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When I was a kid, I asked my dad about them and he said (not in so many words) that it was because sunlight and artificial light have different spectrums and they picked colors that reflect only the wavelengths that are contained in sunlight but not artificial light.

It was a good enough explanation as a kid but I've since learned that when he doesn't know an answer to a science question he makes something up that sounds good, so now I'm not really convinced. If it were a matter of the spectrum of the light, wouldn't some colors appear under incandescent light that don't appear under fluorescent light, or vice-versa? And why is it that there don't seem to be any natural things (that I know of) that exhibit this same behavior?

It seems to me, it would have to be that the ink (or whatever coloring substance) is specifically designed to react to sunlight--maybe chemically?--instead of just being in colors that don't appear in artificial light.

But I'm just guessing. (Like father like son?) Does anybody know how these work?

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    $\begingroup$ In addition to these other answers specifically for del-sol, I've also worked with thermochromic and photochromic pigment that changes color when exposed to heat/light. Many of these kinds are colored when cool and turn clear when warm. It's possible to print a colored design on the shirt and cover it with white photochromic pigment so when in the sun, the pigment clears and allows the colors to be seen. There were lots of clothing in the 90s made with this stuff, always had pink armpits on blue shirts ;p $\endgroup$ – coblr May 24 '16 at 22:12

A quick look at the Del Sol How it Works page shows their explanation,


The Spectrachrome® crystal reveals color upon irradiation by ultraviolet waves; i.e., sunlight. When a flower blooms, the result is the exposure of the inherent color of the flower. A Spectrachrome® crystal is similar in that an energy-shift occurs causing the color of the dye to become visible to the human eye. The shifting or "twisting" of the dye is referred to as a molecular excitation transition. The dye does not actually "change" color; rather, it becomes visible to the human eye. Research shows that some animals; e.g., certain species of bats, can actually see the color of a Spectrachrome® crystal in its inactive state.


Although each Spectrachrome® crystal operates at a slightly different wavelength, the optimal wavelength is 365 nanometers.

which is about reasonable. Note that there doesn't need to be any particular correlation between the wavelength of light that activates a certain ink colour and the colour of that ink. Most artificial lights are pretty thin on the UV (by design - it's not something we want to be around all the time), but black lights are readily available if you want to test this experimentally.

In particular, this means that the explanation in the OP,

sunlight and artificial light have different spectrums and they pick colors that reflect only the wavelengths that are contained in sunlight but not artificial light.

is not quite right. It's not that the spectrum of sunlight is reflected; instead, sunlight contains spectral components that 'activate' the ink so that the dye becomes visible.

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    $\begingroup$ This would mean his father was quite right, sort of. $\endgroup$ – joojaa May 23 '16 at 22:14
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    $\begingroup$ What this seems to imply is that the dye molecule has an energy state about 3.4 eV above the ground state, which it can get excited into by the 365nm "activating" UV, and additional states between 5 eV and 6.5 eV that enable it to absorb visible light once it has been kicked into the 3.4 eV state. But visible photons contain too little energy to reach any of those upper state when the dye is in the ground state. $\endgroup$ – hmakholm left over Monica May 24 '16 at 0:15

There is an explanation on the Del Sol web site but it omits the technical details. This is probably because Del Sol regard it as commercially valuable confidential intellectual property, and I have to concede that they are probably correct.

Anyhow, some Googling has turned up a suggestion for how it works but there is no proof for this idea so treat it as just a hypothesis until such time as Del Sol reveal their technology.

The hypothesis is that colours are due to fluorescence. The dyes use absorb light and re-emit it at a different wavelength giving the colours we see on the clothes. The colour change happens because the dyes can exist in two states, one of which fluoresces at uv wavelengths and the other at visible wavelengths. The change between the two states is triggered by ultra-violet light.

Artificial light contains very little ultraviolet so the dyes revert to the state in which they do not fluoresce in the visible wavelengths. Sunlight contains significant uv and in sunlight the dyes change to the form that fluoresces at visible wavelengths. That's why the colour appears only in sunlight.

