# Is the free energy bounded

Can we say that the free energy of a mixture of a finite number of fluids situated in a bounded container $\Omega$ is bounded?

I was thinking of this as an optional hypothesis for one mathematical theorem I want to prove, and I wondered if the energy of some object can be considered to have an upper bound (maybe very large, but still finite). This seems reasonable if the object is bounded. Still, I was thinking of the black holes as mass concentrated in a point, and their energy is huge (I guess).

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The answer really depends on the physics you put in. If the free energy were unbound, one would have to account for the full physics of GUT scales, which we don't know. Classically, you could think of a GR big bang, which would give positive answer. – Alexey Bobrick Apr 21 '12 at 23:49

## 3 Answers

I will assume you mean the free energy of a classical Newtonian collection of particles at temperature T and volume V, defined as the "free entropy" times the temperature or $T\log(Z)$ where Z is the sum over all configurations of $e^{-\beta E}$. This quantity is bounded for reasonable force-laws (meaning you can't extract too much energy by packing the atoms close) because the volume of the constant E phase-space surface for N particles roughly grows as V^N times the product of N spheres of radius $\sqrt{2mkT}$. So the free energy is bounded above by a small multiple of the ideal gas free energy (the multiple is determined by the detail of the force law).

But at high temperature, the total free entropy is (up to an additive constant) the free gas free entropy

$$\log(Z) = \log({V^N\over N!}) + \log((\sqrt{2mT})^{3N})\approx N\log({N\over V}) + {3\over 2}N \log(T)$$

So that not only is the free energy unbounded, but the free energy divided by the temperature is unbounded at large T.

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Sort of: it depends on whether you take the possibility of black holes seriously.

The energy of "some object", i.e. any particular, bounded system, is always bounded (otherwise you have a perpetual motion machine); the energy a given region can contain is not: you can in principle pack as much energy as you want to in to some fixed volume if you're clever enough.

This is within the limits of reason and non-general-relativistic dynamics, of course. If you pack energy so closely that the associated mass, through $E=mc^2$, exceeds the Scharzschild limit, then whatever it is you've got in there will collapse into a black hole and you will lose that energy behind the event horizon.

I should also mention that while black holes look like they have a lot of energy, in fact they do not, in the following sense: the collapse of a star into a black hole is a spontaneous process (i.e. it does not require any action from outside the system) and as such it has to be exothermic, so it will give out some energy as heat, and the total energy inside any box you place around it will at most remain constant. The energy given off comes from the gravitational potential energy associated with stuff being further apart, which you might not have counted in the first place.

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Only energy differences are measurable, so you would need to define a reference state before you start thinking about what it means to be bounded. That's why most of the time people talk about $\Delta G$, i.e. $G-G_0$, rather than just $G$.

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Yes, but it still can be bounded once you set a datum. The bound depends upon the choice of datum, that's all. – Manishearth♦ Apr 21 '12 at 18:41
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