New Tired Light (NTL)

[tiredlyndon]

Greetings!
New Tired Light provides an explanation of redshifts in a static universe in terms of a photon - electron interaction in the IGM.
NTL first saw light of day in 1996 when it was first proposed on internet sites such as 'bad Astronomy' but since then has been published in around 15 papers in peer reviewed journals and conference papers.
NTL uses known physics to predict numerical outcomes such as the Hubble constant and the CMBR which can be tested.
In 2016, a paper was published predicting a relationship between the dispersion measure (DM) of a fast radio burst (FRB) and the redshift of the host galaxy and can be found here.
The relationship is:
DM(IGM)= (mc/2hr)ln(1+z) +DM(host)
BUT... in papers the DM has been weighted for expansion and so we must remove this weighting to return to the original data ie for a static universe.
m=rest mass of the electron
c=speed of light
h=plank constant
r=classical radius of the electron.
Inserting values for these constants gives:
DM(IGM)=(7.318x10^25)ln(1+z) + DM(host) in SI units
Notice that there are no 'adjustable parameters' as in the BB. A graph of DM(IGM) versus ln(1+z) will have a gradient of (mc/2hr).
That was 2016. In 2023 we have 23 FRB's localised to host galaxies and this gave us an opportunity to test this prediction of 7 years ago.
In the summer of 2024 I published a paper testing this prediction and can be found here
Since then a further ten FRB's have been localised giving us 33 data points (which must be statistically viable??)
I have updated the graph to include all data points.
33 data points.docx

You will notice that the Gradient = 7.33x10^25 m^-2 - a difference of 1.6% from the NTL prediction of 7.318x10^25m^-2 by NTL 7 year ago when none of this was known.
So, questions please - but remember NTL got it right!
Cheers
Lyndon
bringing cosmology back down to Earth

[Francois-Zinserling]

I am as desperate as anyone else for a static universe solution.
How does the solution account for the electrons [not] scattering the photons?
The first paper mentions:

The dispersion of radio waves is known to be produced by the interaction of the photons of the radiation with the electrons in space.

But this is not a good thing. Electrons in space are not the radio transmitter's friends. Very much not. Dispersion here means 'scattering'.

Agreed that the CMB is not an 'image' of anything, but the images of distant galaxies are supposedly real. Then if their light is also 'tired' due to electron dispersion, we would not be able to resolve an image here at earth.

The papers are filled with good logical mathematics and sound reasoning, and the data appears to support the premise.
I need some help with the fundamentals.

[tiredlyndon]

Hi Francois,
Thanks for your interest in NTL.
electrons are the best friends of electromagnetic radiation (radio or light) provided they remember their place and behave!
For instance, in glass, the electrons are fixed in the atoms. Photon comes along and is absorbed by the electron and the system oscillates. There is a delay, before the electron emits a 'new' photon. in the same direction as the original one. This is why the speed of light in glass is less than in a vacuum. Between atoms the photons travel at 3x10^8m/s but due to the delays on absorption/re-emission the average speed is less. Note that there is no recoil here because the electron is fixed in the atom which is fixed in the pane of glass so the whole pane would have to recoil.
The problem is that just about everyone (except me!) is using an old model of the IGM which was shown to be wrong over 50 years ago! A background of hard X-rays had been found indicating a plasma at 10^8K but this was later found to be untrue and due to AGN’s at high redshift. But this is the model people still use – a hot plasma of separate protons and electrons whizzing about disorderly.
See here
A research probe launched in 2012 stated

