0
$\begingroup$

I have a differential equation for an electromagentic wave propogating in the z-direction and oriented alone the x-axis:

$$\frac{d^2 E_x}{d z^2}+\omega^2 \mu \epsilon E_x=0$$

and if I say $k^2= \omega^2 \mu \epsilon$

the solution to the differential equation yield the following solution:

$$E_x=E_{x}^+ e^{-jkz} + E_{x}^- e^{jkz}$$

where $E_{x}^+$ and $E_{x}^-$ are arbitrary constants.

$E_{x}^+ e^{-jkz}$ is called the forward travelling wave and $E_{x}^+ e^{+jkz}$ is called the backward travelling wave.

Why is the exponent with the negative sign considered as the forward travelling wave while the one with the positive sign considered the backward travelling wave?

Does this have anything to do with stability issues?

$\endgroup$
2
  • $\begingroup$ $\frac{\partial^2 E_x}{\partial z^2}+\omega^2 \mu \epsilon E_x=0$ is a partial differential equation (PDE). Is that what you meant to write? $\endgroup$
    – Gert
    Jun 8 at 5:57
  • $\begingroup$ no I made a mistake there, the partial would have been with the curl, but I evaluated the curl and since $E_x$ is dependent only on z, it is total differential $\endgroup$
    – Orpheus
    Jun 8 at 6:16
2
$\begingroup$

Your equation is incomplete and is not an electromagnetic wave equation. Any wave function $f$ has to depend on both position and time: $$f = f(x,t)$$ as well as satisfy the wave equation $$\frac{\partial^2 f}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 f}{\partial t^2}$$ Sinusoidal solutions to the wave equation will, in general, have $(kx \pm \omega t)$ as the argument. It should not be difficult to see that $(kx - \omega t)$ corresponds to a wave traveling in the $+x$ direction, and $(kx + \omega t)$ corresponds to a wave traveling in the $-x$ direction.

$\endgroup$
5
  • $\begingroup$ my equation is in the phasor form, the derivative with respect to time is reflected in the $\omega$ $\endgroup$
    – Orpheus
    Jun 8 at 6:17
  • 1
    $\begingroup$ @Orpheus Then you should've indicated it clearly in your original question. In this case, the argument is the same: when the signs of the coefficients of $x$ and $t$ are opposite, the wave is traveling forward. If they have the same sign, the wave is traveling backward. Therefore if you already have assumed $E_x e^{i \omega t}$, then the forward-traveling wave will have a negative $z$ in the exponent. $\endgroup$ Jun 8 at 6:34
  • $\begingroup$ Yes...I want to know why is that when the signs are opposite...the wave is forward travelling....and when opposite it is backward travelling? and my apologies for not having mentioned that equation was in the phasor form.... $\endgroup$
    – Orpheus
    Jun 8 at 6:38
  • 1
    $\begingroup$ @Orpheus Imagine an ordinary function $f(x)$ plotted in the usual way. If I want to move the function to the right (in the $+x$ direction) by $a$ units, then the new function will be $f(x-a)$. Since I have subtracted $a$, the same point on the graph will now be at a greater value of $x$ (i.e. shifted to the right). So in our example $a$ plays the role of $\omega t$. $\endgroup$ Jun 8 at 6:57
  • $\begingroup$ Thanks a lot @Vincent Thacker! $\endgroup$
    – Orpheus
    Jun 8 at 7:15
1
$\begingroup$

We start with the Electromagnetic wave equations \begin{align} \nabla^2\mathbf E\boldsymbol{-} \dfrac{1}{c^2}\dfrac{\partial^2 \mathbf E}{\partial t^2}& \boldsymbol{=}\boldsymbol{0} \tag{01a}\label{01a}\\ \nabla^2\mathbf B\boldsymbol{-} \dfrac{1}{c^2}\dfrac{\partial^2 \mathbf B}{\partial t^2}& \boldsymbol{=}\boldsymbol{0} \tag{01b}\label{01b} \end{align} where $\mathbf E\boldsymbol{=}\mathbf E\left(x,y,z,t\right)$ , $\mathbf B\boldsymbol{=}\mathbf B\left(x,y,z,t\right)$ are functions of the space-time coordinates and $1/c^2\boldsymbol{=}\mu\epsilon$.

Now for an electromagnetic wave propagating in the $z-$direction and oriented alone the $x-$axis we have \begin{equation} \mathbf E\left(x,y,z,t\right)\boldsymbol{=}\mathrm E_{x}\left(z,t\right)\mathbf i \tag{02}\label{02} \end{equation} where $\mathbf i$ the unit vector along the $x-$axis. Equation \eqref{01a} yields \begin{equation} \dfrac{\partial^2 \mathrm E_{x}}{\partial z^2}\boldsymbol{-} \dfrac{1}{c^2}\dfrac{\partial^2 \mathrm E_{x}}{\partial t^2} \boldsymbol{=}0 \tag{03}\label{03} \end{equation} The general solution of this wave equation is not something new, you could find it in many textbooks on electromagnetics, on the web, here in this site etc. But I'll sketch in summary a proof to realize that what you are missing here is the time dependence part which prevents you from understanding the $''$travelling$''$ of the wave, forward or backward.

