Biting the hand that feeds me here... but 1030 is obviously 10^30. This is a common effect of moving between formats with different capacity for superscripts. The whole page uses this form fairly consistently for powers of ten.

As a crude estimate, the number is not bad. 10^29 would be better, but for a back of the envelope 10^30 is easier to remember.

Diamond has about 1.76e29 atoms per cubic meter. A quick look did not turn up anything with more atoms per unit volume.

Lead has about 3.3e28 atoms per cubic meter. Most solids will be somewhere between these two values.

Cheers -- Sylas

Lets not feel too sorry for Sylas, I am sure that he will agree that
"Those who live by the sword die by the sword"
As for the page on my website, yeh, thanks for pointing that out. I was in the process of updating it and as Sylas said, am swopping over from 'pagebuilder' software to 'frontpage'. The cut and paste lost something in the translation ('pagebuilder is hopeless with mathematical symbols). I will sort it and add some more to the site as soon as I have converted Papageno and Sylas over to Tired Light. Next week.
Cheers,
Lyndon

I just noticed this:

A photon with a frequency of 6Hz has a wavelength of 50,000 kilometers. The collision in this case is indeed with many electrons at once.

There is a sharp change in the nature of the interaction at the plasma frequency. Below this, photons are able to excite plasma oscillations and become absorbed, with great efficiency. The plasma is opaque. Above this, and photons are not able to excite plasma oscillations, and the plasma is transparent.

The visible light photon, with a wavelength of 500 nanometers, does not collide with many electrons at once. The analysis of its interaction can be confined to a single electron. Its frequency is 600 TeraHz, and it does not transfer energy into the plasma, except by Compton scattering from a single electron with the consequent change in direction.

Lyndon disagrees with this, of course, but is unable to give a balanced quantified analysis of energy and momentum showing with numbers where the the momentum and energy goes in his alleged interactions.

Cheers -- Sylas

PS. Added in edit. No need to feel sorry for me at all. I cheerfully acknowledge that I am an airchair physicist. I make errors from time to time, and some subjects are well beyond my ability. I like having my errors pointed out, by anyone. No-one dies, or is even much inconvenienced; unless the errors are not pointed out or addressed of course.

Lyndon is under no obligation to pay the slightest attention to my comments. I have no particular hope or expectation that he'll ever learn a thing from my pointing out his errors. I'll keep doing it, however. I've learned quite a bit about plasmas in these threads, and I suspect that there may be readers who are learning other aspects of physics by trying to follow along and see where the errors really are.

Sylas,
In your fairy tale over the page why did you choose:
What has this got to do with anything?
Cheers,
Lyndon.

You spoke earlier, in this post, of how long the absorption interaction would take, but gave no numbers of your own. So I tried to pick a generous number.

The number I picked is the time it takes a photon to move from one particle to another in the plasma. The order of magnitude of distance between particles is a meter or so. Light travels that distance in 3.33e-9 seconds. It's also a nice round magnitude (1/3) for back of the envelope calculations, which is all we really needed. You've been speaking of an interaction between a photon and a particular individual electron, so this time is a sensible upper bound. More than this, and we can't really be talking about a particular electron any more.

Another way to proceed is to think of the cycle time for the photon. For the 500 nm photon we have been considering, the frequency is 600 teraHz, and the cycle time is1.67e-15 seconds. I've allowed enough time for 2 million cycles.

You tell me. It's your absorption reaction. How long do you want to allow for the absorption of an photon by an electron? Is 2000000 cycles not enough?

In other reactions used in real physics, it's nothing like that big, but hey. You gave no numbers and no analysis of your own, and I knew there was plenty of room to spare. So I picked a large time interval, and showed that even this is nowhere near enough to allow for the transfers of energy and momentum required to make up for the fundamental mismatch that has been pointed out lo these many threads.

Cheers -- Sylas

It's the time it takes for the next particle over to be able to have any reaction to event at the electron we're looking at. EM forces propogate at the speed of light, so the light-travel time is a reasonable limiting timescale to use to look the actions of electrons in your theory, Lyndon.

(I see I've had this page open too long, and Sylas has beat me to the answer, but mine is also valid.)

No, Sylas, this is your calculation here, you tell us. So apart from being:

What physical significance does the time between collisions have to the time for collision?

