The drift velocity is the combined effect of the accelerating electric field and the scattering of electrons, which works like a viscous friction.
An electron is oscillating about a trajectory, but is zipping from scattering event to scattering, drifting slowly in the direction of the accelerating force.

Where?
Definitely not in the last two pages.

Except that your "theory" is based on electron-photon scattering, not on the effect of a macroscopic EM wave on an electron (maybe you still do not realize the difference).

__________________
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 you are saying that the electric field of the photon imparts momentum perpendicular to the direction the photon was travelling in (and recoil velocity)? Why does this cause oscillation rather than an additional scattering component?

Because of the other electrons imparting a restoring force? Assuming that would work (which it doesn't, as has been shown numerous times) you then have the problem of energy being diffused through the system...

How on earth does the electron reclaim that energy from the system to re-emit the photon?

So you should be able to easily show a quantitative analysis, demonstrating that both momentum and energy conserved at the same time. Yet you've failed to do so.

Light is a transverse wave. When you set up your T/V aerial, do you lie the metal rods along the direction in which the wave is travelling ie point the rods towards the transmitter? No you most certainly do not. You put the metal rods perpendicular to the line between yourself and the transmitter so that the elctric fields cause the electrons in the rods to move up and down.

It only gets diffused through the system at resonce 6 Hz in IG space. Otherwise it is contained within a very small number of electrons - as in atom.

The electron does not 'reclaim' the energy - the system of electrons emit a new photon, as in an atom.
Cheers,
Lyndon

My reply Sylas is fine on this.
Cheers,
Lyndon

Sorry, but no. The burden of proof remains on you, lyndon, if you expect to support your idea. You claim that energy and momentum are conserved, yet you're unable to demonstrate it, except by handwaving arguments. You claim that Sylas has done the math incorrectly, yet you're unable to give a quantitative analysis of the situation, and are instead trying to evade the issue.

I don't 'claim' it. I have explained why it is wrong many times and above i have done sums to back this up. Lets wait for Sylas' reply.
Cheers,
Lyndon

Your only quantitative response above dealt only with energy, as far as the transverse motion goes, not with momentum. Your only response to the question of the transverse momentum of the electron has been a vague analogy to radio antennae or television aerials. You have not demonstrated quantitatively that both momentum and energy are conserved. You brought up the simple model with three electrons. You need to actually show that the additional two electrons can provide the necessary restoring forces for the electron to oscillate, and that you can account for the energy lost by the photon while also balancing the momentum. A handwaving argument won't suffice.

An atom where each electron was ~1 meter apart? Anyway, the energy would be diffused to every electron within the deflected electron's field, regardless of resonance. What is the length of time between absorbtion of the photon and re-emission? That will allow you to calculate a rough estimate of how many electrons have become part of the system.

The system emits the electron? I haven't heard that before. How does that work? Where does the photon then emit from, precisely? How on earth does the system remember what direction to send the photon off in?

I guess that due to the oscillation, you have a fifty percent chance of the photon going in the right direction, and a fifty percent chance of it going back where it came from.
Just kidding, but the whole theory is becoming so nonsensical that kidding is all that is left to do...

__________________
Knowledge is a curse, but ignorance is worse

I am quoting this post:
There are two ways to show that photo-absorption by an electron is impossible.

• The simplest is energy momentum conservation. The electron can absorb the momentum, but in doing so it can only account for 0.00024% of the energy. I used this method here.
  
• Another way is the inconsistency of momentum conservation with impulse directions. By conservation of momentum, the impulse to the electron must be along the direction of the photon motion. But by consideration of the wave, the impulse must be perpendicular to the photon direction. The two requirements cannot be reconciled in an alleged absorption of a photon by an electron, and this is another reason it is impossible.

It is Lyndon's error... not anyone else's.... to fail to account for the inevitable transverse impulse and consequent scatter that always arises in photon-electron interactions. In real physics, there is always a transverse component, and therefore always a scatter angle for the photon.

