New Twist on Hubble's Law

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15 years 6 months ago #15172 by JMB
Replied by JMB on topic Reply from Jacques Moret-Bailly
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by lyndonashmore</i>
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But, in Creil, photon - photon collision cross sections are known, photon densities are known. Must be a simple calculation to work out an expression for the Hubble constant and Whallah! test it by putting numbers in.
Have you done this?
I know 'Tired Light' [url]http//www.lyndonashmore.com[/url] has and gets it right
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'Tired light' requires new laws of physics while CREIL uses only regular spectroscopy. There is however a similarity between both effects: they are proportional to the path of light, assuming a constant density of gas for the CREIL.

A computation of the CREIL effect requires lhe knowledge of the Raman polarisability of atomic hydrogen for the quadrupolar spin recoupling transitions at 178, 59 and 24 MHz in the states of principal quantum number n=2.

If somebody knows these polarisabilities or can find them, for instance experimentally or by an ab initio computation, I will be happy.

However, the computation of the Hubble constant by CREIL requires the density of atomic hydrogen in each of the involved levels. The Hubble law corresponds to an assumed constant density of these atoms.

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15 years 6 months ago #15173 by JMB
Replied by JMB on topic Reply from Jacques Moret-Bailly
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by lyndonashmore</i>
I know 'Tired Light' [url]http//www.lyndonashmore.com[/url] has and gets it right
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Unhappily the 'tired light' theory requires new physics because you assume that the COHERENCE of light-matter interactions may be neglected: there is a coherence of phases between an incident wave and the scattered wave.

The study of the propagation of light in matter requires one of the two equivalent theories:
- The old theory of Lord Rayleigh in which the coherence is directly taken into account. If an incoherent scattering occurs, by a collision in a low pressure gas, the frequency of scattered light may be changed, but the images are destroyed. Else, no exchange of energy result from a variation of the translational energy of the atom, all coherent exchanges of energy result from transitions in the atom.
- Einstein's theory which introduces the A and B 'Einstein coefficients'. A corresponds to the spontaneous emission of light and is useless if the 'residual' ( or 'zero point' or 'stochastic') field is not subtracted from the true electromagnetic field. As in Einstein theory the coherence is not explicitely used (only thermodynamics!), it is difficult to say that it is wrong. This theory does not introduce any translational energy.

If an electron (or an other particle) interacts with light, it may be accelerated and absorb energy, but this energy is fully returned to the field if the energy of the electron has not been changed by some interaction. In CREIL, this interaction corresponds to a (virtual) quadrupolar transition. 'virtual' is important because the atom is only 'dressed' by the redshifted radiation, and simultaneously 'dressed' by the thermal radiation, so that the stationary state of the atom is not changed when the pulse of light stops.

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15 years 6 months ago #23693 by lyndonashmore
Nah!
Tired light needs no new Physics. It certainly is so in my case ( www.lyndonashmore.com )tired light uses standard Physics.
Light travels at 3x10^8m/s in a vacuum.
In a medium (glass)it travels slower.
Why?
Cos it is constantly absorbed by electrons in the medium and then re emitted.
Since there is a delay between absorption and re emission the average speed is less in the medium.
Photons travel at 3x10^8 m/s between atoms but the delays slow down the average speed.
In glass the electrons cannot recoil so there is no energy loss but in space where it is squidgy, the electrons recoil. Some of the energy of the photon is lost to the recoiling electon on both absorption and re emission. (Mossbauer effect)
Energy of photon is reduced so frequency is reduced.
Frequency reduced wavelength increases.
redshift.
the Hubble relation becomes;
Photons of light from galaxies twice as far away, travel twice the distance through intergalactic space, make twice as many collisions, lose twice as much energy, undergo twice the loss in frequency and suffer twice the increase in wavelength (redshift).
All good Physics no new stuff. all common sense.
the problem with CREIL is that it requires coincidences ie two photons and so is unlikely - Okams razor and all that.
cheers,
lyndon

lyndon ashmore - bringing cosmology back down to earth.

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15 years 6 months ago #23415 by JMB
Replied by JMB on topic Reply from Jacques Moret-Bailly
I agree with Lyndon's description up to:
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by lyndonashmore</i>
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Photons travel at 3x10^8 m/s between atoms but the delays slow down the average speed.
In glass the electrons cannot recoil so there is no energy loss but in space where it is squidgy, the electrons recoil. Some of the energy of the photon is lost to the recoiling electon on both absorption and re emission. (Mossbauer effect)
Energy of photon is reduced so frequency is reduced.
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If the electron is free, the energy of acceleration of the electron is exactly returned to the light beam.

