# At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces?

London Dispersion forces are forces between non-polar non-ionic molecules that the random fluctuation of their electrons makes them temporarily dipoles.

• What is the frequency that this occurs?
• Every how many milliseconds does a molecule like this becomes a dipole?
• Are there such numbers for some specific molecules out there?
• Good question! I don't have an answer offhand, but I suspect you're off by several orders of magnitude in your time estimate -- I would think the electronic fluctuations occur on a femtosecond or picosecond time scale. Commented Jul 29, 2015 at 12:55
• In my point of view oscillations in time are an artifact of the orbital picture. The time independent Schrödinger equation and its exact solutions account completely for London dispersion interaction. The problem occurs when you build wave functions and interaction energies for multi-electron systems using one-electron wave functions and Fock space and so on. This picture can be patched by London dispersion. But the oscillations are pure mathematical model without physical counterpart. Commented Jul 29, 2023 at 7:52

According to Prof. Loren Williams of Georgia Tech, the fluctuations occur on the femtosecond $\left(10^{-15}\ \mathrm{s}\right)$ time scale:

Figure 13 shows how fluctuating dipoles of liquid Xenon (or Helium or Neon, etc) are coupled. Darker blue indicates higher electron density. The fluctuations are correlated and are very fast, on the femtosecond ($10^{-15}$ second) timescale. Adjacent Xenon atoms experience electrostatic attraction from the transient dipoles. Two different representations of fluctuating dipoles are shown.

This corresponds to a frequency of $1\ \mathrm{PHz}$ (petahertz), which is about five hundred thousand times faster than most modern computer processors.

Another estimate of the time scale (albeit one devised by me and thus of potentially suspect validity), is the SI value of the atomic unit for time, derived by dividing the reduced Planck constant by the atomic unit for energy, the Hartree (values are rounded a bit):

$$\left\{atomic\ time\right\} = \frac{\hbar}{E_\mathrm{h}} = \frac{1.05457\times 10^{-34}\ \mathrm{J\cdot s}}{4.3597\times 10^{-18}\ \mathrm{J}} = 2.41888\times 10^{-17}\ \mathrm{s} \approx 0.024\ \mathrm{fs} = 24\ \mathrm{as}$$

That last value is $24$ attoseconds, which is a well and truly short span of time. For comparison, this corresponds to a frequency of $41\ \mathrm{PHz}$.

Whatever the exact value, I have to believe the fluctuations are fast enough so as to be impossible to measure directly. (I would be fascinated to learn otherwise, though!) Thus, even if different atoms/molecules do actually have different frequencies of London dipole oscillation, we have no way of finding out these values.

• This is a nice description of dipole oscillation, but doesn't answer the posted question. They are asking how often formally nonpolar particles form an instantaneous dipole. The particle is polar for nearly the entire femtosecond it takes to flip polarity. Commented Jul 29, 2015 at 13:47
• @DanBurden Uh-oh, looks like this is going to get out of hand quickly. :-) My answer attempts to address the question on its own premise, that the dipoles do disappear. Yours rejects the question's premise, arguing that the dipoles are never (or nearly never) absent. I leave it to the community to vote their preferences. Commented Jul 29, 2015 at 13:51
• This is speculation on my part, but perhaps the polarization time scale depends on the dimensions of the system. I wonder if sufficiently large electron clouds can be polarized in a measurable amount of time. Maybe this could be done with fullerenes, or Rydberg atoms? Commented Jul 29, 2015 at 14:47
• Is there a kind of spectroscopy based on this oscillation? I.e., would liquid xenon absorb electromagnetic radiation at 41 PHz, thereby driving the oscillation? Just curious. Commented Jul 30, 2015 at 12:33
• There is something screwy about this number 41 petahertz. That corresponds to x radiation at a wavelength of about 7.3 nanometers - hence ionizing radiation. That is a lot more energy than I would expect to be required to cause oscillations of electron density. Commented Jul 30, 2015 at 16:40

While dispersion forces dominate in non-polar, non-ionic systems, they are actually felt by all chemical species that contain electrons. But, for simplicity's sake, let's use the example of a non-polar, non-ionic atom or molecule to describe these forces in this answer.

Surprisingly, these molecules are almost never completely "non-polar". There is almost always a dipole moment, but the strength and direction are random. The electron densities around these atoms and molecules are always shifting, which almost always leaves an uneven distribution of electron density around the nucleus.

The London dispersion forces are simply the sum of interactions between the constantly fluctuating dipoles of one particle, and the induced dipoles within the neighboring particles. The influence of the electrostatic force is directly proportional to the product of polarizabilities for each participating particle.

To summarize: Molecules or atoms with no permanent dipole are almost never in a completely "non-polar" state, due to the constantly fluctuating electron cloud. This negates the question of instantaneous dipole frequency, because it is never not a dipole.

• While this is a good description of London forces, it doesn't address the actual question of the frequency of appearance of the dipoles. Commented Jul 29, 2015 at 13:27
• Thanks! I was trying to answer the question by saying that the molecules are never not dipoles. I apologize if this is unclear. Commented Jul 29, 2015 at 13:29
• Yep, that point was unclear, at least for me. Edit made it much better! :-) Commented Jul 29, 2015 at 13:35
• Hmm... so, perhaps a better question is something like: "Below what time scale is time-averaging of electron positions insufficient to be able to treat a nominally non-polar atom/molecule as lacking a dipole?" Commented Jul 29, 2015 at 13:55
• Maybe - I think it would help to define a nonpolar molecule as having a polarity below some threshold. For example, very very small electron density asymmetries can probably be considered nonpolar. However, I think the threshold would have to be dependent on polarizability, temperature, particle size, etc. It's a tricky problem. Commented Jul 29, 2015 at 14:03

The electron rich/ electron poor regions tend to oscillate at a period of about 1-10 femtoseconds.