# Molecular simulations of proteins: how good an approximation is electrostatics?

This question has to do with molecular dynamics simulation of molecules, especially proteins, using software such as GROMACS.

This type of software uses empirical force fields (eg AMBER) and some constraints on covalent bond length and geometry; the simulation proceeds in steps of typically a few femtoseconds and can cover tens to hundreds of nanoseconds at a time.

The forces considered here are pure electrostatics; there is no magnetic or coupled electric-magnetic component. Of course since there are electric charges in the simulation (eg polar groups, molecular dipoles, ...), and they move, and moving charges produce magnetic fields which exert force on other moving charges, the simulation is going to miss some things.

The question is: how good of an approximation is electrostatics in this context?

This is about developing some intuition and good order of magnitude estimates, not about any kind of precision. I think the pieces of this are: How large are the charges? and How fast do they move? (due to thermal motion, or when transitioning between stable conformations) and consequently: How strong is the resulting magnetic field, and what is the ratio of electrostatic force between charges to the magnetic force between moving charges?

As a biologist, the kinds of proteins I would be interested in especially as model systems are voltage-sensing domains (VSDs) of transmembrane proteins, and tubulin/microtubules. Both of these have large conformational changes between different states, and have mobile subunits with large dipoles. Looking up VSDs, it seems the distance of movement between states can be 10-40 A (1 and 2), the macrodipoles of subunits of the protein can be 20-25 Debye (3), and the motion time scale is nanoseconds to milliseconds (4 and 5).

• 10 angstroms in a nanosecond is sloooooooooooow (from an atomic perspective). For comparison, the 1s electron in hydrogen moves at about 0.5 angstrom per atomic unit of time (24 attoseconds), and that is still 137 ($=1/\alpha$) times slower than the speed of light. The motion you're describing is some seven to thirteen orders of magnitude slower. – Emilio Pisanty Dec 15 '19 at 10:22
• @EmilioPisanty Interesting! I'm trying to build an intuitive picture and this helps. It seems the electron motion "current" in a hydrogen is balanced by an opposing motion 50-100pm away, while the moving protein subunit effectively has a few electrons worth of charges separated by 2-3nm or so (and not necessarily moving in the same direction). So the B field wouldn't cancel nearly as much. Does that sound right to you? – Alex I Dec 15 '19 at 20:40
• Not really, no. In hydrogen the current is circular (or at least it can be) and the magnetic effects don't cancel out - they provide the spin-orbit coupling at fine-structure level, which is on the fifth or sixth significant figure of the eigenenergies. LS coupling can be stronger than that, though, but my gut feeling is that it will be the first magnetic effect you'll see, by a long shot. If it's not there, you can ignore magnetism, I would say. – Emilio Pisanty Dec 15 '19 at 23:27
• Hi @AlexI , great question. There is a proposal for materials modeling SE site currently in the commitment phase. It would be great if you could support the proposal by committing to it. – rashid Feb 27 '20 at 13:45

The magnetic field of a moving charge can be obtained from the expression of the Liénard–Wiechert potential and it can be expressed as a function of the corresponding electric field as $${\bf B}({\bf r},t) = \frac{{\bf n}(t_r)}{c} \times {\bf E}({\bf r},t)$$ where $$t_r = t- \frac{\left|{\bf r}-{\bf r^{\prime}}\right|}{c}$$ is the retarded time, $${\bf n}$$ is the unit vector pointing in the direction from the source and $$c$$ the speed of light.
It is clear from the above formula, that the Lorentz force on a charge moving with speed $$v$$ cannot be larger than $$\frac{v}{c}E$$, therefore the ratio between magnetic and electric force cannot be larger that $$\frac{v}{c}$$.
In a thermalized system $$v\sim \sqrt{k_BT/m}$$, therefore $$\frac{\text{magnetic force}}{\text{electric force}}\sim\frac{v}{c}\sim \sqrt{\frac{k_BT}{mc^2}}\ll 1$$ Even in the case of a proton, the ratio is of the order $$10^{-5}$$, at room temperature. Taking into account that effective force fields used in computer simulations have uncertainties orders of magnitude larger, and that numerical integration algorithms introduce in turn additional, larger inaccuracies, I do not see any reason to be worried about neglecting magnetic forces.
• @AlexI Once the field has been generated by one charge, it is there and any other charge will experience a force $q v \times B$. – GiorgioP Dec 15 '19 at 18:54