Does the absolute value of the speed of EM reflecting from a surface have an effect on radiation pressure?

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  • #1
Neeraj
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Even if a beam of light strikes a reflective surface at an angle 'A', the change in momentum should be 2mc, P=2IcosA/C but I find it P= 2Icos^2(A)/C.
 

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  • #2
hutchphd
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Even if a beam of light strikes a reflective surface at an angle 'A', the change in momentum should be 2mc, P=2IcosA/C but I find it P= 2Icos^2(A)/C.
Where did you see this?...I believe you are correct
 
  • #3
Neeraj
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Where did you see this?...I believe you are correct
Which one? I believe speed of light is constant in every frame we presume. I found it in a book.
 
  • #4
hutchphd
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If the velocity of the surface is large (near c) there will also be an appreciable effect from the Doppler shift, so the result will need modification. If you quote a result it is good form to provide the source (I.e. which book..there are quite a few!).

Also you say p=mc ...what is m? This is not correct.
 
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  • #5
Neeraj
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If the velocity of the surface is large (near c) there will also be an appreciable effect from the Doppler shift, so the result will need modification. If you quote a result it is good form to provide the source (I.e. which book..there are quite a few!).

Also you say p=mc ...what is m? This is not correct.
Here we require the change in momentum to find the force and then the pressure, and since we can't associate mass to photons, mass energy equivalence can be used to replace mass (p.s. It was explained in photoelectric effect to find out radiation pressure)I don't remember the exact source, I just wanted to confirm if I am right or not.
 
  • #6
Neeraj
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Here we require the change in momentum to find the force and then the pressure, and since we can't associate mass to photons, mass energy equivalence can be used to replace mass (p.s. It was explained in photoelectric effect to find out radiation pressure)I don't remember the exact source, I just wanted to confirm if I am right or not.
If you can check, I can send an image on how they derived it.
 
  • #7
jbriggs444
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If you can check, I can send an image on how they derived it.
If you are using relativistic mass (*shudder*) then mc^2 = E = pc. Divide by c to get mc = E/c = p
 
  • #8
Neeraj
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If you are using relativistic mass (*shudder*) then mc^2 = E = pc. Divide by c to get mc = E/c = p
Photons are massless so in order to avoid the term m we can use debroglie's dual nature theory or whatever you just said
 

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