middle term, we have:

dW 1 1 ∂ ~ ~ ∂ ~ ~ 1 ~ ~ = B B + ε0 E E dτ E B n^ dA dt ¡ 2 V µ0 ∂t ¢ ∂t ¢ ¡ S µ0 £ ¢ Z · ³ ´ ³ ´¸ Z ³ ´ Or, for a ﬁxed volume V,

d 1 B2 1 dW ε E2 + dτ = E~ B~ n^ dA dt 2 0 µ ¡ µ £ ¢ ¡ dt Z µ 0 ¶ ZS 0 1 ³ ´ = E~ B~ n^ dA ~j E~ dτ (2) ¡ S µ0 £ ¢ ¡ V ¢ Z ³ ´ Z This equation should be interpreted as follows:

rate of change of em energy stored in the volume = energy ﬂow into the volume - rate at which energy is converted to non-em forms as the ﬁelds do work..

Remember that ^n points outward from the volume. Mathematically: dU dW = S~ ^n dA (3) dt ¡ ¢ ¡ dt Zs where 1 S~ = E~ B~ (4) µ0 £ ³ ´ is the Poynting vector that describes energy flow in the fields. Its units are J/m2 s. S~ n^ is positive when energy flows out of the volume. ¢ ¢

2 We can convert equation (2) to di¤erential form in the usual way, using the divergence theorem. 2 ∂ 1 2 B 1 ε0E + + ~j E~ + ~ E~ B~ dτ = 0 V ∂t 2 µ0 ¢ r ¢ µ0 £ Z · µ ¶ ³ ´¸ for any volume V, so 2 ∂ 1 2 B ~ ~ 1 ~ ~ ε0E + + ~j E + E B = 0 ∂t 2 µ0 ¢ r ¢ µ0 £ µ ¶ ³ ´ or, equivalently, ∂u E M + ~ S~ = ~j E~ (5) ∂t r ¢ ¡ ¢ It is interesting at this point to consider energy flow in a simple circuit with a battery and a resistor. The battery supplies energy to the circuit, and the resistor converts that energy to non-electro-magnetic forms (primarily thermal energy). So energy is transported from the battery to the resistor. But how? First let’s look at the field configuration. There are both electric and magnetic fields. After the switch is closed the battery quickly establishes the necessary charge distribution to create the potential distribution and electric field that drive current through the resistor. The current loop produces a magnetic field, as shown in the diagram. Then there is a non-zero Poynting flux that transmits energy across the empty space between the battery and the resistor. S~ is greatest near (but not in ) the wires, where the fields are greatest. ¯ ¯ ¯ ¯ ¯ ¯

3 1.2 Conservation of momentum 1.2.1 Newton’s 3rd law in E&M It is easy to show that Newton’s third law holds in electrostatics: The force on charge A due to charge B is equal and opposite to the charge on B due to A ¡ this comes directly from Coulomb’s law. But things get more interesting when magnetic ﬁelds are involved. The current density due to a charge q moving with speed ~v is ~j = ρ~v = qδ (~r ~r (t)) ~v ¡ q and this current density produces a magnetic ﬁeld according to the Biot-Savart Law. Provided that v c, we may write ¿ µ ~j R~ B~ (~r) = 0 £ dτ where R~ = ~r ~r 4π R3 0 ¡ 0 Z µ ~v R~ = 0 q δ (~r ~r (t)) £ dτ 4π ¡ q R3 Z ~ µ0 ~v R = q £ where R~ = ~r ~rq (t) 4π R3 ¡ Now consider two charges moving at right angles, as shown:

~ Charge 1 produces a magnetic ﬁeld B1 at the position of charge 2, with a resulting force exerted on 2 by 1

F~ = q ~v B~ on 2 by 1 2 2 £ 1 that points downward. But the magnetic ﬁeld produced by charge 2 is zero at the (instantaneous) position of charge 1 because ~v R~ = 0, and so the force £ exerted on 1 by 2 is zero. This result clearly violates Newtons’ third law. It then appears that momentum is not conserved, as

d~p d~p 2 = F~ ; 1 = F~ = 0 dt on 2 by 1 dt on 1 by 2 But momentum conservation is one of the most fundamental principles in physics. The problem is resolved by realising that there is momentum stored in the ﬁelds that is also changing. It should not surprise us that momentum is stored in the ﬁelds if energy is. 1 2 Consider a particle moving with velocity ~v. It has kinetic energy K = 2 mv

