The same way they do it in a metal.
The electrons cannot leave the volume occupied by the plasma.
If they try, a net positive charge arises, and they are pulled back.


You are wrong.
The electrons do not oscillate about an equilibrium point. They fly around in the plasma.
If they were oscillating about a point, they would be bound to it (like in an atom), and there would not be a plasma.

Do you remember the analogy with sound?
The air molecules have Brownian motion, yet we can have oscillations in the pressure, on length-scales much large than the average distance between the molecules.


What are your apologies for?

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

Plasma clouds in IG space are huge. 99.9% of all space? How could an electron 'try to leave it"?


They don't oscillate about 'a point' but their shm is superimposed on their random thermal motion.
You have yet to explain anything. To be honest this is basically hand waving. How, by looking at the individual electrons, do we get density waves in plasma? That is, if you don't agree with 'my' version.
cheers,
Lyndon


What are your apologies for?[/quote]

Is the density uniform?


Because hand-waving is enough to show that you do not understand basic physics.

How do the electrons oscillate, if they are zipping around?
What makes them oscillate?

By the way, you still have not shown your full calculations about... anything.


At length-scales much larger than the average distance between particles, one sees a density of charges, not single charge carriers.
The density is nearly unifrom, because of the relatively high speed and random motion of the electrons.
An electromagnetic wave with a wavelength comparable to these length-scales, does not interact with one electron at a time, but with a high number of electrons at the same time.

We get density waves in plasma, as we can get sound in air.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

papagenorote

And what do these electrons do when the wave interacts with them?
I mean how do they move?
Cheers,
lyndon

These electrons are mostly ~1 metre apart from each other. Considering that they will be moving at something like 6e6 m/s (to use Sylas's figures), the average electron is going to be passing by an average of 6e6 electrons per second. This means that the net force exerted on any particular electron, including our deflected electron, is going to be changing incredibly fast, and in a totally chaotic manner. There is no way a single electron can perform SHM under these circumstances.

Edit to change wording.

At those length-scales, talking about motions of single electrons is not very helpful, which is why you always find the charge density in your sources.

If you treat the electromagnetic wave as a macroscopic oscillating electric field, the single electron is accelerated.
If you treat it as a collection of photons, you have lots of electron scattering lots of photons.

(An electron oscillates in a high-power laser light, because it is scattering a lot of photons: for each photon it recoils, and adding up all the recoils, you end up with an oscillation. This is Feynman's picture in his QED book, as far as I understand.)

But in your "theory" you try to explain the red-shift as a sum of single-electron/single-photon scattering, which is not the same as macroscopic electormagnetic waves and plasma oscillations.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

But it is helpful to me.

As it is accelerated what does it do, What path does this acceleration cause the electron to follow?
A wave is a predictable thing, so we must be able to determine the effects of the acceleration on the electron.
So how do we get a density wave - a predictable thing with a calculable frequency if the electrons are all scattering the photons randomly?

A laser beam travelling from left to right can only cause the electron to recoil from left to right. How does 'adding up all these recoils' cause it to 'oscillate back and forth'?

Macroscopic effects are the net result of microscopic effects.
Cheers,
Lyndon

Because you do not actually perform proper calculations.

If you treat the electron as a classical particle, the acceleration is in the direction of the force.


As the name says, you use the charge density for the calculations.
If there is an external electric field, the motions are nearly random.
Add up enough nearly random motions and, if they are correlated, you can end up with a not-so-random density oscillation.


There is no reason for the recoil to be limited along the direction of propagation.


If you studied solid state physics, you would know things can be more complicated (see many-body effects).

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

Why 6e6 per second? A second is a long time in the life of a e-m wave. Light has a frequency of about 6x10^14Hz. It takes 1.7x10^-15 secs for one oscillation. During this time the electron will have covered 10^-8 metre thermally.It is nothing.
Consider ac (and despite what others on this boeard say electrons do perform shm in ac currents. It is the basis (page 4?) of radio transmission.
In ac the thermal motion of the electrons is at about 10^5 m/s. One cycle takes 0.02 s. During this time the electron will have travelled 2km thermally - and tet we still see our T/v sets.
Electrons are extremely mobile.
Cheers,
Lyndon

I will accept a 'hands in packets' none hand waving qualititive answer from you here Papageno.

