\(\cfrac { \partial T }{ \partial t } =\cfrac { x }{ g_{ T } } \cfrac { \partial^{ 3 }T }{ \partial^{ 3 }t } \)
\( \int \partial T=\cfrac { x }{ g_{ T } } \int { \cfrac { \partial^{ 3 }T }{ \partial^{ 3 }t } } \partial t\)
\( T(t)T(x)=\cfrac { x }{ g_{ T } } (\cfrac { \partial^{ 2 }T }{ \partial t^{ 2 } } +C(x) +C)\) ---(*)
where \(C(x)\) is a function only in \(x\) and \(C\) is a numerical constant.
If we insist that \(T\) propagate as a wave at velocity \(v\),
\(\cfrac { \partial^{ 2 }T }{ \partial t^{ 2 } }={ v }^{ 2 }\cfrac { \partial^{ 2 }T }{ \partial x^{ 2 } }\)
Substitute the above into (*)
\( T(t)T(x)=\cfrac { x }{ g_{ T } } ({ v }^{ 2 }\cfrac {\partial^{ 2 }T }{ \partial x^{ 2 } } +C(x) +C)\)
when we let \(T=T(t)T(x)\).
If \(T(t)\) has no constant term,
\(C(x)=C=0\),
\( T(t)T(x) ={ v }^{ 2 }\cfrac{ x } { g_{ T } }T(t)\cfrac { d^{ 2 }T(x) }{ d x^{ 2 } } \)
\( g_{ T } =v^2\cfrac{x}{T(x)}\cfrac { d^{ 2 }T(x) }{ d x^{ 2 } } \)
The biggest assumption here is that \(T\) propagate as a wave at velocity \(v\). We know that \(g_T\rightarrow 0\) as \(x\rightarrow \infty\). Thermal gravity is positive, and points away from higher temperature. It is in the same direction as \(x\). \(T(x)\) as a product with \(T(t)\) satisfies the wave equation. Just on the boundary of the hot spot, \(g_T\) is maximum. Thereafter, \(g_T\) decreases monotonously.
Still looking for \(g_T(x)\)...
Because the immediate answer is,
\( T(x)={ e }^{ -Ax }\)
where
\( A=\cfrac { 1 }{ v }\sqrt{\cfrac { { g }_{ T } }{ L}} \)
which leads right back to
\( g_{ T } =v^2x(\cfrac { 1 }{ v }\sqrt{\cfrac { { g }_{ T } }{ L}})^2 \)
\( g_{ T } = x{\cfrac { { g }_{ T } }{ L}} \)
\(x=L\)
which is undeniable. Very Hot...
Because the immediate answer is,
\( T(x)={ e }^{ -Ax }\)
where
\( A=\cfrac { 1 }{ v }\sqrt{\cfrac { { g }_{ T } }{ L}} \)
which leads right back to
\( g_{ T } =v^2x(\cfrac { 1 }{ v }\sqrt{\cfrac { { g }_{ T } }{ L}})^2 \)
\( g_{ T } = x{\cfrac { { g }_{ T } }{ L}} \)
\(x=L\)
which is undeniable. Very Hot...
