Ashmore's paradox

[lyndonashmore]

gritmonger wrote

Well, since I appear to have been brought into the discussion- the conjecture appears to be the fewer electrons the photon hits, the more the redshift- then shouldn't very large very close objects be much more redshifted than smaller or even point-like objects? I mean, their light goes through more electrons, essentially mimicking a dense plasma, if spread over time.

You would have joined the discussion anyway! No the electrons have to be in a plasma where they can perform SHM

Additionally, it appears that there should be redshifting in light going through glass or water: small, as you state, but small does not imply immeasurable. This would be a perfect test of your theory! Where could we get an appropriately large amount of transparent material? Fiber optics?

Amazingly thin strands of glass ... that can be thousands of miles long are used to deliver the optical network’s payload.
http://web10.eppg.com/betabooks/oct01/velte/ch01.html

This same book excerpt mentions attenuation of the light- but not by redshifting: by absorption when hitting impurities. On, or off: no redshift. Very pure optical glass thousands of miles long produces no redshift in a single-frequency laser light passing through it; just attenuation from some absorption.

So far the test does not appear to bear out tired light through photon absorption and re-emmision.

[/quote]
You need to read a book on QED or special relativity. French states "the recoil may to all intents and purposes be completely surpressed for atoms in a crystalline lattice" Ergo no redshift in transparent solids.

[ExpErdMann]

Lyndon, I see in your paper there is a connection to Marmet's work. John Kierein is not mentioned but there are similarities there too. You might want to get John's reaction - he's on this board.

[gritmonger]

You would have joined the discussion anyway!

I'm sorry, is that an indictment of some sort?

No the electrons have to be in a plasma where they can perform SHM

You need to read a book on QED or special relativity. French states "the recoil may to all intents and purposes be completely surpressed for atoms in a crystalline lattice" Ergo no redshift in transparent solids.

Gosh- and you'd said earlier there would be but it would be small. That still doesn't address the redshift getting larger the fewer electrons it encounters... and why electrons of atoms in a disorganized glass state - not a crystal lattice, by the way - would necessarily obey this rule as well.

A noncrystalline material, which has no long-range order, is called an amorphous, vitreous, or glassy material. It is also often referred to as an amorphous solid, although there are distinct differences between solids and glasses: most notably, the process of forming a glass does not release the latent heat of fusion.
- from the Wikipedia entry on crystal

[lyndonashmore]

papageno wrote
As far as I know, Hubble constant is not derived from hr/m.
It might be just a coincidence. Why do you link them?

Because they have the same value. The Hubble constant is derived from hr/m that is the whole point of this discussion topic

Photoabsorption is still referred to electrons bound to nuclei (where the electrons can absorb and emit photon by changing their state).
In a plasma they are not.

In a plasma electrons interact with each other and perform SHM if displaced. An electron that can oscillate can absorb and reemit photons of light i.e photoabsorption

Mössbauer effect is referred to nuclei bound in a lattice.
How do you apply it to electrons in a plasma?

because it is all due to recoil on absorption and emission and the resulting redshift or not.

The atoms in a glass can vibrate.
The recoil is transmitted to the whole mass of the block only if the atom cannot vibrate. Otherwise the recoil just excites lattice vibrations.

No, recoil is often 'neglected' - see the QEd book cited earlier

Why?
According to your picture, a denser plasma means more absorption and emission processes.
And why would the effective mass of the electron be larger?

Because the coulomb forces are stronger and restrict the recoil.

Compton effect deal with scattering of photons on free electrons?
Why can't you apply to electrons in a plasma

Because in Compton the electrons go whizzing one way whilst the photon goes the other. In a plasma the electrons are restricted and perform SHM rather than going zooming off. This is why this effect is nowhere like Compton effect.

[lyndonashmore]

gritmonger
You would have joined the discussion anyway!

I'm sorry, is that an indictment of some sort? [/quote]
Its a compliment - I know someone who likes a good argument.

Gosh- and you'd said earlier there would be but it would be small.

'To all intents and purposes can be neglected' means 'small' to me.

That still doesn't address the redshift getting larger the fewer electrons it encounters... and why electrons of atoms in a disorganized glass state - not a crystal lattice, by the way - would necessarily obey this rule as well.

