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Meaning of the Maxwell Tensor Meaning of the Maxwell Tensor We define the Maxwell Stress Tensor as 1 1 1 T = ε E E − δ E E + B B − δ B B ij 0 i j ij k k µ i j ij k k 2 0 2 While it looks scary, the Maxwell tensor is just a different way of writing the fact that electric fields exert forces on charges, and that magnetic fields exert forces on cur- rents. Dimensionally, Momentum T = (Area)(Time) This is the dimensionality of a pressure: Force = dP dt = Momentum Area Area (Area)(Time) Meaning of the Maxwell Tensor (2) If we have a uniform electric field in the x direction, what are the components of the tensor? E = (E,0,0) ; B = (0,0,0) 2 1 0 0 1 0 0 ε 1 0 0 2 1 1 0 E = ε 0 0 0 − 0 1 0 + ( ) = 0 −1 0 Tij 0 E 0 µ − 0 0 0 2 0 0 1 0 2 0 0 1 This says that there is negative x momentum flowing in the x direction, and positive y momentum flowing in the y direction, and positive z momentum flowing in the z di- rection. But it’s a constant, with no divergence, so if we integrate the flux of any component out of (not through!) a closed volume, it’s zero Meaning of the Maxwell Tensor (3) But what if the Maxwell tensor is not uniform? Consider a parallel-plate capacitor, and do the surface integral with part of the surface between the plates, and part out- side of the plates Now the x-flux integral will be finite on one end face, and zero on the other. So x-momentum is flowing though the ε 2 closed surface, and the pressure is 0 Ex 2. The force per unit area is the charge per unit area times the electric field at the charges. With this field, the = ε charge per unit area on the plates is Q A 0 Ex . ε 2 If we used Ex for that field, we would get 0 Ex , which is twice what we calculated from the Maxwell stress tensor, which is wrong. I wasted an hour last night worrying about this, until I re- membered the solution (Griffiths 3d section 2.5.3). Meaning of the Maxwell Tensor (4) Charges reside at the surface of a conductor. So the electric field is finite on the exterior side, and zero on the interior side. The proper electric field to use turns out to be the average of the full field and the zero field. One way to see this is to make the charge layer a finite thickness, so the electric field ramps up linearly from zero to the full value, and compute the integrated force on the distributed charge. Another way to see it is to consider a uniform charge sheet not in a conductor, which makes equal and op- ε ( ) posite electric fields of 0Q 2A on opposite sides, but with a superposed uniform electric field of the same strength, which doubles the field on one side and cancels it on the other. If we isolate a patch of the charge, the charge sheet exerts no net force on the patch by symme- try, and all the force is from the uniform external field. With either explanation, the actual net force on the ca- ε 2 pacitor plate is 0 Ex 2, in agreement with the Maxwell stress tensor calculation. The moral is, sometimes the equations are smarter than me! Meaning of Field Momentum Density Consider two positive point charges moving at equal speeds along lines that cross at right angles, but timed so they don’t actually collide: V V B F At the instant show, the horizontally moving particle makes a magnetic field that is into the screen, and the V × B force on the vertically moving particle is then hori- zontal and backwards. But the vertically moving particle cannot possibly make an equal and opposite backwards force on the other par- ticle, either electrically or magnetically! So we have a violation of Newton’s Third Law, i.e., momentum is not conserved! It will get even worse later, when we will find that the electric forces between particles in motion aren’t instan- taneously balanced! The solution to these paradoxes is that the fields bet- ween the particles must also carry momentum, and only the sum of the particle momentum and field momentum (and net Maxwell stress flux) is conserved. Angular Momentum in Statics If a wire with charge Q per length is parallel to the axis of a solenoid with magnetic field B, and we move it radi- ally from the edge to the center with velocity