5.3 Transverse waves

Transverse waves

A plane wave solution to the electromagnetic wave equation for the $\vec{E}$ field is $$\vec{E}(\vec{r},t)=\overrightarrow{E_0}{e}^{i\left(\vec{k}\cdot \vec{r}-\omega t\right)}$$ In vacuum with no currents present we know that: $\vec{\nabla }\cdot \vec{E}=0$ . Recall that earlier we showed $$\vec{\nabla }\cdot \vec{E}=i\vec{k}\cdot \overrightarrow{E_0}{e}^{i\left(\vec{k}\cdot \vec{r}-\omega t\right)}$$ So $$\vec{\nabla }\cdot \vec{E}=i\vec{k}\cdot \overrightarrow{E_0}{e}^{i\left(\vec{k}\cdot \vec{r}-\omega t\right)}=0$$ implies that the $\vec{E}$ associated with our plane wave is perpendicular to its direction of motion.

Likewise $\vec{\nabla }\cdot \vec{B}=0$ implies that the $\vec{B}$ field is also perpendicular to the direction of motion Lets pick a specific simple case: $$\vec{E}=\widehat{}{E}_y(x,t)$$ Then Faraday's law $$\vec{\nabla }\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$ tells us that (since $\frac{\partial E_y}{\partial z}=0$ ) $$\frac{\partial E_y}{\partial x}\widehat{k}=-\frac{\partial B_z}{\partial t}\widehat{k}$$ That is the $\vec{B}$ field is at Right angles to the $\vec{E}$ field.Also $${\displaystyle \begin{array}{c}{B}_z=-\int \frac{\partial E_y}{\partial x}dt\\ =-\int \frac{\partial }{\partial x}{E}_0{e}^{i\left(kx-\omega t\right)}dt\\ =-\int \frac{\partial }{\partial x}{E}_0{e}^{ikx}{e}^{-i\omega t}dt\\ =-\frac{\partial }{\partial x}{E}_0{e}^{ikx}\int e^{-i\omega t}dt\\ =-ik{E}_0{e}^{ikx}\int e^{-i\omega t}dt\\ =-ik{E}_0{e}^{ikx}\frac{e^{-i\omega t}}{-i\omega }\\ =\frac{1}{c}{E}_0{e}^{ikx}{e}^{-i\omega t}\\ =\frac{1}{c}{E}_y\\ \end{array}}$$ I leave as an exercise showing $\frac{k}{\omega }=\frac{1}{c}$

A movie demonstrating a plane wave can be seen at

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An applet can be viewed at

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