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\topmatter
\title
Lorentz Transformations And Statistical Mechanics
\endtitle
\rightheadtext{Lorentz Transformations And Statistical Mechanics}
		 

         % Put all authors' names within the next environment.
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Daniel Henry Gottlieb
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\leftheadtext{Daniel Henry Gottlieb} 
	        

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Math. Dept., Purdue University, West Lafayette, Indiana       
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\email
gottlieb@math.purdue.edu      
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exponential map, singularity, electro-magnetism, energy-momentum, vector bundles, 
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\endtopmatter
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\head 1.\ Introduction\endhead

In [Gottlieb (1998)] and [Gottlieb (2000)] we launched a study of Lorentz
transformations.  We find that every Lorentz transformation can be expressed
as an exponential $e^F$ where $F$ is a skew symmetric operator with respect
to the Minkowski metric $\langle \ ,\ \rangle$ of form $- + + +$.  We provided $F$ with the notation of electromagnetism.  Thus we can describe boosts as pure 
$\vec E$ fields and rotations as pure $\vec B$ fields.  An important class
of Lorentz transformations corresponding to radiation made its appearance.  We
have yet to see a description of these ``radiation'' transformations in the
Physics literature.

The complexification of the Lorentz Transformations yield very beautiful 
algebra which is related to Clifford Algebras as well as Lie Algebra theory.

The purpose of this paper is to confront the basic part of Statistical
Mechanics with the Lorentz transformation formalism and see how much of 
Statistical Mechanics can be described in this alternative language.  

Using our Lorentzian picture, we identify the central exponentials of 
Statistical Mechanics, $e^{- \beta \epsilon}$, as the eigenvalues of suitable 
Lorentz transformations.  Since we can have complex eigenvalues, we have a natural scheme which gives Fermions with half integers and Bosons with whole integers and naturally produces the formulas.
$$
   \langle n_k\rangle = \ds{1 \over e^{\beta(\epsilon_k - \mu)} \mp 1}
$$
for the mean state occupations numbers $\langle n_k\rangle$ for bosons, with the $-1$ and fermions with the $+1$.  The same qualities play a role in the description of the Maxwell-Boltzman, Bose-Einstein, and Fermi-Dirac statistics.

Finally, we can also give a count of the energy density, in a manner analogous
to Rayleigh-Jeans, but not counting by squares of integers because of the wave
equation, instead using an argument analogous to the appearance of the sum
of squares using Pauli matrices in spin arguments.


\head 2. Lorentz Transformations \endhead

Here we follow [Gottlieb (1998)] and [Gottlieb (2000)] and recall the notation
for Lorentz transformations.  Let $M$ be Minkowski space with inner product
$\langle \ , \ \rangle$ of the form $- + + +$.  Let $e_0, e_1, e_2, e_3$ be an orthonormal basis
with $e_0$ a time-like vector.  A linear operator $F: M \rightarrow M$ which
is skew symmetric with respect to the inner product $\langle \ , \ \rangle$ has a matrix 
representation of the form
$$
   F = \left(\matrix
             0& \vec E^T\\
        \vec E& \times \vec B\\   
        \endmatrix \right)
$$
where $\times \vec B$ is a $3 \times 3$ matrix such that $(\times \vec B) 
\vec v = 
\vec v 
\times \vec B$, the cross product of $\vec v$ with $\vec B$.  The 
dual $F^*$ of $F$ is given by 
$$
   F^* = \left(\matrix
             0& - \vec B^T\\
       -\vec B& \times \vec E\\
\endmatrix \right)
$$
We complexify $F$ by $cF:= F - i F^*$.  Its matrix representations is
$$
   F= \left(\matrix 0& \vec A^T\\
               \vec A& \times (-i \vec A)\\ 
            \endmatrix \right)
        \hbox{where } \vec A = \vec E + i \vec B
$$
These complexified operators satisfy some remarkable Properties:

\item{a)}$cF_1 cF_2 + cF_2 cF_1 = 2 \langle \vec A_1, \vec A_2 \rangle I$ where $\langle \ , \ \rangle$
denotes the complexification of the usual inner product of $\Bbb R ^3$.
Note, $\langle \ , \ \rangle$ is not the Hermitian form, that is our inner
product satisfies $i \langle \vec v, \vec w \rangle =
\langle i \vec v, \vec w \rangle = \langle \vec v, i \vec w \rangle$.

\item{b)}The same property holds for the complex conjugates $c \ov F_1$
and $c \ov F_2$ of course.  But also, they commute, $c \ov F_1 c \ov F_2
= c \ov F_2 c \ov  F_1$

\item{c)}$c F c \ov F= 2T_F$ where $T_F$ is proportional to the stress-energy 
tensor of electro-magnetic fields.

Now $e^F = I + F + \ds{F^2 \over 2!} + \ds{F^3 \over 3!} + \ldots : M 
\rightarrow M$ is a proper Lorentz transformation.  It satisfies 
$e^F = e^{cF \over 2} e^{c \ov F \over 2 }$.  The algebraic properties of $c F$ give rise 
to a simple expression for its exponential:
$$
   e^{cF} = \cosh (\lambda_{cF})I + \ds{\sinh (\lambda_{cF}) \over \lambda_{cF}}         cF
$$
where $\lambda_{cF}$ is an eigenvalue for $cF$.  There are only, at most, two
values for the eigenvalues of $cF$, namely $\lambda_{cF}$ and $- \lambda_{cF}$.

Now every proper Lorentz transformation $L \in SO  (3, 1) ^+$ is 
equal to
the exponential of some skew symmetric linear operator.  That is, $L= e^F$
where $F \in \frak s \frak o (3,1)$.  This was proved by Mitsuru Nishikawa
in \cite{1983} for $SO (p,1)^+$. 
I will 
prove it below (Theorem 4.1), since the proof is novel even though it only
works for $p=3$.  In general, it is unusual that a non-compact
connected Lie group such as $SO (3,1)^+$ has every element 
expressed 
as an exponential.  But it is true for $SO (3,1)^+$.

On the other hand, $L$ can be the exponential of more than one $F$, that is
$L= e^F = e^G$ where $F \not= G$.  An important example is the fact that the
identity is the exponential of an infinite number of skew operators.  In 
fact, $I = e^F$ for $F \not= 0$ if and only if $\lambda_{cF} = 2 \pi ni$ 
where $n$ is an integer.  Thus we may say that a Lorentz transformation has
``internal states'' corresponding to integers.  We will identify the odd $n$
with fermions and the even $n$ with bosons for reasons seen below.

