Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1880296 times)

WarpTech

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6640 on: 08/07/2015 02:19 AM »
...
I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ...
...
Disclaiming again, I probably broke the math.

The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.
Todd

If F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart.  Specifically, as the size of the waveguide decreases Vp increases.  An increase in Vp decreases F.  That would make the force on the little end --smaller-- than the force on the big end.

Is that right?

Pretty much... It is that the momentum of an EM wave depends on the phase velocity and Planck's constant. It is the phase velocity, Vp = w/k that give us the momentum per photon, p=h*k. So if Vp is increasing, p is decreasing, and if Vp is decreasing, p is increasing. Either way, the force on the frustum is in the same direction in order to conserve momentum. The momentum is "forward rectified" by the gradient. It is only the reflection at the big end that is opposing this force. Anything that reduces the force on the big end, will result in forward thrust.

In order for the big end to 100% cancel these forces, the force on the big end would have to be larger than the force at the small end. Definitely not smaller. If it is smaller by hook or by crook, or by leakage or heat, then the frustum will feel a thrust forward.

The real question is still, how does it get thrust greater than a photon rocket? I see 3 possibilities:

1. A group velocity that is faster than light in free space.
2. A phase velocity that is slower than light in free space.
3. A means to change the speed of light such that Vg*Vp = (c/K)2, in which case the group velocity could be faster than light in a frustum, but slower than light in free space and result in an amplified ratio;

Vg/c2 -> K*Vg/c2, where K > 1

This way, nothing ever exceeds c in free space, but it reduces both the phase velocity and the group velocity to be much less than c. It is consistent with the PV Model of gravity. I just don't know how to derive it yet. I have 2 key pieces to the puzzle. Squeezed light behaves this way, and what Zeng and Fan show for an impedance curve looks almost exactly like a Reissner Nordstrom "charged" metric refractive index. Somehow, this all seems to make sense to me, but not well enough to write down the correct equations.
Todd

Rodal

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6641 on: 08/07/2015 02:47 AM »
I tried to model TheTraveller's EmDrive Mark 2 in 3D, which was a bit difficult since the only dimensions TT provided were end diameters Ds and Db and the length (Rb - Rs)*

Ds = 159 mm
Db = 400 mm
Rb-Rs = 240.7 mm

Rs = Ds = 159 mm
Rb = Db = 400 mm
half-cone angle = 30°

I get slightly different dimensions, on the order of tenths of a millimeter.  Which brings up the question, how tight are the tolerances on the manufacturing process, and exactly how tight they need to be to get good results?

Flux-Capacitor is correct that:

Db = 2 r2 Sin [(30/180) Pi] = 2 r2 Sin [Pi/6]= 2 r1 (1/2) = r2 exactly
Ds = 2 r1 Sin [(30/180) Pi] = 2 r1 Sin [Pi/6]= 2 r1 (1/2) = r1 exactly

An angle of 30 degrees results in  Db = r2 and Ds = r1 exactly (to infinite precision) because Sin[30 degrees] =1/2 exactly, regardless of the value of Db and Ds.

But TheTraveller did NOT give an angle of 30 degrees.  Instead TheTraveller defined the difference between the spherical radii=r2-r1, and TheTraveller defined them such that the angle is NOT 30 degrees

lmbfan  is correct that the numbers given by TheTraveller do NOT agree with this exactly

The reason is that

r2 - r1 = 0.4 - 0.159 = 0.241 exactly

but TheTraveller instead gave the incorrect value of

r2 - r1 = 0.2407 m
(which is weird, since those extra digits of 0.7 mm are unnecessary and end up screwing up the values for r1, r2 and theta, while just giving 0.241 gives an exact 30 degrees)

This value is incompatible with 30 degrees.  That's where the problem comes from.

This value gives:

r1 = (r2 -r1)/(Db/Ds - 1) =  0.2407 /(0.400/0.159 - 1)
= 0.158802 m

r2 = (r2 -r1)/(1 - Ds/Db ) =  0.2407 /(1 - 0.159/0.400)
= 0.399502 m

theta = ArcSin[Db/(2 r2) ]= 30.0412 degrees

So, again

r2 - r1 = 0.2407  is only valid for 30.0412 degrees

For theta = 30 degrees one must have  r2 - r1 = 0.241m

//////////////////////////////////////////

Any way, whether using 30 degrees or 30.0412 degrees this geometry is going to resonate at TE012 at a lower natural frequency than the 2.45 GHz TheTraveller calculated.

