Does entropy increase or decrease when black holes merge?

[bon]

Homework Statement

The Hawking/Bekenstein relation gives entropy S for a black hole as proportional to its area:

S = kc^3 / 4G hbar A

Does the entropy increase or decreatse when two black holes coalese into one?

What is the entropy change for the universe (in equivalent bits of information) when two solar mass black holes coalesce..?

Homework Equations

The Attempt at a Solution

Not sure what happens or how to work out entropy change..

What stays constant before and after? total mass?

Thanks

 

[zhermes]

Not sure what happens or how to work out entropy change..
What stays constant before and after? total mass?

Correct. When black-holes merge, you form a new black hole with a mass equal to the sum of the original black-holes. This is required from conservation of energy.

 

[bon]

Ok Thank you! It now asks me:

The internal energy of the black hole is given by the Einstein relation E= mc^2/ Find the temperature of the black hole in terms of its mass..

Im not sure how to do this..none of my equations contain temperature! How can i relate temperature to the energy of the black hole?

Thanks!

 

[zhermes]

They're (hopefully) referring to the temperature of http://en.wikipedia.org/wiki/Hawking_radiation" . That article gives you the equation to find the temperature. Let me know if that doesn't make sense.

 

[bon]

Ok thanks, but I don\'t think we are meant to use that because it\'s not given..I think we have to derive it somehow?

Also - still stuck on my previous question...

Does the entropy increase or decreatse when two black holes coalese into one?

What is the entropy change for the universe (in equivalent bits of information) when two solar mass black holes coalesce..?

So I see that the entropy will increase because A increases..and I can work out the entropy change..but what does it mean \"in equivalent bits of info\"? THanks!

 

[zhermes]

Ok thanks, but I don\'t think we are meant to use that because it\'s not given..I think we have to derive it somehow?

You definitely don't need to derive the equation for hawking radiation. I can't think of any other temperature they might be referring to...

So I see that the entropy will increase because A increases..and I can work out the entropy change..but what does it mean \"in equivalent bits of info\"? THanks!

Great. 'in bits' just means using a particular unit of measurement of entropy. I'm not sure if you're familiar with any computer-science, but its the same 'bit'---referring to a single piece of information (yes or no; 0 or 1). You'd have to check your formula to see what units its in---I'm not sure off hand. I'm sure google could find out.

 

[bon]

OK i see now. Thanks. I\'m trying to do the bit that says: find the temp of a black hole in terms of its mass.

Surely you can use the first law: dU = Tds - pdV

this implies T = dU/dS at const. Vol

What i don't understnad is how this can be:

U = mc^2 so we are left with dm/DS at const V. But how can S change at const V if S = kA where A is area. If Vol is fixed, so is area, thus so is volume?

THanks!

 

[bon]

You definitely don\'t need to derive the equation for hawking radiation. I can\'t think of any other temperature they might be referring to...


Great. \'in bits\' just means using a particular unit of measurement of entropy. I\'m not sure if you\'re familiar with any computer-science, but its the same \'bit\'---referring to a single piece of information (yes or no; 0 or 1). You\'d have to check your formula to see what units its in---I\'m not sure off hand. I\'m sure google could find out.

Hey. Just thought I\'d quote you so you see my reply.. thanks.

 

[zhermes]

Surely you can use the first law: dU = Tds - pdV
this implies T = dU/dS at const. Vol
...
But how can S change at const V if S = kA where A is area. If Vol is fixed, so is area, thus so is volume?

I'm not sure if that will work (honestly I kind of doubt it), but its a good exercise either way.

You can't treat either the Volume or the Area as a constant (because they're not). You have an expression for the entropy, the volume, the potential energy, and you're looking for the temperature. You also need an expression for the pressure---the only option would be gravitational.

Give it a try, let me know what you get!