Chapter 9: Q12P (page 400)

In the complex notation there is a clever device for finding the time average of a product. Suppose $f (r, t) = A c o s (k \times r - ω t + δ_{a})$ and $g (r, t) = B c o s (k \times r - ω t + δ_{b})$ . Show that $< f g > = (1 / 2) R e (\tilde{f g})$ , where the star denotes complex conjugation. [Note that this only works if the two waves have the same k and $ω$ , but they need not have the same amplitude or phase.] For example,
$< u > = \frac{1}{4} R e (ε_{0} \tilde{E} \times \tilde{E} + \frac{1}{μ_{0}} \tilde{B} \times \tilde{B}) a n d < S > = \frac{1}{2 μ_{0}} R e (\tilde{E} \times \tilde{B) .}$ and .

Short Answer

Expert verified

It is proved that $<f g> = (1 / 2) R e (\tilde{f} \tilde{g})$ .

Step by step solution

Expression for the f(r,t) and g(r,t):

Write the expression for f (r , t).

$f (r, t) = A c o s (k . r - ω t + δ_{a})$ …. (1)

Write the expression for g( r , t).

$g (r, t) = B c o s (k . r - ω t + δ_{b})$ …. (1)

Determine the <fg>:

Find the $<f g>$ as follows.

$<f g> = \frac{1}{T} \int_{g}^{T} A \cos (k . r - ω t + δ_{a}) . B \cos (k . r - ω t + δ_{b}) d t = \frac{A B}{T} \int_{g}^{T} [\cos (2 k . r - ω t + δ_{a} + δ_{b} + \cos (δ_{a} - δ_{b})] d t = \frac{A B}{T} \cos (δ_{a} - δ_{b}) T = \frac{1}{2} A B \cos (δ_{a} - δ_{b})$

Determine (12)Re(fg)^:

Write the equation in the complex notation.

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

Where, $\tilde{A} = a e^{i δ_{a}} a n d \tilde{B} = B e^{- i δ_{b}}$ .

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

$\frac{1}{2} (\tilde{f} \tilde{g^{*}}) = (\tilde{A} e^{i (k . r - ω t)}) (\tilde{B} e^{i (k . r - ω t)}) = \frac{1}{2} \tilde{A} \tilde{B} = \frac{1}{2} A B e^{i (δ_{a} - δ_{b})} = \frac{1}{2} A B (\cos (δ_{a} - δ_{b}) + i \sin (δ_{a} - δ_{b}))$