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Why is it that thermal radiation of a black body usually described by a spectral distribution function rather than an intensity vs frequency curve?

I have a vague explanation for this: Any measured intensity of radiation will contain a spectrum of frequencies which are impossible to separate (even with a very good spectrometer, one can only measure the intensity of an interval). So it is practically impossible to assign a particular intensity to a particular frequency.

Is my reasoning correct?

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um, isn't a "spectral distribution function" the same thing as an "intensity vs frequency curve"? –  Jeremy Jan 18 '11 at 20:40
    
Right, Jeremy. I personally have seen Intensity versus wavelength often. –  Georg Jan 18 '11 at 20:50
    
@Jeremy: In a spectral distribution function, the quantity which is considered is the spectral radiancy $R_T(\nu)$. $R_T(\nu)$ is defined in such a way that $R_T(\nu)d\nu$ gives the intensity of the radiation $dI$ in the interval $(\nu,\nu+d\nu)$ (similarly for wavelength). Integrating the spectral radiancy curve, one would get the total intensity emitted at a particular frequency (which by Stefan's law is proportional to $T^4$). –  Bernhard Heijstek Jan 18 '11 at 20:58
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2 Answers 2

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It's true that you can't talk about intensity without referring to some specific frequency range, but the reason is the definition of intensity itself, not the inherent limits of our measuring devices. Intensity, also called irradiance, is defined as power per unit area. Typically it takes into account radiation at all frequencies (or wavelengths), but you can also talk about the intensity within a particular band like $400\text{ nm}-700\text{ nm}$.

If you wanted to create a graph of intensity versus frequency, you could certainly do so by dividing the frequency axis into bins and plotting one point for each bin representing the total intensity detected within that frequency range. Obviously, the smaller you can make the bins, the more detailed information you get about the spectrum, but the intensity values become correspondingly smaller (and harder to measure accurately) as the bins get smaller. Besides, if everyone did that, different plots from different experiments would be very difficult to compare unless they all decided on a standard bin size.

In order to make data from different experiments comparable, you normalize the intensities by dividing each intensity measurement by the width of the bin. In the limit as the bin width goes to zero, this is just the spectral irradiance.

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Thanks! But I have seen plots of intensity vs frequency (mostly on non-technical sites). So are these incorrect? –  Bernhard Heijstek Jan 18 '11 at 21:39
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@finemann: I couldn't say without knowing how the plots were generated. It could be that they're using the binning procedure I talked about, or it could be that the plots are just mislabeled. –  David Z Jan 18 '11 at 22:10
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Dear Finemann, the "spectral power distribution"

http://en.wikipedia.org/wiki/Spectral_power_distribution

and "intensity vs frequency" contain exactly the same information. They're so easily related that physicists wouldn't even consider them "two different things".

The spectral power distribution is the (infinitesimal) power per unit area and per unit wavelength: $$ M_\lambda = \frac{\delta\Phi}{\delta A\delta \lambda} $$ You may simply write $\lambda$ as $c/f$ and use the right differentials of functions, $\delta\lambda=-(c/f^2)\times \delta f$ and you will get the analogous distribution in terms of the frequencies. The graph will be just flipped from the left to right (that's because of the irrelevant minus sign), and on one side, it will be stretched vertically and squeezed horizontally. There is no international ban on using the frequencies, and indeed, it's true that even top physicists usually prefer to parameterize radiation by its frequency rather than the wavelength. You may have seen "popular" sources that used frequencies instead of the wavelength but it is not because they were incorrect or misleading; they did copy conventions that common in research-level physics.

You may also think that the words "intensity" and "power" are different things. But "intensity" is just a popular term for "irradiance" which is just power per unit area, see

http://en.wikipedia.org/wiki/Radiant_intensity

More properly, "radiant intensity" should be counted as power of a localized source per solid angle (note that if the radiation is radial, the power only depends on the solid angle and not the distance at which you absorb the light).

Black body spectral power distribution is continuous: a black body is indeed able to emit at all frequencies. A usual disclaimer for continuous distributions should be added: the power emitted by a continuous power distribution at a particular exact frequency is always exactly zero. You need a whole interval of frequencies (or wavelengths) to get a nonzero answer. The (infinitesimal) power coming from this interval of frequencies will then be proportional to the (infinitesimal) width of the interval of frequencies. That's why we talk about the ratios of differentials etc.

Cheers LM

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