# Why is the helmholtz free energy minimized?

I don't quite understand the principle of minimum energy despite having read the derivation on Wikipedia.

I think I got lost when the free energy was defined as $A= \max_S{\left(U-TS\right)}$, because I don't know why is the max there.

• The wikipedia page has $\max_S$, not the maximum of the expression.
– user195162
Jun 13, 2018 at 19:30
• @Quantumness I know, but I still don't even know what's that and why is that. Jun 13, 2018 at 19:33
• It is referring to maximum entropy, the second law of thermodynamics.
– user195162
Jun 13, 2018 at 19:43
• @Quantumness I still don't get the derivation :( Jun 13, 2018 at 19:50
• Could you refer to the specific aspect where you are confused? I'm not sure I understand what you are looking for.
– user195162
Jun 13, 2018 at 20:05

Here's an alternative derivation to show that Helmholtz energy will be minimized.

Consider the fact that by the second law of thermodynamics, the total entropy $S$ of the universe must increase. That is, the sum of all entropies has the relationship

$$dS_{\rm universe} = dS_{\rm sys} + dS_{\rm surr} \geq 0,$$

where 'sys' and 'surr' represent ours system and surrounds respectively. Hold this result for the moment. Recall the thermodynamic identity $dU = TdS-PdV+\mu dN$, and recognize that for constant volume and number of particles,

$$dS_{\rm surr} = \frac{dU_{\rm surr}}{T} = - \frac{dU_{\rm sys}}{T}.$$

Note that the last equality comes from the first law: energy is conserved, so $dU_{\rm surr} = -dU_{\rm sys}$. Then, plugging this into our earlier result and multiplying by $T$, we have

$$dS_{\rm universe} = dS_{\rm sys} + dS_{\rm surr} = dS_{\rm sys} - \frac{dU_{\rm sys}}{T}$$ $$\rightarrow TdS_{\rm universe} = TdS_{\rm sys} - dU_{\rm sys}.$$

Now the proper definition for Helmholtz free energy is $F = U - TS$, so for constant temperature, $dF = dU - TdS = -(TdS-dU)$. We can plug this into our last result as

$$T dS_{\rm universe} = - dF_{\rm sys}$$

and finally

$$dS_{\rm universe} = - \frac{dF_{\rm sys}}{T}.$$

Note that in order to maximize the energy of the universe, you have make the Helmholtz free energy as negative as possible (negative, but large magnitude).