Question about entropy of photons

[JesseM]

Someone made the following argument:

Each photon has exactly the entropy of ONE, regardless of the photon's energy, and regardless whether you put the photon in a bigger box or a smaller box. (certainly you can not put a photon into a box smaller than its wavelength.)
To see that, remember entropy times temperature equals to energy:
s*(kT) = E
A photon's temperature equals to the blackbody temperature at which it is emitted, which means kT = E. So S equals ONE.

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.

 

[kanato]

In blackbody radiation, photons obey Bose-Einstein statistics, so the partition function is something like
Z = \prod_{n=1}^\infty \frac{1}{e^{n h/\lambda kT} - 1}

where \lambda 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
S = \left( \frac{\partial (kT \ln Z)}{\partial T} \right)_{V,N}

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 TS = E + PV - \mu N, where \mu 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.

 

[dextercioby]

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.

 

[lanjarote]

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

 

[kanato]

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

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.

 

[dextercioby]

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.

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...

Daniel.

 

[kanato]

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...

Daniel.

Hmm... I'm not actually familiar with QFT, so this is new to me. But interesting, nonetheless.

 

[dextercioby]

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...

Daniel.

 

[JesseM]

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:

In contrast to the molecules in a gas, the different wavelengths of radiation do not collide with each other in the middle of the oven. Nevertheless, energy can shift from one mode to the other through the intermediary of the oscillators in the wall - which is to say, just the atoms and molecules the wall is made of. An electromagnetic wave in the oven can set a charge oscillating in the wall, energy can be transferred to other charges in the wall, which will in general oscillate and radiate at different frequencies. Thus the different frequencies of radiation inside the oven will come to thermal equilibrium.

 

[dextercioby]

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:

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.

 

[JesseM]

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)...?

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?)

 

[JesseM]

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 E=h\upsilon 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 &lt;s&gt; = 1/(1 - e^{\hbar \omega / T}) = 1/(1 - e^{\hbar 2 \pi \upsilon / T}). This gives photon number as a function of temperature T, but the second page also mentions the Stefan-Boltzmann law of radiation, U/V = \pi^2 T^2 / 15 \hbar^3 c^3, which can be rearranged to give temperature as a function of the volume of the box and the total energy it contains, or T = \sqrt{15 \hbar^3 c^3 U / \pi^2 V}...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 &lt;s&gt; = 1/(1 - e^{2 \upsilon / \sqrt{15 \hbar c^3 U / V}}). If the box has edges of equal length L, the volume would be L^3...would the allowable modes all be of the form \lambda = L/n, or \upsilon = nc/L? 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 1/(1 - e^{2n / L \sqrt{15 \hbar c U / V}})? The page above mentions that the sum over positive n in three dimensions is given by 1/8 \int_{0}^{\infty} 4 \pi n^2, 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 \pi n^2 / 2 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 \sqrt{ \pi^2 L^3 U / 15 \hbar^3 c^3}. 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.

 

[JesseM]

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?

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 (4 \pi^2 V T^3 ) / (45 \hbar^3 c^3 ).

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

 

[JesseM]

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. http://www.phys-astro.sonoma.edu/people/faculty/tenn/P314/BlackbodyRadiation.pdf is also equal to a constant times V T^3, so if you keep U constant the entropy would also be proportional to V^{1/4}.

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 \upsilon, so that the number of photons would simply be U/h\upsilon...does anyone know how you'd calculate the entropy in this case?

 

[dextercioby]

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

Daniel.

 

[JesseM]

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

Daniel.

That was what I just calculated, I got (16 \pi Zeta[3] k^3 V T^3 ) / (c^3 h^3)

 

[dextercioby]

Wrong answer,think again.

Daniel.

 

[JesseM]

Wrong answer,think again.

Daniel.

Can you explain which part of my derivation is wrong? Do you agree with this page that the number of photons per mode is 1 / (e^{h\upsilon / kT} - 1)? Do you agree with http://www.phys-astro.sonoma.edu/people/faculty/tenn/P314/BlackbodyRadiation.pdf that the number of modes between frequency \upsilon and \upsilon + d\upsilon is ( 8 \pi V \upsilon^2 \, d\upsilon ) / c^3? If so, wouldn't the total number of photons be equal to the integral (8 \pi V / c^3) \int_{0}^{\infty} \upsilon^2 / (e^{h\upsilon / kT} - 1) \, d\upsilon?

 

[dextercioby]

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

Daniel.

 

[JesseM]

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

Daniel.

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 h\upsilon.

 

[kanato]

The answer given in my statistical mechanics textbook is
\overline{N} = V \frac{2 \zeta(3)(kT)^3}{\pi^2 \hbar^3 c^3} \prop VT^3

It is calculated by taking the Rayleigh expression for the number of normal modes of vibration, ie. the density of states as
g(\omega) d\omega = \frac{V \omega^2 d\omega}{\pi^2 c^3}
and integrating the probability of occupation \langle n_s \rangle = (e^{\hbar \omega/kT} - 1 )^{-1} over all states.

Also calculated from the grand partition function is the pressure volume relation
PV = \frac{8 \pi^5 V}{45 h^3 c^3} (kT)^4 = \frac{1}{3}U

Notice that the pressure is independent of the volume.

So then when we go to calculate the fluctuation in the number of particles, it's given by
\frac{\overline{\Delta N}^2}{\overline{N}^2} = -\frac{kT}{V} \left( \frac{\partial V}{\partial P} \right)_T
we find that the fluctuations in the number of photons are infinite, because (\partial P/\partial V)_T is zero. So it's hard to take the formula for the number of particles at face value.

