Black hole area theorem states that surface area of a black hole does not decrease with time (see page 10 of Introductory Lectures on Black Hole Thermodynamics, Ted Jacobson http://www.physics.umd.edu/grt/taj/776b/lectures.pdf). How then does the surface area of a Black hole decrease via Hawking radiation? Is it because mass of black hole decreases or one of the assumptions of the theorem is violated?


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    – Qmechanic
    Commented Mar 12, 2015 at 15:20

2 Answers 2


I'm surprised that Jacobson's notes, which are great by the way, failed to mention this explicitly, but the reason is that Hawking radiation indeed violates one of the theorems assumptions, namely a positive energy condition.

Positive energy conditions normaly state that the energy density according to all observers is non-negative. When we pass to the context of Quantum Field Theory we must instead talk about the expectation values of energy density. As you can see even in Minkowski space, quantum fields will in general have negative energy density in some points of space for some observers. Outside general relativity energy is invariant under a constant, so one can rescale it, but in GR energy gravitates, and violations of positive energy condition can lead to expansion rather than contraction under gravitational influence, just like a positive cosmological constant.

Note that this is not a quantum vc. classical issue, the classical scalar field also violates energy conditions. But usually one assumes that classical fields are just electromagnetic plus fluids, who all satisfy the positive energy conditions


The fact is that the Hawking area theorem is a classical result, while Hawking radiation is an intrinsically quantum effect.

The Hawking theorem states that in any classical process the area of the Black hole cannot decrease, but radiation emission is not a classical process.


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