New Twist on Hubble's Law

Hi Jacques
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"> Not clear:
You suppose that photons are absorbed and reemitted. Two problems may appear because you do not indicate how the photons are reemitted:
- Is it a spontaneous emission? if yes, light is scattered into all direstions.
- Do you obtain a mixture of incident photons and scattered photons? if yes, the lines are broadened.
The CREIL solves these problems, it is possible that the same process works, but it must be justified.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Yes, there is a delay between absorption and re emission. This is how light travels through any medium and why the speed of light in a medium is less than that in a vacuum. Photons travel at the speed of light between atoms/electrons but the delays at each interaction reduce the average speed. This is why in sparsely populated plasma we get the recoil that leads to the redshift.
The broadening of the lines is not a problem as the effect is less than the thermal broadening.

I didnt get the chance to show the power of this Tired Light Theory at the CCC2 conference as I did it as a poster (having already had most of it published already) so let me do that now.
Lets take the K Line (3.934x10-7 m) A2005 in the Corona Borealis Supercluster
[url] www.atlasoftheuniverse.com/superc/cbo.html [/url].
It is a distance of 1025Mly or 9.7x10^24m away.
I showed here
Page 53 [url] www.i-b-r.org/Incons.GravFinalGED-I.pdf [/url]
that the increase in wavelength on each recoil interaction is h/mc (h = Plank constant, m = rest mass of the electron, c = speed of light in a vacuum).
This has a value of 2.43x10^-12m.
To get the number of collisions we need the collision cross-section sigma for electron photon absorbtion..
These can be found here [url] www-cxro.lbl.gov/optical_constants/intro.html [/url]
For absorption, sigma = 2rf(lamda) Where lambda is the wavelength of the photon concerned and r is the classical radius of the electron (r = 2.82x10^-15m).
For low energy X rays and Hydrogen atoms f has values between 0 and 1 - and f basically shows the probability of the photon being absorbed and not re-emitted. If q is the probability of the photon being re emitted then f + q = 1 as one or the other must happen.
So collision cross section for absorption re emission is qr(lamda) or (1 - f)r(lamda).
When well away from resonance frequencies/wavelengths the tables show f = 0.
Ergo, sigma(absorption with re emission) = 2r(lamda).
For our K Line, sigma = 2.12x10^-21m^2.
The mean free path is 1/[n(sigma)] where n is the average number of electrons per cubic metre of space. Using WMAP data for local space [url] www.aanda.org/index.php?option=article&a...aah4525/aah4525.html [/url]

n has a minimum of 2.2x10^-7 cm^3 or an average of 0.22 electrons per cubic metre of space and increases with redshift.
That gives our mean free path as 2.14^21m (i.e on average, one collision every 230000 ly).
If the total distance to A2005 in the Corona Borealis Supercluster is 9.7x10^24m, the photon makes an average of 4500 collisions and at each interaction the wavelength increase by 2.43x10^-12m.
The total increase in wavelength suffered by our photon on its journey is 4500x2.43x10^-12 or 1.1x10^-8m
Redshift, z= (Increase in wavelength)/(original wavelength) = (1.1x10^-/( 3.934x10^-7) = 0.03 ie.Predicted z by Tired Light = 0.03.
The measured value is 0.076 ([url] www.atlasoftheuniverse.com/superc/cbo.html [/url]) - so our predicted value is pretty close considering that the value used for n was a minimum and contains a large degree of uncertainty.
There is no 'new' Physics here and I believe everything is referenced.
Can you do that with the CREIL theory?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">No. A fraction of the energy at the front of the pulse is used to dress the atoms. As the dressing of matter disappears at the end of the pulse, the available corresponding energy is returned to the wave.

Pay attention: the energy used to dress the atoms is not quantized. Else, how could the weakest light beams dress all atoms of a big prism? (refraction laws work at all energies).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
If the Universe is static and infinite (as you appear to be saying - as you are trying to explain redshift by mechanisms other than expansion as I do) then there must be an infinite number of atoms to be dressed. This would take an infinite amount of energy to dress them and so there would be none left over for the wave. Does that mean that with CREIL, no light would travel anywhere in an infinite universe?
Best Regards,
lyndon


lyndon ashmore - bringing cosmology back down to earth.

- To avoid a blurring of the images, the interaction of light with matter MUST BE COHERENT, as the interaction which produces the refraction.
- To avoid a blurring of the spectra, the interaction must produce A SINGLE FREQUENCY.

