$\tau_o$ Or No $\tau_o$

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.