Clarification on propagation directions of wave equations

  • Thread starter Thread starter Sandbo
  • Start date Start date
  • Tags Tags
    Propagation Wave
AI Thread Summary
The discussion centers on the propagation direction of the wave equation ψ(x) = A{e^{-j(kx - ωt)}}. It is clarified that the wave propagates in the positive x direction due to the form of the equation, where the term (kx - ωt) indicates movement in that direction. The conversation also addresses potential confusion arising from different representations of wave equations, particularly regarding the signs of the terms involved. A key point made is that the phase velocity v is determined by the relationship v = ω/k, which remains consistent regardless of the sign of k and ω. Ultimately, the understanding of wave propagation direction is reinforced through mathematical reasoning and the relationship between time and spatial variables in the wave equation.
Sandbo
Messages
18
Reaction score
0
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.
 
Physics news on Phys.org
\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:
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
Back
Top