The cloud of \(T^{-}\) particles that surrounds the \(T^{+}\) particle orbit are in materials that are good conductor of heat, free to flow. Heat is the result of the flow of negative temperature particles. This flow of \(T^{-}\) particles between two bodies stops, when the attractive forces on \(T^{-}\) due to the \(T^{+}\) particles of the nuclei of the two bodies in thermal contact cancel each other.
\(T^{-}\) effects \(E\), the electric field due to a spinning \(T^{+}\) particle in the nucleus by changing the effective value of \(T^{+}\) and the orbit of \(T^{+}\).
\(E=\cfrac{\tau_{\small{E}o}}{4\pi}\cfrac{T^{+}.v_{\small{T}}\times \hat{r}}{r^2}\)
This \(E\) field effects both the orbit of the proton and electron. When temperature decreases, with increasing \(T^{-}\) the orbital radius of the \(T^{+}\) increases resulting in an decrease in \(E\). The charged particles (proton and electron) are held loosely by the weak field. When temperature increases, the number of \(T^{-}\) particles decreases, \(T^{+}\) falls closer into the nucleus, its orbit decreases. The weak field strengthen and the charged particles are held more tightly to the nucleus.
Electrical sparks from a cold piece of metal? Yes! Colder temperature results in higher conductivity of a metal because of more free electrons? Yes.
When two bodies have the same temperature but one feels hot the other feel cold, we attribute this to the higher thermal conductivity of the latter. Thermal conduction is the conduction of negative temperature particles. When two bodies have the same temperature but one feels hot for a longer time than another, we attribute this to the higher thermal capacity of the former. Thermal capacity is the ability to store negative temperature particles.
But the \(T\) field inside the \(T^{-}\) particle shell is zero, as the \(E\) field inside a charged shell is zero.
How does the \(T^{-}\) particle cloud effects the \(T^{+}\) particle in orbit?
The point is to realize that the positive particle is also in orbit and this orbit is rotating about one of its diameter. So all forces are along the radial line through the sphere transcribed by the positive particle between the two shells.
The capacitance of this shells is,
\(C=\cfrac{Q}{V}\)
\(C=\cfrac{4\pi\varepsilon_o}{\cfrac{1}{r_{\small{+}}}-\cfrac{1}{r_{\small{-}}}}=\cfrac{4\pi\varepsilon_o}{r_{\small{-}}-r_{\small{+}}}.r_{\small{-}}r_{\small{+}}\)
Consider the partial derivatives,
\(\cfrac{\partial C}{\partial\,r_{\small{+}}}=+\cfrac{4\pi\varepsilon_o}{{\left(\cfrac{1}r_{\small{+}}-\cfrac{1}{r_{\small{-}}}\right)^2}}.\cfrac{1}{r^2_{\small{+}}}=+\cfrac{4\pi\varepsilon_o}{\left(r_{\small{-}}-r_{\small{+}}\right)^2}.r^2_{\small{-}}\)
\(\cfrac{\partial C}{\partial\,r_{\small{-}}}=-\cfrac{4\pi\varepsilon_o}{{\left(\cfrac{1}r_{\small{+}}-\cfrac{1}{r_{\small{-}}}\right)^2}}.\cfrac{1}{r^2_{\small{-}}}=-\cfrac{4\pi\varepsilon_o}{\left(r_{\small{-}}-r_{\small{+}}\right)^2}.r^2_{\small{+}}\)
But strictly speacking,
\(\require{cancel}\)
\(\cfrac { dC }{ dQ } =\cfrac { \partial C }{ \partial \, Q } \cancelto{0}{\cfrac { \partial Q }{ \partial \, r_{ { + } } }} +\cfrac { \partial C }{ \partial \, Q } \cfrac { \partial Q }{ \partial \, r_{ { - } } }\)
because the \(T\) field due to \(Q\) inside the \(T^{-}\) cloud shell is zero. Changing \(Q\) should not change \(r_{\small{+}}\).
Stuck!