You have posted a Pseudo scientific 'crank' post which is purporting to tell us something and it is based on estimates chosen because they are If you have no clue about what the quantities involved are, then why do the sums and, more to the point, why post it?
Cheers,
Lyndon

So the 'reaction' is passed on 'one electron at a time?' Is this what you are saying?

Ah, so your answer is based on "nice numbers that fit onto the back of an envelope" too! Very scientific
Cheers,
Lyndon

Shrug. Suit yourself. I told you in the immediately preceding post, but I'll do so one more time.

The time is an absurdly large number, which definitely has no credible significance as the time for Lyndon's reaction. The real time for an absorption has to be far far less than this; and so the number I am using is a valid upper bound.

Because Lyndon does not give quantified values for various essential features in his interaction, and because his interaction is not something covered in any real physics reference, I propose to give the bounds within which Lyndon will have to work, should he ever decide to quantify the time of the interaction himself.

Lyndon has never provided a quantified analysis of energy momentum in which the distributions of energy and momentum are identified, quantified and balanced, excepting only the time that he introduced a variable electron rest mass.

I did not use the time between collisions. I used the travel time between particles, which is about the duration for which a photon is closest to one particular particle of the plasma. The alleged absorption interaction has to resolve itself well within this time, given the need to conserve momentum and energy. It's also time enough for 2000000 cycles of the photon frequency. That's a huge upper bound for the time it takes a photon to be absorbed.

Because it suffices to show that even if Lyndon allows a very large amount of time for the alleged duration of the electron-photon collision, it still will not be enough time to transfer any significant amount of energy and momentum to other particles in the plasma.

The point is that photon absorption by electrons in rarefied plasma is not physically possible. The energy momentum budget cannot match.

To review, the various proposals Lyndon has given in an attempt to find some other energy or momentum sink are as follows:

• Variable electron rest mass. The only one quantified, and a simple error in particle physics.
• ”Steadying” of the electron with emission of a microwave photon. Fails by a bit over three orders of magnitude.
• An oscillation of individual electrons. Does not actually give any extra energy term; the energy of any induced oscillation of the particle is the same as the kinetic energy given in a collision, and which has already been taken into account.
• A transfer of energy to other particles in the plasma as density oscillations or phonons. This fails by many orders of magnitude as well, because the forces between particles in the plasma, and the duration of the collision, is far below what is required to makeup the mismatch in the energy momentum budget. This was the point of the most recent analysis.

By the way, in my view the really ridiculous part of my analysis is where I summed the magnitudes of forces. Sometime I'll give a stronger bound, taking an upper bound on the forces by having all positive particles in one half of the Debye sphere, and all negative particles in the other. This will also give a huge upper bound; but that's all we need to show that Lyndon's absorption reaction is physically impossible.

Does anyone have a physically sensible model for the forces that will be felt on a particle from its neighbours within the Debye sphere? It should be a continuously changing force as nearby particles are moving with the termal energy, a bit like a Brownian motion, but from electrical interactions.

Cheers -- Sylas

PS. This is certainly back of the envelope stuff, which is far better than vague and dubious analogies with no numbers. It really only takes quite elementary physics to see the errors in Lyndon's physics. Some of it, like the notorious errors in dimensional analysis, is early high school level. Other errors involve more advanced concepts, but can be resolved without getting into very difficult analysis. All in all, a good learning opportunity for people who want to try their hand and finding errors in crank physics.

I was expecting you to misinterpret that. As Sylas says in his most recent post, it is a valid upper bound. That makes it very back-of-the-envelope. But it is okay to test the water temperature with your toe instead of doing a cannonball into the deep end. It could save you some trauma in the end, as well.

Nope. What I'm saying is that the change in the properties of an electron can only propogate at the speed of light (unless your paradox here has some "proof" that this aspect of GR is wrong). It's postion, velocity, etc, will be transmitted and responded to at this speed. So your "restoring forces" and "SHM" won't start until the light-travel time (between the electron in question and any other particle) has passed after the collision. That's what I'm saying.

It's too late, I'm too tired, I know this will come back and bite me, but I'm going to post it.

Actually, I just got a really interesting idea that seems to answer the point of my original intended post. Two-photon emission. That's the only solution I see to the energy budget problem. Even if it were an inelastic collision and energy were dissipated by something other than the electron and CMB photon, another photon seems to be the only viable option.

