# Principle of minimum energy

1. Feb 2, 2014

### aaaa202

Confuses me. In which case is it the helmholtz free energy, the gibbs free energy, the energy that gets minimized and why? Also is it consistent with energy conservation and how is that possible if you use it on two systems exchanging energy. Can we know the total energy of the total system for the principle to apply or only the average?

2. Feb 3, 2014

### Chronos

The laws of thermodynamics insist energy is exchanged. 'Negative' energy cannot be exchanged. That would be like a credit card that refunds you more than you charge. It doesn't happen, otherwise the universe would already be broke.

3. Feb 3, 2014

### stevendaryl

Staff Emeritus
The reason for the different kinds of energy is because there are different kinds of constraints affecting a system.

Suppose you have a small system that is free to exchange energy with a much larger system (say, a reservoir of gas or liquid). In that case, it doesn't make sense to say that the small system should minimize its energy, because any energy that leaves the small system must go to the large system. What it does make sense to do is to maximize the total entropy, $S$, which is the sum of the entropy $S_1$ of the small system and the entropy $S_2$ of the big system.

Maximizing the total entropy turns out to be the same thing as minimizing the quantity

$\mathcal{E} - S_1(\mathcal{E}) T$

where $\mathcal{E}$ is the energy of the small system, subject to the constraint that $T =$ constant.

If the volume of the small system is allowed to change, as well, (imagine a balloon in a much larger room--the balloon can increase its volume, but only by decreasing the volume available in the room outside of the balloon), then maximizing the total entropy in this case is the same thing as minimizing the quantity

$\mathcal{E} - S_1(\mathcal{E}) T + P V_1$

where $V_1$ is the volume of the small system, subject to the constraints that $T =$ constant and $P =$ constant.

The appropriate values for $P$ and $T$ are determined by the entropy of the large system:

$\frac{1}{T} = \frac{\partial S_2}{\partial E_2}$
$\frac{P}{T} = \frac{\partial S_2}{\partial V_2}$

where $E_2$ is the energy of the large system and $V_2$ is its volume.

Last edited: Feb 3, 2014
4. Feb 3, 2014

### aaaa202

Okay can you explain in more detail why it is that for a system at constant pressure and temperature that it is the Gibbs free energy that gets minimized and not the Helmholtz energy for instance? (maybe they both get minimized but G deals with the relevant variables)

G=U-TS+pV

dG= dU-TdS+SdT+pdV+Vdp = Vdp = 0

If one uses dU = TdS - pdV and dT=dP=0

This doesn't seem right - shouldn't it decrease? - when should I use dU = dQ+dW and when should I use TdS - pdV - for some reason I don't like to equate those two expressions even though my book does.

Also since F=U-TS and G=U-TS+pV will F not be minimized whenever G is minimized.

5. Feb 3, 2014

### stevendaryl

Staff Emeritus
That's not correct. Since
$dU = T dS - P dV$
$dG = T dS - P dV - (T dS + S dT) + (P dV + V dP) = V dP - S dT$

So $dG = 0$ when $P$ and $T$ are held constant.

$dS$ is defined to be $\frac{dQ}{T}$ and $dW$ is defined to be $-P dV$, so it doesn't matter which form you use, as long as you are talking about infinitesimal, reversible changes.

If you have a system that is free to radiate away energy to the environment, then at equilibrium, then energy will be at a minimum. So you minimize $U$.

If the temperature is being held constant, then instead you minimize:

$F = U - ST$

(Helmholtz free energy)

$dF = -P dV - S dT$

So $dF = 0$ if $V$ and $T$ are held constant.

6. Feb 3, 2014

### aaaa202

But why will U (in the case where F=U-TS is minimized) not also be minimized not also be minized? Is it because our system can get energy for free from the environment? (I guess that's the idea behind introducing F) But either way, will the system not still also have minimum in U when F is minimal?

7. Feb 3, 2014

### stevendaryl

Staff Emeritus
If the small system is free to exchange energy with a reservoir, then as I said, any energy that leaves the small system must be transferred to the reservoir. So minimizing the energy of the small system is maximizing the energy of the reservoir. So what actually happens is determined by the temperatures. If the small system has a higher temperature than the reservoir, then energy will flow from the small system to the reservoir. If the small system has a lower temperature, then energy will flow from the reservoir to the small system. Eventually, they will achieve the same temperature. At that point, neither system will be at a minimum energy, but will be at the minimum value of $U - S T$, subject to the constraint that the temperature is the equilibrium temperature.

8. Feb 4, 2014

### aaaa202

Let me just try to understand again:

Maximum entropy principle: We know the average energy of a system interacting with a resevoir and find the energy distribution by maximizing S(U).

Minimum energy principle: We know the average entropy of a system interacting with a resevoir and find the energy distribution by minimizing U(S)? Or do we know the exact entropy? I don't see how the latter case can be possible since we can't know the entropy of a system completely if it can exchange energy (and thus entropy) with a resevoir.

Minimum free energy principle: We know ... ?