This multiplicity of internal states holds for all proper Lorentz 
transformations $L$, except for the ``radiative'' $L$.  That is those $L = e^N$ 
where $N$ is a radiative, or what we shall now call a {\it null} skew 
symmetric operator.  That is, $e^F = e^N$ implies $F=N$ if $N$ is null.

\head 3. Occupation Numbers \endhead


We will choose [Tolman (1938)] as our text.  It is very carefully written 
classic exposition of Statistical Mechanics.  In that text some equations
(141.1-3) give the equations for a key concept of Statistical Mechanics:
The mean numbers of elements $\langle n_k\rangle$ in a given energy state $k$ with energy $\epsilon_k$ at equilibrium.  They read:
$$
\align
   \langle n_k\rangle &= \ds{1 \over e^{\alpha + \beta \epsilon_k}} 
            \quad \hbox{(Maxwell-Boltzmann) }\\
   \langle n_k\rangle &= \ds{1 \over e^{\alpha + \beta \epsilon_k} -1} 
            \quad \hbox{(Bose-Einstein) }\\
   \langle n_k\rangle &= \ds{1 \over e^{\alpha + \beta \epsilon_k} +1}
            \quad \hbox{(Fermi-Dirac) }\\ 
\endalign
$$
Here $\alpha$ is a constant depending on the chemical potential of the 
particles, and \break
$\beta = \ds{1 \over kT}$ where $k$ is Boltzmann's constant
and $T$ is the temperature of the equilibrium.

Let us consider the postulate that the exponentials in these equations should
be represented as the eigenvalue of a proper Lorentz transformation.  Then a
first guess would be that $L= e^F$ where $\lambda_F = \alpha + \beta \epsilon$
and $\lambda_{F^*} = 2 \pi n$.  Thus 
$$
    \lambda_{cF} = \lambda_F - i 
    \lambda_{F^*} = (\alpha + \beta \epsilon_k) - 2 \pi ni
$$
Let $s$ be a null
eigenvector of $L$.  There are only two of them generically.  Let $s$ be the 
eigenvector corresponding to $\lambda_F = \alpha + \beta \epsilon_k > 0$.  Then
$$
   L^{-1}s = e^{-F}s = \ds{1 \over e^{\alpha + \beta \epsilon_k}} s = \langle n_k\rangle s
$$
for Maxwell-Boltzmann particles, and
$$
   (L - I)^{-1}s = (e^F - I)^{-1}s = \ds{1 \over 
   e^{\alpha + \beta \epsilon_k}-1} s = \langle n_k\rangle s
$$
   for Bose-Einstein particles, and
$$
   (L + I)^{-1}s = (e^F + I)s = \ds{1 \over 
   e^{\alpha + \beta \epsilon_k}+1}
   s = \langle n_k\rangle s 
$$
for Fermi-Dirac particles. 

Now to present ``Physical arguments'' that such a choice is reasonable (aside 
from the fact that it gives the right answer), we follow the history of the
``derivation'' the three formulas for $\langle n_k\rangle$.

Boltzmann generalizing Maxwell's distribution of velocities in a gas, counted
the number of distributions of $n$ particles in $k$ states.  Then he used the 
method of Lagrangian multipliers to find the distribution of maximum 
probability.  This method produces the constants $\alpha$ and $\beta$ arising 
in the exponentials as Langrangian multipliers.

Gibbs realized that identical particles should be counted as 
indistinguishable elements in order to get the right answer to what is called
the Gibbs' Paradox.  Then to obtain Bose-Einstein and the Fermi-Dirac 
occupation numbers, one used Boltzmann's procedures based on two different
methods of counting identical particles.  These methods of counting are called
the Einstein-Bose and the Fermi-Dirac statistics.  Though they gave the right 
answers, they did violence to the idea of particle.  Eventually with the
discovery of Schrodinger's Equation, it was asserted that what was really
being counted were the numbers of different combinations of basic eigen-solutions 
of Schrodinger's equation whose eigenvalues added up to the energy
$\epsilon_k$.  For Fermi-Dirac statistics, one counts only anti-symmetric 
combinations of basic solutions, whereas for Bose-Einstein statistics, one
counts only the symmetric combinations.

We can give a count of appropriate Lorentz Transformations which will naturally
give the same statistics.  Let us observe that the three types of statistics
arise from the following methods of counting.  Suppose $G$ is a free group
with $k$ generators $\alpha_1, \ldots, a_k$ and we want to count the number of
words of length $N$ which can be formed out of these generators such that we
have $n_i$ copies of $a_i$ for each $i$ between $1, \ldots, N$.  Thus we have
$n_1 + \ldots + n_k = N$.  For example, if we have $2$ generators $a$ 
and 
$b$ and we want to look at the words of length $N=4$ made out of $3= n_1 \ \ 
a$'s
and $1 = n_2 \ \ b$'s we get $\{ba^3, ab a^2, a^2 b a, a^3 b\}$.  This count 
will give us Maxwell-Boltzmann Statistics.

For Bose-Einstein Statistics, we let $G$ be a free abelian group on $k$
generators $b_1, \ldots, b_k$ and we count the number of words of length $N$
for which there are $n_i$ of the $b_i$.  Thus if $a$ and $b$ commute, the set 
of words with $N=4$ and $n_1 =3$ and $n_2 = 1$ is $\{a^3 b\}$.

For Fermi-Dirac Statistics, we consider the elementary abelian $2$-group
$G \cong \Bbb Z_2 \oplus \Bbb Z_2 \oplus \ldots \oplus \Bbb Z _k$ with 
generators 
$c_1, \ldots, c_k$ and we count the number of words of length $n$.  
Since $c^2_i = 1$ we see that $n_i$ is either $0$ or $1$.

We take advantage of this group theoretical count to use the fact that the 
Lorentz transformations form a group.  To each particle of energy $E_i$, we 
assign a Lorentz transformation $e^F$ where $F$ has eigenvalue $E_i$.
If we insist that one of the two null-eigenvectors of $F$ coincide with a
fixed null-vector $s$, then $e^F = e^{F_1} e^{F_2} \ldots, e^{F_k}$ and so the
eigenvalue of $e^F$ for $s$ is $e^{E_1 + \ldots + E_k}$, so $E = E_1 + \ldots
+ E_k$.  This corresponds to the fact that the product of eigenfunctions of
basic particle states give rise to the sum of eigenvalues under a suitable
Hamiltonian, as in [Tolman (1938), equation 87.7].

