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Can energy be lost from a Photon?

  1. Jul 22, 2009 #1
    Is energy lost from photons when travelling through the air (eg: friction) to an object from a source such as the sun?

    If so or not, please explain it me simply. Thank you.
     
  2. jcsd
  3. Jul 22, 2009 #2
    The energy of photons is absorbed by matter in a manner VERY different from friction, and since you asked for a simple explanation, I won't go into what actually happens.
    But it's not like a projectile going through the air. Photons don't get slowed down by traveling through the air, that's not how they lose energy. (Photons are light, and travel at the speed of light)
    They have a certain amount of energy "stored" in each photon, and it can only be imparted completely. A photon can't lose just part of its energy, it's all or nothing. (Please correct me if I'm misleading the TS, since I have only a basic understanding of the subject)

    But yes, the photons originate from the high-energy reactions within the sun. They themselves do have energy, and they can impart that energy onto particles. Originally, it came from the thermo-nuclear reactions in the sun.

    There are many different sources for photons, but whenever one is "created" from matter, that matter ends up being less energetic. Again, this is all very simplistic, but I hope that answers your question.
     
  4. Jul 22, 2009 #3
    So photons don't lose their energy because they travel at the speed of light and keep their momentum?

    So during the photoelectric effect, photons are emmitted from an EM source with a high frequency and these photons collide with electrons on a charged object giving them energy to move either away from the atom or further in. (Crompton effect occurs I think).
     
  5. Jul 22, 2009 #4
    Photons can gain or lose energy travelling in a gravitational field, but they maintain a speed of C.
    Photons from a star lose energy struggling up through the star's gravitational field, they tend towards lower frequencies ( to the red)
    Photons from space descending through the Earth's gravitational field gain energy, move towards the blue.
    A Black Hole is the situation where the photon has lost all energy in the extreme gravitational field.
     
  6. Jul 22, 2009 #5

    diazona

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    I like that description. (with the caveat that it only applies for collisions, not redshifts)
     
  7. Jul 22, 2009 #6
    "They have a certain amount of energy "stored" in each photon, and it can only be imparted completely. A photon can't lose just part of its energy, it's all or nothing"

    The energy E = hf so if the frequency changes, the energy changes. But the energy is stored as momentum. p = hf/c So if momentum changes, energy changes and frequency changes.
    Seen in the Hubble red shift and gravitational red and blue shifts.
     
  8. Jul 22, 2009 #7
    Photons from nuclear isomeric transitions (very narrow linewidths) traveling uphill (22.5 meters uphill from a gravitational source) lose energy, as proved in the Pound Rebka experiment (in 1959) using the Mossbauer Effect in iron-57.
    See
    http://en.wikipedia.org/wiki/Falling_photon
    The effect was a few parts in 10^15 energy "loss", basically just a red shift.
    At one time, I used a Bragg-crystal-diffraction spectrometer to look for energy loss of a 84.2 KeV nuclear transition gamma (with very narrow linewidth) by putting 1" of aluminum in front of the source. I could not see any broadening of the gamma line, nor any energy shift of the order of 1-eV or more. I could not detect any forward scattered Compton photons either.
    [Edit] 1" of aluminum represents about 1.35 photon attenuation lengths at 84.261 KeV.
     
    Last edited: Jul 22, 2009
  9. Jul 23, 2009 #8
    i do not agree with map19's response. to the best of my knowledge, photons do not lose energy "struggling" to escape a gravitational field, and they do not "lose all energy" in a BH - photons cannot escape a BH because the escape velocity within the EH is > C. a discussion of this can be found in bruce harvey's paper here:
    http://users.powernet.co.uk/bearsoft/new_site/phys/grav-photons.pdf
    which states:
    "The energy content of photons is unaffected by gravity. A photon emitted from the sun appears red shifted
    because it had less energy when it was emitted than it would have had if the atom which emitted it was on
    earth."

    photons can, however, lose energy over long periods of time traveling through space as they become red-shifted due to the expansion of the universe. this effect can be seen clearly in CMBR.
     
    Last edited by a moderator: Apr 24, 2017
  10. Jul 23, 2009 #9
    How do you explain the Pound & Rebka Mossbauer Effect experiment on the effect of gravity on the 14.7 KeV iron-57 line.
    Photons from nuclear isomeric transitions (very narrow linewidths) traveling uphill (22.5 meters uphill from a gravitational source) lose energy, as proved in the Pound Rebka experiment (in 1959) using the Mossbauer Effect in iron-57.
    See
    http://en.wikipedia.org/wiki/Falling_photon
     
  11. Jul 23, 2009 #10
  12. Jul 23, 2009 #11
    this is plainly wrong: photons of high frequency (one color, say ultraviolt) have high energy and photons of low frequency (another color, rad infrared) have low energy.


    The energy content IS affected (as noted in the prior two posts) and can be considered due to time dilation accompanying increasing gravitational potential: a photon emitted from a gravitational source IS of lower frequency, is less energetic, than it would have been if emitted from a lesser gravitational source. In fact all chemical and nuclear processes are slowed in such a gravitational potential as reflected in general relativity.
     
  13. Jul 24, 2009 #12
    i stand corrected - thanks. however, can you please explain to me where this lost energy goes? it isnt radiated, so how is the energy conserved?
     
  14. Jul 24, 2009 #13
    Example 1: Reflection. A photon reflects off a moving mirror.


    Example 2: Coherent forward scattering in media yielding an effective index of refraction. A photon enters a medium travelling in one direction and exits while moving in another direction.
     
  15. Jul 24, 2009 #14

    dlgoff

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    Yep. I remember doing an Fe-57 Mossbauer Effect experiment in a undergraduate lab where we put the source on a model RR car and ran it down an inclined track toward the detector. Just for reference: http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/mossfe.html#c1"
     
    Last edited by a moderator: Apr 24, 2017
  16. Jul 25, 2009 #15
    "please explain to me where this lost energy goes?"
    Yes, well, you picked a slightly contentious issue.
    If the frequency and energy is reduced by red shift that is caused by time dilation then the total energy is preserved. You see a lower energy over a longer time.
    However, I recently read a paper arguing for a new look at the energy balance. Sorry I don't have a reference handy.
     
  17. Jul 28, 2009 #16
    We know from classical physics that light waves entering a transparent material with index of refraction n>1 that the light wave will be partially refracted and partially reflected. Light with normal incidence on glass will be 96% refracted (transmitted), and 4% reflected. Does this mean that 96% of the energy of a single photon is refracted, and 4% of the energy reflected? No, photons are refracted 96% of the time with 100% of the energy, and reflected 4% of the time with 100% of the energy.
     
  18. Jul 28, 2009 #17
    Bob S is, of course, correct. This is covered in detail in Feynman's book QED.
    What may confuse students is because of the slower speed in glass (and any medium ) the energy and frequency remain the same for refracted light, but because of v = frequency x wavelength, the wavelength is reduced.
     
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