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The wave equation is: $$\nabla^2 \mathbf{E} + k^2 \mathbf{E} = 0$$

Using separation of variables, we get a solution of $E = a(x)b(y)c(z)d(t)$. Say for the $x$-direction we get a solution of:

$$ a(x) = C_1 e^{-j k_x x} + C_2 e^{+j k_x x}$$

and the time solution is

$$ d(t) = C_3 e^{-j \omega t} + C_4 e^{+j \omega t}$$

But I noticed that for physics books, they apply boundary conditions that yield the following scenario

$$ a(x) = 0 + C_2 e^{+j k_x x}$$

and

$$ d(t) = C_3 e^{-j \omega t} + 0$$

Hence, we get

$$ a(x)d(t) = C_5 e^{\mathbf{+}j k_x x \mathbf{-} j \omega t}$$

For engineering books, they apply boundary conditions that yield the following scenario

$$ a(x) = C_1 e^{-j k_x x} + 0$$

and

$$ d(t) = 0 + C_4 e^{+j \omega t}$$

Hence, we get

$$ a(x)d(t) = C_5 e^{\mathbf{-}j k_x x \mathbf{+} j \omega t}$$

Question: which set of boundary conditions is the "correct" one for wave propagation? Or what does the engineering book assume over the physics book (and vice-versa)

Background:

  1. The physics book (e.g. Griffith, chapter 9) claims $+k_x$ is for the incident/transmitted wave, while $-k_x$ is for the reflected wave. Moreover, He states $\pm \omega$ controls the wave direction (i.e. moving to the left or right).
  2. The engineering book (Pozar, chapter 1) states "terms with + signs result in waves traveling in the negative $x,y,z$ direction, while terms with - signs result in waves traveling in the positive direction" when talking about the sign of $k_x$ and Pozar slaps a $e^{+j \omega t}$ onto the final solution.
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    $\begingroup$ I just want to mention that there is a certain small advantage of using engineering equation with $e^{-j \omega t}$, because a Fourier transform of derivative will give $\frac{df}{dt} \doteqdot j \omega \hat{f}(\omega)$. Thus the Fourier transform will change a diff equation into polynomial, and no negative sign will appear. In this case it is similar to Laplace transform where it transform partial derivative into polynomials and no negative sign appear. It seems that mnemonically it is more simple this way. $\endgroup$ Jan 25, 2023 at 13:51

1 Answer 1

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It all goes back to the ancient habit of defining what a Fourier transform is. By tradition long lost in the mist of history, physicists define a Fourier transform of a function $f(t)$ as $$F_p(\omega)=\int_{-\infty}^{+\infty}dt f(t) e^{\mathfrak i \omega t}$$ but engineers define it as $$F_e(\omega)=\int_{-\infty}^{+\infty}dt f(t) e^{-\mathfrak j \omega t}$$ so that we have $F_p(-\omega) = F_e(\omega)$. There is some reason in the engineers definition because we, at least I do, intuitively think in terms of positive frequencies and then since the Inverse transforms is $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}d\omega F_e(\omega) e^{\mathfrak j \omega t}$$ you can say that you assemble an arbitrary $f(t)$ from its frequency components represented at $\omega$ by the complex sinusoids $e^{\mathfrak j \omega t}$ with amplitudes $F_e(\omega)$. Of course a physicist would say his frequency contents are equally well represented at positive frequencies by the sinusoids $e^{-\mathfrak i \omega t}$. Don't worry, you get used to it, just mentally change every $\mathfrak j$ to $-\mathfrak i$ or $\mathfrak i$ to $-\mathfrak j$.

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    $\begingroup$ It seems to depend on physicist. I have more often seen the symmetrical version normalized with $\frac{1}{\sqrt{2 \pi}}$. $\endgroup$
    – Triatticus
    Jan 25, 2023 at 14:36
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    $\begingroup$ Right. I’ve also talked about this difference in conventions in this answer, when referring to phasors: physics.stackexchange.com/a/687491/111969 Another point worth noting is that frequently engineers define the Fourier transform as a function of the frequency rather than the angular frequency, because frequency is what instruments indicate and you can also get rid of the $2\pi$ factors in front of the integrals. $\endgroup$ Jan 25, 2023 at 21:39
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    $\begingroup$ Thank you for this amazing answer! I do have one more question though: f(x) should be replaced with f(t) in $F_{p}(\omega)$? If no, why is this the case? $\endgroup$ Jan 26, 2023 at 9:37
  • $\begingroup$ TYPO ALERT! Thanks for noticing, $\endgroup$
    – hyportnex
    Jan 26, 2023 at 9:46

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