Now I conjecture (but I have not proved mathematically) that a generic set of
Lorentz transformations $e^{F_i}$ all of whose positive eigenvalue's 
eigenvector coincide with the null vector $s$, but whose other eigenvectors
are distributed randomly, will look like a free group.  These should correspond
to Maxwell-Boltzmann statistics.

If our particles are not distinguishable, they cohere in a way, so we will let
them be represented by Lorentz transformations both of whose null-eigenvectors
agree.  This implies that the Lorentz transformation $e^{F_i}$ as well as 
their exponents $F_i$, commute.  For $\lambda_{F_i} > 0$, the set $\{e^{F_1}
\ldots, e^{F_k}\}$ should behave like the generators of an abelian free group
and thus give rise to Bose-Einstein statistics.

For Fermi-Dirac statistics, we let the particles be represented by $e^{F_i}$ 
where $\lambda_{cF}: = e^{(2ni + 1)F_i}$.  This represents a rotation
around $180^{\circ}$, so $(e^{F_i})^2 = I$.  So we have at least a central
element in the count of words for $\Bbb Z_2 \oplus \Bbb Z_2 \oplus \ldots
\oplus \Bbb Z_2$.
 

Boltsmann's method arose from an argument by Maxwell in which particles are
moving and colliding at statistical equilibrium.  We can use Lorentz 
transformations to describe some features of a kinetic equilibrium which will 
make the choice of $e^F - I$ natural.

We consider particles at equilibrium in Minkowski space.  They are at 
equilibrium with respect to an observer $u$.  We let $u$ denote a time-like
vector of unit length $<u, u> = -1$.  We consider a particle at rest which its
excited into motion.  Its new four-velocity will be denoted by $u^{\prime}$.
Then $u^{\prime} - u$ represent a space-like component of the velocity.  We
consider more particles at rest.  Then the average length of the 
vectors $||u^{\prime}_i - u||$ must be a descriptor of the equilibrium process.
Now we can choose a skew operator $F_i$, so that $e^{F_i}$ is a Lorentz
transformation, so that $e^{F_i} u = u^{\prime}_i$.  Thus we can describe the
$u^{\prime}_i - u$ as $(e^{F_i} - I)u$.  If we restrict ourselves to those
$F_i$, all of which have the null vector $s$ as an eigenvector corresponding to
$\lambda_{F_i} > 0$, then we should expect properties of the average $\ds{1
\over N} (F_1 + \ldots, + F_N)$ to be a descriptor of the equilibrium.  An
example would be the average of the eigenvalues $\lambda_{cF_i}$ of the $F_i$.
Thus we should expect an average $F$ so that $(e^F - I)u$ has average length.

We may do the same argument with $4$-momenta $p_i$ for appropriate classes of
particles.  Now $e^{F_i} p_i = p^{\prime}_i$ has the same length.  So we should
get $\langle e^{F_i} p_0 - p_0, u\rangle=$ energy difference for $i^{th}$ particle.  Then
$\langle \sum (e^{F_i} - I) p_0, u\rangle=$ total energy difference
Then $\langle \sum \ds{(e^{F_i} - I) \over N} p_0, u\rangle=$ Average energy difference.

Let $e^F$ be Lorentz transformation so that $e^{\lambda_F}= 
\ds{e^{\lambda_{F_1} + \ldots + \lambda _{F_N}} \over N}$ and so that $s$ is
a null-eigenvector of $F$.  Then $(e^F - I) s = (e^{\lambda_F} - 1)s = \ov s$
where $\ov s$ is the average energy times $s$.  So 
$(e^{\lambda F} -1)^{-1} \ov s = \langle n_k\rangle \ov s$.

\head 4. Mathematics\endhead
\proclaim{Theorem 4.1}The exponential map Exp: $\frak s \frak o (3,1) 
\rightarrow SO(3,1)^+$
given by $F \rightarrow e^F$ is onto.  That is, for every proper Lorentz 
transformation $L$, there exists an $F  \in  \frak s \frak o (3,1)$ so that 
$L = e^F$.
\endproclaim

To prove the above theorem, we need to consider the complexification $S: =
\frak s \frak o (3,1) \otimes \Bbb C$ operating on $\Bbb R^{3,1} \otimes 
\Bbb C$. This is isomorphic to $\Bbb C ^4$ and has an inner
product which is of the type $- + + +$ on $\Bbb R^{3,1}$ and extends to the complex 
vectors by $\langle i \vec v, \vec w\rangle = \langle \vec v, i \vec w\rangle = i \langle \vec v, \vec w\rangle$.
See [Gottlieb (2000), p 2] for more details.

Now let $c: \frak s \frak o (3,1) \rightarrow S$ by $cF= F - i F^*$.  The image
of $c$,
denoted $cS$, is a three dimensional complex vector space.  The set of 
operators of the form $aI + b cF$ will be denoted by $D$.  Note that $D$ is
a vector space isomorphic to $\Bbb R ^{3,1} \otimes \Bbb C$, and that $D$ is 
closed under multiplication.

\proclaim {Lemma 4.2}Let $F$ and $G \in  cS$ denote $cF$ and $cG$.  Then
$(a I + \beta F) (\alpha I +  G) = (a \alpha + b \beta \langle F,G\rangle) I + (b \alpha F + a \beta G + \ds{b \beta \over 2} [F,G])$
\endproclaim
Now we say that $L \ \in \ D$ is a {\it complex Lorentz transformation}
if $\langle Lu, Lv\rangle = \langle u, v\rangle$.  Any complex Lorentz transformation $L$ must have the
form $L= a I + bF$, where $F  \in  cS$, such that $a^2 - b^2 
\lambda^2_F =1$.

That is, $L^{-1} = a I - bF$.