Now, before Lyndon leaps at the chance to use this, there is a definitive way to predict the energy probability distribution of the second photon based on the energy of the incident photon. I'm just way to tired to hash through the details right now. This could be an interesting way to quantify redshift through each collision as well. I'll pursue this more when I wake up, if someone hasn't pointed out the hole in this that I'm too tired to see. ( )

Oh, heres a potential hole: there may be a spin conservation issue. :-k Hmm. I'll sleep on it. 8-[

This has got me confused, ok, most of this thread confuses me, but, I really don't understand what you mean here.

Are you saying that the photon simutaneously can collide with many electrons over a span of 50,000km or that it hits them over the course of one wavelength.

Be gentle, I'm no physisist

__________________
MrObvious

Yes... I considered this also, in this post. There are some other conservation laws and principles that may apply, and I'm not good enough at quantum physics to identify them.

You can calculate what is permitted by energy budgets as follows. First, split the original photon into the red-shifted photon you want to continue, and another one carrying the difference in energy and moving in exactly the same path. Then take the second photon, and do a Compton interaction with the electron. End result: one redshifted photon on the original trajectory, plus a photon and a recoiling electron.

Electrons don't split up photons in this way, but the energy budget works.

It does not match Lyndon's descriptions, however. There is no absorption; a free electron cannot absorb a photon. There is also no basis for the "calculations" Lyndon gives for the expected redshift. This method works with any redshift you like.

Cheers -- Sylas

Tobin's right, you'll run into problems with spin. A photon has a spin of 1, so there's no way to conserve spin in an interaction that starts with one photon and one electron and ends up with two photons and one electron.

Sylas, nice work. Lyndon, for all your criticism of Sylas's back of the envelope calculation, you seem to have missed an important point. In every case where he made a rough estimate, he gave your model the benefit of the doubt, and estimated in the direction that would make it most likely to come out physically acceptable. In some cases, he did so by a few orders of magnitude (such as assuming that all of the interacting particles in the plasma were electrons). And your idea still fails conservation of energy and momentum by several orders of magnitude more. So if it doesn't come close to working with the most generous estimates that are reasonable, it can't work if you put in better estimates (which will be less generous).

And do you really still think it's meaningful to compare H and hr/m directly, even though they do not have the same dimensions?

Grey, thankyou for confiruming what I was thinking about Sylas's estimations; that while Sylas's figures for things such as the interaction time are 'back of the envelope', they are bounds which, when tightened around the real figures, just cause Lyndon's theory to be more and more implausible.

So the good thing about posting tired is that my follow-up will be posted before I get to it. :wink:

This is exactly what I was going to say about the whole situation. Most of it had entered my head last night, but wasn't quite coherent enough at that hour to post it. As Grey said, spin conservation also kills this before it leaves the gate. Well, good. Thanks, guys.

So it looks like we are back to Lyndon's Tired Light theory!
I understand that there are 'back of the envelope calculations to show that bees can't fly, that supertankers couldn't bebuilt. I had a set of encyclopedias when I was little that said a 'back of the envelope calculation' had said that man would not get into space let alone onto the Moon.
In the end what matters is what happens, not what is written on the back of second class mail.
Cheers,
Lyndon

You have an amazing ability to draw conclusions wich have nothing to do with the previous debate, and also to mock people who at least take the effort to put some numbers to your handwaving.
We (well, they) are showing you that 'what happens' is not what you describe, but for some reason, that doesn't matter to you. I don't know where you have written your theory on, but it is wrong, and that has been shown to you numerous times in the previous thread and this one.

__________________
Knowledge is a curse, but ignorance is worse

Um, no. The point (which perhaps you missed) is that your model can't possibly work, because it violates basic conservation laws. Unless you'd like to propose that conservation of momentum and energy aren't valid, but I'd want to see some hard evidence to back up such a claim.

Light behaves in some ways like a wave, and other other ways like a particle. Generally, the more energetic it is, the more it seems like a particle; but the longer the wave length the more it seems like a wave.

There are some intriguing experiments, the results of which cannot be understood except by recognizing that light is both a wave and a particle. One of these is the single photon/double slit experiment. Light is directed through two slits, and we find an interference pattern. This is a wave-like phenomenon, and most easily explained in terms of a wave passing through both slits, and interfering with itself to give the pattern. But we can make it particle-like as well, by using light of such low intensity that there can only be one photon in transit at any time, or by using photoelectric detectors to see where the light arrives. Such detectors reply on particle-like properties of light.

