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Pressure of Photon Gas 
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#1
Jan2514, 10:03 PM

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1. The problem statement, all variables and given/known data
Consider a photon gas (particlelike nature) with N photons of monochromatic light in a box that has a volume V. You can assume everything is perfectly reflecting. What is the pressure of the photon gas based on the ideal gas law derivation? 2. Relevant equations N/A. 3. The attempt at a solution 1) [itex]P=F_{\text{on molecule}}/A[/itex] 2) [itex]F = \frac{dp}{dt}= \frac{2h\nu}{c\Delta t}[/itex] 3) Plugging 2 into 1 yields [itex]P = \frac{2h\nu}{Ac\Delta t}[/itex] 4) Define Δt as the time it takes for the photons to undergo one roundtrip in the box. So, [itex]\Delta t = \frac{2L}{c}[/itex] 5) Plugging 4 into 3 yields [itex]P = \frac{h\nu}{V}[/itex] 6) This can be rearranged to yield [itex]PV=h\nu[/itex] 7) For multiple photons, [itex]PV = Nh\nu_{\text{avg}}[/itex] (since the photon gas is uniform in frequency, [itex]\nu_{\text{avg}} = \nu[/itex]) I believe it's supposed to be [itex]PV=\frac{1}{3}Nhν[/itex], but I can't figure out why! I'm also not sure if that's the right answer, so any clues would be appreciated. 4. Variables P = Pressure, F = Force, A = Area, L = Length, V = Volume (V = A*L), ν = Frequency, t = Time, p = Momentum, c = Speed of light, h = Planck's constant, N = number of photon 


#2
Jan2514, 10:33 PM

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Hi argon. Welcome to PF!
Are all the photons moving back and forth between just two walls of the box? What about the other 4 walls? 


#3
Jan2514, 10:43 PM

P: 5

Hello! Why thank you (I've actually been here before, but I couldn't be bothered resetting my other account's password!)
That is a very good question. You know, the problem statement doesn't specify. That being said, in class I believe we only considered gases going between the 2 walls of the box when deriving the ideal gas law. Edit: In fact, we did only consider the x direction. 


#4
Jan2514, 11:06 PM

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Pressure of Photon Gas
A simple model is to assume that a certain fraction of the total number of particles contributes to the pressure between 2 opposing walls. 


#5
Jan2514, 11:10 PM

P: 5

Perhaps it is because we derived the ideal gas equation based on a pistoncylinder system (and not a box) where the piston was in the horizontal direction. In this case, it would be impossible for a pressure to be in the y or z direction, as particles can only collide with the piston in the x direction. I guess the system is fundamentally different than the one proposed in the problem statement.



#6
Jan2514, 11:31 PM

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There are different levels of sophistication for deriving the pressure. The particles (or photons) are moving randomly in all directions and there are particles striking all of the walls. A complete derivation would take into account all the possible directions of motion of the particles.
But you get the right result if you take a simplified approach and assume that the particles move only in the [itex]\pm[/itex]x direction, the [itex]\pm[/itex]y direction, or the [itex]\pm[/itex]z direction and that each particle can travel round trips between walls without interference from other particles. This is a drastic oversimplification of the true gas! But you get the right result. Based on your derivation that you gave in your original post, it looks to me that you are probably expected to use this oversimplified model. But I'm only guessing. For this simplified model, what fraction of the total number of particles would you expect to travel in the [itex]\pm[/itex]x direction and contribute to pressure on the two walls that you were considering? How does this affect your answer? 


#7
Jan2514, 11:44 PM

P: 5

Thank you for your help and patience.
I believe that we are, indeed, expected to consider vastly simplified cases. I think as the course progresses we'll be addressing these further. Assuming truly random motion, I would expect 1/3 of the particles to travel in the x direction and contribute to the system in consideration, which would then bring in that factor of 1/3 I was pondering about. Thank you. 


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