\proclaim {Theorem 4.3}Every complex Lorentz transformation $L$ is an exponential, 
that is $L= e^F$ for some $F \ \in \ cS$, except for $L = -I + N$ where 
$N  \in  cS$ is null, that is $N^2 = 0$.
\endproclaim

\demo{Proof}Recall [Gottlieb (2000), Theorem 8.5] where $F  \in  cS$ 
that
$$
    e^F = \cosh (\lambda_F) I + \ds{\sinh (\lambda_F) \over \lambda_F} F
\tag {**}
$$
Now $L= aI + H$ where $H  \in  cS$ and $a^2 - \lambda^2_H = 1$.  
So the first obstruction to showing that $L$ is an exponential is solving the 
equation
$\cosh (\lambda)= a$.  We shall show below that such a $\lambda$ always 
exists.  Next, if \ \ $\ds{\sinh (\lambda) \over \lambda} \not= 0$, then
$$
   L= aI + H = \cosh(\lambda) I + \ds{\sinh  \lambda \over \lambda} 
                \left(\ds{\lambda \over \sinh \lambda} H\right)
             = \cosh (\lambda) I + \ds{\sinh \lambda \over \lambda} D = e^D
$$  
Hence $L$ may not be an exponential if \ \ $\ds{\sinh (\lambda) \over \lambda}=0$.

Now $\ds{\sinh \lambda \over \lambda} = 0$ exactly when $\lambda= \pi ni$ for
$n$ a non-zero integer. (Note that $\ds{\sinh (0) \over 0} =1$).  Then 
$$
   a = \cosh (\lambda) = \cosh (\pi ni)= \cos (\pi n) = (-1)^n.
$$
If $n$ is even, then $L= I + N = e^N$ where $N$ must be null.

If $n$ is odd, then $a= (-1)^n = -1$, so $L= -I + N$ where $N$ must be null
or zero.  Now $e^{B}= -I$ where $B  \in  cS$ has eigenvalue 
$(2k + 1) \pi i$.  But $-I + N = -e^{-N}$ cannot be an exponential, however $N$
must be null.  This proves Theorem 4.3 except for the following lemma.
\enddemo

\proclaim{Lemma 4.4}
\item{a)}$\cosh(\lambda) = a$ always has a solution over the complex numbers.

\item{b)}$\sinh(\lambda) = 0$ if and only if $\lambda= \pi ni$.
\endproclaim

\demo{Proof}First we show b).  $\sinh (\lambda) = \ds{e^{\lambda} - 
e^{- \lambda} \over 2} = 0$

Thus $e^{2 \lambda} = 1$, hence $2 \lambda = 2 \pi ni$ so $\lambda= \pi ni$.

Next we show a).
$\cosh(\lambda) = \ds{e^{\lambda} + e^{- \lambda} \over 2} = a$.  
Hence $(e^{\lambda})^2 - 2 ae^{\lambda} + 1 = 0$

Hence $e^{\lambda} = \ds{2a \pm \sqrt{4a^2 - 4} \over 2}= a \pm \sqrt{a^2 - 1}$.

Now $e^{\lambda} = b$ has a solution for all $b$ except $b=0$.  But $a \pm 
\sqrt{a^2 -1}$ cannot equal zero, hence we have shown there is a solution
for each $a$.
\enddemo

\demo{Proof of Theorem 4.1}We show the exponential map is onto $SO (3,1)^+$ 
by showing the products of two exponentials is an exponential.  That is 
$e^F e^G= e^D$ for $F, G, D \in \frak so (3,1)$.  Now $e^F = 
e^{{1 \over 2}cF} e^{{1 \over 2} \ov cF}$ where $\ov c F= F + i F^*$.  This follows
since $cF$ and $\ov c F$ commute.  Also for this reason, $e^{cF}$ and $ e^{\ov c G}$
commute.  Thus $e^F e^G = e^{{1 \over 2} cF} e^{{1 \over 2} cG} 
e^{{1 \over 2}\ov c F} e^{{1 \over 2} \ov c G}$.  Now $e^{{1 \over 2} cF} 
e^{{1 \over 2} cG}$ is a complex Lorentz transformation in $D$.  So either it 
is an exponential 
$e^{cD}$, or it has the form $-I + cN= - e^{cN}$ by Theorem 4.3.  Now Theorem
4.3 also holds for $\ov D \in \ov c S \otimes \Bbb C$.  Hence we have 
$e^F e^G = e^{2D}$ or $e^F e^G= (-e^{cN}) (-e^{\ov c N})= e^{2N}$.
\enddemo


\proclaim{Corollary 4.5} The exponential map $\text{Exp}:S\to SO(\Bbb R^{3,1}\otimes\Bbb C)$ is not onto.  If $N\in \frak s \frak o(3,1)$ is null, then $-e^N$ is
not an exponential even though $-e^{cN}$ is an exponential.
\endproclaim

\demo{Proof} As explained in [Gottlieb (2000)], we can extend duality $F^*$ to
skew symmetric matrices $\pmatrix 0 & \vec E\\  \vec E & \times \vec B\endpmatrix$
where $\vec E$ and $\vec B$ are complex vectors.  Then $cF=F-iF^*$ and
$\ov cF=F+iF^*$ satisfy the same properties as in the complexification of the 
real case.  Now consider $e^Fe^G$ where $F$, $G\in S$.  Then $cF={1\over 2}\,cF+{1\over 2}\,\ov c F$, so $e^Fe^G=e^{{1\over 2}cF}
e^{{1\over 2}\ov cF}e^{{1\over 2}cG}e^{{1\over 2}\ov cG}$.  Now $cF=cA$ for some
$A\in \frak s \frak o(3,1)$, and $\ov c F= \ov cA'$ for $A' \in \frak s \frak o(3,1)$
$$
e^Fe^G=e^{cA} e^{\ov cA'} e^{cB}e^{\ov cB'}=(e^{cA}e^{cB})(e^{\ov cA'}e^{\ov cB'})
$$
and so $e^{cA}e^{cB}$ equals either $e^{cD}$ or $-e^{cN}$.  But $(-I)e^{cN}=e^{(2n+1)\pi i\ov cE}e^{cN}=e^{(2n+i)\pi i\ov c E+cN}$ where $E$ has eigenvalue
equal to $1$.  So in both cases $e^{cA}e^{cB}$ is an
exponential.