Such a set up still shows an interference pattern. Somehow, even if considered one photon at a time, the photon path depends on both the slits. Sometimes people speak of the photon passing through both slits at once. This is an instance of wave particle duality, which is at the heart of quantum physics. Quantum physics is (IMO) one of the hardest parts of physics to grasp. Even if you can crunch all the numbers and formulae correctly, finding an intuition for what you are calculating is difficult and tends to defy some very basic assumptions we have about how the world works.

Returning to the photon with a wavelength of 50,000 km. Such a photon does not have a clearly identified location. Its trajectory will be impacted by many electrons over a wide area, in much the same way as a photon trajectory may be impacted by having a range of alternative slits that interfere with each other. Put another way, such a photon does not "see" individual particles in the plasma. It sees the combined fields, and interacts with the plasma as a whole.

However, if the photon is highly energetic, then it is more localised. A visible light photon, with a wavelength of 500 nm, can only "see" one electron at a time in the rarefied intergalactic plasma.

In reading the notes on plasma physics that have been cited in the thread, and also further notes from the same sites and sources, the analysis of index of refraction and speed of propagation for light in a plasma is treated with wave analysis. The more energetic the light, the less propagation speed is affected. One of the terms in the formulae is sqrt(f^2-p^2), where f is the frequency of the electromagnetic wave and p is the plasma frequency. This becomes imaginary with f is less than the plasma frequency, which corresponds to damping of the wave. But above the plasma frequency, and you have transmission with reduced velocity but no energy loss. For a frequency far in excess of the plasma frequency, the difference in speed of propagation is negligible, and there is effectively no refraction.

Cheers -- Sylas

PS. The stuff on plasma is based on what I have been learning just in the last week. Don't take it as expert opinion. Comments or corrections welcome.

Lyndon, why do you refer to the underlying process as the Mossbauer effect? I thought that that was effectively all about coupling the recoiling nucleus to the far greater inertia of the crystal lattice, and hence reducing the energy loss of the emitted photon due to recoil. If you're not coupling to a greater mass (which your equation in terms of the electron mass indicates that you are not) then it surely isn't the Mossbauer effect.

As you say that your paper has been accepted for publication in a peer reviewed journal, I would assume that they would have spotted this, so perhaps I have misunderstood why you are calling it the Mossbauer effect.

Bad assumption. The journal in question is Galilean Electrodynamics. These clowns are not doing Lyndon any favours.

Having errors pointed out may be painful for some people, but it is an essential part of developing useful ideas. The journal reviewers appear to be incompetant, and this merely allows one to go public with all the errors flapping in the breeze. No additional prestige or credibility results from such publication.

Cheers -- Sylas

Thanks Sylas, that makes it much clearer for me. The photon at a time with the double slit experiment cleared things up for me greatly.

Much appreciated mate

__________________
MrObvious

I would suggest that you edit your last post Sylas. To refer to a named third party in such a way is unfair to them and sounds like sour grapes on your part, rather than informed discussion. This said, I will not address any points of this nature again.
Cheers,
Lyndon

I identified the journal that published Lyndon's paper, and expressed a very low opinion of its quality, and the quality of its review process.

It is hard to understand this comment by Lyndon as anything but a refusal to allow negative comment to be made about a journal. Balderdash. There is nothing remotely unfair about public expression of a low opinion of a named journal, or the breathtaking incompetance of unnamed reviewers. It is perfectly standard to evaluate any journal by the quality of any paper that get through its review process. That's exactly what I have done, and to call this unfair is just bizarre.

It is also easy to check over some sample papers accepted in the past through the link I provided to their own home page.

Lyndon's comments are getting just getting more and more out of touch.

Cheers -- Sylas

Yes and no. When one gives an effect, 'good science' requires one to give both sides.
For instance, if we say "how does a mercury thermometer work". Most books will say "because mercury expands as its thermal energy increases". Which is true but doesn't answer anything. The 'good science' answer is mercury expands but the glass does not hence the mercury climbs up the tube".
Same with the Mossbauer effect , to say "nuclei 'coupled' to a far greater mass do not recoil when they emit gamma rays" is meaningless unless one also states that "those that are only 'loosely coupled' actually do recoil". Meaningless because without the second statement all nucleii might not recoil, coupled or uncoupled. And then there is no effect.
However, the same effect occurs with electrons and atoms - they only look at nuclei and gamma rays experimentally because for electrons and atoms their thermal motion produces doppler effects that 'mask' the Mossbauer effect - so we don't see it. It is still there.
As a result it is correct to use the term Mossbauer effect for my interaction.


