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Thread: Why the Compton Effect does not cause Cosmological Redshift

  1. #31
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    On 2003-01-29 21:25, Tim Thompson wrote:
    JK: Where did you get the idea that photons with less energy than the rest energy of an electron don't scatter from electrons?

    From Classical Electrodynamics, J.D. Jackson, John Wiley & Sons, 3rd edition, 1999. See section 14.8, pp.694-697.
    Quote: "The classical Thomson formula is valid only at low frequencies where the momentum of the incident photon can be ignored. When the photon's momentum hbar*omega/c becomes comparable to or larger than mc, modifications occur. These can be called quantum mechanical effects, since the concept of photons as massless particles with momentum and energy is certainly quantum mechanical (pace, Newton!), but granting that, most of the modifications are purely kinematical. The most important change is the one observed experimentally by Compton. The energy or momentum of the scattered photon is les than the incident energy because the charged particle recoils during the collision." (pp. 695-696).

    The same discussion occurs in section 14.7 of the 2nd edition, which I have from my student days and is probably more common on bookshelves.

    From Radiative Processes in Astrophysics, G.B. Rybicki & A.P. Lightman, John Wiley & sons, 1979. See chapter 7, p.195, "Compton Scattering".
    Quote: "For low photon energies, h*nu very much less than mc^2, the scattering of radiation from free charges reduces to the classical Thomson scattering, discussed in chapter 4." The authors go on to develop the theory of Compton scattering & inverse Compton scattering in great detail.

    JK: The Klein-Nishina formula gives the cross section and it reduces to the Thomson cross section for a free electron.

    Correct. However, in the Klein-Nishina formula, the cross section is energy dependent. Compton scattering becomes less efficient at higher energies (see Rybicki & Lightman, p. 197). This in turn means that the optical depth for Compton scattering is also energy dependent. That's why the Compton effect cannot produce a wavelength-independent redshift, because the scattering optical depth is wavelength (energy) dependent.

    It may not be exactly correct to say that Compton scattering does not occur at all for low energy photons, but it is correct to say that the scattering optical depth will drop so close to zero that Compton scattering ceases to be a physically significant process at such low photon energies. Compton himself discovered the effect with X-rays, where photon energies are high enough to make the process work.

    <font size=-1>[ This Message was edited by: Tim Thompson on 2003-01-29 21:26 ]</font>
    Even for cases in which E_ph < m_c^2 (i.e., lower photon energies than Klein-Nishina cross-section regime), Compton scattering does apply, but the SHIFT will be wavelength dependent. This is because as Tim points out, the shift is in proportion to the incident photon wavelength to the compton wavelength of an electron. So as the wavelength of the incident photon becomes very long compared to the Compton wavelength (0.0242631 Angstroms), then the wavelength of the outgoing scattered photon tends toward that of the incident -- i.e., NO CHANGE. So it's not the Compton cross section that goes to zero for low energies (i.e., UV, optical, etc), it's the shift itself. Only at the very high energies do further quantum mechanical effects reduce the cross section in some proportion to the incident photon energy.

    As JS Princeton said, all we need to do is go to the paper. The explanation is quite simple.

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-29 23:30 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-30 08:49 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-30 08:50 ]</font>

  2. #32
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    The Compton effect does NOT occur for energies of the photon which are LARGER than twice the rest energy of the electron. (i.e. the wavelengths are LESS than 1/2 the compton wavelength of the electron) When such a high energy photon hits an electron then there is electron-positron pair formation. This is the mechanism for conservation of mass-energy in this regime for exceedingly high energy gamma rays. However, for all lower energies, conservation of energy requires that the Compton effect cause a recoil of the free electron. Indeed, as Spiff points out, the shift PER Collision is small compared to the wavelength of the initial photon, but the number of collisions is very large, being proportional to the wavelength when the photon is traversing a very long distance through the extensive electron cloud. The Thomson scattering ignores the momentum change not because it isn't there, but because it is small per collision. This small size is easily seen in that it takes all the electrons along the sight to sun for a very small Compton effect red shift to be seen (Z on the order of 10^-6). But it can't be ignored for the cosmological red shift where the number of interactions is huge and the distance of travel is so far. It is generally assumed that the wavefront is not reconstructed which is why it is thought that a large optical depth will not happen, but blurring will occur. This is true for mechanisms where the photon is absorbed and re-emitted or where the ExH vector is perturbed as in small mirrors or fog droplets which are large in comparison to the photon wavelength, but not so for Compton scattering from small free particles the size of electrons and positrons or anything smaller than the 1/2 the wavelength of the photon. See my paper for my arguments:
    http://www.geocities.com/CapeCanaver...5/compton.html

