Electromagnetic waves

Electromagnetic (EM) waves are produced by accelerating charges.  They are changing electric and magnetic fields, carrying energy through space.  EM waves require no medium, they can travel through empty space.  Sinusoidal plane waves are one type of electromagnetic waves.  Not all EM waves are sinusoidal plane waves, but all electromagnetic waves can be viewed as a linear superposition of sinusoidal plane waves traveling in arbitrary directions.  A plane EM wave traveling in the x-direction is of the form

E(x,t) = Emaxcos(kx-wt+f),

B(x,t) = Bmaxcos(kx-wt+f).

E is the electric field vector and B is the magnetic field vector of the EM wave.  For electromagnetic waves E and B are always perpendicular to each other, and perpendicular to the direction of propagation.  The direction of propagation is the direction of E´B.

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If, for a wave traveling in the x-direction E = Ej, then B = Bk and j´k = iElectromagnetic waves are transverse waves.

The wave vector k points into the direction of propagation, and its magnitude k = 2p/l, where l is the wavelength of the wave.  The frequency f of the wave is f = w/2pw is the angular frequency.  The speed of any sinusoidal wave is the product of its wavelength and frequency.

v = lf.

The speed of any electromagnetic waves in free space is the speed of light c = 3´108m/s.  Electromagnetic waves can have any wavelength l or frequency f as long as lf = c.

When electromagnetic waves travel through a medium, the speed of the waves in the medium is v = c/n, where n is the index of refraction of the medium.  When an EM wave travels from one medium with index of refraction n1 into another medium with a different index of refraction n2, then its frequency remains the same, but its speed and wavelength change.  For air n is nearly equal to 1.

The electromagnetic spectrum

Electromagnetic waves are categorized according to their frequency f or, equivalently, according to their wavelength l = c/f.  Visible light has a wavelength range from ~400nm to ~700nm.  Violet light has a wavelength of ~400nm, and a frequency of ~7.5´1014 Hz.  Red light has a wavelength of ~700nm, and a frequency of ~4.3´1014 Hz.

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Visible light makes up just a small part of the full electromagnetic spectrum.  Electromagnetic waves with shorter wavelengths and higher frequencies include ultraviolet light, X-rays, and gamma rays.  Electromagnetic waves with longer wavelengths and lower frequencies include infrared light, microwaves, and radio and television waves.

Type of Radiation

Frequency Range (Hz)

Wavelength Range

gamma-rays 1020 - 1024 <10-12m
x-rays 1017 - 1020 1nm - 1pm
ultraviolet 1015 - 1017 400nm - 1nm
visible 4-7.5´1014 750nm - 400nm
near-infrared 1´1014 - 4´1014 2.5mm - 750nm
infrared 1013 - 1014 25mm - 2.5mm
microwaves 3´1011 - 1013 1mm - 25mm
radio waves <3´1011 >1mm

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Electromagnetic waves

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

S = EB/(m0) = E2/(m0c),

since for electromagnetic waves B = E/c.  The units of S are J/(m2s).  m0 is a constant called the permeability of free space, m0 = 410-7 N/A2.

The Poynting vector is the energy flux vector.  It is named after John Henry Poynting.   Its direction is the direction of propagation of the wave, i.e. the direction in which the energy is transported.

S = (1/m0)E´B.

Energy per unit area per unit time is power per unit area.  S represents the power per unit area in an electromagnetic wave.   If an electromagnetic wave falls onto an area A where it is absorbed, then the power delivered to that area is P= S×A.

Electromagnetic waves transport energy.  EM wave also transport momentum.  The momentum flux is S/c.  S/c 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 falls onto an area A where it is absorbed, the momentum delivered to that area in a direction perpendicular to the area per unit time is dp^/dt = (1/c)S×A.

The momentum of the object absorbing the radiation therefore changes.  The rate of change is dp^/dt = (1/c)SA^, where A^ is the cross-sectional area of the object perpendicular to the direction of propagation of the electromagnetic wave.  The momentum of an object changes if a force is acting on it.

F^ = dp^/dt = (1/c)SA^

is the force exerted by the radiation on the object that is absorbing the radiation.  Dividing both sides of this equation by A^, we find the radiation pressure (force per unit area) on an area perpendicular to the flux, P = (1/c)S.   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.

Problems:

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 1000W/m2?
Solution:
The intensity S is the energy per unit area per unit time or the power per unit area. Multiply S by the time Dt yield the energy that falls on an unit area in the time Dt.  This energy is contained in a cylindrical volume of unit area and height h, where h = cDt.  Light travels with speed c, so it travels a distance of h = cDt in a time interval Dt.  If we set h = 1m, then  Dt = (1/3)´10-8s.  So E = (S/(3´108s))(1m2) is the energy that is contained in a cubic meter.  E = ((1000W/m2)/(3´108s))(1m2) = 3.33´10-6J.
A plane electromagnetic wave of intensity 6W/m2 strikes a small pocket mirror of area 40cm2 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.
dp/dt = 2(1/c)S×A.
S×A = (6W/m2)(4´10-3m2) = 0.024W.
dp/dt = 2´0.024W/(3´108m/s) = 1.6´ 10-10kgm/s2.
Each second the wave transfers 1.6´ 10-10kgm/s of momentum to the mirror.
(b)  F= dp/dt = 1.6´10-10N.
The eye is most sensitive to light having a wavelength of 5.5´10-7m., which is in the green-yellow region of the electromagnetic spectrum.  what is the frequency of this light?
Solution:
f = v/l = (3´108m/s)/(5.5´10-7m) = 5.455´1014Hz.

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