Calculation of Poynting vector of standing electromagnetic wave

Question

I have this question in my homework where I have the following phasor of the electric field (we assume that all waves have $\omega$ frequency): $$\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}\left[\hat{x}+\left(2+1.5j\right)\hat{y}+\left(2+3j\right)\hat{z}\right]e^{j\left(4x-4y+2z\right)}$$ and we are asked to find another plane wave $\overline{\mathbf{E}}_{2}\left(x,y,z\right)$ s.t their sum will carry power only in the $\hat{y}$ direction. I suggested the following plane wave: $$\overline{\mathbf{E}}_{2}\left(x,y,z\right)=E_{0}\left[-\hat{x}+\left(2+1.5j\right)\hat{y}-\left(2+3j\right)\hat{z}\right]e^{-j\left(4x+4y+2z\right)}$$ and I tried to calculate the average power by calculating: $$\vec{\boldsymbol{S}}_{tot} =\frac{1}{2}\Re\{{\overline{\mathbf{E}}_{tot}\times\overline{\mathbf{H}}_{tot}^{*}}\}$$and show that it has only a $\hat{y}$ non-zero component. I know this must be true but I probably have an arithmetic mistake (or something sillier). I don't know if it's legit to post my calculation here in hope that someone can catch my mistake but after $5$ hours of trying to figure out where's my mistake I am simply desperate. I'm sorry if it's too messy or too elaborate but I wrote it in LyX and converted it to Latex and paste it here. so here it is: \begin{align} \overline{\mathbf{E}}_{tot}\left(x,y,z\right) & =\overline{\mathbf{E}}_{1}\left(x,y,z\right)+\overline{\mathbf{E}}_{2}\left(x,y,z\right)= \\ & =E_{0}\left[\hat{x}+\left(2+1.5j\right)\hat{y}+\left(2+3j\right)\hat{z}\right]e^{j\left(4x-4y+2z\right)} \\ & \quad +E_{0}\left[-\hat{x}+\left(2+1.5j\right)\hat{y}-\left(2+3j\right)\hat{z}\right]e^{-j\left(4x+4y+2z\right)}=\\ & =E_{0}\left[\hat{x}\left(e^{j\left(4x+2z\right)}-e^{-j\left(4x+2z\right)}\right)+\left(2+1.5j\right)\hat{y}\left[e^{j\left(4x+2z\right)}+e^{-j\left(4x+2z\right)}\right] \right.\\ & \qquad \left. +\left(2+3j\right)\hat{z}\left(e^{j\left(4x+2z\right)}-e^{-j\left(4x+2z\right)}\right)\right]e^{-j4y}=\\ & =E_{0}\left[2j\sin\left(4x+2z\right)\hat{x}+\left(4+3j\right)\cos\left(4x+2z\right)\hat{y}+\left(-6+4j\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y} \end{align} The fields themselves: \begin{align} \overline{E}_{tot,x} & =2jE_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \overline{E}_{tot,y} &=\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \overline{E}_{tot,z} &=\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \end{align}

Their derivatives:

\begin{align} \partial_{y}\overline{E}_{tot,z}&=\partial_{y}\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=\left(16+24j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{z}\overline{E}_{tot,y}&=\partial_{z}\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)=-\left(8+6j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{z}\overline{E}_{tot,x}&=\partial_{z}\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=4jE_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \partial_{x}\overline{E}_{tot,z}&=\partial_{x}\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=-\left(24-16j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y} \\ \partial_{x}\overline{E}_{tot,y}&=\partial_{x}\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)=-\left(16+12j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y} \\ \partial_{y}\overline{E}_{tot,x}&=\partial_{y}\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)=8E_{0}\sin\left(4x+2z\right)e^{-j4y} \end{align}

