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In physics, when we solve an PDE or ODE, the solution usually has the form of \begin{equation} f=C_+e^{i\lambda x}+C_-e^{-i\lambda x} \end{equation} and the "causility" will eliminate one term as it violates the "physics".
I am wondering how the "causility" is defined here. In detail, I will focus on the time harmonic term as it directly reflects the "causility".
As for time harmonic, the harmonic term usually takes the form \begin{equation} T(t)\sim e^{-i\omega t} \end{equation} However, in some books or papers, there seems to be another set of notations, they replace the imaginary unit $i$ with $-j$, which reads \begin{equation} T(t)\sim e^{j\omega t} \end{equation} I've heard some explanations on this topic like:

  1. Both $i$ and $j$ are square roots of $-1$, but "$i$" is "$+\sqrt{-1}$" while "$j$" stands for "$-\sqrt{-1}$".
  2. In engineering, especially electronic engineering, "$i(t)$" is preseved as the transient current. So those people will use "$j$" as the imaginary unit.

Both notion sets actually work for me. However, during review tasks, I saw a few (very few) papers were parepared with notation \begin{equation} T(t)\sim e^{i\omega t} \end{equation} With this notion set, the harmonic phase development will be quite confusing, the phase term can be any combination of: \begin{equation} \pm kr \pm \omega t \end{equation}

Here, my question is:
Why is time harmonic term $e^{-i\omega t}$ rather than $e^{i\omega t}$, is there any reason for this or is it idiomatic since the first guy (who's that guy?)

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    $\begingroup$ Hi Tippsie. Welcome to Phys.SE. Please only ask 1 question per post. $\endgroup$
    – Qmechanic
    Commented Mar 2, 2023 at 17:18
  • $\begingroup$ Thanks Qmechanic, just removed the 2nd question. $\endgroup$
    – Tippsie
    Commented Mar 2, 2023 at 17:34
  • $\begingroup$ "I am wondering how the 'causility' [sic] is defined here." Causality generally means that the past affects the future, and not vice versa. I kick a dog, the dog yelps. My action (kick) caused the dog's response (yelp). The "Caus-" part of "causality" comes from the word "cause." $\endgroup$
    – hft
    Commented Mar 2, 2023 at 18:37
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    $\begingroup$ What book uses $j=-\sqrt{-1}$? $\endgroup$
    – Ghoster
    Commented Mar 2, 2023 at 18:44

2 Answers 2

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  1. $j$ is the engineering notation. There is no difference except in notation between $e^{i\omega t}$ and $e^{j\omega t}$.
  2. what matters is the relative phase. $kx-\omega t$ or $-kx+\omega t$ both describe waves moving toward the $+x$ direction, whereas $kx+\omega t$ describes a wave moving in the $-x$ direction. The rest is convention, which may be different in different textbook. I’m a little more used to $kx-\omega t$ because the point of phasors is to remove the explicit time-dependence and it $e^{ikx}$ avoids using a $-$ sign all the time.
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  • $\begingroup$ Thanks ZeroTheHere, but I would argue $e^{-i\omega t}$ is identical with $e^{j\omega t}$, i.e. $i=-j$ $\endgroup$
    – Tippsie
    Commented Mar 2, 2023 at 17:37
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    $\begingroup$ @Tippsie I have no idea why you would think this, and I’ve never seen this in use in 20 years of teaching E&M to engineers. $\endgroup$ Commented Mar 2, 2023 at 17:49
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    $\begingroup$ I first heard that as a joke in an undergrad lecture. "You know, the electrical engineers chose the other root of $\sqrt{-1}$ and called it $j$." [Stares out the window pensively.] "Sometimes I worry that we chose the wrong root and they chose the right root." $\endgroup$ Commented Mar 2, 2023 at 19:18
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assume you want to solve this ODE

$$\ddot x+\omega^2\,x=0$$

you make the Ansatz $~x(t)=A\,\rm e^{\lambda\,t}$ and obtain

$$\underbrace{A\,\rm e^{\lambda\,t}}_{\ne 0}\,\left(\lambda^2+\omega^2\right)=0$$ from here $~\lambda=\pm i\,\omega~$ thus our solution is

$$x(t)=A\rm e^{+i\omega\,t}\quad,A\rm e^{-i\omega\,t}$$ or because we have linear ODE, the superposition of both solution
$$x(t)=A\rm e^{+i\omega\,t}+B \rm e^{-i\omega\,t}$$

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