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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.

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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).

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

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