# Can we show that the helmholtz free energy at equilibrium is minimized from its second derivative?

I can show that $$dF = 0$$ at equilibrium, where F is Helmholtz free energy. But mathematically, starting from $$dF = d(U-TS)$$ I want to show that its second derivative, $$d^2F > 0$$ at equilibrium and thus I can prove that the equilibrium state has minimum Gibbs free energy. I am starting like $$d^2F=d(d(U-TS))$$ $$d^2F = d(dU-TdS-SdT)$$$$d^2F=d(TdS-PdV-TdS-SdT)$$$$d^2F=d(-PdV-SdT)$$ For a system at constant volume, $$dV =0$$ and at equilibrium state, $$dT =0$$, so clearly I would get $$d^2F=0$$ which is obviously incorrect, so am I doing something wrong in here, or my approach is totally wrong. Please do suggest. Thanks.

The first derivative at an extremum point $$x_0$$ is $$0$$ only when evaluated at $$x_0$$. Same goes with the second derivative. So you first have to evaluate the second derivative, then you plug in the specific $$x=x_0$$.

For example, take $$y = x^2$$.
$$\frac{\mathrm{d}y}{\mathrm{d}x}= 2x$$ and $$\frac{\mathrm{d}^2y}{\mathrm{d^2}x}= 2$$.
At $$x = 0$$, the minimum, $$\frac{\mathrm{d}y}{\mathrm{d}x} |_0 = 0$$, and $$\frac{\mathrm{d}^2y}{\mathrm{d^2}x}|_0= 2$$.
If we were to apply your logic however, we would have $$\frac{\mathrm{d}^2y}{\mathrm{d^2}x} = \frac{\mathrm{d}}{\mathrm{d}x} \underbrace{\left (\frac{\mathrm{d}y}{\mathrm{d}x} \right)}_{=0}=0.$$

So, you have to expand your last line.
$$\mathrm{d}^2F = -\mathrm{d}P\mathrm{d}V - P\mathrm{d}^2V-\mathrm{d}S\mathrm{d}T-S\mathrm{d}^2T.$$ Now you apply the condition that you are at equilibrium, so $$\mathrm{d}T|_{\mathrm{eq}} = \mathrm{d}V|_{\mathrm{eq}} = \mathrm{d}P|_{\mathrm{eq}} = \mathrm{d}S|_{\mathrm{eq}} = 0$$, so that: $$\mathrm{d}^2F|_{\mathrm{eq}} = - P\mathrm{d}^2V|_{\mathrm{eq}} -S\mathrm{d}^2T|_{\mathrm{eq}} .$$

Then I guess that if pressure is positive, then the volume is a maximum so $$\mathrm{d}^2V<0$$ so that the first term is positive. Probably a similar argument for the second term...?

• Thanks, but I am finding it hard to understand how $d^2V < 0$ and $d^2T<0$. I know that both P and entropy S is positive, but how $d^2V$ and $d^2T$ are negative. I would be very grateful of you could elaborate it a bit further. Thanks. – Roshan Shrestha Apr 28 at 3:39
• I am kinda guessing here, I was just pointing out the maths. I gave a plausible reason for the first term to be positive in the answer. For the second term, Wikipedia says "at constant temperature, the Helmholtz free energy is minimized at equilibrium." (on the Helmholtz free energy page) so I guess the temperature change is 0, which eliminates the term. – SuperCiocia Apr 28 at 3:46
• Ok, thanks, but then even though $dT=0$, it should not mean that $d^2T=0$ right, also I have not been able to grip this idea why $d^2V<0$ or shall I say how Volume is minimum at equilibrium. Thanks. – Roshan Shrestha Apr 28 at 3:50
• Well if temperature is constant, then $T=c$, $\mathrm{d}^mT =0$. For the volume: My interpretation is that, if the pressure $P$ is positive, then the gas occupies the maximum volume it can by pushing on all its boundaries. So I say the volume should be a maximum, hence $\mathrm{d}^2 V$ @ eq $< 0$. – SuperCiocia Apr 28 at 4:39

Start with the 1st and 2nd Laws written as $$dU=\delta Q + \delta W$$, $$F=U-TS$$ and $$\delta Q \le TdS$$, where $$\delta W$$ is the work done by external forces on the system whose internal energy is $$U$$, and $$\delta Q$$ is the heat transferred from the environment to the system.

Then we have for any process that $$d(F+TS)=dF+TdS+SdT \le TdS + \delta W$$ and $$dF+SdT \le \delta W \tag{1}\label{1}$$ where in $$\eqref{1}$$ equality holds iff the process is reversible.

Now assume that the process is such that the external work done is zero $$\delta W=0$$ then you have $$dF\vert_{\delta W=0} \le -SdT$$, and if the process is also isothermal, that is $$T=const,\; dT=0$$, then you must have $$dF \le 0 \tag{2}\label{2}$$

What does it mean that the free energy cannot increase $$\eqref{2}$$? If a reversible process then $$dF=0$$ and it is not changing. If it is an irreversible process then $$dF<0$$ and $$F$$ must decrease. As $$F$$ is bounded from below when decreases it must reach a minimum eventually. Being a minimum its 2nd derivative, if exists, must be positive, that is equilibrium with the given constraints.

Important: Actually, we do not have to assume that the whole process is isothermal. Instead, it is sufficient to assume that the heat (entropy) exchange with the environment is always at the same temperature, $$dT=0$$. For example, the system's internal temperature during equilibration may change, only its interaction with the outside world must be at a fixed temperature.

• Well, I was not looking for this answer particularly, as I was more into Mathematical understanding of it, I mean is there any way I can prove $d^2F > 0$ at equilibrium. But, thanks for a great explanation of why F is bounded from below and will have minimum as it is decreasing. – Roshan Shrestha Apr 28 at 3:55
• If you accept the entropy maximum principle and its consequence that $S$ is a concave function of its variables from below then it follows that $F$ and $U$ and $G$, etc., (the various potentials by Legendre transformations of $U$) are convex functions from below. You can find it in Callen: THERMODYNAMICS AND AN INTRODUCTION TO THERMOSTATISTICS chapters 5 and 8. (This is an excellent book, strongly recommended. Here is an older edition archive.org/details/thermodynamicsin00call/page/16 that you can read online) – hyportnex Apr 28 at 13:56