# Importance of energy relation in Landau & Lifschitz "Statistical Physics"?

At the end of section 11 in "Statistical Physics" by Landau & Lifschitz, a supposedly important equation is derived.

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Comparing (11.1) with (11.2) we find: $$\frac{\overline{\partial E (p,q;\lambda )}}{\partial \lambda}=\left(\frac{\partial E}{\partial \lambda}\right)_S \tag{11.3}$$ This is the required formula. It allows us to calculate, by thermodynamic methods, the mean values of quantities of the type $$\partial E(p,q;\lambda)/\partial\lambda$$ (over an equilibrium statistical distribution).

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In the equation above, $$E$$ represents the energy of the Hamiltonian/thermodynamic system (assumed to be classical), $$p$$ and $$q$$ the phase space coordinates, and $$\lambda$$ a constrained variable characterizing the boundary/external conditions on the system. The subscripted $$S$$ on the right-hand side means that the derivative is taken at constant entropy (this is done because the slowly changing boundary/external conditions is assumed to be done adiabatically, so the entropy is constant). Also, the overbar on the left-hand side means the derivative is acting on the energy averaged over the phase-space (i.e. statistically averaged energy for the system).

I really cannot see how this equation is important. In face, I don't even see what this means. Could you help elucidate the meaning of this equation which the authors claim to be important?

This formula links thermodynamic (macroscopic) quantities like E, S, and $$\lambda$$ that you can measure, and that describe a thermodynamical system, to microscopical quantities.

Namely: Given a System that is in a macrostate described by the entropy S and the parameter $$\lambda$$, the (average, depending on the ensemble) Energy E ($$E = \langle H \rangle$$) will be a function of these state variables: $$E = E(S, \lambda)$$.

Up until here we don't make any statement about the internal structure of the system. But we know that our system has Microstates (with certain probabilities) and that each of this microstates ($$\Gamma$$) yields Observables, like the Energy $$H(\Gamma, \lambda)$$ or the derivative of the Energy $$\frac{\partial H(\Gamma,\lambda)}{\partial \lambda}$$.

The asked about relation now relates those: \begin{align} \langle \frac{ \partial H(\Gamma,\lambda)}{\partial \lambda} \rangle = \frac{\partial E(S, \lambda)}{\partial \lambda} \end{align}

This is a remarkable result, as gives a microscopic meaning to the rhs: If you change the parameter $$\lambda$$ during a process that leaves the entropy invariant, the amount of energy you gain is exactly the same as the average of the change in energy of the microscopic states. It is by no means self-evident that this should hold: For example, it might be that Different microstates have the same Energy, but not the same derivative, or that the distributions of both are not the same.

Just to be clear, the left-side $$E(p,q;\lambda)$$ represents the Hamiltonian, whereas the right-side $$E$$ the thermodynamic energy. Importance of this equation is obvious if you look at the first law for a fixed number of particles

$$dE = TdS - X_i d\lambda_i$$

where $$X_i=(\partial E/\partial \lambda_i)_S$$ is the generalized (thermodynamic) force corresponding to the external parameter $$\lambda_i$$.

• Could you be more explicit please, in particular regarding the Hamiltonian? Coz it seems to mean that the Hamiltonian resembles in some way the internal energy of a system. Which is... not surprising at all? Commented Aug 26, 2022 at 9:43
• It is not the same. Hamiltonian is the microscopic description of the system, whereas the internal energy (a macroscopic description) is just the average value (calculated with respect to some distribution) of the Hamiltonian. Commented Aug 26, 2022 at 11:09