W9_parallel_resonance.eps

Week 10: Maxwell’s Equations and Light 363

direction of the field as one expects from the relationJ~=σE~ (a form of Ohm’s

Law, recall, from our discussion of conduction and resistance). Themagnetic fieldalso

penetrates a short distance into the surface and exerts a forceon this surface current.

As you can see from the figure, this force is expected to bein the direction of the

waveand will be spread out on the entire conducting surface.

This simple picture demonstrates that just as the electromagneticwave carriesenergy

(per unit time), it carrieslinear momentum(per unit time) and exerts a force on any

conducting surface it collides with. From our previous discussion of dielectrics, which

alsodevelop a (bound) surface charge density that reduces the electric field, we expect

a dielectric surface to also have a (much weaker) surface currentparallel to the electric

field and to still experience a force when impacted by an electromagnetic wave in direct

proportion to the energy absorbed by the surface per unit time.

Indeed, the transfer of momentum to the surface follows the same general rules we

learned in the first half of this course when discussing momentum transfer by things like

basketballs hitting a floor and bouncing off versus baseballs being caught. If any surface

absorbs the energy transmitted by radiation, it also absorbs the momentum transmitted

by the radiation (like a baseball beingcaughtby an ice skater). If the surface reflects the

energy of the radiation, it picks uptwicethe momentum transmitted by the radiation

(less a small amount needed to balance energy and momentum simultaneously), like a

baseball caught by an ice skater who thenthrows it back(almost) as fast as it was moving

when it was caught.

We will idealize these two rules and assume that absorption transfers exactly the momen-

tum of wave in the direction of the Poynting vector, and that reflectio of a wave transfers

twice thecomponentof the momentum of the wave perpendicular to the surface.

The remaining question is, how much momentum does a wave carry, and how can we

compute the force exerted by the wave on any given surface? Theanswer to these two

questions – well beyond the scope of this course toderive– is that themomentum density

of an electromagnetic waves is:

~g=

1

c^2

S~ (893)

The magnitude of the momentum ∆ptransferred to a surface areaAthat absorbs an

electromagnetic wave and that is normal to the wave direction, perunit time, in time

∆t, is then all of the momentum in the box of volumeAc∆tas usual (we’ve used this

argument many times before) or:

∆p=

1

c^2

|S~|Ac∆t (894)

If we divide bothAand ∆tto the left, we get the force per unit area exerted on the

surface:

Pr=

1

A

∆p

∆t

=

|S~|

c

(895)

This is called theradiation pressureexerted on the surface by the electromagnetic

field, assuming normal incidence and complete absorption of the wave. One then finds

the total force the usual way:

F~=AS~

c

(896)

in the direction of the wave (the Poynting vector direction itself).