Many BABB contributors are inclined to think these discussions are a waste of time. I don't agree. The point is not to teach Lyndon anything... that is plainly impossible. But looking into each new error and distraction is a chance to learn something, even if Lyndon himself never avails himself of the opportunity.

So how does photo-absorption occur in real physics? Don't physicists speak of electrons absorbing photons? Can't the objections I am raising carry over to photo-absorption or electron-photon interactions proposed in text books?

• In real physics, the absorption of a photon is never with an electron a meter away from other particles. It is with an electron in very close association with a positive nucleus; an atom. In this case, the fields push the electron one way, and the nucleus the other, and these transverse momentum contributions cancel. The two particles are given an increased separation, which corresponds to an increase in potential energy. In quantum physics the electron jumps to an outer orbital which means that the atom is in an excited state.
  
• In interactions between a photon and an electron considered classically as a wave interaction, motion of the particle sets up new fields that combine with fields of the incident wave. The end result is a new direction for the photon. This is essential to conserve momentum, and it also feeds back into motions of the particle while the interaction proceeds. Lyndon's unquantified comments about the wave neglect to consider direction change in the wave itself.
  
• In interaction with an electromagnetic wave, the particle is initially accelerated perpendicular to the photon's line of motion. Given this motion, the particle is then able to interact also with the magnetic fields associated with a photon, and this acts perpendicular to the field, and perpendicular to the particle motion, to contribute a further force parallel to the photon's line of motion. The effect is small, but it exists; but Lyndon has ignored magnetic fields entirely.

This serves to demonstrate that in real physics, you should get the same answer, even if you apply different kinds of analysis to the problem. No matter which way you look at it, Lyndon's interaction is impossible.

• It violates conservation of energy/momentum.
• It ignores the transverse impulse that scatters the photon.
• It uses photo-absorption calculations for particles that don't absorb photons.
• It uses unquantified hand waving to transfer most the incoming energy into the plasma; in conflict with the maximum possible transfer through the calculated forces.
• It brings all that energy back from the plasma, by means unexplained, to power re-emission of a redshifted photon.
• It confuses oscillatory motions of a particle with density waves in billions of randomly moving particles.
• It uses the wrong cross section. No published source; no set of tabulated results, no discussion or theoretical calculation, ever uses a cross section of 2rλ.

ROFL!

This oscillation only occurs while the electron is still interacting with the photon's electromagnetic fields. Once the photon is gone (either absorbed or emitted) this oscillation is over, and all that is left is a small residual impulse on the electron. Lyndon proposes that a photon is absorbed. So where does the energy go? He can't appeal to the oscillations that occur while the photon is still involved. It's gone mate, by his own description.

Lyndon should use his own figures. What I actually say is that 99.99976% of the energy is unaccounted for. Lyndon can try and put that into oscillations if he likes; but it is deliberately dishonest to pretend that I say the energy goes into oscillations. Oscillations are physically unable to get that much energy from a 500 nm photon; and I have backed that up by quantified consideration of the available forces.

If you want to involve other particles a meter or more away, be my guest, but quantify the forces involved. If the energy ends up in motions of any other particles, then you also have to get it all back to power the emission of a redshifted photon!

Here is Lyndon's attempt to manage the energy momentum budget. I have highlighted in blue one particularly reprehensible bit of deliberate dishonesty; and I shall show some trivial errors that make this unacceptable even for high school level.

What is this "according to Sylas" nonsense? I don't use oscillations. I point out that you CAN'T get oscillations that have the requisite energy. I point out that the ACTUAL energy of any subsequent oscillation for the electron is the kinetic energy it gets in the collision, which is more than five orders of magnitude LESS than what Lyndon quotes here. I've done this repeatedly, in post after post after post, and suddenly Lyndon pops up with an oscillation in which the kinetic energy is that of the photon; NOT the energy of the electron.

This is dishonest.

The above still fails to be an energy-momentum budget, because it never mentions the momentum!
Momentum of electron = sqrt(2Em) = 8.51e-25 kg m/s.
Initial momentum of the 500 nm photon = 1.33e-27 kg m/s.
Now the momentum does not balance. The magnitudes differ by a factor of 642!