“We do not find any direct evidence of the so-called ‘warm hot intergalactic medium’”

here
That is, an IGM of hot ionised H atoms is wrong! But workers still use this model to discredit Tired Light and ignore the model when looking at FRB’s. To localise the FRB they look down the line of sight – these images are not blurred and neither are Tired light images blurred.
So, what is in the IGM? Dust particles (silicon and graphite) are sputtered out of galaxies and these become charged by the photoelectric effect. They reach an equilibrium potential of about +20ev where the rate of release of electrons by the PE effect is equal to the rate of recombination
Ref
The dust particles have at equilibrium between 10 – 1000 protons on them depending on particle size. The question these workers did not ask was what happened to the 10 – 1000 electrons that sped off into the IGM?
Well when you have a group of electrons without the protons (which are on the dust particles) they form onto a Wigner – Seitz BCC crystal lattice held there by their mutual repulsion. They can and do perform SHM about their lattice positions and so they can and do absorb and re-emit photons. But unlike the electrons in glass, they recoil on absorption and re-emission. Some of the energy of the absorbed and emitted photon is transferred to the recoiling electron (emitted in the form of two low energy CMBR photons). But unlike Compton scatter, recoil takes place along the line of site so there is no blurring of the image with NTL.
In Dispersion Measure (DM) longer wavelengths arrive later than shorter ones – well where have they been? The answer is it is the same process as NTL and differing frequencies suffer different time delays/Number of interactions along the way.
Hope this helps.
Cheers
Lyndon
Bringing cosmology back down to Earth

[Francois-Zinserling]

Dear Lyndon

I am 99% with you on the TL effect, but 0.1% on electrons being the cause.

For instance, in glass, the electrons are fixed in the atoms. Photon comes along and is absorbed by the electron and the system oscillates. There is a delay, before the electron emits a 'new' photon. in the same direction as the original one. This is why the speed of light in glass is less than in a vacuum

Visible light through glass does not have the required photon energy to excite the electrons. By excite, meaning getting absorbed and re-emitted. If they were absorbed and re-emitted, two things would happen: 1) All the light coming out would be one colour (of the particular electron band - e.g. infrared) and 2) the light would be scattered.
Thus visible light goes through glass without passing through electrons.
Since light is an EM wave, there is an overall E/B effect which slows down the wave, but the only way to keep the image coherent is to have no direct electron interaction. As soon as interaction happens, as it might for UV light, the photons are scattered and the image is diffused.

Another misnomer is 'conservation of momentum', which cannot be applied to an electron attached in an atom, since momentum is shared and the photon is still scattered.

Because the photons are not absorbed and re-emitted in glass, when the photon eventually exits the glass, it regains its entire momentum. No redshift occurs.

The Wigner – Seitz BCC crystal lattice may be proposed as a similar structure like glass, but the above would still hold true - direct interaction with electrons would diffuse the image, and if it doesn't, if it is just slowed in transit, then momentum is regained upon exit.

[RedshiftDrift]

@Francois-Zinserling

Visible light through glass does not have the required photon energy to excite (getting absorbed and re-emitted) the electrons.

Correct. But absorption-emission is not the only type of interaction of light with electrons!

Thus visible light goes through glass without passing through electrons.

Incorrect! The index of refraction of glass is caused by a polarization interaction of light on the electrons. Your argument fails to account for the index of refraction (among other things).

[Francois-Zinserling]

"Passing through electrons" may have been a poor choice of words on my part.
If you mean by 'interaction of light with electrons' that light is affected by the E/B fields of electrons, yes.

The photon does not lose energy though unless work was done and moving an electron back and forth does not qualify as 'work'.

[tiredlyndon]

Dear Francois

Visible light through glass does not have the required photon energy to excite the electrons. By excite, meaning getting absorbed and re-emitted.

We don't want the photon to excite the electron in an atom as this leads to scatter and absorption lines. We want the energy of the photons to be well away from the excitation energy so that all he photons are absorbed and emitted as a new photons.
As Feynman wrote in his book QED,

The transmission and reflection of light is nothing more than an electron picking up a photon, scratching its head so to speak, and emitting a 'new' photon

As for glass,

For instance, in glass, the electrons are fixed in the atoms. Photon comes along and is absorbed by the electron and the system oscillates. There is a delay, before the electron emits a 'new' photon. in the same direction as the original one. This is why the speed of light in glass is less than in a vacuum. Between atoms the photons travel at 3x10^8m/s but due to the delays on absorption/re-emission the average speed is less. Note that there is no recoil here because the electron is fixed in the atom which is fixed in the pane of glass so the whole pane would have to recoil.

