What Thermal Gravity? Too Hot To Handle...

From the post "Heat Wave"

$\cfrac { \partial T }{ \partial t } =\cfrac { x }{ g_{ T } } \cfrac { \partial^{ 3 }T }{ \partial^{ 3 }t }$

$\int  \partial T=\cfrac { x }{ g_{ T } } \int { \cfrac { \partial^{ 3 }T }{ \partial^{ 3 }t }  } \partial t$

$T(t)T(x)=\cfrac { x }{ g_{ T } } (\cfrac { \partial^{ 2 }T }{ \partial t^{ 2 } } +C(x) +C)$ ---(*)

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...