Derivation of $\beta$

If we take the entropy definition as (see [8] §38, [7] Ch.9B, and [4] Ch.1,§16 )

$$\mbox{d}S ~=~-\mbox{d}\big{(}\sum_iP_i\ln P_i\big{)}~=\sum_i\ln P_i\mbox{d} P_i~+~\sum_ i\mbox{d}P_i$$ (24)

where the entropy is measured in bit. The last term in the above equation is seen to be zero by differentiating the normalization constant $\sum_iP_i=1$.

Taking the logarithm of equation (18) we get $\ln P_i=-\beta E_i-\ln Z$ , and by using this value in equation (24), we obtain

$$-\mbox{d}\big{(}\sum_iP_i\ln P_i\big{)}=-\beta\sum_iE_i\mbox{d}P_i-\ln Z\sum_i\mbox{d}P_i~~.$$ (25)

Again the last term vanishes. Now, rewriting the second law $\mbox{d}E=T\mbox{d}S-P\mbox{d}V$ as $\mbox{d}S=\frac{1}{T}(\mbox{d}E+P\mbox{d}V)$. If we fix the volume then the last equation would become $\mbox{d}S=\frac{\mbox{d}E}{T}$ .   Combining this equation with equation (25) as follows

$$\frac{1}{T}(\mbox{d}E)~=\mbox{d}S~=~\beta\sum_iE_i\mbox{d}P_i~~.$$ (26)

It is evident that $\beta$ must be equal to the inverse of temperature

$$\beta~=~\frac{1}{T}~~.$$ (27)

In the above sections we have proven that the dissipation is equal to the energy fluctation and that both are proportional to the heat capacity. It remains to be seen what the energy dissipation is in thermodynamics.