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Question about entropy of photons

  1. Feb 3, 2005 #1

    JesseM

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    Someone made the following argument:
    Aside from the questionable idea of talking about the entropy or temperature of a single photon, is he correct that if you had a large number of photons with a black body spectrum trapped inside a box, the entropy of this collection of photons would be independent of the volume of the box?

    edit: I think the main error here is that if you have some radiation in a box with a blackbody spectrum, the temperature of the radiation will decrease as the volume increases, since temperature basically measures thermal energy per degree of freedom, so larger volume=lower temperature if the energy is constant...for example, the cosmic microwave background radiation has a blackbody spectrum, and its temperature decreases as the universe expands.
     
    Last edited: Feb 3, 2005
  2. jcsd
  3. Feb 3, 2005 #2
    In blackbody radiation, photons obey Bose-Einstein statistics, so the partition function is something like
    [tex] Z = \prod_{n=1}^\infty \frac{1}{e^{n h/\lambda kT} - 1} [/tex]

    where [tex]\lambda[/tex] is the characteristic wavelength that meets the boundary conditions (twice the box size). You can put a photon in a box smaller than the wavelength, as long as it's exactly half as big.

    Anyway, I don't know exactly what that evaluates to, but the entropy of the system can be calculated by
    [tex] S = \left( \frac{\partial (kT \ln Z)}{\partial T} \right)_{V,N} [/tex]

    and I would most certainly bet that it's not a constant. Btw, I think that person is invoking the Gibbs-Duhem relation, which is [tex] TS = E + PV - \mu N [/tex], where [tex] \mu [/tex] is zero, because particle number is not conserved, but V certainly isn't zero, and P wouldn't be zero because there would be a flux of photons escaping from the box if a hole was drilled into it.
     
  4. Feb 3, 2005 #3

    dextercioby

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    There's no such thing as temperature,pressure,internal energy or entropy of a single photon,simply because the the the quantities forementioned are defined for macroscopic states & systems...

    Daniel.
     
  5. Feb 6, 2005 #4
    Entropy of a photon

    The entropy of a photon is dependent of how much do you know about it, if you know its propagation direction , and you know that it is a single particle then you have two posible states of spin (+/-1). So the entropy is

    S= log2(2) bits = k ln(2) juls/ºK
     
  6. Feb 6, 2005 #5
    Btw, spin-1 particles such as photons have three possible spin states: +1, 0, -1. That's a knowable state of a particle, but whether or not it's knowable is irrelevant. This definition of entropy does not make sense, because if we take N*k ln(2) that does not give us the entropy of a system composed of many photons.
     
  7. Feb 6, 2005 #6

    dextercioby

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    Incorrect,photons are massless spin 1 particles and that's why they have only 2 degrees of freedom which in this case are the only 2 (+1 & -1) eigenvalues of the Helicity operator...
    Read more into it... :wink:

    Daniel.
     
  8. Feb 7, 2005 #7
    Hmm... I'm not actually familiar with QFT, so this is new to me. But interesting, nonetheless.
     
  9. Feb 7, 2005 #8

    dextercioby

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    Yes,indeed,this is a QFT EFFECT,since applying the (quantum) theory of (spin) angular momentum to photons needs to be done after:
    1.Discussing the classical em.field in the Lagrangian formalism.
    2.Making the Hamiltonian analysis of the classical theory.
    3.Identifyng the algebra of observables for the EM field.
    4.Quantizing these observables in agreement with the second principles (either BRST or Batalin-Vilkoviski are the most elegant).

    Just then u can wonder about the spin of the EM field/photon.

    A totally different approach would involve group theory.It's really elegant... :wink:

    Daniel.
     
  10. Feb 7, 2005 #9

    JesseM

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    One other question I have about this situation--if you have a box filled with blackbody radiation, and you increase the size of the box, will the number of photons change? This page says that blackbody radiation reaches equilibrium by interacting with the walls of the chamber, so it seems possible that the number of photons absorbed by the wall wouldn't always match the number emitted:
     
  11. Feb 7, 2005 #10

    dextercioby

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    What do you mean "change".Question:You have a (closed) box of volume V at temperature T filled with photons.What is the no.of photons...?If u increase the V (keeping the T constant),how does th # of photons vary(if so)...?

