# Why is time harmonic follow the form of $e^{-i\omega t}$, not $e^{i\omega t}$? [closed]

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?)

• Hi Tippsie. Welcome to Phys.SE. Please only ask 1 question per post. Mar 2 at 17:18
• Thanks Qmechanic, just removed the 2nd question. Mar 2 at 17:34
• "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."
– hft
Mar 2 at 18:37
• What book uses $j=-\sqrt{-1}$? Mar 2 at 18:44

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.
• Thanks ZeroTheHere, but I would argue $e^{-i\omega t}$ is identical with $e^{j\omega t}$, i.e. $i=-j$ Mar 2 at 17:37
• @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. Mar 2 at 17:49
• 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." Mar 2 at 19:18

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
$$x(t)=A\rm e^{+i\omega\,t}+B \rm e^{-i\omega\,t}$$