# Proof of thermodynamic stability condition

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• chimay
chimay
TL;DR Summary
Question about a mathematical detail in the proof of the thermodynamic stability relation
I am watching Kardar's Statistical Mechanics course in my spare time and I am struggling to understand a mathematical detail in the proof of the thermodynamic stability condition. See Eq. I.62 here.

The author considers a homogeneous system at equilibrium with intensive and extensive variables $(T,J,\mu)$ and $(E,x,N)$, respectively. Then, He imagines that the latter system is divided into two subsystems (A and B) that can exhange energy. Under the assumption that $E,x$ and $N$ are conserved, we have $\delta E_A=-\delta E_B$, $\delta x_A=-\delta x_B$ and $\delta N_A=-\delta N_B$. What is not clear to me is the statement "Since the intensive variables are themselves functions of the extensive coordinates, to first order in the variations of $(E, x, N)$, we have $\delta T_A=-\delta T_B$, $\delta J_A=-\delta J_B$ and $\delta \mu_A=-\delta \mu_B$."

Can anyone explain to me the previous statement more in detail?

Thank you!

If, for example, ##T=f(E)##, then to the first order, ##\delta T=f'\, \delta E##.
Then, ##\delta E_A = -\delta E_B## implies ##\delta T_A = -\delta T_B##.

chimay
$\delta T_A = \frac{\partial f}{\partial E}\bigg\rvert_{T_A} \delta E_A$ and
$\delta T_B = \frac{\partial f}{\partial E}\bigg\rvert_{T_B} \delta E_B$.

However, since the two subsystems are not necessarily equal, even though $\delta E_A = - \delta E_B$, $\delta T_A \ne \delta T_B$ because $\frac{\partial f}{\partial E}\bigg\rvert_{T_A} \ne \frac{\partial f}{\partial E}\bigg\rvert_{T_B}$.

Where am I wrong?

chimay said:
However, since the two subsystems are not necessarily equal, even though δEA=−δEB, δTA≠δTB because ∂f∂E|TA≠∂f∂E|TB.
They say, "The two subsystems, A and B, initially have the same values for the intensive variables". That is, ##T_A=T_B##. Then, ##\frac{\partial f}{\partial E}\bigg\rvert_{T_A} = \frac{\partial f}{\partial E}\bigg\rvert_{T_B}##.

chimay
You are right! I read it so many times but I missed that. Indeed, since the overall system is at equilibrium, it makes sense that the two "subsystems" have the same values for the intensive variables. Kardar's book is very dense of information and I find it hard to follow sometimes...

Hill

## What is thermodynamic stability?

Thermodynamic stability refers to the condition where a system is in a state of equilibrium and any small perturbations or changes in the system's conditions do not lead to a spontaneous change in its state. This implies that the system's free energy is at a minimum, making it stable against fluctuations.

## Why is the second derivative of free energy important for stability?

The second derivative of the free energy with respect to an extensive variable (like volume or number of particles) is crucial because it indicates the curvature of the free energy function. For a system to be thermodynamically stable, this second derivative must be positive, ensuring that the free energy is at a local minimum and the system resists small perturbations.

## How do you determine the stability of a multi-component system?

For a multi-component system, stability is assessed using the Hessian matrix of second derivatives of the free energy with respect to various extensive variables. The matrix must be positive definite, meaning all its eigenvalues are positive. This ensures that the free energy is at a local minimum in all directions of the variable space.

## What role does entropy play in thermodynamic stability?

Entropy is a measure of the disorder or randomness in a system. For a system to be thermodynamically stable, its entropy should be maximized at equilibrium. This means that any small perturbation should not decrease the entropy, ensuring the system returns to its equilibrium state, which corresponds to the maximum entropy condition.

## Can a system be stable if it is not in equilibrium?

No, a system cannot be thermodynamically stable if it is not in equilibrium. Stability requires that the system be in a state where its free energy is minimized and entropy is maximized. If the system is not in equilibrium, it means there are still processes occurring to reduce free energy and increase entropy, indicating that the system is not yet stable.

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