# Susceptibilities and response functions

It is often confusing whether a susceptibility is the same as a response function, specially that often they are used interchangeably, in the context of statistical mechanics and thermodynamics. Very generally:

Response function:

For response functions, typical examples would be thermal expansivity $\alpha,$ isothermal compressibility $\kappa_T,$ specific heats $C_v$, $C_p,$ at least for these examples they seem all to be given by first derivatives of either a system parameter or a potential:

$$\alpha = \frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_{P,N}, \, \kappa_T = -\frac{1}{V} \left(\frac{\partial V}{\partial P}\right)_{T,N}, \, C_v = \left(\frac{\partial E}{\partial T}\right)_{V,N}$$

1. So can one define response functions as first derivatives (I guess talking of first derivatives already assumes linear responses) of a system's observables (e.g. $V$) and potentials (e.g. $E$) with respect to system parameters (e.g. $T,$ $P$) without loss of generality?

Susceptibilities:

Wikipedia definition:

In physics, the susceptibility of a material or substance describes its response to an applied field. More general, a susceptibility is a quantification for the change of an extensive property under variation of an intensive property.

Typical quantities we refer to as susceptibilities are magnetic and electric susceptibilities, describe the change of magnetization and polarisation with respect to changes of the magnetic field $h$ and electric field $E$ respectively. So one writes, for the magnetic susceptibility e.g.:

$$\chi = \left(\frac{\partial M}{\partial h}\right)_T$$ But the magnetization itself seems to be a response function given by: $$M = \left(\frac{\partial F}{\partial h}\right)_T$$ Where $F$ is the Helmholtz free energy. Combining the two expression we can write the susceptibility as the second derivative of $F$: $$\chi = \left(\frac{\partial^2 F}{\partial h^2}\right)_T$$

1. The above in mind, was it correct to call the magnetization a response function? As it would be well in line with the given definition of response functions in first part.
2. From the final expression of $\chi,$ can one conclude that susceptibilities are usually given by second order derivatives of thermodynamic potentials with respect to a system parameter or an external field?

1. Closing remark: All of this seems to rather point at the fact that response functions and susceptibilities cannot actually be used interchangeably. Anyhow, I really hope someone can resolve such confusions by giving more consistent or complete definitions of response functions and susceptibilities.

response function = susceptibility = (pure or mixed) second derivative of a (Helmholtz, Gibbs, etc.) free energy.

Magnetization (a first, not second derivative of a free energy) is not a response function as the free energy is not observable, so one cannot observe its response to a change of some variable.

• So short and so concise, thanks for this. Is it correct to regard first derivatives of free energy as observables and the 2nd derivatives as the rate of change of those observable? Finally, what type of derivatives do we mean exactly? I mean is it assumed to be derivatives (always) with respect to external fields, e.g. the magnetic field, or can one also define response functions in terms of free energy derivatives with respect to intensive/extensive system variables? – user929304 Apr 16 '15 at 8:33
• @user929304: There are different kinds of free energy, each one depending on a distinguished set of variables. The first derivatives of a free energy are the observables conjugate to those with respect to one the derivatives are taken. Which pairs of observables are meaningful depends on the system under study. For a magnetic system, the free energy is a function of temperature and magnetic field, and the conjugate observables are entropy and magnetization. Their changes gives three different susceptibilities. – Arnold Neumaier Apr 17 '15 at 12:13
• For a pure chemical system, the Helmholtz free energy is a function of temperature and volume, and the conjugate observables are entropy and pressure. Alternatively, the Gibbs free energy is a function of temperature and pressure, and the conjugate observables are entropy and volume. – Arnold Neumaier Apr 17 '15 at 12:13

I might be able to answer your question in the context of linear response theory:

Response function: the power series expansion of the applied field generated by a weak external perturbation. Mathematically speaking, we can relate the average value of an observable $X$_i to the response function $\chi$ via \begin{align} \langle X_i(t)\rangle=\int_0^t dt'' \sum_j \chi_{ij}(t,\,t'')f_i(t'') \end{align} where $f_i(t)$ is the external perturbation. We can also express it purely in terms of the known observable of the system: \begin{align} \chi_{ij}(t,\,t')=\beta X_i(t)\dot{X}_j(t')su \end{align}

The generalized susceptibility: define this as $\chi(\omega)$. This is the ratio of the response of an average observable to an external force $F(\omega)$: \begin{align} \chi(\omega)=\frac{\Delta \langle X(\omega)\rangle}{F(\omega)} \end{align}

Furthermore, the susceptibility is the Laplace-Fourier transform of the linear response function--that is, \begin{align} \chi(\omega)=\int_0^\infty dt \chi(t)\exp(-i\omega t) \end{align} Many texts (on non-equilibrium statistical mechanics, at least) use a very liberal definition of response function--that is, one that is synonymous with the susceptibility. For a non-equilibrium stat mech perspective, look in Pottier's 2012 text.

• Thanks a lot for your answer. I hope it's ok if I ask 1-2 questions. 1) $\chi_{ij}$ in the first equation, do the indices mean that the response function is a 2nd rank tensor? 2) What's the intuition behind the 2nd equation? I mean the observable times its own time derivative? (what is $su$ at the end of that eq.?) 3) Why did we not care for the frequency of perturbation $\omega$ when defining the response function? (in lin. response theory, are the response functions taken as $\omega$-independent?) Thanks you very much in advance for further clarifications. – user929304 Apr 14 '15 at 6:43