The inaccuracy of TheTraveller and Shawyer's code is more evident for this case because of the higher cone angle 30 degrees (compared with other EM Drives that range from 15 degrees to 20 degrees), and therefore the more inaccurate their approximate formulas.

It has a very high theoretical Q though, close to 100,000 much higher than the Fligh Thruster

Flux Capacitor:

I don't use COMSOL for these calculations.  I prefer to use the exact solution because... it is exact, while numerical solutions are approximate.
« Last Edit: 08/07/2015 04:29 AM by Rodal »

ElizabethGreene

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6642 on: 08/07/2015 04:33 AM »
...
I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ...
...
Disclaiming again, I probably broke the math.

The equation would be F = (Vg/c^2)*P = P/Vp.
<snipp>

If F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart.  Specifically, as the size of the waveguide decreases Vp increases.  An increase in Vp decreases F.  That would make the force on the little end --smaller-- than the force on the big end.

Is that right?

Thank you both for the kind replies.  I looked back in my notebook from yesterday and I did manage to derive 2P/Vp (the expanded form) on my first attempt, but I wrote "that can't be right" next to it and swapped in Vg.  I"ll be more careful in the future.

Closing the loop:
My math regarding fine tuning the small end dimensions for maximum thrust was wrong.

The corrected equation is F=2P/V_p.  Playing with the numbers for this reveals approximately a 2 order of magnitude decrease in the small end force for small end sizes of 10 microns above the calculated non-tapered waveguide cutoff.

These results do not explain -at all- why the device appears to thrust in the wrong direction in the Boeing video.

I'll look at side forces next.

deltaMass

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6643 on: 08/07/2015 05:11 AM »
The 3D integration of all photonic forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell.
The 3D integration of all Lorentz forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell too.

So if it moves, you have chosen to reject Maxwell.
And when it moves, you have a further set of choices: reject Noether or reject Einstein.

I hope you know what you're doing, whoever you are.

tleach

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6644 on: 08/07/2015 05:15 AM »
As to how you can reconcile that the force being larger on the Big End results in motion towards the Small End

That's why I prefer McCulloch's theory: according to MiHsC, as they travel towards the big end, photons gain momentum. When they bounce back towards the small end, they loose momentum. So to obey CoM, the cavity has to move from the big end towards the small end.

So I had a little extra time today and I actually took a look at McCulloch's equations.  I don't know much about equations, but what I did realize was that they were actually pretty simple.  "I could probably program this into a spreadsheet," I thought aloud.  He's only using the frustum dimensions (big end, small end and length), the power, and the Q (which he defines on his blog as the "number of bounces of a typical photon inside the cavity").

So, I tried (see attached .xls).  I'm not entirely certain that I translated things correctly, but I'm sure someone here can point out my errors...

I included drive specs I was playing around with (we'll see if I can talk my wife into letting me do a build) as well as some specs and Df I stole from TheTraveller's spreadsheet.  There are also specs, predicted force and observed force for 2 of Shawyer's drives (I got those numbers from McCulloch's blog) and Tajmar's drive (again, the specifications came from McCulloch's blog).

Would a spreadsheet like this (only with several more of the various theories/equations programed in) be useful to have over on the wiki?  I think it could be helpful for the builders to plug their specifications in and get back several different thrust predictions, each based on a different theory.

EDIT: http://physicsfromtheedge.blogspot.co.uk/2015/02/mihsc-vs-emdrive-data-3d.html

Original
F = PQL/c * (1/wb - 1/ws)

3D
F = 6PQL/c * ( 1/(L+4wb) - 1/(L+4ws) )
« Last Edit: 08/07/2015 02:07 PM by tleach »
T. Thor Leach

lmbfan

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6645 on: 08/07/2015 05:26 AM »

r2 - r1 = 0.4 - 0.159 = 0.241 exactly

This agrees with my analysis.  The horizontal dimension should be 241 mm, or the small end should be 159.3 mm.  If the manufacturing tolerance is greater than a few tenths, it doesn't matter too much, hence my question.  If, on the other hand, the tolerance is less than that (which is not uncommon for precision components), it does matter.  I know that many microwave devices require tolerances on the order of .001", which works out to ~.03 mm, but this is a novel application, and my experience is less useful.