    <font size=-1>[ This Message was edited by: John Kierein on 2003-01-30 04:03 ]</font>

    <font size=-1>[ This Message was edited by: John Kierein on 2003-01-30 04:05 ]</font>

    <font size=-1>[ This Message was edited by: John Kierein on 2003-01-30 04:19 ]</font>

    <font size=-1>[ This Message was edited by: John Kierein on 2003-01-30 04:24 ]</font>

  3. #33
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    JK, well, that's a start. Now you admit there is an upperthreshhold for redshift (albeit ridiculously high at twice the mass of the electron). Of course, a lot of your assumptions about the interstellar/intergalactic plasma come from convenient theorizing and are not backed up by complementary observations. I've been calculated column densities of electrons for any number of different objects at different redshifts and let me tell you there is no such thing as uniformity across the sky. Add to that the wavelength dependence of Compton's Scattering (well documented in the paper I cited) is simply ignored by you. And I say we have excellent reasons not to give your theory a second look.

  4. #34
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    Ok folks. Note that JK has yet to confront the problem raised here by this very topic -- that Compton scattering induces a wavelength dependent redshift. Nevermind his suspect arguments that objects through his scattering fog wouldn't look blurry, the fact remains that the Compton effect produces redshifts that are wavelength dependent.

    Here is the Compton scattering formula:

    z_compton = lambda_c/lambda_i * (1 - cos (phi))

    per scattering, where lambda_c is the compton wavelength of the electron = 0.02426 Angstroms, lambda_i is the initial wavelength of the electron (assumed here to have a sub-relativistic momentum), and phi is the scattering angle of the photon relative to its initial direction. In JK's universe, this is always zero.

    So even if all the other magic happens (as in there are enough electrons per m^3 along the sight line to cause the number of scatterings JK needs, as in the light is never spatially blurred, as in the light is never spectrally blurred, as in the plasma he proposes has never been observed by the light IT SHOULD EMIT ON ITS OWN...), why is it that we do not observe a wavelength dependent redshift? Because it's not caused by the Compton effect, that's why.


  5. #35
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    It is NOT zero. See my paper. There is an average scattering angle for the photon travelling at c. But the group velocity is less than c because, although each individual travels at c, the path length is slightly longer due to the multiple interactions and the group wavefront is reconstructed to travel at the speed in the medium given by the index of refraction of this transparent medium.
    As an aside, the reason a cloud is opaque to visible light is that the fog droplets are larger than wavelength, so the ExH vector is reflected like off a mirror. But clouds are transparent to radar because the radar wavelengths are larger than the fog droplets. But the clouds do have an index of refraction.

  6. #36
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    Does anyone else see how we're asking a straightforward question about the implied wavelength dependence on redshift and there is no hint of a response offerred?

  7. #37
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    On 2003-01-30 11:11, John Kierein wrote:
    It is NOT zero. See my paper. There is an average scattering angle for the photon travelling at c. But the group velocity is less than c because, although each individual travels at c, the path length is slightly longer due to the multiple interactions and the group wavefront is reconstructed to travel at the speed in the medium given by the index of refraction of this transparent medium.
    By "It" I assume you mean the angle of the photon deflection. Ok, but....

    What you describe is a medium whose index of refraction is large, compared to that in a vacuum. This is an optical effect that comes about via the passage of light through dense media, such as glass. What index of refraction (as a function of wavelength) are you assuming? Remember too, that we're not just speaking of the interaction of radio waves with electrons where TINY effects such as you describe (wavelength dependent delays, if I understood you correctly) are observed in strong Galactic radio emitters -- but ALSO infrared, optical, UV and X-rays with spectral features within them that ALL show the same redshift of the quasar.

    And I stand by my statement that you still have not explained why we don't observe wavelength dependent redshifts, of the kind specifically predicted by Compton.

    As an aside, the reason a cloud is opaque to visible light is that the fog droplets are larger than wavelength, so the ExH vector is reflected like off a mirror. But clouds are transparent to radar because the radar wavelengths are larger than the fog droplets. But the clouds do have an index of refraction.
    I have no argument there.



