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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 = ½d¿ = ~j ^n dA dt dt ¡ ¢ Z Z and di¤erential @½ + ~ ~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 = "0E + d¿ 2 2 ¹ 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 = ½d¿ : The work done on this charge is dW = ½~v E~ d¿ = ~j E~ d¿ dt ¢ ¢ Now we sum up over the whole volume to find the total rate of doing work: dW = ~j E~ d¿ 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~ "0 (1) ¹0 r £ ¡ @t 1 Thus dW 1 @E~ = E~ ~ B~ " d¿ dt ¢ ¹ r £ ¡ 0 @t ZV à 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 Faradayzs law for ~ E~ to get: r £ @B~ E~ ~ B~ = B~ ~ E~ B~ ¢ r £ ¢ á @t ! ¡ r ¢ £ ³ ´ ³ ´ Then ~ ~ dW 1 ~ @B 1 ~ ~ ~ ~ @E = B + E B + E "0 d¿ dt ¡ V "¹0 ¢ @t ¹0 r ¢ £ ¢ @t # Z ³ ´ Using the divergence theorem on the middle term, we have: dW 1 1 @ ~ ~ @ ~ ~ 1 ~ ~ = B B + "0 E E d¿ E B n^ dA dt ¡ 2 V ¹0 @t ¢ @t ¢ ¡ S ¹0 £ ¢ Z · ³ ´ ³ ´¸ Z ³ ´ Or, for a fixed volume V; d 1 B2 1 dW " E2 + d¿ = E~ B~ n^ dA dt 2 0 ¹ ¡ ¹ £ ¢ ¡ dt Z µ 0 ¶ ZS 0 1 ³ ´ = E~ B~ n^ dA ~j E~ d¿ (2) ¡ S ¹0 £ ¢ ¡ 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) ¹0 £ 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 "0E + + ~j E~ + ~ E~ B~ d¿ = 0 V @t 2 ¹0 ¢ r ¢ ¹0 £ Z · µ ¶ ³ ´¸ for any volume V; so 2 @ 1 2 B ~ ~ 1 ~ ~ "0E + + ~j E + E B = 0 @t 2 ¹0 ¢ r ¢ ¹0 £ 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 Newtonzs 3rd law in E&M It is easy to show that Newtonzs 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 Coulombzs 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 = ½~v = q± (~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 ± (~r ~r (t)) £ d¿ 4π ¡ q R3 Z ~ ¹0 ~v R = q £ where R~ = ~r ~rq (t) 4¼ 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 particlezs 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 à ¡ 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 ½ and ~j : ½ = " ~ E~ 0r ¢ and ~j from equation (1): ~ ~ ~ ~ ~ 1 ~ ~ @E ~ f = "0 E E + B "0 B r ¢ à ¹0 r £ ¡ @t ! £ ³ ´ Letzs 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~ (Faradayzs 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~ = "ijkEj"klm Em £ r £ i @xl h ³ ´i @ = (±il±jm ±im±jl) 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 = "0 ~ E~ E~ ~ B B~ ~ B~ "0 ~ E E~ ~ E~ "0 E~ B~ r ¢ ¡ ¹0 2 r ¡ ¢ r ¡ 2r ¡ ¢ r ¡ @t £ ³ ´ · ³ ´ ¸ · ³ ´ ¸ ³ ´ @ B2 = " E~ B~ ~ + " E2 ¡ 0 @t £ ¡ r ¹ 0 µ 0 ¶ ³ ´ 1 +"0 ~ E~ E~ + E~ ~ E~ + B~ ~ B~ + B~ ~ B~ r ¢ ¢ r ¹0 ¢ 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 = "0¹0 u+"0 E E + E E + B B + B B ¡ @t ¡r r ¢ ¢ r ¹0 ¢ 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 = "0 + + "0 ~ E~ Ei + E~ ~ Ei + B~ ~ Bi + Bi ~ B~ @xj ¡@xi 2 ¹0 r ¢ ¢ r ¹0 ¢ r r ¢ · ¸ h³ ´ ³ ´ i h³ ´ ³ ´i @ E2 B2 @E @ 1 @ @B = ± " + + " j E + E E + B B + B j ¡ ij @x 0 2 ¹ 0 @x i j @x i ¹ j @x i i @x j · 0 ¸ · j j ¸ 0 · j j ¸ @ E2 B2 @ @ 1 = ± " + + (" E E ) + B B ¡ ij @x 0 2 ¹ @x 0 j i @x ¹ j i j · 0 ¸ j j µ 0 ¶ @ E2 B2 1 = ± " + + " E E + B B (7) @x ¡ ij 0 2 ¹ 0 j i ¹ j i j · · 0 ¸ 0 ¸ We may write the components conveniently as a matrix.
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