4 and momentum ~p = m~v. The ratio of the magnitudes is K v = ~p 2 j j Later this semester, we will see that as the particle’s speed approaches the speed of light, the ratio becomes

K γ 1 c v2 c = c ¡ = c 1 1 2 c as v c ~p γ v Ã ¡ r ¡ c ! v ! ! j j µ ¶ ³ ´ We have been using a classical field model of electricity and magnetism, but we could also model the fields using a particle picture. The photon is a massless particle that travels at speed c, and so for the photon K/p = c. Thus we might guess that for the EM field, the momentum flux is given by

S~ 1 = E~ B~ c cµ0 £ ³ ´ and then the momentum density

f lux S~ ~ = = = ε E~ B~ (6) PEM c c2 0 £ ³ ´ Our next task is to prove this conjecture.

1.2.2 The Maxwell Stress tensor We begin by looking at the force per unit volume f~ in a system of charged particles and ﬁelds. The charge in a volume dτ is dq = ρdτ, and then

f~dτ = dF~ = ρdτ E~ + ~v B~ £ f~ = ρE~ + ~j B~³ ´ £ Now we use Maxwell’s equations to eliminate ρ and ~j : ρ = ε ~ E~ 0r ¢ and ~j from equation (1): ~ ~ ~ ~ ~ 1 ~ ~ ∂E ~ f = ε0 E E + B ε0 B r ¢ Ã µ0 r £ ¡ ∂t ! £ ³ ´ Let’s work on the last term. Remember: the order of the vectors in a cross product matters!

∂ ∂E~ ∂B~ E~ B~ = B~ + E~ ∂t £ ∂t £ £ ∂t ³ ´ ∂E~ = B~ + E~ ~ E~ (Faraday’s law) ∂t £ £ ¡r £ ³ ´ 5 Thus ∂E~ ∂ B~ = E~ B~ + E~ ~ E~ ∂t £ ∂t £ £ r £ Now we may expand the triple cro³ss prod´uct, to ge³t ´

∂ E~ ~ E~ = εijkEjεklm Em £ r £ i ∂xl h ³ ´i ∂ = (δilδjm δimδjl) Ej Em ¡ ∂xl ∂ ∂ = Ej Ej Ej Ei ∂xi ¡ ∂xj 1 ∂ 2 ~ ~ = E E Ei 2 ∂xi ¡ ¢ r ³ ´ with a similar expansion for ~ B~ B~ = B~ ~ B~ . Putting all this r £ £ ¡ £ r £ back into f~, we get ³ ´ ³ ´

~ 1 1 2 1 2 ∂ f = ε0 ~ E~ E~ ~ B B~ ~ B~ ε0 ~ E E~ ~ E~ ε0 E~ B~ r ¢ ¡ µ0 2 r ¡ ¢ r ¡ 2r ¡ ¢ r ¡ ∂t £ ³ ´ · ³ ´ ¸ · ³ ´ ¸ ³ ´ ∂ B2 = ε E~ B~ ~ + ε E2 ¡ 0 ∂t £ ¡ r µ 0 µ 0 ¶ ³ ´ 1 +ε0 ~ E~ E~ + E~ ~ E~ + B~ ~ B~ + B~ ~ B~ r ¢ ¢ r µ0 ¢ r r ¢ h³ ´ ³ ´ i h³ ´ ³ ´i In the last line I added the term B~ ~ B~ , which is identically zero, to make r ¢ the last two terms symmetrical in E~³and B~´. Tidying up a bit, we get

~ ~ ∂S ~ ~ ~ ~ ~ ~ ~ 1 ~ ~ ~ ~ ~ ~ f = ε0µ0 u+ε0 E E + E E + B B + B B ¡ ∂t ¡r r ¢ ¢ r µ0 ¢ r r ¢ h³ ´ ³ ´ i h³ ´ ³ ´i The ﬁrst term is, from equation (6),

1 ∂S~ ∂ ~ = PEM ¡ c2 ∂t ¡ ∂t

The remaining terms may be written as the divergence of a tensor Tij . A tensor as a total of 3 3 = 9 components, labelled with the two indices i and j, each £ of which ranges over the values 1-3, 1 = x, 2 = y, 3 = z. It is much easier to

