I am given the following problem about fluctuation dissipation theorem:

Consider an external force $f(t)= \frac{f_0}{2}(e^{i\omega_0 t}+e^{-i\omega_0 t})$ acting on a particle with momentum $p=mv$ in some environment. The total Hamiltonian reads:

$$H = \frac{p^2}{2m}+U(\{q,p\}_{env})-\frac{p f(t)}{m},$$

where $env$ stands for the environment positions and momenta.

I am asked to find the average dissipation rate. And I am given that $\langle v(\omega)\rangle=\chi_v(\omega)f(\omega)$, where $\chi_v(\omega)$ is the velocity response function. The answer should be as follows:

$$\Big\langle \frac{dE}{dt}\Big\rangle=\frac{1}{2}f_0^2 \chi_v^{\prime\prime}(\omega_0)\omega_0$$ where $\chi^{\prime\prime}_v(\omega_0)$ denotes the imaginary part of $\chi_v(\omega_0)$.

My trial:

We can find $\langle v(t)\rangle$ by using the inverse Fourier transform:

$$\langle v(t)\rangle=\frac{1}{2\pi}\int^{\infty}_{-\infty}\chi_v(\omega)f(\omega) e^{-i\omega t} d\omega.$$

We note that the Fourier transform of the force $f(t)$ is given by:

$$f(\omega) = f_0 \pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)].$$

We therefore find that:

$$\langle v(t) \rangle = \frac{f_0}{2}\int^{\infty}_{-\infty}\chi_v(\omega)[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)] e^{-i \omega t}d\omega.$$

From this I find that $\langle v(t) \rangle$ is given by:

$$\langle v(t) \rangle = \frac{f_0}{2} (\chi_v(\omega_0)e^{-i\omega_0 t}+\chi_v(-\omega_0)e^{i\omega_0 t}).$$

Now I used the following to find the dissipation of the energy of the particle:

$$\frac{dE}{dt} = f(t)\langle v(t) \rangle = \frac{f^2_0}{4}(e^{i\omega_0 t}+e^{-i\omega_0 t})(\chi_v(\omega_0)e^{-i\omega_0 t}+\chi_v(-\omega_0)e^{i\omega_0 t}).$$

We can average this over some time period $T$ as follows:

$$\Big\langle \frac{dE}{dt} \Big\rangle = \frac{1}{T}\int^T_0 \frac{dE}{dt} dt = \frac{f_0^2}{4T} \int^T_0 (\chi_v (\omega_0 )+\chi_v(-\omega_0)+e^{2i \omega_0 t}\chi_v(-\omega_0)+e^{-2i \omega_0 t}\chi_v (\omega_0))dt.$$

Since we take $T = 2\pi/\omega_0$ the last two terms give a vanishing contribution to the integral. We therefore find that:

$$\Big\langle \frac{dE}{dt} \Big\rangle = \frac{f^2_0}{4}(\chi_v(\omega_0)+\chi_v(-\omega_0)).$$

Then we see that because of $\chi_v(\omega_0) = \chi^\prime_v(\omega_0)+i\chi^{\prime \prime}_v (\omega_0)$ we can write:

$$\chi_v(\omega_0)+\chi_v(-\omega_0) = 2 \chi^\prime_v(\omega_0),$$ where $\chi^\prime$ denotes the real part of $\chi$, we can do this because the real part is even and the imaginary part is odd in $\omega_0$. But then I find the following:

$$\Big\langle \frac{dE}{dt} \Big \rangle = \frac{f^2_0}{2}\chi_v^\prime(\omega_0),$$

but this is different from what I have to proof.

Question: can somebody spot the difference or help me with this problem?

  • $\begingroup$ Check your formula for $dE/dt$. On the right, $f(t)\langle v(t)\rangle$ must be wrong (has units of energy, not energy/time). $\endgroup$
    – user197851
    Commented Nov 1, 2018 at 18:20
  • $\begingroup$ Hi thank you for the answer. One question: velocity times a force just gives a power right so not an energy? $\endgroup$ Commented Nov 1, 2018 at 18:50
  • $\begingroup$ You are only calling $f$ a force! It appears in your perturbed hamiltonian as a term $-(p/m)f(t)=-vf(t)$. So its units are the same as momentum. A real force would appear as a term $-xf(t)$ or similar. $\endgroup$
    – user197851
    Commented Nov 1, 2018 at 18:54
  • $\begingroup$ You are right, but the problem is that it is stated in the question that way literally. I think they might be mistaken and that $f$ indeed should be interpreted as external momentum.... Thank you professor! $\endgroup$ Commented Nov 1, 2018 at 19:01
  • 1
    $\begingroup$ Often one refers to a "generalized force". I'm not saying the problem lies in the definition of the perturbation term in the hamiltonian; I'm saying, check your formula for the rate of change of energy. I think it needs another time derivative somewhere. More than that, I don't want to say! $\endgroup$
    – user197851
    Commented Nov 1, 2018 at 19:03

2 Answers 2


Since you've set out your solution, I hope it should be OK to offer some context.