ElizabethGreene

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6646 on: 08/07/2015 06:26 AM »
So if it moves, you have chosen to reject Maxwell.
And when it moves, you have a further set of choices: reject Noether or reject Einstein.

I choose to reject none at this point.  Instead I will, following your example, read the papers critically (and slowly).  Part of that means, to the limited extent my pidgin math allows, playing with numbers to see what pops out the other side.

It hasn't been unproductive.  Two days of spare cpu cycles noodling with small end parameters taught me that Mr. S's equations do not describe the observed behavior.  That was worth the time invested.  Much better than watching reruns of Law & Order.

ThinkerX

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6647 on: 08/07/2015 06:40 AM »
Quote

The 3D integration of all photonic forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell.
The 3D integration of all Lorentz forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell too.

So if it moves, you have chosen to reject Maxwell.
And when it moves, you have a further set of choices: reject Noether or reject Einstein.

I hope you know what you're doing, whoever you are.

Key word in this to me is 'averaged.'

Lately, the thought has crossed my mind more than once that maybe something about this device breaks or twists the averages somehow.    There are such things as loaded dice.  And coins do land on edge now and again.  Maybe the EM Drive represents one such?

TheTraveller

Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6648 on: 08/07/2015 07:20 AM »
The inaccuracy of TheTraveller and Shawyer's code is more evident for this case because of the higher cone angle 30 degrees (compared with other EM Drives that range from 15 degrees to 20 degrees), and therefore the more inaccurate their approximate formulas.

Thanks for the good laugh.

You who have no working EMDrives created with your exact solution, derate Shawyer's solution where he has shown 3 working EMDrives and in the case of my estimated Flight Thruster dimensions, produced the quoted resonance at 3.9003GHz, which says my estimated length was 0.6mm too short for resonance at 3.85GHz.

The reality is you have no real idea if your numbers are correct or not, but continually push them as if they represent reality, which as your resonance is so far off the Shawyer solution says they are no correct. Not even close.

BTW how often does your private and not for publication exact solution vary?
« Last Edit: 08/07/2015 07:21 AM by TheTraveller »
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TheTraveller

Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6649 on: 08/07/2015 07:29 AM »
...
I played with the numbers yesterday for F=2P/Vg -> F=2*P*Sqrt(1-(c/(2*(a+x)*f))^2) and found ...
...
Disclaiming again, I probably broke the math.

The equation would be F = (Vg/c^2)*P = P/Vp. IF you use standard waveguide physics that is. In the case of the EM Drive, this may be, F = (Vg*K/c^2)*P, where K is a function TBD by experiment.
Todd

If F=P/Vp is correct for traditional physics, then Mr. Shawyer's theory falls apart.  Specifically, as the size of the waveguide decreases Vp increases.  An increase in Vp decreases F.  That would make the force on the little end --smaller-- than the force on the big end.

Is that right?

Pretty much... It is that the momentum of an EM wave depends on the phase velocity and Planck's constant. It is the phase velocity, Vp = w/k that give us the momentum per photon, p=h*k. So if Vp is increasing, p is decreasing, and if Vp is decreasing, p is increasing. Either way, the force on the frustum is in the same direction in order to conserve momentum. The momentum is "forward rectified" by the gradient. It is only the reflection at the big end that is opposing this force. Anything that reduces the force on the big end, will result in forward thrust.

In order for the big end to 100% cancel these forces, the force on the big end would have to be larger than the force at the small end. Definitely not smaller. If it is smaller by hook or by crook, or by leakage or heat, then the frustum will feel a thrust forward.

