I think partial of this question being asked before but I have some other doubts.As shown in this post how to add two plane waves if they are propagating in different direction?, reading the third reply by Wouter. If we have two waves

$$ f_1 = \sin(\vec{k}\cdot\vec{r} + \omega t), f_2 = \sin(\vec{q}\cdot\vec{r} - \omega t) $$

1) my first question, in one text, I saw that to create the standing wave, we should have the two sinusoidal waves propagate in counter direction. I am confusing with above function of wave. I know that the group velocity is defined as $$\vec{v}_p = \hat{k}\omega/|\vec{k}|$$ so if we want two waves propagate in opposite direction, we could have one $\omega$ and the other $-\omega$ or we have have one $\vec{k}$ positive and the other one $-\vec{k}$. But starting from above waves, $\omega$ is opposite, so we should make sure $\vec{k}=\vec{q}$. But what's the physical significance to make $-\omega$? To me, it makes more sense to

$$ f_1 = \sin(\vec{k}\cdot\vec{r} + \omega t), f_2 = \sin(\vec{q}\cdot\vec{r} + \omega t) \quad \mbox{with} \quad \vec{q}=-\vec{k} $$

2) Let start from the last formulas, let $\vec{q}=-\vec{k}$ so to have the sum in the form

$$ f_1+f_2 = \sin(\vec{k}\cdot\vec{r})\cos(\omega t) $$

what I understand from the text is standing wave should be stationary in space. But there is a modulated amplitude $\cos(\omega t)$ there so does it really stationary (or standing wave)?

3) If I have two such "standing waves" propagating along different direction, says $\cos^{-1}(\vec{p}\cdot\vec{q})=\pi/6$

$$ s_1 = \sin(\vec{p}\cdot\vec{r})\cos(\omega t), \quad s_2 = \sin(\vec{q}\cdot\vec{r})\cos(\omega t) $$

so will these two wave makes a two-dimensional standing wave if $|q|=|p|$?


1 Answer 1

  1. It's not very useful to think of the frequency as negative. The wave vector $\vec{k}$ should be interpreted as the indicating the direction of the wave, so if the wave is going in the opposite direction, then $\vec{k}$ should be made negative, not $\omega$.

  2. Here is an animation that may help you. It shows the two propagating waves, and the resulting standing wave. It's called a "standing" wave in the sense that the nodes (places where the wave function has a value of zero) of the wave do not move.

  3. It's a little odd to say you have standing waves that are propagating, but yes, you will have a two dimensional standing wave, with a different wavevector in each dimension. Here is a visualization of it frozen in time (I used $|\vec{p}| = |\vec{q}| = 1$ for simplicity).

  • $\begingroup$ Thanks Brionius. That helps to clarify my question. So the amplitude of the "standing wave" is not constant as stated in the text, is it only stationary in space but the amplitude is still changing in time, right? For 3, I should not use the word 'propagate', I mean there are two standing waves along two direction making 30 degree. I wonder how to visualize that 2D standing wave then? $\endgroup$ Jan 3, 2015 at 21:37
  • $\begingroup$ As per your amplitude question, I know what you mean, and yes - your idea is correct. However, I would be careful about the word "amplitude" - traditionally the "amplitude" of a standing wave is interpreted as the maximum value of the wave function over time, which doesn't change. $\endgroup$
    – Brionius
    Jan 3, 2015 at 22:15
  • $\begingroup$ As for the 2D question, here is what it would look like (frozen in time). I used $|\vec{p}| = |\vec{q}| = 1$ for simplicity. Added that to the answer too. $\endgroup$
    – Brionius
    Jan 3, 2015 at 22:18
  • $\begingroup$ Thanks a lot. It is clear now. But another question just comes up to me. If people use the laser to create the optical standing wave, so the term $\cos\omega t$ that modulate the amplitude is related to the frequency of the laser? Since the frequency is so high, what do we see for optical standing wave then? $\endgroup$ Jan 4, 2015 at 0:25
  • $\begingroup$ Can you give an example of "seeing an optical standing wave"? That could mean a few different things, and I'm not sure what you mean. In general, though, our eyes are equipped with molecules that have resonances in the visible frequency range (100-1000 THz, roughly). Those molecules absorb electromagnetic waves with frequencies in that range, which causes a neural stimulus to travel to the brain, resulting in the experience of "seeing light". $\endgroup$
    – Brionius
    Jan 4, 2015 at 2:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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