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Wien's Displacement Law frequency vs wavelength

  1. Jan 16, 2009 #1
    I hope this is the right place. Since the law is derived from statistical mechanics, I assumed this was the place to post. I understand how the Wien's Displacement law is derived. What I don't understand, other than mathematically, is why there is a difference between the peak intensity as measured by wavelength and frequency. Wikipedia does as good a job as anyone in providing the background: http://en.wikipedia.org/wiki/Wien's_displacement_law

    Seeing as each wavelength has a one-to-one correspondence with a particular frequency, I would expect that the peak would occur at the same place when plotted against either. Again, I understand how, from the math, we find that it doesn't. What is wrong with my thinking?

    If I were to go do an experiment and, let's say, filter out all light except for a particular frequency/wavelength, and do this repeatedly across the spectrum, which plot would I expect and why?

    Thanks.
     
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  3. Jan 16, 2009 #2

    mgb_phys

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    There isn't a difference, you can just divide C by the curve to get the curve for frequency.
    The only reason it's quoted in terms of wavelength is that for reasonable temperature objects you are in a region where you normally work in wavelengths (UV-IR) rather than energy (x-ray) or frequency (radio)
     
  4. Jan 16, 2009 #3
    Can you please elaborate? I'm certain that the formula [tex]c=\lambda\nu[/tex], does not relate the peak of the frequency curve to the peak of the wavelength curve. You can check by attempting to calculate the temperature of the sun given the form found using wavelength and the form using frequency. If that formula worked, the temperatures would agree, but they do not.

    Start with any arbitrary temperature and you will find that they don't relate by that equation.

    For peak intensity: (excuse the units)

    [tex] \nu_{max} = 5.879\times10^{10} T [/tex]

    and

    [tex] \lambda_{max} T = 2.898\times10^{6} [/tex]

    Maybe I'm misinterpreting what you mean by "divide C by the curve."
     
  5. Jan 16, 2009 #4

    mgb_phys

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    Whatever peak wavelength you get from wien's law, you must get the corresponding peak frequency from [tex]c=\lambda\nu[/tex]
    (sorry lambda isn't showing up for me ?)
     
  6. Jan 16, 2009 #5
    maximum of the blackbody spectrum

    I agree with buttersrocks, I also derived the two displacement laws with exactly the same constants of proportionality. Apparently these are not the only relations and if you look at the bottom of this page

    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/wien3.html

    you should know what to look for in other sources. Heald is mentioned but I just looked through his "Classical Electromagnetic Radiation" and I couldn't find anything related to this problem. His other book, "Physics of Waves", also doesn't seem to cover it but I only look at the table of contents so I'm not sure.
     
  7. Jan 16, 2009 #6

    Cthugha

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    No, you don't. The scales are pretty nonlinear. You know that [tex]\nu=\frac{c}{\lambda}[/tex] so you can easily see that the range of lambdas you span by going one unit step in [tex]\nu[/tex] further depends strongly on which [tex]\nu[/tex] you are at. The number of lambdas will be very large for small [tex]\nu[/tex] and will be rather small for large [tex]\nu[/tex]. If you go one unit step in [tex]\lambda[/tex] you will see something similar. This is just an application of the chain rule and has already been discussed here: https://www.physicsforums.com/showthread.php?t=132258.
     
  8. Jan 16, 2009 #7

    mgb_phys

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    Sorry am I misunderstanding the question?
    If Wein's law predicts a peak wavelength for a blackbody at a certain temperature then a photon of that wavelength has a frequncy given by c=\lambda \nu
    The relationship between the wavelength and frequency of a photon doesn't depend on what created it.
    Similairly a peak wavelength - temperature plot and a peak frequency - temperature plot are going to have different shapes ( one is the reciprocal of the other) but you can convert between them with the above equation.
     
  9. Jan 16, 2009 #8

    Cthugha

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    Well, you can convert the peak wavelength of Wien's law to a frequency, but the peak frequency of Wien's law will NOT give the same result. This is a result of the chain rule.