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  • $\begingroup$ +1: First thing I thought when I read the Del Sol explanation was "isn't that just an obtuse description of fluorescence?" $\endgroup$ – RBarryYoung May 24 '16 at 13:43
  • $\begingroup$ Also, don't fluorescent lights emit UV? $\endgroup$ – RBarryYoung May 24 '16 at 13:44
  • $\begingroup$ @RBarryYoung fluorescent lights produce UV internally, which is converted to visible light by the phosphorous coating on the inside of the glass tube. Also, glass is not usually very transparant to UV, and it would make sense for fluorescent lights to be made with UV-opaque glass. As far as I know, virtually no UV makes it out of the tube. $\endgroup$ – marcelm May 24 '16 at 19:42
  • $\begingroup$ UV-C fluorescent tubes have clear glass and when lit they have a strange fuzzy blue color that is not pleasant to look at ("So don't do that!"). Obviously some UV-C makes it through the extremely thin glass or they would not make these, for use in Bug Zappers and sanitizing lamps. $\endgroup$ – user95006 May 25 '16 at 0:29
  • $\begingroup$ Ordinary fluorescence does not seems to address the point that the fabrics appear (based on the images here) to be darker in sunlight -- that is they start absorbing more of the visible spectrum, not emitting additional colors! $\endgroup$ – hmakholm left over Monica May 25 '16 at 9:02

To discriminate between the other two answers. Lifetime spectroscopy would be capable of distinguishing between photochromism (in this case, reversible ultraviolet-switchable reflectance, example: hexaarylbiimidazole where the transition time is milliseconds to seconds) and ultraviolet fluorescence (where the "transition time" is zero, although the excited fluorscent state probably has a lifetime of nanoseconds or shorter).

One way to implement this is with a small black light and a large high speed box fan. The fan acts as an optical chopper. Shining the light through the blades of the fan at the shirt, if the effect is fluorescence, the color will immediately appear every time the shirt is illuminated. If the effect is photochromism, the color will appear if the fan is set to a slow enough rotation speed (perhaps stopped, if the transition time is "long").

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  • $\begingroup$ Fluorescence can be slow enough that this wouldn't work, but if you turned the ambient light off that would make it obvious. $\endgroup$ – Chris H May 24 '16 at 8:27
  • $\begingroup$ @ChrisH : Are you sure you're not conflating fluorescence with phosphorescence? Only once have I had to sample a fluorescence lifetime slower than 1 MHz (i.e., lifetime greater than 1 $\mu\text{s}$) and that turned out to be a sample prep error that included a phosphorescent taggant. (This is not my work, but it is a typical example.) $\endgroup$ – Eric Towers May 24 '16 at 8:35
  • $\begingroup$ Yes, semi-deliberately if sloppily (from reading too much about (fluorescent+phosphorescent) dyes this morning, plus work involving long-lifetime PL out to ms). $\endgroup$ – Chris H May 24 '16 at 9:23

I believe the dyes behave similar to glow in the dark materials; They fluoresce a certain wavelength over time after being radiated.
In the case of these shirts you'll have different chemicals for different colours, but which are all "activated" by UV radiation.

There is a pretty nice explanation on Wikipedia, but basically this:

You can excite an electron by providing it with a high enough packet of energy - i.e UV rays for example. This means that the electron has more energy, and moves to a higher energy state (often thought of as a higher orbital). What usually happens is that that electron drops down to its usual energy state pretty much immediately, which gives off light (a vibrating electron generates photons). A nice example of this are those camping gas-lights: put in heat, electron excites and de-excites immediately, but constantly, giving you a nice constant bright light)

However, for phosphorescence, the electron only drops down some time later, which means that if you put in energy now, it gives off light until later, albeit not so bright.

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  • $\begingroup$ Ordinary fluorescence does not seems to address the point that the fabrics appear (based on the images here) to be darker in sunlight -- that is they start absorbing more of the visible spectrum, not emitting additional colors! $\endgroup$ – hmakholm left over Monica May 25 '16 at 9:06

As has been commented by fractalspawn this is almost certainly a photochromic effect (https://en.wikipedia.org/wiki/Photochromism). In this process a molecule in form A can absorb a visible or UV photon and be isomerized into another form B. This will have a different absorption spectrum than the starting material and if this absorbs in the visible part of the spectrum a color change is observed.

In photochromic molecules, a different coloured photon can change B back to A. It might also be the case that thermally B can make the transition back to A. Without knowing more about the compounds in the T-shirt one cannot be certain. Similarly the reaction rate constants cannot be known, but in many of these types of compounds isomerisation rate constants can be $>10^9 s^{-1}$.

It is always possible for fluorescence to occur from species A or B but this has to compete with reaction (making the yield small) and would be very hard to observe in bright light whereas a color change due to absorption is easily observable.

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