# [Physics] Why is there a negative sign in front of the optical wave

conventionsopticswaves

In undergrad I lost (a lot) of marks in my optics class for writing:

$$A(t) = \exp(i(\omega t + \phi))$$

$$A(t) = \exp(i(-\omega t + \phi))$$

In a derivation where I must have needed a plane wave. At the time I thought the TA was being pedantic. Both forms represent a plane wave and are mathematically equivalent within some aesthetic changes (namely $\omega -> -\omega$). Now I realize this has huge implications if we need to take the derivative of the signal.

Physically why is it negative omega? Why do I see a negative sign pop up in every time it involves $t$, but the more standard for of the Fourier series is used otherwise?

I don't think it's very likely, but one other thing I can think of: when the sign before the $\omega$ is a minus then the wave represents a wave travelling to the "right" - positive x direction - and maybe your TA wants only waves travelling to the right. ^^

In case you don't know why the minus sign represents a wave travelling in the positive x direction:

Normally I write a plane wave like this (I consider only the x component here):

$$exp[i(k_xx\pm\omega t)]$$

The phase of the plane wave is obviously the term in the exponential. Let's consider a point of constant phase, let's say zero. The term in the brackets must then give

$$k_xx\pm\omega t=0$$

When the wave and hence this point of constant phase is travelling to the right then $x$ increases. Since time also increases then the only possibility to obtain a wave travelling in the positive x-direction is to put a minus sign before the omega, such that

$$k_xx-\omega t=0$$

for all times for this point of constant phase.

This convention of course depends on the way he writes the plane wave. If your write the plane wave as

$$exp[i(\omega t\pm k_xx)]$$

Then there would be no minus sign befor the omega.