6 write equations in index notation when we have tensors. So ∂ ∂ E2 B2 1 Tij = ε0 + + ε0 ~ E~ Ei + E~ ~ Ei + B~ ~ Bi + Bi ~ B~ ∂xj ¡∂xi 2 µ0 r ¢ ¢ r µ0 ¢ r r ¢ · ¸ h³ ´ ³ ´ i h³ ´ ³ ´i ∂ E2 B2 ∂E ∂ 1 ∂ ∂B = δ ε + + ε j E + E E + B B + B j ¡ ij ∂x 0 2 µ 0 ∂x i j ∂x i µ j ∂x i i ∂x j · 0 ¸ · j j ¸ 0 · j j ¸ ∂ E2 B2 ∂ ∂ 1 = δ ε + + (ε E E ) + B B ¡ ij ∂x 0 2 µ ∂x 0 j i ∂x µ j i j · 0 ¸ j j µ 0 ¶ ∂ E2 B2 1 = δ ε + + ε E E + B B (7) ∂x ¡ ij 0 2 µ 0 j i µ j i j · · 0 ¸ 0 ¸ We may write the components conveniently as a matrix. The ﬁrst term con- tributes only when i = j, ie on the diagonal of the matrix.

ε0 2 2 2 2 Ex Ey Ez 1 1 1 2¡ 2¡ 2 ε0ExEy + µ BxBy ε0ExEz + µ BxBz + 2µ Bx By Bz 0 0 0 0¡ ¡ ¡ ¢ 1 ε0 E2 E2 E2 ¡ 1 ¢ 2 y x z 1 Ã!T = B ε0ExEy + µ BxBy 1 2¡ 2¡ 2 ε0EyEz + µ By Bz C 0 + By Bx Bz 0 B 2µ 0¡ ¡ ¡ ¢ C B ε0 E2 E2 E2 C B 1 ¡ 1 ¢ 2 z x y C B ε0ExEz + µ BxBz ε0EyEz + µ By Bz 1 2¡ 2¡ 2 C B 0 0 + Bz Bx By C B 2µ0¡ ¡ ¡ ¢ C (8) @ ¡ ¢ A Note that this tensor is symmetric, that is:

Tij = Tji Finally we write the force density equation as: ~ ∂ E M , i ∂Tij fi = P + (9) ¡ ∂t ∂xj and integrating over the volume:

∂ ~ , i ∂T f dτ = PEM dτ + ij dτ i ¡ ∂t ∂x ZV ZV ZV j Using the divergence theorem, we convert the last term to a surface integral:

∂ ~ , i F = PEM dτ + T n dA (10) i ¡ ∂t ij j ZV ZS or equivalently d F~ = ~ dτ + ÃT! dA~ (11) ¡ dt PEM ¢ Z ZS or d ~ dτ = Ã!T dA~ F~ dt PEM ¢ ¡ Z ZS We may interpret this as:

7 Rate of change of stored EM momentum in the volume=-force exerted by ﬁelds in the volume on charges in the volume - force exerted by ﬁelds in the volume on the surface of the volume.

The physical interpretation of the tensor component Tij is:

Tij is the force per unit area in the ith direction exerted on an area element with normal in the jth direction.

The diagonal elements such as Txx are pressures (normal force per unit area) while the o¤-diagonal components are shears. Let’s see what happens to equation 11 when there is no time dependence. We have F~ = ÃT! dA~ ¢ ZS This means that we can calculate the force equally well by calculating the force per unit volume, and integrating over the volume, or the force per unit area and integrating over the area. For example, Suppose we have two point charges separated by a distance D. The volume integral is just Coulomb’s law: q q F~ = 1 2 r^ 2 4πε0D But we should be able to get the result by integrating over any surface com- pletely surrounding one of the charges.We need the total electric ﬁeld due to both charges: 1 q q E~ = E~ + E~ = 1 r^ + 2 r^ 1 2 4πε r2 1 r2 2 0 µ 1 2 ¶ Let’s put the x axis along the line joining the two charges, with the origin half ¡ way between them.

