# Problem with setting Dielectric

Dear Lammps users,
As suggested by @Germain in this post, I’m creating a new thread to discuss the problem.

Currently, I’m trying to reproduce a paper titled " Self-assembly of amphiphilic polyelectrolytes in trivalent salt solution". In the paper the authors have use truncated Lennard Jones Potential and Coulomb Potential as their non-bonded interactions for their polyelectrolytes. Further, they have explicitly mentioned in the SI of the paper that:

In my version of LAMMPS script I’m using a pair_style lj/cut/coul/long, pair style with respective cutoffs. Now, As per my understanding of electrostatic interactions. It depends on the charges and dielectric of the medium one is considering. The equivalent value of charges in the LJ units can be set by choosing some unit scale. As given in the units. In my case the value of charge corresponding to 1 e comes to be 15 q^* in LJ units ( q^* = q/\sqrt{4 \pi \epsilon_0 \sigma \epsilon}, q= 1e, \sigma= 0.24 nm, \epsilon=4.11 \times 10^{-21} J)

I believe the value of desired dielectric can be set in three different ways, either one fixes their charges or one fix their bjerrum length values or Temperature. Given that \lambda_B = 3 \sigma, q=15 4 \pi \epsilon_0=1 in LJ units and kBT=1, at 298K (that’s the \epsilon). So won’t the formula reduces to \lambda_B = (q^*)^2/\epsilon_r ?

However, if I use this formula and evaluate \epsilon_r, it comes to be 75. The value seems to be a bit odd. Because as per the lammps documentation of the dielectric command “a value of 4.0 reduces the Coulombic interactions to 25% of their default strength” so if I’m using the value of 75 as the dielectric that means the coulombic interactions are reduced to 1.3% of their default strength. This seems a bit unphysical to me.

So, somewhere my understanding is flawed. It’ll be really great if anyone can explain it or maybe point me in the right direction.

Thanks

Yours Sincerely
Akshay

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There’s a sentence in your very screenshot stating a relative permittivity (\epsilon_r) of 80 (unitless). This is the experimentally measured value for water. Water is in fact extremely polar and shields interactions between ions very well (if it did not, ionic substances would not dissolve in water), and a straightforward molecular dynamics simulation of ions in water will show that those ions are very sparingly attracted to / repelled by each other relative to their vacuum interactions.

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As stated elsewhere, I have good reasons to think this reasoning is incorrect.

You are actually taking the problem in reverse. The paper takes k_BT=1 with T=298K as an energy scale and the Bjerrum length as a length scale. You don’t have to recompute it. \lambda_B=3\sigma is already in reduced units. As stated in the paper, the relative permittivity of your medium has already be taken into account. You shouldn’t have to modify it. This is part of the reason you charge value is higher.

Another way to put is is that the formula in you screenshot can be rewritten:

U=k_BT\frac{e^2}{r_{ij}}\frac{q_iq_j}{4\pi\epsilon_0\epsilon_r{}}

Where you can clearly see the reduced charges on the right. You can now see (I hope) why k_BT is an energy scale and \lambda_B is actually a length scale.

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You meany the reduced charge values I mentioned in my original post? For a given energy scale and length scale I used q^* = \frac{q}{\sqrt{4 \pi \epsilon_0 \sigma \epsilon}} where q= electronic charge. q* came to be 15. And for trivalent salt, trivalent cation I used 3 \times q* value and monovalent anion I used 1 \times q* value. Further for a given bjerrum length I was trying to evaluate thw dielectric value.

I’ll set dielectric value to be 80 as it’ll take the said bjerrum length into account.

I apologise for my naïveté @Germain.

Don’t apologise, there is no stupid question.

However, you miss the point: the dielectric constant is taken into account in computing the Bjerrum length, which is used as a scaling value.

With the settings from the paper, you are already doing simulations taking into account a medium with a relative permittivity of 80. If you want to take into account a different permittivity, you would have to recompute the Bjerrum length and derive the \sigma value from it.

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Now I understand it. Thank you so much @Germain.

You mentioned this in your earlier post. I didn’t understand what made you say this. As stated earlier, i used the given(from the paper) length and energy scale to evaluate the charge in reduced units.

I think I might understand what @Germain is concerned about, so here’s some advice.

Firstly, if you are not confident using LJ units, then don’t use LJ units. Convert lengths into angstroms and energies into kcal/mol using the stated energy scales, use temperatures in kelvins, set the ionic charges in multiples of e (so a trivalent cation is just +3), set dielectric 80, and run. You don’t have to worry about Bjerrum lengths and reduced charges any more.

Secondly, you should realize that there are two completely equivalent ways to include a relative dielectric constant:

• You can use “effective charges”, where you reduce every charge from its original value q to the screened value q/\sqrt{\varepsilon_r}, so that when evaluated with an unmodified dielectric \varepsilon_r = 1 the forces are reduced as expected.
• Or you can set dielectric XX`, so that when evaluated with unmodified charges the forces are reduced as expected.

But if you change both the dielectric and the charges then you are double-counting the effect (resulting in an effective dielectric of \varepsilon_r^2) and that will be very wrong.

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Thank you so much @srtee, @Germain for your kind responses. I understood where was I going wrong.