A noncrystalline material, which has no long-range order, is called an amorphous, vitreous, or glassy material. It is also often referred to as an amorphous solid, although there are distinct differences between solids and glasses: most notably, the process of forming a glass does not release the latent heat of fusion.
- from the Wikipedia entry on crystal

[/quote]
a redshift of h/mc on each interaction is as big as it gets. This is the value I use. As the number of electrons per cubic metre increases the redshift reduces on each interaction

[Normandy6644]

papageno wrote
As far as I know, Hubble constant is not derived from hr/m.
It might be just a coincidence. Why do you link them?

Because they have the same value. The Hubble constant is derived from hr/m that is the whole point of this discussion topic

As I've said, you derived nothing. Not only is your argument without any logical framework, the math is incorrect. I don't know how many times I need to keep saying it.

How about everyone here who knows calculus try to do the integral he set up (incorrectly, but ignore that for now) and see if you get the same answer as he did. I'm curious now, because so far I've confirmed this with only 2 people.

[lyndonashmore]

Lyndon, I see in your paper there is a connection to Marmet's work. John Kierein is not mentioned but there are similarities there too. You might want to get John's reaction - he's on this board.

Thanks, the more the merrier.
I haven't seen John's work but will look for it. I only saw Marmet's work last year! As I have said before, I follow JJ Thompson ideas - don't do any research until after you have come up with your own ideas!

[lyndonashmore]

Normandy6644

As I've said, you derived nothing. Not only is your argument without any logical framework, the math is incorrect. I don't know how many times I need to keep saying it.

Where is the flaw in the 'logical argument"?

How about everyone here who knows calculus try to do the integral he set up (incorrectly, but ignore that for now) and see if you get the same answer as he did. I'm curious now, because so far I've confirmed this with only 2 people

Sounds like a good idea to me

[snowflakeuniverse]

Hi Lyndon

I am looking over your work, very interesting.

A critical parameter in your equation is the electron density, ne. (e normally subscripted). Double ne and you double Ho, according to your model.

Ne is experimentally derived, I would presume. I tried to check your source for the density of electrons in free space ( you referenced "Principles of Physical Cosmology" by Peebles, 1993) and I could not find the figure. There is no listing under electron density nor did I find anything in the section on interstellar medium. (In fact the density in the interstellar medium, for Peebles, all tend to include the effects of an expanding space-time field.)

I tried an internet search but all I was able to find were electron densities within galaxies, just outside galaxies, around the sun and above the atmosphere.

Please site a source for ne . If you give me the page number for peebles, that would work, we probably have the same text if yours is softbound.

Snowflake

[lyndonashmore]

[quote]

I'm going through more carefully now and I have two mathematical problems so far:

1) On page 5, you say that


Sum(x=0..N-1) [lambda + x(h/mc)]^-1=2nrd


(I'm dropping all subscripts right now to make things a bit easier to read, and type)

You then say that this is equal to an integral of [lambda + x(h/mc)]^-1 dx. My problem is, where is the dx coming from? You can't just stick it in there, because an integral is defined as the limit of a Riemann sum, and your first expression is just a sum without any infinitessimally small area element to integrate over. If the left side has a dx (or any vanishingly small element), the right side must have this also.

Try the Euler-Maclaurin formula http://encyclopedia.thefreedictionar...urin%20formula
Does this help?

[lyndonashmore]

Hi Lyndon

I am looking over your work, very interesting.

A critical parameter in your equation is the electron density, ne. (e normally subscripted). Double ne and you double Ho, according to your model.

Ne is experimentally derived, I would presume. I tried to check your source for the density of electrons in free space ( you referenced "Principles of Physical Cosmology" by Peebles, 1993) and I could not find the figure. There is no listing under electron density nor did I find anything in the section on interstellar medium. (In fact the density in the interstellar medium, for Peebles, all tend to include the effects of an expanding space-time field.)

I tried an internet search but all I was able to find were electron densities within galaxies, just outside galaxies, around the sun and above the atmosphere.

Please site a source for ne . If you give me the page number for peebles, that would work, we probably have the same text if yours is softbound.

Snowflake

Hi Snowflake,
Probably the best ref for electron density in IG space is C. Deffayet, D. harari. JP Uzan M. Zaldarriaga Phs Rev D 66d3517D 6 pages 2002. I say best because it is recent 2002. Phys rev is a pain cos you have to pay (at least I have) but a copy is available free at
http://xxx.lanl.gov/pdf/hep-ph/0112118
This gives the value of n as being approx 0.1 per metre cubed.
Zombeck is older (1990) "handbook of Space Astronomy and astrophysics"
http://ads.harvard.edu/cgi-bin/bbrow...a&page=286 (its at the ADS data base - may have typed the ref in wrong, getting late here)
This gives n as being approx or less than 10 per cubic metre.
Surprisingly enough you can have more electrons per cubic metre in IG space than you might have elsewhere.
Hope this helps.
Zombeck

