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426 THE QUANTUM THEORY

momenta between p* and p* + dp'. Supply a further factor 2 to count polariza-

tions. This produces the quantity STT V(ps)^2 dps, which equals A^3 ZS by virtue of the

relation ps = hv'/c. Hence Zs is the number of cells of size A^3 contained in the

particle phase space region being considered. How innocent it looks, yet how new

it was. Recall that the kinematics of the Compton effect had been written down

only about a year and a half earlier. Here was a new application of p — hv/c\

Before I turn to the rest of Bose's derivation, I shall comment briefly on the

subject of photon spin. When Bose introduced his polarization factor of 2, he noted

that 'it seems required' to do so. This slight hesitation is understandable. Who in

1924 had ever heard of a particle with two states of polarization? For some time,

this remained a rather obscure issue. After the discovery of the electron spin,

Ehrenfest asked Einstein 'to tell [him] how the analogous hypothesis is to be stated

for light-corpuscles, in a relativistically correct way' [E3]. As is well known, this

is a delicate problem since there exists, of course, no rest frame definition of spin

in this instance. Moreover, gauge invariance renders ambiguous the separation

into orbital and intrinsic angular momentum (see, e.g., [Jl]). It is not surprising,

therefore, that in 1926 the question of photon spin seemed quite confusing to Ein-

stein. In fact, he went so far as to say that he was 'inclined to doubt whether the

angular momentum law can be maintained in the quantum theory. At any rate,

its significance is much less deep than that of the momentum law' [E4]. I believe

that this is an interesting comment on the state of the art some fifty years ago and

that otherwise not too much should be made of it.

Let us return to Bose. His new interpretation of Zs was in terms of 'number of

cells,' not 'number of particles.' This must have led him to follow Boltzmann's

counting but to replace everywhere 'particles' by 'cells,' a procedure he neither did

nor could justify—but which gave the right answer. It may help to understand

Bose's remark that he did not know that he was 'doing something which was

really different from what Boltzmann would have done, from Boltzmann statis-

tics,' if I recall at this point Boltzmann's coarse-grained counting, which is dis-

cussed at more length in Section 4b.

Boltzmann. Partition N particles with total energy E over the one-particle

phase space cells «,, w 2 ,... There are NA particles in <OA. Their mean energy is

£A. We have

The relative probability W of this coarse-grained state is

The equilibrium entropy S is given by

(23.5)

(23.6)

(23.7)