Cosmic Microwave Background and Black Holes

[mathematic]

Black holes are described as losing mass (very slowly - unless hole is microscopic) by Hawking radiation.

 

The CMB is all over, so black holes will gain mass from CMB radiation impinging on them. Presumably bigger black holes will gain more.

 

Question - at what size will these mass changes balance?

[J.C.MacSwell]

Black holes are described as losing mass (very slowly - unless hole is microscopic) by Hawking radiation.

 

The CMB is all over, so black holes will gain mass from CMB radiation impinging on them. Presumably bigger black holes will gain more.

 

Question - at what size will these mass changes balance?

From Wikipedia on Hawking Radiation

 

The black hole radiation temperature will balance the CMBR when the black hole is approximately the mass of the moon.

 

https://en.wikipedia.org/wiki/Hawking_radiation

[imatfaal]

Black holes are described as losing mass (very slowly - unless hole is microscopic) by Hawking radiation.

 

The CMB is all over, so black holes will gain mass from CMB radiation impinging on them. Presumably bigger black holes will gain more.

 

Question - at what size will these mass changes balance?

 

I am not quite sure from your OP whether you know why larger black holes gain mass and smaller (very small) lose mass. Apologies if I have misread

 

Simplistically it is a temperature thing - large black holes are very very cold, they are so cold that the vacuum of space "warmed" by the CMBR is actually warmer. Large black holes are warmed up by their surroundings - they gain energy, increase their mass, and get colder! Small black holes (smaller than the Moon as JCMcS said above) are hotter than their environment, they give off energy, decrease their mass, and get hotter. Adding energy to the black hole seems to make it colder.

 

The temperature of the black hole (the thermodynamic temperature which corresponds to a black body which emits the same radiation spectrum as the black hole emits hawking radiation) is inversely proportional to the mass of the black hole

 

[latex]T = \frac{1}{M} \cdot \frac{\hbar c^3}{8 k \pi G}[/latex]

[Airbrush]

When you say large, or supermassive, black holes are very cold, does that mean the average temperature inside the event horizon including the singularity? Is gravity so great that it freezes the motion of atoms inside the singularity? So any heat generated by matter entering the black hole is instantly frozen as it slams into the singularity?

Edited June 22, 2015 by Airbrush

[Mordred]

No it's the blackbody temperature at the EH. Not the Temperature inside the EH. The blackbody temperature in this case is loosely defined as the ability of one particle of the particle/anti particle pair to escape. The pair forms just outside the EH. One of the pair falls into the EH. The other escapes.

Inside the EH. We simply don't know but according to the ideal gas laws the temperature of the singularity should be in excess of several million degrees.

 

Based upon Pv=nRT. Higher density equals higher temperatures.

Edited June 22, 2015 by Mordred

[imatfaal]

When you say large, or supermassive, black holes are very cold, does that mean the average temperature inside the event horizon including the singularity? Is gravity so great that it freezes the motion of atoms inside the singularity? So any heat generated by matter entering the black hole is instantly frozen as it slams into the singularity?

 

I said "Simplistically ... black holes are very very cold" - and you should note that lower down I explained more what I meant by the temperature of a black hole - the thermodynamic temperature which corresponds to a black body which emits the same radiation spectrum as the black hole emits hawking radiation. That is to say that we can calculate what temperature a black body would need to be to emit the same spectrum of radiation as the black hole emits. We just do not know - and to an extent cannot know - what goes on inside the event horizon; there is a disconnect and if we knew about stuff happening inside then it would no longer be an event horizon because there would be an information flow.

 

nb - in case you do not know the term black body - it has nothing to do with black holes. A black body is a theoretical ideal within thermodynamics - black body radiation is radiation emitted dependent purely on the temperature of the black body in a state of thermal equilibrium. Every termperature has a corresponding spectrum of radiation that would be emitted as blackbody radiation

Edited June 22, 2015 by imatfaal
xposted with Mordred

[MigL]

Properties of a BH, like mass, entropy and temperature, are defined by the event horizon.

They have little to do with what is actually 'inside' the event horizon.