In his own paper, Lyndon calculates kinetic energy for the electron as 9.65e-25, using conservation of momentum. [[See formula Q^2/2mc^2 on the bottom of page 3, and plug in the numbers.]] That is the energy for the electron that belongs in the above picture.

A high school physics student should be able to figure out how to give a quantified energy momentum budget. You have to calculate numbers for energy, and numbers for momentum, and show that they balance. Lyndon has never done this.

Cheers -- Sylas

PS. I had not heard of the Mössbauer Effect before reading Lyndon's stuff. Initially I accepted his description of straight line transmission with no scatter, and repeated it myself in some posts. I now suspect this is wrong, and that photons can be scatted in all directions. The distinguishing feature is not a lack of scatter, but negligible recoil in the atom, and hence negligible change in wavelength. I'll be checking this further to confirm or retract; whether I'm right or wrong I'll have learned something.

[[Edit to fix a typo in one formula; numbers unchanged.]]

Ashmore called his "effect" double Mössbauer, because he does not actually understand what characterizes the real Mössbauer effect.

This is the typical situation for Mössbauer effect.
Take a crystal lattice containing radioactive (gamma emitting) Iron isotopes.
The excited Iron nuclei emit gamma photons. If the energy of that photon is lower than the energy necessary to excite a phonon (quantum of lattice vibrations), the emission of the gamma photon is effectively recoil-less. The recoil is transferred to the whole crystal (10^23 atoms), rather than to the emitting atom (which would make the atom vibrate in the lattice).
The spectral line corresponding to this recoil-less emission, is very narrow, and makes some nice experiments possible.
For example, it has been used to measure gravitational red-shift on Earth, between the basement and the top of a tower.
In a teaching lab, I used it to measure the hyperfine field in Iron.

__________________
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)

This much I think I understand.

As a test of my understanding... it seems to me that the recoil-less emission should mean that photons can be absorbed by an atom, and then re-emitted with exactly the same frequency to all intents and purposes, with no energy loss to the lattice. The re-emitted photon can continue to pass through the lattice, possibly being absorbed and re-emitted a number of times, all with no loss of energy. But so far I see nothing to constrain it to remain in a straight line.

I've seen references compare this with "resonance fluorescence of the yellow D lines of sodium in sodium vapour". See, for example Recoilless nuclear resonance absorption of gamma radiation, which is Rudolf Mössbauer's Nobel lecture. In this case, the fluorescence means that the whole gas glows as photons bounce around within the gas. I guess the same thing happens for gamma photons in a crystal lattice at the excitation frequency of the Mössbauer effect.

If a beam of gamma ray photons of exactly the right frequency are directed into the lattice, I would expect to see photons emerging from the lattice at the same frequency and in every possible direction.

Have I got this right? When I first read Lyndon's paper, I got the impression that Mössbauer effect meant that the photons had no scatter angle, but now I think this is just another error on Lyndon's part. Or perhaps it is my error and I have misunderstood Mössbauer's lecture.

Unless I am badly mistaken, Lyndon's effect is completely the opposite of the Mössbauer effect in all the essential respects:

• The Mössbauer effect works only for very very tightly constrained frequencies. The Lyndon effect allegedly works right across the spectrum.
• The Mössbauer effect is distinguished by no recoil and no redshift. The Lyndon effect allegedly involves a redshift to the photon and two recoils to an electron.
• The Mössbauer effect involves something comparable to fluorescence, in which a lattice "glows" with gamma ray photons emerging in all directions. The Lyndon effect allegedly involves no scattering at all, with the photons emerging with direction unchanged.

It is the last point on which I am least certain.

Thanks for any help -- Sylas

When you jump, the Earth recoils.
However, the Earth's mass is so much larger than yours, that the recoil is negligibly small.