    Just consider you have two distributions and want to change variables [tex]\nu[/tex]=[tex]\nu(\lambda)[/tex]:

    f([tex] \lambda [/tex]) d[tex] \lambda [/tex]=g([tex] \nu [/tex]) d[tex] \nu [/tex]

    Now g([tex] \nu [/tex])=f([tex] \lambda [/tex]) [tex]\frac{d \lambda}{d \nu}[/tex]

    Differentiate this with respect to [tex]\lambda[/tex] to find the peak for [tex]\lambda[/tex]:

    [tex]\frac{d g(\nu)}{d \lambda}=\frac{d f(\lambda)}{d \lambda} \frac{d \lambda}{d \nu} + f(\lambda) \frac{d^2 \lambda}{d \nu^2}[/tex]

    At the peak wavelength one of the terms vanishes:

    [tex]\frac{d g(\nu(\lambda_{PEAK}))}{d \lambda}=f(\lambda_{PEAK}) \frac{d^2 \lambda}{d \nu^2}[/tex]

    So if both distributions are supposed to have the peak at the same wavelength, the term [tex]\frac{d^2 \lambda}{d \nu^2 }[/tex] must vanish. This is clearly not the case here, so both distributions will peak at different wavelengths.
     
  10. Jan 17, 2009 #9
    To get back to the OP's question:

    First the part I think you already know:
    The blackbody spectrum is a plot of power distribution throughout wavelengths/frequency. So if you want to find the power for some band of wavelengths [tex]\Delta \lambda[/tex], then you would integrate. Say we measure wavelength bandwidths of 100 nm; e.g. 600-700 nm, 700-800 nm, 800-900 nm, etc. In all cases, we have kept [tex]\Delta \lambda=100 nm[/tex] constant. Now convert each of these numbers into frequency. You will find now that [tex]\Delta \nu[/tex] does not remain constant. This is the reason for the different curves depending if you are working in the wavelength domain or the frequency domain.

    It all comes down to conservation of energy. An integration over a certain range in the wavelength domain must give the same value for the corresponding range in the frequency domain. However, holding the wavelength range constant and moving throughout the spectrum has the effect of varying the frequency range and vice-versa. This is due to the inverse relation between frequency and wavelength.

    Suppose I gave you a bag of marbles with different colors/weight: 2 x red/2 g, 5 x orange/2 g, 3 x yellow/4 g, 3 x green/4 g, and 2 x blue/6 g. Which type of marble do we have the most off?

    Orange marbles!
    No!
    4 g marbles!
    um....

    Because we don't have a direct relationship between color and weight, we have different peak positions.

    Depends on your experiment. Are you measuring wavelength (e.g. using diffraction gratings) or frequency/energy (e.g. using atomic transitions)?
     
  11. Jan 17, 2009 #10
    Bear with me for a moment...

    Okay, I understand the combinatorial example, but does it apply? For each particular frequency there is one and only one wavelength that it relates to. You don't have 100hz and 1000hz both corresponding to the same wavelength. Hence, my confusion.

    Also, how do you distinguish whether you are measuring wavelength or frequency, since they are directly related. You can typically express atomic transitions in terms of wavelength as well. Don't we know that there IS a direct relationship between frequency and wavelength? ([tex]c=\nu\lambda[/tex]) Or is that relationship not as fundamental as I thought? (I can't see how this would be the case.) Perhaps I just need a bit more elaboration.
     
  12. Jan 17, 2009 #11

    Cthugha

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    Just try something: Take some random pairs of equally spaced frequencies, for example 100 Hz and 1100 Hz, 3100 Hz and 4100 Hz and 1000000 Hz and 1001000 Hz. Now calculate the corresponding wavelengths. You will see that although the differences in frequency are constant, the differences in wavelength will be very different. You can also convert some constant wavelength differences into frequencies to see that it works the other way round, too. Note, that this works for ANY difference, no matter, whether you choose 1000 Hz, 100000000Hz or 0.1 Hz so any linear scale in frequency will give you a different peak than using a linear scale in wavelength.
     
  13. Jan 17, 2009 #12
    Okay, I see where it is coming from now. Thanks.
     
  14. Jan 17, 2009 #13
    True, the combinatorial example isn't exactly what's going on with the blackbody example, the point I was trying to make is that in both cases, you do not have a direct relationship between the two domains you are examining. Note that [tex]\nu=c/\lambda[/tex] is not a "direct" relationship. A direct relation would be energy and frequency. In that case the Wien peaks match up.

    As for how you know what you are measuring; you have to think about it. You will never measure precisely x, but rather some [tex]x\pm\Delta x[/tex]. So in the laboratory, it is wise to make your experiment hold [tex]\Delta x[/tex] constant. So, for example a diffraction grating, the diffraction properties are directly related to wavelength ([tex]m\lambda=d sin \theta[/tex]), therefore you hold [tex]\Delta \lambda[/tex] constant and get the blackbody spectrum in the wavelength domain.
     
  15. Jan 17, 2009 #14
    Thanks cmos. That last part of the last post really solidified it for me. You are correct, I was not using "direct relationship" in the mathematically rigorous sense; I should have seen that. I think that this discussion really helped to destroy a few subconscious misconceptions. Much appreciated.
     
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