A Penrose process and area theorem

  • Thread starter Sumarna
  • Start date
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 :(
 
25,832
6,649
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.
 
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?
 
25,832
6,649
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.
 
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

Science Advisor
1,505
381
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##.
 
O yes now i get it
 

Want to reply to this thread?

"Penrose process and area theorem" You must log in or register to reply here.

Related Threads for: Penrose process and area theorem

  • Posted
Replies
3
Views
780
Replies
2
Views
2K
Replies
3
Views
832
  • Posted
Replies
1
Views
2K
  • Posted
Replies
2
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top