In e-m waves the force varies periodically so are you now saying that this electron performs SHM in a classical sense?


What is 'nearly random' is it something superimposed on top of their random motion to make their motion nearly random?


True but can the photons 'suck' the electrons backwards?


I thought we had already agrred on this - I did.
Cheers
Lyndon

So, there is not enough time for the electron to interact with other electrons and lose energy.

Why don't you cite a proper textbook about soild state physics, or metals?
Why do you have to rely on googling "SHM"?

Others on this board have explained why electrons in a wire are not oscillating about a point.
But you come up with a reference to a 6-pages long document: at least you could cite Feynman Lectures (but he probably does not say what you would like to hear).

Yes, but in a wire the electrons scatters on all sorts of things: what is the average scattering time (picoseconds, maybe?)?
High-Electron-Mobility-Transistors at low temperatures (1 K) have an average scattering time of 100s of picoseconds (10^-10 secs). In those 0.02 secs, an electron has scattered an average of 10^8 (100 milion) times.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

You have been asked many times to provide your full calculations, but you never complied.
Pot calling the kettle black: the burden of proof is still yours.


A classical EM wave and classical charged particle: forced oscillation.


Mathematically, you could treat it as superimposed. But it is not an oscillation, because the other particles are not arranged periodically.


What exactly is your question? ("suck"?)


You did what?
Study solid state physics? If so, you wasted your time, because you don't even understand the simplest concepts of conduction in metals.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

papageno wrote
So that any one can follow the link - not everyone has access to a library, or wants to.

This makes it worse. In IG space the electrons do not make any collisions as they oscillate. If electrons in ac can do shm and collide all these times then there is no problem with my theory.
cheers,
lyndon

papageno wrote
And on that thought Papageno, me old mate, I will leave you. The weekend has started, quiz time at the 'Alamo' along with a few bevies. I will get back to this post tomorrow (late)
Cheers,
Lyndon

Here: Solid State Physics, Brief Review, slides, 321 pages!.

Again obfuscation.
You were talking about wires, and I addressed wires.
Yet again you show that you do not understand that the electrons are not oscillating with an AC current: the charge density is.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

And that scattering shows up as heat - which means for an infinitely old universe and your theory to work, we should be piping hot (incredibly intense microwave background...). Else the energy simply disappears, and conservation of energy does not work in your universe.

(emphasis mine)

But that's exactly it: they don't! The individual electrons are zipping about all over the place. It is only when you take the collective behaviour of many electrons, that you get your SHM.

Your model seems to think of electrons as cars on a busy motorway, all trying to keep an even safe distance from each other. One car briefly slows down beforing speeding back up, and you get a 'wave' of close-together cars travelling back down the line. If that car at the front repeatedly speeds up and slows down, you'd get a continous 'wave' going down the line. The mechanics are different, but the principal is identical to what you are suggesting.

But that is incorrect. Its more like dodgems, everyone trying to maintain a safe distance but essentially driving in random directions (ok dodgems for scaredy cats, just bear with me ). Any one dodgem goes in a random direction, but looking from a distance you would still see waves, ebbs and flows of 'dodgem density'.

I am inclined to agree here. lyndonashmore, revising history is a very big no-no here. Editing is okay under limited conditions, but deleting a post when caught in a mistake?

There is enough confusion here (barely) to keep me from banning you outright. Instead I will give you a very stern warning. Don't do it again.

I also am of the opinion that you are not addressing points made very clearly, but I am not about to dredge through 22 pages of this argument to pick out every point you've dodged. But I would advise against that as well.

__________________
Phil Plait
The Bad Astronomer
http://www.badastronomy.com
badastro@badastronomy.com

Here is an example of a quantified energy momentum analysis.

Values for speed of light, Planck's constant, electron charge and mass:
c = 3.00e8 m/s
h = 6.63e-34 kg.m^2/s
e = 1.60e-19 C
m = 9.11e-31 kg

Initial photon, wavelength, energy, and momentum:
λ = 5.00e-7 m
Q = 3.97e-19 J (Q = hc/λ)
p = 1.325e-27 kg.m/s (p = h/λ)
This is the energy and momentum that must remain balanced. If numbers are not given adding up to these values, both for energy and momentum, then the balance has not been shown.

