# Can anybody help with group velocity simulations?

1. Jul 17, 2014

### mysearch

On first reading, the description of ‘group velocity [vg]’ appears to be quite straightforward. However, I also found a number of speculative explanations as to ‘how’ and ‘why’ the group velocity may exceed the ‘phase velocity [vp]’. Therefore, in order to get a better intuitive understanding of the issues, I am ‘attempting’ to simulate 4 different types of beat wave configurations:

• 1-way, non-dispersive phase waves
• 2-way, non-dispersive phase waves – see 2ND.gif attachment
• 1-way, dispersive phase waves
• 2-way, dispersive phase waves – see 2D.gif attachment

Each simulation uses the same basic approach in which the amplitude of two individual phase waves is added, and calculated, for all values of [x], which is then displayed as a single frame in the animation. The process is then repeated for incrementing values of time [t] to create the next frame within the animation. While the first simulation in the list above produced ‘sensible’ results, i.e. [vg=vp], all the subsequent permutations, based on the same algorithms, lead to ‘unexplained’ results. For example, the second simulation changes only the direction of one of the phase waves, but appears to suggest a group velocity that is 10x the phase velocity, although the actual value calculated was [vg=1]. My initial lines of thoughts:

• There is an mistake in the simulation. If so, it is not obvious to me, but see next bullets.

• The equation used to calculate [vg=dw/dk] might have to accommodate the direction of the phase wave velocity [vp]. For example, given that [vp=w/k], such that [k=w/vp], then treating [vp] as a vector having magnitude and direction might suggest that the sign of [k] must also be direction dependent. However, it is unclear whether this would explain the anomalous result associated with the 1-way simulations.

• The equation [vg=dw/dk] would be greater than [vp], when [dw>dk], which appears possible in dispersive media – see equation [1] in attached pdf for details. However, it is not clear this would explain the non-dispersive cases.

• The simulations are only showing an 'pattern shift'. For example, when the phase waves propagate with velocity [c] in opposite directions, the phase relationship between these waves constantly changes with time, i.e. on every frame. So while the ‘shift’ in the beat waveform might suggest a propagation velocity greater than [c], it has nothing to do with anything propagating through the media, rather it simply reflects the rate of change in phase shift between the phase waves.

I have attached ‘simulations.pdf’ in order to give some more details of the simulation results for anybody who might be interested in this topic. I have also attempted to attach two gif files to give ‘examples’ of the animations, although the upload size required 500 frames to be reduced to just 25 and so are they very jerky.

Anyway, I would appreciate any help on offer as I am sure somebody must have already resolve these issues. Thanks

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2. Jul 20, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Jul 21, 2014

### olivermsun

Since nobody seems to be picking up your questions, let me point out a couple things that I noticed about your simulation and derivation:

Yes, the phase vector $k$ is signed. Alternatively, you can view the back-propagating wave as having frequency $-\omega_2$, which would explain why you are getting an envelope velocity which looks like $\dfrac{\omega_1+\omega_2}{k_1-k_2}$ instead of $\dfrac{\omega_1-\omega_2}{k_1-k_2}.$

I believe so. Your wave phases are propagating in opposite directions, so there is no sense of any actual "group" propagating through the medium.

4. Jul 22, 2014

### mysearch

I think we have to accept that the PF is not a guaranteed service, but sometimes it is worth a shot. Often I simply post some details for my own backup and cross-reference; plus it might help somebody else who is treading the same path at some later time.

While I agree with you, working on some updates to the simulation seems to suggest that the equations above require the $\pm$ sign to be assigned to $\omega$, not k; otherwise the equations below do not seem to work for waves travelling in the same and opposite direction:

$ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{ω_p}{k_p}$

$ω_g = \dfrac{ω_1 - ω_2}{2}; k_g = \dfrac{k_1 - k_2}{2}; v_g = \dfrac{ω_g}{k_g}$

Why this is so is not clear to me, as I would argue that frequency in the form of $\omega$ is essentially direction independent, if you ignore negative time [t]. As such, it 'should' be the wavelength in the form of the wave number [k] that must change to match the requirements of the propagation media and the direction of the travelling waves in order to comply with the following equation; although the simulation suggests otherwise(!):

$v=f \lambda = \dfrac{\omega}{k} \Rightarrow \pm \omega = \pm v*k$

Anyway, I have attached some updated animations, which now show the relative phase and group velocity on the beat waves, see bottom two red and black waveforms, for the following configurations.

