Heat Wave

From the post "Temperature",

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

where the distance between the hot spots is made to vary in $x$, and $T$ has a dependence on $x$.  ie $T\rightarrow T(x,t)$.   But immediately we see that as $x\rightarrow\infty$,  when the two hot spots are set far apart and we set $T\rightarrow0$ at the far end, ie only one hot spot,

$\cfrac{\partial T}{\partial t}\rightarrow\infty$.

As such we know that for this single hot spot to radiate wave,  we have to consider thermal gravity, specifically the term $\cfrac{x}{g_T}$ such that

$\cfrac{\partial T}{\partial t}\nrightarrow\infty$.

Intuitively this make sense, because to feel the heat from the hot spot at large distances, the hot spot must be delivering large amount of $T$.

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

as  $x$  does not change with time, and   $g_T$   given   $x$   does not change with time.

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

If we insist that $T$ propagate as a wave at velocity $v$,

$\cfrac { \partial^{ 2 }T }{ \partial t^{ 2 } }={ v }^{ 2 }\cfrac { \partial^{ 2 }T }{ \partial x^{ 2 } }$

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

If we consider $T(x,t)$ to be of the form $T(x)T(t)$,

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

and $T(t)$ is without a DC value,

$C(x)=0$

$T(t)T(x)=\cfrac { x }{ g_{ T } } { v }^{ 2 }\cfrac {\partial^{ 2 }T }{ \partial x^{ 2 } }$

$T(t)T(x) =\cfrac { x }{ g_{ T } } { v }^{ 2 }T(t)\cfrac { d^{ 2 }T(x) }{d x^{ 2 } }$

$T(x)=\cfrac { x }{ g_{ T } } { v }^{ 2 }\cfrac {d^{ 2 }T(x) }{ d x^{ 2 } }$

Using an analogy from gravity, where This is wrong. Gravity cannot be of the same form as around a solid mass as when analogously the mass is being poured out.

${ g }_{ T }=\frac { { H }_{ o } }{ { x }^{ 2 } }$

where $H_o$ is the thermal equivalent of $G_o$ and

${ g }_{ To }=\frac { { H }_{ o } }{ { x_{ o } }^{ 2 } }$.

where $x≥x_o$,   $x_o$ is the extend of the hot spot.

Thermal gravity, $g_{T}$ is positive as it is in the direction of $x$, but in our analogy gravity pulls $T$ away and has a negative sign.  We will retain the negative sign because it is consistent with $T$ moving outwards.  So,

${ g }_{ T }=-\frac { { H }_{ o } }{ { x }^{ 2 } }$

$\cfrac { { -H }_{ o } }{ { x }^{ 3 }{ v }^{ 2 } }T(x) =\cfrac { \partial^{ 2 }T(x) }{ \partial x^{ 2 } }$

$\iint {\cfrac{1}{T(x)}}\partial^2T(x)=\iint { \frac { { -H }_{ o } }{ { x }^{ 3 }{ c }^{ 2 } } (\partial x)^{ 2 } }$

$\int{ln(T(x))}\partial T(x) =\frac { { -H }_{ o } }{ { v }^{ 2 } } \int { -\frac { 2 }{ x^{ 2 } }  } +C \partial x$

$T(x)(ln(T(x))-1) =\frac { { -H }_{ o } }{ { v }^{ 2 } } (\frac { 2 }{ x } +Cx+B)$

when $x\rightarrow \infty$,   $T(x)\rightarrow 0$,    $B=0$,    $C=0$

Let,  $B=\cfrac {-2 { H }_{ o } }{ {v }^{ 2 } }$,

$T(x)(ln(T(x))-1)  = \cfrac { B}{ x }$

$ln(T(x))-1  =\cfrac{B}{xT(x)}$

$T(x)=e^{\left\{\cfrac{B}{xT(x)}+1\right\}}$

$T(x)=ee^{\left\{ \cfrac { B }{ xT(x) } \right\}}$

If we consider the power series expansion of $e^x$,

$T(x)=e(1+\cfrac { B }{ xT(x) } +\frac { (\cfrac { B }{ xT(x) } )^{ 2 } }{ 2 } +\frac { (\cfrac { B }{ xT(x) } )^{ 3 } }{ 6 } +\frac { { (\cfrac { B }{ xT(x) } ) }^{ 4 } }{ 24 } +..)$

$T(x)=e(1+\cfrac { B }{ xT(x) } +\frac { B^{ 2 } }{ 2x^{ 2 }T^{ 2 }(x) } +\frac { B^{ 3 } }{ 6x^{ 3 }T^{ 3 }(x) } +\frac { { B }^{ 4 } }{ 24x^{ 3 }T^{ 3 }(x) } +..)$

As a first approximation, let $x$ be large that all higher term of $\cfrac{1}{x}$ can be ignored,

$xT^{ 2 }(x)-exT(x)-eB=0$

$T(x)=\cfrac { ex\pm \sqrt { e^{ 2 }x^{ 2 }+4exB }  }{ 2x }$

Because $T(x\rightarrow\infty)=0$,

$T(x)=\cfrac { e }{ 2 } -\sqrt { \cfrac { e^{ 2 } }{ 4 } +\cfrac { eB }{ x }  }$

$T(x)=\cfrac { e }{ 2 } -\sqrt { \cfrac { e^{ 2 } }{ 4 } -\cfrac { 2eH_o }{ xv^2 }  }$

$T(x)=\cfrac { e }{ 2 } -\sqrt { \cfrac { e^{ 2 } }{ 4 } -\cfrac { e2g_{To}x^2_o}{ v^2 }\cfrac{1}{x}  }$

At $x=x_o$,

$T(x_o)=\cfrac { e }{ 2 } -\sqrt { \cfrac { e^{ 2 } }{ 4 } -\cfrac { e2g_{To}x_o}{ v^2 }}$

$T(x_o)=\cfrac { e }{ 2 } -\sqrt { \cfrac { e^{ 2 } }{ 4 } -\cfrac{e}{2}\cfrac { 4g_{To}x_o}{ v^2 }}$

$T(x_o)=T_{max}=\cfrac { e }{ 2 }$    when,

$\cfrac { 4g_{To}x_o}{ v^2 }=\cfrac{e}{2}$

$g_{To}=e\cfrac{v^2}{8x_o}$

$T(x)$ is of course appropriately scaled in practical use,

$T_{sun}=AT(x)$

where A is a constant, to reflect the surface temperature of the hot spot, like the Sun.

 $T_{surface}=AT(x=x_o)=A\cfrac{e}{2}$

$A=\cfrac{2T_{surface}}{e}$

$T_{sun}=\cfrac{2T_{surface}}{e}T(x)$ and

$x_o$  =  radius of the Sun.

A plot of e/2-(e^2/4-100*e/x)^(1/2) is shown below,

This plot starts at $\cfrac{e^2}{4}=\cfrac{100e}{x}$,   $x$ = 147.1, decreases almost exponentially to zero at infinity.

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