Question 13.9.Q2: A microwave travelling in a uniform EM waveguide can be char......

(a) As shown in Prob. 277, components {E}_z\ and\ \mathcal{B}_z of the electric and magnetic fields E and B, respectively, are determined from wave equations for {E}_z\ and\ \mathcal{B}_z. For TE_{mn} modes \left({E}_z = 0\right) everywhere in the waveguide core) propagating in a uniform rectangular EM waveguide the results for {E}_z\ and\ \mathcal{B}_z are as follows

{E}_z(x, y, z, t)=0 \quad \text { and } \quad \mathcal{B}_z(x, y, z, t)=\mathcal{B}_0 \cos \left(\frac{m \pi}{a} x\right) \cos \left(\frac{n \pi}{b} y\right) e^{i \varphi} \text {, }          (13.281)

where m and n are integers, and φ = k_zz − ωt is the phase of the microwave. When deriving the other components for the electric and magnetic fields \left({E}_x ,{E}_y ,\ \mathcal{B}_x\ and\ \mathcal{B}_y \right) we will need derivatives ∂\mathcal{B}_z/∂x\ and\ ∂\mathcal{B}_z/∂y, so we state them here

\frac{\partial \mathcal{B}_z}{\partial x}=-\frac{m \pi}{a} \mathcal{B}_0 \sin \left(\frac{m \pi}{a} x\right) \cos \left(\frac{n \pi}{b} y\right) e^{i \varphi}           (13.282)

and

\frac{\partial \mathcal{B}_z}{\partial y}=-\frac{n \pi}{b} \mathcal{B}_0 \cos \left(\frac{m \pi}{a} x\right) \sin \left(\frac{n \pi}{b} y\right) e^{i \varphi} .          (13.283)

In (13.130) it was shown that components {E}_x\ and  {E}_y as well as \mathcal{B}_x and \mathcal{B}_y can be expressed simply as function of axial components and for special modes (TE and TM) the situation is even simpler with transverse components for TE modes expressed as a function of \mathcal{B}_z and for TM modes as a function of {E}_z. For the TE_{mn} modes we thus have the following expressions for components {E}_x ,  {E}_y ,\ \mathcal{B}_x ,\ and\ \mathcal{B}_y derived from Maxwell equations

{E}_x=\frac{i \omega}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial y}=-\frac{i \omega}{\gamma^2} \frac{n \pi}{b} \mathcal{B}_0 \cos \left(\frac{m \pi}{a} x\right) \sin \left(\frac{n \pi}{b} y\right) e^{i \varphi},          (13.284)

{E}_y=-\frac{i \omega}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial x}=\frac{i \omega}{\gamma^2} \frac{m \pi}{a} \mathcal{B}_0 \sin \left(\frac{m \pi}{a} x\right) \cos \left(\frac{n \pi}{b} y\right) e^{i \varphi}           (13.285)

\mathcal{B}_x=\frac{i k_z}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial x}=-\frac{i k_z}{\gamma^2} \frac{m \pi}{a} \mathcal{B}_0 \sin \left(\frac{m \pi}{a} x\right) \cos \left(\frac{n \pi}{b} y\right) e^{i \varphi}          (13.286)

and

\mathcal{B}_y=\frac{i k_z}{\gamma^2} \frac{\partial \mathcal{B}_z}{\partial y}=-\frac{i k_z}{\gamma^2} \frac{n \pi}{b} \mathcal{B}_0 \cos \left(\frac{m \pi}{a} x\right) \sin \left(\frac{n \pi}{b} y\right) e^{i \varphi},           (13.287)

where parameter γ is defined as

γ^2 = k^2  –  k^2_z = \frac{\omega^2}{c^2}  –  k^2_z = \left( \frac{m \pi}{a}\right)^2 + \left( \frac{n \pi}{b}\right)^2 = \frac{\omega^2_c}{c^2}            (13.288)

with ω_c the cutoff frequency for the uniform rectangular EM waveguide and TE_{mn} mode.

(b) The Poynting vector S is in general defined as the energy flow per unit time (power) per unit area A and is given by the vector product S =  {E}\times \mathcal{B}/μ_0, where μ_0 is the magnetic constant. For periodic sinusoidal electromagnetic (EM) fields of more interest is the mean Poynting vector \bar{S} averaged over time and determined by treating the electric and magnetic field vectors E and B as complex vectors to get

\overline{\mathbf{S}}=\frac{1}{2 \mu_0} \boldsymbol{E} \times \boldsymbol{B}^*         (13.289)

with B^∗ the complex conjugate of B and the factor 1/2 arising from the time average of the sinusoidal function over one time period.
In matrix format we now express the mean Poynting vector \bar{S} as