[Normandy6644]

I'm going through more carefully now and I have two mathematical problems so far:

1) On page 5, you say that


Sum(x=0..N-1) [lambda + x(h/mc)]^-1=2nrd


(I'm dropping all subscripts right now to make things a bit easier to read, and type)

You then say that this is equal to an integral of [lambda + x(h/mc)]^-1 dx. My problem is, where is the dx coming from? You can't just stick it in there, because an integral is defined as the limit of a Riemann sum, and your first expression is just a sum without any infinitessimally small area element to integrate over. If the left side has a dx (or any vanishingly small element), the right side must have this also.

Try the Euler-Maclaurin formula http://encyclopedia.thefreedictionar...urin%20formula
Does this help?

It makes no difference (I'm not new to math), since you have no infinitessimal. Integrals are limits of Riemann sums, taken over a vanishingly small division. You can't just set a discrete quantity like that sum to an integral, it doesn't work that way.

[lyndonashmore]

Normandy6644
It makes no difference (I'm not new to math), since you have no infinitessimal. Integrals are limits of Riemann sums, taken over a vanishingly small division. You can't just set a discrete quantity like that sum to an integral, it doesn't work that way

but what do Euler and Maclaurin have to say about it? It is a recognised mathematical process. If you don't like it do it the long way, sum the arithmetic progression and do it that way. One gets the same answer, all roads lead to Rome, but some are longer than others!
Lyndon

[Normandy6644]

Normandy6644
It makes no difference (I'm not new to math), since you have no infinitessimal. Integrals are limits of Riemann sums, taken over a vanishingly small division. You can't just set a discrete quantity like that sum to an integral, it doesn't work that way

but what do Euler and Maclaurin have to say about it? It is a recognised mathematical process. If you don't like it do it the long way, sum the arithmetic progression and do it that way. One gets the same answer, all roads lead to Rome, but some are longer than others!
Lyndon

You seem to have a very loose grasp on the mathematics here. Either way, even if you can do the integral (which I still don't believe you can), your answer that you give as a result is incorrect. There should be no lambda left, since it cancels after you exponentiate both sides.

[lyndonashmore]

Normandy6644
You seem to have a very loose grasp on the mathematics here.

true, I meant to say Arithmetic series.

Either way, even if you can do the integral (which I still don't believe you can), your answer that you give as a result is incorrect. There should be no lambda left, since it cancels after you exponentiate both sides

There is no 'either way'. You complained about my sums. I showed you that all was well since I had applied the Euler-Maclaurin formula and gave you a quote. Now, does this apply or not?
Lets sort this little problem out and then we will move on to the integral. Why do you believe that the Euler-Maclauren formula does not apply in this case? It swings either way, Integrals can be approximated to summations and summations can be approximated to integrals.

[gritmonger]

gritmonger
You would have joined the discussion anyway!

I'm sorry, is that an indictment of some sort?

Its a compliment - I know someone who likes a good argument.
[/quote]

Gad! I already have a reputation!

'To all intents and purposes can be neglected' means 'small' to me.

But not nonexistant. It's important, because if you could show this effect in a local context, it would go a long way towards proving your theories. This is the same reason the Gravity Probe B was launched: despite the incredibly small effect predicted, finding it would still go a long way towards providing a proof to relativity predictions.

a redshift of h/mc on each interaction is as big as it gets. This is the value I use. As the number of electrons per cubic metre increases the redshift reduces on each interaction

So, larger redshift objects are closer. A very dim object of redshift 10 is closer than the Andromeda galaxy?

[Normandy6644]

Normandy6644
You seem to have a very loose grasp on the mathematics here.

true, I meant to say Arithmetic series.

Either way, even if you can do the integral (which I still don't believe you can), your answer that you give as a result is incorrect. There should be no lambda left, since it cancels after you exponentiate both sides

There is no 'either way'. You complained about my sums. I showed you that all was well since I had applied the Euler-Maclaurin formula and gave you a quote. Now, does this apply or not?
Lets sort this little problem out and then we will move on to the integral. Why do you believe that the Euler-Maclauren formula does not apply in this case? It swings either way, Integrals can be approximated to summations and summations can be approximated to integrals.

The formula does not seem to apply to this situation. You're calculating a sum and converting it into an integral, but according to the formula you should have a series that deals with the Bernoulli numbers and periodic series. Note the form given for the formula here and you'll see what I mean. It does not say you can just make the sum an integral; there are other complicated terms left over.

