Wednesday, October 8, 2014

My Very Own Partition...

Consider a function so constructed such that,

\(Z_{s}=\cfrac { d\, \beta e^{ f(E_{ s }) } }{ d\, E_{ s } } =\beta f^{ ' }(E_{ s })e^{ f(E_{ s }) }\)

where  \(\beta\)  is a constant,  and  \(f(E_{s})\)  is a function in  \(E_s\)  with roots  \(E_{si}\)  \(i=1,2,3...\).  ie.

\( f(E_{ s }=E_{ { si } })=0\quad \quad \quad i=1,2,3...\)

then

\( e^{ f(E_{ s }=E_{ { si } }) }=e^{ 0 }=1\)

and,

\( \left[{ \cfrac { d\, \beta e^{ f(E_{ s }) } }{ d\, E_{ s } }  }\right]_{ E_{ si } }=\beta f^{ ' }(E_{ s }=E_{ { si } }).1=Z_{ si }\)

Since differentiation by product rule leaves behind one term without each root factor, we define,

\( f^{ ' }(E_{ si })=p(Z_{ si }).(E_{ si }-E_{ s1 })(E_{ si }-E_{ s2 })..\)

except for the factor of  \((E_{si}-E_{si})\) such that,

\( f^{ ' }(E_{ si })=p(Z_{ si }).\prod \limits_{\substack{ j=1 \\ j\ne i }}^n (E_{ si }-E_{ { sj} })=\cfrac{Z_{ si }}{\beta}\)

where  \(p(Z_{si})\)  is defined by,

\( p(Z_{ si })=\cfrac { Z_{ si } }{ \beta .\prod\limits _{\substack{ j=1 \\ j\ne i }}^n (E_{ si }-E_{ { sj} }) } \)

where we insist that  \(E_{si}\ne E_{sj}\)  for  \(i\ne j\), each root of  \(f(E_s)\)  is unique.  \(p(Z_{si})\)  can also be defined as  \(p(E_{si})\)  as

\(p'(E_{s})\prod_i(E_s-E_{si})\)    is zero for all  \(E_s=E_{si}\).  We will still have,

\( p(E_{ si })=\cfrac { Z_{ si } }{ \beta .\prod\limits _{\substack{ j=1 \\ j\ne i }}^n (E_{ si }-E_{ { sj} }) } \)

\(\beta\)  is defined such that,

\( \int _{ 0 }^{ E_s max }{ Z_{ s } } d\, E_{ s }=Z=\left[ \beta e^{ f(E_{ s }) } \right] ^{ E_s max  }_{ 0 }=\sum\limits _{ i }^{ all\, i } Z_{ si }\)

\(\beta=\cfrac{Z}{\left[ e^{ f(E_{ s }) } \right] ^{ E_s max  }_{ 0 }}\)

If  \(e^{ f(E_{ s }max)}\rightarrow0\)    for    \(E_{ s }max\rightarrow \infty\)

\(\beta=\cfrac{Z}{0-e^{ f(0) }}=-\cfrac{Z}{e^{ f(0) }}\)

at  \(e^{f(E_s=0)}\).

Notice the expression for  \(Z\)  is always zero integrating from root to root of  \(f(E_s)\).  This is because  \(e^{ f(E_{ s })}\)  intersect  \(y=1\)  repeatedly such that the area above and below the x-axis of   \(e^{ f(E_{ s })}\)  is equal, from root to root of  \(f(E_s)\).  This means, the value of  \(Z\)  is determined by that part of the curve from zero to the first root of \(f(E_s)\)  and, the last root of  \(f(E_s)\)  to the tail end of  \(e^{ f(E_{ s })}\)  towards infinity.

If the first root is zero then  \(\beta\)  can be found from,

\(\beta=\cfrac{Z}{0-e^{ f(E_s\,last) }}=-\cfrac{Z}{e^{ f(E_s\,last) }}=-Z\)

because we know that the integration of  \(Z_s\)  with boundaries between roots of  \(f(E_s)\)  sums to zero.

No need to involve  \(T\),  which ever way temperature is defined.