- #1

MisterX

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## Homework Statement

Condensed/simplified problem statement

[itex]\vec{E} = f_{y}(x-ct)\hat{y} + f_{z}(x-ct)\hat{z} \\

\vec{B} = g_{y}(x-ct)\hat{y} + g_{z}(x-ct)\hat{z} \\[/itex]

All the f and g functions go to zero as their parameters go to ±∞.

Show that g

_{y}= f

_{z}and g

_{z}= -f

_{y}

## Homework Equations

[itex]\nabla \cdot \vec{E} = 0 \\

\nabla \cdot \vec{B} = 0 [/itex]

[itex]\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} [/itex]

[itex]\nabla \times \vec{B} = \frac{1}{c^{2}}\frac{\partial \vec{E}}{\partial t}

[/itex]

## The Attempt at a Solution

[itex]\nabla \times \vec{E} = -\frac{\partial f_{z}(x-ct)}{\partial x}\hat{y} + \frac{\partial f_{y}(x-ct)}{\partial x}\hat{z} = -\frac{\partial \vec{B}}{\partial t} = \\-(g^{'}_{y}(x-ct)(-c)\hat{y} + g^{'}_{z}(x-ct)(-c)\hat{z}) = c(g^{'}_{y}(x-ct)\hat{y} + g^{'}_{z}(x-ct)\hat{z}) [/itex]

The -c factors come from the chain rule when differentiating B with respect to t. So then I have the following.

[itex]-\frac{\partial f_{z}(x-ct)}{\partial x} = -f^{'}_{z}(x-ct) = cg^{'}_{y}(x-ct) \\

\frac{\partial f_{y}(x-ct)}{\partial x} = f^{'}_{y}(x-ct) = cg^{'}_{z}(x-ct)[/itex]

The problem is to relate the functions directly and not by their derivatives. We are given that the functions all go to zero as the parameter goes to +/- infinity. So maybe it would be okay to just integrate without concern for a constant added on. But supposedly the c factor isn't there. The answer is supposed to have[itex]g_{y} = -f_{z}[/itex]. How do I get there?

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