Clarification on propagation directions of wave equations

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Discussion Overview

The discussion revolves around the propagation directions of wave equations, specifically analyzing the mathematical forms of wave functions and their implications for directionality. Participants explore the relationship between the wave equation parameters and the resulting wave propagation direction, engaging in both conceptual clarification and technical reasoning.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the propagation direction of the wave described by the equation \(\psi (x) = A{e^{-j(kx - \omega t)}}\), suggesting it propagates in the positive x direction based on the form of the cosine function derived from the exponential.
  • Another participant points out that the phase velocity of the wave is given by \(\omega/k\) and notes that this can also be expressed as \((- \omega)/(-k)\), implying a relationship between the signs of \(\omega\) and \(k\) and the direction of propagation.
  • Concerns are raised about a potential typo in lecture notes regarding the general solution of wave equations, with a suggestion that the correct form should represent waves traveling in both directions.
  • A participant expresses confusion about the relationship between the negative sign in the term \(-\omega t\) and the resulting positive direction of wave propagation, seeking clarification on how this leads to a positive velocity.
  • One participant attempts to reason through their own question, suggesting that for a wave described by \(A \cos(kx - \omega t)\), an increase in time implies a shift of the wave to the right, supporting the idea of positive x direction propagation.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the wave equation forms and their propagation directions. There is no consensus on the interpretation of the signs in the equations and their effects on directionality, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants reference various forms of wave equations and their solutions, but there is uncertainty regarding the implications of specific terms and signs. The discussion highlights the complexity of wave propagation concepts and the need for careful consideration of mathematical expressions.

Sandbo
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I would like to ask some questions to confirm if what I have acquired so far are correct, please point out my faults if any.
If all the below statements are too bulky to read, my question is actually as short as:
\psi (x) = A{e^{-j(kx - \omega t)}}
What is the propagation direction of this wave? Positive x or negative x? Why(which terms tell this)?



Regarding the propagation direction of wave equations, to the best of my understanding, in the following wave equation:
\psi (x) = A{e^{ - j(kx - \omega t)}}, let B = j(kx - \omega t)
The direction of the wave propagation is towards the positive x direction (+ve), for the following reasons:
1. when we consider the exponential, we are actually taking only the real part of it, which is \cos (kx - \omega t).
As \cos x = \cos ( - x), therefore no matter the sign before B is + or - (i.e. \psi (x) = A{e^{j(kx - \omega t)}}{\rm{ }}or{\rm{ }}\psi (x) = A{e^{ - j(kx - \omega t)}}), the propagation direction will not change, remains as positive x direction.

However, from a lecture note I recently read, I saw something kind of contradicting to the above:
\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ - j(kx - \omega t)}}
"Consider a free particle that travels only in the positive direction of x. Let the arbitrary constant B be zero."
It seems the negative sign on the latter term told us that it's propagating towards negative x direction and that is omitted, which is not consistent with my previous understanding of wave equations.

I am sorry if I have put too much redundancy here, simply trying to make it as clearly as I can.
Thanks for reading.
 
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\psi (x) = A{e^{j(kx - \omega t)}}

The [phase] velocity of the wave given by the equation above is ω/k. You may notice that ω/k = (-ω)/(-k).
 
Last edited:
Sandbo said:
However, from a lecture note I recently read, I saw something kind of contradicting to the above:
\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ - j(kx - \omega t)}}

I think there is a typo here. The general solution is of the form f(x-ct) + g(x+ct) for arbitrary functions f and g, representing waves traveling in each directiion. Your notes should be
\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ j(kx + \omega t)}}
or something equivalent to that.
 
MisterX said:
\psi (x) = A{e^{j(kx - \omega t)}}

The [phase] velocity of the wave given by the equation above is ω/k. You may notice that ω/k = (-ω)/(-k).

Thanks for the clarification,
from the phase velocity then it means the propagation direction is determined by the sign difference between k and \omega.

AlephZero said:
I think there is a typo here. The general solution is of the form f(x-ct) + g(x+ct) for arbitrary functions f and g, representing waves traveling in each directiion. Your notes should be
\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ j(kx + \omega t)}}
or something equivalent to that.

Also, yes I think that is a typo after reading more, thanks for pointing out:biggrin:
 
Last edited:
MisterX said:
\psi (x) = A{e^{j(kx - \omega t)}}

The [phase] velocity of the wave given by the equation above is ω/k. You may notice that ω/k = (-ω)/(-k).

Allow me to have one follow up question,
why it should be kx-wt ending up with a relationship of v = (positive w) / (positive k)?
The sign before w is a negative sign(-).:confused:

And in the same way when we consider a wave Acos(kx+wt), we say v = k/-w and the wave propagate in -x direction.

Would you share more on where does the conclusion [(kx-wt) = positive x direction] stem from?
Thank you.
 
Sandbo said:
Allow me to have one follow up question,
why it should be kx-wt ending up with a relationship of v = (positive w) / (positive k)?
The sign before w is a negative sign(-).:confused:

And in the same way when we consider a wave Acos(kx+wt), we say v = k/-w and the wave propagate in -x direction.

Would you share more on where does the conclusion [(kx-wt) = positive x direction] stem from?
Thank you.

Trying to ans own question:
for a wave A = Ao cos(kx-wt),
in case of an increased t, from the relation (kx-wt), the particle possessing same amplitude should have been shifted to the right(in (kx-wt), x is also increased to oppose the change of t making (kx-wt) a constant to keep the amplitude).
With all particles having the same, the waveform will therefore move in +x direction.

This is phenomenal explanation I can come up with.
Please let me know in case there is a concrete mathematics deviation which I will be happier with.:biggrin:
 

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