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I Radiation Pressure and Conservation of Momentum

  1. May 30, 2017 #1
    How do you account for the conservation of momentum for a photon? Specifically, if you have light travelling in a medium where the refractive index is not constant. For example, a graded index multi mode fiber optic.

    So here is another diagram to help articulate my question. Inside this graded index fiber, light curves off it's trajectory. If light has momentum, what is the opposite effect of this radial acceleration?
    light.png
     
  2. jcsd
  3. May 30, 2017 #2

    Dale

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    In a medium you cannot uniquely separate out the EM momentum and the material momentum. The total momentum of the wave and the medium is conserved.
     
  4. May 30, 2017 #3
    Does that mean the medium(fiber optic), experiences equal and opposite accelerations as the photons?
     
  5. May 30, 2017 #4

    Dale

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    Yes (equal and opposite changes in momentum), although this is a classical phenomenon so you can just talk about the classical electromagnetic waves and don't need to actually deal with quantized photons.
     
  6. May 31, 2017 #5

    sophiecentaur

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    Radiation pressure doesn't need photons to account for it. There is a force which results from a classical EM wave encountering a medium. Very text book stuff long before QM came along.
     
  7. May 31, 2017 #6
    I'm intrigued by the momentum em waves exert on the medium. Doesn't this also mean that sunlight curving into the earth is pushing the medium(atmosphere) in the opposite direction?
     
  8. May 31, 2017 #7

    Nugatory

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    Ever wondered why the tails of comets always point away from the sun?
     
  9. May 31, 2017 #8

    jbriggs444

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    The standard explanation has little to do with radiation pressure. Similarly, the rotation direction of a Crookes radiometer has little to do with radiation pressure.
     
  10. May 31, 2017 #9

    Nugatory

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    Ah - you're right, there's the solar wind which is an outgoing stream of charged particles. I've always understood that both radiation pressure and solar wind contribute to the formation oif cometary tails, but I've tried looking into the relative contribution of each.
     
  11. May 31, 2017 #10

    sophiecentaur

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    I can only reply "Why not?"
    But the actual value of this pressure is what counts. It has visible effects.
     
  12. May 31, 2017 #11

    sophiecentaur

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    I know that a standard Crooke's Radiometer works on temperature of the surfaces and local gas pressure. I have always assured students that, if you had a deep enough vacuum, it would go the other way.
    Does anyone know of a machine that works 'properly' in the presence of just sunlight? Was I guilty of Teacher BS?
     
  13. May 31, 2017 #12

    jbriggs444

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    You are right, of course and taught your students correctly. I was only going after the simple misconception that a comet or a Crookes radiometer is an immediate proof of radiation pressure.
     
  14. May 31, 2017 #13

    sophiecentaur

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    But is there a 'proper' one in some museum somewhere? It nags my conscience sometimes. I would never, knowingly lie to students. :biggrin:
     
  15. May 31, 2017 #14
    I can't help you with that but I do have a question.
    Why doesn't it take the same amount of energy to stop a photon as it does to create it? A 680 nm photon has 1.8232936424298 ev. It has 9.2058823529411764705882352941176e-28 kg m s of momentum. Or 9.027886617647058e-27 watts or 5.634763867393764823e-8 ev. Why is this the case? If a photon stops, how would it still have energy?
     
  16. May 31, 2017 #15

    sophiecentaur

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    If it "stops" then it doesn't;t exist, Its energy and momentum will have been transferred elsewhere. It's not like a billiard ball that hits something and then stays there. It has no mass so it can quite happily (for us) disappear.
    Edit: What's with the mention of Watts?I don't understand that.
     
  17. May 31, 2017 #16
    Just to convert to ev from watts. I just don't understand why the momentum is .00000003 of the initial energy. If this photon hit a medium with a refraction index of 2, would half of this momentum be transferred to the medium? When it exits this medium, does the medium again experience acceleration opposite of the photon?
     
  18. Jun 1, 2017 #17

    sophiecentaur

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    eV is Energy. Watts is Power (= Energy / time). There is no meaningful 'conversion'.
    How can you compare them? They are different quantities and different units. That number you quote is just 1/c.
    What you are doing is Numerology, which is not Science. Work in a different Unit system and the two above 'conversions' give different numbers.
     
  19. Jun 1, 2017 #18
    I looked at it from a different angle this morning. Since I know that the photon has 1.8233 ev of energy. I took an electron and calculated its velocity in vacuum after accelerating across 1.8233 volts. Then I found the momentum it would have. The answer was 7.2977e -25 newt m/s. I converted this to kg m/s, 7.444153e-26.

    So a photon has 2 orders of magnitude less momentum for the same energy as an electron. I am unsure of the math though.
     
  20. Jun 1, 2017 #19
    Looks like it was actually 7.246e-25 kg m/s so it is 3 orders of magnitude more than a photon of the same energy.

    Does them seem correct or atleast a feasible method to compare the two?
     
  21. Jun 1, 2017 #20

    jbriggs444

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    You were challenged on a conversion between energy units and power units. You responded with a way to start with the energy of an electron and end with its corresponding momentum. No, that does not give you a conversion factor between power and energy.

    It does not even give you a conversion factor between energy and momentum. The conversion is not linear. Momentum does not scale linearly with energy except in the high energy limit.
     
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