W9_parallel_resonance.eps

356 Week 10: Maxwell’s Equations and Light

Maxwell Displacement Current – this is where Maxwell’s contribution shines!

∮

B~·dℓ = μ 0 ǫ 0 d

dt

∫

∆A

E~·ˆndA

By∆y+ 0·∆z−By(z+ ∆z)∆y− 0 ·∆z = μ 0 ǫ 0

d

dt

(Ex∆A)

−(By(z+ ∆z)−By(z)) ∆y = μ 0 ǫ 0

dEx

dt

∆y∆z

(By(z+ ∆z)−By(z))

∆z = −μ^0 ǫ^0

dEx

dt

dBy

dz

= −μ 0 ǫ 0

dEx

dt

where we have taken the limit ∆z→0 as before in the last step^100.

Since we’re going to use these two results alot, let’s write them down right next to each

other:

dEx

dz

= −

dBy

dt

(848)

dBy

dz

= −μ 0 ǫ 0 dEx

dt

(849)

Although they don’tlookmuch like it, these are both still Faraday’s Law and Ampere’s

Law (with the MDC) respectively, although expressed only for two particular components

of the electric and magnetic field.

Well, we could have had (say) ay-directed electric dipole instead, or (since our coordinate

system was arbitrary) we could just rotate it byπ/2 around thezaxis to makeExinto

EyandByinto−Bxin the new coordinate system (imagine lifting they-axisupand

pushx-back into the page as you mentally rotate figure 141). In that case one expects

to get:

dEy

dz

= dBx

dt

(850)

dBx

dz

= μ 0 ǫ 0

dEy

dt

(851)

from an identical argument to the one above, something you can verify by completely

recapitulating the derivation above as part of your homework^101.

This is all very well, but so far it is still not spectacular. To make it spectacular, we

(say) differentiate the first of these equations with respect toz:

d

dz

dEx

dz

=

d^2 Ex

dz^2

=−

d

dz

dBy

dt

=−

d

dt

dBy

dz

(852)

(^100) Once again, this should be∂B∂zy=−μ 0 ǫ 0 ∂E∂tx, but in this one dimensional, non-relativistic treatment it doesn’t
matter.
(^101) Sure, sure, they should all be partials. In fact, you are basically deriving:
∇~×E~ = −∂B~
∂t
∇~×B~ = μ 0 ǫ 0 ∂E~
∂t,
the grown-up way of writing the source free Faraday’s and Ampere’s Laws in terms of thecurl, a component pair at
a time. You can actually get all six terms in these two equations from our one original result by mentally rotating
the arbitrary right-handed coordinate system into all six indepedent orientations. Or you can use Stokes Theorem,
which we basically just derived. Since advanced students derived the partial differential form for Gauss’s Law in the
second week, we have now derived the partial differential form for the whole set of Maxwell’s Equations, at least once
the source terms are put back in...