1
$\begingroup$

A paper was passed around by our professor. The paper has a derivation of noise in gene expression from Langevin equations. The actual context is not so important, it's just that I think there might be a simple error in the derivation. Here is the part of the derivation:

$$\frac{d \delta r}{dt} + \gamma_R \delta r = \eta_R \, .$$

Fourier transforming these equations by setting $x(t)=\int x(\omega) \exp(i \omega t) d\omega/2\pi$,

$$\frac{\delta r(\omega)}{\eta_R(\omega)} = \frac{1}{\gamma_R + i \omega} \, \quad \left \langle |\eta_R |^2 \right \rangle = q_R \, ,$$

so that the steady state value of the fluctuation is

$$\langle \delta r \rangle = \int \frac{d \omega}{2\pi} \frac{1}{\gamma_R^2 + \omega^2} q_R = \frac{q_R}{2 \gamma_R} $$

$\eta$ is the random noise term. I think there is a square root missing in the derivation, or I just do not see how it went from $\eta$ to $\langle | \eta^2 | \rangle$

$\endgroup$
4
+50
$\begingroup$

There are a couple of mistakes.

Though

$$ \delta r(\omega) = \frac{\eta(\omega)}{\gamma + i \omega} $$

is correct. However, if we assume the noise is $\delta$-correlated in the time domain,

$$\langle \eta(t)\eta(t') \rangle = q\delta(t-t')$$

this leads to similarly $\delta$-correlated noise in the frequency domain

$$\langle \eta(\omega)\eta(\omega')\rangle = 2 \pi q \delta(\omega+\omega')$$

contrary to your notes.

With this the final result is obtained step-by-step as

$$ \begin{split} \langle (\delta r(t))^2\rangle &= \frac{1}{4 \pi^2}\int_{-\infty}^\infty d \omega \int_{-\infty}^\infty d \omega' \,\mathrm{e}^{i (\omega+\omega')t} \langle \delta r(\omega) \delta r(\omega') \rangle \\& = \frac{1}{4 \pi^2}\int_{-\infty}^\infty d \omega \int_{-\infty}^\infty d \omega' \, \mathrm{e}^{i (\omega+\omega')t}\,\frac{\langle \eta(\omega) \eta(\omega') \rangle}{(\gamma+i \omega)(\gamma+i \omega')} \\ &= \frac{1}{2 \pi}\int_{-\infty}^\infty d \omega \frac{q}{\gamma^2+\omega^2} \\ & = \frac{q}{2 \gamma} \end{split} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I might be lacking in knowledge about correlation function, but why does delta function correlation in time domain give $2\pi q\delta(\omega+\omega')$ correlation in frequency domain? $\endgroup$ – Ilya Lapan Dec 20 '16 at 11:01
  • 1
    $\begingroup$ The Fourier transform is pretty much defined by the fact that the Fourier transform of a single $\delta$ function is a plane wave and vice versa. Given this, one way to see the result you are asking about is as follows: You are Fourier transforming in both coordinates, $t$ and $t'$. One Fourier transform turns a $\delta$-function into a plane wave, and the second transform turns that plane wave back into a $\delta$-function. $\endgroup$ – ComptonScattering Dec 20 '16 at 12:03
1
$\begingroup$

There clearly is a square missing. You can see that dimensionally. Moreover you expect $\langle \delta r \rangle $ to be vanishing and only its squared value to be finite.

| cite | improve this answer | |
$\endgroup$

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.