• 1-way, non-dispersive phase waves – see 1.gif
• 2-way, non-dispersive phase waves – see 2.gif
• 1-way, dispersive phase waves – see 3.gif
• 2-way, dispersive phase waves – see 4.gif

Given the upload limit I have restricted the animation to the first 50 frames of 500. Despite the reservations above, these animations seem to make sense, but maybe I am still missing some explanation of the issue above. Anyway, as indicated, the information posted might help somebody at a later time.

Yes, I still hold to this view. Appreciated your feedback

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5. Jul 22, 2014

### olivermsun

I don't think it should matter where you put the sign. For positive wavenumber $k$, then the wave should have the form
$$\cos(k(x-vt))$$ if it's forward propagating with phase velocity $v$, and
$$\cos(k(x+vt))$$ if it's backward propagating with phase velocity $-v$.

Then $\omega = kv$ in the forward case and $\omega = -kv$ in the backward case.

If you choose the convention that $\omega$ is positive, then the back-propagating wave can be written as
$$\cos(kx+kvt) = \cos(-kx-kvt),$$
which is the same as a forward-propagating wave with negative wavenumber $-k$ and positive frequency $\omega = -kv$.

The signs for the sum and difference waves should all work out as long as you choose one convention and stick to it.

Last edited: Jul 22, 2014
6. Jul 22, 2014

### mysearch

Again, I do not disagree with your reasoning, but my simulation results gave up 2 conflicting sets of results depending on whether using [w=v*k], which worked, as opposed to [k=w/v], which didn’t. The following equations are used in both cases below, but seem to give up different results based on the sign of $[ \omega ]$ or [k] - see results below as an example of the 2-way non-dispersive case using both approaches.

$ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{ω_p}{k_p}$

$ω_g = \dfrac{ω_1 - ω_2}{2}; k_g = \dfrac{k_1 - k_2}{2}; v_g = \dfrac{ω_g}{k_g}$

‘ Common parameters
‘-------------------------
c = 1
w0 = 0.2
k0 = w0/c = 0.2
dw0 = w0*0.1 = 0.02
dk0 = dw0/c = 0.02

‘calculate [w] from [k]
‘this appears to work
------------------------
'wave-1:
k1 = k0+dk0 = 0.22
v1 = c*1 =1
w1 = v1*k1 = 0.22
'wave-2:
k2 = k0-dk0 = 0.18
v2 = -c*1 = -1
w2 = v2*k2 = -0.18
' deltas:
wp = (w1+w2)/2 = 0.02
kp = (k1+k2)/2 = 0.2
vp = wp/kp = 0.1
wg = (w1-w2)/2 = 0.2
kg = (k1-k2)/2 = 0.02
vg = wg/kg = 10

‘calculate [k] from [w]
‘this does not appear to work
-------------------------
'wave-1:
w1 = w0+dw0 = 0.22
v1 = c*1 =1
k1 = w1/v1 = 0.22
'wave-2:
w2 = w0-dw0 = 0.18
v2 = -c*1 = -1
k2 = w2/v2 = -0.18
' deltas:
wp = (w1+w2)/2 = 0.2
kp = (k1+k2)/2 = 0.02
vp = wp/kp = 10
wg = (w1-w2)/2 = 0.02
kg = (k1-k2)/2 = 0.2
vg = wg/kg = 0.1

Maybe there a mistake somebody can see. I argued for $A=A_0cos( \omega t ± kx)$ rather than $A=A_0cos( kx ± \omega t)$ because an oscillating charged particle might be said to define the frequency, independent of the media, which then propagates through a given media with a wavelength defined by the refractive index [n] in order to maintain $[v= \omega/k]$. In this context, the only way [ $\omega$ t] is negative is when you are calculating the amplitude at some earlier point in time, which is not the issue in this simulation. Again, if somebody can spot an error it would be much appreciated as it is bugging me.

