Derive electromagnetic wave equation in free space
Answers
Explanation:
Derivation of the Electromagnetic Wave Equation in a Vacuum from Maxwell's Law's
In free space q = 0 and I = 0. That fact greatly simplifies Maxwell's equations. We begin with Faraday's law:
ò E×ds = -dΦB/dt.
We assume we are dealing with a wave moving in the +x direction at a point where E is in the
positive y-direction and B is in the positive z-direction. To apply Faraday's law we may
evaluate the line integral around the infinitesimal rectangle pictured at the right. Going around
the rectangle in the direction shown, since ds is perpendicular to E along the top and bottom
of the rectangle, the result is
ò E×ds = E(x+dx, t)l- E(x, t)l » (¶E/¶x) dx l.
Evaluating the right-hand side of Faraday's law gives us, since ΦB = Bl dx,
-dΦB/dt = l dx (¶B/¶t).
Finally, plugging the last two results into Faraday's law yields
¶E/¶x = -¶B/¶t.
Next we use the Ampere-Maxwell law, evaluated in free space where I = 0,
ò B×ds = ε0μ0 (dΦE/dt).
Again we assume we are dealing with a wave moving in the +x direction, still at a point where
E is in the positive y-direction and B is in the positive z-direction. To apply the Ampere-
Maxwell law we may evaluate the line integral around the infinitesimal rectangle pictured at the
right. Going around the rectangle in the direction shown, since ds is perpendicular to B along the
short sides of the rectangle, the result is
ò B×ds = B(x, t)l- B(x+dx, t)l » -(¶B/¶x) dx l.
Evaluating the right-hand side of the Ampere-Maxwell law gives us, since ΦE= El dx,
dΦE/dt = l dx (¶E/¶t).
Finally, plugging the last two results into the Ampere-Maxwell law yields
¶B/¶x = - μ0ε0 ¶E/¶t.
Now, taking the partial derivative with respect to x of the first highlighted equation, then plugging in the second highlighted equation,
¶2E/¶x2 = -¶(¶B/¶t)/¶x = -¶(¶B/¶x)/¶t = -¶(- μ0ε0 ¶E/¶t)/¶t,
or
¶2E/¶x2 = μ0ε0 ¶2E/¶t2.
Similarly, taking the partial derivative with respect to x of the second highlighted equation, then plugging in the first highlighted equation, and simplifying,
¶2B/¶x2 = μ0ε0 ¶2B/¶t2.
If we compare these two equations with the generalized wave equation, ¶2y/¶t2 = (1/v2) ¶2y/¶x2, we see that both the equation for E and that for B have exactly the form of the wave equation with the speed of the wave being v = 1/Ö(μ0ε0). We call this electormagnetic wave speed c, i.e.,
c = 1/Ö(μ0ε0).
We check by calculation that
c = 1/Ö[(4π ´ 10-7 T×m/A)(8.854187817 ´ 10-12 C2/N×m2) = 2.99792458 ´ 108 m/s!
The simplest solutions to the two aqua-highlighted differential equations above are
E = Emax cos (kx-ωt),
and
B = Bmax cos (kx-ωt).
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