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Another thread on this site established that the photon has a gravitational field of its own. Then the photon must lose energy to the gravitational wave the photon makes when it travels through space, I assume? But I have never seen a calculation of the redshift due to this effect by mainstream astronomers. But prof. Jian-Miin Liu (Nanking University) has some interesting papers on this subject on the internet, using the GTR as his theoretical background. I would be very glad to hear your views on this.

Urb
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Göran Rosander
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    Could you reference/cite the other link you mention in your first line, thanks –  Jan 20 '17 at 23:26
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    Possible duplicate of Do photons and cosmic rays radiate energy through gravitational waves? If not, why not? – ProfRob Jan 23 '17 at 07:42
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    Please note that the expansion of the universe is seen not just in the redshifts of photons, but in the dilation of time intervals between events occurring at high redshifts, like typeIa supernova lightcurve decays. The latter cannot possibly be attributed to GWs (or any of the other tired light ideas). – ProfRob Jan 23 '17 at 07:47
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    Also http://physics.stackexchange.com/q/240440/ – ProfRob Jan 23 '17 at 07:52
  • Gedanken experiment: A laser at great distance from Earth with its beam directed against the Earth. Between the laser and the Earth there is a tired light mechanism. The laser light is consequently red shifted. A timer placed on the laser opens and closes a light valve, e.g. one sec. on - one sec. off and so on. Every period of light has a fixed number of light maxima, not changed by the mechanism. But we receive a longer laser beam because the number of maximas x the longer wavelength = give a longer period than one sec. with a constant velocity of light. Nova curves add nothing to red shift. – Göran Rosander Jan 25 '17 at 05:44
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    @Countto10 It was a time ago so I don't remember clearly,but I think an experiment with two antiparallell laser beams draw my attention. – Göran Rosander Jan 25 '17 at 12:40
  • @GöranRosander Your gedanken experiment makes no sense. Tired light has to reduce the "number of peaks" per second - that is what redshift is. It removes energy from the wavepacket, it does not smear it out over a longer period of time. You appear to be asking for a dispersion relation which makes light cover the distance between source and observer in a time that depends whether it is at the front or back of the wave packet. – ProfRob Jan 25 '17 at 21:35
  • Linked: http://physics.stackexchange.com/questions/156618/tired-light-red-shift-hypothesis?rq=1 – ProfRob Jan 25 '17 at 21:48
  • And by the way - the neutron stars in your deleted "answer" had a time-dependent quadrupole moment - as per John Rennie's answer. – ProfRob Jan 25 '17 at 21:50
  • @Rob Jeffries I'm sorry to say that you have not understood what I have written. "Tired light has to reduce the "number of peaks""/sec. Exactly. Next step of thought: The (redshifted) wavelength X number (not reduced) of peaks => and you get a distance LARGER than 300,000 km. But the light train was exactly 300,000 km when transmitted. Consequently, you have a time dilation of light (redshift) and also of the one (to more than one sec) sec. timer. => You don't add any fact by measuring nova curves if you have measured the redshift. – Göran Rosander Jan 27 '17 at 12:37
  • @GöranRosander There is no law of "conservation of number of peaks"; you are quite misguided. – ProfRob Jan 27 '17 at 17:28
  • @Rob Jeffries Please enlighten me about the mechanism you consider being able to change the number o peaks: – Göran Rosander Jan 27 '17 at 21:04
  • @GöranRosander Anything that changes the energy of light in the rest frame of an observer. e.g. Compton scattering; any other tired light mechanism. What preserves the number is space expansion or Doppler shift. – ProfRob Jan 27 '17 at 22:48
  • @Rob Jeffries If you have an mechanism as the one I have written about above, based on the GTR, that smoothly acts on every part of the laser beam, no way that the number of peaks are going to change. You just make a statement with no physical logic or reason. "You are quite misguided." – Göran Rosander Jan 28 '17 at 00:02
  • @GöranRosander You haven't written about anything, or even provided a reference, let alone to one in a peer-reviewed journal. – ProfRob Jan 28 '17 at 07:41
  • @Rob Jeffries That's correct. My 'gedanken' experiment is my own, I have not seen it anywhere else. – Göran Rosander Jan 28 '17 at 12:30
  • @Rob Jeffries I mentioned prof. Jian Miin Liu. The papers I downloaded 15 years ago can still probably be found on Internet; I just googled him. If mainstream cosmologists had been more interested of discussing the weaker points of the big bang theory it certainly had been easier for me to give a reference. – Göran Rosander Jan 28 '17 at 12:48

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Not everything that has a gravitational field emits gravitational waves. Gravitational waves are emitted if the gravitational field is time dependent and has a time dependent quadrupole moment.

The gravitational field of a propagating photon, or indeed any propagating particle, does not have a time dependent quadrupole moment so it does not lose any energy by emission of gravitational waves.

John Rennie
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When a photon is absorbed, then one does expect a ringing of the gravitational field. Some of the energy of the photon will dissipate as a gravitational wave under those conditions.

Continuous
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  • I don't believe this is accurate. Can you provide any references or explanation? – DilithiumMatrix Jan 23 '17 at 01:23
  • As an example of a propagating particle let us take a neutron star - a massive particle indeed but well inside the definition "any propagating particle". Scientists have been awarded the Nobel Prize for their scientific work on two such particles and with precision calculated the energy loss when the 'particles' propagate in a gravitational field. Do you really mean "any propagating particle? – Göran Rosander Jan 25 '17 at 05:23