[snowflakeuniverse]

Hi Lyndon

I checked your reference for the electron density, and that reference gave the Peebles reference that you cited, also with no page number or chapter number etc. So I cannot find the observational justification for an electron density of .1 electrons/metercubed.

Also the second source you mentioned has a suggested electron density of 100 electrons/metercubed. This is 1000 times the value you use, and it is 2000 times the necessary value for your relationship to work.

It appears that if the relationship you propose is valid, there currently is no experimental evidence to strongly support it.

I do not mean to say it is wrong, it is just that the proof appears to be lacking.

Snowflake

[lyndonashmore]

snowflakeuniverse
I checked your reference for the electron density, and that reference gave the Peebles reference that you cited, also with no page number or chapter number etc. So I cannot find the observational justification for an electron density of .1 electrons/metercubed.

So you have a published reference there hence it is valid

Also the second source you mentioned has a suggested electron density of 100 electrons/metercubed. This is 1000 times the value you use, and it is 2000 times the necessary value for your relationship to work

Might I respectively suggest that you visit the optician! The reference gives n as approx or less than 10^-5 electrons per cubic centimetre. There are 10^6 cubic centimetres in a cubic metre so that puts n at around or less than 10 in 1990.
or try this one;
http://www.coseti.org/9301-020.htm
" For communications within interplanetary space, Ne = 106 to 1010 m-3, while in intergalactic space, Ne < 10 m-3. The sign of the dispersion is such as to cause the lower frequency to be delayed with respect to the higher frequency, i.e., low frequencies move slower."

It appears that if the relationship you propose is valid, there currently is no experimental evidence to strongly support it.

I do not mean to say it is wrong, it is just that the proof appears to be lacking.

Snowflake

One can only say this after dismissing all the evidence!
The published value of n in 2002 is approximately 0.1 electrons per cubic metre. I do not know where you got the 1000 from but all values are within range and support this theory and consequently there is strong evidence for it.
cheers Lyndon

[papageno]

papageno wrote
As far as I know, Hubble constant is not derived from hr/m.
It might be just a coincidence. Why do you link them?

Because they have the same value. The Hubble constant is derived from hr/m that is the whole point of this discussion topic.

Are you basing a physical link on a numerical coincidence?
What would happen if we used different units?

You can argue that there is a physical connection if the numerical coincidence does not depend on the units chosen: have you checked?
(In your paper you do not discuss it.)

Photoabsorption is still referred to electrons bound to nuclei (where the electrons can absorb and emit photon by changing their state).
In a plasma they are not.

In a plasma electrons interact with each other and perform SHM if displaced. An electron that can oscillate can absorb and reemit photons of light i.e photoabsorption

If an electron is not bound to a nucleus, why would it oscillate?
Without a nucleus, what keeps an electron from "flying" around randomly?

Mössbauer effect is referred to nuclei bound in a lattice.
How do you apply it to electrons in a plasma?

because it is all due to recoil on absorption and emission and the resulting redshift or not.

"Mössbauer effect" has a precise meaning: it deals with absorption and emission of photons by nuclei, not electrons, in a solid.

The atoms in a glass can vibrate.
The recoil is transmitted to the whole mass of the block only if the atom cannot vibrate. Otherwise the recoil just excites lattice vibrations.

No, recoil is often 'neglected' - see the QEd book cited earlier

The effect of the recoil exciting lattice vibrations is to broaden the spectral lines.
The Mössbauer effect is nice because there is no such broadening and the lines can be very sharp. This allows, for example, to probe the hyperfine interaction in atoms.

Why?
According to your picture, a denser plasma means more absorption and emission processes.
And why would the effective mass of the electron be larger?

Because the coulomb forces are stronger and restrict the recoil.

Even in a plasma, the electrons are still fermions.
In a Fermi gas (such as the plasma you are talking about) a higher density means that the electrons have a higher average kinetic energy. As a consequence, the potential energy (due to Coulomb interaction) is less effective.
And still, in your picture, a higher density means a higher chance for a photon to be absorbed and re-emitted.

About effective mass, how exactly does it work in your theory?

Compton effect deal with scattering of photons on free electrons?
Why can't you apply to electrons in a plasma

Because in Compton the electrons go whizzing one way whilst the photon goes the other. In a plasma the electrons are restricted and perform SHM rather than going zooming off. This is why this effect is nowhere like Compton effect.