The gamma photons are not constrained on a line.
Whether the emission (absorption does not seem very likely, since the gamma photon needs to hit a nucleus) is recoil-less, depends on the lattice.
The phonon spectrum is not necessarily isotropic, so you might have absence of recoil in one direction, but recoil in another direction even though the energy is the same.
The emission by the nucleus is isotropic.*


There is definitely similarity between transitions between atomic levels, and transitions between nuclear levels: the difference is the energy scale.
However, I do not expect absence of recoil in gas atoms, because the atom are not constrained. So I would expect broadening of the spectral line in the fluorescent emission.


You have to consider the probability for a nucleus to absorb a gamma photon.
Then you have to consider the spectrum of the lattice vibrations, to see when recoil is absent.


Don't worry. Ashmore does not understand the Mössbauer, hence his use of the name is misguided.


The main point is that the recoil is absent because there is not enough energy to excite lattice vibrations.
Normal emission: jumping from a spring-board.
Mössbauer emission: jumping from a hard concrete wall.



* Well, it depends on magnetic field and temperature.
For anisotropic gamma emission, search for low temperature nuclear orientation.

__________________
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)

Hi Sylas,
I am still awaiting a reply to my post. The long 'epistle according to Sylas' above answers nothing but repeats all your wrong sums.
You have spent the last 20 odd pages treating light as a longitudinal wave. This has been pointed out to you time and time again. We would like a response.
In using mv^2/2 to represent the whole of the photon energy, using v as the recoil velocity, you are saying light is longitudinal.
Can we have a response to this point please, and then you can go on to proving Mr Mossbauer wrong.
Cheers,
Lyndon

There is no point anyone even considering answering that question unless you somehow manage to resolve the numerous issues surrounding your electron 'oscillating'.

I will repeat the last issue I had...
An atom where each electron was ~1 meter apart? Anyway, the energy would be diffused to every electron within the deflected electron's field, regardless of resonance. What is the length of time between absorbtion of the photon and re-emission? That will allow you to calculate a rough estimate of how many electrons have become part of the system.

The system emits the electron? I haven't heard that before. How does that work? Where does the photon then emit from, precisely? How on earth does the system remember what direction to send the photon off in?

And I'm still awaiting a response to my post. You know, one where you actually show mathematically that your system of three electrons that you were using to model this interaction can absorb a photon and still conserve both momentum and energy, rather than just making vague references to radio antennae.

Vague, vague?
It isn't vague at all.
momentum is only conserved if there are no external forces acting. In "physics according to Sylas volume 1" he expects momentum to be conserved, but we don't.
The momentum of the electron in a direction perpendicular to that in which the photon was originally travelling is not conserved because there is an external force acting on it - due to the oscillating electric fields of the photon. This is why it gains energy and momentum in this direction.
If one wants momentum to be conserved then one must look at the universe as a whole - including when the photon was transmitted.
A wibbly wobbly when it was created is cancelled by a wibbly wobbly when it was absorbed.
Cheers,
Lyndon
If one momentum

This is standard physics.
The frequency of oscillation of the electrons about 1m apart in plasma of IG space is a published result, about 30 or less hertz.
In any case, this is how light travels through glass. I am not inventing anything new here on this point, I am just applying known physics of transparent materials to space.
And getting the correct answers - H = 2nhr/m.
Cheers,
Lyndon.

Sure. The response is that you are still deliberately lying about my treatment. It does not use or depend on the wave at all.

My analysis has been simply conservation of momentum. Invoking the wave directions does not absolve you from the responsibility of balancing the initial momentum with the momentum of the final state.

You have a photon with momentum in a certain direction. That momentum must be conserved, and it is not conserved in your reaction. You can do an energy momentum budget just by adding up the energy and momentum of all the particles involved in the reaction.

I take it back. High school students can do energy momentum budgets; but we have yet to see a complete balanced budget from Lyndon, that takes both energy and momentum into account and shows them conserved.

But let's talk about waves, if you prefer that approach. This gives another proof that the photo-absorption by an electron is impossible.

