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Suppose I have two waves travelling along the positive and negative $x$ axis, and are given by : $$y_1=A\sin(kx-\omega t)\,\,\,\,,\,\,\,y_2=A\sin(kx+\omega t)$$

What would be the phase difference between these two waves at a particular point ?

If I define the phase difference as the difference between the arguments, then I get :

$$\Delta \phi=kx+\omega t-(kx-\omega t)=2\omega t$$

But, I could have easily defined the waves, by keeping a positive sign in front of $\omega t$ instead of $kx$. So in that case, my arguments would have become $\omega t-kx$ and $\omega t+kx$ instead. In this case, the phase difference at any point comes out to be :

$$\Delta\phi=\omega t+kx-(\omega t-kx)=2kx$$

At any value of $x=x_0$, this phase difference is constant.

So, I get two contradictory answers here. In the previous case, the phase difference at any point, varied over time. In the second case, this phase difference was constant at a given point, and varied from point to point.

Which one is correct, and how should I know, which one to choose, in situations such as these ?

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    $\begingroup$ Some musings: Are you sure it makes sense to talk about phase differences for waves traveling in opposite directions? I am struggling to think of a meaningful definition in this case. Maybe if you looked at $y_1+y_2=2A\sin(kx)\cos(\omega t)$? From this expression it looks like some "phase" would vary independently in space and time. $\endgroup$
    – Michael M
    Jan 21 at 13:07
  • $\begingroup$ @MichaelM yeah, I understand. The doubt comes from a type of problem in interference, where there are two hypothetical coherent sources on either side of a flat line, and a detector moves along this line to note the interference. Now, the argument used here, is that, if the distance between two sources is $l$, then at a distance $x$ from one of the sources, the light from one source travels $x$, while light from the other source travels $l-x$, and so there is a path difference. If this difference is $n\lambda$, we see a bright fringe, else a dark one. $\endgroup$
    – RayPalmer
    Jan 21 at 13:36
  • $\begingroup$ @MichaelM this argument seemed too simplistic and somewhat wrong, because, first of all, they are ignoring the fact that the phase changes over time, as well as distance. That is why I tried to derive a general phase difference at a point, using the two wave equations separately. However, as you can see, I'm struggling with this quite a bit, as I'm unable to find a proper definition of the phase as well. $\endgroup$
    – RayPalmer
    Jan 21 at 13:37

3 Answers 3

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Phase difference as a constant, independent on time, can be defined only between two waves with the same wave vector and frequency, which is not the case in the example given in the OP, where the waves propagate in the opposite directions.

More generally, the phase difference is defined between two points in space and time. E.g., if we have waves $$y_1(\mathbf{x},t)=\cos\phi_1(\mathbf{x},t), y_2(\mathbf{x},t)=\cos\phi_2(\mathbf{x},t),$$ we could define a phase difference between points $\mathbf{x}_1,t_1$ and $\mathbf{x}_2,t_2$ as $$\Delta \phi(\mathbf{x}_1,t_1;\mathbf{x}_2,t_2 )=\phi_1(\mathbf{x}_1,t_1)-\phi_2(\mathbf{x}_2,t_2).$$

Thus, the phase differences defined in the Op correspond to two different cases:

  • same space point, but different time
  • same time point, but different locations in space

Remark One also has to agree about what is considered as a positive/negative frequency and the phase - the paradox in the OP might be simply due to exploiting even symmetry of the cosine function.

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  • $\begingroup$ Are you saying that, in a sum, if I choose a certain wave travelling in a given direction to have $\cos(kx-\omega t)$ form, then all waves in that given direction must have a positive $kx$ and negative $\omega t$ in order for us to calculate some phase difference or something ? $\endgroup$
    – RayPalmer
    Jan 23 at 16:35
  • $\begingroup$ For example, if I define a wave as $\cos(kx-\omega t)$, then I can also write this as $\cos(\omega t-kx)$ by exploiting the symmetry. Now if I try to find the phase difference between these two waves, I'd get $2kx$ or $2\omega t$, which should both be wrong, since it is the same wave, and must therefore have $0$ or $2n\pi$ phase difference. $\endgroup$
    – RayPalmer
    Jan 23 at 16:37
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    $\begingroup$ @RayPalmer to find the phase difference you need to define phase: $kx -\omega t +\phi$ or otherwise. But if you use two different definitions for the same qyantity, you risk obtaining misleading results. $\endgroup$ Jan 23 at 16:51
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I could have easily defined the waves, by keeping a positive sign in front of ωt instead of kx.