The first step is to consider that $\mathrm E_{x}\left(z,t\right)$ is the product of a function of space coordinates, here $z$, and a function of time $t$ as follows \begin{equation} \mathrm E_{x}\left(z,t\right) \boldsymbol{=}\mathcal E\left(z\right)\mathcal T\left(t\right) \tag{04}\label{04} \end{equation} Then from equation \eqref{03} we have \begin{equation} T\left(t\right)\dfrac{\mathrm d^2 \mathcal E\left(z\right)}{\mathrm dz^2}\boldsymbol{-}\dfrac{1}{c^2}\mathcal E\left(z\right)\dfrac{\mathrm d^2 \mathcal T\left(t\right)}{\mathrm dt^2}\boldsymbol{=}0 \nonumber \end{equation} that is \begin{equation} c^2\dfrac{1}{\mathcal E\left(z\right)}\dfrac{\mathrm d^2 \mathcal E\left(z\right)}{\mathrm dz^2}\boldsymbol{=}\dfrac{1}{T\left(t\right)}\dfrac{\mathrm d^2 \mathcal T\left(t\right)}{\mathrm dt^2} \tag{05}\label{05} \end{equation} Since the lhs is a function of $z$ only and the rhs is a function of $t$ only then both must be a constant \begin{equation} c^2\dfrac{1}{\mathcal E\left(z\right)}\dfrac{\mathrm d^2 \mathcal E\left(z\right)}{\mathrm dz^2}\boldsymbol{=}\boldsymbol{-}\omega^2\boldsymbol{=}\dfrac{1}{T\left(t\right)}\dfrac{\mathrm d^2 \mathcal T\left(t\right)}{\mathrm dt^2}\,,\qquad \omega\in \mathbb C \tag{06}\label{06} \end{equation} yielding a system of two linear differential equations of 2nd order \begin{align} \dfrac{\mathrm d^2 \mathcal T\left(t\right)}{\mathrm dt^2}\boldsymbol{+}\omega^2 \mathcal T\left(t\right)& \boldsymbol{=}0 \tag{07a}\label{07a}\\ \dfrac{\mathrm d^2 \mathcal E\left(z\right)}{\mathrm dz^2}\boldsymbol{+}\left(\dfrac{\omega}{c}\right)^2\mathcal E\left(z\right)& \boldsymbol{=}0 \tag{07b}\label{07b} \end{align} and defining \begin{equation} \mathrm k\boldsymbol{\equiv}\dfrac{\omega}{c} \tag{08}\label{08} \end{equation} we have \begin{align} \dfrac{\mathrm d^2 \mathcal T\left(t\right)}{\mathrm dt^2}\boldsymbol{+}\omega^2 \mathcal T\left(t\right)& \boldsymbol{=}0 \tag{09a}\label{09a}\\ \dfrac{\mathrm d^2 \mathcal E\left(z\right)}{\mathrm dz^2}\boldsymbol{+}\mathrm k^2\mathcal E\left(z\right)& \boldsymbol{=}0 \tag{09b}\label{09b} \end{align} with general solutions respectively(1) \begin{align} \mathcal T\left(t\right) & \boldsymbol{=} \mathrm A_1\, e^{\boldsymbol{+}i\,\omega\,t}\boldsymbol{+}\mathrm A_2\, e^{\boldsymbol{-}i\,\omega\,t} \tag{10a}\label{10a}\\ \mathcal E\left(z\right) & \boldsymbol{=}\mathrm B_1\, e^{\boldsymbol{+}i\,\mathrm k\,z}\boldsymbol{+}\mathrm B_2\, e^{\boldsymbol{-}i\,\mathrm k\,z} \tag{10b}\label{10b} \end{align} giving the following general solution of \eqref{03} \begin{equation} \mathrm E_{x}\left(z,t\right) \boldsymbol{=}\left[\mathrm C_1\, e^{i\left(\mathrm k\,z\boldsymbol{+}\omega\,t\right)}\boldsymbol{+}\mathrm C_2\, e^{\boldsymbol{-}i\left(\mathrm k\,z\boldsymbol{+}\omega\,t\right)}\right]\boldsymbol{+}\left[\mathrm D_1\, e^{i\left(\mathrm k\,z\boldsymbol{-}\omega\,t\right)}\boldsymbol{+}\mathrm D_2\, e^{\boldsymbol{-}i\left(\mathrm k\,z\boldsymbol{-}\omega\,t\right)}\right] \tag{11}\label{11} \end{equation} If $\,\mathrm k \in \mathbb R\,$ then in order to be $\,\mathrm E_{x}\left(z,t\right) \in \mathbb R\,$ we must have \begin{equation} \mathrm C_1\boldsymbol{=}\mathrm C_2\boldsymbol{=}\tfrac12\mathrm E^{\boldsymbol{+}}_{0z} \in \mathbb R\,,\qquad \mathrm D_1\boldsymbol{=}\mathrm D_2\boldsymbol{=}\tfrac12\mathrm E^{\boldsymbol{-}}_{0z}\in \mathbb R \tag{12}\label{12} \end{equation} giving \begin{equation} \mathrm E_{x}\left(z,t\right)\boldsymbol{=}\mathrm E^{\boldsymbol{+}}_{0z}\cos\left(\mathrm k\,z\boldsymbol{+}\omega\,t\right)\boldsymbol{+} \mathrm E^{\boldsymbol{-}}_{0z}\cos\left(\mathrm k\,z\boldsymbol{-}\omega\,t\right) \tag{13}\label{13} \end{equation} Based on this solution the answer to your question about the travelling waves is given very well by @Vincent Thacker.

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$

(1) If you want to see how to solve this kind of differential equations without assuming a priori that the solution would be exponential or sinusoidal take a look in my answer here Need help understanding an equation of motion for a pendulum.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.