Recoil electron with kinetic energy K after absorbing momentum p
v = 1.46e3 m/s (v = p/m)
K = 9.64e-25 J (K = mv^2/2)

At this point in the analysis, we can balance the momentum, but (1-K/Q) = 99.99976% of the energy is still unaccounted for.

If this energy transfers to the rest of the plasma, then it will require over 400,000 more electrons with the same amount of energy to make up the balance (Q/K = 4.12e5). Energy transfers to the rest of the plasma occur as an electron moves through the electric fields; but the electron cannot transfer more than the energy K of its own motion. The photon is allegedly absorbed; it can't go on and interact with 400,000 more electrons itself.

Thus there is no possibility of energy balance at this point in the interactions.

In real physics, the energy of photo-absorption is actually taken up by excitation of an atom to a new energy level. This is how it occurs in French, and in every published source we have considered on photo-absorption. This is why photo-absorption in real science is for electrons bound to atoms; never for ionized electrons a meter or so away from any other particle.

----------------------------------

In his paper Lyndon then describes the recoiling electron as decelerating by interactions with the rest of the plasma and emitting a photon with energy K, by bremsstrahlung.

Emission of a CMB photon.
CMBλ = 2.06e-1 m (CMBλ = hc/K)

This is the only quantified interaction Lyndon gives with the rest of the plasma. It could occur, for example, in a close encounter with a heavy positive ion. Energy momentum can balance for the CMB emission, with energy K of the electron becoming energy of the CMB photon, and with momentum sqrt(2Km) of the electron absorbed into the heavy ion with a negligible change in energy.

This is no help for the 99.99975% of the initial energy Q still unaccounted.

-------------------------------------------------------

In his paper Lyndon then describes the electron as emitting a new redshifted photon, and recoiling again.

This energy is going to let the energy budget balance, but where did the energy come from? The electron only got energy K, and that's already gone into a CMB photon.

Basically, the original energy has to be stored somewhere, to power this emission. Lyndon has spoken of "storing" it in the rest of the plasma, and we've seen above that this is impossible. But even if it was stored in the rest of the plasma, that involves dissipation into enormous numbers of other particles. How is it then all brought back together into the original electron, so as to make physical sense of the electron emitting the photon? This is now also a violation of basic thermodynamics.

In real physics, the absorbing particle is an atom. The energy of photo-absorption gets stored in one of the well-defined fixed energy levels of the atom. This is why all the real absorption and emission reactions occur at very well defined energies; not spread out over the spectrum like the alleged Lyndon effect.

Another curious property of the emitted redshifted photon is that somehow the electron "remembers" the direction of the original photon, and emits it again with zero scatter angle!

--------------------------------------------------------

We can almost balance energy by combining absorption and emission into a single interaction. Lyndon hints at this on the top of page 4, when he speaks of the form factor f2 as follows: "One meaning that the photon has been absorbed and the electron remaining in an excited state and zero meaning that the photon was absorbed and an identical photon reemitted."

Actually, f2 being zero simply means no absorption at all; but let it pass. Also, Lyndon does not reemit an identical photon but one that is redshifted. Let that pass too. The question is if we can balance energy momentum if we consider the redshifted photon to be emitted again immediately.

The amount of redshift Lyndon invokes corresponds to loss of energy of 2K. Furthermore, the photon has no scatter angle. This is crucial to the impossibility of balancing energy momentum. If scatter was allowed, we could have simply used a Compton effect.

If we combined an incoming photon with energy Q, and an outgoing photon with energy Q-2K, then there is still 2K of energy to balance. Because Lyndon insists on no scatter angle, both momentum vectors are along the same line, and so there is still 2K/c momentum, in the direction of the original photon, still to be accounted for.

This is, in fact, exactly equivalent to absorbing a photon of energy 2K! The same analysis as above shows that an electron cannot absorb that amount of energy and momentum to balance the books. It can absorb momentum, but most of the energy 2K remains unaccounted.

How about if we include the CMB photons in the interaction? They have energy K, so two of them balances the energy budget, and also allows for spin-parity to be preserved.