By what are the electrons restricted?
If they are not bound to nuclei, what keeps them from zooming off?

[Gerbil94]

When I evaluate the integral at the limits, I get (L=lambda)

L+(h/mc)(N-1)-L=exp(2nrdh/mc)

or N=(mc/h)[exp(2nrhd/mc)+(h/mc)], which has no lambda dependence, which in turn is the basic for the rest of your discussion.

I might have made a mistake which is why I'm asking others to double check me and see if I made an error, as well as check my reasoning for why I don't think you can do the integral in the first place.

I think you have made a mistake in the evaluation. Integrating, I get

2nrdh/mc = ln(L + (N-1)(h/mc)) - ln(L)

If I exponentiate both sides I get

exp(2nrdh/mc) = 1 + (N-1)(h/mcL)

which still depends on lambda. It looks to me like you've done

exp(ln a + ln b) = exp(ln a) + exp(ln b) = a + b

But this is not the case. Instead,

exp(ln a + ln b) = exp(ln ab) = ab.

The MacLaurin formula is misapplied in the paper, although fortunately for the author even the crippled approximation is very good at the sort of N he probably needs for significant redshifts:

Lambda = 500e-9, h/mc = 2.4264e-12:

N=1: z=5e-6; Sum = 2.000e6, Integral = 0.0
N=10: z=5e-5; Sum = 1.999e7, Integral = 1.799e7
N=100: z=5e-4; Sum = 2.999e8, Integral = 1.979e8
N=1e3: z=5e-3; Sum = 1.995e9, Integral = 1.993e9
N=1e4: z=0.05; Sum = 1.953e10, Integral = 1.953e10
N=1e5: z=0.5; Sum = 1.630e11, Integral = 1.630e11
N=1e6: z=5; Sum = 7.282e11, Integral = 7.282e11

I have not read the rest of the paper yet.

[Normandy6644]

Yes, you are correct (I think) about the integral. However, i still am very uneasy about the sum/integral problem, especially because it is used to derive a more general formula. As of now I'm still not convinced about the validity, but I will work on it some more. it won't change my opinion that the results are contrived though.

[snowflakeuniverse]

Hi Lyndon

You are right, I read the chart wrong, the density in the second reference you gave is 100 electrons per cubic meter for intergalactic space, not 1000. (On my computer, the chart shows up vertically and when I tip my head over it gets a bit difficult to read it.)

So instead of your relationship being off from observation by a factor of 2000, we are now down to 200. Things are getting better. Also the chart does indicate that the number of electrons in space should be less than this. Things are getting even better. But it is still fairly far off from being convincing.

The electron density of .1 electrons per cubic meter is closer to the value you need, which is reputed to be somewhere in Peebles, “Principles of Physical Cosmology”. I have not been able to find it, and I spent almost an hour looking thorough the text. My fear is that the original source that you used to lead to your value may have been a made up the figure and you may be the victim of someone else’s error. Maybe they had to turn their head on the side to read a chart and they got it wrong. As far as I can tell there is no evidence of an electron density in space of .1 electrons in Peebles 1993 text and until someone finds it, or explains how they derived it from assumptions from Peebles, I would tend to not have much faith in the number. Also, Peebles would have this figure as a reference, it really would be good to have the original source. If we do not, there is the real risk that a faulty figure could become some kind of universally accepted figure when it should not.

The third, and latest reference you provided, lists an electron density of 10 per meter cubed, or less. This is within a factor of 20 of your desired relationship, which is not too bad, considering the guesswork involved in establishing electron density. (Could there be a very dense electron cloud around a galaxy and between it there is “empty” space? (photons and neutrinos fill space) Are there studies of signal/frequency spread detected when the signal passes through the edge of a galaxy?). It would be good if he provided the source for his estimates. I got the impression that he that did not determine the delay in signal based upon frequency but someone else did. Also, the date of the source is from 1978, seems we should have a better figure by now. (Hopefully a .05 electrons per meter cubed value)

Until we pin down the .1 electron density from a confirmed source, I do not believe it should be used. , (There is no evidence of it in Peebles, as far as I can tell). The only figures we have are between 20 to 200 times the necessary densities to conform to your relationship.

This again is not too bad, but it is not that good either.

Snowflake

[lyndonashmore]

papageno "]

papageno wrote
As far as I know, Hubble constant is not derived from hr/m.
It might be just a coincidence. Why do you link them?

Because they have the same value. The Hubble constant is derived from hr/m that is the whole point of this discussion topic.

