A Penrose process and area theorem

1. Nov 9, 2016

Sumarna

Hawking area theorem says that area of black hole generally never decrease. Penrose process says that energy can be extracted from black hole. Energy extraction will decrease mass? if yes then if mass is decreased then will area also decrease?
I am confusing things here :(

2. Nov 9, 2016

Staff: Mentor

Yes. (Note that this theorem excludes any quantum effects.)

Yes, but this only happens with a rotating black hole, and the hole also loses angular momentum in this process.

No, because for a rotating hole the area does not just depend on the mass. It depends on, heuristically, $\sqrt{M^2 - a^2}$, where $a$ is the hole's angular momentum per unit mass. In the Penrose process, $M$ decreases, but $a$ also decreases, in such a way that the horizon area ends up larger.

3. Nov 9, 2016

Sumarna

This has confused me more.. if both mass and angular momentum are decreasing then area must also decrease. How it end up increasing?

4. Nov 9, 2016

Staff: Mentor

No. Look at the minus sign in front of $a^2$ in the heuristic formula I gave. If angular momentum decreases, the area increases.

5. Nov 9, 2016

Sumarna

Consider $A=8\pi M(M+\sqrt{M^2-a^2})$ which is area of rotating black hole. So if both mass M and angular momentum a are decreasing then area will increase? when i am decreasing these two terms, area is also decreasing.

6. Nov 9, 2016

martinbn

If $a$ is decreased enough the area will increase. For example start with $M$ and $a$ equal, then the area is $A=8\pi M^2$. Now decrease $M$ to $\frac34 M$ and $a$ to zero, then the area will be $A=9\pi M^2$.

7. Nov 9, 2016

Sumarna

O yes now i get it