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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!

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  • $\begingroup$ Use $\vec{k}\cdot \vec{r} = |k| |r| \cos \theta$ where $\theta$ is angle between the waves. $\endgroup$ – ja72 Jun 30 '13 at 12:35
  • $\begingroup$ I think the 2 waves should be $y_1=A\sin(\vec{k}\cdot\vec{r} + \omega t)\vec{e}_{\vec{k}}$ and $y_2 = A\sin(\vec{q}\cdot\vec{r} + \omega t+\phi)\vec{e}_{\vec{q}}$ $\endgroup$ – Paracosmiste Jun 30 '13 at 15:26
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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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  • $\begingroup$ Hi David, thanks for the detail explanation. I know that question might be vague since it is pretty confusing to me when I type the question. Anyway, your math looks pretty straightforward to combine the y component to the amplitude. Since you are using complex number, I think in the real physical case, we only look at the real part, right? So the wave should look like $A*\cos(k_yy)[\cos(k_x x-\omega t) + \cos(k_x x + \omega t)]$, so it is a standing wave along the x direction but the amplitude is y-dependent instead of constant. Am I understanding it correctly? $\endgroup$ – user1285419 Jul 2 '13 at 23:03
  • $\begingroup$ So if I two sine waves (with same magnitude of wavevector) propagating on x-y plane and making 60 degree with each other i.e. first $\vec{k}$ making 60 degree with the positive x, another $\vec{k}$ making 120 with the positive x, what should I see if I visualize it? Should I see a standing wave on x axis with the amplitude is A (since on x-axis, y=0, so $\cos(k_yy)=1$. So is that's the case, what's the $\cos(k_yy)$ really affecting? $\endgroup$ – user1285419 Jul 2 '13 at 23:17
  • $\begingroup$ You need to be careful how and when you take the real portion. If you're using a complex function to model real phenomena, and have an equation like (complex function) x (other complex function) = (complex answer), you have to take the real portion of the final answer, and not any of the intermediate steps. When you multiply complex exponentials, the only effect is a change in phase. It can't create a change in amplitude like you've described. $\endgroup$ – David Jul 2 '13 at 23:26
  • $\begingroup$ It is pretty confusing when explain things in complex. So I just want to know if I have two waves like what described above, should I see one horizontal standing wave (along x and y=0)? $\endgroup$ – user1285419 Jul 2 '13 at 23:41
  • $\begingroup$ Sorry, my original conclusion was wrong. I've fixed it. You will actually see something like a standing wave, as long as the angle between the two waves isn't 0. $\endgroup$ – David Jul 3 '13 at 4:39
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The other answer are good, but this might help you visualise the result. It is easy to produce visualizations if you have access to a package like Mathematica (you could also do this with python+matplotlib, gnuplot or Matlab or just about anything really). I've generated plots of two waves in 2D, one going in the positive $x$ direction and the other going at an angle $\theta$ relative to the $x$ axis. The amplitudes, wavelengths and frequencies are the same. Here is the code:

wave1[x_, y_, t_] := Sin[x - t];
wave2[x_, y_, t_, \[Theta]_] := Sin[Cos[\[Theta]] x + Sin[\[Theta]] y - t];
frames[\[Theta]_] := frames[\[Theta]] = Table[Plot3D[wave1[x, y, t] + wave2[x, y, t, \[Theta]],
    {x, -10, 10}, {y, -10, 10}, PlotLabel -> "\[Theta] = " <> ToString[\[Theta]]], {t, 0, 10}];
Table[Export["twowaves_\[Theta]_" <> ToString[\[Theta]] <> ".gif", frames[\[Theta]]], {\[Theta], 0, 2 \[Pi], 0.5}]

Selected plots shown below. Note that the sum of waves simplifies to

$$ \sin(x-t)+\sin(\cos(\theta)x+\sin(\theta)y-t)=2 \cos\left(\frac{1}{2} x \left(\cos\theta-1\right)+\frac{1}{2} y \sin\theta\right) \sin\left(\frac{1}{2} x \left(\cos\theta+1\right)+\frac{1}{2} y \sin\theta-t\right). $$