Not everybody realizes that there are two ways of deriving this result, and if they have only seen one way, there is a temptation to regard the other way as "wrong". In the following I'll consider the case where the perturbed energy is $E=E_0-f(t)A(x,p)$, and $A(x,p)$ is some function of the system coordinates and momenta. $A(t)$ is a convenient shorthand for $A(x(t),p(t))$. Angle brackets represent averages in the perturbed ensemble.

In A Modern Course in Statistical Physics by LE Reichl, for instance, 4th edition, p262, eqns (7.123)-(7.130) (these same equations appear in differently numbered chapters in earlier editions) you will see dissipation expressed as the rate of doing work on the surrounding medium. I'll change the sign of this, and express it as the rate of doing work on the system of interest (power received) $$ P(t) = \frac{dW}{dt} = f(t)\frac{d\langle A(t)\rangle}{dt} $$ This is the answer you have given, with $A=p/m$.

In Statistical Mechanics by DA McQuarrie, p540, Problem 21-58, and in other places such as HC Andersen's course notes, the dissipation is calculated in terms of the rate of change of energy of the system. In this case, $$ P'(t) = \frac{d\langle E\rangle}{dt} = -\frac{df}{dt}\langle A(t)\rangle $$ Andersen's notes are quite good in giving a simple derivation of this, and in explaining whether we should be considering $E$ or $E_0$ here.

These two definitions, $P$ and $P'$, of the instantaneous power are different from each other.

There are a few places where both expressions are given and compared: for example, Introduction to Modern Statistical Mechanics by D Chandler, p258, section 8.7, and Nonequilibrium Statistical Physics by N Pottier, p390, eqns (14.1.1)-(14.1.11). Chandler points out that, if the two expressions are averaged over a time period, one can integrate by parts, assuming that the boundary terms at $t=0$ and $t=T$ are negligible, which should be true if $T$ is long enough, $$ \overline{P(t)} =\frac{1}{T}\int_0^T dt \, f(t)\frac{d\langle A(t)\rangle}{dt} =-\frac{1}{T}\int_0^T dt \, \frac{df}{dt}\langle A(t)\rangle =\overline{P'(t)} $$ so the averages turn out to be equal. Something similar happens when you integrate over a cycle for an oscillatory perturbation. And Pottier actually says

As shown by formulae (14.1.4) and (14.1.10), the instantaneous power received by the system is not equal to the instantaneous evolution rate of the energy of the system coupled to the field. However, as displayed by formulas (14.1.5) and (14.1.11), these quantities are equal on average. This is why the average power dissipated within the system may also be obtained from the average evolution rate of the total energy of the system coupled to the field.


The mistake in the calculation lies in the fact that $f(t)$ is not a force but a generalized force, meaning in this case it has units of momentum. And therefore the equation for the dissipation rate is not quite right. The average dissipation rate is therefore given by the following:

$$\Big \langle \frac{dE}{dt} \Big\rangle = \frac{1}{T}\int^T_0f(t)\frac{d\langle v(t)\rangle}{dt} dt,$$

(or the other form as stated by the professor above).

From this we find that:

$$\frac{dE}{dt} = \frac{f^2_0}{4}(i\omega_0 \chi_v(-\omega_0) e^{i\omega_0 t}-i\omega_0 \chi_v(\omega_0)e^{-i\omega_0 t})(e^{i\omega_0 t}+e^{-i\omega_0 t}).$$

Averaging over time gives:

$$\Big\langle \frac{dE}{dt} \Big\rangle = \frac{f_0^2}{4i}\omega_0 (\chi_v(\omega_0)-\chi_v(-\omega_0)).$$

Since $$\chi_v(\omega_0)-\chi_v(-\omega_0) = 2i \chi_v^{\prime\prime}(\omega_0),$$

we can write the previous expression as follows:

$$\Big \langle \frac{dE}{dt} \Big \rangle = \frac{f^2_0}{2}\omega_0 \chi_v^{\prime\prime}(\omega_0).$$

Credits to the professor helping me with this question.


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