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Can anybody help with group velocity simulations? 
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#1
Jul1714, 04:17 AM

PF Gold
P: 521

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:
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:
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 


#2
Jul2014, 10:35 PM

Admin
P: 9,311

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
Jul2114, 09:32 PM

P: 629

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



#4
Jul2214, 03:35 AM

PF Gold
P: 521

Can anybody help with group velocity simulations?
[itex] ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{ω_p}{k_p}[/itex] [itex] ω_g = \dfrac{ω_1  ω_2}{2}; k_g = \dfrac{k_1  k_2}{2}; v_g = \dfrac{ω_g}{k_g}[/itex] 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.
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. 


#5
Jul2214, 08:27 AM

P: 629

$$\cos(k(xvt))$$ 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 backpropagating wave can be written as $$\cos(kx+kvt) = \cos(kxkvt),$$ which is the same as a forwardpropagating 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. 


#6
Jul2214, 09:38 AM

PF Gold
P: 521

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 2way nondispersive case using both approaches.
[itex] ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{ω_p}{k_p}[/itex] [itex] ω_g = \dfrac{ω_1  ω_2}{2}; k_g = \dfrac{k_1  k_2}{2}; v_g = \dfrac{ω_g}{k_g}[/itex] ‘ 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  'wave1: k1 = k0+dk0 = 0.22 v1 = c*1 =1 w1 = v1*k1 = 0.22 'wave2: k2 = k0dk0 = 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 = (w1w2)/2 = 0.2 kg = (k1k2)/2 = 0.02 vg = wg/kg = 10 ‘calculate [k] from [w] ‘this does not appear to work  'wave1: w1 = w0+dw0 = 0.22 v1 = c*1 =1 k1 = w1/v1 = 0.22 'wave2: w2 = w0dw0 = 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 = (w1w2)/2 = 0.02 kg = (k1k2)/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
Jul2214, 09:55 AM

P: 629




#8
Jul2314, 04:01 AM

PF Gold
P: 521




#9
Jul2314, 09:34 AM

P: 629

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
Jul2414, 04:35 AM

PF Gold
P: 521

Again, appreciate the feedback as it is making me review all my assumptions. Some initial thoughts in a somewhat random order:
[1] [itex]A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)[/itex] I started from the form [wtkx] for the reasons explained above: [2] [itex]A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ][/itex] However, it seemed equally valid to start from [3] [itex]A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t + k_2 x) ][/itex] 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. Inline 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: [itex] ω_p = \dfrac{ω_1 + ω_2}{2}; k_p = \dfrac{k_1 + (k_2)}{2}; v_p = \dfrac{ω_p}{k_p}[/itex] [itex] ω_g = \dfrac{ω_1  ω_2}{2}; k_g = \dfrac{k_1 – (k_2)}{2}; v_g = \dfrac{ω_g}{k_g}[/itex] 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]. “In this specific nondispersive 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 spacetime, as it more accurately reflects the phase shift between waves1 and wave2 occurring at all value of [x] simultaneously in time [t]. This phase shift effect, leading to the perception of [vg>1] is higher when wave1 and wave1 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 'snowblind' by looking at these equations for too long, which is why a second pair of 'fresh eyes' is helpful. Thanks 


#11
Jul2414, 09:42 AM

P: 629

##\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##. 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. However, when you use [3], you don't "doublecount" 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
Jul2514, 04:09 AM

PF Gold
P: 521

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 nondispersive case, where the two plane waves propagate in opposite directions.
[1] [itex]Wave1: \omega_1=0.22; k_1=0.22; v_1=+1[/itex] [2] [itex]Wave2: \omega_2=0.18; k_2=0.18; v_2=1[/itex] I believe we agree that the following equation is able to represent the beat superposition: [2] [itex]A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)[/itex] Where the value of [ ## \omega_p, k_p, v_p##] and [ ## \omega_g, k_g, v_g##] were defined by: [3] [itex] \omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}[/itex] [4] [itex] \omega_g = \dfrac{\omega _1  ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}[/itex] However, [3] and [4] were derived from [5] below, where the waves are travelling in the same direction: [5] [itex]A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ][/itex] As such, it would seem that if [3] and [4] are to work, the direction of wave1 and wave2 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(xvt \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 longwinded, but I wanted to try to explain why I was interested in the physical interpretation of these equations as much as their mathematical consistency. 


#13
Jul2514, 09:28 AM

P: 629

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


#14
Jul2614, 03:54 AM

PF Gold
P: 521

[1] [itex]A=A_0 [cos( \omega_1 t – k_1 x) + cos( \omega_2 t – k_2 x) ][/itex] This leads to the following equations used in the simulations: [2] [itex]A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)[/itex] Where the value of [ ## \omega_p, k_p, v_p##] and [ ## \omega_g, k_g, v_g##] were defined by: [3] [itex] \omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}[/itex] [4] [itex] \omega_g = \dfrac{\omega _1  ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}[/itex] As they stand, these equations seem to give the correct results, when the 2 waves propagate in the same direction, i.e. the 1way configurations. [3a] [itex] \omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + [k_2]}{2}; v_p = \dfrac{\omega _p}{k_p}[/itex] [4a] [itex] \omega_g = \dfrac{\omega _1  ω_2}{2}; k_g = \dfrac{k_1 – [k_2]}{2}; v_g = \dfrac{\omega _g}{k_g}[/itex] While the suggestion appears to required the reversing of the ##p## and ##g## definition as follows: [4b] [itex] \omega _g = \dfrac{\omega _1 + ω_2}{2}; k_g = \dfrac{k_1 + [k_2]}{2}; v_g = \dfrac{\omega _g}{k_g}[/itex] [3b] [itex] \omega_p = \dfrac{\omega _1  ω_2}{2}; k_p = \dfrac{k_1 – [k_2]}{2}; v_p = \dfrac{\omega _p}{k_p}[/itex] The simulation of the 2way 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] [itex] \omega _p = \dfrac{\omega _1 + [ω_2]}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}[/itex] [4c] [itex] \omega_g = \dfrac{\omega _1 – [ω_2]}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}[/itex] 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 


#15
Jul2614, 09:26 AM

P: 629

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


#16
Jul2714, 05:12 AM

PF Gold
P: 521

[1] [itex]A=2 A_0 cos( \omega_p t – k_p x) cos( \omega_g t – k_g x)[/itex] [2] [itex] \omega _p = \dfrac{\omega _1 + ω_2}{2}; k_p = \dfrac{k_1 + k_2}{2}; v_p = \dfrac{\omega _p}{k_p}[/itex] [3] [itex] \omega_g = \dfrac{\omega _1  ω_2}{2}; k_g = \dfrac{k_1 – k_2}{2}; v_g = \dfrac{\omega _g}{k_g}[/itex] 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] [itex]A=2 A_0 cos( \omega_p t – k_p x)[/itex] [5] [itex]A=2 A_0 cos( \omega_g t – k_g x)[/itex] 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: [6] [itex] \dfrac{ \partial A^2}{\partial t^2} = v^2 \dfrac{ \partial A^2}{\partial x^2}; \ where \ A=A_0sin(\omega t  kx) [/itex] 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. 


#17
Jul2714, 06:39 AM

P: 629

## \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. 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. I'm glad you got your questions sorted out. Cheers! 


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