This allows a balanced budget at last. The final energy budget is as follows. There is Q-2K for the redshifted photon, and 2K for the two CMB photons, and a negligible quantity of energy left over for the electron.

We still need to account for momentum, and the initial momentum is Q/c.

The red-shifted photon is (Q-2K)/c along the same line.
The CMB photons contribute K/c, twice, along any two lines we like

The total momentum of the three photons is a maximum when they are all along the same line, giving Q/c with nothing more required. It is a minumum when the CMB photons are backscattered, effectively subtracting from the momentum of the redshifted photon, giving a total of (Q-4K)/c. Other values for the photon momentum contribution can have magnitudes with anything between these two values, depending on the directions.

The momentum remaining to be balanced is thus anything from 0 to 4K/c, and this can be taken up by the electron, with a negligible difference to the energy.

That is, the books can balance if the interaction involves immediate re-emission of a redshifted photon with no scatter, and with the remaining energy taken up immediately by more photons, and with a small impulse to the electron of no more than 4K/c.

The velocity given to the electron with this impluse is 4K/mc, so the maximum allowed energy and velocity for the electron is
Electron maximum velocity: 4K/mc = 1.4e-2 m/s
Electron maximum energy: 8K^2/mc^2 = 9.1e-35 J

There is no possibility of a balanced energy momentum budget for Lyndon's reaction if the stationary electron ever exceeds a velocity of 14 millimeters per second. Any intermediate state with more velocity than this in the electron is bound to fail a simple energy momentum conservation test. All photons have to be emitted at once in the initial collision.

Lyndon won't accept this. He may accuse me of "forgetting" some contribution, but in reality I have considered and quantified the other sinks for energy or momentum and shown that they are not able to balance the books. He may accuse me of getting my "sums" wrong, but you can be sure he won't offer any quantified corrections, or balanced energy momentum analysis of his own.

Cheers -- Sylas

[[ Minor grammatical edits applied, and converted some formulae to simpler classical approximations; all numbers remain as original. ]]

Sylas, we have been through this many times before and time and time again I have shown you where you have gone wrong.
You ignore my responses but keep generating random numbers in an attempt to discredit a perfectly good theory.
Let me remind you once again, Light is a transverse wave. In transverse waves the energy is stored in oscillating electric and magnetic fields that oscillate in a direction perpendicular to that in which the wave travels. A light wave travelling from left to right has momentum from left to right but the electric fields oscillate up and down. Now when our electron absorbs this chunk of light it recoils from left to right but it oscillates up and down – in the direction of the field and this is where the energy is.
Do we understand that Sylas? If not say so and we will go over it again. If yes then let us look at your first paragraph.

[quote]You have a recoil velocity of 1.46e3m/s and recoil energy of 9.64e-25J. Which is fine and this is the amount emitted as a photon of CMB. But you then come up with a strange idea that somehow, there is a great deal of energy ‘missing’ and somehow the only way that this energy can be transferred to other electrons is by a repetition of the recoil shown above and thus involving 400,000 other electrons – ie longitudinal waves. Utter rubbish!
The ‘missing’ energy is stored in the oscillations of the electron up and down and this is caused by the electric fields of the light wave driving it. This is the energy that is re-emitted as a new photon.

Let’s do this once again, The photon comes in and sets the electron and it neighbours oscillating. This is where most (99.99976%) of the energy goes (to use your figures). The electron recoils in picking up the momentum of the photon and collects 0.00024% of the energy which is radiated as a photon of CMB. On re-emission the energy of the electron oscillations is given out as a new photon with, again, a little lost as CMB in recoil.
How can such sparse plasma involve such an amount of energy, says Sylas? Well let’s do some sums. Instead of involving a cast of millions as Sylas does let’s use just three electrons and give them the whole energy of the photon for good measure. If it works here then it will work anywhere.
The effect we are looking at is a linear one so, to simplify the situation slightly, let’s nail the outer two electrons down and let the middle one absorb the photon and oscillate between them.
Photon comes in from left to right and is absorbed, our middle electron is set into oscillation up and down between our two neighbours above and below it (whilst our electron recoils from left to right). We have done the recoil so let us look at the oscillations. According to Sylas the energy stored in this oscillation is equal to the energy of the photon which is 3.97e-19J. At the center of the oscillation all this energy takes the form of KE so what is the velocity of the electron? Answer is 9.5x10^5 m/s. Not a problem there is there since the random thermal energy gives the electron a velocity of 2.1x10^6m/s? Hardly an exceptional result is it? So there is no problem with the KE, let’s look at the PE.
As the electron oscillates upwards towards the electron above it, that electron repels our oscillating electron and slows it down. KE of electron is converted to electrical potential energy. Let’s work out how far our electrons will be apart when it is brought to rest and all of our photon energy, having been firstly converted to KE is now converted to electrical PE. We will ignore the effects of the lower electron as being negligible at this point.


