Wave energy and momentum

Electromagnetic waves transport energy through space.  In free space this energy is transported by the wave with speed c.  The magnitude of the energy flux S is the amount of energy that crosses a unit area perpendicular to the direction of propagation of the wave per unit time.  It is given by the energy density times the speed c.
The energy per unit volume in the electric field is uE = ½ε0E2,
the energy per unit volume in the magnetic field is uB = B2/(2μ0), and B = E/c.
We also have μ0ε0 = 1/c2.  We can therefore write

S = c(uE + uB) = ½(cε0E2 + E2/(cμ0)) =   ½(cε0E2 + cε0E2)  = cε0E2 = EB/μ0.

The units of the energy flux S are J/(m2s) = W/m2.

Note:
The energy transported by an electromagnetic wave is proportional to the square of the amplitude, Emax2, of the wave.
For all waves, the energy transported is proportional to the square of the amplitude.

The energy flux vector is S.  It has magnitude S and points in the direction of propagation, i.e. in the direction of E × B.
It is called the Poynting vector and can also be written as

S
= (E × B)/μ0.

For a sinusoidal plane wave both E and B vary as a function of time as sin(kx - ωt + φ), and S varies as sin2(kx - ωt + φ).
To find the magnitude of the average energy flux or intensity (I) of the wave we average sin2(kx - ωt + φ) over each period, which yields ½.  Then

I = <S> = ½(cε0Emax2).

Problem:

How much electromagnetic energy per cubic meter is contained in sunlight if the intensity of the sunlight at the earth's surface under a fairly clear sky is 1000 W/m2?

Solution:
The intensity <S> is the energy per unit area per unit time or the power per unit area.  Multiplying <S> by the time ∆t yields the energy that falls on an unit area in the time ∆t.  This energy is contained in a volume of unit area and height h, where h = c∆t.  Light travels with speed c, so it travels a distance of h = c∆t in a time interval ∆t.  If we set h = 1 m, then ∆t = (1/3)*10-8 s.
So Energy = <S>*(1/3)*10-8 s *(1 m2) is the energy that is contained in a cubic meter.
Energy = (1000 W/m2)*(1/3)*10-8 s*(1 m2) = 3.33*10-6 J.

Problem:

If the intensity of the sunlight at the earth's surface is 1000 W/m2, what is the rate P at which the sun radiates light energy in W?

Solution:
The average Earth-Sun distance is R =1.496*108 km.  At a distance r from the Sun, the light energy emitted by the Sun has spread over an area of 4πr2, the surface area of a sphere with radius r, centered at the Sun.
So at the earth distance R, the intensity is I = P/(4πR2), where P is the power of the sun, (the rate at which it emits light energy.)
We have P = I*4πR2 = (1000 W/m2)*4π(1.496*1011 m)2 = 2.81*1026 W.
(The total energy output (not just visible light) of the Sun is 3.86*1026 W.)

Problem:

Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.5 mW.
(a)  If such a laser beam is projected onto a circular spot 1 mm in radius, what is its intensity?
(b)  Find the peak electric field strength.
(c)  Find the peak magnetic field strength.

Solution:
(a)  Intensity I = energy/(area*time) = power/area = 0.5*10-3 W/(π*10-6 m2) = 1.59*103 W/m2.
(b)  I = ½(cε0Emax2). 
Emax2 = 2I/(cε0) = (2*1.59*103 W/m2)/(3*108 m/s *8.854*10-12 C/(V m) =1 .2*106 N2/C2.
Emax = 1095 N/C.
(c)  Bmax = Emax/c = (1095 N/C)/(3*108 m/s) = 3.65*10-6 T.


Electromagnetic waves transport energy.  EM wave also transport momentum.  The momentum flux is <S>/c.  It is the amount of momentum that crosses a unit area perpendicular to the direction of propagation of the wave per unit time.  If an electromagnetic wave is absorbed, momentum conservation requires that the object acquires momentum.  The object is therefore acted on by a force, F = ∆p /∆t.  The radiation exerts radiation pressure P = F/A on the object.  If the radiation is reflected instead of absorbed, then its momentum changes direction.  The radiation pressure on an object that reflects the radiation is therefore twice the radiation pressure on an object that absorbs the radiation.

Problem:

A plane electromagnetic wave of intensity 6 W/m2 strikes a small pocket mirror of area 40 cm2 held perpendicular to the approaching wave.
(a)  What momentum does the wave transfer to the mirror each second?
(b)  Find the force that the wave exerts on the mirror.

Solution:
(a)  The wave is reflected by the mirror. The momentum transferred to the mirror per unit area per second is twice the momentum of the light striking the mirror per unit area per second.
∆p/∆t = 2(1/c)<S>A.
<S>A = (6 W/m2)(4*10-3 m2) = 0.024 W.
∆p/∆t = 2*0.024 W/(3*108 m/s) = 1.6*10-10 kgm/s2.
Each second the wave transfers 1.6*10-10 kgm/s of momentum to the mirror.
(b)  F = ∆p/∆t = 1.6*10-10 N.