    Daniel.
     
  12. Feb 7, 2005 #11

    JesseM

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    I don't know anything about quantum electrodynamics--are you implying the question is meaningless? I thought particle number was one of the things you could measure in quantum field theories (even if it can change over time through creation and annihilation, would there not be an average expected number of photons in a box filled with radiation at equilibrium?)
     
  13. Feb 8, 2005 #12

    JesseM

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    OK, I did a little more research and found out you don't need QED to get the average number of photons in the box, you can just use the fact that [tex]E=h\upsilon[/tex] for a photon, so if you know the energy of a given mode of radiation in the box (formula given here), you can divide by h times the frequency of that mode to get the average number of photons in that mode, which is given on this page as [tex]<s> = 1/(1 - e^{\hbar \omega / T}) = 1/(1 - e^{\hbar 2 \pi \upsilon / T})[/tex]. This gives photon number as a function of temperature T, but the second page also mentions the Stefan-Boltzmann law of radiation, [tex]U/V = \pi^2 T^2 / 15 \hbar^3 c^3[/tex], which can be rearranged to give temperature as a function of the volume of the box and the total energy it contains, or [tex]T = \sqrt{15 \hbar^3 c^3 U / \pi^2 V}[/tex]...then you should be able to plug this into the equation for <s> above to get the average number of photons in a mode as a function of volume and energy, or [tex]<s> = 1/(1 - e^{2 \upsilon / \sqrt{15 \hbar c^3 U / V}})[/tex]. If the box has edges of equal length L, the volume would be [tex]L^3[/tex]...would the allowable modes all be of the form [tex]\lambda = L/n[/tex], or [tex]\upsilon = nc/L[/tex]? If so, will the average number of photons in a box with sides L and containing total energy U be the sum-over-n of [tex]1/(1 - e^{2n / L \sqrt{15 \hbar c U / V}})[/tex]? The page above mentions that the sum over positive n in three dimensions is given by [tex]1/8 \int_{0}^{\infty} 4 \pi n^2[/tex], I'm not quite sure how you'd apply that to the expression for the number of photons I got (would you just replace n by [tex]\pi n^2 / 2[/tex] and integrate from 0 to infinity?)

    Finally, if the energy in the box is at an equilibrium, can you assume that the entropy is just equal to the total energy U divided by the temperature T? If so, the entropy for a box with sides L would be [tex]\sqrt{ \pi^2 L^3 U / 15 \hbar^3 c^3}[/tex]. If this is correct, then clearly the entropy will change if you change L while keeping U constant, and it doesn't appear that his claim that the entropy is proportional to the number of photons would be correct.
     
  14. Feb 8, 2005 #13

    JesseM

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    Never mind, I got confused about the difference between the total energy U of a system and the total "heat content" Q in the equation S = Q/T...I realized "heat content" is really another word for enthalpy, which is equal to U + PV. Anyway, I don't need to derive the total entropy of the box filled with blackbody radiation because that page I referenced above already gives it as [tex](4 \pi^2 V T^3 ) / (45 \hbar^3 c^3 )[/tex].

    How about my other question, though? If you know the number of photons in a mode with frequency [tex]\upsilon[/tex] is [tex]1/(1 - e^{\hbar 2 \pi \upsilon / T})[/tex], and you know the allowable frequencies are all of the form [tex]\upsilon = n c / L[/tex] (where n is a postive integer), and that page says that "The sum over positive n in 3 dimensions becomes [tex]1/8 \int_{0}^{\infty} 4\pi n^2 \, dn[/tex]", what integral do you have to do to find the total number of photons in the box?
     
