2
$\begingroup$

I am having a lot of trouble understanding the influence of volume on relative permittivity/static dielectric constant. I'm not a chemist, but am using molecular dynamics to calculate relative permittivity for use in other work.

A predecessor used this equation for relative permittivity $ε_\mathrm{r}$:

$$ε_\mathrm{r} = 1 + \frac{4π\sigma_M}{3k_\mathrm{B}TV}$$

where $V$ is the volume of the simulation, $T$ is temperature, $\sigma_M$ is the variance of the total dipole moment, and $k_\mathrm{B}$ is the Boltzmann constant.

Here is what I do not understand: If I use a different $V $(i.e., a different number of atoms) for my simulation compared to my predecessor, then my calculated relative permittivity will be different from his. I have done this to make sure, and indeed, the difference is quite significant.

Is there something silly I am misunderstanding here? Like "$V$" is not the average box/cell volume of the simulation, but something that is not size-dependent?

|improve this question
$\endgroup$
  • 1
    $\begingroup$ Your equation looks right. It is the same as can be found in this paper, arxiv.org/ftp/arxiv/papers/1009/1009.5958.pdf. I am not certain of this, but I would guess that the variance of the total dipole moment would change when you change the total volume, and thus your result should be more or less volume independent. $\endgroup$ – Erik Kjellgren Oct 20 '19 at 9:05
  • $\begingroup$ Thanks very much, Erik. I wondered if that were meant to be the case. So then does it seem likely to you that when two calculations (each with a different number of atoms) lead to different calculated values, that the system volume is too small for one of the calculations? $\endgroup$ – Antst Oct 20 '19 at 9:07
  • $\begingroup$ And a follow-up question to that for anyone who is interested: would it make sense to run the smaller system for longer to achieve the same relative permittivity value as the larger system? Or is there something more fundamental that's a problem about using a smaller system? $\endgroup$ – Antst Oct 20 '19 at 9:20
  • $\begingroup$ It's not quite clear how you determine the variance of total dipole moment. Is it just a mean square? Could you probably elaborate on this a bit further and add the exact math expression you are using in the equation? $\endgroup$ – andselisk♦ Oct 20 '19 at 10:31
  • 1
    $\begingroup$ Thank you very much for asking, andselisk, and thank you for your edits. The molecular dynamics software I'm using outputs the total dipole moment and I'm using Python/numpty's "variance" function (e.g., "np.var") to calculate the variance. Please feel free to ask if you would like me to add any more detail about this. $\endgroup$ – Antst Oct 20 '19 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.