# Carnot Cycle for the Photon Gas

Light is something like raindrops. Each little lump of light is called a photon and if the light is all one color, all the “raindrops” are the same.

The photon gas, a box filled with little light lumps, plays a very important role in physics. As a source of black body radiation — collect the photons coming from a small hole in the box and sort them by energy — it put Planck on the road to quantum mechanics, and although his hope that electromagnetic radiation was the origin of irreversibility did not work out as he intended, it is well worth studying this idea.  Here, however, I investigate the photon gas as a medium for a Carnot engine.

I already devoted one post to some of the thermodynamic properties of the photon gas, and here I want to use those properties. First to test an important statement by Carnot:

The motive power of heat is independent of the agents employed to realize it; its quantity is fixed solely by the temperatures of the bodies between which is effected, finally, the transfer of the caloric.

and secondly to show that a number of statements Carnot makes about his systems are not only completely wrong (some of them are also wrong for non ideal classical gases), but also irrelevant to his derivation.

The properties of the photon gas are very different from those of an ideal (particle) gas, but still simple enough for analytical treatment. The number of particles is not fixed, photons disappear by being absorbed by the walls, and others (keeping track of individual photons is fundamentally impossible) appear by being emitted. If I change the temperature of the wall the photons do not move faster as classical particles would do, but change color[1]. Photons cannot feel each other, just like the atoms or molecules in an ideal gas, but nevertheless the “photon gas law” (relation between temperature, pressure and volume) is very different from that of the classical ideal gas. Properties of  the ideal particle gas and the corresponding Carnot cycle can be found in almost every textbook on thermodynamics, and I use those occasionally for comparison without further explanation.

The Carnot cycle consists of isotherms and adiabats. Along an isotherm the temperature is constant, and if I let the gas expand, in this case from point A to point B, heat enters the system and is converted to work. The isotherm for the photon gas is given by the expression[2]:

$p = \frac{1}{3}bT^4\quad\mbox{with} \quad b = 7.57\times10^{-21}\,{\rm bar\, K^{-4}}$

The value of $$b$$ is very small, which means that the pressure resulting from the ever present photons (there are photons everywhere, always) is usually completely negligible. At room temperature ($$T=300\,$$K) the pressure of the air around us is about 1 bar, and the photons give an extra contribution of  $$\approx 10^{-11}\,$$bar, not something to worry about in daily life.

The amount of work done on the gas is by definition given by:

$w_{\rm A\to B}= -\int_A^B pdV = -p_A(V_{\rm B}-V_{\rm A}) = -\frac{1}{3}b T_h^4(V_{\rm B}-V_{\rm A})$

where $$T_h$$ is the temperature of the hot reservoir. Since the final volume ($$V_B$$) is larger than the initial volume  ($$V_A$$) this quantity is negative, meaning that the gas does the work. This would mean that the energy of the gas decreases, but that is compensated by an amount of heat entering the gas. But in this case that is not enough.

The energy of the photon gas is given by $$U=bVT^4$$, and this allows us to calculate the amount of  heat entering the gas when going from point A to point B. This is one of the properties where the photon gas differs greatly from a particle gas. If I let a particle gas expand, the density decreases. But in the photon gas, the walls start producing photons to keep the density the same, and that costs energy[2]. According to the first law of thermodynamics $$\Delta U = q +w$$, so that $$q = \Delta U -w$$, and a quick calculation shows that the amount of heat absorbed from the hot reservoir is equal to

$q_{\rm A\to B} = \frac{4}{3}b T_h^4(V_{\rm B}-V_{\rm A})$

which shows that only 1/4 of the heat entering the system in an isothermal expansion is converted to work. The rest is needed to create the photons for the larger volume. This is in contrast with the ideal particle gas, where all the heat is converted to work, something that is implicitly used by Carnot.

The second step in the cycle is the adiabatic expansion, necessary to change the temperature from that of the hot reservoir ($$T_h$$) to that of the cold one ($$T_l$$), in accordance with Carnot’s requirement that temperature changes should only be due to volume changes. I derived in the photon gas post that an adiabatic curve is given by the relation $$VT^3=\,$$ constant. If I halve the temperature and at the same time increase the volume by a factor eight no heat is absorbed or disposed of. Along this curve, from B to C, no heat enters the system, by definition, and so the work performed on the gas is[3]:

$w_{\rm B\to C} = \Delta U_{\rm BC} = bV_{\rm C}T_l^4 – b V_{\rm B}T_h^4 = bV_{\rm B}T_h^3(T_l-T_h)$

where I used that $$V_{\rm B}T_h^3 = V_{\rm C}T_l^3$$ (both points are on the same adiabat). Again this work is negative, since the photon gas gives it to us, thereby loosing energy. Since the energy is proportional to $$VT^4$$ halving the temperature and increasing the volume eightfold takes half of the energy away.