    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-30 13:39 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-31 12:22 ]</font>

  8. #38
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    Spiff: The answer to your ? is in this paper as I have given above. http://www.geocities.com/CapeCanaver...5/compton.html
    Want me to copy it here? It's a little long for that.
    The cosmological red shifts are wavelength dependent for a given distance. delta L = HDL. The shift is delta L, and it is a function of wavelength. Maybe I don't understand your question?

  9. #39
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    On 2003-01-30 14:39, John Kierein wrote:
    Spiff: The answer to your ? is in this paper as I have given above. http://www.geocities.com/CapeCanaver...5/compton.html
    Want me to copy it here? It's a little long for that.
    The cosmological red shifts are wavelength dependent for a given distance. delta L = HDL. The shift is delta L, and it is a function of wavelength. Maybe I don't understand your question?
    A few of questions about this paper:
    (1) What value do you use for rho?
    (2) What probability function do you use (P(theta)) in evaluating the average value of cos(theta)?
    (3) How do you account for the change in the distance D due to the scattering?
    (4) Do you have calculations which show the change in apparent angular size of an object as function of d (since the angular size would not go as 1/d because of the random walk caused by the scattering)?
    (5) You mentioned earlier that you did not think the blurring would occur. Do you have anything more quantitative on this point?

  10. #40
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    On 2003-01-30 14:39, John Kierein wrote:
    Spiff: The answer to your ? is in this paper as I have given above. http://www.geocities.com/CapeCanaver...5/compton.html
    Want me to copy it here? It's a little long for that.
    The cosmological red shifts are wavelength dependent for a given distance. delta L = HDL. The shift is delta L, and it is a function of wavelength. Maybe I don't understand your question?
    If you are quoting the Hubble law, it is:

    v = H x d, and so

    cz = H x d Does anyone see a wavelength dependence here in the expanding universe paradigm redshift z? There is none.

    The redshift z that we measure from galaxy and quasar spectra does NOT (can I say it any louder?) DOES NOT depend upon the wavelength of the emitted photon. You've confused the redshift z with the DIFFERENCE in wavelength (observed - emitted) that is indeed wavelength dependent.

    Look, here is an explicit example. Here I compute the redshift from four of the strongest emission lines in quasar spectra:

    for a z = 2 quasar, here are the typical measured wavelengths (lambda_obs) in Angstroms:
    *****************************
    Table
    Lyman alpha: lambda_em = 1216
    lambda_obs = 3648

    C IV: lambda_em = 1550
    lambda_obs = 4650

    Mg II: lambda_em = 2800
    lambda_obs = 8400

    H alpha: lambda_em = 6563
    lambda_obs = 19689 (1.9689m)
    *****************************

    where lambda_em is the rest (lab) wavelength of the emission line transition.

    So when I compute z I will find the same value, whether for Lyman alpha or for H alpha.

    Here is how astronomers measure the redshift from the quasar's or galaxy's spectrum:
    z = (lambda_obs - lambda_em)/lambda_em

    e.g.,
    z(Lya) = (3648-1216)/1216 = 2.00
    z(Halpha) = (19,689 - 6563)/6563 = 2.00

    SO AGAIN -- we measure the same value of the redshift z (or nearly so) for all emission lines from the quasar, and for that matter the same redshift for stellar spectral features in high redshift normal galaxies. We can also measure stellar spectral features from the normal galaxy surrounding the quasar (lying in the galaxy's nucleus) and they are the same (very near so) as those from the emission lines of the quasar.

    What will Compton scattering find?

    Look at its expression:
    z_c = lambda_c / lambda_em (1-cos(theta))

    lambda_c is a constant -- and so IT IS PLAIN TO SEE THAT z_c IS INVERSELY PROPORTIONAL TO THE EMITTED WAVELENGTH OF LIGHT. This guarantees that every emission line will have its own compton redshift. THIS IS NOT A N OBSERVED FEATURE OF OUR UNIVERSE.

    e.g.,
    It will find a redshift z_c that is 6563/1216 = 5.4 TIMES LARGER for Lyman alpha than for Halpha, and the redshifts we would measure from each and every emission line from X-rays all the way into the infrared (which span a factor of 10,000 in wavlength) would be different! and by huge factors.

    For instance, the iron K-alpha line near 6.4 keV (1.9 Angstroms) would suffer (from one scattering) a Compton redshift of (ignoring the cos(theta) term):

    z_c = lambda_c / lambda_em = 0.02426 / 1.9 = 0.0128, where lambda_c is the Compton wavelength.

