A Penrose Process & Hawking Area Theorem Explained

[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 :(

 

[PeterDonis]

Hawking area theorem says that area of black hole generally never decrease.

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

Penrose process says that energy can be extracted from black hole. Energy extraction will decrease mass?

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

if mass is decreased then will area also decrease?

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.

 

[Sumarna]

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.

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

 

[PeterDonis]

if both mass and angular momentum are decreasing then area must also decrease.

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

 

[Sumarna]

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

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.

 

[martinbn]

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.

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 $A=9\pi M^2$.

 

[Sumarna]

O yes now i get it