The real question is still, how does it get thrust greater than a photon rocket? I see 3 possibilities:

1. A group velocity that is faster than light in free space.
2. A phase velocity that is slower than light in free space.
3. A means to change the speed of light such that Vg*Vp = (c/K)2, in which case the group velocity could be faster than light in a frustum, but slower than light in free space and result in an amplified ratio;

Vg/c2 -> K*Vg/c2, where K > 1

This way, nothing ever exceeds c in free space, but it reduces both the phase velocity and the group velocity to be much less than c. It is consistent with the PV Model of gravity. I just don't know how to derive it yet. I have 2 key pieces to the puzzle. Squeezed light behaves this way, and what Zeng and Fan show for an impedance curve looks almost exactly like a Reissner Nordstrom "charged" metric refractive index. Somehow, this all seems to make sense to me, but not well enough to write down the correct equations.
Todd

Why do you and others keep referring to phase velocity inside a waveguide as it is above c. Inside a waveguide energy moves at group velocity, which is determined from the guide wavelength, which is determined from the cutoff wavelength. The momentum in the wave varies as the guide wavelength varies, longer at the small end and shorter at the big end.

http://www.microwaves101.com/encyclopedias/waveguide-mathematics#velocity

Cullen 15, attached, is based on this and proved that it is correct, so please stop talking above phase velocity inside a waveguide as it is above c and nothing travels above c.

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deltaMass

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6650 on: 08/07/2015 07:32 AM »
Quote

The 3D integration of all photonic forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell.
The 3D integration of all Lorentz forces over the entire internal surface area, averaged over one complete cycle, is zero. That's Maxwell too.

So if it moves, you have chosen to reject Maxwell.
And when it moves, you have a further set of choices: reject Noether or reject Einstein.

I hope you know what you're doing, whoever you are.

Key word in this to me is 'averaged.'

Lately, the thought has crossed my mind more than once that maybe something about this device breaks or twists the averages somehow.    There are such things as loaded dice.  And coins do land on edge now and again.  Maybe the EM Drive represents one such?
I think I understand your comment, but if what you suggest were to be true, then one cycle of RF energy from time t0 to t1 would be in some way different to another cycle from t1 to t2. That does seem rather unlikely, absent an explanatory mechanism, don't you think?

TheTraveller

Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6651 on: 08/07/2015 07:37 AM »
As the wave moves from the small end toward the big end it expands, but it's wavelength is getting shorter because the phase velocity is getting slower. So the momentum of the wave is increasing when moving from the small end toward the big end. Therefore, to conserve momentum, the frustum must move the other way. It feels the force that moves it in direction toward the small end. This is thrust, just like a rocket.

When the wave reflects from the big end, the amount of momentum it transfers depends on the impedance match and the angle of reflection. If the impedance doesn't match exactly, or the angle of reflection is not perpendicular, then the force will vary too. Don't ask me for the equations please, I don't have them quite right yet. So this is just hand waving, based on my understanding.

When the wave is traveling toward the small end, its momentum is decreasing and therefore, to conserve momentum the frustum must again feel a force in the "forward" direction, like a sail on a boat. This is due to the gradient, which is unique to the cone geometry. According to Zeng & Fan's paper,  if the phase constant Beta goes to zero, there is a perfect impedance match and all the power and momentum is attenuated into the frustum with no reflection.

So regardless if the wave is traveling forward or backwards inside, the frustum will be pushed "forward" toward the small end. The only component that is in the other direction, opposing this thrust is the reflection from the big end. From what I've seen in the meep models, the wave is not perpendicular to the big end. The angle of reflection produces a glancing blow. Anything that reduces the momentum transferred to the big end plate, results in thrust forward.
Todd

Nicely explained, except it is group velocity and not phase velocity. Additionally the group velocity is faster at the big end and slower at the small end.

The bad bounce introduces phase distortion into the wave and is why spherical end plates are so important.
« Last Edit: 08/07/2015 07:41 AM by TheTraveller »
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flux_capacitor

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6652 on: 08/07/2015 08:21 AM »
Any way, whether using 30 degrees or 30.0412 degrees this geometry is going to resonate at TE012 at a lower natural frequency than the 2.45 GHz TheTraveller calculated.

So what do your exact solutions give for each value of the possible angles we found (30° and 30.0412°) for TE012 and TE013? Are they really different with such a slight difference?

I think TheTraveller may have accounted for that lower natural frequency, as he also noted on his drawing: "Min operational frequency 2.3 GHz".

TheTraveller

Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6653 on: 08/07/2015 08:53 AM »
Any way, whether using 30 degrees or 30.0412 degrees this geometry is going to resonate at TE012 at a lower natural frequency than the 2.45 GHz TheTraveller calculated.

So what do your exact solutions give for each value of the possible angles we found (30° and 30.0412°) for TE012 and TE013? Are they really different with such a slight difference?

