in an online video lecture,(around 36min, where the exactly statement is at 36min33secs.) i got one question, suppose we have a system of $N$ particles, $\left\{ {{{\vec r}_i}(t)} \right\}i = 1, \cdot \cdot \cdot ,N$ are the position vectors of the particles. I was told in the lecture that the so-called self intermediate scattering function is defined as. $${F_s}(k,t) = \frac{1}{N}\left\langle {\sum\limits_{i = 1}^N {{e^{i\vec k \cdot [{{\vec r}_i}(t) - {{\vec r}_i}(0)]}}} } \right\rangle.$$ (for homogeneous system, it only depends on the absolute value of $\vec k$.)

Furthrmore, it is said by the lecturer that when $k \to 0$, $${F_s}(k,t) \to \frac{1}{N}\left\langle {\sum\limits_{i = 1}^N {{{[{{\vec r}_i}(t) - {{\vec r}_i}(0)]}^2}} } \right\rangle$$

but i can't see why. Could anybody give me some help on it.


1 Answer 1


Start from this expression :

$$e^{ikx}=\cos kx + i \sin kx$$

As $kx \to 0$ we get $\sin kx \to kx$ and $\cos kx \to 1-\frac {(kx)^2} 2$

Now from listening to that section of the video I do not think that the lecturer is saying $e^{ix}\to x^2$ but is saying that there is a contribution in the low $kx$ range that includes a squared factor, which is coming from the expansion of $\cos kx$ in the low $kx$ range.

The squared quantity expectation value is referred by him as the mean squared displacement, but it is not equated to $F_s(k,t)$ (at least not in the section of video you refer to).

  • $\begingroup$ Thanks StephenG! I think I misunderstood what the lecturer said. $\endgroup$
    – FaDA
    Oct 4, 2019 at 9:53

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.