The value for \(\tau_o\) is too high,
\(\tau_o\ne\cfrac{c^4}{4\pi G}\)
There is no reason to equate gravitational force will the thermal force. How then to obtain \(\tau_o\)?
From the post "Cold Temperature Requires Thick Skin, Very Thick Too",
\( T^{ 2 }_{ c }=4\pi \tau _{ o }\cfrac { (m_{ e }+m_{ p })^{ 2 } } { m_{ e }m_{ p } }(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { 1 }{ 2 } m_{ e }v^{ 2 }r_{ e })\)
Let's assume that the element Silver has a double root, and that it is this double root that is the reason for its high electrical and thermal conductivity. Since Silver has a atomic number of 47, and \(r_e=144 pm\)
\( T^{ 2 }_{ c }=4\pi \tau _{ o }\cfrac { (m_{ e }+47m_{ p })^{ 2 } }{ 47m_{ e }m_{ p } } (\cfrac { 47q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { 1 }{ 2 } m_{ e }v^{ 2 }r_{ e })\)
\(\tau _{ o }={ 47m_{ e }m_{ p }T^{ 2 }_{ c } }/\left\{ { (m_{ e }+47m_{ p })^{ 2 }(\cfrac { 47q^{ 2 } }{ \varepsilon _{ o } } -2\pi m_{ e }v^{ 2 }r_{ e }) } \right\} \)
Both electrical and thermal conductivity of Silver peak around 10 K. Which is equivalent to,
\(T_c\) = 10/11600 = 8.62e-4 eV
Therefore,
\(\tau_o\) =( 47*9.10938e-31*1.67262e-27)*(8.62e-4)^2/(( 9.10938e-31+47*1.67262e-27)^2*(47*(1.602176565e-19)^2/(8.8541878176e-12)-2*pi*9.10938291e-31*(2^(1/2)*299792458)^2*144e-12))
\(\tau_o\) =-5.817e10
negative!? This is because the term,
\(\cfrac { 47q^{ 2 } }{ \varepsilon _{ o } } -2\pi m_{ e }v^{ 2 }r_{ e }\) = 47*(1.602176565e-19)^2/(8.8541878176e-12)-2*pi*9.10938291e-31*((2)^(1/2)*299792458)^2*1.44e-10=-1.480e^-22
is very small.
If however, we were to use \(r_{pe}\) = 3.41e-15 m from the post "Hydrogen, And Everyone Has His Spins", I know hydrogen is not silver, it is the order of things that counts.
\(\tau_o\) =( 47*9.10938e-31*1.67262e-27)*(8.62e-4)^2/(( 9.10938e-31+47*1.67262e-27)^2*(47*(1.602176565e-19)^2/(8.8541878176e-12)-2*pi*9.10938291e-31*(2^(1/2)*299792458)^2*3.41e-15))
\(\tau_o\) = 6.486e13 N-1(eV)2m-2
There is noway to check this value as yet.