This is based on the book, 'special relativity' by French so I am happy with this

[HanDeBruijn]

This entry is spotted in the list of References as contained in your first paper:
[3] Kiernein, J and Marmet, L. (2016) Private Emails

It reminds me of an old thread: "Big Bang: Tweak it or throw it out? ". The following formula is found in that post by @RedshiftDrift .

$$ H_0 = \zeta\frac{\alpha h^2}{m_e^2 c}n_e \qquad \mbox{(:STz)} $$

Compare this with a formula in your paper, namely

$$ H = 2 n_e h r_e / m_e \qquad (2) $$

where it is known that

$$ r_e = \alpha \frac{\hbar}{m_e c} $$

and so

$$ H = 2 n_e h r_e / m_e = 2 n_e h \left(\alpha \frac{\hbar}{m_e c}\right)/ m_e = \zeta\frac{\alpha h^2}{m_e^2 c}n_e = H_0 $$

with equality for $\zeta=1/\pi$ . The two formulas are similar but not exactly the same.

[tiredlyndon]

p.s. you may get a faster reply if you could direct me to where the equations are!

[tiredlyndon]

BTW
As a taster - from the internet

do you see a σ=2rλ? σ=collision cross section
I just need to explain f(2)

[HanDeBruijn]

@tiredlyndon .

p.s. you may get a faster reply if you could direct me to where the equations are!

Oh well, now I see that you have them almost everywhere, with better explanation than where I was looking first. I'm sorry.

[tiredlyndon]

not forgotten you - I have two NTL projects on the go at present. reply soon

[tiredlyndon]

Hi,
So lets look at collision cross sections for the NTL.
Firstly we need to look at all the waffling that goes on with light.
There are bulk properties but it all boils down to an individual photon interacting with a single electron.
That is it and all that matters.
NTL uses standard physics when a photon travels through glass it is absorbed and re-emitted by the atoms (basically the electrons in those atoms) absorbing and re-emitting photons.
There is a delay between absorption and re-emission hence the speed of light in glass is less than it is in a vacuum.
Photons travel at the speed of light between interactions but on average their speed in glass is less because the total time is greater due to the delays suffered by absorption re-emission.
Same in the plasma of the IGM.
How else can dispersion measure happen?
Photons of long wavelength arrive after photons of shorter wavelength - so where have they been?
Is einstein wrong? why do photons of long wavelength arrive after photons of shorter wavelength ?
It is to do with the photon interactions with individual electrons, they are held over ie delayed. But to be delayed the electron has to store the photon energy as potential energy ie oscillate.
The problem is 'scientists' are repeating a model of the IGM that was shown to be wrong over 50 year ago!
The electrons are separated from the protons by the PE effect - UV etc acting on dust particles remove electrons. The protons are held fast on the dust particles but the electrons whiz off to fill the voids of the IGM. Once there they form Wigner Seitz BCC crystals held there by their mutual repulsion.
They perform SHM about their equilibrium positions and absorb and re-emit photons with a delay.
Photon comes in, sets the electron into oscillation (hf transferred to vibrational energy), bit of recoil and this energy is 'lost' to the photon but emitted as a secondary photon which forms the CMBR.
Oscillating electron emits a new photon (again a bit of recoil and hence another CMBR photon) and overall the photon has been redshifted.
The transmission of light or any other phenomena is just a case of a single photon interacting with a single electron. Any other theory is wrong!
So, lets talk about collision cross-sections for NTL (or the transmission of photons with matter).
Collision cross-sections are usually worked out theoretically which means one cannot trust them!
I submitted NTL to several journals and I think it was 1998? (certainly after APJ said "never ever submit a paper to us again" and long before certain sites give a birthday of 2015 !) that I submitted to A&A. I received a nice thank you and a statement saying they would not publish the paper unless it had official, referenced collision cross-sections. Fair enough
So, to be honest, I knew that for NTL to work the collision cross-section had to be 2rλ
Hope I am not boring you with the history of NTL, but one day you can tell your grandchildren, Lyndon Ashmore told me this....
Why,
Because redshift is independent of wavelength!
There are over 100 tired light theories we are told and this single requirement sorts them out. Do they predict the same redshift for all wavelengths from a particular galaxy?
This is the acid test.
NTL passes it since the collision cross-section is dependent on the wavelength.
Hence, a photon with twice the wavelength from a particular galaxy has twice the collision cross section, makes twice as many collisions, undergoes twice the loss of energy, experiences twice the drop in frequency and hence twice the increase in wavelength ,
The redshift = increase in wavelength / original wavelength is constant ie twice the original wavelength, twice the increase in wavelength gives the same value for all wavelengths.
If anyone tries to sell you a tired light theory ask this question.
But A&A wanted published collision cross sections.
Anyway (for your grandchildren) , mad about astronomy and cosmology, I had set up an astronomy evening at school, over 100 kids parents etc coming , planetarium set up to give them a preview for what they were going to see. Three telescopes 8/9 inch reflectors with anyone else bringing their own.
All set up and I had 2 hours to kill.
So, thinks me, what if I typed in to a search engine 'sigma = 2rλ? (it was yahoo - just for your grandkids) ie a reverse search. What would it give?
and it led me here
here
So, this is an indirect route but gets us there.
Here at the Lawrence Berkely national laboratory they had had experimentally measured not the cross-section for transmission but for absorption for low energy X rays interacting with matter!.
So here was a way in
here
So we don't have the collision cross-section for transmission ie NTL but we do have it for absorption.
Collision cross-sections are basically probabilities and you will see that the collision cross section has two components f(2) and 2rλ
2rλ is fixed and is the probability of the photon being absorbed in the first place and f(2) is the probability of it being retained ie absorbed. If you look at the data for f(2), f(2) has a value of 1 if the energy of the photon is equal to the energy of an energy level in a hydrogen atom and '0' if it is well away.
ie
If energy of photon = an energy level then that photon it always absorbed.
BUT if the energy of the photon is well away from any internal energy levels it is always re-emitted with a new photon.
There are only two outcomes, either the photon is absorbed or it is re-emitted (NTL) . so, if the probability of the photon is being absorbed is f(2) then the probability of it being re-emitted (NTL) must be (1-f(2))
Radio, light is well away from the resonant frequency of the IGM (15Hz?) so f(2) is always '0' s the probability of it being re-emitted as a new photon is '1' and hence the overall collision cross-section for NTL is 2rλ!!!Hope this helps/
Lyndon
BTW what did A&A say when I presented published collision cross-sections?
'we only accept papers from professional ....'

