\(\cfrac{\partial T}{\partial t}=\cfrac{x}{g_T}\cfrac{\partial^3 T}{\partial t^3}\)
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,
\({ g }_{ T }=-\frac { { H }_{ o } }{ { x }^{ 2 } }\)
\( \iint {\cfrac{1}{T(x)}}\partial^2T(x)=\iint { \frac { { -H }_{ o } }{ { x }^{ 3 }{ c }^{ 2 } } (\partial x)^{ 2 } } \)
\( \int{ln(T(x))}\partial T(x) =\frac { { -H }_{ o } }{ { v }^{ 2 } } \int { -\frac { 2 }{ x^{ 2 } } } +C \partial x\)
\( T(x)(ln(T(x))-1) =\frac { { -H }_{ o } }{ { v }^{ 2 } } (\frac { 2 }{ x } +Cx+B)\)
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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