Apparent Paradox?

We demand rigidly defined areas of doubt and uncertainty.

Douglas Adams



In 2001 Bartell published a paper [1] in which he posed, among others, the following question:


Suppose \(N\) photons of frequency \(\nu\) are emitted by a laser aimed at a stationary black body. The black body absorbs the energy \(nh\nu\) and converts it to thermal energy (heat). Compare this with the energy \(N h\nu’\) absorbed if the black body is moving away from the source. By the Doppler effect, \(\nu’\) is less than \(\nu\) and consequently the absorber sees and absorbs photons of lower energy than emitted by the laser. Where did the extra energy go? The answer to this simple question eludes many physical chemistry professors. It does, however, yield some important results

I am not a physical chemistry professor, so the answer did not elude me. The question rather reminded me of an old riddle:

Three friends want to stay in a hotel and are told the room is $30 (I said is was an old riddle). They go up to their room, and then the receptionist discovers there is actually a discount for that night and the room is only $25. So he sends up the bellboy with five dollars. The bellboy has no idea how to split five dollars three ways, so he decides to give each of the guests one dollar, and pockets the other two. This means that each friend now paid $9, and the bellboy has $2, a total of $29. What happened to the remaining dollar?

You can probably easily figure out the answer to this riddle, you don’t have to be a professor for that either. The answer Bartell gives to his own question, however, is very confusing and leaves much to be desired. Here is his answer:

Where the remainder of the energy goes is in work done on the moving object by the radiation pressure of the photons acting over a distance. By applying the first law of thermodynamics it is quite easy to show that this pressure is just the energy density $E/V$ of the radiation. An interesting application is to operate a Carnot engine using a “photon gas” to exert pressure on the piston. From the efficiency of this Carnot engine comes the Stefan-Boltzmann radiation law. It is only a small skip from the concept of pressure exerted by radiation to illustrate Einstein’s favorite way of demonstrating that \(E=mc^2\)

There are in fact quite a few things completely wrong about this answer. The first statement implies that the pressure should increase if the body is moving away faster from the radiation source. In addition it is very questionable if thermodynamics, which holds for equilibrium systems like black body radiation, holds for photons radiated by a laser. In fact, of course, it does not. The pressure, by the way, is not \(E/V\) in a thermodynamical system, but \(E/3V\). The efficiency of Carnot cycles should be calculated using reversible processes where the system goes through equilibrium states. Although some of the statements in the answer are correct, these have nothing to do with this problem.

To the left is a pictorial representation of what is happening. A photon approaches a block, which is already warm, and is absorbed by it [2]. The block moves with velocity \(V\), and the frequency of the photon is \(\nu\). After the collision the photon is absorbed, making the block slightly hotter, and, since momentum is also added to the block, it moves slightly faster as well. There are two fundamental laws relevant here: conservation of momentum, and conservation of energy. The energy of the photon is $h\nu$, and its momentum is \(h\nu/c\). Therefore we can write these laws for \(N\) photons being absorbed as:

\[ \frac{Nh\nu}{c} +MV = MV’ \] and \[ Nh\nu + \frac{1}{2}MV^2 + U_i = \frac{1}{2} M{V’}^2 + U_f \]

where \(M\) is the mass of the block, and \(U\) its internal energy which changes from \(U_i\) initially to \(U_f\) in the final state. The difference in internal energies is equal to the heat generated in absorption: \(Q = U_f-U_i\). I also assumed that the velocity of the body is much smaller than \(c\).

I did make a small simplification here, neglecting the increase in mass due to absorption of energy. This is completely justified [3]. And, as Bartell does, we ignore the fact that black bodies also radiate, which is, however, essential in understanding the thermodynamic expressions alluded to. 

First of all we note that momentum conservation implies that the change in momentum of the body: 

\[ \Delta P = MV’-MV = \frac{Nh\nu}{c}\]

does not depend on the initial velocity \(V\). This is somewhat counterintuitive, especially since the second equation shows that the change in kinetic energy, \(\Delta K\), does depend on the initial velocity.  That equation can be written as

\[ \Delta K +Q = Nh\nu\]

which shows that part of the energy is dissipated as heat, and part is converted to kinetic energy. Solving the conservation of momentum equation for \(V’\), and substituting the answer into the conservation of energy equation yields an expression for \(Q\):

\[ Q = Nh\nu\left(1-\frac{V}{c}\right) = Nh\nu’\]

which shows that the heat is indeed related to the absorption of a Doppler shifted photon. At the same time the change in kinetic energy is given by

\[ \Delta K = \frac{(Nh\nu)^2}{2Mc^2} + \frac{Nh\nu}{c} V\]

The first term must be neglected in view of the approximations already made and the second is a product of the total momentum of the photons and the initial velocity of the body. I have no good intuitive explanation for these results. From the frame of reference of the moving block the frequency of the photon is reduced as the Doppler effect requires, and conservation of energy implies that the kinetic energy increases. Why the change in momentum (and thus the force due to the photons) is independent of the velocity, and the change in kinetic energy increases with it, is a mystery to me. Whatever the resolution of this problem is, the results show that the original question is ill posed: you should always take all contributions to the energy into account. The change in kinetic energy can indeed be interpreted as the work done on the system, so the first law of thermodynamics is satisfied. But it should be noted that this is not so much a thermodynamical law as a very general statement about all processes in this universe.

