Zangwill derives electromagnetic planes waves by:

  1. Showing that wave-like solutions to the Maxwell equations exist.
  2. Supposing that plane wave solutions to the wave equation depend on one spatial variable:

$$\vec{w} = \vec{w}(z,t) \text{ (where w is a plane wave solution)}.$$

From (1.) and (2.), it follows that $\vec{E}(z,t)$ and its associated $\vec{B}(z,t)$ constitute components of a valid electromagnetic plane wave. Zangwill starts with $\vec{E}(z,t)$ and derives $\vec{B}(z,t)$ via Faraday's law (in free space).

It is shown via Gauss's law and Ampere's law that the $\hat{z}$ component of $\vec{E}(z,t)$ is at most a constant. For simplicity, I think, we then "safely" set it to zero.

Because $\vec{E}(z,t)$ is also be a valid solution to the wave equation, it takes on the form:

$$\vec{E}_+(z,t) = \vec{f}_\perp(z+ct) \text{ and } \vec{E}_-(z,t) = \vec{g}_\perp(z+ct).$$

Because $E_z = 0$ (the $\hat{z}$ component of the electric field) and using Faraday's law in free space:

$$\frac{\partial}{\partial t} \vec{B}_+(z,t) = -\hat{z} \text{ }\times \text{ } \frac{\partial}{\partial z}\vec{f}_\perp(z+ct) = -\hat{z} \text{ }\times \text{ } \frac{1}{c}\frac{\partial}{\partial t}\vec{f}_\perp(z+ct).$$

I've ommitted the (-) component. My questions are:

  1. How do you get to the right hand side of the above equation?
  2. Zangwill states:

"Integration of the first and last terms in (16.26) gives the magnetic field up to a function of z alone. The latter may be dropped because we are only interested in time-varying fields. Therefore,

$$ c\vec{B}_+(z,t) = -\hat{z} \text{ } \times \vec{f}_\perp(z+ct)." $$

What does he mean by this quote? Why do we not just equate the left most and right most term and show that the integrands are equal, resulting in the same result.


1 Answer 1


The answer to 1) is chain rule.


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