Actually, in this case you cannot do that. Here you have defined the waves in terms of $\sin$ functions. So $\sin(kx-\omega t)$ is not the same wave as $\sin(\omega t - k x)$.

However, you could have asked the question in terms of $\cos$, and in that case $\cos(kx-\omega t)$ is the same wave as $\cos(\omega t - k x)$. In that case you would indeed get your two different scenarios.

I get two contradictory answers here. In the previous case, the phase difference at any point, varied over time. In the second case, this phase difference was constant at a given point, and varied from point to point.

So, using $\cos$ waves you do get two different answers, but they are not contradictory, they are completely equivalent. There is no difference between a wave with spatially varying phase whose amplitude changes in time and a wave with temporally varying phase whose amplitude changes in space. Those are just two equivalent ways of describing the same wave pattern.

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  • $\begingroup$ yes, but suppose I take two waves $\cos(kx-\omega t)$ and $\cos(\omega t-kx)$. These are the same wave, and if I substract the phases, I'd get either $2kx$ or $2\omega t$ depending on which one I'm substracting from which one. However, according to you, both are equivalent, but aren't they both wrong ? Since these are the same wave, shouldn't I get $0$ or $2n\pi$ phase difference ? $\endgroup$
    – RayPalmer
    Jan 23 at 16:34
  • $\begingroup$ @RayPalmer you could say that they are both wrong or both right. It doesn’t particularly matter. They sum to the same thing either way, and either way the phase difference is simply not a particularly important feature of this combined wave. $\endgroup$
    – Dale
    Jan 23 at 16:58
  • $\begingroup$ can I then say something like, in order for phase difference to make sense, I have to define the phase in the same exact way for all the waves in question i.e. either $kx-\omega t +\theta$ or $\omega t-kx+\theta$. Only then it would make sense. In case of two opposite waves, this is not true as the two waves a difference definiton of phase. Is that the source of this paradox ? $\endgroup$
    – RayPalmer
    Jan 23 at 17:01
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I believe that both answers are correct and they actually have the same meaning/interpretation.

The second one is more intuitive in the sense that most people in the field of acoustics and/or waves in general, are quite familiar with standing waves, where the phase difference between the two (monochromatic) waves is space/distance dependent.

Now, the first case may be somewhat unintuitive. There is no notion of space here but there is time. This seems like the phase difference between two monochromatic waves changes with time, which for static sources (which they can also be positioned at $+\infty$ or $-\infty$ as long as they don't move) doesn't make sense. Yet, we haven't considered the fact that we are talking about progressive waves. This means that as time passes, the spatial coordinate also changes. In this case it does make sense for the phase difference to vary with time.

This may be more obvious if we look at the argument as presented in D'Alembert's solution which is

$$u(x, t) = f(x - ct) + g(x + ct)$$

where $f$ and $g$ are arbitrary functions with their argument having this specified form $x - ct$ and $x + ct$.

This solution describes one dimensional progressive waves, traveling with speed $c$, as does the function you provided (the $\sin$ function). Even when we talk about standing waves, we "compose" them with progressive waves.

The key word here is progressive, which actually intertwines the spatial and the temporal dimensions through speed. Thus for static conditions, either in space or time (in essence they both have the same effect) both solutions are valid since when you stop time you get both $t = t_{1}$ and $x = x_{1}$, where $t_{1}$ and $x_{1}$ are arbitrary time and point in space.

I believe that without specifying initial and/or boundary conditions you won't be able to get a specific phase difference formula. This is the case for the formula provided by Michael M in their comment.

So, to conclude the answer, I believe that the correct answer for your case is neither of them! What you seek is found in the formula provided by Michael M in their answer. The equation providing the sum of the two progressive waves is (using Michael M' formula directly and adapting the notation a bit)

$$ y_{1}(x, t) + y_{2}(x, t) =2 A \sin(kx) \cos(ωt) $$

Where you can see that, as Michael M states, the phase difference is related to both time and space, so this equation does not describe a progressive wave. Please not in the above equation that this is the resulting equation in the case both progressive waves that create this pattern (standing wave) have equal amplitude $A$.

Thus, using this equation you can find the phase difference both in space and time.

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