The easiest surface to use is a rectangular box, with ﬂat surfaces at x = 0, x , y and z . This box encloses charge 2, but if we choose ! 1 ! §1 ! §1

8 x , that box encloses charge 1. The fields go to zero fast enough that the ! ¡1 only non-zero contribution is from the surface at finite x. (The fields decrease like 1/distance2, so the product of two field components goes like 1/distance4, but the area only increases as distance2, so the product 0 as 1/distance2. ) Since the box enclosing charge one has outward normal!n^ = x^ while the box enclosing charge 2 has outward normal ^n = ^x, we can see immediately that the ¡ forces each charge exerts on the other are equal and opposite. Let’s calculate the force on charge 2. On the plane

1 q D q D E~ = 1 x^ + yy^ + zz^ + 2 ¡ x^ + yy^ + z^z 4πε 0 2 3/2 2 2 3/2 2 1 0 D + y2 + z2 µ ¶ D + y2 + z2 µ ¶ B 2 2 C @ ³¡ ¢ ´ ³¡ ¢ ´ A + + 1 1 F = dy dzT x ¡ xx Z¡1 Z¡1 + + 1 1 ε = dy dz 0 E2 E2 E2 ¡ 2 x ¡ y ¡ z Z¡1 Z¡1 + + ¡ E2 ¢ = 1 dy 1 dzε E2 ¡ 0 x ¡ 2 Z¡1 Z¡1 µ ¶ 2 + + q1 D/2 q2 D/2 1 1 1 (D 2/4+y2 +z2) 3/2 ¡ (D 2/4+y 2+z2) 3/2 = 2 dy dz 0 2 1 ¡ (4π) ε0 h 1 q1r^1 q1r^2 i Z¡1 Z¡1 2 D 2/4+y 2+z2 + D 2/4+y 2+z 2 B ¡ C @ h i A We may use polar coordinates on the plane, where y2 + z2 = s2

2π + 2 2 2 2 + 2 1 q D /4 πq D 1 du D sdsdθ 1 = 1 where u = + s2 2 2 3 3 0 0 (D /4 + s ) 4 D 2/4 u 4 Z Z Z 2 2 2 πq D 1 1 2πq = 1 = 1 4 ¡ 2u2 D2 µ ¶¯D 2/4 ¯ ¯ and ¯ 2π + q2 1 1 4πq2 1 sdsdθ 1 = πq2 = 1 2 1 2 2 2 2 ¡ u 2 D Z0 Z0 (D /4 + s ) µ ¶¯D /4 ¯ ¯ Thus the ﬁrst term in each of the brackets sum to give ¯ 2πq2 1 4πq2 1 1 = 0 D2 ¡ 2 D2 Of course we expect this because the result should involve only the product of the two di¤erent charges. The same thing happens with the last term in each

9 of the two brackets. Thus we are left with the cross terms.

2π + 2 2 2 1 q1q2D /2 D 1 du D 2 sdsdθ ¡ = πq1q2 where u = + s 2 2 3 3 0 0 (D /4 + s ) ¡ 2 D 2/4 u 4 Z Z Z 2 D 1 1 = πq q ¡ 1 2 2 ¡ 2u2 µ ¶¯D 2/4 q q ¯ = 4π 1 2 ¯ ¡ D2 ¯ and

2π + 2π + 2 2 2 1 2q q ^r r^ 1 q1q2 r + r r sdsdθ 1 2 1 ¢ 2 = 2sdsdθ y z ¡ x 2 2 2 2 2 2 2 0 0 (D /4 + s ) 0 0 (D /4 + s ) r Z Z Z Z ¡ ¢ 2π + 2 2 1 s D /4 = q1q2 2sdsdθ ¡ 2 2 3 0 0 (D /4 + s ) Z Z ¡ ¢ 2π + 2 2 2 1 s + D /4 D /2 = q1q2 2sdsdθ ¡ 2 2 3 0 0 (D /4 + s ) Z Z ¡ ¢ 2π + 2 1 1 D /2 = q1q2 2sdsdθ 2 2 2 2 2 3 0 0 (D /4 + s ) ¡ (D /4 + s ) Z Z " # We havre already done the second integral, so let’s work on the ﬁrst:

2π + 1 1 1 1 8π 2sdsdθ = 2π du = 2 2 2 2 2 0 0 (D /4 + s ) D 2/4 u D Z Z Z Thus q q 4π 1 8π 8π F = 1 2 x ¡ 2 ¡ D2 ¡ 2 D2 ¡ D2 (4π) ε0 µ · ¸¶ q1q2 = 2 4πε0D as expected. Finally let’s look at Fy .