Now $-e^{cN}$ is an exponential since $-e^{cN}=e^{\pi i\ov cE}e^{cN}=
e^{\pi i\ov cE+cN}$ where $ E$ has eigenvalue $\lambda_{c E}=1$.
On the other hand $-e^N$, where $N$ is null and real, cannot be an 
exponential, since if
$-e^N=e^F$, then  $s$, the unique eigenvector for $e^N$, applied to this
equation gives 
$-s=e^Fs=e^{\lambda_F}s$, so $\lambda_F=(2n+1)\pi i$ for some $n$.  Thus $F$
has another eigenvector, which contradicts $-e^N$ having only one.
\qed \enddemo

\head 5.  Lorentz transformations sharing eigenvectors\endhead

Suppose that we have a set of complex Lorentz transformations $L_i\in cS$ so
that there is a null vector $s$ which is an eigenvector for all the $L_i$.
Let $N$ be a null operator in $cS$.  Then each $F_i$ can be expressed as
$F_i=\lambda_i E+\alpha_iN$ where $E$ and $N$ share the eigenvector $s$ and has
$\lambda_E=1$, and $L_i=e^{F_i}$; or else $L_i=-e^{\alpha_iN}$.

Now $F_iN=\lambda_iN$.  So $e^{F_i} N = e^{\lambda_i}N$ and $Ne^{F_i}=e^{-\lambda_i}N$.  This follows since $N$ and $F_i$  share an
eigenvector, so $NF_i=-F_iN$.

\proclaim{Lemma 5.1} If $F$, $N\in cS$ and $N^2=0$ and $FN=\lambda N$,  then
$e^Fe^N=e^F+e^\lambda N=e^H$ where $\ds H=F+{\lambda_Fe^{\lambda_F}\over \sinh \lambda_F}\,N$.  The last equality holds only if $\lambda_F\not= n\pi i$ for
non zero integers $n$.
\endproclaim

\demo{Proof} 
$$
\aligned
e^Fe^N &= e^F(I+N)=e^F+e^FN=e^F+e^\lambda N=\cosh \lambda I+ {\sinh \lambda\over \lambda}\,F+e^\lambda N\\
&= \cosh \lambda I+{\sinh \lambda\over \lambda}\,H,
\endaligned
$$
provided $\ds{\sinh \lambda\over\lambda}\not=0$.
\enddemo

\proclaim{Theorem 5.2} Let $L_1 L_2\dots L_k$ be a product of complex Lorentz
transformations $L_i\in cS$ all sharing a common eigenvector $s$.  Assume that
none of the $L_i$ have eigenvalues of the form $\lambda_i=n\pi i$ for
$n\not=0$, and that the sum of the eigenvalues $\sum_1^k\, \lambda_i $ is not equal to $n\pi i$ for nonzero $n$.
  So each $L_i=e^{\lambda_iE+\alpha_iN}$ or else $-e^{\alpha_iN}$.
Then $\ds L_1\dots L_k=\pm\left(e^{(\sum_1^k\, \lambda_i)E}+\left(\underset {i=1}\to{\overset k\to\sum} e^{\lambda_1+\dots+\lambda_{i-1}-\lambda_{i+1}+\dots+\lambda_k}
\alpha_i {\sinh \lambda_i\over \lambda_i}\right)N\right)$.
\endproclaim

\demo{Proof}
$$
\aligned
L_1\dots L_k&= \pm \prod_1^k e^{\lambda_iE+\alpha_iN}=\pm \prod_1^k (e^{\lambda_iE}+\alpha_i{\sinh \lambda_i\over \lambda_i} N)\\
& = \pm\left(e^{\left({\sum_1^k} \lambda_i \right)E}+\sum_{j=1}^k e^{\left(\sum_1^{j-1}\lambda_i \right)} \alpha_j{\sinh \lambda_j\over \lambda_j}N e^{(\sum_{j+1}^k\lambda_i)E}+0+0+\dots+ 0\right)\\
&=\pm\left(e^{\left(\sum_1^k \lambda_i\right)E} +\left(\sum_{j=1}^k e^{(\sum_1^{j-1}\lambda_i-\sum_{j+1}^k \lambda_i)}{\sinh \lambda_j\over\lambda_j}\alpha_j\right)N\right]\\
& = \pm e^H
\endaligned
$$
where
$$
H := {\sum_1^k \lambda_i\over \sinh(\sum_1^k \lambda_i)}
\left(\left(\sum_1^k \lambda_i\right)E+\sum_{j=1}^k \left(e^{\sum_1^{j-1} \lambda_j-\sum_{j+1}^k \lambda_j}\right){\sinh \lambda_j\over \lambda_j} \alpha_j N\right)
$$
assuming $\ds\sum_1^k \lambda_i\not= n\pi i$ for $n\not= 0$.
\enddemo


\proclaim{Corollary 5.3} Suppose the $L_i$ commute.  Then $L_i=e^{\lambda_i(E+N)}$,  or else $L_i=\pm e^{\alpha_i N}$ for all $i$.  Then
$\prod_1^k L_i=e^{aE}+bN$ where $a=\sum_i^k \lambda_i$ and\newline
$b=\sum_{1}^k \exp\left[\sum_{1}^{j-1} \lambda_i+\sum_{j+1}^k (-\lambda_i)\right]{\sinh \lambda_j\over \lambda_j}$.  Hence $\prod L_i=e^{(\sum_1^k \lambda_i)(E+N)}$.
\endproclaim

Let $F\in cS$ and let suppose $N^+$ and $N^-$ are null operators in $cS$ so
that $FN^+=\lambda_FN^+$ and $FN^-= -\lambda_FN^-$.  Then $F+aN^+$ has the same
eigenvalue as $F$ with $N^+$ as the same ``eigenvector''.  Here $a\in\bC$.  The
question we pose is:  What is the ``eigenvector'' $N'$ corresponding to $-\lambda_F$?

\proclaim{Theorem 5.4} 

\noindent a) Suppose $(F+aN^+)N'=-\lambda N'$.  Then
$\ds a={-\langle F,N'
\rangle\over \langle N^+,N'\rangle}$.

