Question

Confirm that the energy in the$\mathit{T}{\mathit{E}}_{\mathbf{m}\mathbf{n}}$mode travels at the group velocity. [Hint: Find the time-averaged Poynting vector $\mathbf{<}\mathbf{S}\mathbf{>}$and the energy density $\mathbf{<}\mathbf{u}\mathbf{>}$(use Prob. 9.12 if you wish). Integrate over the cross-section of the waveguide to get the energy per unit time and per unit length carried by the wave, and take their ratio.]

Short answer

Answer

The velocity of energy carried by the electromagnetic wave is confirmed to be the group velocity of the wave.

Step-by-step solution

Expression for the velocity of the energy carried by the electromagnetic wave traveling in a waveguide:

Write the expression for the velocity of the energy carried by the electromagnetic wave traveling in a waveguide.

$\mathit{V}\mathbf{=}\frac{\mathbf{\int }\mathbf{<}\overrightarrow{\mathbf{S}}\mathbf{>}\mathbf{d}\overrightarrow{\mathbf{A}}}{\mathbf{\int }\mathbf{<}\overrightarrow{\mathbf{u}}\mathbf{>}\mathbf{d}\overrightarrow{\mathbf{A}}}$ …… (1)

Here, v is the velocity of electromagnetic wave, is the time-averaged pointing vector for the electromagnetic wave, is the energy per unit volume of the electromagnetic wave, and $\overrightarrow{A}$is the cross-section of the waveguide.

Write the expression for the time-averaged pointing vector for the electromagnetic wave.

…… (2)

Write the expression for the energy per unit volume of the electromagnetic wave.

…… (3)

Here, $*$represents the complex conjugate of the quantity.

Determine the time-averaged pointing vector for the electromagnetic wave:

Write the value of $\left(\overrightarrow{E}\times \overrightarrow{B}\right)$.

$\left(\overrightarrow{B}\times \overrightarrow{B}\right)\mathbf{=}\mathbf{\left(}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{y}}\mathbf{\right)}\hat{\mathbf{x}}\mathbf{-}\mathbf{\left(}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{x}}\mathbf{\right)}\hat{\mathbf{x}}\mathbf{+}\mathbf{(}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{x}}\mathbf{-}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{y}}\mathbf{)}\hat{\mathbf{z}}$

As $\mathbf{\left(}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{y}}\mathbf{\right)}$and $\left(B_z^{*}{E}_x\right)$are imaginary components, the value of localid="1658406246663" $\mathit{R}\mathit{e}\left(\overrightarrow{E}\times \overrightarrow{B}\right)$will be,

localid="1658406255214" $\mathit{R}\mathit{e}\left(\overrightarrow{E}\times \overrightarrow{B}\right)\mathbf{=}\mathbf{(}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{x}}\mathbf{-}{\mathbf{B}}_{\mathbf{z}}^{\mathbf{*}}{\mathbf{E}}_{\mathbf{y}}\mathbf{)}\hat{\mathbf{z}}$

Using equation 9.180 and 9.186, write the value of ${\mathit{B}}_{\mathbf{y}}^{\mathbf{*}}\mathbf{,}{\mathit{E}}_{\mathbf{x}}\mathbf{,}{\mathit{B}}_{\mathbf{x}}^{\mathbf{*}}$and ${\mathit{E}}_{\mathbf{y}}$.

$$\begin{gathered} {\mathit{B}}_{\mathbf{y}}^{\mathbf{*}}\mathbf{=}\frac{\mathbf{-}\mathbf{i}\mathbf{k}}{{\left({\displaystyle \frac{\omega }{c}}\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}\left(-\frac{n\pi }{b}\right){\mathit{B}}_{\mathbf{0}}\mathbf{}\mathbf{cos}\left(\frac{m\pi x}{a}\right)\mathbf{sin}\left(\frac{n\pi y}{b}\right) \\ {\mathit{E}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{-}\mathbf{i\omega }}{{\mathbf{\left(}{\displaystyle \frac{\omega }{c}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}\mathbf{\left(}\frac{\mathbf{-}\mathbf{n\pi }}{\mathbf{b}}\mathbf{\right)}{\mathit{B}}_{\mathbf{0}}\mathbf{}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{m\pi x}}{\mathbf{a}}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{n\pi y}}{\mathbf{b}}\mathbf{\right)} \\ {\mathit{B}}_{\mathbf{x}}^{\mathbf{*}}\mathbf{=}\frac{\mathbf{-}\mathbf{ik}}{{\mathbf{\left(}{\displaystyle \frac{\omega }{c}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}\mathbf{\left(}\frac{\mathbf{m\pi }}{\mathbf{a}}\mathbf{\right)}{\mathit{B}}_{\mathbf{0}}\mathbf{}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{m\pi x}}{\mathbf{a}}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{n\pi y}}{\mathbf{b}}\mathbf{\right)} \\ {\mathit{E}}_{\mathbf{y}}\mathbf{=}\frac{\mathbf{-}\mathbf{ik}}{{\mathbf{\left(}{\displaystyle \frac{\omega }{c}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}\mathbf{\left(}\frac{\mathbf{-}\mathbf{m\pi }}{\mathbf{a}}\mathbf{\right)}{\mathit{B}}_{\mathbf{0}}\mathbf{}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{m\pi x}}{\mathbf{a}}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{n\pi y}}{\mathbf{b}}\mathbf{\right)} \end{gathered}$$

