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Does quantization mean there are only so many colors?

  1. May 8, 2013 #1
    Given that a photon of known wavelength is emitted when an electron goes to a lower energy state, if there are only so many different types of atoms, and only so many electron orbits around those nuclei, the number of wavelengths possible is finite: If true, then the EM spectrum is quantised, not continuous.

    ...or does Heisenberg's uncertainty blur the actual wavelengths that are measured? (enough to blur the frequency with its neighbors and provide a continuous spectrum?


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  3. May 8, 2013 #2


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    Red shift alone assures that wavelength is NOT quantized. That MIGHT have to be based on an assumption that spacetime is continuous and not quantized, but I'm not sure.
  4. May 8, 2013 #3


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    There are many other ways that photons can be produced besides the atomic energy levels. Synchrotron radiation, Cherenkov radiation, bremsstrahlung, Compton scattering, and so on. These all produce light of continuous wavelengths. Any time a charged particle accelerates, or collides with another, a photon can be produced.
  5. May 8, 2013 #4

    I probably should have been more specific - I figure red/blueshift could make any wavelength appear to be any other, but I'm wondering more about everyday experience; the light from the sun and from tungsten filaments where relativistic effects are negligible.

    So Bill, what I understand from you is that essentially any time kinetic energy is released as a photon, it depends upon the speed of the original particles, which could be anything - thus giving us a continuous spectrum.

    So, if have understand that correctly, that leads me to two thoughts;

    A) That black body radiation would be quantised to certain wavelengths depending on the composition of the black body because of the nature of the way it emits (ie; tungsten light would be quantised to certain wavelengths and therefore not continuous). But light from the sun would have many particle collisions at random velocities, creating a continuous spectrum.

    B) If the continuous spectrum is made so by the transfer of kinetic energy into light, could a study of photon wavelengths from what should be a continuous spectrum source prove whether or not time is quantised?
  6. May 9, 2013 #5
    They are both continuous because of the broadening mechanisms described above. You only need to look into natural broadening to find that any single transition is still a continuous spectrum because of the necessarily finite lifetimes of both lower and upper states in the transition. These finite lifetimes limits the precision to which their energies can be known, which limits the precision to which the transition frequency is known. So even a sodium lamp emission, which is extremely narrowband, is still a continuous function over frequency space.

    A tungsten lamp has an astronomical number of vibration modes (it's effectively one huge molecule), so its spectrum will contain an astronomical number of transitions, each one broadened by the above mechanism (if not others). The result is a true continuous spectrum. Now it might not be a true blackbody, but it still is continuous.

    This calculation is probably way WAY off, but in general, the number of vibration modes of a molecule of N particles is 3N. So if you take a 1g tungsten filament, that contains NA/183 tungsten atoms with (3*NA/183) vibration modes. If you give each mode a quantum number from 1...10 then there are (3*NA/183)^10 different quantum vibration states. If you (crudely) ignore forbidden transitions then there are ((3*NA/183)^10)*((3*NA/183)^10 -1)/2 different transitions.

    That's 10^439 vibration transitions alone.

    So there are quite a lot of discrete transitions; so many that if you pack these onto a spectrum with uniform spacing, I'm sure you can show that the spectral distance between any two adjacent transitions is far FAR smaller than the half-width of the natural broadening curve- the spectrum is certainly continuous, and furthermore, it is likely to be fairly smooth.
  7. May 9, 2013 #6
    Thanks Mike,

    Makes sense. So if I have you right, the uncertainty principle means that no two photons emitted from an electron decaying to a lower energy state are the same wavelength exactly. And then there are other processes that can modify that wavelength even more.

    After I wrote this I figured another reason why time couldn't be quantised - it would create a nyquist frequency above which no wavelength could be coherently represented by time quanta.
  8. May 9, 2013 #7


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    Whenever I see a question such as this (and we HAVE seen many of them on here), I think that there is a complete misunderstanding that light can only be generated via atomic transition. This, of course, is severely wrong, as has been mentioned above.

    When I used to conduct undergrad labs, one of the things I like to do when students were using a spectrometer to look at the emission lines from gasses, is to also have them look at the spectrum from an ordinary, incandescent light bulb. Here's an example.

    This is a typical spectrum from a hydrogen discharge tube showing the Balmer lines:


    This is an example of atomic transition, where you see discrete spectral lines. Now, compare this to the spectrum from an incandescent light bulb:


    One can immediately see a clear difference between the two. There is more of a "continuous" blend of colors or spectrum, meaning that the light emission here covers many, many wavelengths that are practically continuous.

    {Credit: http://www1.assumption.edu/users/bniece/spectra/lightsources.html"

    This is because the emission mechanism for these two light sources are very difference. Light just doesn't HAVE to be emitted via atomic transition. It can also be emitted via vibration, agitation, etc. of charged particles. That is how we generate light at synchrotron light sources, for example.

    So this misleading idea needs to be corrected.

    Edit: Argh! The images are so big, it is messing up the formatting. I'll copy those images, resize them, and then will link back to it when I have some time after work today. Sorry about that, folks.

  9. May 9, 2013 #8


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    Nema problema Zz. In the excellent new PF layout it only affects your post (and not as before when the whole page was ‘messed up’). On my screen it looks fairly decent both in IE & Firefox.

    Thanks for interesting/beautiful pictures!

    Maybe this is slightly OT... and it looks like everyone are aware, anyhow we should maybe point out that color & wavelength are two different things, where the former is generated in our brains as combination of the second.

    And not only that, we can’t always trust our gray clump. The color pink (magenta) is something solely produced in the brain as ‘stand in’ for the combination of the ‘extremes’ red + violet, which would normally produce something halfway between red and violet in the spectrum, which would be green.

    The Brain however does not fancy the green solution, so it invents a new color, PINK!

    Try to quantize that little tuber... :smile:
  10. May 9, 2013 #9
    I don't know anything about quantisation of time, but you have it right about natural broadening. Doppler mechanisms are important for gaseous absorbers even at room temperature (the gases move pretty fast and even small Doppler shifts can cause significant broadening, particularly at high frequencies).

    It's also important to remember that any practical measurement of a spectrum is a convolution of the spectral irradiance on the measurement aperture with some spectral responsivity curve of the measuring device itself. So even if you did have a stupidly narrowband source, you would still measure a continuous spectrum because the device itself will output a (small) signal even if it is calibrated to an off-linecentre frequency.

    In other words you never measure the absorption at a single frequency, you always measure its average over a narrow range. It's the same as how you'd never infer a probability simply by taking the height of a probability density function- you need to integrate it over x+dx to get an finite probability.
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