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I Diffuse reflection

  1. Jun 3, 2018 #1

    pervect

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    In a previous thread, we discussed specular reflection from a stationary and moving mirror, and argued that the momentum transfer was always normal to the mirrors surface.

    I became curious as if this was true in the case of diffuse reflection. But I'm not sure how to calculate that. It's probably sufficient to calculate this in the rest frame of the mirror, though.

    (I marked the thread as I level, but if an A-level response would be better, that's OK too.)
     
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  3. Jun 6, 2018 #2

    JMz

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    By diffuse, you presumably mean that the mirror consists of a distribution of randomly orientated, small reflecting regions (facets), whose mean normals are the same as the normal of the mirror as a whole, right? In that case, each facet's momentum transfer is along its own normal. (Note that this is working with geometrical optics, so we are assuming the wavelengths are short compared to the facets' diameters.)
     
  4. Jun 8, 2018 #3

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    A quote from Wikipedia on diffuse reflection might clarify the mechanism a bit:

    So I know the process involves scattering, but I'm not sure of the details. Perhaps the question could use some refining - if so, the "reflection from a polished marble surface" as opposed to "reflection from a mirror" would be a specific inspirational example for why I asked the question. However, it's possible some other example would shed some insight on the issue and be easier to talk about and illustrate the physics more.
     
  5. Jun 11, 2018 #4

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    My technical background on the practical optical physics includes (ancient) techniques for producing diffuse reflectants, as for calibrations. Those relied on a crystalline powder -- likely TiO2, though I don't recall. The crystals therefore perform as phase-randomizers: both tiny prisms and tiny mirrors, albeit mirrors that, as often as not, send their reflected photons further into the powder, just because of the random orientation of the facets, rather than directly back out of it. (In certain ways, the mathematics is similar to diffuse scattering in particle physics or gravitation.)

    Whether this is what marble effectively does or not, I do not know. But I do know that that powder was used for the very purpose of producing an easily reproducible, very diffuse (Lambertian) reflector. So I suggest that this may serve as a tractable model for your question.

    The key for your question, I believe, is the word random. If the material were either an absorber or a retro-reflector, momentum transfer would obviously be along the direction of incidence, rather than normal to the surface. It's the randomization that leaves the photon flux with no other preferred direction than the surface normal.

    For single-facet reflectors, i.e., for random tiny mirrors, this involves an integration over either all the facet orientations or, equivalently, over the outbound photons' directions. The nice thing (for perfect non-absorbers) is that, even for multiple reflections and for refraction, all the photons ultimately come out again, away from the surface. So you can perform the very same integration over the outbound photons' directions, without regard to their complicated paths in the middle. If the reflection is indeed Lambertian, the integration is symmetric with respect to the surface normal, so that's the direction of momentum transfer.

    In fact, even if the surface only behaves as a linear combination of a Lambertian and a specular reflector, that can obviously hold as well, because it holds for both components. To take your example, perhaps that applies to polished marble.
     
  6. Jun 11, 2018 #5

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    BTW, I am assuming here that the surface is nearly flat. If it's significantly concave, the result might be the same, but the proof modestly more complicated. If it's not concave everywhere, my guess is that (a) it's still true provided that there is a well-defined normal (untrue for a sphere, for instance), but (b) it's a significantly more complicated proof. However, I don't know that.
     
  7. Jun 12, 2018 #6

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    Thank you for your help, it's appreciated. I'm going to go on a bit and "think aloud" as to what I think your remarks imply.

    Everything I've been reading suggests that we consider diffuse reflection in general to be described as scattering (with Lambert scattering being an idealization that is specific enough to compute).

    For instance, if we take a dust cloud, there is the Poynting-Robertson effect, which slowly causes a dust cloud orbiting a star to spiral in. Aw Wiki epxlains, in the frame of the dust, tellar aberration of the light from the star causes it to appear to come at an angle slightly "ahead" of the star. The momentum of the light is essentially absorbed and re-radiated uniformly in all directions, so the net momentum transfer is as if the light were absorbed.

    I don't think a mirror with specular reflection would suffer any Poynting-Robertson drag, the light would still be aberrated, but the reflected light would cancel out the drag effect.

    If we consider a flat reflector (the original problem), I'd think that the radiation can only occur in a 180 degree sollid angle, and this also allows for a net momentum transfer that's not normal to the mirror's surface whenever the light does not strike the mirror at a normal angle. So the dust-cloud model wouldn't really apply. I'm thinking that limiting the solid angle of the re-emitted radiation would actually fight the Poynting-Robertson effect, but I'm ont really sure yet.
     
  8. Jun 12, 2018 #7

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    Ah - yes, if you are thinking of dust, then of course thermal emission can occur in all directions (if the dust is either thermally conductive or spinning rapidly), including "out the back", away from the light source. And I believe that emission would indeed be considered diffuse. However, it would not be considered "reflective", because the original (say, visible) photons would have been converted to much lower energies (infrared), not merely bounced about and eventually released from the surface.
     
  9. Jun 13, 2018 #8

    JMz

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    As I, too, think aloud here, it seems to me that -
    1. The diffuse output from a flat surface will transfer momentum into the surface normal (average across all outgoing photons), as stated above.
    2. However, the light source will transfer momentum into it in the direction of the source, as you said.

    The two energies will obviously be equal, and the magnitude of momentum = energy/c for photons. So do we conclude that the net momentum transfer will occur in the direction (i.e., angle) exactly 1/2 way between the input direction and the surface normal?
     
  10. Jun 13, 2018 #9

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    I'm not sure - not without doing more detailed calculations and a lot more thinking. It looks like Wiki has a formula for the reflection, Lambert's cosine law.
     
  11. Jun 14, 2018 #10

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    Yes, but it didn't seem to provide a direction answer to your question. That's where my 2-part question came in.
     
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