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?


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.

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$
    – user929304
    Apr 16, 2015 at 8:33
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 17, 2015 at 12:13
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 17, 2015 at 12:13

It's been a while since this question was asked, but I think there's a big picture missing in these answers. The connection to probability theory will provide a robust framework to understand why @Arnold's first statement makes sense. Further, linear response theory (I discuss here Physics Stack 20797) is a formal way to work backward up the derivative chain. (i.e. predict magnetization $m$ given your response function/susceptibility $\chi$).

Part 1 - Average Magnetization The average magnetization is defined as, $$\left< m \right> = \sum_i m_i \times p_i $$. But what is $p_i$?? Let's use our thermodynamics class and consider the partition function. Which is the sum of all the Boltzmann factors and defines the probability measure for a system in thermal equilibrium, $$\mathcal{Z} = \sum_i e^{-\beta E(s_i)}$$ where, $\beta = \frac{1}{T}$, and $E(s_i)=E_i$ is the energy of the $i^{th}$ state. This is just an isolated system. The probability of having energy $E_i$ is defined by this probability measure as, $$p_i=\frac{e^{-\beta E_i}}{\mathcal{Z}}$$ Now let's turn on some external magnetic field $H_j$ which interacts with each of the $j$ particles in state $i$ via their magnetization $m_j(s_i)$. Here the magnetic field interacts with each particle magnetization so we use $m$. $$\mathcal{Z} = \sum_i e^{-\beta \left(-E_i - \sum_j H_j m_j(s_i) \right)}$$ Note, that we pick this sign convention because $-(E + mH)$ looks like a negative Enthalpy and matches the Ising model definition. We'll suppress the state dependence in $m_j(s_i)=m_j$ cause it's not relevant to our upcoming math. Anywho, the probability of being in state $s_i$ is, $$p_i=\frac{e^{\beta E_i + \beta \sum_j H_j m_j}}{\mathcal{Z}}$$ NOW we can compute the average magnetization. $$\left< m \right> = \sum_i m_i \times p_i = \sum_i m_i \times \left(\frac{e^{\beta E_i + \beta \sum_j H_j m_j}}{\mathcal{Z}} \right)$$ Rearrange terms, $$ \left< m \right> = \frac{1}{\mathcal{Z}} \sum_i m_i e^{\beta E_i + \beta \sum_j H_j m_j} = \frac{T}{\mathcal{Z}} \frac{\partial}{\partial H_i} \sum_i e^{\beta E_i + \beta \sum_j H_j m_j} = \frac{T}{\mathcal{Z}} \frac{\partial}{\partial H_i} \mathcal{Z}$$ So in this case we recognize that the partition function is the Moment Generating Function (MGF) of probability distribution function. The average value of the magnetization $\left< m \right>$ is the first moment (i.e the first derivate of $\mathcal{Z}$).

Part 2 - The Free Energy We can got one step further by using $\frac{1}{f[x]}\partial_x f(x) = \partial_x \log[f(x)]$. $$ \left< m \right> = T \frac{\partial}{\partial H_k} \log[\mathcal{Z}]$$ Recognize that $F=-T \log [Z]$ and all spins have the same coupling $H_j = H, \, \, \forall j$ $$ \left< m \right> = - \frac{\partial F}{\partial H} $$ If we plug in this value of $m$ into your final equation we recover that $$ \chi = -\frac{\partial}{\partial H} \left< m \right> = \frac{\partial}{\partial H} \frac{\partial F}{\partial H} = \frac{\partial^2 F}{\left( \partial H \right)^2} $$

This is not surprising!... "Why Dude!? Cause it seems pretty cool!!"... well I'll tell you! The partition function is the Moment Generating Function. The log of the MGF is the Cumulant Generating Function (CGF). This tells us that the second derivative of $F$ is the variance of the distribution / correlation in the system. $$\frac{\partial^2 F}{\left( \partial H \right)^2} = \left< m^2 \right> - \left< m_i \right>^2 $$ The susceptibility is by definition the correlation in the system.

$$\frac{\partial^2 F}{\left( \partial H \right)^2} = \left< m^2 \right> - \left< m_i \right>^2 = \chi$$

In total, I have just shown that (1) the partition function $\mathcal{Z}$ is the moment generating function and (2) the free energy, $F$ is the cumulant generating function. And this explains why the derivative look the way they do! Furthermore, an open question here is "how to interpret @Arnold's second statement", because a physicist can certainly measure the free energy in their simulations and experiments. A quick example from my field is that we use the Helmholtz energy (a free energy like the Enthalpy) to study a phase transition in the nuclear strong force. While that heuristic may work for him, I'm not sure that's a broadly applicable.

  • $\begingroup$ Many thanks Thomas for this highly elaborate answer! I will spend the proper time it deserves to be read, and get back to you with potential follow-up questions :p $\endgroup$
    – user929304
    Dec 15, 2020 at 12:51
  • $\begingroup$ "Note, that this is the right sign because E−mH looks like the first law of thermo U−pV" The logic doesn't follow. The minus sign of the term -pV is unusual and arises because pressure tends to decrease volume. Other intensive variables tend to increase their conjugate extensive variable, and this is the case with mH (i.e., a magnetic field tends to increase magnetization). Furthermore, "U-pV" is not a law at all. I suggest instead that the subtraction in E-mH constitutes a Legendre transform. $\endgroup$ May 3 at 17:47
  • $\begingroup$ I refer you to Wikipedia's "First law of thermodynamics" subsection sign conventions. en.wikipedia.org/wiki/… There it is discussed that physicists generally follow my convention and chemists normally follow your convention. I am a physicist so I follow physics conventions. $\endgroup$
    – ThomasTuna
    May 3 at 19:46
  • $\begingroup$ Additionally, you could just check out the DEFINITON of the Ising model (en.wikipedia.org/wiki/Ising_model#Definition) where my $E(s_i)$ matches the $-\sum_{<i,j>} J_{ij} \sigma_i \sigma_j$ and my $-\sum_j H_j m_j$ matches the $-\mu \sum_j h_j \sigma_j$. $\endgroup$
    – ThomasTuna
    May 3 at 20:05
  • $\begingroup$ That being said, I agree with your desire to have the Enthalpy emerge via a Legendre Transform, that would be a cool connection to make. So I'll switch around some stuff with my E_i to include your suggestion. $\endgroup$
    – ThomasTuna
    May 3 at 20:08

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.

  • $\begingroup$ 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. $\endgroup$
    – user929304
    Apr 14, 2015 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.