I think TheTraveller may have accounted for that lower natural frequency, as he also noted on his drawing: "Min operational frequency 2.3 GHz".

I redesigned the frustum to have a lower operational freq of 2.30GHz, which increased the small end diameter and altered the length so as to give me good headroom when operating at 2.45GHz and to allow wider manufacturing tolerance.

Screenshot of my design spreadsheet is attached.

As you can see the TE013 frustum length is 0.05mm too short where length is defined as per the attached.
« Last Edit: 08/07/2015 09:12 AM by TheTraveller »
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Mulletron

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6654 on: 08/07/2015 09:08 AM »
It doesn't matter if there is a more momentum at one vs the other. This theory of operation will not result in the frustum moving because....

The small end and the large end are physically connected, so the total momentum is ZERO.

This mechanism of action is a no go. It is just like pushing your car from the inside.

Now that I've poopood theories which rely on the above (more momentum transfer to one end plate vs the other), I'll offer a down to earth idea, which is similar and may be equally abysmal, but doesn't rely on electromagnetic vacuum field fluctuations as I'm accustomed to doing.

So we know that friction is a non conservative force. I'd rather shake a stick at the friction from the air against the cavity walls while it is circulating within the cavity. I can see this happening as long as thermal equilibrium hasn't been achieved. Is it useful? I don't know.

This should be super easy to test. A simple heat sink or water jacket attached to the frustum would make or break thrust with this mechanism. I think any mechanical engineer could blow this idea up quite easily too.

I don't have much confidence in the internal air friction idea yet, but I'm forced to look at more mundane explanations than momentum from the QV, which I believe is happening but isn't the dominant force here.
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TheTraveller

Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6655 on: 08/07/2015 09:24 AM »
It doesn't matter if there is a more momentum at one vs the other. This theory of operation will not result in the frustum moving because....

The small end and the large end are physically connected, so the total momentum is ZERO.

This mechanism of action is a no go. It is just like pushing your car from the inside.

This is not like pushing your car from the inside. This is about the EM wave gaining and losing momentum inside the tapered waveguide and how that momentum gain and loss affects the frustum.

As the EM wave moves from the small end to the big end, the EM waves momentum increases as the wavelength shortens and the group velocity increases. In needs to balance that momentum increase by pushing the frustum the other way, ie toward the small end.

When the EM wave moves big end to small end, it needs to dump EM wave momentum because the EM wave's momentum is decreasing as the wavelength increases and the group velocity decreases. The dumped momentum pushes the frustum toward the small end.

The bounce at the small end adds to the big to small end momentum gain but offset by the small to big end bounce momentum gain at the big end.

Shawyer has already said the end plate forces versus the frustum forces, from just the bounce action will not results in any external Force being generated.
« Last Edit: 08/07/2015 09:44 AM by TheTraveller »
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deltaMass

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6656 on: 08/07/2015 10:09 AM »
Oh, I see. That was supposed to be physics.

Carry on.

TheTraveller

Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6657 on: 08/07/2015 10:26 AM »
Oh, I see. That was supposed to be physics.

Carry on.

So as the EM wave moves from the small to the big end end and the momentum in the EM wave increases, there is no effect on the frustum?

Likewise as the EM wave moves big end to small end and the momentum in the EM wave decreases, there is no effect on the frustum?

Yup then according to you, momentum changes in the EM wave just happen totally isolated from the real world and without any corresponding momentum changes in the frustum.

Nice physics you believe in there.

Oh yes I see, you believe the energy / momentum inside the tapered waveguide (frustum) moves at superluminal above c phase velocity and the group velocity / guide wavelength doesn't change as the EM wave moves from one end of the frustum to the other, so there is no momentum change in the EM wave.

What shall we call this? Denier physics?
« Last Edit: 08/07/2015 10:29 AM by TheTraveller »
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deltaMass

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6658 on: 08/07/2015 10:48 AM »
I'm unfamiliar with the term. Was it supposed to be French?

Mulletron

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Re: EM Drive Developments - related to space flight applications - Thread 3
« Reply #6659 on: 08/07/2015 10:54 AM »
This is useful for those still arguing over group vs phase velocity.

http://www-math.mit.edu/~stevenj/papers/PovinelliIb04.pdf