    However, Halpha (Balmer alpha) would suffer (from one scattering) Compton redshift of (again ignoring the cos(theta) term):

    z_c = 0.02426 / 6563 = 3.70 x 10^-6.

    These redshifts z are NOWHERE NEAR EACH OTHER, and in fact as we've been saying all along, the Compton redshift z_c scales inversely with wavelength of the incident photon, which can be easily seen above (but I'll spell it out):

    z_c(Fe Kalpha) / z_c(Balmer alpha) = 0.0128 / 3.70 x 10^-6 = 3460 which is precisely (within round-off error) the ratio of 6563A / 1.9A.

    WE DON'T OBSERVE WAVELENGTH DEPENDENT REDSHIFTS, z. I'll say it again: You've confused the redshift z with the DIFFERENCE in wavelength (observed - emitted) that is indeed wavelength dependent (see the table above).

    We don't observe a wavelength dependent redshift, z.

    I am sorry about the tone of this note, but I've been telling JK this for over 2 years now, and either JK ignores it hoping it will go away, or JK cannot do the (straightfoward) maths, or I don't know what.


    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-31 12:19 ]</font>

  11. #41
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    Orion38: Isn`t this aspect than Marmet resolve when he introduce the bremsstrahlung in his calculations? (A New Non-Doppler Redshift)

    Not that I can see. Marmet is incorrect in saying that bremsstrahlung is ignored (easily seen by just looking at the chapter on bremsstrahlung in Rybicki & Lightman). When a photon Compton scatters off of an electron, the electron will emit bremsstrahlung radiation, but at an extremely long wavelength, since the acceleration of the electron is small in most cases. The result will be a Comptonized photon and a bremsstrahlung photon. You still have to deal with the fact of the Comptonized photon, and its lack of ability to generate a wavelength independent redshift.

    Also note that Marmet's treatment is not very convincing. He argues in section 3 of his online paper that photon interactions with neutral hydrogen produces a redshift, and even computes the expected scattering cross section for the hydrogen atom. But neutral atoms and charged free electrons are different kinds of beasts.

    I think Marmet seriously overestimates the liklihood that a photon of any given wavelength will interact with a neutral hydrogen atom. After all, his computation of the cross section in appendix B is nothing more than the volume of the atom, cut to a cross sectional area. He never considers the energetics allowed by quantum mechanics, which is what controls the probability that a photon will interact with a hydrogen atom of that size. I think the real probability is much smaller.

    Note also that he derives a required minimum density of neutral hydrogen, 2.5x10<sup>4</sup> atoms/m<sup>3</sup>, or 0.025 atoms/cm<sup>3</sup> (which he oddly approximates as 0.01). But we know, by virtue of the absence of continuum absorption, that the intergalactic density can't be more than 10<sup>-12</sup> atoms/cm<sup>3</sup> (Astrophysical Concepts, Martin Harwit, Springer, 1998 (3rd edition), chapter 9, "Cosmic Gas and Dust", page 352). So we need about 10 more orders of magnitude of hydrogen to make Marmet's model work, even if it is not based on questionable basic physics. And, if I am right about the energetics, then 0.025 atoms/cm<sup>3</sup> is a serious underestimate of the density Marmet really needs.

  12. #42
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    Compton effect and CMB

    Since we are talking about the Compton effect & redshifts, it might also be worth the effort to include a note about the cosmic microwave background (CMB). The FIRAS instrument on COBE measured the spectrum to be thermal, with a temperature 2.72500.0010 Kelvins. If this background radiation were subject to Compton scattering, the spectrum would be distorted away from a pure thermal spectrum.

    The possibility of distortion by Comptonization has been considered. the integrated Compton effect is shown by the compton y parameter:

    y = integral( k(T<sub>e</sub>-T<sub>p</sub>)/M<sub>e</sub>c<sup>2</sup> )dtau<sub>e</sub>

    where T<sub>e</sub> is the electron temperature, T<sub>p</sub> is the photon temperature, M<sub>e</sub>c<sup>2</sup> is the electron rest energy, and tau<sub>e</sub> is the optical depth to electrom Compton scattering.

    The value of the Compton y parameter is derived from the shape of the FIRAS spectrum, and shown to be no more than 1.5x10<sup>-5</sup> at the 95% confidence level (Fixsen et al., 1996; Fixsen et al., 2002). This implies that the CMB is exactly what it appears to be, a thermal background, and not the result of any Compton effect.

