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Reflective Radiation Pressure

  1. Aug 16, 2013 #1

    I am just trying to understand the basis of radiation pressure. I understand radiation pressure due to absorption, but I am having a hard time understanding the radiation pressure due to reflection.

    From what I understand there will be an incoming photon with momentum p = E/c. The normal momentum will be imparted into the reflecting surface upon impact, resulting in a transfer of energy of Ef/c * cos(a) for incident angle a.

    Now the photon will also be reflected in a random direction. I think that there should be an integral over the solid angle to capture all possible reflection angles, but I am not sure what to do exactly. I know Wiki says Preflec=2Ef/c * cos2a for an incident angle a. I am not sure where the cos squared comes from.
  2. jcsd
  3. Aug 16, 2013 #2


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    Hello rodriguez1gv and welcome to PF!

    The wiki article assumes that the angle of reflection is the same as the angle of incidence rather than random reflection direction.

    Note this statement from http://en.wikipedia.org/wiki/Radiation_pressure#Radiation_pressure_by_particle_model:_photons :

    "The orientation of a reflector determines the component of momentum normal to its surface, and also affects the frontal area of the surface facing the energy source..."

    The article is considering a beam of light in which the cross sectional area of the beam is larger than the area of the reflecting surface (like holding a mirror in sunlight.) Think about what happens as you alter the angle of tilt of the reflecting surface in the beam. Will the same number of photons strike the surface per unit time?
  4. Aug 16, 2013 #3


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    hi rodriguez1gv! welcome to pf! :smile:
    imagine that you have a 1 sq cm tube of light falling on the reflector

    if it falls perpendicuarly, it it spread over 1 sq cm of the reflector

    but if it falls at an angle θ, it falls on 1/cosθ sq cm, so it is more spread out, and is diluted by a factor cosθ

    furthermore, only the component of momentum perpendicular to the reflector is reversed, so that's another factor of cosθ :wink:

    (wikipedia talks about a perfect reflector, so the calculation will be slightly different if the surface is matt)
  5. Aug 17, 2013 #4
    So, to reiterate, the cos squared is accounting for the perpendicular energy and a reduced incident flux due to the tilt of a surface? That makes sense I think, Thanks for the help!
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