# Fluctuation dissipation theorem : how to identify the response variable and the force in general?

I fully re-edited my question

I have a super basic question. Note that I am just beginning to learn linear response theory.

## General context:

If I consider a linear, time invariant, causal system, relating an input $$E$$ to an output $$S$$, I know that the relationship between $$E$$ and $$S$$ respect the following relationship:

$$S(t)=\int_{-\infty}^{+\infty} \chi(t-t') E(t') dt' + S(-\infty)$$

Where $$\chi(u<0)=0$$ to respect causality. $$\chi$$ is called the linear response function, its Fourier transform is called the susceptibility.

Usually we consider the output being $$\widetilde{S}(t)=S(t)-S(-\infty)$$ to avoid having this extra term on the r.h.s. From now on I will assume $$S(-\infty)=0$$.

All this is math. Now we apply this theory to describe physical systems.

The imaginary part of the susceptibility is supposed to represent the dissipation occuring in a linear system. The susceptibility is the Fourier transform of the linear response function.

However in principle, for it to represent dissipation, $$E(t)$$ and $$S(t)$$ must be specific variables and not "any". I take the example of the relationship voltage-current arround a resistor. For me $$E(t)=I(t)$$ and $$S(t)=U(t)$$. The system being linear, causal and time invariant, I can write:

$$U(t)=\int_{-\infty}^t \chi(t-t') I(t') dt'$$

In practice, here: $$\chi(t-t')=R \delta(t-t')$$.

And we see $$\chi(\omega)=R$$ which is purely real. Then it is here the real part that represent dissipation and not the imaginary one.

## My question

How can I know which quantity should represent the input (generalized force) and output (response variable) so that in fits in the usual framework in which it is applied in physics. Basically we expect that the physical interpretation of dissipation holds for the susceptibility. But maybe there are other physical interpretation that must hold as well. I would like an answer general enough.

Indeed up to my understanding, all the theorem in linear response theory are "simply" mathematical derivations. It is when we do physics that we say "this represents dissipation". Thus I expect in principle that we can take any variable as force and any variable as response (as long as the system is linear,causaul,time invariant). It is only at the physical interpretation level that one must be careful.

Do you also confirm this statement I make ?

• You'll get a clearer picture, if you start with an RC circuit driven by a time-dependent emf. Then it will be clear what are the force, the response variable, and the susceptibility. And do not forget the $\theta(t-t')$ to account for causality, since otherwise you'll not get real $and$ imaginary parts of $\chi$. – Roger Vadim Apr 25 '20 at 19:54
• @Vadim thank you. So I guess my question is then how do I know what is the force and the response variable indeed. – StarBucK Apr 25 '20 at 20:16

How can I know which quantity should represent the input (generalized force) and output (response variable) so that in fits in the usual framework in which it is applied in physics.

In the standard setup of linear response theory, the Hamiltonian contains the product of the input $$F$$ and the output $$x$$, $$H_{\text{int}} \supset F(t) x.$$ Examples of pairs of this form include force and position, pressure and volume, and external magnetic field and magnetization. This is similar to the definition of conjugate variables in thermodynamics, since differentially we have $$dU = F \, dx$$.

Indeed up to my understanding, all the theorem in linear response theory are "simply" mathematical derivations. It is when we do physics that we say "this represents dissipation". Thus I expect in principle that we can take any variable as force and any variable as response

I wouldn't agree with that at all. The trivial parts of linear response theory are indeed independent of what you choose to be the input and output, since they follow from the symmetries alone. But statements such as the fluctuation dissipation theorem are proven starting from the assumption I made above. Of course, you cannot say anything whatsoever about energy dissipation unless you assume something about the Hamiltonian.

• Thank you for your answer. Do you have a good reference to learn this ? I basically only learnt it from wikipedia or some bad links. physics.stackexchange.com/questions/552468/… – StarBucK May 16 '20 at 19:19
• @StarBucK All I know about linear response theory is from David Tong's brief notes. I agree that a lot of sources are not very good -- in particular Wikipedia almost actively tries to hide the fact that I mentioned. – knzhou May 16 '20 at 19:20
• Incidentally, I think Tong's linear response pdf is the only one (from him) that seems to have major erroneous claims. Tong is fantastic hands down, compared to even proper books otherwise! – Vivek May 17 '20 at 2:22
• @Vivek Well, you can't just say that and leave! What were the errors? – knzhou May 17 '20 at 3:58
• @knzhou If memory serves me right, the version of fluctuation dissipation-theorem he quotes with arbitrary observables $O_i$ and $O_j$ is not true. – Vivek May 18 '20 at 14:10

The Fourier Transform is a linear operator with respect to the "outside" multiplications, i.e. the physical entities like voltage, current, susceptibility, inductance etc. A linear replacement of the amplitude I with I - I" will yield a Fourier transform with a factor I - I".

But the FT is non-linear with respect to the time parameter. If the time parameter t in the time domain is replaced by t - t`, the FT will be multiplied by a complex exponential factor, according to the displacement or time shifting rule.

• Im sorry but I dont understand what you mean. I just applied the convolution product propertie of the FFT here. – StarBucK Apr 25 '20 at 21:03
• Also sorry - maybe I did misunderstand something. Could you please elaborate what your definitions are for all values like R, suscept., I both in time and frequency domain? – xeeka Apr 25 '20 at 21:31
• My point is that in linear response theory, we can relate a force to a response variable via a reponse function which corresponds to the $\chi$ I wrote in my message. The fluctuation dissipation theorem says that the imaginary par of the Fourier transform of $\chi$ corresponds to the dissipation. Here I have a pure resistor and the fourier transform of $\chi$ is purely real, thus I don't understand. I guess my problem is that I do not use the theorem in an appropriate way because I did not used the proper force and response variable. I would like to understand my mistake. – StarBucK Apr 25 '20 at 22:13