Hope this makes some sense as I am rushing at the end of my day. Thanks

7. Jul 22, 2014

### olivermsun

In both cases the analysis shows that you have a product of two waves, a short wave with $k = 0.2, w = 0.02, v = 0.1$ and a long wave $k= 0.02, w = 0.2, v = 10$. The only difference I can see is which wave you call $g$ and which wave you call $p$. Maybe I'm missing something.

8. Jul 23, 2014

### mysearch

Apologises if my rushed outline in post #6 was confusing. In an attempt to provide some greater clarity to the issue I am trying to resolve, I have attached a pdf file to this post that provides more detail. Some limited animations produced by the simulation are attached to post #4 with 2.gif showing the specific case detailed in the pdf file. While it is realised that nobody may be interested in working through all this detail, should anybody be able to resolve the apparent anomaly, it would be appreciated if they could let me know via the PF forum. Many thanks.

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9. Jul 23, 2014

### olivermsun

In your pdf you mention several times that you apply your values to 2 sets of equations and get different values of $v_p$ and $v_g$. The 2 sets of equations describe different situations, so they should get different values.

You have to choose either

a) a single equation and signed $k$ (or $\omega$), to signify the direction of propagation, or

b) or two different equations depending on propagation direction and unsigned $k, \omega$.

Also, you mention at the top of page 2 that both values appear to be wrong. Just looking at your simulation #2, it seems consistent with what I said in post #7, which is that you have a wave with $k=0.2, v=\text{small}$, modulated by a wave with $k=0.02, v=10$. What you've computed in the box is indeed choice a) above: using the single equation $\cos(\omega_1 t - k_1 x) + \cos(\omega_2 t - k_2 x)$ with a signed $k_2$, and you seem to have the correct result for when the $k=0.18$ wave is propagating to the left.

Finally, the choice of the form $\cos(\omega t\pm\ kx)$ vs. the form $\cos(kx \pm \omega t)$ makes no difference at all because
$$\cos(\omega t- \ kx) = \cos(-(\omega t - \ kx)) = \cos(kx - \omega t).$$

10. Jul 24, 2014

### mysearch

Again, appreciate the feedback as it is making me review all my assumptions. Some initial thoughts in a somewhat random order:

I do not disagree with the mathematically equivalence of the cosine functions shown. However, in isolation of the cosine, (wt-kx) is not equivalent to (kx-wt); especially when considered from the physical interpretation of a propagating wave. This, in part, was one of the points initially raised because the simulation only considers incrementing time [t]. If frequency [w] is always a positive, then [wt] must always be positive in the cases being considered. Therefore, [wt $\pm$ kx] was assumed to be the appropriate starting point. On this basis, it was assumed that [ ($\pm$ k) = w/($\pm$ v) ] not [ ($\pm$ w) = ($\pm$ v)*k ], although this appeared to lead to problems when calculating [wp,kp] and [wg,kg] as I tried to illustrate.

When I first did the derivation of the superposition beat equation, i.e.

[1] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$

I started from the form [wt-kx] for the reasons explained above:

[2] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ]$

However, it seemed equally valid to start from

[3] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t + k_2 x) ]$

Although the derivations based on [2] or [3] both lead to [1], the definitions of [kp,kg] changes in sign depending on whether [2] or [3] is used. In-line with your advice in (a), the simulation has tried to use only the equations from one derivation, i.e. as derived from [2], where it was assumed that changing the sign of [k] would change the direction of the second plane wave [w2, -k2], which it does. The results I previously gave for the second option based on [3] were erroneous as they end up only reflecting the waves propagating in the same direction. However, based on option a) and [ ($\pm$ k) = w/($\pm$ v)], the following equations were assumed to reflect the values of [wp,kp] and [wg,kg] when [-k2] reflects the second wave propagating in the opposite direction:

$ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + (-k_2)}{2}; v_p = \dfrac{ω_p}{k_p}$

$ω_g = \dfrac{ω_1 - ω_2}{2}; k_g = \dfrac{k_1 – (-k_2)}{2}; v_g = \dfrac{ω_g}{k_g}$

As stated, this did not seem to provide consistent results. However, when I changed the sign dependency to [w], not [k], i.e. [($\pm$ w)=($\pm$v)*k] everything appears to work and I believe I can physically explain the values of [vp] and [vg], see my quote below, but not necessarily [ $\pm$w].

