Phase difference between two waves in opposite directions

Question

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 ?

Answer 1

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.

Answer 2

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.

Answer 3

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.