I was playing around with a comb, my hair, and pieces of paper when it occurred to me that inducing a dipole on the pieces of paper meant that many valence electrons previously on the side closer to the negatively charged comb were now on the side farther from the comb. This implies that the ionization energy and the energy needed to change an electron's energy level changed on every ion that was previously an atom.

This further would imply that the wavelengths of light absorbed by the newly made ions is now different, which means that the wavelengths most reflected towards my eyes should be different as well.

So this then begs the question: Why don't I see any difference in color when electric dipoles are induced?

The easy answer is that it does but not enough to be noticeable, but I haven't seen this effect in anything I've induced a dipole in.


You haven't seen these effects because your eyes are rather insensitive to wavelength changes. A moderately resolving spectrometer can detect these changes quite easily. It's called "Stark effect" and it can be observed in atomic spectroscopy: https://en.wikipedia.org/wiki/Stark_effect. For magnetic fields the analog is called "Zeeman effect": https://en.wikipedia.org/wiki/Zeeman_effect.

If the objects are conducting, these atomic physics effects would be masked by electronic conduction effects, which are different but still measurable. If you want to see what local fields can do to metals, look no further than "surface plasmon resonance": https://en.wikipedia.org/wiki/Surface_plasmon_resonance, which is now a widely used biochemical analysis technique because of its ease of application, its sensitivity and the ability to modify it to many different molecular detection systems.

  • $\begingroup$ Thank you! Would it be observable in something that is reflecting a wavelength of light that is at the threshold between what is absorbed and what isn't absorbed in an atom's spectrum? i.e. shining a laser of red light at a surface that gets reflected by the atom but absorbed by the ion? $\endgroup$ – Striker May 14 '16 at 20:40
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    $\begingroup$ @KennyDuran: You mean intensity modulation by a steep optical filter, like what they are using to detect the Doppler changes in light of stars with planets? I believe that would be plenty sensitive to show these changes. I would have to look at the order of the coefficient in both effects, but I am fairly sure they can be seen rather easily with much less resolving equipment. $\endgroup$ – CuriousOne May 14 '16 at 20:45
  • $\begingroup$ I believe that would work, but what I'm suggesting is more for a cheap home experiment to detect the difference with the human eye. The example would ideally have a single wavelength of light shined onto a surface that would then scatter and be visible with the human eye. Next a dipole is established, and at least one of the two ions in the surface would absorb that wavelength rather than reflect it. Thus the Stark effect is observed by a lack of light because the ion's emission spectrum absorbs wavelengths that the atom's didn't. $\endgroup$ – Striker May 14 '16 at 21:11
  • $\begingroup$ @KennyDuran: Why does it matter to you if you can detect something with your own eyes? There is no particular merit in that. An eye is an optical detector of so-so sensitivity and resolution. It doesn't give you any philosophical legs up over a machine that converts light with better sensitivity and frequency resolution. After looking at some of the experimental descriptions it doesn't seem like an easy experiment. $\endgroup$ – CuriousOne May 14 '16 at 21:25
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    $\begingroup$ Oh, I see. :-) The difficulty lies in other parts of the experiment like temperature stabilization and one has to compensate for the magnetic fields because the magnetic Zeeman effect is much stronger, I believe. Getting the electric field high enough is a problem and the lines are broadened, so that the effective separation is more of a dip in the spectrum, it seems. I haven't seen this experiment since I was in university, so forgive me that I am rusty about the details. In any case, it's not a spectacular return for the amount of hardware that's needed, I'd say. $\endgroup$ – CuriousOne May 14 '16 at 22:11

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