\noindent
b)\ \ The $N'$ for $F+aN^+$ is given by any complex multiple of
$$\ds N'={-a\over 2\lambda^2}F+{1\over 2} N^- -{a^2\over 2\lambda^2} N^+$$  We are
choosing $N^-$ so that $\langle N^+,N^-\rangle=2$.
\endproclaim

\demo{Proof of a)}  Expand $(F+aN^+)N'=-\lambda N'$ and use the equation
$FG=\langle F,G\rangle I+{1\over 2}[F,G]$ found in [Gottlieb (2000), Lemma 5.5]
to arrive at the equation
$$
(\langle F,N'\rangle +a\langle N^+,N'\rangle)I +{1\over 2}[F+aN_+,N']=-\lambda N'.
$$
The coefficient of $I$ must be zero since skew operators have zero trace.  The
unique $a$ so that the coefficient is zero is the required result.
\enddemo

\demo{Proof of b)}  Let $u$ be an observer.  Then $Fu$ completely determines
$F\in cS$.  So choose a basis in $\bR^{3,1}\otimes \bC$ so that
$Fu=(0,0,\lambda)$, $N^+u=(1,i,0)$ and $N^-u=(1,-i,0)$.  Let $N'u=(r,s,t)$.
We can choose $N'$ so that $\langle N^+,N'\rangle=1$.  Then $\langle F,N'\rangle=-a$.  Also $\langle N',N'\rangle=0$.  From these three equations
we obtain three equations
$$
\aligned
r+si &=1\\
\lambda t &= -a\\
r^2+s^2+t^2 &= 0.\endaligned
$$
The solution is $\ds N'u=\left({\lambda^2-a^2\over 2\lambda^2},
-{(\lambda^2+a^2)i\over 2\lambda^2}, -{a\over\lambda}\right)$.  We can write
this as
$$
\aligned
N'u &= -{a\over 2\lambda}(0,0,1)+{1\over 2}(1,-i,0)-{a^2\over 2\lambda^2}(1,i,0)\\
&= -{a\over 2\lambda^2} Fu +{1\over 2} N^-u-{a^2\over 2\lambda^2} N^+u.
\endaligned
$$
This gives us the general equation since it holds for all vectors if it holds
for $u$. \qed
\enddemo

A simple method to calculate the real eigenvector $s'$ is to calculate the
complex 3 vector $N'u=: \vec E+i\vec B$.  Then $s'=E^2u+\vec E\times \vec B$.

We see from this process that $F+aN^+$ has a different eigenvectors $s'$
corresponding to $-\lambda_F$ and they are in $1-1$ correspondence with the
points in $\bC$, assuming $F$ is not null.

Another way to calculate the other eigenvector is to note that
 $F+N^+=e^{-{N^+\over 2}}Fe^{N^+\over 2}$.  Now
$e^{-{N^+\over 2}}N^-$ is an eigen-operator for $F+N^+$ associated with
$-\lambda_F$.  It follows that a real eigenvector can be obtained by
$s^-=e^{-N}s^-$ where the $N$ denotes the real null operator of $N^+$ .

Now there is a cononical equation for an eigenvector $s$ corresponding to the
$\lm_F$ eigenvalue of $F\in {\frak so}(3,1)$.  For a given observer $u$, $F$ is
determined by $\vec E$ and $\vec B$.  Then the canonical eigenvector $s$ is
given by the following equation, [Gottlieb (1998), Corollary 6.8].
$$
\aligned
s &:= {1\over 2} (\lm_{cF}I+cF)(\ov{\lm_{cF}I-cF})\\
&=\left(\lm_T +{E^2+B^2\over 2}\right)u+\vec E\times \vec B+\lm_F\vec E-
\lm_{F^*}\vec B.
\endaligned
$$

\proclaim{Theorem 5.5}  Let $F_0+N=F$, where $F_0$, $N$, and $F\in  
\frak s \frak o(3,1)$ and
suppose $N$ is null whose eigenvector, $s_1$, lies along the eigenvector, $s_0$,
of  $F_0$ corresponding to the eigenvalue $\lm_{cF_0}$.  Then $s=s_0+s_1$,
where $s$ is the canonical eigenvector of $F$, if we assume that $F_0$ has
linearly dependent $\vec E$ and $\vec B$ for observer $u$.

\endproclaim

\demo{Proof} Now in $u$'s rest space, $F_0$ is composed out of $\vec E_0$ and
$\vec B_0$ where $\lm_{F_0}\vec E_0-\lm_{F*} \vec B_0=: \vec k$, and
$\vec E_0\times \vec B_0=0$.  $N_1$ is determined by $\vec E_1$ and $\vec B_1$
where $E_1=B_1$, $\vec E_1\cdot \vec B_1=0$ and $\vec E_1\cdot\vec k=\vec B_1\cdot \vec k=0$.  Then $F$ is determined by $\vec E=\vec E_0+\vec E_1$ and
$\vec B=\vec B_0+\vec B_1$.  Under these conditions,
$$
s=\left(\lm_T+{E^2+B^2\over 2}\right)u+\vec E\times \vec B+\lm_F\vec E-\lm_{F*}\vec B
\tag a
$$
$$
s_0 =\left(\lm_T+{E_0^2+B_0^2\over 2}\right)u+ \lm_F\vec E_0-\lm_{F*}\vec B_0\tag b
$$
$$
s_1= E_1^2u+\vec E_1\times \vec E_2. \tag c
$$

Now we eliminate $\vec E$ and $\vec B$ in equation (a), use the fact that
$$
{E^2+B^2\over 2} = {E_0^2+B_0^2\over 2} +{E_1^2+B_1^2\over 2}=\lm_T+E_1^2
$$
and then that $\vec E\times\vec B=\vec E_1\times\vec B_1+\vec E_0\times\vec B_1+
\vec E_1\times \vec B_0$ and equations b) and c) to obtain $s=s_0+ s_1+
\text{space-like vectors}$.  But the space-like vectors must add up to a light-light vector which implies that they must add up to $\vec 0$.  This implies that $s_0+s_1=s$.
\enddemo

\head 6. Bosons and Fermions\endhead

Now recall that the mean occupation number $\langle n_k\rangle$ of states of a
particle at equilibrium at a temperature $T$ is given by 
$$
\langle n_k\rangle= {1\over e^{\beta (\epsilon_k-\mu)}\mp 1}
$$
depending on  whether the particle is a boson or a fermion.  Here $\epsilon_k$
is the energy of the $k$th state, $\beta ={1\over kT}$ and $\mu$ is the chemical
potential.

Our first approximation to describing this was to postulate that there was a
Lorentz transformation $e^F$ so that $\lm_F=\beta(\epsilon_k-\mu)$.  However,
there are different $F$'s which give rise to this Lorentz transformation.  So
we can consider $F$ with eigenvalues $\lm_{cF}=\beta(\epsilon_k-\mu)+2\pi si$
where $s$ is a half integer ${n\over 2}$.  Then
$$
{1\over e^{\lm_{cF}}-1}=(-1)^{2s}\langle n_k\rangle.
$$
We then should correspond the integers $s$ to bosons and the half integers
 $s$ to
fermions in order to get the correct formulas.

