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If I have an $\mathbf{E}$ field:

$$ \mathbf{E}_1 = x E_0 e^{-j(y-z) } $$

I think I can find its wave vector direction by finding the $\mathbf{H}$ field and then solving for the Poynting vector $\mathbf{E} \times \mathbf{H}$, but how do I find the wavenumber $K$? I know $K = 2\pi/\lambda$, but is this all I can say about it or can I actually solve for it with this information?

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  • $\begingroup$ Ya it's a constant and j is sqrt(-1) $\endgroup$
    – Austin
    Jun 11, 2015 at 19:25
  • $\begingroup$ So this is like a complex exponential form of a plane wave, except it's also scaled by x? $\endgroup$
    – Alan
    Jun 11, 2015 at 20:08
  • $\begingroup$ X is the unit vector for it so it should be x hat, but the poynting vector is in the (-y +z) direction I think $\endgroup$
    – Austin
    Jun 11, 2015 at 20:12
  • $\begingroup$ Ah, so $xE_0$ is the vector constant. You are right about direction of the wave. What else does the exponent $j(0,-1,1)\cdot(x,y,z)$ tell you? $\endgroup$
    – Alan
    Jun 11, 2015 at 20:19
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    $\begingroup$ so can you do sqrt(0^2 + (-1)^2 + 1^2) so K = sqrt(2)? $\endgroup$
    – Austin
    Jun 11, 2015 at 23:07

2 Answers 2

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Remember that $\vec{E}=E_{0}e^{-i\vec{k}\cdot\vec{r}}\hat{x}$, so that, just by identifying the variables $\vec{k}\cdot\vec{r}=k_{x}x + k_{y}y + k_{z}z$ ; from that you should be able to get the wave vector $\vec{k}$.

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  • $\begingroup$ So does $\vec{k}$ = (-$k_{y}y + k_{z}z$) and K(wavenumber) = magnitude of wave vector = sqrt(0^2 + (-1)^2 + 1^2) = sqrt(2)? $\endgroup$
    – Austin
    Jun 12, 2015 at 0:41
  • $\begingroup$ Careful $\vec{k}$ is a vector, so that: $\vec{k}=\hat{y}-\hat{z}$; the magnitude is, of course, $\sqrt{2}$, with $k_{y}=1$ and $k_{z}=-1$, according to the notation used. $\endgroup$
    – user82635
    Jun 12, 2015 at 0:57
  • $\begingroup$ Hmm sorry still a bit confused.. why isn't it k⃗ = -ŷ +ẑ and why is magnitude 1/sqrt(2) instead of just sqrt(2)? $\endgroup$
    – Austin
    Jun 12, 2015 at 1:02
  • $\begingroup$ Look at the answer, I corrected it.However, since you are using $-j$ and not $j$, then the sign for your $\vec{k}$ is wrong. $\endgroup$
    – user82635
    Jun 12, 2015 at 1:04
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Both

\begin{align} \tilde{\bf E}&=\hat{x} E_0 \exp(-i(\pm\frac{\sqrt{2}}{\mathrm{m}} (\hat{x}\frac{\sqrt{2}}{2}+\hat{y}\frac{\sqrt{2}}{2})\cdot(x,y,z)-\omega t))\\ &=\hat{x}E_0 \exp(-i(\vec{k}_\pm\cdot \vec{r}-\omega t)) \end{align} are part of a valid electrodynamic field travelling in opposite directions. (1/m is just the unit 1/meters) with

$$\vec{k}_{\pm}=\pm \sqrt{2}(\hat{x}\frac{\sqrt{2}}{2}+\hat{y}\frac{\sqrt{2}}{2}) \frac{1}{\mathrm{m}}$$

So your problem of determining wave vector and Poynting vector doesn't have a unique solution. But the wave number $k$ is $\sqrt{2}\mathrm{m}^{-1}$. Do you see how i determined that? i just made it so the exponent was in the form $|\vec{k}|\hat{k}\cdot \vec{r}$

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