W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 361

No energy in that region of space before, yes energy after, therefore energy is carried by

the field from the source to the region of space. Simple!

Naturally, we’d like to be able to compute howmuchenergy is being carried along by the

field. To find out, we resort to what should now be a very familiar argument. In a time

∆t, all of the energy in a box of lengthc∆twill be carried through the cross-sectional

areaAof it’s end. The amount of energy is:

∆U=

1

2

(

ǫ 0 E^2 +

1

μ 0

B^2

)

c∆tA (886)

The power per unit area per unit time that is carried throughAis a quantity we define

to be theintensityof the light wave:

I=

P

A

=

∆U

A∆t

=

1

2

(

ǫ 0 E^2 +

1

μ 0

B^2

)

c (887)

Let’s do a bit of algebra. For the moment, let’s once again concentrate on our familiar

harmonic pairEx(z, t) andBy(z, t). ThenEx^2 =Ex(cBy) andB^2 y=By(Ex/c), so if we

multiply this out we get:

I=

1

2

(

ǫ 0 ExByc^2 +

1

μ 0

ExBy

)

(888)

Butc^2 =ǫ 01 μ 0 so that:

I=

1

μ 0

ExBy (889)

Note as well, thatagainby a hopefully familiar argument, we derived the above for the

“special” case of a surface ∆Athat is perpendicular to the direction of propagation. By

now we should easily be able to see that if we tip this surface into ∆A′at some angleθ, we

will increase its area by 1/cos(θ) and will need to compensate by multiplying it by cos(θ).

This makes the power through the surface not justP=I∆AbutP =I∆A′cos(θ) or

more generally, if we define the “vector intensity” in the direction by:

S~=^1

μ 0

E~×B~ (890)

• a quantity eponymously named thePoynting vector(yes, itpoyntsin the direction
  that the wave propagates, har har) then the power through anysurfaceSisflux of the
  Poynting vectorthrough that surface:

P=

∫

S

S~·nˆdA (891)

Again, I’m hoping that I don’t have to do much more than this – sketchout one more

example of how the flow of a vector fieldthrougha surface is conserved and correctly

accounted for by the flux integral.

The intensity is thus themagnitude of the Poynting vector:

I=|S~| (892)

and is still a very useful quantity in its own right.

The Poynting vector is actually pretty much magical. For example, it doesn’t just work

with dynamic electromagnetic waves – it works forstatic fieldsas well. In fact, for your

homework you will prove that the flux of the Poynting vector into a resistor, and inductor

and a capacitor all precisely equalV I–I^2 R,LIdI/dtandQI/Crespectively. This seems