In the undergraduate course about the wave, there stated for two harmonic waves propagating in opposite direction, then the resulting wave will be a standing wave. In math, it is like
$$y_1 = A\sin(kx + \omega t), y_2 = A\sin(kx - \omega t)$$
so $$ y = y_1+y_2 = A\sin(kx + \omega t) + A\sin(kx - \omega t) = 2A \sin(kx)\cos(\omega t) $$
I am thinking what happen if we have the two plane waves propagating along two different direction (says making angle 60 degree, i.e. the two wave are making ). I know that if that's the case, we cannot write $kx$ but we need to consider the $k$ is a vector such that
$$y_1 = A\sin(\vec{k}\cdot\vec{r} + \omega t), y_2 = A\sin(\vec{k}\cdot\vec{r} - \omega t)$$
But if we look at the horizontal direction (i.e. x) and vertical direction (i.e. y), what can we tell about the resulting wave along x and along y? I am thinking from physical point of view, if we look at the horizontal direction, should the waves still added up to a standing wave because the x components of waves are propagating in opposite direction. But along the vertical direction, the y components of waves are propagating in the same direction so there is no standing wave. Is that correct? If so, how to prove that in math? The term $\vec{k}\cdot\vec{r}$ is very confusing!
Answer
The question is unclear, but I believe can be summarized as "can standing waves form from plane waves that propagate at some arbitrary angle to each other?"
A standing wave is most easily understood in one dimension, and can be described by the equation. $$ u = A\cos(k x)\cos(\omega t) $$
It's a simple product-sum trig identity, which can be found on this page that relates the standing wave to the waves propagating in opposite directions.
$$ 2A\cos(k x)\cos(\omega t)= A[\cos(kx -\omega t)+\cos(-kx - \omega t)] $$
In a scalar formulation, it's convenient to define the positive and negative direction of propogation through the negative $\omega t$. Since we'll be working in a vector formulation with $k$, it's easier to show the direction through the sign on $k$. There could also be an arbitrary phase.
Now to show that something like a standing wave could occur in two dimensions (easily generalized to 3 dimensions), it's easier to use complex exponentials to represent the waves. Adding the two waves (q is the wavenumber of the second wave): $$ Ae^{i(\mathbf{k \cdot r}-\omega t)} + Ae^{i(\mathbf{q \cdot r} - \omega t)} = Ae^{ik_{y}y}e^{i(k_{x}x-\omega t)}+Ae^{iq_{y}y}e^{i(q_{x}x-\omega t)} $$ If the y component of the wavenumber is identical for both waves, then the y component can be combined with the amplitude to form a complex amplitude common between both waves, with a phase that depends on y. $$ Ae^{ik_{y}y}(e^{i(k_{x}x-\omega t)}+e^{i(q_{x}x-\omega t)}) $$ Reverting back to a trig representation, and ignoring the y dependent phase: $$ A[\cos(k_{x}x-\omega t)+cos(q_{x}x-\omega t)] $$ You should be able to recognize that this is a standing wave if the x components of the wavenumber are equal magnitude, but opposite direction. This suggests that two plane waves with common phase, that have the same amplitude of wavenumber, will in fact make a standing wave when viewed within a particular plane.
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