Clarification on propagation directions of wave equations

  • Thread starter Thread starter Sandbo
  • Start date Start date
  • Tags Tags
    Propagation Wave
Click For Summary
SUMMARY

The discussion centers on the propagation direction of wave equations, specifically the equation ψ(x) = A{e^{-j(kx - ωt)}}, which propagates in the positive x direction. The participants clarify that the sign of the wave number k and angular frequency ω determines the direction of wave propagation. The general solution for wave equations is expressed as f(x - ct) + g(x + ct), indicating waves traveling in both directions. The phase velocity is defined as ω/k, reinforcing the relationship between the signs of k and ω and the direction of wave travel.

PREREQUISITES
  • Understanding of wave equations and their mathematical representations
  • Familiarity with complex exponentials in wave mechanics
  • Knowledge of phase velocity and its calculation
  • Basic principles of Fourier analysis and wave propagation
NEXT STEPS
  • Study the derivation of wave equations in classical mechanics
  • Learn about the implications of wave number (k) and angular frequency (ω) on wave behavior
  • Explore the concept of phase velocity in different media
  • Investigate the mathematical formulation of wave superposition and interference
USEFUL FOR

Students and professionals in physics, particularly those focusing on wave mechanics, signal processing, and applied mathematics. This discussion is beneficial for anyone seeking to deepen their understanding of wave propagation and its mathematical foundations.

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:
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
19
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
27
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K