All I can say is that the paper has been peer reviewed several times - including by journals that even Sylas would not libel, and whilst they have whinged at this or that, no one has ever complained about the use of Mosssbauer.
That said, we seem to spend so much time arguing over the name, I am coming to the conclusion that it may be prudent to drop it to concentrate peoples attention on the real issues.
Cheers,
Lyndon

It has been peer reviewed and rejected several times, I assume. Can you give us the reasons they did? That may be interesting.
And when has it been published? I've only seen the mention that it has been accepted (in september, I believe), but no mention of the actual publication yet.
Finally, what Sylas said was not libel, it was an opinion (and an informed opinion at that, as it was based on articles actually published by the magazine).

__________________
Knowledge is a curse, but ignorance is worse

Back of the envelope calculations are fine if the estimates used have some meaning. I have pointed out such areas in Sylas' calculations already - time of impact for one. Remember, there are 'sticky' patches on the back of every envelope and this is one for Sylas. As a result, one can't do that here until one has a good physical reason for coming up with a ball park figure.
What to do then if back of envelopes are out?
Simple, suck it and see!
That is try it and see what you get. If your final answer is way out then you know the effect does not work.
So, lets do a "proof of the pudding is in the eating" analysis.
I say the effect works, Sylas says it does not so lets assume it does for a moment and try it.
Increase in wavelength on each absorption and re-emission is h/mc - the compton constant.
The published result for the photoabsorption collision cross section is 2rλ.
Putting this into our calculation the total shift in the wavelength for a photon travelling a distance, d, can be found and hence the redshift.
For astronomically 'small' distances (z less than 0.2) we get H = 2nhr/m.
We know h, r and m for the electron already so we will look up what other scientists are using for n and that is 0.1 to 10 electrons per cubic metre. For H = 72 km/s per Mpc, n should be about 0.6 per cubic metre. This result is certainly consistent if not even pretty good. Notice that the only thing we have assumed is that the electron oscillates and tales up the energy - everything else is 'known' physics and published by scientists other than I. Once you accept that it works everything falls into place.
The proof of the pudding is in the eating.
But cry the cynics, it could be a coincidence? Maybe, so what to do?
The proof of the pudding is in the eating.
What about large distances? Well since the wavelength of the photon redshifts as it travels, the collision cross section increases as it travels and this gives us an exponential hubble diagram. That is, accept the electron takes up the energy and one not only get a good match with the value of H but one gets the exponential nature of the Hubble diagram too. I had this long before it was 'accepted' that the Hubble diagram has an exponential nature - the Bb had to drag in inflation.
The proof of the pudding is in the eating.
But, Cry the cynics, that could be coincidence too.
What do we do?
The proof of the pudding .....
Lets have a look at the energy transferred to the recoiling electron. How much energy is transferred there. The plasma will reach thermal equilibrium eventually and re radiate this. What will the wavelength of the photons be?
Shock horror, its in the microwave and here we have all this CMB. That is, accept the electron takes up the energy and one not only get a good match with the value of H and the exponential nature of the Hubble diagram but one gets the CMB too!
and on and on till we are totally sickof pudding.
This is the difference. Sylas trumps up 'estimates' with no physical meaning, does a calculation and says the theory is wrong. But is he generating random numbers?
I try it and find it works excedingly well.
The proof of the pudding is in the eating.


Yes, because I compare H and 'hr/m in each cubic metre of space'
That is, just why is the measured value of the Hubble constant equal to 'this much (hr/m) of an electron in each cubic metre of space? This statement reaks of Tired Light. I say that redshift is due to photons interacting with electrons in IG space as they travel and then we find
"the measured value of the Hubble constant is equal to 'this much (hr/m) of an electron in each cubic metre of space".

The question you should be asking is
"If the hubble constant has nothing to do with the electron why is the measured value of the Hubble constant equal to 'this much (hr/m) of an electron in each cubic metre of space?"
Cheers,
Lyndon

What makes the cubic metre of space so special? Why not a cubic kilometre? After all, the Hubble constant is only a number per second, so cubic metres have nothing to do with it anymore than cubic inches or cubic lightyears.

__________________
Knowledge is a curse, but ignorance is worse

Then use the full form of "why is H = 2nhr/m", it is fine by me.
Cheers,
lyndon