Consider this,
\(\cfrac{d kg}{dt}=A{ \begin{matrix} lim \\ \Delta t\rightarrow0 \end{matrix} } \left\{\cfrac{\Delta kg}{\Delta t} \right\}=A\cfrac{(kg_1-kg_2)}{\Delta t}\)
Assuming that \(g\), gravity is a constant,
\(\cfrac{d kg}{dt}=\cfrac{A}{g}{ \begin{matrix} lim \\ \Delta t\rightarrow0 \end{matrix} } \left\{\cfrac{(kg_1-kg_2)g}{\Delta t} \right\}\)
\(\cfrac{d kg}{dt}=\cfrac{A}{g}{ \begin{matrix} lim \\ \Delta t\rightarrow0 \end{matrix} } \left\{\cfrac{\Delta F}{\Delta t} \right\}\)
\(\cfrac{d kg}{dt}=\cfrac{mA}{g}{ \begin{matrix} lim \\ \Delta t\rightarrow0 \end{matrix} } \left\{\cfrac{\Delta a}{\Delta t} \right\}\)
where \(m\) is the fluid inside the connecting tube.
\(\cfrac{d kg}{dt}=\cfrac{mA}{g}\cfrac{d^3x }{d t^3}\)
\(\Delta kg = \rho X_A \Delta x\), \(m= \rho X_A L\)
where \(X_A\) is the cross-sectional area of the connecting tube, \(L\) the length of the tube and \(\rho\) the density of the fluid.
\(\Delta x = \cfrac{\Delta kg}{\rho X_A}\)
\(\cfrac{\Delta x}{\Delta t} = \cfrac{1}{\rho X_A}\cfrac{\Delta kg}{\Delta t}\)
When we apply \(\lim{\Delta t\to0}\),
\(\cfrac{dx}{dt}= \cfrac{1}{\rho X_A}\cfrac{dkg}{dt}\)
\(\cfrac{d^3x}{dt^3} = \cfrac{1}{\rho X_A}\cfrac{d^3 kg}{dt^3}\) --- (**)
So,
\(\cfrac{d kg}{dt}=A\cfrac{m}{g}\cfrac{1}{\rho X_A}\cfrac{d^3 kg}{dt^3}\)
\(\cfrac{d kg}{dt}=A\cfrac{\rho X_A L}{g}\cfrac{1}{\rho X_A}\cfrac{d^3 kg}{dt^3}\)
\(\cfrac{d kg}{dt}=A\cfrac{L}{g}\cfrac{d^3 kg}{dt^3}\)
where A is the constant of proportionality that can be 1.
Now comes the hard part,
Consider an analogy,
\(kg:)_\supset T\)
where the relational symbol, \(:)_\supset \) means "analogous to"
where \(T\) is temperature, \(A\) is a proportionality constant, \(L\) is the distance between the two hot spots and \(g_T\) an analogous gravity term that is here responsible for the flow of T. A hot spot thins out space around it. A gravity pull develops towards the direction of denser space, as a result of time speed slowing down in the denser space region. This gravity pulls \(T\) into the region of denser and colder space. Thermal gravity, \(g_T\) develops as a result of conservation of energy across time and space,
\(v^2_t+v^2_s=c^2\) from which we obtained \(g\), gravity.
\((v^2_t+v^2_s)+(v^2_{tc}+v^2_c)=c^2\)
from which we obtained \(g\), gravity and electrostatic field acceleration. And
\((v^2_t+v^2_s+v^2_{rg})+(v^2_{tc}+v^2_{rc}+v^2_c)=c^2\)
from which we obtained in addition, two temperature components. These temperature components are the results of rotation \(v^2_{rg}\) and \(v^2_{rc}\), they do not have independent time components. Their time components are respectively, in compliance to the right hand rule, \(v_t\) and \(v_{tc}\).
We note that from (*),
\(\cfrac{d T}{dt}=A\cfrac{L}{g_T}\cfrac{d^3 }{dt^3}\left\{ \cfrac{1}{2}mv^2\right\}=A\cfrac{L}{2g_T}\cfrac{d^3 }{d^3t}\left\{ mv^2\right\}\)
where both \(m\) and \(v\) can change with time.
In general when \(g\) is not considered time invariant (case of transient before steady state), we start from
\(\cfrac{d T}{dt}=mA{ \begin{matrix} lim \\ \Delta t\rightarrow0 \end{matrix} } \left\{\cfrac{\Delta }{\Delta t}\left\{\cfrac{a}{g}\right\} \right\}\)
\(\cfrac{d T}{dt}=mA\cfrac{d}{dt}\left\{\cfrac{1}{g}\cfrac{d^2x }{d t^2}\right\}\)
\(\cfrac{d T}{dt}=mA\left\{\cfrac{1}{g}\cfrac{d^3x }{d t^3}-\cfrac{1}{g^2}\cfrac{dg}{dt}\cfrac{d^2x }{d t^2}\right\}\)
Substitute (**)
\(\cfrac{d T}{dt}=AL\left\{\cfrac{1}{g}\cfrac{d^3kg }{d t^3}-\cfrac{1}{g^2}\cfrac{dg}{dt}\cfrac{d^2kg }{d t^2}\right\}\)
and apply the analgy, \(kg\) with \(T\),
\(\cfrac{d T}{dt}=AL\left\{\cfrac{1}{g}\cfrac{d^3T }{d t^3}-\cfrac{1}{g^2}\cfrac{dg}{dt}\cfrac{d^2T }{d t^2}\right\}\)
And when we consider other time components,
\(\cfrac{d T}{dt}=A\cfrac{L}{2}\left\{\cfrac{1}{g}\cfrac{d^3}{d t^3}\left\{\sqrt{q^2v^4_{rc}+m^2v^4_{rg}}\right\}-\cfrac{1}{g^2}\cfrac{dg}{dt}\cfrac{d^2}{dt^2}\left\{\sqrt{q^2v^4_{rc}+m^2v^4_{rg}}\right\}\right\}\)
\(g^2=g^2_{rc}+g^2_{rg}\)
where \(g_{rc}\) and \(g_{rg}\) are the result of time slowing down in the respective time dimensions. And of course we can have,
\(\cfrac{d T}{dt}\cfrac{dt}{dx}=\cfrac{d T}{dx}=A\cfrac{L}{2v_x}\left\{\cfrac{1}{g}\cfrac{d^3}{d t^3}\left\{\sqrt{q^2v^4_{rc}+m^2v^4_{rg}}\right\}-\cfrac{1}{g^2}\cfrac{dg}{dt}\cfrac{d^2}{dt^2}\left\{\sqrt{q^2v^4_{rc}+m^2v^4_{rg}}\right\}\right\}\)
where \(v_x=\cfrac{dx}{dt}\). And, for steady state, \(t\rightarrow\infty\)
\(\int^{T(x)}_0{dT}=T(x)=A\cfrac{L}{2}\int^{\infty}_0{\left\{\cfrac{1}{g}\cfrac{d^3}{d t^3}\left\{\sqrt{q^2v^4_{rc}+m^2v^4_{rg}}\right\}-\cfrac{1}{g^2}\cfrac{dg}{dt}\cfrac{d^2}{dt^2}\left\{\sqrt{q^2v^4_{rc}+m^2v^4_{rg}}\right\}\right\}dt}\)
And so, out of the kitchen we go.