Now we need to get back to point A. For this we use the trajectory C$$\to$$D$$\to$$A: isothermal compression followed by adiabatic compression. Calculations very similar to the ones above give for the C$$\to$$D isotherm:

$w_{\rm C\to D}=-\frac{1}{3}b T_l^4(V_{\rm D}-V_{\rm C})\quad\mbox{and}\quad q_{\rm C\to D} = \frac{4}{3}b T_l^4(V_{\rm D}-V_{\rm C})$

Now heat flows out of the system, four times as much as the work needed to perform the compression, but since the volume decreases, we need to destroy a lot of photons in the process. and that energy is just lost. Well, not really lost. The energy goes into the low temperature heat bath, but since that has infinite heat capacity it won’t notice a few photons more or less. Using that along an adiabat $$VT^3$$ is constant, $$w_{\rm C\to D}$$ can also be written as:

$w_{\rm C\to D}=-\frac{1}{3}b T_l T_h^3l(V_{\rm A}-V_{\rm B})$

The final adiabatic step also costs us work which is gained by the system (which again looses a few photons since the volume decreases):

$w_{\rm D\to A} = bV_{\rm A}T_h^3(T_h – T_l)$

Again: this is positive, since we now perform work on the system.

In the first two steps I obtained work from the system, and in the final two steps I needed to perform work to bring the photon gas back to its original state. If I obtained more work in the first two steps than needed in steps three and four, I gained something in the process that I can use for putting things on top of other things. In the first step heat was absorbed from the hot environment, and in the third step heat was dumped in the cold environment. If I obtained useful work (and I did as it turns out) less heat is dumped in the low temperature than was gained from the high temperature bath. The difference is converted to work by my engine. I will not derive it, but carefully adding up all heat and work contributions will show that the total amount of energy is conserved. The proof is straightforward and left to the reader. In fact it is only a test for the correctness of the expressions, since I made use of energy conservation, so if I don’t get that in the end I made a mistake. I didn’t.

My goal was a different one, however. I want the efficiency: how much useful work can I get from a given quantity of heat entering the gas at the high temperature.  The total amount of work I get can be calculated by adding all contributions. The final expression can be written as:

$w_{\rm tot} = \frac{4}{3} T_h^3 (V_{\rm B}-V_{\rm A})(T_h-T_l)$

By definition the efficiency is the amount of work obtained divided by the amount of heat absorbed:

$\eta = \frac{w_{\rm tot}}{q_{\rm A\to B}} = \frac{\frac{4}{3} T_h^3 (V_{\rm B}-V_{\rm A})(T_h-T_l) }{\frac{4}{3}b T_h^4(V_{\rm B}-V_{\rm A})}= 1-\frac{T_l}{T_h}$

which shows[4] that the photon gas based engine has exactly the same efficiency as the Carnot engine based on the ideal classical particle gas[5]. It is not possible to create a perpetual motion machine (pmm)  of the second kind by combining a classical heat engine with a photon engine. Alas. Although this should be the case (otherwise I’m sure someone or something would have already constructed a pmm) but nevertheless it is somewhat surprising, in view of the completely different properties of the two gases.  And, as I will show in a separate post, almost all the assumptions Carnot makes in his proof are not satisfied by the photon gas.

Could we build an engine based on photons? I don’t see why not. It will be as efficient as any other engine, but the amount of heat per stroke we can convert is rather limited[6]. Steam engines work so well since the water can take in a massive amount of heat when converted to steam, and although only a much smaller part than the maximum of that is converted to useful work, it is still a workable solution. But who knows what options the future will bring. My interest currently is a different one: understanding entropy. And Carnot’s cycle was later shown to lead to that concept directly[7]. Long before people were thinking about microstates. Had he been a little more mathematically inclined he could have found the inherent contradiction in his assumption about conservation of caloric, and have obtained entropy as the “falling substance” from high to low temperature, in the process.

[1] Individual photons cannot do such a thing, what happens is that at higher temperatures the walls of the black box emit more photons with higher energy, and as a consequence the spectrum shifts to the blue.

[2] Note that this does not depend on the volume, if the volume increases at the same temperature the pressure remains the same. The walls just make more photons. This also implies that for the photon gas volume is not a state function. A topic for a later post, in which I show that Carnot could have derived that caloric is not a conserved quantity.

[3] It is common practice in thermodynamics to refer always to the system, work done by the system is just the negative of work done on the system. Same for heat. Of course as humans we are interested in how much useful work we can obtain, so in the first two steps we get useful work. Also standard in thermodynamics: changes in quantities  are always calculated as final state – initial state.

[4] Again, straightforward if you make use of the fact that A and D, as well as B and C are connected by adiabats.

[5] The properties of the photon gas derive from the quantum nature of this gas. There are a few other systems that can be used for similar calculations as the one I present here, for instance the quantum particle in a box (not photons, but actual particles, but now with a wave function) or a system of spins in a magnetic field. It should not come as a surprise that also these systems lead to the same efficiency. It is apparently a property of our universe that all systems have the same efficiency, and intimately linked to the fact that there are no pmm’s. However, the search is still on.

[6] If we expand the volume of the photon gas from 1m3 to 2m3 at room temperature, about 0.1 nJ of heat is absorbed. This does not look very promising for a heat engine based on photons. But that was not the point.

[7] In this context you may want to convince yourself that $\frac{q_{\rm A\to B}}{T_h} + \frac{q_{\rm C\to D}}{T_l} = 0$.

### 2 thoughts on “Carnot Cycle for the Photon Gas”

• September 12, 2017 at 4:16 am

For the work done from C to D, what happened to the (Vd – Vc) term? It doesn’t show up in your expression for total work. Thanks!

• September 12, 2017 at 8:06 am

I used that VT3 is constant along the adiabatic curve to express everything in the change of volume along the high temperature isotherm. I’ll add a line in the text to make that more clear. Thanks.