Understanding Reflective Radiation Pressure

[rodriguez1gv]

Hi,

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 E_f/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 P_reflec=2E_f/c * cos²a for an incident angle a. I am not sure where the cos squared comes from.

 

[TSny]

Hello rodriguez1gv and welcome to PF!

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 E_f/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 P_reflec=2E_f/c * cos²a for an incident angle a.

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

I am not sure where the cos squared comes from.

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?

 

[tiny-tim]

hi rodriguez1gv! welcome to pf!

. I am not sure where the cos squared comes from.

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

(wikipedia talks about a perfect reflector, so the calculation will be slightly different if the surface is matt)

 

[rodriguez1gv]

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!

 

[Nickie]

Hi there,

Thank you for your inquiry about reflective radiation pressure. Radiation pressure is a phenomenon that occurs when electromagnetic radiation, such as light, exerts a force on a surface. This force is caused by the transfer of momentum from the photons of the radiation to the surface.

When it comes to reflective radiation pressure, there are a few key factors to consider. First, as you mentioned, the momentum of a photon is given by p = E/c, where E is the energy of the photon and c is the speed of light. This momentum is transferred to the surface upon impact, resulting in a force being exerted on the surface.

Now, in the case of reflection, the photon will be reflected in a random direction, as you mentioned. However, this random direction is not completely random. The angle of reflection will depend on the angle of incidence and the properties of the surface, such as its reflectivity. This is where the cos squared term comes in.

The cos squared term represents the cosine of the angle of reflection, squared. This is because the force exerted by the photon is proportional to the cosine of the angle of reflection. This means that for a given incident angle, the force exerted by the photon will be greater if the angle of reflection is closer to 90 degrees (perpendicular to the surface) rather than closer to 0 degrees (parallel to the surface).

To capture all possible reflection angles, an integral over the solid angle is necessary. This integral will take into account the different orientations of the surface and the resulting variations in the angle of reflection.

I hope this helps to clarify the concept of reflective radiation pressure. If you have any further questions, please don't hesitate to ask. Thank you for your interest in this topic.

Best regards,