 

[dextercioby]

The bottom line is that in the cavity ANY NUMBER OF PHOTONS FROM 0 TO INFINITY CAN EXIST AT THE SAME TIME.That is an assumption made at the very beginning:chemical potential of the photon gas is ZERO,meaning that 'N' is not a thermodynamical variable.
Of course this fact would indeed reflect in the infinite fluctuations of N...

Daniel.

 

[JesseM]

According to quantum electrodynamics, is the number of photons N in a given region of space a variable that can actually be measured? If so, then if we performed a series of measurements on a box filled with blackbody radiation, what statistical distribution would we observe for N?

 

[dextercioby]

It's not QED that states what I've said earlier/in my previous post.

What do you mean statistical distribution...?Wrt what?

Daniel.

 

[JesseM]

It's not QED that states what I've said earlier/in my previous post.

I know, but I'm wondering if the lack of ability to say anything about the number of photons in the box is just showing that nonrelativistic QM is insufficient for answering this question.

What do you mean statistical distribution...?Wrt what?

I just mean the probability distribution that tells you the likelihood that a given measurement will yield a given number of photons N (or a number in some range N to N'). If you make a sufficiently large number of measurements and graph the relative frequencies of different outcomes, the result will have to approach some probability distribution, no?

 

[dextercioby]

For the second part i wouldn't know what kinda measuements one may do...That number,as i said,it's basically undetermined,it can be any number from 0 to infinity.
No,it doesn't have to do with "insufficiency" of nonrelativistic QM...

Daniel.

 

[JesseM]

For the second part i wouldn't know what kinda measuements one may do...

But particle number is one of the variables you can measure in quantum field theories, right? Or is there some problem with measuring the number of photons that makes it different than measuring the number of some other particle like electrons?

That number,as i said,it's basically undetermined,it can be any number from 0 to infinity.

But is that because there's no way to measure it, or is it because you can measure it but the result can be anything from zero to infinity? If you can measure it, it can't be equally probable that you'll get any value from zero to infinity even if any value is possible, can it? I don't think the idea of a uniform probability distribution on all the numbers from zero to infinity even makes sense--wouldn't that mean that whatever number you guess beforehand, you'll have a 100% chance of getting a larger number than that when you make your measurement, since there are an infinite number of integers larger than your guess but only a finite number of integers smaller than or equal to your guess?

 

[dextercioby]

But particle number is one of the variables you can measure in quantum field theories, right? Or is there some problem with measuring the number of photons that makes it different than measuring the number of some other particle like electrons?

You probably didn't understand why in this specific case,the # of photons is allowed to vary over this huge interval:the atoms which make up the walls of the box are in thermodynamic equilibrium with the radiation within,which means that these atoms continuously emit & absorb radiation.That fact makes impossible the determination of the # of photons at a certain moment of time in the box...

Daniel.

 

[JesseM]

You probably didn't understand why in this specific case,the # of photons is allowed to vary over this huge interval:the atoms which make up the walls of the box are in thermodynamic equilibrium with the radiation within,which means that these atoms continuously emit & absorb radiation.That fact makes impossible the determination of the # of photons at a certain moment of time in the box...

Impossible in principle, or just impossible in practice? Couldn't you also imagine some high-energy system where electrons are constantly being annihilated and created, but electron number is still a well-defined measurable property of the system (even though the uncertainty in this number might be high for short time intervals)?

Also, I was thinking about what it means to say the average size of fluctuations is infinite, and it seems to me it doesn't automatically mean that talking about the average is physically meaningless. For example, imagine a situation where there's a 1/2 chance the fluctuation will deviate from the average by 2, a 1/4 chance it will deviate from the average by 4, a 1/8 chance it will deviate from the average by 8, etc...in this case the average fluctuation would be 2*(1/2) + 4*(1/4) + 8*(1/8) + ... = 1 + 1 + 1 + ... = infinite, but it's still true that in 3/4 of all cases you'll get within 4 of the average. Now, I'm not suggesting the typical fluctuations in photon number would really be so small, but the point is that in principle an infinite average fluctuation size in the value of a quantity doesn't mean that it's meaningless to talk about the average value of that quantity.

 

[JesseM]

The answer given in my statistical mechanics textbook is
\overline{N} = V \frac{2 \zeta(3)(kT)^3}{\pi^2 \hbar^3 c^3} \prop VT^3

It is calculated by taking the Rayleigh expression for the number of normal modes of vibration, ie. the density of states as
g(\omega) d\omega = \frac{V \omega^2 d\omega}{\pi^2 c^3}
and integrating the probability of occupation \langle n_s \rangle = (e^{\hbar \omega/kT} - 1 )^{-1} over all states.

I tried doing this integral and I didn't get the answer you give above...are you sure it's not just \overline{N} = V \frac{2 \zeta(3)(kT)^3}{\pi^2 \hbar^3 c^3}?

So then when we go to calculate the fluctuation in the number of particles, it's given by
\frac{\overline{\Delta N}^2}{\overline{N}^2} = -\frac{kT}{V} \left( \frac{\partial V}{\partial P} \right)_T
we find that the fluctuations in the number of photons are infinite, because (\partial P/\partial V)_T is zero. So it's hard to take the formula for the number of particles at face value.

Well, see the second paragraph of my last post above, I don't think an infinite average fluctuation size necessarily means the average itself is physically meaningless. If it was meaningless, why would your textbook give it at all? Does it make any comments about the significance of the infinite average fluctuation result?