From the post "Stress When The Heat Is On, Thermal Stress",
\(\cfrac{\partial ({d}_{s})}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}+\cfrac{\partial }{\partial T}\left\{\cfrac{\partial (d_s)}{\partial x}\right\}.T\)
\(\left\{1-T.\cfrac{\partial }{\partial T}\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}\)
\(\cfrac{\partial (d_s)}{\partial x}=\left\{1-T.\cfrac{\partial }{\partial T}\right\}^{-1}\left\{\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}\right\}\)
where
\(\left\{1-T.\cfrac{\partial }{\partial T}\right\}^{-1}\) is the inverse of the operator \(\left\{1-T.\cfrac{\partial }{\partial T}\right\}\)
This turns out to be very difficult.
But consider this,
\(\cfrac { \partial }{ \partial ln(T) } \equiv\cfrac { \partial T }{ \partial ln(T) } \cfrac { \partial }{ \partial T } \equiv T\cfrac { \partial }{ \partial T } \)
So,
\(\left\{1-\cfrac{\partial }{\partial ln(T) }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial T}.\cfrac{d T(x)}{d x}\)
\(\left\{1-\cfrac{\partial }{\partial ln(T) }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial x}\cfrac{\partial x }{\partial T}.\cfrac{d T(x)}{d T}\cfrac{d T}{d x}\)
that is to say,
\(\left\{1-\cfrac{\partial }{\partial ln(T) }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial x}.\cfrac{d T(x)}{d T}\)
\(\left\{1-\cfrac{\partial }{\partial ln(T) }\right\}\cfrac{\partial (d_s)}{\partial x}=\cfrac{\partial (d_s)}{\partial x}\)
which suggests (just mathematical trickery),
\(\cfrac{\partial }{\partial ln(T) }\left\{\cfrac{\partial (d_s)}{\partial x}\right\}=0\)
that \(\cfrac{\partial (d_s)}{\partial x}\) is independent of \({ ln(T) } \) or
\(\cfrac{\partial }{\partial x}\left\{\cfrac{\partial (d_s)}{\partial { ln(T) } }\right\}=0\)
that \(\cfrac{\partial (d_s)}{\partial ln(T) } \) is independent of \(x \)
One way this could happen is that \(d_s\) is a linear combination of functions in \(x\) and \({ ln(T) } \). This is to say,
\(d_s=A.f(x)+B.h({ ln(T) }) \)
where \(A\) and \(B\) are constants and \(f(x)\) is a function only of \(x\) and \(h({ln(T)})\) is a function only of \({ ln(T) }\).
That space density \(d_s\) is the superposition of space distribution itself, expressed as \(A.f(x)\), and the effect of thermal energy in the form of \(B.h({ ln(T) })\).
In the case of uniformly distributed space,
\(f(x)=C\) where \(C\) is a constant
\(d_s=d_n+B.h({ ln(T) })\)
where \(A.C=d_n\) is the normal space density and,
\(h({ ln(T) })|_{lim\,T\rightarrow0}=0\)
such that when \(T=0\), \(d_s=d_n\)
The points here are the \( ln(T) \) dependence of \(d_s\) and superposition \(d_s=A.f(x)+B.h({ ln(T) }) \).