    Last edited: Feb 8, 2005
  15. Feb 11, 2005 #14

    JesseM

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    Well, I figured out how to calculate the number of photons in a box filled with blackbody radiation, and it turns out he's correct in the sense that the entropy is directly proportional to the number of photons. This page gives a derivation of the Stefan-Boltzmann law for the energy density of blackbody radiation in a box...the trick is that the number of modes between frequency [tex]\upsilon[/tex] and [tex]\upsilon + d\upsilon[/tex] is equal to [tex]( 8 \pi V \upsilon^2 \, d\upsilon ) / c^3[/tex], where V is the volume of the box...so, if you know the energy per mode is [tex]h\upsilon / (e^{h\upsilon / kT} - 1)[/tex], then you just integrate [tex](8 \pi h V / c^3) \int_{0}^{\infty} \upsilon^3 / (e^{h\upsilon / kT} - 1) \, d\upsilon[/tex] to get the total energy...so by the same token, if we know the number of photons per mode is equal to [tex]1 / (e^{h\upsilon / kT} - 1)[/tex], then you'd do the integral [tex](8 \pi V / c^3) \int_{0}^{\infty} \upsilon^2 / (e^{h\upsilon / kT} - 1) \, d\upsilon[/tex], which by my calculations works out to [tex](16 \pi Zeta[3] k^3 V T^3 ) / (c^3 h^3)[/tex]. So since the Stefan-Boltzmann law says that [tex]U/V = (8 \pi^5 k^4 T^4 / (15 h^3 c^3 )[/tex], we know that [tex]T = ((15 U h^3 c^3) / (8 \pi^5 V k^4))^{1/4}[/tex], so plugging that into the formula for the number of photons we can conclude that if you vary V while keeping the total energy U constant, the number of photons is proportional to [tex]V^{1/4}[/tex]. Meanwhile, the entropy formula given at the bottom of this page is also equal to a constant times [tex]V T^3[/tex], so if you keep U constant the entropy would also be proportional to [tex]V^{1/4}[/tex].

    So is there any way to poke a hole in his "every photon contributes the same amount to the total entropy of a system" argument? What if we consider a cavity filled with non-blackbody radiation--say, one where all the system's energy U is concentrated in a single mode with frequency [tex]\upsilon[/tex], so that the number of photons would simply be [tex]U/h\upsilon[/tex]...does anyone know how you'd calculate the entropy in this case?
     
    Last edited: Feb 11, 2005
  16. Feb 11, 2005 #15

    dextercioby

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    And how many photons are there in a box with blackbody radiation of volume V at temperature T...?

    Daniel.
     
  17. Feb 11, 2005 #16

    JesseM

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    That was what I just calculated, I got [tex](16 \pi Zeta[3] k^3 V T^3 ) / (c^3 h^3)[/tex]
     
  18. Feb 11, 2005 #17

    dextercioby

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    Wrong answer,think again.

    Daniel.
     
  19. Feb 11, 2005 #18

    JesseM

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    Can you explain which part of my derivation is wrong? Do you agree with this page that the number of photons per mode is [tex]1 / (e^{h\upsilon / kT} - 1)[/tex]? Do you agree with this page that the number of modes between frequency [tex]\upsilon[/tex] and [tex]\upsilon + d\upsilon[/tex] is [tex]( 8 \pi V \upsilon^2 \, d\upsilon ) / c^3[/tex]? If so, wouldn't the total number of photons be equal to the integral [tex](8 \pi V / c^3) \int_{0}^{\infty} \upsilon^2 / (e^{h\upsilon / kT} - 1) \, d\upsilon[/tex]?
     
  20. Feb 11, 2005 #19

    dextercioby

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    What is the statisctical theory behind blackbody radiation...?This is a question of principles,not of numbers/formulas...

    Daniel.
     
  21. Feb 11, 2005 #20

    JesseM

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    Well, I'm not sure of the exact derivation from statistical mechanics, but I think the idea is that the different modes can exchange energy by interacting with oscillating charges in the wall, and you look for the equilibrium distribution of energy among modes. And the idea of "number of photons in a mode" is that a mode can't have an arbitrary energy, but can only have an energy that's some integer multiple of [tex]h\upsilon[/tex].
     
    Last edited: Feb 11, 2005
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