For the reasons I think we both agree, the simulation should be consistent, i.e. based on ‘choice a)’. However, I do not believe any of the simulation results on page 2 are right, when the direction of the second plane wave was defined by [-k]. Only the configuration based on [ ($\pm$w) = ($\pm$v)*k] at the bottom on page 2 and the results on page 3 appear consistent with my interpretation of the beat superposition, i.e.
“In this specific non-dispersive case, where [v1=+1, v2=-1], the rationale for [vp=0.1] can be explained as being analogous to a standing superposition waves. For as [Δω] approaches zero, the phase velocity of the superposition wave, i.e. the higher frequency component of the beat waveform must also approach zero. While a group velocity [vg>1] appears anomalous, it is argued that the group wave envelope is not actually propagating through space-time, as it more accurately reflects the phase shift between waves-1 and wave-2 occurring at all value of [x] simultaneously in time [t]. This phase shift effect, leading to the perception of [vg>1] is higher when wave-1 and wave-1 propagate in opposite directions. However, while now having to argue for the results above, it is still unclear why these equations require the directional [±] sign to be assigned to [ω] not [k].”​

So while I have always had a simulation that works, based on option a), my main issue is that I cannot physically explain why the sign of second plane wave [w2, k2] has to be assigned to [w2] not [k2] to get the correct values of [vp] and [vg]. Maybe there is another [$\pm$] I am missing - I will keep checking but I have possibly become 'snow-blind' by looking at these equations for too long, which is why a second pair of 'fresh eyes' is helpful. Thanks

11. Jul 24, 2014

### olivermsun

I guess I still don't understand why there's any difference in physical interpretation.

$\cos(kx-\omega t) = \cos\left(k\left(x-\dfrac{\omega}{k}t \right)\right)$ is a (spatial) waveform with shape $\cos(kx)$ which is translating to the right at speed $\omega/k$ per unit time.

$\cos(\omega t - kx) = \cos \left(\omega\left(t - \dfrac{k}{\omega} x\right)\right)$ is a (time) oscillation $\cos(\omega t)$ that is translating forward in time at $\dfrac{k}{\omega}$ per unit distance.

By mathematical equality,
$\cos(\omega t - kx) = \cos(-kx + \omega t) = \cos \left(-k\left(x - \dfrac{\omega}{k} t\right)\right) = \cos \left( k \left(x - \dfrac{\omega}{k} t \right)\right)$, which is exactly a wave $\cos(kx)$ that is translating to the right at speed $\omega/k$.

You don't need to carry the $\pm$ signs everywhere. You just pick a convention, and then the signs work out.

Look at $A(x, t) = \cos(\omega t - kx)$: when $\omega$ and $k$ are both positive, then $v = \omega/k$ is also positive, so you have rightward phase propagation. If $\omega$ and $k$ have different signs, then you will have leftward phase propagation. If you decide that $\omega$ is always positive, then the sign of $k$ is enough to know whether the wave is propagating to the right or the left.

As they should. In [3] the second wave is $\cos( \omega_2 t + k_2 x) = \cos\left( k_2 x - \left(-\dfrac{\omega_2}{k_2}t \right)\right)$. This means that for $\omega_2, k_2$ both positive, the second wave is propagating to the left at $\omega_2/k_2$. This approach is fine, but then you have to use different equations for different combinations of propagation directions.

However, when you use [3], you don't "double-count" the propagation direction by also using a negative $k_2$. You use $k_1, k_2$ both positive and the leftward propagation is built into the equation.

So for example, on your page 2 in the box under
$$\cos(\omega_1 t - k_1 x) + cos(\omega_2 t + k_2 x)$$
you should have
\begin{align} k_p &= \dfrac{0.22 - 0.18}{2} = 0.02; \quad v_p = \dfrac{0.2}{0.02} = 10 \\ k_g &= \dfrac{0.22 + 0.18}{2} = 0.2; \quad v_g = \dfrac{0.02}{0.2} = 0.1 \end{align}
which would make the results consistent.

