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S.N.~Bose, ZP, {\bf 26,} 178 \hfill {\large \bf 1924}\\
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{\Large \bf Planck's Law and Light Quantum Hypothesis.}\\
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S.N.~Bose\\
~~~\\
(Received 1924)
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Planck's formula for the distribution of energy in the radiation from a black
body was the starting point of the quantum theory, which has been developed
during the last 20 years and has borne a wealth of fruit in energy domain of
physics. Since its publication in 1901 many methods for deriving this law have
been proposed. It is recognized that basic assumptions of the quantum theory are
irreconcilable with the laws of classical electrodynamics. All derivations up to
now use the relation
$$
\rho_{\nu} d \nu = \frac{8 \pi \nu^2 d \nu}{c^3}~ E,
$$
that is, the relation between the radiation density and the mean energy of an
oscillator, and they make assumptions about the number of degrees of freedom
of the ether, which appear in the above formula (the first factor on the right--
hand side). This factor, however, can be derived only from classical theory.
This is the unsatisfactory feature in all derivations and it is therefore no
wonder that attempts are being made to obtain a derivation that is free of this
logical flaw.
Einstein has given a remarkably elegant derivation. He recognized the logical
defect of all previous derivations and tried to deduce the formula independently
of classical theory. From very simple assumptions about the energy exchange
between molecules and a radiation field he found the relation
$$
\rho_{\nu} = \frac{\alpha_{mn}}{e^\frac{\epsilon_m - \epsilon_n}{kT} -1}.
$$
To make this formula agree with Planck's he had to use Wien's displacement law
and Bohr's correspondence principle. Wien's law is based on classical theory and
the correspondence principle assumes that the quantum theory and the classical
theory coincide in centrum limits.
In all cases it appears to me that the derivations have not been sufficiently
justified from a logical point of view. As opposed to these the light quantum
hypothesis combined with statistical mechanics (as it was formulated to meet the
needs of the quantum theory) appears sufficient for the derivation of the law
independent of classical theory. In the following I shall sketch the method
briefly.
Let the radiation be enclosed in the volume $V$ and let its total
energy be $E$. Let
various types of quanta be present of abundances $N_8$ and energy
$h \nu_8(s = 0 \quad \mbox{to} \quad s = \infty).$
The total energy is then
\begin{equation}
E = \sum_{s} N_s h \nu_s = V \int \rho_{\nu} d \nu
\end{equation}
The solution of the problem therefore requires the determination
of the $N_S$,
which, in turn, determine $\rho_{\nu}$.
If we can give the probability for each distribution characterized
by arbitrary values of $N_s$ then the solution is given by the
condition that
this probability is to be a maximum, keeping in mind the condition (1) which is
a constraint on the problem. We now seek this probability.
The quantum has the momentum $\frac{h \nu_s}{c}$ in the direction of its motion.
The momentary state of the quantum is characterized by its coordinates $x, y, z$
and the corresponding components of the momentum $p_x, p_y, p_z$. These six
quantities can be considered as point coordinates in a six--dimensional space,
where we have the relation
$$
p_{x^2} + p_{y^2} + p_{z^2} = \frac{h^2\nu^2}{c^2},
$$
in virtue of which point representing the quantum in our six--dimensional space
is forced to lie on a cylindrical surface determined by the frequency. To the
frequency range $d_{\nu_s}$ there belongs in this sense the phase space
$$
\int dx~ dy~ dz~ dp_x~ dp_y ~dp_z = V \cdot 4 \pi(h_{\nu}/c)^2 hd_{\nu}/c
= 4 \pi \cdot h^3 \nu^3/c^3 \cdot V \cdot d \nu
$$
If we divide the total phase volume into cells of size $h^3$, there are then
$4 \pi \cdot \nu^2/c^3 \cdot d_{\nu}$ cells in the frequency range $d_{\nu}$.
Nothing definite can be said about the method of dividing the phase space in
this manner. However, the total number of cells must be considered as equal to
the number of possible ways of placing a quantum in this volume. To take into
account polarization it appears necessary to multiply this number by 2 so that
we obtain $8 \pi V \nu^2 d \nu/c^3$ as the number of cells belonging to
$d_{\nu}$.
