Blackhole mass = Hawking temperature x Entropy?

[johne1618]

According to Blackhole thermodynamics a black hole has an entropy that is proportional to its surface area and a temperature called the Hawking temperature.

If one multiplies the Hawking temperature by the entropy one gets a thermal energy.

Would this energy be equal to the mass/energy of the black hole itself?

John

 

[johne1618]

Actually I answered my own question.

By integrating the thermodynamic relation:

dU = T dS

where T is the Hawking temperature and S is the black hole entropy one obtains the relation:

G M / R = c^2 / 2

which is the Scwartzchild radius relation for a black hole.

Thus the whole of the mass of a black hole can be considered as the thermal energy of its event horizon. This is in accord with the holographic principle that the internal state of a black hole is completely described by the state of its surface area.

 

[johne1618]

If the holographic principle also applies to the Universe as a whole then the Universe is a maximum entropy/information object like a black hole with its entropy determined by the surface area of its expanding event horizon.

If we assume that this event horizon has a Hawking temperature and that its surface thermal energy equals the mass/energy of the Universe then the above analysis again applies so that we again get the relationship:

G * M / R = c^2 / 2

where now M and R is the mass and radius of the Universe.

If you plug this relation into the Friedman equations one finds that it implies a spacially flat linearly expanding model of the Universe that exactly obeys the Hubble equation. This model is both simple and elegant and also fits the current Universe expansion data pretty well.