The magnetic field: \begin{align*} \Longrightarrow\,\,\,\overline{\mathbf{H}}_{tot}\left(x,y,z\right) & =-\frac{1}{j\omega\mu}\nabla\times\overline{\mathbf{E}}_{tot}=\frac{j}{\omega\mu}\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ \partial_{x} & \partial_{y} & \partial_{z}\\ \overline{E}_{tot,x} & \overline{E}_{tot,y} & \overline{E}_{tot,z} \end{vmatrix}\\ & =\frac{j}{\omega\mu}\left[ \hat{x}\left(\partial_{y}\overline{E}_{tot,z}-\partial_{z}\overline{E}_{tot,y}\right)+\hat{y}\left(\partial_{z}\overline{E}_{tot,x}-\partial_{x}\overline{E}_{tot,z}\right) \right.\\ & \qquad \qquad \left. +\hat{z}\left(\partial_{x}\overline{E}_{tot,y}-\partial_{y}\overline{E}_{tot,x}\right) \right] =\\ & =\frac{jE_{0}}{\omega\mu}\left[\left(16+24j+8+6j\right)\sin\left(4x+2z\right)\hat{x}+\left(4j+24-16j\right)\cos\left(4x+2z\right)\hat{y} \right.\\ & \qquad \qquad \left. -\left(16+12j+8\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y}=\\ & =\frac{jE_{0}}{\omega\mu}\left[\left(24+30j\right)\sin\left(4x+2z\right)\hat{x}+\left(24-12j\right)\cos\left(4x+2z\right)\hat{y} \right.\\ & \qquad \qquad \left. -\left(24+12j\right)\sin\left(4x+2z\right)\hat{z}\right]e^{-j4y} \end{align*}

and its components,

\begin{align} \overline{H}_{tot,x}&=-\frac{E_{0}}{\omega\mu}\left(30-24j\right)\sin\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,y}&=\frac{E_{0}}{\omega\mu}\left(12+24j\right)\cos\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,z}&=\frac{E_{0}}{\omega\mu}\left(12-24j\right)\sin\left(4x+2z\right)e^{-j4y} \\ \overline{H}_{tot,x}^{*}&=-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y} \\ \overline{H}_{tot,y}^{*}&=\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y} \\ \overline{H}_{tot,z}^{*}&=\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y} \end{align}

giving the Poynting vector

\begin{align*} \vec{\boldsymbol{S}}_{tot} & =\frac{1}{2}\overline{\mathbf{E}}_{tot}\times\overline{\mathbf{H}}_{tot}^{*}=\frac{1}{2}\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ \overline{E}_{tot,x} & \overline{E}_{tot,y} & \overline{E}_{tot,z}\\ \overline{H}_{tot,x}^{*} & \overline{H}_{tot,y}^{*} & \overline{H}_{tot,z}^{*} \end{vmatrix}=\\ & =\frac{1}{2}\left[\hat{x}\left(\overline{E}_{tot,y}\cdot\overline{H}_{tot,z}^{*}-\overline{E}_{tot,z}\cdot\overline{H}_{tot,y}^{*}\right)+\hat{y}\left(\overline{E}_{tot,z}\cdot\overline{H}_{tot,x}^{*}-\overline{E}_{tot,x}\cdot\overline{H}_{tot,z}^{*}\right) \right. \\ & \qquad \left. +\hat{z}\left(\overline{E}_{tot,x}\cdot\overline{H}_{tot,y}^{*}-\overline{E}_{tot,y}\cdot\overline{H}_{tot,x}^{*}\right)\right] \end{align*}

with components

\begin{align*} S_{tot,x} & =\frac{1}{2}\left(\overline{E}_{tot,y}\cdot\overline{H}_{tot,z}^{*}-\overline{E}_{tot,z}\cdot\overline{H}_{tot,y}^{*}\right)=\\ & =\frac{1}{2}\left(\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y}\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(\left(4+3j\right)\cdot\left(12+24j\right)-\left(-6+4j\right)\cdot\left(12-24j\right)\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(-24+132j-24-192j\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =-\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(24+60j\right)\sin\left(8x+4z\right) \end{align*}