We can associate a $cF$ to an equilibrium as easily as $F$.  The $(e^{cF}-I)$
will give us a real vector which describes a mean change of direction, and a
complex vector which describes some sort of mean rotation or spinning.  Then
$\lm_{cF}=\beta(\epsilon_k-\mu)+2\pi si$.  Hence
$$
(e^{cF}-I)^{-1}cN= (-1)^{2s}\langle n_k\rangle cN
$$
where $cN$ is the null eigen-operator for $cF$, that is $cFcN=\lm_{cF}cN$.

Now suppose $e^{cF}$ represents some mean description of a system in
equilibrium at temperature $T$.  We expect $\lm_{cF}$ to remain constant in
time.  But $e^{cF}$ distinguishes different directions, so we would expect
$e^{cF}$ to vary in time so that $e^{\lm_{cF}}$ remains constant.

There is a beautiful formula, due to Helgason, which gives the differential of
the exponential map.  See [Gottlieb (2000), \S4] or [Helgason (1978), p.105,
Theorem 7.1].

\proclaim{Theorem 6.1. (Helgason)} Let $\frak g$ be the Lie Algebra for a Lie group 
$\bG$.
Then $\nabla_G e^F\colon={d\over dt}(e^{F+tG})|_{t=0}=e^F\cdot (g[ad F](G))$ where $ad\ F\colon \frak g\to \frak g$ sends $ X \longmapsto [F,X]$ and $g[\xi]$ is the power series of the function $g(\xi)=\ds{e^{-\xi}-1\over -\xi}$.
\endproclaim


Now $cF$ has two eigen-operators, $cN^+$ and $cN^-$ so that
$cFcN^+=\lm_{ cF}cN^+$ and $cFcN^-=-\lm_{ cF}cN^-$.  Hence  we get
$$
\nabla_{cN}e^{cF} =e^{\lm_{cF}}\left({e^{-2\lm_{cF}}-1\over -2\lm_{cF}}\right)
cN = {e^{\lm_{cF}}-e^{-\lm_{cF}}\over 2\lm_{cF}}=
{\sinh(\lm_{cF})\over\lm_{cF}}cN.
$$
This also holds for $cN^-$ and so it is true for any linear combination of
$cN$ and $cN^-$.  This subspace is characterized by $cG$ such that
$\langle cG, cF\rangle=0$.  This $G\longmapsto \nabla_Ge^F$ is completely
characterized by the following theorem.

\proclaim{Theorem 6.2} $\ds\nabla_{cG}e^{cF}={\sinh \lm_{cF}\over \lm_{cF}}cG$ if
$\langle cF, cG\rangle =0$, and $\nabla_{cF}e^{cF}=e^F\circ F$.
\endproclaim

Now if $e^F\in SO(3,1)^+$ represents the process in equilibrium, we saw
that $\lm_{cF}=\beta(\epsilon_k-\mu)+2n\pi i$ was a reasonable choice of
eigenvalue.  Since $e^F=e^{{1\over 2}c F} e^{{1\over 2}\ov cF}$, the choice for
a complexified $cF$ representing the process should have eigenvalue
$\lm_{cF}={1\over 2} \beta(\epsilon_k-\mu)+n\pi i$.  Hence
$$
\aligned
{\sinh \lm_{cF}\over \lm_{cF}}&= e^{-\lm_{cF}}\left[{e^{+2\lm_{cF}}-1\over
+2\lm_{cF}}\right] = {e^{-{1\over 2}\beta(\epsilon_k-\mu)-n\pi i}\left({1\over\langle n_k\rangle}\right)\over \beta(\epsilon_k-\mu)+2n\pi i}\\
&={1\over\langle n_k\rangle} \cdot {(\beta(\epsilon_k-\mu)-2n\pi i)\over
\beta^2(\epsilon_k-\mu)^2+4n^2\pi^2}\cdot (-1)^ne^{-{1\over 2}\beta(\epsilon_k-\mu)}.
\endaligned\tag *
$$


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

{\bf Scholium 6.3}  Let $cF$ represent an equilibrium process then

\item{a)}  $\ds\nabla_G e^{cF} = {(-1)^n\over \langle n_k\rangle}
e^{-{1\over 2}\beta(\epsilon_k-\mu)} {(\beta(\epsilon_k-\mu)-2n\pi i)\over
\beta^2(\epsilon_k-\mu)^2+4n^2\pi^2} cG \text{ if } \langle cG, cF\rangle=0$.

\item{b)} $\ds e^{cF} = \cosh({1\over 2}\beta(\epsilon_k-\mu)+n\pi i)I+
{(-1)^n\over \langle n_k\rangle}
e^{-{1\over 2}\beta(\epsilon_k-\mu)}{(\beta(\epsilon_k-\mu)-2n\pi i)\over
\beta^2(\epsilon_k-\mu)^2+4n^2\pi^2}cF$.

\demo{Proof} We prove a) by substituting (*) into the previous theorem.  We
prove b) by substituting (*) into equation (**) in proof of Theorem 4.3.

Now we take the differential of $e^F$ for $F\in \frak{so}(3,1)$ and $N$ null
with a common eigenvalue.
\enddemo

\proclaim{Theorem 6.4}  If $N$, $F\in\frak{so}(3,1)$ Where $N$ and $F$ both
share an eigenvector and $N$ is null, then
$\ds e^{-F}\nabla_Ne^F={e^{-\lm_{cF}}-1\over -\lm_{cF}}\cdot N$ where the dot
is electromagnetic duality.
\endproclaim

\demo{Proof} 
$$
\aligned
[F,N] &= \left[{1\over 2} cF+{1\over 2} cF, {1\over 2} cN+{1\over 2} cN\right]
={1\over 4}\left([cF, cN], [\ov c F, \ov c N]\right)\\
&= {2\over 4} \text{Re}[cF,cN]=\text{ Re}\left({1\over 2}(2\lm_{cF}cN)\right)=
\lm_{cF}\cdot N.
\endaligned
$$
Now $\ds e^{-F}\circ \nabla_Ne^F = g([F,N])={e^{-\lm_{cF}}-1\over -\lm_{cF}}\cdot N$.
\enddemo