+ + 1 1 F = dy dzT y ¡ yx Z¡1 Z¡1 + + 1 1 = dy dzε E E ¡ 0 x y Z¡1 Z¡1 + + 1 1 1 q D/2 q D/2 = dy dz 1 2 2 2 2 2 3/2 2 2 2 3/2 ¡ (4π) ε0 "(D /4 + y + z ) ¡ (D /4 + y + z ) # Z¡1 Z¡1

q1y q2y 3/2 3/2 £ "(D2/4 + y2 + z2) ¡ (D2/4 + y2 + z2) #

10 The terms look like

+ + 2 1 1 q yD/2 dy dz 1 = 0 (D2/4 + y2 + z2)3 Z¡1 Z¡1 because the integral over y has an odd integrand integrated over an even interval. The cross terms are zero for the same reason.

+ + 1 1 q q yD/2 dy dz 1 2 = 0 (D2/4 + y2 + z2)3 Z¡1 Z¡1 Thus the force only has an x component, as expected. ¡

1.2.3 Conservation of momentum Let’s return to the full force law (11), Since the force acting on the charges in the volume changes their mechanical momentum, we may rewrite the equation as: d ~EM + ~M ech dτ = ÃT! dA~ dt P P S ¢ Z ³ ´ Z or, in di¤erential form,

∂ ~ + ~ = ~ ÃT! (12) ∂t PEM PMech r ¢ ³ ´ 1.3 Angular momentum If the ﬁelds carry momentum, they must also carry angular momentum:

~` = ~r ~ EM £ PEM = ε ~r E~ B~ 0 £ £ ³ ´ As an example, consider a conﬁguration of charges and currents as follows.

We have two concentric cylindrical shells of charge with radii a and c > a. The

11 cylinders have equal but opposite charges, so the surface charge densities are σ and σc/a respectively. There is an electric field ¡ σa E~ = s^ ε0s for c > s > a. A solenoid with radius b, c > b > a, carries azimuthal current I, so a magnetic field ~ B = µ0nIz^ exists in the region s < b. Thus for a < s < b we have both electric and magnetic fields. 2 ~ σa σa ^ EM = ε0 ^s µ0nIz^ = µ0nIθ P ε0s £ ¡ 2s and the angular momentum density about the origin is: σa `~ = ~r µ nIθ^ EM £ ¡ s 0 σa³ ´ = µ nI (s^s + zz^) θ^ ¡ s 0 £ z = σaµ nI z^ s^ ¡ 0 ¡ s ³ ´ The total angular momentum contained in a cylinder of height h is ~ L~ EM = ÈM dτ Z h/2 2π b z = dz dφ sds σaµ0nI z^ ^s h/2 0 a ¡ s Z¡ Z Z n ³ ó = πσaµ nI b2 a2 h z^ ¡ 0 ¡ Now if we turn the current o¤, B~¡ 0 an¢d L~ 0 also. Since there are no ! E M ! external torques, L~ should be conserved. Aha, but the changing B~ produces an azimuthal E~ , by Faraday’s law, and the resulting torque on the cylinders causes them to rotate. We have

d 2 2πsEθ = πs B ¡ dt s dI E = µ n θ ¡ 2 0 dt and the torque on a height h of the cylinder of radius a < b is

~τ = aσ (2πah) Eθ^z a dI dL~ = 2πσa2h µ n z^ = cylinder ¡ 2 0 dt dt Thus the inner cylinder gains angular momentum:

t2 ~ ~ dLcylinder 3 t2 Linner = dt = πσa hµ n I z^ dt ¡ 0 jt1 Zt1 3 = πσa hµ0nI z^

12 The initial ﬂux through a circle of radius c is πb2B, since B is zero for b < s < c, so we have b2 dI E (c) = µ n θ ¡ 2c 0 dt and, since the outer cylinder has a negative charge density,

a b2 dI ~τ = 2π σ c2h µ n z^ c 2c 0 dt ³ ´ giving L~ = πachb2µ nI z^ outer ¡ 0 The total angular momentum gained by the two cylinders is

L~ = πσaµ nIhz^ b2 a2 z^ = L~ m ech ¡ 0 ¡ EM The total electromagnetic angular mome¡ntum th¢at disappears exactly equals the total mechanical angular momentum gained by the cylinders. Angular momentum is conserved.