12. Jul 25, 2014

### mysearch

Many thanks for post #11, it helped me focus on what I was trying to convey when referring to the mathematical and physical interpretation of these wave equations. Of course, you may not agree! However, before commenting on the last post, can I raise a direct question regarding the result of just one specific simulation previously cited as it might also help me understand where any discrepancy lies. The example is the non-dispersive case, where the two plane waves propagate in opposite directions.

[1] $Wave-1: \omega_1=0.22; k_1=0.22; v_1=+1$
[2] $Wave-2: \omega_2=0.18; k_2=0.18; v_2=-1$

I believe we agree that the following equation is able to represent the beat superposition:

[2] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$

Where the value of [ $\omega_p, k_p, v_p$] and [ $\omega_g, k_g, v_g$] were defined by:

[3] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$

[4] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$

However, [3] and [4] were derived from [5] below, where the waves are travelling in the same direction:

[5] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ]$

As such, it would seem that if [3] and [4] are to work, the direction of wave-1 and wave-2 has to be accounted somewhere.

So what values of [$v_p, v_g$] do you think results in this specific case?
Can you also identify whether you assumed [$\omega_2$] or [$k_2$] to be negative or some other permutation?

Anyway, turning to the issue of physical interpretation.

As indicated, I found these equations to be really useful in reviewing the physical assumptions underpinning them for they explicitly help identify where the wave velocity [v] appears in these equations, e.g.

[6] $\cos(kx-\omega t) = \cos\left(k\left(x-\dfrac{\omega}{k}t \right)\right) = cos\left(k\left(x-vt \right)\right)$

[7] $\cos(\omega t - kx) = \cos \left(\omega\left(t - \dfrac{k}{\omega} x\right)\right) =\cos \left(\omega\left(t - \dfrac{x}{v} \right)\right)$

Now, in the context of [6] and [7], the vector quantity of velocity [v] that carries the direction [$\pm$] sign can be more clearly seen; although it does not necessarily explain how it should be inferred on [$\omega$] or [k], i.e.

[8] $\pm v= \pm f λ = \pm \dfrac{\omega}{k} = \pm \dfrac{x}{t}$

Mathematically, based on [8], it would seem that either [f or $\lambda$] or [ $\omega$ or k] or [x or t] may assume the [$\pm$] sign. However, I have argued that the idea of ‘negative’ frequency [f, $\omega$] has no physical meaning, while the simulation has also constrained time [t] to only positive values. As a composite quantity with the restriction of [+t], negative velocity [v] seems to only makes sense in terms of [-x/+t], but which then implies that the negative sign has to extend to [ $\lambda$ and k], although by a somewhat circular argument, it might be said that the idea of negative wavelength is only inferred from the directional velocity.

Sorry that this is a bit long-winded, but I wanted to try to explain why I was interested in the physical interpretation of these equations as much as their mathematical consistency.

I entirely agree and I believe my initial simulation, based on [1] thru [5] was doing just this, except when applying the directional sign to [k] not [$ω$] I didn’t seem to get the right values of [$v_p, v_g$] from [3] and [4], hence the specific example/question at this begin of this post to cross-check what values you think result in this case. Thanks

Last edited: Jul 25, 2014
13. Jul 25, 2014

### olivermsun

If you are going to use this equation and account for propagation direction using signed values of $\omega$ and/or $k$, then your Wave-2 is inconsistent as written.

\begin{alignat}{3} \text{Wave 1:}\quad \lvert\omega_1\rvert &= 0.22; \quad \lvert k_1\rvert &= 0.22; \quad v_1 &= +1 \\ \text{Wave 2:}\quad \lvert\omega_2\rvert &= 0.18; \quad \lvert k_2\rvert &= 0.18; \quad v_2 &= -1 \end{alignat}
which implies that exactly one of $\omega_2$ or $k_2$ is negative.

If signed values of $\omega, k$ are allowed, then the waves in [5] do not necessarily travel in the same direction. Instead, $v_p, v_g$ in [3] and [4] have signs which are determined by the signs of $\omega_p, k_p$ and $\omega_g, k_g$.

I assumed $k_2$ to be negative $(k_2 = -\lvert k_2 \rvert)$. As I explained in my previous post, you get the same results either way, except that your $p$ and $g$ labels need to be swapped.