and

\begin{align*} S_{tot,y} & =\frac{1}{2}\left(\overline{E}_{tot,z}\cdot\overline{H}_{tot,x}^{*}-\overline{E}_{tot,x}\cdot\overline{H}_{tot,z}^{*}\right)=\\ & =\frac{1}{2}\left(\left(\left(-6+4j\right)E_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12+24j\right)\sin\left(4x+2z\right)e^{j4y}\right)\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(-\left(-6+4j\right)\cdot\left(30+24j\right)-2j\cdot\left(12+24j\right)\right)\sin^{2}\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(276+{24j}+48-{24j}\right)\sin^{2}\left(4x+2z\right)=\\ & =\frac{162\left|E_{0}\right|^{2}}{\omega\mu}\sin^{2}\left(4x+2z\right) \end{align*}

and

\begin{align*} S_{tot,z} & =\frac{1}{2}\left(\overline{E}_{tot,x}\cdot\overline{H}_{tot,y}^{*}-\overline{E}_{tot,y}\cdot\overline{H}_{tot,x}^{*}\right)=\\ & =\frac{1}{2}\left(\left(2jE_{0}\sin\left(4x+2z\right)e^{-j4y}\right)\cdot\left(\frac{E_{0}^{*}}{\omega\mu}\left(12-24j\right)\cos\left(4x+2z\right)e^{j4y}\right) \right. \\ & \qquad \left. -\left(\left(4+3j\right)E_{0}\cos\left(4x+2z\right)e^{-j4y}\right)\cdot\left(-\frac{E_{0}^{*}}{\omega\mu}\left(30+24j\right)\sin\left(4x+2z\right)e^{j4y}\right)\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(2j\cdot\left(12-24j\right)+\left(4+3j\right)\cdot\left(30+24j\right)\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(48+24j+48+186j\right)\sin\left(4x+2z\right)\cos\left(4x+2z\right)=\\ & =\frac{\left|E_{0}\right|^{2}}{2\omega\mu}\left(48+105j\right)\sin\left(8x+4z\right) \end{align*}

So, as you can see the real parts of the $\hat{x}$ and $\hat{z}$ components are not $0$.

Any help would be much appriciated.

Answer 1

I wrote the matlab code below. The values of $E_1$ and $E_2$ are assumed to be at $t=0$ and at $\vec{r}=0$.

The result is that $P=0$ for your choice of $E_2$. If you change the sign of $E_2$ the result is a momentum / energy transport density in the -y direction. A result of P along the positive y-direction seems hard to achieve since the initial wave travels in the -y direction.

r = [0,0,0];

t = 0;

Amp1 = [1,2+1.5j,2+3j]; 

k1 = [4,-4,2];

ph1 = exp(j*(sqrt(sum(k1.*k1))*t+sum(k1.*r)));

E1 = Amp1*ph1;

Amp2 = [-1,2+1.5j,-(2+3j)]; 

k2 = [-4,-4,-2];

ph2 = exp(j*(sqrt(sum(k2.*k2))*t+sum(k2.*r)));

E2 = Amp2*ph2;

Poy = real(cross(E1+E2,cross(k1,conj(E1))+cross(k2,conj(E2))));

display(Poy);

Answer 2

This is almost certainly caused by the minus sign in the exponent in $$ e^{-j\left(4x+4y+2z\right)} = e^{j\left(-4x+4y-2z\right)} $$ in your second wave; the wave that results is precisely the clear answer if you were asked to take the wave $e^{j\left(4x-4y+2z\right)}$ and cancel out the power flow in the $x$ and $z$ directions, but not the $y$ direction, by turning the first two into standing waves.

You then also need to check that the polarizations work out correctly (though they look OK to me at a first glance), and goodness help whoever tries to wade into the thicket of formulas you've posted (which, frankly, should not be done in terms of the total fields ─ calculate everything in terms of $\mathbf E_1$ and $\mathbf E_2$ all the way through, and only put in their explicit expressions when they're absolutely crucial). But the clear place to start is with the correct exponent.