Are you basing a physical link on a numerical coincidence?
What would happen if we used different units?

You can argue that there is a physical connection if the numerical coincidence does not depend on the units chosen: have you checked?
(In your paper you do not discuss it.)

The whole point is that in an expanding universe there is no physical link between H and hr/m so why are they the same – because the Bb is wrong, it is tired light. In tired light H = 2nhr/m so it is no big deal if they just happen to be the same. I.e. what is the probability of them being the same in a) BB where there is no link b) tired light where h = 2nhr/m? ”Scientists should be suspicious of things that have the same value but do not appear to be linked” R. P. Kirshner

If an electron is not bound to a nucleus, why would it oscillate?
Without a nucleus, what keeps an electron from "flying" around randomly?

An electron in a plasma oscillates if it is displaced. The frequency of oscillation is called the langmuir frequency. The langmuir frequency depends upon the plasma density. Basically, on the macroscopic a plasma is neutral. When an electron absorbs a photon it recoils. electrostatic forces from the other charges in the plasma try to restore it to its original position to make the plasma neutral again etc etc it performs SHM.

"Mössbauer effect" has a precise meaning: it deals with absorption and emission of photons by nuclei, not electrons, in a solid

They are always treated in the same chapter in the book, call it the Ashmore – Mossbauer effect if you like.

Even in a plasma, the electrons are still fermions.
In a Fermi gas (such as the plasma you are talking about) a higher density means that the electrons have a higher average kinetic energy. As a consequence, the potential energy (due to Coulomb interaction) is less effective.
And still, in your picture, a higher density means a higher chance for a photon to be absorbed and re-emitted.

About effective mass, how exactly does it work in your theory

The greater n the stronger the restoring forces. This means that the same external force produces less acceleration of the plasma electron. Its effective mass (m* = F/a) is greater. i.e. http://urap.gsfc.nasa.gov/www/abstra...ng_1993_1.html
“During the Ulysses flyby of Jupiter, the spacecraft crossed the outer part of the lo plasma torus along a basically North-to-South trajectory at a Jovicentric distance of about 8 R(J)…. assuming an effective mass of about 20 proton masses”

Compton effect deal with scattering of photons on free electrons?
Why can't you apply to electrons in a plasma

Because in Compton the electrons go whizzing one way whilst the photon goes the other. In a plasma the electrons are restricted and perform SHM rather than going zooming off. This is why this effect is nowhere like Compton effect.

By what are the electrons restricted?
If they are not bound to nuclei, what keeps them from zooming off?

[/quote]
The other charges in the plasma. see above

[lyndonashmore]

Gerbil94 wrote

although fortunately for the author even the crippled approximation is very good at the sort of N he probably needs for significant redshifts:

Thanks Gerbil94, but when you say "fortunately for the author" do you mean that I was correct all along?!!!!
No need to answer that

Lyndon

[lyndonashmore]

Hi Lyndon

You are right, I read the chart wrong, the density in the second reference you gave is 100 electrons per cubic meter for intergalactic space, not 1000. (On my computer, the chart shows up vertically and when I tip my head over it gets a bit difficult to read it.)

So instead of your relationship being off from observation by a factor of 2000, we are now down to 200. Things are getting better. Also the chart does indicate that the number of electrons in space should be less than this. Things are getting even better. But it is still fairly far off from being convincing.

The electron density of .1 electrons per cubic meter is closer to the value you need, which is reputed to be somewhere in Peebles, “Principles of Physical Cosmology”. I have not been able to find it, and I spent almost an hour looking thorough the text. My fear is that the original source that you used to lead to your value may have been a made up the figure and you may be the victim of someone else’s error. Maybe they had to turn their head on the side to read a chart and they got it wrong. As far as I can tell there is no evidence of an electron density in space of .1 electrons in Peebles 1993 text and until someone finds it, or explains how they derived it from assumptions from Peebles, I would tend to not have much faith in the number. Also, Peebles would have this figure as a reference, it really would be good to have the original source. If we do not, there is the real risk that a faulty figure could become some kind of universally accepted figure when it should not.

The third, and latest reference you provided, lists an electron density of 10 per meter cubed, or less. This is within a factor of 20 of your desired relationship, which is not too bad, considering the guesswork involved in establishing electron density. (Could there be a very dense electron cloud around a galaxy and between it there is “empty” space? (photons and neutrinos fill space) Are there studies of signal/frequency spread detected when the signal passes through the edge of a galaxy?). It would be good if he provided the source for his estimates. I got the impression that he that did not determine the delay in signal based upon frequency but someone else did. Also, the date of the source is from 1978, seems we should have a better figure by now. (Hopefully a .05 electrons per meter cubed value)

Until we pin down the .1 electron density from a confirmed source, I do not believe it should be used. , (There is no evidence of it in Peebles, as far as I can tell). The only figures we have are between 20 to 200 times the necessary densities to conform to your relationship.