    References

    Fixsen, et al., 1996: The Cosmic Microwave Background Spectrum from the Full COBE/FIRAS Data Set, D.J. Fixsen et al., Astrophysical Journal 473:576-587, December 20, 1996.

    Fixsen, et al., 2002: The Spectral Results of the Far Infrared Absolute Spectrophotometer Instrument on COBE, Astrophysical Journal 581:817-822, December 20, 2002.

  13. #43
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    Well, Spaceman spelled it out clear and plainly as the nose on any person's face who has a nose. I await anxiously the answer to this quandary that has been the spectre for the idea for the last two pages of non-answering on the part of JK.

  14. #44
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    Since I believe the Compton effect to be indiscernible from the doppler effect, I think the authors were expecting some other difference. They expect the doppler effect from a big bang. The shape of the curve deviates from the black body curve at the hectometric wavengths. If the CMBR is thermal, it is just the background temperature of a static universe, as shown by Max Born.

  15. #45
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    While the thread is on the CMB, let me ask this, what is the CMB? That meaning, what is the substance of the CMB, is it strictly in the background, at the edge of the perceivable universe, or is it pervasive throughout the universe? If the CMB is thermal, then would it not be detectable everywhere, since even local space would have a thermal signature to detect? If its background what exactly is it that's emitting this radiation?

  16. #46
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    On 2003-01-31 09:32, John Kierein wrote:
    Since I believe the Compton effect to be indiscernible from the doppler effect, I think the authors were expecting some other difference. They expect the doppler effect from a big bang. The shape of the curve deviates from the black body curve at the hectometric wavengths. If the CMBR is thermal, it is just the background temperature of a static universe, as shown by Max Born.
    One can believe what one wants.

    "I believe I can fly, and so I can jump off this 100 story building and I'll soar through the skies." "I believe that pi is a rational number; it must be so because circles are so perfect in their symmetry. pi must be 3. I will now design airliners, buildings and bridges, believing that pi is 3.00."

    However, nature's laws will correct one's beliefs, if those beliefs run counter those laws. Science isn't about belief, and supposedly we are discussing science on this board, not beliefs.

    And the simple facts are:

    1) WE DON'T OBSERVE A WAVELENGTH DEPENDENT REDSHIFT, z.

    2) COMPTON SCATTERING BY ITS VERY NATURE PRODUCES REDSHIFTS, z, THAT ARE WAVELENGTH DEPENDENT.(nevermind how such enormous redshifts are actually generated by this mechanism without breaking/misapplying nature's laws, contradicting what we observe in the universe, or both)

    And to correct another misconception....
    The redshift due to expansion is not actually Doppler. They share similarities (especially at low redshifts), but they ARE NOT the same. In fact it is incorrect to use the special relativistic Doppler formula (even though some elementary astronomy texts still do this - aargghh!). General relativity and the expansion of space-time are not described by special relativity or the Doppler formalism. The latter describe what happens when things move through space-time. They do not apply to describing phenomena associated with the expansion (or other distortions) of space-time itself.



    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-01-31 12:09 ]</font>

  17. #47
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    Doodler: ... what is the CMB?

    The CMB (Cosmic Microwave Background) is a radiation that distributed throughout the universe, and has an energy spectrum consistent with a thermal black body at 2.72500.0010 Kelvins. It is observable everywhere.

    It is called the "cosmic microwave background because it is observed in the "microwave" region of the electromagnetic spectrum, its peak intensity at a wavelength of about 0.19 cm. Kierein's comment that "The shape of the curve deviates from the black body curve at the hectometric wavengths." is a meaningless irrelevancy. We can see the CMB at those wavelengths because there are no other competing sources of emission. There would never be any reason to expect a thermal spectrum at such long wavelengths, so the observation that we don't only confirms elementary expectations.

    It is called the "cosmic microwave background because it is observed to come from all directions, and has essentially the same energy spectrum in all directions (with minor deviations in temperature but not in thermal shape). It is not generated by discrete sources, but rather appears to fill space uniformly as a photon gas. we observe it as a "background" behind other, discrete sources, such as galaxies & galaxy clusters.

    And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.