\proclaim{Scholium 6.5} Let $F$ represent an equilibrium at temperature $T$ with null $N$ sharing eigenvector and $\lm_{cF}=\beta(\epsilon_k-\mu)+2\pi ni$.  Then
$\ds e^{-F}\circ\nabla_N e^F = {e^{-\lm_{cf}}\over \langle n_k\rangle \lm_{cF}}
\cdot N$.
\endproclaim

\head \ 7. The Rayleigh-Jeans Factor \endhead

The main use of the mean occupations numbers is in the establishment of useful
differential formulas.  These comprise the {\it Planck radiation law},
[Tolman (1938), (93.6)]
$$
du = {8 \pi hv^3\over c^3} {1\over e^{hv/kT}-1}dv.
$$
The number of particles in a Bose-Einsten gas [Tolman (1938), 93.11]
$$
dn ={4\pi vg\over h^3} m\sqrt{2m} {\epsilon^{1\over 2}d\epsilon\over 
e^{\alpha+\beta\epsilon}-1}
$$
The number of particles in a Fermi-Dirac gas [Tolman (1938),94.6]
$$
dn ={4\pi vg\over h^3} m\sqrt{2m} {\epsilon^{1\over 2}d\epsilon\over e^{\alpha+\beta\epsilon}+1}
$$
The coefficients $\ds{4\pi vg\over h^3} m  \sqrt{2m\epsilon}$ of the last two
equations are derived in [Tolman (1938), \S7].  They arise from solutions of
the stationary Schrodinger's equation
$$
{\partial^2 u\over\partial x^2} + {\partial^2u\over\partial y^2} +
{\partial^2u\over\partial z^2} = -{8\pi^2m\over h^2}E.
$$
The appropriate solutions are characterized by triples of non-negative integers
$(n_1,n_2,n_3)$ which lie in a small spherical shell in the first octant.
This gives the number of eigen-solutions as $\ds G={4\pi v\over h^3} m  \sqrt{2m E}\Delta E$, which is [Tolman (1938), (71.16)].

We may produce the same lattice of integers by using skew symmetric operators.
Let $E_x$, $E_y$, $E_z$ be skew symmetric real operators such that
$E_{x}u=\vec e_x$, the unit vector in the $x$ direction.  Similarly for $E_y$
and $E_z$.  Let $cE_x$, $cE_y$, $cE_z$ be their complexifications.  Then
$cE_x^2=cE_y^2=cE_z^2=I$ and they all anticommute with each other, i.e.
$cE_xcE_y=-cE_ycE_x$.

Now consider $cE=n_1cE_x +n_2cE_y+n_3cE_z$.  Then
$\lm_{cE}^2I=cE^2=(n_1^2+n_2^2+n_3^2)I$.  Now $\lm_{cE}$ is real, so
$\lm_{cE}^2 =|\lm_{cE}\ov\lm_{cE}|=2\lm_T$.  We can identify $\lm_T$ with energy
since $T={1\over 2} cE\ov cE$ corresponds to the stress energy tensor and so
$2EV$ (where $V$ is volume) should be the energy.  This necessitates letting
$\ds{h\over 2m}\,cE_x$, $\ds{h\over 2m}cE_y$, $\ds{h\over 2m}cE_z$ be
our basic set of skew operators.  Counting the number of allowable triples
$(n_1,n_2,n_3)$ as in the usual argument gives us the same formula as in the
standard argument.

If we are in a situation where the opposite eigenvector $s$-corresponding to
$\ds\lm_{cE_x}=-{h\over 2m}$ is distinguished, as by an opposite spin, then we
want to count $\ds{h\over 2m}cE_x$ twice and so we multiply by two the
factor to get the number of eigenstates $G$ corresponding to the energy range
$E$ to $E+\Delta E$.  That is
$$
G={8\pi V\over h^3} m\sqrt{2m\,E}\Delta E
$$
for spin.

Multiplying the mean occupation numbers by the number of particles gives us the
formulas above .

For the black body radiation formula, we have
$$
du = 2\cdot {4\pi V^2\over c^3}\,d\nu\cdot kT\cdot {{h\nu\over hT}\over
e^{h\nu/kT}-1}.
$$
Now the factor $\ds{8\pi\nu^2\over c^2} kTd\nu$ gives the Rayeigh-Jeans Law.
So
$$
{e^{h\nu/kT}-1\over {h\nu\over kT}}={8\pi\nu^2\over c^3} kT{d\nu\over du}.
$$

Now let $cN^-$ be the eigen-operator corresponding to $cFcN^-=-\lm_{cF}=
-{h\nu\over kT}$.  Then
$$
e^{-cF} \nabla_{cN^-} e^{cF} = {8\pi V^2\over c^3} kT {d\nu\over du} cN^-.
$$
\bigskip

\Refs
\widestnumber\key {ON}

\ref
\manyby
Daniel H. Gottlieb (1998)
\paper Skew Symmetric Bundle Maps
\jour   Contemporary Mathematics
\vol. 220
\pages pp. 117 - 141
\endref

\ref
\manyby
Daniel H. Gottlieb (2000)
\paper Fields of Lorentz transformations on space-time
\jour   (to appear)
\vol
\pages 
\endref

\ref
\by Daniel H. Gottlieb (1986)
\paper Robots and fibre bundles
\jour Bulletin de la Societe Mathematique de Belgique
\vol 38
\yr 
\pages 219--223
\endref

\ref
\by Daniel H. Gottlieb (1988)
\paper Topology and the Robot Arm
\jour 
Applicandae Mathematicae
\vol 11
\yr 
\pages 117--121
\endref

\ref\by Helgason (1978), Sigurdur\book Differentiable Geometry, Lie Groups, and
Symmetric Spaces\publ Academic Press\publaddr New York\yr 
\endref

\ref
\by Mitsuru Nishikawa (1983)
\paper On the exponential map of the group $O(p,q)_0$
\jour Memoirs of the Faculty of Science, Kyushu Univ.
\vol 37, ser. A
\pages 63--69
\endref

\ref 
\manyby Stephen Parrott (1987)
\paper Relativistic Electrodynamics and Differential Geometry
\jour  Springer-Verlag, New York
\vol
\yr 
\pages 
\endref





\bye