Either the electron is free and it returns the recoil energy to the EM field, either it is bound to an atom, and we must know the aim of its energy.

It is very dangerous to use the photon representation of electromagnetic fields, because there is a TEMPTATION TO NEGLECT THE PHASE of the pilot wave (Schroedinger wave, de Broglie wave, electromagnetic wave ? ). Applying your method, you necessarily find the same conclusion than the majority of physicists who said: "Townes maser will not work".

The coherence must be taken into account, the simplest way being the use of 1917 Einstein theory.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">
All good Physics no new stuff. all common sense.
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What does the energy of your electron become? In Mossbauer effect, we know what the energy becomes.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">
the problem with CREIL is that it requires coincidences ie two photons and so is unlikely
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No, the CREIL effect is a parametric effect: when light interacts with an atom to be refracted, infinitesimal energies polarize all atoms of the refracting medium (the atoms are in a "dressed state" non-stationary combination of stationary states), and this energy is completely returned to the light beam when the pulse of light stops.
In CREIL, several beams dress the atoms, so that, while after the light pulses, the atom return to its stationary state, the exchanges of energy with the various beam may be changed, to increase the entropy.

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15 years 6 months ago #23601 by lyndonashmore
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If the electron is free, the energy of acceleration of the electron is exactly returned to the light beam.

Either the electron is free and it returns the recoil energy to the EM field, either it is bound to an atom, and we must know the aim of its energy.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
But electrons in plasma are not free. They act collectiveley. They even oscillate and this is how they absorb and reemit photons.
However, the natural frequency is much less than that of the photons of light and so resonance absorption does not take place. The photons are re emitted. But since the electrons recoiled on absorption and re emission some energy is lost. E = hf so frequency is less wavelength longer. Redshifted.
see [url] ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&page=286 [/url]
In IG sace the natural frequency is less than 30Hz.


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">No, the CREIL effect is a parametric effect: when light interacts with an atom to be refracted, infinitesimal energies polarize all atoms of the refracting medium <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
All the atoms in the refracting medium? Including the ones in front of the light beam? Does that mean energy travels faster than light?
Cheers,
lyndon

lyndon ashmore - bringing cosmology back down to earth.

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15 years 6 months ago #15184 by JMB
Replied by JMB on topic Reply from Jacques Moret-Bailly
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by lyndonashmore</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">
But electrons in plasma are not free. They act collectiveley. They even oscillate and this is how they absorb and reemit photons.
However, the natural frequency is much less than that of the photons of light and so resonance absorption does not take place. The photons are re emitted. But since the electrons recoiled on absorption and re emission some energy is lost. E = hf so frequency is less wavelength longer. Redshifted.
see [url] ads.harvard.edu/cgi-bin/bbrowse?book=hsaa&page=286 [/url]
In IG sace the natural frequency is less than 30Hz.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
There is an ambiguity in your explanations: With a resonance frequency lower than 30Hz, you are in the conditions of the CREIL effect, replacing the spin flip resonance of the atom of hydrogen by the plasma resonance. Thus energy may be lost, it may be a redshift.
This redshift is however lower than the CREIL redshift because as long as the resonance frequency is lower than around 1 GHz, the exchanged energy is proportional to the frequency.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">

No, the CREIL effect is a parametric effect: when light interacts with an atom to be refracted, infinitesimal energies polarize all atoms of the refracting medium <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
All the atoms in the refracting medium? Including the ones in front of the light beam? Does that mean energy travels faster than light?
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> All atoms of the refracting medium are polarized (or dressed by the field in quantum language).
In front of the light beam, the atoms absorb energy to be polarized, this energy being integrally returned to the field when its intensity decreases. The energy does not travel faster than light:
- on the contrary, the small fraction of the energy absorbed at the front of a pulse waits that the pulse decreases to restart its travel.
- during the pulse, the polarized atoms radiate (scatter) a field delayed of pi/2, this radiation requiring no energy because the scattered field is much lower than the exciting field, so that, by interference, the emission of this field requires no energy.

With best regards,
Jacques

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