    The CMB was predicted by early Big Bang cosmologists, who expected the thermal radiation of the "bang" to cool as the universe expanded. But they could not accurately predict the temperature, because they lacked a unique model for the early universe (their idea of the Big Bang was actually rather different from the ideas that are common now). Anti-Bangers often argue that Eddington predicted the background before the Bangers did, and they are half right. He did predict a background dut to thermalized starlight, but explicitly says that the spectrum could not be thermal. Instead, he predicted an effective temperature, that matches the peak intensity but not the spectral distribution of energy. The observed CMB is radically different from Eddington's predicted background, so it is inappropriate to argue that we are only observing what Eddington predicted.

    Here are a couple of websites which may help to understand what the CMB is in more detail.

    <ul>[*]Cosmic Microwave Background (by Tim Thompson, includes a history of its prediction & discovery, also numerous links to papers and other websites, notably the outstanding pages by Wayne Hu & Max Tegmark)[*]The Cosmic Microwave Background (by Arthur Kosowsky, hosted by Caltech's Level5: A Knowledgebase for Extragalactic Astronomy and Cosmology[/list]

  18. #48
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    Tim Thompson wrote:
    And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.
    --
    Marmet`s version of CMB
    http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
    It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
    However, in the standard model of the universe, the simple fundamental Planck's law has been ignored.

  19. #49
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    On 2003-01-31 16:50, Orion38 wrote:
    Tim Thompson wrote:
    And we call it the "cosmic microwave background, because it appears to be a general property of the universe, and not emission from any sources embedded in the universe.
    --
    Marmet`s version of CMB
    http://www.newtonphysics.on.ca/COSMIC/Cosmic.html
    It is recalled that one of the most fundamental laws of physics leads to the prediction that all matter emits electromagnetic radiation. That radiation, called Planck's radiation, covers an electromagnetic spectrum that is characterized by the absolute temperature of the emitting matter. From astronomical observations we observe that most matter in the universe is in the gas phase at 3 K. Stars of course are much hotter. The characteristic Planck's spectrum, corresponding to 3 K, is actually observed in the universe exactly as required.
    However, in the standard model of the universe, the simple fundamental Planck's law has been ignored.
    It's easy to ignore when one realizes that an optically thin (i.e. transparent) body of gas which is constantly being hit with EM radiation does not have a temperature at all. Temperature implies a thermal radiation pattern, and these gases can't have that. It's interesting that it used the term 'planck spectrum,' rather than the more common term 'blackbody spectrum.' It's called 'blackbody radiation' for a reason.

    <font size=-1>[ This Message was edited by: Zathras on 2003-01-31 17:09 ]</font>

  20. #50
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    On 2003-01-31 09:32, John Kierein wrote:
    Since I believe the Compton effect to be indiscernible from the doppler effect, I think the authors were expecting some other difference. They expect the doppler effect from a big bang. The shape of the curve deviates from the black body curve at the hectometric wavengths. If the CMBR is thermal, it is just the background temperature of a static universe, as shown by Max Born.
    This has been proven false and it is basically tantamount to lying if you report that you believe in such a thing without also including the fact that the smoothness and isotropy is not explained by this model.

    This goes for Marmet's ramblings about the subject as well. Just look at the spectrum and decide for yourself, is it a consistent black body? If so then how on Earth can it be a background of something that isn't coherent? The time of last scatter is the ONLY coherent signal that has been proposed by any theory I have seen. All other theories rely on integrated starlight or other source-based radiation that would not give you the incredibly precise Blackbody Spectrum that we witness.

    Still waiting for an explanation on how JK will reconcile the fact that the Compton Effect has a wavelength dependece for any associated redshift.

  21. #51
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    Spiff, you just can't have a single interaction, you must have a number of them. The shift from a single interaction is INDEPENDENT of wavelength; however the number of interactions is huge! The number of interactions is proportional to the wavelength; the longer wavelength photons interact with more of the intervening electrons in direct proportion to the wavelength. So the total red shift from the total nimber of interactions is wavelength dependent just like in the doppler effect. Have you read this or not?????? The red shift is proportional to the wavelength delta lambda ~ lambda for a given distance.
    As I've said many times before, photons interacting with all the electrons in the sun's atmosphere only produce a shift on the order of delta lambda/lambda of about 10^-6
    http://www.geocities.com/CapeCanaver...5/compton.html

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    This does NOT address the fundamental issue that there is a wavelength dependence for coherent sources.


    Look at it simply: you have a source (like a galaxy) that has a specific energy across the spectrum. Sometimes the most energetic are the lower wavelengths, sometimes the most energetic are the higher wavelengths. REGARDLESS of how many photons are of individual wavelength there is NO WAVELENGTH DEPENDENCE OF REDSHIFT.