!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!

Papageno,
We are still awaiting a description of what you mean by Nearly random motion of electrons in ac.
Cheers,
Lyndon

Now, what's the upward momentum of this electron? And where did that momentum come from, since we've already used the entire momentum of the incoming photon? Or, more precisely, the incoming photon had no net upward momentum. So if the electron now has momentum in that direction, it must be balanced by an equal momentum downward, so that the total remains zero, just as it was before absorption. Again, you fail to show conservation of both energy and momentum.

It seems that you were not paying attention.
What makes the motion only nearly random, is the effect of external field.
Add nearly random motions of many electrons, and you end up with a charge density that oscillates, while the lectrons still zip around nearly randomly.

It was addressed here, but you did not reply to it.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

Grey wrote
Not a problem Grey,
I trust that you do accept that the momentum of a photon is in the direction in which it is traveling?
Ah! You do. I trust that you also accept that the light is transverse and that the electric field causes the electron to oscillate up and down. Since this is accepted Physics I trust that you do.
Now, if you remember, I nailed the other two electrons down for simplicity. When your T/V aerial picks up radio signals the electrons move up and down either individually or in 'bulk' depending upon your view point. Where is momentum conserved here? Same thing.
Cheers,
lyndon

Where are your full calculations?
I suggest you take the BA's warning very seriously.


Apparently you do not understand what Sylas is saying.
Unlike you, he justifies the formulae and the assumptions he uses for is quantitative estimates.
You give only unjustified hand-waving.


Again confusing an macroscopic EM wave with a single photon.


You do not grasp the idea that a macroscopic EM wave forcing a charged particle in an oscillating motion, requires the partgicle to scatter many photons (many recoils, adding up to a oscillation).

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

OK missed that,
Papageno wrote
So what are you superimposing on your 'random motion' to make it 'nearly random' and I still don't understand what 'nearly random' is.
I think that you are just trying to avoid agreeing with me.
Cheers
Lyndon

There is nothing to agree with.
Random motion is for electrons that do not interact with other electrons.
Nearly random is for electrons whose motion is dominated by kinetic energy (thermal motion), with an external field acting on them.
How strong the effect is, depends on the scattering rate.
In a typical conductor, the scattering rate is many orders of magnitude higher than the frequency of typical AC voltages, which means that within one period of the electric field, an electron has scattered bilions or milions of times and the motion is effectively nearly random (with DC voltages you end up with a drift velocity, which is orders of magnitude smaller than the "actual" velocity of the electrons).
Add all the nearly random motions of the electrons, and, since these "nearlies" are correlated because they are produced by the same external AC electric field, you end up with an oscillation in the charge density.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

So what is the 'drift' velocity in ac. You have given it for DC but resorted back to this 'charge density term for ac. An ac currect switched on for 0.00001s must be like a dc so there must be an equivalent ac drift velocity.
Cheers,
lyndon

There is no drift velocit in AC, because the external electric field oscillates.
Unless, of course, you observe the system of time-scale much longer that the average scattering time, but much shorter that the period of the AC oscillation.

Now, let's get back to IG plasma, and explain why you do not provide your full calcuations to show us that you are right.

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

OK I agree,
You know, we know, the 'drift velocity' in ac is time dependent and shm you just won't admit it.
I gave my answer numerically to Sylas and showed where he went wrong in detail. As I always do.
The point I have just made with yourself is that it is what happens between collisions that matters because in IG space the electrons dion't make any during one cycle.
Cheers,
Lyndon