If you use the photon representation, you must take into account the PHASE OF ITS PILOT WAVE. The computation is difficult and useless, because Einstein (1917) gave the exact solution of the interaction of a light beam with matter (Einstein coefficients). Combining this result with Planck's result (1911) which gives the lowest value of the energy in a mode (hf/2), only Einstein's B coefficient is useful.
Einstein considered only the cases in which the state of matter is changed, but his theory works if matter is simply polarized (dressed), provided that it returns to its initial state after the interaction. To have a change of energy, at least two beams are necessary. There are conditions of coherence :
- in a crystal, the wavelengths of two beams may be equal although their frequencies are different, so that a lot of experiments combine the frequencies of lasers, multiply them, ...).
- in a gas, the conditions were written by G. L. Lamb Jr.
This can work with plasma resonances, but, as the frequencies are low, the result is weaker than with the spin-flip resonances in excited states (not in the ground state, 1420 MHz does not verify Lamb's condition).

Look at a good book of optics the theory of refraction. You will see that refraction works at any energy, in extremely large volumes: the polarization requires, for each atom an infinitesimal energy which is returned to the wave at the end of the pulse.

I am careful: the use of the photon representation is very dangerous, it requires much attention and big computations. The semi-classical theory has no failure, no paradox (such as EPR); the experiments done to show that the semiclassical theory is wrong use a wrong classical computation (neglecting a part of the light energy:hf/2) .On the contrary, the start up of the lasers is not explained by quantum electrodynamics.
I agree that it may be a redshift produced by a plasma. But, it requires a resonance at a low frequency

Hi Jacques,
I posted a 'scientific' post. It was fully referenced. So if you have an argument with it then lets please cut out the 'hand waving' and give scientific, referenced replies.



<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If you use the photon representation, you must take into account the PHASE OF ITS PILOT WAVE. The computation is difficult and useless, because Einstein (1917) gave the exact solution of the interaction of a light beam with matter (Einstein coefficients). <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

In 1917 we had the russian revolution and the battle of Ypres. Why should science from then be revelant to us now?




<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Combining this result with Planck's result (1911) which gives the lowest value of the energy in a mode (hf/2), only Einstein's B coefficient is useful.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In 1912 the Titanic sank. does that mean that ships cannot sail across the Atlantic?
With respect, science has moved on. Please refer to my post with RECENT references and lets move on from there - to a time when we have been to the Moon and back and not from a time when man could not fly or even ride a car.
cheers,
lyndon

lyndon ashmore - bringing cosmology back down to earth.

Hi Jacques, earlier you asked
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">
What does the energy of your electron become? In Mossbauer effect, we know what the energy becomes.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

So....
Lets take the power of this theory one step further and derive the CMB.
In this theory, the energy transferred to the recoiling electron is re radiated as a CMB photon.
Momentum of photon = hf/c and this is transferred to our electron which gains momentum mv (m=mass of electron and v = recoil velocity)
By conservation of momentum
Mv=hf/c so recoil velocity, v = hf/mc
Kinetic Energy of recoiling electron = (1/2)mv^2 = (h^2f^2)/(2mc^2)
This energy is radiated as a secondary photon of frequency f where f is given by:
F = (hf^2)/(2mc^2)
So lets look at an example and take the Lyman Alpha emission line (must be a lot of these photons since it is Hydrogen the building block of the universe).
F = 2.47x10^15Hz en.wikipedia.org/wiki/Lyman-alpha_line
The secondary photon given off by a photon electron interaction with these photons has a frequency of 2.45x10^10Hz .
The wavelength of these secondary photons is 1.21x10^-2m
Ie in the microwave region.
The CMB peaks at 0.21x10^-2m and so the radiation from the recoiling electrons in this Tired light Theory is a contender for the CMB. To get the peak photons we need original photons with a wavelength of 5x10^-8m ie UV.
NB in this theory there are Two CMB photons per interaction one on absorption and one on re emission.
Cheers,
Lyndon


lyndon ashmore - bringing cosmology back down to earth.

<i>Originally posted by lyndonashmore</i>
<br />
A small effort and you will get the CREIL. It remains two fundamental problems solved by the CREIL:
1- How do the scattered photon gets the same path than the exciting one?
2- How do all photons exchange the same energy to avoid a blurring of the light?
Best regards

Hi Jacques,

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by JMB</i>
<br />

<i>Originally posted by lyndonashmore</i>
<br />
A small effort and you will get the CREIL. It remains two fundamental problems solved by the CREIL:
1- How do the scattered photon gets the same path than the exciting one?
2- How do all photons exchange the same energy to avoid a blurring of the light?
Best regards

<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
As I have asked before ,
the small effort that i need to get the CREIL is numbers.
Please predict a value for the Hubble constant and get it correct as I have done with Tired Light.
Then please predict a value for the CMB and get it correct as I have done with Tired Light.
In Tired Light and plasma the electrons are not totally 'free' but interact with all the other charges in the Debye sphere. This keeps the photons travelling in a straight line (as in glass) and are so not scattered.
Regards,
lyndon
(travelling at present but will try to keep in touch on a fairly regular basis).
Cheers,
Lyndon

lyndon ashmore - bringing cosmology back down to earth.