of thermodynamics $$\label{eq:mq}
  \delta {m_{bh}}=  \frac{\kappa}{8 \pi} \delta A + \Omega_H \delta J,$$ where $$\label{eq:7}
  \kappa= \frac{1}{4{m_{bh}}} \left(1-\frac{(8\pi J)^2 }{A^2} \right),\quad
  \Omega_H=\frac{4\pi J}{A \,{m_{bh}}}.$$ In equation (\[eq:mq\]) we have followed the standard notation for the formulation of the first law; we emphasize, however, that in our context this equation is a trivial consequence of . In axial symmetry $\delta
J=0$ and hence we obtain $$\delta  {m_{bh}}=  \frac{\kappa}{8 \pi} \delta A.$$ By the area theorem we have $$\delta A \geq 0.$$ Then $\delta {m_{bh}}\geq 0$ if and only if $\kappa \geq 0$, that is $\delta {m_{bh}}\geq
0$ if and only if the inequality (\[eq:JA\]) holds. Then, it is natural to conjecture that this inequality should be satisfied for any axially symmetric black hole. If the horizon violates (\[eq:JA\]), then in the evolution the area will increase but the mass ${m_{bh}}$ will decrease. This will indicate that the quantity ${m_{bh}}$ does not have the desired physical meaning. Also, a rigidity statement is expected. Namely, the equality in (\[eq:JA\]) is reached only by the extreme Kerr black hole where $\kappa=0$.

This inequality provides a remarkable quasi-local measure of how far a dynamical black hole is from the extreme case, namely an ‘extremality criteria’ 