Answer 3

After a lot of effort trying to find a condition for the wave I'm adding ($\overline{\mathbf{E}}_{2}$) it turned out that my math was fine but I chose the wrong relative phases for the different components of $\overline{\mathbf{E}}_{2}$ because I was under the impression that the relative phases of the components of $\overline{\mathbf{E}}_{2}$ doesn't matter (which means that I have more than one d.o.f) but that is wrong in the general case (it is true for example for a wave with linear polarization in the form: $\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}e^{-j\left(k_xx+k_zz\right)}\hat{y}$)

Anyway, I finally managed to find necessary and sufficient conditions for $\overline{\mathbf{E}}_{2}$ s.t. $\overline{\mathbf{E}}_{1}+\overline{\mathbf{E}}_{2}$ will carry power only in the $\hat{y}$ direction. let's make it general. the most general form of a phasor of planar wave is: $$\overline{\mathbf{E}}_{1}(x,y,z)=[A\hat{x}+B\hat{y}+C\hat{z}]e^{-j(k_xx+k_yy+k_zz)}$$ where: $$A=|A|e^{j\varphi_A}\,\,,B=|B|e^{j\varphi_B}\,\,,C=|C|e^{j\varphi_C}$$ if we want to add another wave $\overline{\mathbf{E}}_{2}$ s.t. their sum will carry power only in the $\hat{y}$ direction we have to choose the following $\mathbf{k}_2$ vector of $\overline{\mathbf{E}}_{2}$: $$\mathbf{k}_2=(-k_x,k_y,-k_z)$$ and we also need to choose the amplitudes of the components of $\overline{\mathbf{E}}_{2}$ such that they equal in absolute value to the corresponding amplitudes of $\overline{\mathbf{E}}_{1}$ because we need to create sinusoidal dependence in the $x$ and $z$ coordinates. thus, the most general form of $\overline{\mathbf{E}}_{2}$ we can offer is: $$\overline{\mathbf{E}}_{2}(x,y,z)=[Ae^{j\varphi_1}\hat{x}+Be^{j\varphi_2}\hat{y}+Ce^{j\varphi_3}\hat{z}]e^{-j(-k_xx+k_yy-k_zz)}$$ now, all that left is to do the (tedious) math and calculate the complex poynting vector of the sum of the 2 waves and finally require that the real parts of it's $x$ and $z$ components will vanish. after doing that I found that $\varphi_1, \varphi_2, \varphi_3$ must hold the following equations: $$\varphi_1 = \varphi_2 + 2(\varphi_B-\varphi_A)+\pi$$ $$\varphi_3 = \varphi_2 + 2(\varphi_B-\varphi_C)+\pi$$ and $\varphi_2$ stays free to choose (which means that we only left with one d.o.f). The above condition doesn't need to hold in the case that $B=0$ or $k_y=0$ or {$A=0$ and $C=0$} (which for example, corresponds to the simple case of $\overline{\mathbf{E}}_{1}\left(x,y,z\right)=E_{0}e^{-j\left(k_xx+k_zz\right)}\hat{y}$ because in that case we have $k_y=0$ and also $A=0$ and $C=0$).

In the specific example of this question we have: $$\overline{\mathbf{E}}_{1}(x,y,z)=E_{0}[\hat{x}+(2+1.5j)\hat{y}+(2+3j)\hat{z}]e^{-j(-4x+4y-2z)}$$ So, if we follow the above condition we get the following wave: $$\overline{\mathbf{E}}_{2}(x,y,z)=E_{0}e^{j\varphi}[-\hat{x}+(2-1.5j)\hat{y}+(-2+3j)\hat{z}]e^{-j(4x+4y+2z)}$$ where $\varphi$ can be any phase (this is our d.o.f) and under that choice we get that $\overline{\mathbf{E}}_{1}+\overline{\mathbf{E}}_{2}$ carries power only in the $\hat{y}$ direction.