The wave is longitudinal. It therefore gives the electron a push to one side. Therefore the photon cannot be absorbed, since it must go off to the opposite side to counterbalance. Equal and opposite reactions -- but violated if the photon is absorbed. Therefore absorption is impossible.

If you EVER get around to doing a complete energy momentum budget, you will have to consider that the electron DOES get pushed sideways, and you have to find SOMETHING ELSE pushed the other way to balance.

In real physics of photoabsorption, it is the positive nucleus.

In real physics of Compton effect, it is a scattered photon.

In the Lyndon effect, there is nothing to compensate the sideways movement of the electron, and so the effect is a violation of trivial school level physics.

Cheers -- Sylas

The 'we' you are talking about, is that a pluralis majestatem, or do you seriously think that the rest of us is waiting for Sylas to respond as well? I can't speak for everyone here, but I for one expect better answers from you, while I'm quite satisfied with the efforts of Sylas thus far.
You are always going on about what you see as his wrong sums and random numbers, but you don't give anything better or even a good reason why we should doubt Sylas.

__________________
Knowledge is a curse, but ignorance is worse

You've never given a quantitative analysis of the total momentum and energy conservation of the system, so, yes, I call that vague.

Momentum is conserved in any closed system. So for example, if we consider the system of the incoming photon and the plasma, and ignore outside interactions, we should be able to show conservation of both momentum and energy. In the broadest sense, yes, perhaps the only truly closed system is the entire universe, but I'm sure you'll agree that we can assume conservation of mometum and energy when working with orbital dynamics in the solar system, say, without needing to take into account interactions with the Andromeda galaxy. Or will you agree to that? Let me know, if you don't.

Similarly, if we have a photon interacting with a plasma, we can consider just the plasma and the photon as a closed system. If there are other outside forces involved with the photon and the plasma that prevent conservation of momentum and energy, you should be able to identify and calculate them. If not, you should be able to work out the details, assuming that only the photon and the plasma are involved, yet you've failed, again, to give a quantitative analysis of the matter.

If only including three electrons (the one doing the absorption and its nearest neighbors, presumably) won't allow you to do a quantitative analysis, then first I'll ask why you brought up that model in response to questions about conservation of energy and momentum. And second, I'd expect you to expand your simple model, so that you can show it. Remember, burden of proof here is on you, if you expect to convince anyone to take your ideas seriously.

The electric field of the photon isn't an external force, it's part of the system in question. The incoming photon has a definite momentum, the electrons with which it's interacting have a definite momentum. As with any interaction, you can add those all up, and it has to be the same before and after the interaction takes place. Or are you actually saying that your theory violates conservation of momentum?

Sylas, you're letting Lyndon get to you. I'm sure you meant to say that the wave is transverse here (especially since you then speak about the wave being deflected to the side, which wouldn't be the case for a longitudinal wave, so it looks like a mistake of words rather than one of logic). Perhaps having Lyndon continually saying that you're treating the wave as longitudinal left you with that word on your brain.

Sorry, you didn't answer a single one of my questions there.

I didn't ask for the oscillation frequency, I asked for the amount of time between photon-absorbtion and photon emission. I'm not the first to ask this. So how long is it?

You also failed to answer the question of how a system of electrons produces a single photon. How does a system of electrons suddenly behave as one to produce a photon? Where is the photon emitted from, exactly? By what process is it emitted? How does the system know where to fire it to prevent scatter?

Edited to clarify questions

I am not getting to Sylas at all. He is continuing to show his true colours here. Throughout this thread he has done his sums on light being longitudinal.
He has worked out the recoil velocity of the electron and, when he finds that he cannot equate the energy of the photon with that picked up in recoil, he says "where has the energy gone" . In transwerse waves they are separate entities. One deducts the energy picked up in recoil from the energy of the photon and it is this that produces the redshift.
Cheers,
Lyndon

You know, you could settle this real easy. Instead of a lot of word salad, why don't you just produce the quantitative analysis showing the conservation of momentum and energy in your process? Sylas has produced his work, you haven't. Sylas has the calculations to back up his claims, you don't, it's that simple. Until you produce the analysis, your rebuttals to Sylas, without showing the calculations, are not very convincing and make it appear your main effort is to avoid having to provide the calculations. Which makes me think you either are unable to produce the calculations or you could, but don’t dare because you have done them and know they show you to be wrong.

