Question

In the complex notation there is a clever device for finding the time average of a product. Suppose $\mathit{f}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{k}\mathbf{\times }\mathit{r}\mathbf{-}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{\delta }}_{\mathbf{a}}\mathbf{)}$and $g\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathit{B}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{k}\mathbf{\times }\mathit{r}\mathbf{-}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{\delta }}_{\mathbf{b}}\mathbf{)}$. Show that $<fg>\mathbf{=}\left(1/2\right)\mathit{R}\mathit{e}\left(\widetilde{fg}\right)$, where the star denotes complex conjugation. [Note that this only works if the two waves have the same k and$\mathit{\omega }$, but they need not have the same amplitude or phase.] For example,

$<u>\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathit{R}\mathit{e}\left({\epsilon }_0\tilde{E}\times \tilde{E}+\frac{1}{{\mu }_0}\tilde{B}\times \tilde{B}\right)\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}<S>\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}}\mathit{R}\mathit{e}\mathbf{(}\stackrel{\mathbf{~}}{\mathbf{E}}\mathbf{\times }\stackrel{\mathbf{~}}{\mathbf{B}\mathbf{)}\mathbf{.}}$ and .

Short answer

It is proved that .

Step-by-step solution

Expression for the f(r,t) and g(r,t):

Write the expression for f (r , t).

$\mathit{f}\mathbf{}\mathbf{(}\mathbf{}\mathit{r}\mathbf{}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{k}\mathbf{.}\mathit{r}\mathbf{-}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{\delta }}_{\mathbf{a}}\mathbf{)}$ …. (1)

Write the expression for g( r , t).

$\mathit{g}\mathbf{}\mathbf{(}\mathbf{}\mathit{r}\mathbf{}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{B}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{k}\mathbf{.}\mathit{r}\mathbf{-}\mathit{\omega }\mathit{t}\mathbf{+}{\mathit{\delta }}_{\mathbf{b}}\mathbf{)}$ …. (1)

Determine the <fg>:

Find the as follows.

Determine (12)Re(fg)^:

Write the equation in the complex notation.

$$\begin{gathered} \tilde{f}=\tilde{A}{e}^{i(k.r-\omega t)} \\ \tilde{g}=\tilde{B}{e}^{-i(k.r-\omega t)} \end{gathered}$$

Where, $\tilde{A}=a{e}^{i{\delta }_a}and\tilde{B}=B{e}^{-i{\delta }_b}$.

Determine$\frac{1}{2}\left(\tilde{f}\widetilde{g^{*}}\right)$ as follows.

$$\begin{gathered} \frac{1}{2}\left(\tilde{f}\widetilde{g^{*}}\right)=\left(\tilde{A}{e}^{i(k.r-\omega t)}\right)\left(\tilde{B}{e}^{i(k.r-\omega t)}\right) \\ =\frac{1}{2}\tilde{A}\tilde{B} \\ =\frac{1}{2}AB{e}^{i({\delta }_a-{\delta }_b)} \\ =\frac{1}{2}AB\left(\cos \right({\delta }_a-{\delta }_b)+i\sin ({\delta }_a-{\delta }_b\left)\right) \end{gathered}$$

Consider the real term.

Therefore, it is proved that .