Substitute all the above values in equation (2).

$$\begin{gathered} <S>\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}}\left(B_y^{*}{E}_x-B_x^{*}{E}_y\right)\hat{\mathbf{z}} \\ <S>\mathbf{=}\frac{\mathbf{\omega }\mathbf{k}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}}{{\left({\displaystyle \frac{\omega }{c}}\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}\left[\begin{array}{c}{\left(\frac{n}{b}\right)}^{2}co{s}^2\left(\frac{m\pi x}{a}\right)si{n}^2\left(\frac{n\pi y}{b}\right)+\\ {\left(\frac{m}{a}\right)}^{2}si{n}^2\left(\frac{m\pi x}{a}\right)co{s}^2\left(\frac{n\pi y}{b}\right)\hat{z}\end{array}\right] \\ \mathbf{\int }<S>\mathbf{·}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}}\frac{\mathbf{\omega }\mathbf{k}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}}{{\left({\displaystyle \frac{\omega }{c}}\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}}\mathit{a}\mathit{b}\left[{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2\right] \\ \mathbf{\int }<S>\mathbf{·}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}}\frac{\mathbf{\omega }\mathbf{k}{\mathbf{c}}^{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{a}\mathbf{b}}{\left({\displaystyle {\omega }_{mn}^2}\right)} \end{gathered}$$

Determine the energy per unit volume of the electromagnetic wave:

Using the equation9.176, write the value of $\overrightarrow{E}$and $\overrightarrow{B}$.

$$\begin{gathered} \overrightarrow{\mathbf{E}}\mathbf{=}{\overrightarrow{\mathbf{E}}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{i}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega t}\mathbf{)}} \\ {\overrightarrow{B}}^{*}\mathbf{=}{{\overrightarrow{\mathbf{B}}}^{\mathbf{*}}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{i}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega t}\mathbf{)}} \end{gathered}$$

Similarly, using equation 9.180 and 9.186, write the value of ${\mathit{B}}_{\mathbf{z}}^{\mathbf{*}}$and ${\mathit{E}}_{\mathbf{z}}$.

$$\begin{gathered} {\mathit{B}}_{\mathbf{z}}^{\mathbf{*}}\mathbf{=}{\mathit{B}}_{\mathbf{0}}\mathbf{}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{m\pi x}}{\mathbf{a}}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{n\pi y}}{\mathbf{b}}\mathbf{\right)} \\ {\mathit{E}}_{\mathbf{z}}\mathbf{=}\mathbf{0} \end{gathered}$$

Substitute the value of localid="1658406358582" $\overrightarrow{E}$and localid="1658406367803" $\overrightarrow{B}$in equation (3).

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Determine the group velocity:

Substitute $\mathbf{\int }<S>\mathbf{·}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}}\frac{\mathbf{\omega }\mathbf{k}{\mathbf{c}}^{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}}{\left({\omega }_{mn}^2\right)}$and $\mathbf{\int }<u>\mathbf{·}\mathit{d}\mathit{a}\mathbf{=}\frac{{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{a}\mathbf{b}{\mathbf{\omega }}^{\mathbf{2}}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{\omega }}_{\mathbf{m}\mathbf{n}}^{\mathbf{2}}}$in equation (1).

$$\begin{gathered} \mathit{V}\frac{{\displaystyle \frac{\mathbf{1}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}}\frac{\mathbf{\omega }\mathbf{k}{\mathbf{c}}^{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{a}\mathbf{b}}{\left({\omega }_{mn}^3\right)}}}{{\displaystyle \frac{{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{a}\mathbf{b}{\mathbf{\omega }}^{\mathbf{2}}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{\omega }}_{\mathbf{m}\mathbf{n}}^{\mathbf{2}}}}} \\ \mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}}\frac{\mathbf{\omega }\mathbf{k}{\mathbf{c}}^{\mathbf{2}}{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{a}\mathbf{b}}{\left({\omega }_{mn}^2\right)}\mathbf{\times }\frac{\mathbf{8}{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{\omega }}_{\mathbf{m}\mathbf{n}}^{\mathbf{2}}}{{\mathbf{B}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{a}\mathbf{b}\mathbf{\omega }} \\ \mathit{V}\mathbf{=}\frac{\mathbf{k}{\mathbf{c}}^{\mathbf{2}}}{\mathbf{\omega }} \\ \mathit{V}\mathbf{=}\frac{\mathbf{c}}{\mathbf{\omega }}\sqrt{{\mathbf{\omega }}^{\mathbf{2}}\mathbf{-}{\mathbf{\omega }}_{\mathbf{m}\mathbf{n}}^{\mathbf{2}}} \end{gathered}$$

Therefore, the velocity of energy carried by the electromagnetic wave is confirmed to be the group velocity of the wave.