    In other words, JK refuses to answer the criticism. His house of cards has fallen unless he can address this.

  23. #53
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    The redshift depends on the number of electrons each photon sees. There is only a given density of electrons. Photons of longer wavelength are bigger than photons of shorter wavelength. Look at your microwave oven when it is on. You can see inside it through the holes in the metal shield in the door, but the longer wavelength microwaves can't escape. This is because the visible light is smaller than the longer wavelength micrwaves. So when the longer wavelength light sweeps through the elecrons between galaxies, it sees more of the electrons because they are bigger than the shorter wavelength photons. This results in a red shift proportional to the wavelength just like the Doppler effect. I don't see why you should expect anything different to the first order. I guess I don't understand the question, if you are talking about something different. The original number of photons across the spectrum of the source will all be shifted without any deference; the photons are not absorbed; just because the spectrum is more energetic in one part of the spectrum has nothing to do with it. They all see the same electron density. I don't see what you find as a "problem". The only part of the spectrum where the Compton effect doesn't occur is for very energetic gamma rays with an energy greater than twice the rest energy of an electron where the electron-positron pair formation mechanism dominates over the Compton effect for energy conservation. Some will say that the Compton effect doesn't occur for low energy photons, but they are just saying that the shift is very small per photon interaction, not that it doesn't occur at all, and that the Thomson mechanism can approximate it. For the case of the low energy photons, you must realize the size of the photon is very large and has many more interactions so that you can't ignore the cumulative total red shift.

    <font size=-1>[ This Message was edited by: John Kierein on 2003-02-01 04:59 ]</font>

    <font size=-1>[ This Message was edited by: John Kierein on 2003-02-01 05:08 ]</font>

  24. #54
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    On 2003-02-01 04:54, John Kierein wrote:
    So when the longer wavelength light sweeps through the elecrons between galaxies, it sees more of the electrons because they are bigger than the shorter wavelength photons.
    John,

    Da boyz here are used to think that photons and electrons interact only in the way of scattering of point-like particles. They have no idea of interactions of ensembles of photons and electrons where they should've used simple classic electrodynamics. The problem is, they are never taught that science these days...

    I've been through that with them earlier here. They just can't perceive it...

  25. #55
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    Well, first of all we know that Agora's criticism can't be right because JK is using the Compton Effect which is in the beautiful quantum mechanical limit.

    Secondly, JK, consider what Spiff wrote on the last page. The Compton redshift is dependent on wavelength. The redshift we observe is INDEPENDENT of wavelength. Therein lies your problem.

  26. #56
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    What do you mean? The change in wavelength for the doppler effect and the cosmological red shift is dependent on wavelength. Please clarify what you mean? The multiple compton effect and the doppler effect are the same form. Did you read the paper or not?

  27. #57
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    I have read your paper, John. We have explained it to you in plain English and using the proper mathematics. I LINKED to a paper that explains it plainly. If you cannot access said paper, then please tell me. I'm not going to play this game. The evidence has been put on the table. There is a dependence of redshift on wavelength in your model that isn't seen. Repeat: isn't seen. Remember to give the proper definition of redshift which is the one that is basically "normalized". It is not simply the change in wavelength. Look at some basic astronomy text if you're still equivocal.

  28. #58
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    On 2003-02-01 14:42, John Kierein wrote:
    What do you mean? The change in wavelength for the doppler effect and the cosmological red shift is dependent on wavelength. Please clarify what you mean? The multiple compton effect and the doppler effect are the same form. Did you read the paper or not?
    As I said up top, you confuse "change in wavelength" with redshift, z. The former is wavelength dependent in the current paradigm, that latter is observed not to be. The change in wavelength is not the same thing as the redshift, which is (delta lambda) / lambda. In the expanding universe paradigm, this quantity does not depend upon the wavelength. If you really mean redshift, and not change in wavelength, then you need to be more careful what you say.

    Second, you state here and in your pages that longer wavelength photons interact with the electrons more frequently. Where the heck does that come from? There is no wavelength dependence in the free electron cross section except at energies in the Klein-Nishina regime (energies of keV), roughly the same order of magnitude the wavelength of light becomes comparable to the compton wavelength of an electron. Your microwave oven analogy is just hand waving, and as it stands does not apply here.