$\omega$ isn't just a measure of "number of crests passing by per time." It's actually telling you the direction and rate of phase progression, in units of (radians)/time. Similarly, $k$ tells you the rate of phase progression in units of (radians)/length. Physically, the phase progression means the wave is rising or falling between crests and troughs, whether you measure the wave in time at a fixed point in space or whether you look down the x (or -x) direction at a snapshot in time.

I think the easiest way to understand this is to think of plane waves in more than 1 dimension, where $\vec{k}$ is clearly a vector pointing in the direction of positive phase propagation in time. The wavelength is ${\lambda} = {2\pi}/{\lVert \vec{k} \rVert}$, which is always positive. That wave with positive phase propagation in the direction of $\vec{k}$ is exactly equivalent to the wave with the same wavelength but negative phase propagation in the direction of $-\vec{k}$. This happens for basically the same reason why a vector of given length and direction is equal to the vector with negative length pointing in the opposite direction.

Last edited: Jul 25, 2014
14. Jul 26, 2014

### mysearch

Sorry to belabour the issue, but your statement above seems to be the key point that I am missing. For it does seem that if I simply reverse the $p$ and $g$ labels, the 2-way simulations work, but I am not certain why just changing the sign of $[ k_2]$ changes the definition? Simply to recap, I am using following 1-way equation as the basis of the derivation:

[1] $A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ]$

This leads to the following equations used in the simulations:

[2] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$

Where the value of [ $\omega_p, k_p, v_p$] and [ $\omega_g, k_g, v_g$] were defined by:

[3] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$

[4] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$

As they stand, these equations seem to give the correct results, when the 2 waves propagate in the same direction, i.e. the 1-way configurations.

Agreed. Therefore, in the case where the 2 waves propagate in different directions, I have argued that it should be $(k_2 = -\lvert k_2 \rvert)$ that is signed for reasons previously outlined. However, what is not clear to me is why just changing the sign of $[k_2]$ requires the definition of [3] and [4] reversed from $p$ to $g$? For I assumed that changing the sign of $[-k_2]$ would simply lead to:

[3a] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + [-k_2]}{2}; v_p = \dfrac{\omega _p}{k_p}$

[4a] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – [-k_2]}{2}; v_g = \dfrac{\omega _g}{k_g}$

While the suggestion appears to required the reversing of the $p$ and $g$ definition as follows:

[4b] $\omega _g = \dfrac{\omega _1 + ω_2}{2}; k_g = \dfrac{k_1 + [-k_2]}{2}; v_g = \dfrac{\omega _g}{k_g}$

[3b] $\omega_p = \dfrac{\omega _1 - ω_2}{2}; k_p = \dfrac{k_1 – [-k_2]}{2}; v_p = \dfrac{\omega _p}{k_p}$

The simulation of the 2-way cases, where $[-k_2]$ seems to support [3b, 4b]. As this is the central issue underpinning the simulations, I would really like to understand whether the equations, as stated, are correct and why. Just for the record, my original simulation appeared to work for all cases when I changed the sign of [$\omega_2$], i.e.

[3c] $\omega _p = \dfrac{\omega _1 + [-ω_2]}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$

[4c] $\omega_g = \dfrac{\omega _1 – [-ω_2]}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$

While I will try to work through the issues myself, it would be helpful if you indicate whether you support the (a), (b) or (c) form or some other variant. Thanks

Last edited: Jul 26, 2014
15. Jul 26, 2014

### olivermsun

Let me change the way you label your definitions so you can see why the $v$ and $p$ labels get reversed:
Now which of $\lvert k_+\rvert, \lvert k_-\rvert$ is smaller if $k_1, k_2 > 0$? That would normally be the envelope $k_g$, since smaller $\lvert k\rvert$ means longer wavelength.

What if $k_1 > 0, k_2 < 0$?