You only get a standing wave if the space and time dependence seperates. So you need the $x$ and $y$ terms in the $\sin$ to vanish. This requires $\cos\theta=-1$ and $\sin\theta=0$, which has the unique (up to $2\pi$) solution $\theta=\pi$. So you only get standing waves if the two waves are counter propagating. Every other case gives you a travelling wave (the $\sin$ term) modulated by a space-dependent amplitude (the $\cos$ term).

Both waves in the positive $x$ direction:

enter image description here

Wave 2 going slightly up and to the right:

enter image description here

Wave 2 going nearly 90 degrees to wave 1:

enter image description here

Wave 2 nearly opposite wave 1:

enter image description here

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  • $\begingroup$ I like your visualizations but as an answer it doesn't really stand for itself, which is a shame. Some additional explanations of what is going on and how it relates to the question could really make this a great answer. $\endgroup$ – Thriveth Jul 3 '13 at 9:46
  • $\begingroup$ I also make a simulation with matlab but why this simulation based on mathematica is not a standing wave? $\endgroup$ – user1285419 Jul 3 '13 at 16:02
  • $\begingroup$ @user1285419 See the edits. Hope that helps. $\endgroup$ – Michael Brown Jul 4 '13 at 0:37
  • $\begingroup$ Thanks Michael. It works now. May I ask what version of mathematica are you using? The code doesn't really work in my mathematica 9. But I try to use Animate instead, it works but doesn't look smooth. $\endgroup$ – user1285419 Jul 4 '13 at 5:07
  • $\begingroup$ I'm using version 9 on a mid range laptop. I used ListAnimate before exporting the animated gifs. The display in the notebook wasn't very smooth but the exported gifs were fine. Mathematica probably just has an uber heavy rendering front end for animations so I prefer to export. $\endgroup$ – Michael Brown Jul 4 '13 at 5:14
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The first thing you should probably do, just to avoid confusion, is to change the names of your functions. With $y_1$ and $y_2$ there's a possibility to start mixing things up. Furthermore, the $\vec{k}$ and $\vec{r}$ vectors of the first function should be distinct from those of the second function. So let's define

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

where I've given the $\vec{k}$ vector of the second function the symbol $\vec{q}$, just to avoid having to use double subscripts later on. Note also that I didn't distinguish between the $\omega$'s of both functions, so we assume that $|\vec{k}|=|\vec{q}|$. Lastly, I've dropped the amplitude for notational simplicity.

Written differently, the above equations read (assume 2D)

$$\begin{align} f_1(x,y,t) &= \sin{(k_xx+k_yy+\omega t)} \\ f_2(x,y,t) &= \sin{(q_xx+q_yy-\omega t)} \\ \end{align}$$

The sum of these functions gives

$$\begin{align} f_1+f_2 &= \sin{(k_xx+k_yy+\omega t)} + \sin{(q_xx+q_yy-\omega t)} \\ &= 2\sin{\left(\frac{k_xx+k_yy+q_xx+q_yy}{2}\right)}\cos{\left(\frac{k_xx+k_yy+\omega t-q_xx-q_yy+\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(k_x+q_x)x+(k_y+q_y)y}{2}\right)}\cos{\left(\frac{(k_x-q_x)x+(k_y-q_y)y+2\omega t}{2}\right)} \\ &= 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}+\omega t\right)} \end{align}$$

Note that this reduces to the case of simple standing waves if $\vec{k} = \vec{q}$. You can rewrite this using the sum, yielding (with $f=f_1+f_2$)

$$f = 2\sin{\left(\frac{(\vec{k}+\vec{q})\cdot\vec{r}}{2}\right)}\cos{\left(\omega t\right)} \left[\cos{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}-\tan{(\omega t)}\sin{\left(\frac{(\vec{k}-\vec{q})\cdot\vec{r}}{2}\right)}\right]$$

