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Spectral emission by elements occurs in discrete wavelengths. There are only about 118 elements. My question is whether there are pure colors either disfavored in nature, or even disallowed?

The sun (via prism) appears to generate a continuum of visible frequencies, but of course that would probably appear to be the case because the eye would mix adjacent spectral colors and the resultant (near-spectral) color would seem to fill in the gap (note: that is a question, not a statement).

It is often recited that "black bodies" (like the sun) emit a spectrum of radiation, but how would we notice small discontinuities, if they existed?

Red-shift might create visible frequencies that otherwise do not occur. But do we have some basis other than (prism + visual intuition) that all pure visible frequencies do in fact correspond to some possible natural or artificial source?

Apologies for the naivete of the question, if that is the case, and the likelihood that the answer may not have much practical significance. There were related question on the site and on Google but nothing directly on point.

Thanks.

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  • $\begingroup$ Theoretically, any frequency (at least, up to some upper bound that's much larger than the frequency of visible light) can be produced with synchrotron radiation. $\endgroup$ – user27578 Mar 5 '14 at 2:30
  • $\begingroup$ @dgh: Thanks. Looks quite theoretical. I see that it is a light source...will read further. $\endgroup$ – daniel Mar 5 '14 at 2:32
  • $\begingroup$ @daniel Its not 'quite theoretical' as it is more quite real. We've been using it for centuries. However, a more prominent light source, we have blackbody radiation. The light from the Sun, stars, and light bulbs are all from black body radiation. And blackbody radiation produces continuous spectrum. $\endgroup$ – resgh Mar 5 '14 at 3:09
  • $\begingroup$ @namehere: I see it recited that blackbodies emit spectra--and the sun is one. So my question is how can we be sure there is not at least one missing frequency. How would we notice tiny discontinuities? I think dgh might have an answer but I don't know enough about synchrotrons to form a judgment. $\endgroup$ – daniel Mar 5 '14 at 3:31
  • $\begingroup$ @daniel I see. Maybe you could make that clearer in the question. And I'm sorry, the 'centuries' should've been 'decades', referring to synchrotron radiation. $\endgroup$ – resgh Mar 5 '14 at 3:37
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You are right that the sun appears to emit a continuous spectrum, and that's due to the fact that it's a black body to reasonable approximation. We don't even have to consider the spectral response of our eyes, or the color mixing in our brains. Measurements with 'cameras' of one type or another are routinely made to produce plots like this. Discrete spectral emissions from atoms like you mention can also be seen in these plots as spikes that add to the continuous black body curve.

Interestingly enough, you ask about 'discontinuities'. We do notice dim spots when we look at spectra. However, these occur because somewhere between where the light was created and the observer, atoms are getting in the way that are selectively absorbing those wavelengths. In fact, this spectral absorption is useful for telling us about the atomic composition of interstellar space. The point here is that it's not that these wavelengths are missing from nature, it's that something is absorbing them.

Another hint that there is no special wavelength of light that doesn't exist in nature is synchrotron radiation. The spectrum of synchrotron radiation depends on the strength of a background magnetic field, and the energy of the charged particles that are emitting the light. Both of these are continuous quantities, so one would not expect to see a specific wavelength in the middle of a synchrotron spectrum which is dark.

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  • $\begingroup$ Well-written, +1. The plot is interesting. But saying, "black bodies emit a spectrum, the sun is a black body, therefore the sun emits a spectrum" only shifts the question to black bodies; and can't a camera miss a discontinuity? I cannot judge your argument about the synchrotron. If you can flesh out your arguments to avoid circularity I'd probably accept this. $\endgroup$ – daniel Mar 6 '14 at 14:06
  • $\begingroup$ Like most things involving temperature, the answer to your question lies in statistical physics. One way to ask your question is, "What is the probability that I wan't see a photon between energies E and E + dE?" Then to answer the question you have to go back to your quantum mechanics and statistical physics. Since spectral imagers all have some non-zero energy resolution dE, this is also how one would think of experiments. Granted, I haven't done this calculation, but I'm willing to bet this probability is vanishingly small since the spectra that we image are continuous. $\endgroup$ – kristofor Mar 6 '14 at 19:43
  • $\begingroup$ Thanks for this additional comment. The first sentence sort of says it all. $\endgroup$ – daniel Mar 7 '14 at 1:13
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In theory, there are not any disallowed wavelengths from a perfect blackbody. In practice, at the current resolution, spectra that have been observed in nature do not show any gaps. There are absorption and emission lines on most spectrum, of course, but they are due to deviations from black body idealizations, such as gas in front (or within) of the light source (that is not in equilibrium with it).

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  • $\begingroup$ Can you explain the phrase "current resolution?" Thanks. $\endgroup$ – daniel Mar 5 '14 at 21:12
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    $\begingroup$ I mean whatever is the minimum amount of detail that can be observed with the better spectrometer. That is, the frequency difference that corresponds to two adjacent pixels in a spectrum (but I do not know how much that amount currently is). $\endgroup$ – user16007 Mar 6 '14 at 20:17
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In addition to the points that others have brought up about blackbodies, it should be noted that:

  1. with the Doppler shift, you can create any frequency you'd like just by changing the velocity of the light source

  2. Generally, transitions between quantum states are only approximately restricted to discrete energies. Even an isolated atom in an excited state will be able to emit light at energies slightly different from those between the quantum states, with a linewidth that depends on the lifetime of the transition.

For more information about line broadening, see http://en.wikipedia.org/wiki/Line_broadening#Line_Broadening_and_shift.

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  • $\begingroup$ I think "red shift" covers the Doppler effect but +1 for line broadening. Thanks. $\endgroup$ – daniel May 5 '14 at 9:30
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Is your question do colors exist in nature? Would spectral colors not exist only light, and therefore not in "nature"? The terminology for "nature" is confusing, do mean in material existence or matter? Eyes contain cells to decipher wavelengths. The sun is black because the body of the star itself doesn't emit light, the chemical reaction does. Colors are subjected by us, perceived and measured only through human capacity. Photons are a carrier of electromagnetism, so wouldn't it be impossible for us to conclude how this energy "dyes" the light, since even the origin of its existence is unknowable? Plants are green because of the chlorophyll of the chloroplasts, and our skin is pigmented by melanin relative to chemical reaction of ultraviolet light. Any failure to visually perceive anything does not deny its existence. I don't know a whole lot about physics, but I do know about philosophy and my science here is not wrong; I just thought it might be nice to have a broader view to step back into.

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  • $\begingroup$ I mean, sure there are coordinated correlations between an electron (or even an object) and the light perceived from it, but color can only be a property of light $\endgroup$ – C_____ May 5 '14 at 3:28

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