__________________
Some try to tell me, thoughts they cannot defend,... - Moody Blues.

Neptune- The original Dark Matter.

The author feels that this technique of deliberately lying will actually make it easier for you to learn the ideas. - Donald Knuth

Tensor wrote

Its already settled.
I have shown my working many times over. Conservation of energy and momentum is BUILT IN TO MY CALCULATIONS I work out how much momentum and hence energy the electron gains assuming conservation of momentum. I assume conservation of energy and deduct the energy gained by the recoiling electron from that of the incident photon to get the redshift. They have to apply because I have built them in to my calaculations.
Additionally, Sylas has shown us his sums and I have shown him time and time over where his sums are wrong.
The matter is settled, lets move on. Light is transverse - trust me.
Cheers,
Lyndon

For the sake of argument, let's assume that you are not confusing -- again -- macroscopic EM waves with a single photon.
Those "oscillating electric fields of the photon" would not be an external force: their would be the internal force responsible for the electron-photon scattering.
The only external force in the scattering of the electron and the photon is the Coulomb interaction with the other charged particles in the plasma.
You have mto show that the time interval in which the electron-photon scattering occurs, is long enough for the external forces to have any effect on the system.

I urge you again to take the BA's warning seriously, and address the point: provide your full calculations in support of the basic mechanism for your "theory".
Stop hand-waving and tap-dancing around the issue. Your broken-record tactic does not equal addressing the point.

__________________
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)

Not once in this thread have you shown a full treatment showing conservation of both energy and momentum.

You use conservation of momentum to calculate how much energy the electron gains in the direction of the photon's travel. But then you just assume that the rest of the energy is taken up by transverse oscillations and you completely ignore the momentum involved in those transverse oscillations, so your treatment does not conserve momentum.

The issue is not whether light is a transverse oscillation. The issue is that you have, in spite of repeated requests, failed to show conservation of energy and momentum. The matter won't be settled until you give a quantitative treatment showing that.

This is just perfectly normal absorption of a photon. Are you saying that photons are never absorbed?
But lets look at things in more detail. I seem to be cast as the 'bad guy' in all this (which is totally untrue 'cos I am quite nice) so lets have a look at some of Sylas' sums that you all seem to agree with.
On page 20 Sylas said:
Do you agree with this piece of Mathematics? Here Sylas works out the escape velocity for two electrons. Do you really expect me to respond to this sort of nonsense? However, I did point out his mistake and received no response from Sylas. Anyone can make a silly mistake but to actually go ahead and calculate an escape velocity between two repulsive electrons is a a bit more than a ‘silly mistake’

On page 22 we have,
This looks impressive doesn’t it? This is the sort of thing you are all asking me to do as well! Do we all agree with this piece of mathematics? Hands up those who think it is correct? Well it is total nonsense. Here Sylas says that to absorb a photon of light of wavelength 5e-7m will require 400,000 electrons to balance his energy budget. I did it with three electrons and two nails.
But wait a minute, 400,000 electrons to absorb a single photon of light? Hydrogen atoms can not only absorb certain wavelengths of light of this order but they can absorb photons with much more momentum and energy in the UV - and do it individually as atoms in a gas. Hydrogen atom has a mass of about 2000 electrons so why does Sylas need the equivalent of 400,000 electrons to absorb a single photon of light. Rubbish. He is saying that unless you have 200 Hydrogen atoms the momentum/ energy ‘budget’ doesn’t work when a hydrogen atom is excited by a single photon.
I have explained were he is going wrong but the guy won't listen.
I have no intention of trying to repeat this sort of nonsense.


Ah! I see your plan. If you can't prove the theory wrong....
Cheers,
Lyndon