v, there is a Lorentz force per unit length given by Q Q QB F = v × B = (−rˆ v) × (zˆ B) = vφˆ To keep the path radial, we have to apply a torque. Rela- tive to the center of the solenoid, the torque is QBr QBr τ = r × F = v(rˆ × φˆ ) = vzˆ A torque causes a change in angular momentum dL dL dr dL τ = = = v dt dr dt dr Equating these we have QBr dL τ = vzˆ = v dr which we can solve for QBr dL = zˆ dr Angular Momentum in Statics (2) The integral of the angular momentum change from ra- dius r = R to r = 0 is 0 0 2 0 = QBr = QB = QB r L zˆ ∫ rdr zˆ ∫ rdr zˆ R R 2 R QB R2 QBR2 = zˆ 0 − = − zˆ 2 2 When we are done moving the charge to the center of the solenoid, we have applied a torque sufficient to trans- mit this much angular momentum to something. Where does the angular momentum go? Angular Momentum in Statics (3) The charged wire has a radial electric field, so E × B is non-zero. The linear momentum density stored in the fields is p = ε E × B volume 0 The angular momentum density then should be L r × S = = ε r × (E × B ) volume c2 0 The electric field around a line charge by Gauss’ Law is Q 1 E = rˆ πε 2 0r When the charge is at the center of the solenoid, the angular momentum density is L Q 1 QB = ε r × rˆ × zˆ B = −zˆ 0 πε π volume 2 0 r 2 The total field angular momentum per length is thus L QB QBR2 = −zˆ πR2 = −zˆ 2π 2 So the the integrated mechanical torque from the Lor- entz force on the charged wire agrees exactly with the angular momentum density stored in the E × B fields. Angular Momentum in Dynamics Consider two cylinders of radius a and b with charges +Q and −Q, concentric with each other and with a solenoid of with magnetic field B and radius R, a < R < b. If we turn off the solenoid in time T , we generate electric fields, inside and outside, in the cirumferential direction. By the integral form of Faraday’s Law −B Br 2πrE = −πr 2 ⇒ E = φˆ inside T inside 2T −B BR2 2πrE = −πR2 ⇒ E = φˆ outside T outside 2rT These fields generate torques on the charged cylinders for time T and impart angular momenta L = (r × QE )T Ba QBa2 L = aQ Tzˆ = zˆ inside 2T 2 BR2 QBR2 L = b(−Q) Tzˆ = − zˆ outside 2bT 2 QB L = (a 2 − R2 )zˆ total 2 Where did the angular momentum come from? Angular Momentum in Dynamics (2) If the two cylinders have length , the electric field bet- ween them is Q rˆ Q 2πr E = ⇒ E = ε πε 0 2 0 r The angular momentum density is L rˆ Q −1 QB = r × (ε E × B ) = r × ε × B = zˆ 0 0 πε π volume 2 0 r 2 If we multiply by volume we have −1 QB QB L = zˆ [π (R2 − a2 )] = (a2 − R2 )zˆ 2π 2 So again, the angular momentum in the fields equals the mechanical angular momentum. There’s something fishy, though. The electric field doesn’t actually depend on the outer cylinder radius. If we have no outer cylinder, we still get the same electric field, and the same angular momen- tum stored in the field. But angular momentum doesn’t balance if we only have the inner cylinder. Angular Momentum in Dynamics (3) When we turn off the solenoid we generate a circum- ferential electric field outside, which is what exerts the torque on the outer cylinder. That electric field is itself time-varying, so it will generate its own magnetic fields, and propagate as a wave. The wave carries angular momentum to the outer cylinder. If the outer cylinder is 10 light years away, for 10 years the angular momentum will be in the wave. It will eventually be absorbed by the outer cylinder. But it could just go on forever... But what about the case of no inner cylinder, just the out- er one. Now there is zero electric field inside the sole- noid, so there is no E × B momentum stored, so no angu- lar momentum stored. But when we turn the solenoid off, there is still an external electric field that makes the outer cylinder start to rotate. Where did the angular momentum come from this time? Angular Momentum in Dynamics (4) Well, instead of starting with the solenoid on and turning it off, start with it off and turn it on. And instead of mak- ing it a coil of wire, make it another charged cylinder. Rotating the cylinder generates a magnetic field just like a current does.
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