3
$\begingroup$

For a given protein, I know that the NMR spectrometer magnet generates a field $B_0$ and that the interactions with the spins in the local environment generates a much smaller field $B_\mathrm{loc}$ (not necessary aligned with $B_0$).

Owing to Brownian collisions between solvent molecules and the protein, the atoms associated with these spins move, hence making $B_\mathrm{loc}$ a function of time ($B_0$ is constant).

We define a time correlation function$$G(t,\tau) = \overline{B_\mathrm{loc}(t)B_\mathrm{loc}(t+\tau)},$$ where $G(t,\tau)$ is a stationary random function (see Time evolution of correlation functions (specifically Onsager's hypothesis) in time correlation link)
Hence $G(t,\tau)$ only depends on $\tau$, the delay in measuring $B_\mathrm{loc}$.

So, for simplicity we set $t = 0$ and note that

$$G(t,\tau) = \overline{B_\mathrm{loc}(0)^2} \mathrm{e}^{-\tau/\tau_\mathrm{c}},$$

where $\tau_\mathrm{c}$ is the correlation time, the time it takes for the whole molecule to rotate by 1 radian in a process called rotational diffusion.

How do I derive the second equation from the first?

$\endgroup$
1
  • $\begingroup$ You can't prove this, simply because it is not generally true. But it still makes sense to assume this functional form because it seems to fit to many systems. $\endgroup$
    – Nino
    Mar 24, 2017 at 15:34

1 Answer 1

4
$\begingroup$

Your function $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle$ is an autocorrelation function. If it is only a function of $\tau$, the time delay, and not t, this is called the stationary assumption.

The fluctuating field has zero average $\langle B_x(t) \rangle =0$. The bracket implies averaging over a long time for a single spin or average over many spins at a particular time, which is the ergodic hypothesis.

The magnitude of the fluctuating fields is the mean square fluctuation $\langle B_x^2(t)\rangle \ne 0$ and is positive thus the mean square is not zero.

It is also necessary to know how rapidly the fields fluctuate and this is where the autocorrelation $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle \ne 0$ comes in. We compare the field at one point t with another point $t+\tau$ some time later. If $\tau$ is small then the value of $B_x(t)$ and $B_x (t+ \tau )$ are similar and $B_x(t)B_x(t+\tau)$ is large and positive, by comparison if $\tau$ is large the product $B_x(t)B_x(t+\tau)$ is small approaching zero as $\tau$ increases.

Thus the general shape is that $G(\tau)$ is large at small $\tau$ and decays away to zero with increase in $\tau$. Often we assume that the decay is exponential thus $$G(\tau)= \langle B_x^2 \rangle \mathrm{e}^{-|\tau|/\tau_\mathrm{c}}$$ where $\tau_\mathrm{c}$ is a correlation time. This has the correct qualitative form but is difficult to justify on the basis of proper theory.

You state that $\tau_\mathrm{c}$ is caused by rotational diffusion thus it will, in general, be small in fluid solution and large in viscous ones. Of course $\tau_\mathrm{c}$ also depends on the temperature (smaller at larger $T$) and size of the whole molecule's rotation (as a prolate or oblate ellipsoid) or of a mobile group therein, depending upon which property you are observing. Groups in a protein can undergo 'wobbling in a cone' type motion in addition to rotation of the whole protein. Note also that if the molecule is not spherical then it will have more than one rotational relaxation time so $G(\tau)$ may decay with more than one $\tau_\mathrm{c}$. Hope that this helps!

$\endgroup$
6
  • $\begingroup$ Thanks, so no way to prove that assumption? I am assuming spheric molecule and there is an average $\tau_c$ for whole protein. $\endgroup$
    – HighEnergy
    Jul 21, 2016 at 20:04
  • $\begingroup$ And if you could explain the real significance of the spectral density function I get when I Fourier Transform $G(\tau)$ that would be great. It is of the form $J(\omega) = \frac{2\tau_c}{1+\omega^2\tau_c^2}$ $\endgroup$
    – HighEnergy
    Jul 21, 2016 at 20:06
  • $\begingroup$ I am trying to prove the exponential quality of $G(\tau)$, so all help would be appreciated. $\endgroup$
    – HighEnergy
    Jul 21, 2016 at 20:13
  • $\begingroup$ @HighEnergy The Fourier transform of an exponential is the Lorentzian function $J(\omega)$ so its the same thing as $G$ but in the frequency domain. It may make it easier to interpret to look at it this way. You should also probably look at the Weiner-Kninchin relations as these allow you to find the power spectrum (or spectral density) of an autocorrelation. $\endgroup$
    – porphyrin
    Jul 22, 2016 at 7:47
  • $\begingroup$ You gave me information that I needed (and knew somewhat since I had looked at the Weiner-Kninchin relations). I am curious for a given frequency what $J(\omega)$, the spectral density function, means. Also, could I use the fluctuation-dissipation theorem to show that because NMR relaxation (dissipation) is exponentially decaying the time correlation function of $\mathrm{B_{loc}}$ (fluctuation owing to Brownian collisions) should as well? That is my ultimate goal. $\endgroup$
    – HighEnergy
    Jul 22, 2016 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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