    Next, I understand precisely that you require millions of interactions to get the redshifts that are observed. And I am just supposed to believe you that your wavelength compensating cross section mechanism after all of these interactions will precisely lead to a wavelength independent redshift, z? I am just supposed to believe that no spatial smearing occurs? I am just supposed to believe that no spectral smearing occurs? I am just supposed to believe that some coherent wavefront forms after photons scatter millions, billions of times in medium that is millions or even billions of times less dense than some of the best laboratory vacuums, and sends the light straight through this electron scattering medium as if it never interacted at all (except to provide your magic redshift) - and so with effectively zero optical depth (no obscuration)? Just because you state such on a webpage? Saying something does not make it so.

    I've mentioned to you that the problem of photons scattering in Compton thick media (i.e., multiple scattering) is a subfield in astrophysics. I've looked at these papers, and I see nothing that supports your mechanism at all. You are the one that must demonstrate the physical validity (1st) then observational viability (2nd) (i.e., does it predict events that are contrary to observations?). Not the other way around.

    Where are your simulations demonstrating quantitatively what happens to a spectrum of photons passing through a Compton thick medium? What predictions (other than redshift) are made by such a model, and are they consistent or not with the observational data? Quanatitative results, not just waving of hands and saying in effect "because I say so", this is what is demanded by science.



    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-03 15:42 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-03 15:43 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-04 10:25 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-04 12:50 ]</font>

  29. #59
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    Again, JK, what you describe in general, as best as I can decipher, is an optical effect that might occur in macroscopic media of high density and significant indeces of refraction for the wavelengths in question (and I'm talking about X-rays through infrared). However, I doubt that Kierein scattering in detail has any application anywhere.



    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-04 10:17 ]</font>

    <font size=-1>[ This Message was edited by: Spaceman Spiff on 2003-02-04 10:20 ]</font>

  30. #60
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    JK: The shift from a single interaction is INDEPENDENT of wavelength; ...

    Yes and no. The shift in the sense of dw (where w = wavelength and dw = change in wavelength or delta wavelength) is independent of wavelength. So is the astronomical redshift independent of wavelength. However, the two quantities are not the same.

    Astronomical redshift: dw/w<sub>rest</sub>, where dw is the delta wavelength and w<sub>rest</sub> is the rest frame wavelength of the spectral feature.

    Now look at the Compton scattering formula.

    Compton: dw = w<sub>c</sub>*(1-cos{theta}), where dw is once again the delta wavelength, w<sub>c</sub> is the Compton wavelength (a constant), and theta is the photon scattering angle.

    So, if we take the Compton formula and divide both sides by w<sub>rest</sub> we get the astronomical redshift on the left, and we get w<sub>c</sub>*(1-cos{theta})/w<sub>rest</sub> on the right. Now we see that the Compton effect analog of the astronomical redshift is inversly proportional to the rest frame wavelength, an effect which would be very obvious if it were in fact the case. However, we know that there is no such dependence in astronomical redshifts, which implies that the Compton effect cannot be the source of an astronomical redshift.

    JK: Photons of longer wavelength are bigger than photons of shorter wavelength.

    As stated, that is not true, although it is likely to be the case where the photon energies are very different. Photons, being quantum beasties, do not have "length" in the classical sense. However, they do have a coherence length, which is determined by the Heisenberg uncertainty in the photon energy, dE*dT .GE. *hbar, where dE is the uncertainty in the photon energy, dT is the uncertainty in the time it takes to emit the photon, and hbar is Planck's constant divided by 2*pi. The uncertainty dT times the speed of light c is probably a good approximation of the coherence length, or "length" of the photon (I don't know the correct, quantum optical definition). Since dT is process related, and the determinant of the coherence length, then in principle, even photons of widly differing wavelengths could easily have the same "length".

    One particular mistake to avoid is viewing the photon as an extended "wave packet", which is definitely not proper QED (although it was common in the early days of quantum mechanics).

    JK: ...the longer wavelength photons interact with more of the intervening electrons in direct proportion to the wavelength.

    I can think of no physical justification for this statement. We have already seen that the "length" of a photon is a slippery concept, and it certainly is not true that "longer" photons have a larger interaction probability. This can be shown, even if we admit classical "lengths" for quantum photons, because the speed of the photon is, in the vast majority of cases, far in excess of the speed of the electrons. So, the electrons hardly have an opportunity to interact with any other property of the photon than the scattering cross sections found in the various scattering equations or the Klein-Nishina equation.


    <font size=-1>[ This Message was edited by: Tim Thompson on 2003-02-03 18:01 ]</font>

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