The Metropolis Algorithm and Detailed Balance

Nothing hinders us obviously in choosing another acceptance ratio, like a weighting of the two energies via $$ \begin{equation*} A(j\rightarrow i)=\exp{(-\frac{1}{2}\beta(E_i-E_j))}. \end{equation*} $$ However, it is easy to see that such an acceptance ratio would result in fewer accepted moves.