High School Understanding Phase Terminology in Wave Functions

[etotheipi]

Almost everywhere I've looked uses the term phase in different contexts, so I was wondering if someone could let me know whether I've used the right terminology for the different concepts.

For a wave $y = A\sin(kx-\omega t + \varphi)$, I was under the impression that

• The phase, $\phi = (kx-\omega t + \varphi)$ or equivalently its principal value, $(kx-\omega t + \varphi) \text{ mod } 2\pi$. That is, the phase is the angle within the periodic function, so we could say $y=Af(\phi(x,t))$ if $f$ is the periodic function.
• The phase shift is $\varphi$, and is an increment in the phase from a specified reference
• The phase difference $\Delta \phi$ is the difference in phase between two waves, i.e. $\Delta \phi = \phi_{2} - \phi_{1}$. That is, the difference between the angles inside the periodic function.

It's a little confusing since some refer to $\varphi$ as a phase (which can, I suppose, be understood from context) however it seems more consistent to call it a phase shift. Often the symbol $\phi$ is often used to represent what I've called the phase shift, though I'm more concerned about the names than the arbitrary symbol designations.

Is this vaguely correct? Thank you.

 

[Dale]

What you wrote sounds good. However, usage is not very consistent, so don’t get too locked into your specific meaning. In my field, for instance, we always have a phase coherent reference signal. So when we say “phase” we usually mean what you have called “phase difference” with respect to the reference signal.

 

[etotheipi]

What you wrote sounds good. However, usage is not very consistent, so don’t get too locked into your specific meaning. In my field, for instance, we always have a phase coherent reference signal. So when we say “phase” we usually mean what you have called “phase difference” with respect to the reference signal.

Yeah, it appears some mental flexibility is required.

Which makes it a little more difficult for learners like myself, since I need to worry about the actual physics at hand in addition to the convention being employed by the author!

Luckily it shouldn't be too difficult to translate back and forth since the context is usually fairly evident, though it would be nice if everyone could agree! But such is life...

 

[etotheipi]

@Dale I wonder whether you could help with one other clarification.

If two rays interfere at a particular point, we can compute the difference of the phases of the two waves at that particular location, $\Delta \phi$, by considering optical path lengths & initial phases etc., even if the two waves are not parallel or along the same axis. Even if the waves are traveling in the opposite direction but meet at a point $P$, we can calculate $\phi_{1}(x_{p}, t)$ and $\phi_{2}(x_{p},t)$ and then determine what type of interference results.

If, however, two rays of equal wavelength are traveling along the same axis in the same direction, we will often say the waves themselves have a phase difference of $\varphi$ (i.e. if $y_{1} = A\sin{(kx-\omega t)}$ and $y_{2} = A\sin{(kx-\omega t + \varphi)}$). But is saying that two waves have a phase difference of $\varphi$ shorthand for saying $\phi_{2}(x, t) - \phi_{1}(x, t) = \varphi, \forall x$? That is, we're comparing the phase of points on either wave at the same location in space?

Apologies for the wall of text! And thank you for your help!

 

[anorlunda]

we can only refer to the phase difference between two waves if they have the same wavelength

If they have different frequencies, then the phase shift is a function of time. radians/second has the units for rate of change of phase, and the units of frequency.

 

[etotheipi]

If they have different frequencies, then the phase shift is a function of time. radians/second has the units for rate of change of phase, and the units of frequency.

Though wouldn’t the phase difference for two waves of a different frequency along the same axis also depend on the position along the axis, as well as time? In that sense we can’t say the two waves have a phase difference, only at specific positions and times.

 

[anorlunda]

Think about this whole thread. Why does science and engineering speak in mathematics rather than natural language? Your math is correct, but the language confounds you and everyone else. Remember what @Dale said about consistency. If you want sharp definitions, use math, not language.

 

[etotheipi]

Think about this whole thread. Why does science and engineering speak in mathematics rather than natural language? Your math is correct, but the language confounds you and everyone else. Remember what @Dale said about consistency. If you want sharp definitions, use math, not language.

Very true, I'm developing a bit of a habit of being a little pedantic over inconsequential things but it's mainly because I often find that it's easier to internalise something if I can explain it in a coherent way.

But you're right, everyone agrees about the maths.

 

[Dale]

If, however, two rays of equal wavelength are traveling along the same axis in the same direction, we will often say the waves themselves have a phase difference of $\varphi$ (i.e. if $y_{1} = A\sin{(kx-\omega t)}$ and $y_{2} = A\sin{(kx-\omega t + \varphi)}$). But is saying that two waves have a phase difference of $\varphi$ shorthand for saying $\phi_{2}(x, t) - \phi_{1}(x, t) = \varphi, \forall x$?

I would say yes. Although maybe restricting $x$ to some understood region of interest.

 

[etotheipi]

I would say yes. Although maybe restricting $x$ to some understood region of interest.

Yes that's a good point, we could have two beams split and then only recombine later on before they feed into a detector, such that we're only interested the $x$ values within this final interval.