This again is not too bad, but it is not that good either.

Snowflake

Hi SnowflakeUniverse. Thanks for the reply but I don't think we are quite there yet. Zombeck gives a value of n as 10 per cubic metre and not 100 as you say above. Nowhere as I see it gives the value of 100. Zombeck says "n is about or less than 10" - I can't do approx/equal signs here. This is less than 20 off, not 200. Try turning your head the other way around.
As for Peebles the honest answer is that I can't find the book (still looking) but I seem to remeber it being based upon half of all baryons being equally spread out and ionised.
Lets not argue. Why not split the diference, 0.1, 10 .....OK 1 electron per cubic metre. Makes me pretty close to the 0.6!

[lyndonashmore]

gritmonger wrote
But not nonexistant. It's important, because if you could show this effect in a local context, it would go a long way towards proving your theories. This is the same reason the Gravity Probe B was launched: despite the incredibly small effect predicted, finding it would still go a long way towards providing a proof to relativity predictions.

Maybe but interestingly enough, some say that redshifts in supernova Ia's vary with galactic latitude and put this down to redshifting in the plasma around the milky way. I am looking into this at present as my redshift could give the 'cosmological' redshift and the plasma around ours and host galaxies could give an intrinsic redshift.

So, larger redshift objects are closer. A very dim object of redshift 10 is closer than the Andromeda galaxy

No resdhift z is given by z = exp(Hd/c) - 1 where H = 2nhr/m.
Cheers Lyndon

[papageno]

papageno wrote
As far as I know, Hubble constant is not derived from hr/m.
It might be just a coincidence. Why do you link them?

Because they have the same value. The Hubble constant is derived from hr/m that is the whole point of this discussion topic.

Are you basing a physical link on a numerical coincidence?
What would happen if we used different units?

You can argue that there is a physical connection if the numerical coincidence does not depend on the units chosen: have you checked?
(In your paper you do not discuss it.)

The whole point is that in an expanding universe there is no physical link between H and hr/m so why are they the same – because the Bb is wrong, it is tired light. In tired light H = 2nhr/m so it is no big deal if they just happen to be the same. I.e. what is the probability of them being the same in a) BB where there is no link b) tired light where h = 2nhr/m? ”Scientists should be suspicious of things that have the same value but do not appear to be linked” R. P. Kirshner

Show us that H = 2n hr/m is valid even for other units.

By the way, in your derivation have you considered that photons can be scattered in other directions?

If an electron is not bound to a nucleus, why would it oscillate?
Without a nucleus, what keeps an electron from "flying" around randomly?

An electron in a plasma oscillates if it is displaced. The frequency of oscillation is called the langmuir frequency. The langmuir frequency depends upon the plasma density. Basically, on the macroscopic a plasma is neutral. When an electron absorbs a photon it recoils. electrostatic forces from the other charges in the plasma try to restore it to its original position to make the plasma neutral again etc etc it performs SHM.

To form a plasma, electrons are separated from atoms (which get ionized).
If the electrons are not separated from the ions, it is called "gas", not "plasma".
To keep the electrons separated from the ions, the kinetic energy of a single electron must be larger than the Coulomb interaction with any ion.
This means that an electron is nearly freely moving, as long as it stays within the volume occupied by the plasma.

The only "restoring" force is the electric interaction that keeps electrons and ions within the same volume. But it is a force that is important over lengthscales comparable to the volume of the plasma.

If you look at one electron scattering a photon, the electron is effectively free, and the appropriate treament would be Compton scattering. But only as long as the wavelength of the photon is small enough (otherwise you get a response of the plasma as a whole).

"Mössbauer effect" has a precise meaning: it deals with absorption and emission of photons by nuclei, not electrons, in a solid

They are always treated in the same chapter in the book, call it the Ashmore – Mossbauer effect if you like.

Compton: scattering of a photon by electron.
Mössbauer: scattering of a photon by nucleus.

If you want to use one name, you should use "Compton" because you are talking about electrons.

By the way, why don't you consider the effect of ions on the photons?
If your plasma is made of electrons and ions, both should interact with electromagnetic radiation. Why are you considering only electrons?