16. Jul 27, 2014

### mysearch

Well, it took me a bit longer than it should have, but I have finally worked through the permutations, which explains the reversal of the $p$ and $g$ and why this notation can be misleading in terms of the phase and group waves, when the waves propagate in opposite directions. For when deriving the following equations, it was not obvious, at least to me, that reversing the sign of [$k_2$] would cause the inference of $p$ and $g$ to also be reversed in terms of phase and group, as few references seem to cover this point:

[1] $A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)$

[2] $\omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}$

[3] $\omega_g = \dfrac{\omega _1 - ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}$

Purely, by way of reference should anybody else come down this road, the attached file shows the results when [$\omega=v*k$], which works for all values of [ $\pm v$], and [$k=\omega/v$], which works for [$v_1=v_2$] but not [$v_1=-v_2$]. It is also highlighted that the normal definition of [$v_g=\Delta \omega / \Delta k$] looks suspect when [$v_2$] is negative and [$k=\omega /v$], but maybe this definition doesn't support the inclusion of direction? Anyway, splitting the definition of [1] into its two components:

[4] $A=2 A_0 cos( \omega_p t – k_p x)$
[5] $A=2 A_0 cos( \omega_g t – k_g x)$

Based on the argument that [$k=ω/v$], then [4] does describe the phase wave and [5] the group wave, but only when [$v_1=v_2$]. If [$v_1=-v_2$], then the descriptions are reversed, such that [4] now describes the group waves and [5] the phase wave. However, if you ignore the physical implications of negative time and allow [$\omega=v*k$], then [4] describes the phase wave and [5] the group wave, irrespective of the value of [$\pm v$]. In part, because this last approach worked and the notation the $p$ and $g$ suggested phase and group, I didn’t think that just changing the sign of [$k_2$] would change the definition as described, but it does when you sit down and work through the numbers.

While the initial response to this thread didn’t look too promising, I really appreciate the time and effort in helping me resolve the problems I was having with the simulations. While I should really leave matters here, there was one other issue that was touched on throughout this thread, which was summarised in post #13:

I think this is a really interesting summary. The normal assumption is that the relationship between the amplitude [A] and time [t] and space [x] can be anchored in the standard 1D solution of the wave equation:

[6] $\dfrac{ \partial A^2}{\partial t^2} = v^2 \dfrac{ \partial A^2}{\partial x^2}; \ where \ A=A_0sin(\omega t - kx)$

Again, I have deliberately put positive [$\omega t$] before [$\pm kx$] because I have argued that waves only physically propagate through time [t] in the positive direction. As such, I am arguing that negative time [-t] is only a mathematical concept for calculating the amplitude [A] of a wave at some earlier point in time, i.e. it is not a physical propagation mode. Therefore, while I agree with your description of $\vec{k}$ as a vector, especially when extended to 3 dimensions, the scope of frequency [f] or angular frequency [$\omega$] might be open to some debate, when considered in terms of a physical wave system. In this context, [f] or [$\omega$] might simply define the oscillations per unit time [t] without any specific inference on the subsequent propagation of a wave through a given ‘media’, which might be affected by both the refractive index of the media and Doppler effects due to relative motion of the observer, i.e. source and/or receiver. However, the point I am trying to make/clarify is that [f] or [$\omega$] is essentially a scalar quantity without direction, while some might argue that time [t] is a vector quantity, as it may appear to have a forwards and backwards direction, physical systems only appear to go in one direction, i.e. forwards. Therefore, the scope of time as a vector is fairly constrained. I realise this is possibly beyond the scope of this thread, but I would be interested in any alternative perspectives.

Again, really appreciated the help in sorting out my confusion over [$\pm k$]. Many thanks.

#### Attached Files:

• ###### Results.pdf
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17. Jul 27, 2014

### olivermsun

The 1-d wave equation is second order, so it has 2 solutions which have opposite sign relationship between space and time:
$\frac{1}{2}f_0(x - vt)$ and $\frac{1}{2} f_0(x + vt)$.
Once you realize that there are these two independent wave solutions, you can choose whatever convention you like for interpreting the solutions as sums of sine waves in $k, \omega$ with $\pm kv = \omega > 0$ if it's convenient for you.

Choosing to write $\omega t - kx$ instead of $kx - \omega t$ (or any other permutation) makes no difference at all to either the mathematics or the physics. The notation doesn't mean that time is progressing forward in the first case and backward in the second. You still put in the same positive $t, \omega$ and signed $k$ if you like, and get the same thing out.