[sferios]

Hi Lyndon. I've got a question. How are electrons spaced out in the IGM and what are they doing out there? Just sitting around? You mention BCC lattices, but I have a difficult time imagining them. In my understanding, plasma containing both electrons and ions is constantly moving due to the attractive and repulsive forces. But if I read your theory correctly, moving electrons don't contribute to redshift. To absorb and reemit photons with recoil in the same direction, the electrons have to be chilling out with just SHM going on. But how do electrons get so far away from ions? If there are electron clouds just floating around in the IGM, how did they get there? Is this just a fundamental fact of the IGM? Or is it an assumption? Do we have evidence for relatively stationary electron clouds (other than redshift, that is)? I'm trying to picture gigantic electron clouds locked in a repulsive lattice with a single electron every two cubic meters? But when I get to the edges, I see the cloud falling apart as the repulsive force goes on forever. - Emanuel

[sferios]

Zoing! Ok so what's the density of electrons in these clouds?

[tiredlyndon]

Zoing! Ok so what's the density of electrons in these clouds?

0.5 per m^3 gives the Hubble constant and everything else.

[tiredlyndon]

No! Chat GPT looks at old data. It does not re imagine the future.
Put the chairs in BCC format

[tiredlyndon]

Actually, this is a serious scientific site. Pls don't post things like this again
Cheers
Lyndon

[sferios]

Ok no problem.