Even in a plasma, the electrons are still fermions.
In a Fermi gas (such as the plasma you are talking about) a higher density means that the electrons have a higher average kinetic energy. As a consequence, the potential energy (due to Coulomb interaction) is less effective.
And still, in your picture, a higher density means a higher chance for a photon to be absorbed and re-emitted.

About effective mass, how exactly does it work in your theory

The greater n the stronger the restoring forces. This means that the same external force produces less acceleration of the plasma electron. Its effective mass (m* = F/a) is greater. i.e. http://urap.gsfc.nasa.gov/www/abstra...ng_1993_1.html
“During the Ulysses flyby of Jupiter, the spacecraft crossed the outer part of the lo plasma torus along a basically North-to-South trajectory at a Jovicentric distance of about 8 R(J)…. assuming an effective mass of about 20 proton masses”

So it comes up because of the ions in the plasma?

Compton effect deal with scattering of photons on free electrons?
Why can't you apply to electrons in a plasma

Because in Compton the electrons go whizzing one way whilst the photon goes the other. In a plasma the electrons are restricted and perform SHM rather than going zooming off. This is why this effect is nowhere like Compton effect.

By what are the electrons restricted?
If they are not bound to nuclei, what keeps them from zooming off?

The other charges in the plasma. see above

[Normandy6644]

The whole point is that in an expanding universe there is no physical link between H and hr/m so why are they the same – because the Bb is wrong, it is tired light

This is what I mean by your theory lacking logical formalism. You take a numerical coincidence and use it to justify the abandonment of a well established theory, a theory that explains already everything yours does. I see no reason to give up a well established and justified theory just because the Hubble constant is numerically equal to some constants of the electron. It's just not real science.

[lyndonashmore]

papageno wrote
Show us that H = 2n hr/m is valid even for other units.

OK, But later, I need more grape juice for that.

By the way, in your derivation have you considered that photons can be scattered in other directions?

Don't need to as this only dims the light from distant galaxies

To form a plasma, electrons are separated from atoms (which get ionized).
If the electrons are not separated from the ions, it is called "gas", not "plasma".
To keep the electrons separated from the ions, the kinetic energy of a single electron must be larger than the Coulomb interaction with any ion.
This means that an electron is nearly freely moving, as long as it stays within the volume occupied by the plasma.

No, this is incorrect in the intergalactic medium. The plasma here is formed by collisions. I put this in the original paper but I dropped it to make the paper shorter (and hence more publishable) and just quoted the result from other published papers instead. What happens is, in order to create a plasma, energy is needed to produce ions from the neutral atoms of IG space (otherwise the degree of ionization will reduce as the positive and negative particles recombine). This energy can be provided by thermal effects or by high-energy collisions which produce secondary particles. Since cosmic rays are predominantly made up of protons and the IG matter is principally made up of hydrogen atoms the collisions are mainly proton-proton (p + p) collisions. These hadronic collisions always result in copious pion production
The particle density of IG space is extremely small and so the decay length of the secondary particles is much shorter than their interaction length. The secondary particles will decay rather than lose their energy in collisions. The decay chain is: π → μ + ν μ, μ → e + ν μ + ν e. The stable decay products are therefore electrons, positrons and neutrinos. Hope the symbols come out ok! This is why I said earlier, one can have a 'denser' plasma in IG space cos the particles are left to decay by themselves. In denser atmospheres, the particles created bash in to other particles before they decay and come up with all sorts of weird combinations.

The only "restoring" force is the electric interaction that keeps electrons and ions within the same volume. But it is a force that is important over lengthscales comparable to the volume of the plasma.

displaced particles in a plasma perform SHM. This is not new but standard plasma physics in any book on plasmas. Hence langmuir frequency etc.

If you look at one electron scattering a photon, the electron is effectively free, and the appropriate treament would be Compton scattering. But only as long as the wavelength of the photon is small enough (otherwise you get a response of the plasma as a whole).

Compton scatter has NOTHING to do with this. It is standard light travelling through a transparent medium. "transmission of light in a transparent medium is nothing more than an electron picking up a photon, scratching its head and emitting a new photon" R.P.Feynmann Light does not travel through glass by compton scatter.

Compton: scattering of a photon by electron.
Mössbauer: scattering of a photon by nucleususe one name, you should use "Compton" because you are talking about electrons.

By the way, why don't you consider the effect of ions on the photons?
If your plasma is made of electrons and ions, both should interact with electromagnetic radiation. Why are you considering only electrons?

See how plasma is produced above



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