\overline{\mathbf{S}}=\frac{1}{2 \mu_0}\left[\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ {E}_x &  {E}_y &  {E}_z \\ \mathcal{B}_x^* & \mathcal{B}_y^* & \mathcal{B}_z^*\end{array}\right]=\left( {E}_y \mathcal{B}_z^*-  {E}_z \mathcal{B}_y^*\right) \hat{\mathbf{i}}-\left(  {E}_x \mathcal{B}_z^*-  {E}_z \mathcal{B}_x^*\right) \hat{\mathbf{j}}+\left( {E}_x \mathcal{B}_y^*-  {E}_y \mathcal{B}_x^*\right) \hat{\mathbf{k}},           (13.290)

resulting in the following expressions for components \bar{S}_x ,\ \bar{S}_y ,\ and\ \bar{S}_z of the Poynting vector S

and

(c) Mean energy density \bar{ρ}_{en} for the microwave is given by (note: the extra factor \frac{1}{2} again accounts for average over time period)

(d) Mean power \bar{P} flowing through a transverse cross section A of the waveguide core is given by an integral of the axial component of the mean Poynting vector S over A

\bar{P}=\int \overline{\mathbf{S}} \cdot \mathrm{d} \mathbf{A}=\int_0^a \int_0^b \bar{S}_z \mathrm{~d} x \mathrm{~d} y=\frac{\omega k_z \mathcal{B}_0^2 a b}{8 \mu_0 \gamma^4}\left[\left(\frac{n \pi}{b}\right)^2+\left(\frac{m \pi}{a}\right)^2\right],           (13.295)

where we used the following definite integrals

\int_0^a \sin ^2\left(\frac{m \pi}{a} x\right) \mathrm{d} x=\int_0^a \cos ^2\left(\frac{m \pi}{a} x\right) \mathrm{d} x=\frac{a}{2}             (13.296)

and

\int_0^b \sin ^2\left(\frac{n \pi}{b} y\right) \mathrm{d} y=\int_0^b \cos ^2\left(\frac{n \pi}{b} y\right) \mathrm{d} y=\frac{b}{2} .             (13.297)

Recalling the identity (13.288), the mean power \bar{P} flowing through cross section A given in (13.295) simplifies to read

\bar{P}=\frac{\omega k_z \mathcal{B}_0^2 a b}{8 \mu_0 \gamma^4}\left[\left(\frac{n \pi}{b}\right)^2+\left(\frac{m \pi}{a}\right)\right]=\frac{\omega k_z \mathcal{B}_0^2 a b}{8 \mu_0 \omega_{\mathrm{c}}^2}            (13.298)

(e) The time average of total stored energy W_{tot} per unit length of waveguide core is given by the integral of mean energy density ρ_{en} over the cross section of the waveguide core A

W_{\text {tot }}=\int_A \bar{\rho}_{\text {en }} \mathrm{d} A=\int_0^a \int_0^b \bar{\rho}_{\text {en }} \mathrm{d} x \mathrm{~d} y .       (13.299)

Integrating (13.294) in conjunction with (13.288) and the well-known identity c^2 = \left(ε_0μ_0\right)^{−1} we obtain

(f) The velocity υ_{en} of energy propagation in the waveguide core is determined from the ratio between \bar{P} of (13.298) and W_{tot} of (13.300)

v_{\mathrm{en}}=\frac{\int_0^a \int_0^b \bar{S}_z \mathrm{~d} x \mathrm{~d} y}{\int_0^a \int_0^b \bar{\rho}_{\mathrm{en}} \mathrm{d} x \mathrm{~d} y}=\frac{\frac{\omega k c^2 \mathcal{B}_0^2 a b}{8 \mu_0 \omega_{\mathrm{c}}^2}}{\frac{\omega^2  {B}_0^2 a b}{8 \mu_0 \omega_{\mathrm{c}}^2}}=\frac{k_z c^2}{\omega}        (13.301)

Several observations are now possible after a closer look at the result υ_{en} = k_gc^2/ω of (13.301). We note that (13.301) obtained for a rectangular uniform waveguide is identical to (13.280) obtained for a cylindrical uniform waveguide. Therefore, the same conclusions we reached in Prob. 284 for cylindrical waveguide will apply to a rectangular waveguide of interest in this problem. The following points can be made:

(1) Since by definition v_{\mathrm{ph}}=\omega / k_{\mathrm{g}} \text { and } v_{\mathrm{ph}} \geq c \text {, we note that } v_{\mathrm{en}}=c^2 / v_{\mathrm{ph}} \leq c \text {. }

(2) Since v_{\mathrm{ph}} v_{\mathrm{gr}}=c^2 \text {, we note from (13.301) that } v_{\mathrm{en}}=v_{\mathrm{gr}} \text {. }

(3) Since k_{\mathrm{gr}}=\sqrt{\omega^2-\omega_{\mathrm{c}}^2} / c \text {, we note that } v_{\mathrm{en}}=c \sqrt{1-\omega_{\mathrm{c}}^2 / \omega^2}=v_{\mathrm{gr}} \text {. }

Thus, the velocity υ_{en} of energy flow through the waveguide is equal to the group velocity υ_{gr} of microwave propagation in the waveguide.