It is now easy to calculate the thermodynamic probability of a (macroscopically
defined) state. Let $N^s$ be number of quanta belonging to the frequency range
$d \nu^s$. In how many ways can these be distributed among the cells that belong
to $d \nu^s$? Let $p^s_0$ be number of empty cells., $p^s_1$ the number
containing 1 quantum, $p^s_2$ the number containing 2 quanta, and so on. The
number of possible distributions is then
$$
\frac{A^s !}{p^s_0 ! p^s_1 ! \ldots} \qquad \mbox{where} \qquad A^s =
\frac{8 \pi \nu^2}{c^3} \cdot V d \nu^s
$$
and where
$$
N^s = 0 \cdot p^s_0 + 1 \cdot p^s_1 + 2 p^s_2 + \ldots
$$
is the number of quanta belonging to $d \nu^s.$
The probability $W$ of the state defined by all $p^s_r$ is clearly
$$
\Pi \frac{A^s !}{_s p^s_0 ! p^s_1 \ldots}
$$
Taking into account that the $p^s_r$ are large numbers we have
$$
\log W = \sum_s A^s \log A^s - \sum_s \sum_r p^s_r \log p^s_r
$$
where
$$
A^s = \sum_r p^s_r.
$$
This expression must be a maximum under the constraints
$$
E = \sum_s N^s h\nu^s; ~~ N^s = \sum_r = r p^s_r.
$$
Carrying through the variations we obtain the conditions
$$
\sum_s \sum_r \delta p^s_r(1 + \log p^s_r) = 0, ~\sum \delta N^s h \nu^s = 0
$$
$$
\sum_r \delta p^s_r = 0 \qquad \qquad
\delta N^s = \sum_r r \delta p^s_r.
$$
From this we obtain
$$
\sum_r \sum_s \delta p^s_r(1 + \log p^s_r + \lambda^s) +
\frac{1}{\beta} ~\sum_s r \delta p^s_r = 0
$$
From this we first see that
$$
p^s_r = B^s e^{- \frac{\displaystyle rh\nu^s}{\displaystyle \beta}}.
$$
Since, however,
$$
A^s = \sum_r B^s e^{- \frac{\displaystyle rh \nu^s}{\displaystyle \beta}} =
B^s(1 - e^{- \frac{\displaystyle h \nu^s}{\displaystyle \beta}})^{-1}
$$
then
$$
B^s = A^s(1 - e^{- \frac{\displaystyle h \nu^s}{\displaystyle \beta}}).
$$
We further have
$$
N^s = \sum_r r p^s_r = \sum_r r A^s(1 - e^{- \frac{\displaystyle h
\nu^s}{\displaystyle \beta}})
e^{- \frac{\displaystyle rh \nu^s}{\displaystyle \beta}} =
\frac{A^s e^{- \frac{\displaystyle h \nu^s}{\displaystyle \beta}}}
{1 - e^\frac{\displaystyle h \nu^s}{\displaystyle \beta}}
$$
Taking into account the value of $A^s$ found above, we have
$$
E = \sum_S \frac{8 \pi h \nu^{s3} d \nu^s}{c^3}
~V \frac{e^{- \frac{\displaystyle h \nu^s}{\displaystyle \beta}}}
{1 - e^{h\nu^s/\beta}}
$$
Using the result obtained previously
$$
S = k\left [\frac{E}{\beta} - \sum_s A^s \log(1 - e^{h \nu^s/\beta})\right]
$$
and nothing that
$$
\frac{\partial S}{\partial E} = \frac{1}{T}
$$
we obtain
$$
\beta = k T
$$
Hence
$$
E = \sum_S \frac{8 \pi h \nu^{s3}}{c^3}~ V~ \frac{1}{e^{h \nu^s/kT} -1}
~d \nu^s
$$
which is Planck's formula.
{\it Comment of translator.} Bose's derivation of Planck's formula appears to me
to be an important step forward. The method used here gives also the quantum
theory of an ideal gas, as I shall show elsewhere. [A. Einstein]
%SB. = ENCODED 17 JUL 2000 BY NIS;
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