[helpfulgent]

Hello Lyndon,
You write "The IGM plasma is overall neutral. The charge of the IGM as a whole is neutral. "
What evidence do you have for that?
Thank you
David

[tiredlyndon]

Hi David,
In a word, none!
I was presenting a paper on NTL at a general conference in Prague (Vigier) 2018? and as soon as I started hands started to go up (ignored by me till the end) as folk, as always, want to say 'can't be right as IGM ... plasma of electrons protons ... blur.. etc and switch off. Usual stuff to reject tired light out of hand.
Afterwards, one guy, John, came to me and said 'I want to talk to you,'
OK, said I, lets go for a pint and I will buy.
Instantly I had half a dozen professors of solid state physics (John), particle physicists, Cern physicists coming along in a positive way and not the usual negative way (for free beer!)
The end result was John said 'get rid of the protons and everything you say will work.' He's referenced in the paper but would not let me use his job title!!
I said 'you can't do that!'
Why not.' says he, 'all you are doing is changing the potential of the universe.'
50 years of teaching static electricity and emphasising 'you don't create charge, you separate it' meant I wanted to keep the universe neutral with the protons 'fixed' on dust particles or around galaxies and the electrons able to fill the spaces in between.
To get the delay between absorption and re-emission and with no blurring you need the electrons to crystallise and this they will do if you take away the protons (or separate them as in NTL)
An IGM of hot overall neutral plasma will not allow this and would blur.
This idea of the IGM was rejected by the particle physicists as crazy since the the electrons and protons 'would simply, recombine.'
On researching this point I found the actual idea of an IGM consisting of a hot neural plasma had been rejected in the 1970's
So, no evidence just me liking to separate charges and I think a universe with an overall charge would lead to problems.
But NTL would still work as long as there is an excess of electrons
Cheers
Lyndon

[LifeIsStrange]

off topic but @tiredlyndon since you have made a paper titled:

A Relationship between Dispersion Measure and Redshift Derived in Terms of New Tired Light
https://www.semanticscholar.org/paper/A-Relationship-between-Dispersion-Measure-and-in-of-Ashmore/76df669b6c564a272804bb727e22397d92b1fa58

As it has recently been released "the most accurate" local dispersion measure
https://arxiv.org/abs/2410.22098

I was wondering how they result impacts your calculations on NTL

[tiredlyndon]

Not really, the problem with local FRB's is that the uncertainties in the contribution from the MW and halo, plus the host contribution are around as big as the actual measured FRB!
For instance in the paper you quote it says the simulation for the DM MW and halo vary from 18 to 166 pc m^-3 and yet the total DM's measured vary from 87 to 283.
so this implies the range of DM (IGM) varies from -75 to 265!
BUT... we then have the contribution from the host galaxy which they have varying from 17 to 200 and this has to be deducted too!
Enough said ? You need FRB's to be at a decent distance so the uncertainties due to MW and halo and host have less effect.
There are two models of the MW+halo and these can differ by 200 pc m^-3 in the value for the same FRB (workers seem to take the average)
It was also good to see they estimate the electron number density of the IGM as approx 1 per cubic metre which compares well with NTL prediction of 0.5
Interesting though and thanks. It was quite exciting to see one paper in the refs saying they had 43 FRB's but on closer inspection only 31 in the list had redshifts for the hosts,
Always happy to receive data!

[LifeIsStrange]

It was quite exciting to see one paper in the refs saying they had 43 FRB's but

Here is a new repeating FRB with associated redshift
https://phys.org/news/2024-11-astronomers-fast-radio.html

[tiredlyndon]

Thanks NTL predicted = 11x10^24 actual (with weighting for expansion removed ) =4.2x10^24 m^-2
In the ball park of DM =(2hr/mc)ln(1+z)
see attached DM versus ln(1+z) and not too far out (red dot)
But if you look at fig 4 localisation pic the FRB is nowhere near the suggested host galaxy!
Doc14.docx
They say

FRB 20240209A has a projected physical offset of 40 ± 5 kpc from the
center of its host galaxy, making it the FRB with the largest host galaxy offset to date.

So I take it this is not the host galaxy as it is too far away?
If I see other workers using this as a 'well localised FRB then I will use it, if not ....
But thanks again
Keep going
Lyndon