Page 357 - Intro to Tensor Calculus

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I 25. In a vacuum show that E and B satisfy
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1 ∂ E 1 ∂ B
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∇ E = 2 2 ∇ B = 2 2 ∇· E =0 ∇B =0
c ∂t c ∂t
I 26.
(a) Show that the wave equations in problem 25 have solutions in the form of waves traveling in the
x- direction given by

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E = E(x, t)= E 0 e i(kx±ωt) and B = B(x, t)= B 0 e i(kx±ωt)
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where E 0 and B 0 are constants. Note that wave functions of the form u = Ae i(kx±ωt) are called plane
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harmonic waves. Sometimes they are called monochromatic waves. Here i = −1 is an imaginary unit.
Euler’s identity shows that the real and imaginary parts of these type wave functions have the form
A cos(kx ± ωt) and A sin(kx ± ωt).
These represent plane waves. The constant A is the amplitude of the wave , ω is the angular frequency,
and k/2π is called the wave number. The motion is a simple harmonic motion both in time and space.
That is, at a fixed point x the motion is simple harmonic in time and at a fixed time t, the motion is
harmonic in space. By examining each term in the sine and cosine terms we find that x has dimensions of
length, k has dimension of reciprocal length, t has dimensions of time and ω has dimensions of reciprocal
time or angular velocity. The quantity c = ω/k is the wave velocity. The value λ =2π/k has dimension
of length and is called the wavelength and 1/λ is called the wave number. The wave number represents
the number of waves per unit of distance along the x-axis. The period of the wave is T = λ/c =2π/ω
and the frequency is f =1/T. The frequency represents the number of waves which pass a fixed point
in a unit of time.
(b) Show that ω =2πf
(c) Show that c = fλ
(d) Is the wave motion u = sin(kx − ωt) + sin(kx + ωt) a traveling wave? Explain.
2
1 ∂ φ
2
(e) Show that in general the wave equation ∇ φ = have solutions in the form of waves traveling in
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c ∂t 2
either the +x or −x direction given by
φ = φ(x, t)= f(x + ct)+ g(x − ct)
where f and g are arbitrary twice differentiable functions.
(f) Assume a plane electromagnetic wave is moving in the +x direction. Show that the electric field is in
the xy−plane and the magnetic field is in the xz−plane.
Hint: Assume solutions E x = g 1 (x − ct), E y = g 2 (x − ct),E z = g 3 (x − ct),B x = g 4 (x − ct),
B y = g 5(x − ct),B z = g 6 (x − ct)where g i ,i =1, ..., 6 are arbitrary functions. Then show that E x
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does not satisfy ∇· E = 0 which implies g 1 must be independent of x and so not a wave function. Do
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the same for the components of B. Since both ∇· E = ∇· B =0 then E x = B x = 0. Such waves
are called transverse waves because the electric and magnetic fields are perpendicular to the direction
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of propagation. Faraday’s law implies that the E and B waves must be in phase and be mutually
perpendicular to each other.