In fact, the sinusoidal wave itself is just a mathematical concept. The physics are all right there in the wave equation and its solutions, and we just interpret them in various ways. It's similar to the way we interpret certain waveforms as "carriers" modulated by an "envelope," or we can interpret them as a superposition of (unrelated) sine waves; it doesn't matter one bit to the physics.

Okay.

Last edited: Jul 27, 2014
18. Jun 7, 2015

### mysearch

Hi,

I originally raised some questions in this thread last year regarding phase and group velocity, but wanted to return to the subject in order to try to resolve any ‘physical’ interpretation associated with these velocities. In part, my issue relates to the maths and interpretation linked to the simulation attached. This simulation attempts to shows two waves (blue and green) propagating through a ‘non-dispersive’ medium, such that both waves have equal velocity magnitude [v=1], but different angular velocities [w] and wave numbers [k], but where [w/k=v=1]. Within the simulation, 2 different approaches were used to produce the superposition red and black traces, which takes the shape of a beat waveform. The red trace is produced by simply adding the 2 blue and green waveforms, while the black trace is calculated using the following formula:

$A=2 A_0 cos( w_p t – k_p x) cos( w_g t – k_g x)$

The values of $w_p, k_p, v_p$ and $w_g, k_g, v_g$ are determined from the parameters assigned to waves 1 & 2:

$w_p = \dfrac{w_1 + w_2}{2}; \quad k_p = \dfrac{k_1 + (\pm k_2)}{2}; \quad v_p = \dfrac{w_p}{k_p}$

$w_g = \dfrac{w_1 – w_2}{2}; \quad k_p = \dfrac{k_1 - (\pm k_2)}{2}; \quad v_p = \dfrac{w_p}{k_p}$

When blue and green waves propagate in the same direction, the sign of $k_1, k_2$ are the same, such that the $\pm$ can be ignored. However, when these waves propagate in opposite directions, as in this specific case, the wave number [k] effectively acts a vector quantity, which reflects the $\pm$ direction of propagation. Hence the following values for the simulation attached.

$w_p = \dfrac{0.22+0.18}{2}=0.2; \quad k_p = \dfrac{0.22 +(-0.18)}{2} = 0.02; \quad v_p = \dfrac{0.2}{0.02} = 10$

$w_g = \dfrac{0.22-0.18}{2}=0.02; \quad k_g = \dfrac{0.22 – (-0.18)}{2} = 0.2; \quad v_g = \dfrac{0.02}{0.2} = 0.1$

While the red and black outputs are identical, which appears to supports the maths used in both approaches, the values of the phase $v_p$ and group $v_g$ velocities appears difficult to reconcile within the simulation, which is reduced to the first 50 frames. The black dot on both the blue and green traces corresponds to the propagation velocity of these waves, e.g. [v=1]. Superimposed on the bottom black trace is a black and red dot that should correspond to the phase velocity $v_p=10$ and group velocity $v_g=0.1$ respectively. However, the faster black dot appears to tracking the wave packet envelope, which is normally associated with the group velocity, while slower red dot appears to be tracking successive peaks within the underlying phase wave.

Can anybody explain this anomaly?

My initial interpretation is that ‘information’ flowing between [A-B] and [B-A] is physically constrained by [v=1]. As such, any implied ‘superluminal’ inference of [v=10] simply reflects the change in the superposition amplitude at each value of [x] at any instance in time. Therefore, this is not a ‘real’ velocity, but rather a wave pattern that is changing at every position [x] on every tick of the clock. However, I assumed that the group velocity $v_g=0.1$ would have some physical interpretation, i.e. the velocity of the group wave packet, which does appear to be the case, at least, in this simulation. Would be grateful for any knowledgeable insights as to what might be going on. Thanks

P.S. I haven't used the forum in a while. Can you still reverse order the posts within a thread, i.e. latest first? Plus, its seems to take a while for the scrolling within a page to settle down - is this now normal?

Last edited: Jun 7, 2015
19. Jun 7, 2015

### Staff: Mentor

You might get a quicker response to these sorts of questions in the "Forum Feedback and Announcements" forum instead of buried in a highly technical thread on a completely different topic.