From both expressions, it is clear that standing waves are only possible if the wavevectors are in fact equal. If not, there will be a modulation of the standing wave given by the factor between bracket in the last equation, which is a function of both $\vec{r}$ and $t$. The reason why your physical reasoning failed is because of our constraint $|\vec{k}| = |\vec{q}|$. Indeed, the only way in which a standing wave could arise along $x$ e.g. would be if $k_x = q_x$, but because $|\vec{k}| = |\vec{q}|$ this must also mean $k_y = \pm q_y$ and therefore $\vec{k} = \vec{q}$ or $\vec{k} = \vec{p}$ where $\vec{p}$ corresponds to the minus sign. This $\vec{p}$-vector has the same length as $\vec{q}$ but it makes an angle of $\pi-\theta$ with the positive $x$-axis, if $\vec{q}$ makes an angle of $\theta$.

If we hadn't put in that constraint, we would have had to consider distinct frequencies $\nu \neq \omega$ because $\omega/|\vec{k}| = c_1 = c = c_2 = \nu/|\vec{q}|$ must hold. This would have yielded a dependence upon $t$ for the sine in the second-to-last equation as well, making standing waves impossible again, unless $\omega = \nu$, dropping us back into our constraint. The only way out seems to be if $c_1 \neq c_2$, but that's not a physical situation.

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  • $\begingroup$ hi Wouter, thanks for the detail explanation, it takes me sometimes to repeat the math but it is good enough. I have one question: if we have a wave vector directed at some angle with x axis on x-y plane, we will have two component $k_x$ and $k_y$ as you shown in the math. I am wondering in term of component on x and y axis, will the wave still be sinusoidal? If so, will the frequency of each component wave is $\omega$ also? $\endgroup$ – user1285419 Jul 1 '13 at 3:23
  • $\begingroup$ A useful thing to remember when trying to figure out what any function looks like along the $x$- or $y$-axis, is that $y=0$ on the $x$-axis and $x=0$ on the $y$-axis. If you put $y=0$ for example in $f_1$ or $f_2$, you see it is indeed still a sinusoidal function with frequency $\omega$. $\endgroup$ – Wouter Jul 1 '13 at 8:37
  • $\begingroup$ Thanks for explaining the doubt on frequency. I am working out the math you show above. There you stated when $\vec{k}=\vec{q}$, it reduces to a simple standing wave. I am thinking two wavevectors the same doesn't mean they propagate the same direction, because it is $\vec{k}/\omega$ tell the velocity not the $\vec{k}$, right? So what is the physical meaning when we said $\vec{k}=\vec{q}$? $\endgroup$ – user1285419 Jul 1 '13 at 18:14
  • $\begingroup$ Sorry to bother again, it is a bit confusing from the math. As my understanding, if we assume $|\vec{k}|=|\vec{q}|$, we concludes that $\sqrt{k_x^2+k_y^2} = \sqrt{k_y^2 + q_y^2}$ but does it imply $k_x=q_x$ and $k_y=q_y$? If not, how do we know that the resulting wave is simple standing? $\endgroup$ – user1285419 Jul 1 '13 at 18:30
  • $\begingroup$ Well, the physical meaning of the wavevector $\vec{k}$ is that it tells us the direction the wave is travelling in and gives us information about the wavelength ($|\vec{k}|=2\pi/\lambda$). The speed of the wave is given by $\omega/|\vec{k}|$ but its direction is determined by $\vec{k}$. So when we say $\vec{k}=\vec{q}$ it does indeed mean the waves are travelling in the same direction and moreover they have the same wavelength. I've just noticed a little mistake I made by the way. I have $|\vec{k}|\omega=c$ while it should be $\omega/|\vec{k}|$. I'll change that. $\endgroup$ – Wouter Jul 2 '13 at 1:42

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