So, what about the other  parts of the answer? Some of the statements refer to  an idealized system in which photons are absorbed and emitted inside a cavity with temperature \(T\). The radiation coming from a small hole in such a cavity is called black body radiation, the properties of which played a crucial role in the development of physics in the late 19th and early 20th century [4]. It was experimentally determined (Stefan) and rationalized theoretically (Boltzmann) that the internal energy density in such a box of photons is \(u = bT^4\). From this it can be derived that for a system in equilibrium \(p = u/3\). This derivation makes use of the fact that the entropy \(S\) is a state function, and therefore \(dS\) is an exact differential. The fundamental law of thermodynamics \(dU = TdS -pdV\) can also be written as

\[dS = \frac{1}{T}dU +\frac{p}{T} dV = \frac{1}{T}d(uV) + \frac{p}{T} dV = 4bVT^2 dT + \frac{p+bT^4}{T} dV \]

where I used the explicit expression for \(u\). Since this is an exact differential, the following relation must be valid:

\[\left(\frac{\partial 4bVT^2}{\partial V}\right)_T = \left(\frac{\partial (p/T + bT^3)}{\partial T}\right)_V\]

which leads to a differential equation for \(p/T\):

\[\frac{d p/T}{dT} = bT^2\]

Solving this equation gives the pressure as a function of the temperature

\[ p = \frac{1}{3} bT^4 = \frac{1}{3} u \]

which shows dat the pressure is indeed one third of the internal energy density, and not equal to \(U/V\) as Bartell stated. But again, we are talking of two completely different systems. A laser is not an equilibrium system. 

The statement that from the efficiency of the Carnot cycle the expression for the energy density can be derived is simply incorrect. It makes the argument circular, and suggests that we could derive expressions for pressure and internal energy without any prior knowledge. However, to evaluate the amount of heat and work in the steps of the Carnot cycle you need expressions for the isotherms and the adiabats in the \(p,V\) plane. You have to start somewhere. Carnot himself had to know properties of gases to get to his expression for the efficiency [5]. Note that I implicitly used one of Carnot’s results (\(S\) is a state function) to derive the relation between \(p\) and \(u\). Since I treat the Carnot cycle for a box of photons elsewhere on these pages, I will not go into the details here.

I do not know what Einstein’s favorite way of showing \(E=mc^2\) was. Googling the question does not lead to an answer. In [6] Einstein shows that emission of light leads to a decrease in kinetic energy of a body, which can be interpreted as loss of mass in the appropriate frame, and in which he neatly avoids all issues related to pressure by having two photons emitted simultaneously in opposite directions. I doubt if there is a prettier derivation. It has nothing to do with the effect mentioned in the question, since all effects due to changing mass can be neglected for the problem at hand. 

Why do I care? I have no idea. Maybe one of the other of Bartell’s questions (4) gives me a clue:

Can heat be converted completely into work? Such a conversion is not forbidden by the first law. Is it forbidden by the second law? Show how it can be done.

Again the question is irrelevant. It is trivial to convert heat completely into work, or work completely into heat. In fact this happens in all individual steps of the Carnot cycle for an ideal gas. The reason Carnot’s cycle gives the highest possible efficiency is the complete conversion of heat to work in the isothermal expansion step. I think it irritates me that there are many deep and interesting questions related to all of these things, but the ones presented here are riddles and not even good ones at that. It is like counting the money in the hotel puzzle: the presentation starts you off on the wrong foot. There may be a point to that, but I don’t see it. The only lesson I can get from it is that you have to phrase your questions carefully. Although for the first question it was more the answer that irritated me.

[1] L.S. Bartell, Apparent Paradoxes and Instructive Puzzles in Physical Chemistry, J. Chem. Educ., 78, {2001}, 1067-1069.

[2] A red body is black for a blue photon.

[3] One joule of energy corresponds to \(1.5\times10^{18}\) 300 nm photons and a mass of approximately \(10^{-14}\) g. It can also heat 1 g of water by 0.25 degree centigrade if all the energy is absorbed. A laser easily produces that many photons. 

[4] An excellent book about the trials and tribulations this research caused, is T.S. Kuhn’s Black-Body Theory and the Quantum Discontinuity, 1894-1912. University of Chicago Press, 1978.

[5] S. Carnot, Reflections on the Motive Power of Fire. Dover Publications, 1968.

[6] A. Einstein, Ist die Trägheit eines Körpers von seinem Energieinhalt Abhängig?, Annalen der Physik18, (1905), 639. Reprinted in Einstein et al., The Principle of Relativity, Dover Publications.

Leave a Reply

Your email address will not be published. Required fields are marked *