# Talk:Planck's law

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## a devilish plot

The two formulas incrementally plot exactly the same law as functions of respectively frequency and wavelength. They look inexplicably different as formulas, they have different shapes when plotted, and they peak at very different points in the spectrum. This is the result of the interaction of two things: (i) wavelength varies inversely with frequency, and (ii) the formulas are incremental in the sense that they give the radiated intensity only within an infinitesimal frequency band and wavelength band respectively (the meaning of spectral radiance). This is spelled out in the section Planck's_law#Correspondence_between_spectral_variable_forms.

It's too bad the tradition for Planck's law is to plot the x-axis of the law linearly instead of logarithmically as traditionally done in depictions of the EM spectrum (e.g. the figures in that article). The latter would have eliminated all three of the above differences (the peak for both wavelength and frequency would then be at what the article calls the wavelength-frequency-neutral peak intermediate between the wavelength and frequency peaks), as well as allowing the distributions of the Sun and the Earth to be plotted side by side on a common x-axis (covering about four octaves) without one dwarfing the other horizontally (the vertical scales require adjusting). Vaughan Pratt (talk) 01:54, 11 November 2014 (UTC)

Perhaps User:Vaughan Pratt has facile access to three similarly, nearly uniformly, plotted graphs that could be posted here, beside one another? On the left, with a linear wavelength abscissa, in the middle with a twice-labeled logarithmic abscissa, on the right, with a linear frequency abscissa?Chjoaygame (talk) 06:02, 11 November 2014 (UTC)
I pointed out the virtues of a logarithmic abscissa in the article back in November 2011, but it was promptly deleted. Feel free to put it back if you disagree with the deleter. At about that time I also wrote User talk:Vaughan Pratt/Planck's Law as a basis for discussion, which included a graph with two of those curves superimposed, but this went nowhere despite much discussion. I have the third logarithmic plot somewhere, I'll dig it up.
In the meantime you've just now written "That peak location depends on variable choice is not physically significant, but is only an appearance. It can be made to disappear simply by plotting the spectral graph with a logarithmic abscissa scale." That's wrong on two counts.
1. The peak doesn't disappear, it simply moves to a point intermediate between the wavelength or Wien peak and the frequency peak.
2. None of those three peaks have any physical significance that I'm aware of, unlike the median or 50% peak, whose physical significance is that half the radiance (total, not spectral) in W/m2 lies above it and half below.
All four peaks are bold-faced in the section on percentiles. Vaughan Pratt (talk) 01:22, 12 November 2014 (UTC)
I recall that you proposed the logarithmic plot some time ago, but I don't recall the details. I have now spent some time trying to find your logarithmic plots of that time, but not succeeded in finding them. Perhaps you will very kindly show exactly where I can see them?
I like the idea of three separate plots side by side. Just to make it easy to read. Linear frequency, log for both, linear for wavelength.
Sorry about the unresolved anaphoric 'it'. I have made a grammatical fix. You want to point out further that there is no physical significance in the peaks.
If you can find suitable plots I think it would be good. I reserve the right to an opinion about suitability.Chjoaygame (talk) 03:00, 12 November 2014 (UTC)
Thinking it over, I have changed my mind. I have removed my hastily added chat item from the article. I now think this matter needs more conceptual structure. As we now have it, it hardly has physical content, and is just chat. There is potential physical content in it, but to justify a place in the article, the potential needs to be made actual. Something well thought-out about Wien's structural resolution of the Kirchhoff function. Without that it would just be chatter, I think, with no place in the article.Chjoaygame (talk) 08:08, 12 November 2014 (UTC)
Sorry, not following. If you mean you've changed your mind about your "right to an opinion about suitability" then you have my whole-hearted support. Likewise if you mean the inadvisability of hastily added chat items: EBBOM. Vaughan Pratt (talk) 11:31, 12 November 2014 (UTC)
I mean that I think that the logarithmic scale thing, by itself, is lacking in physical conceptual structure.Chjoaygame (talk) 12:20, 12 November 2014 (UTC)
Very interesting. Does a linear scale have more "physical conceptual structure" than a logarithmic scale? Likewise for a scale in 1/x instead of log(x). Vaughan Pratt (talk) 12:56, 12 November 2014 (UTC)
The keyboard of a piano realizes a logarithmic scale for frequency. Surely a keyboard is physical. Vaughan Pratt (talk) 07:51, 25 November 2014 (UTC)
I will let that one pass to the wicket-keeper.Chjoaygame (talk) 10:42, 25 November 2014 (UTC)

## a problem

Yes there is a problem with graphs plotted from the formula because even a professor of physics at a prominent US university has missed the stated fact that the two formula do in fact calulate different values and for this reason peak at different values and not because the absissa for one graph is expressed as wavelength instead of frequency.

I can send you his graphs and comment as proof without identifying him.posted 2014 NCite error: There are <ref> tags on this page without content in them (see the help page). ovember 12

(I have moved the above unsigned comment to the customary place here at the bottom of the page, with the customary new header for a new comment.Chjoaygame (talk) 07:51, 12 November 2014 (UTC)Chjoaygame (talk) 15:29, 12 November 2014 (UTC))

Perhaps you need to take into account that the quantities are not just numbers. They have very different units, as you may check for youselfChjoaygame (talk) 08:12, 12 November 2014 (UTC)
Perhaps you need to take into account that the more relevant quantities ${\displaystyle \nu B_{\nu }(T)}$ and ${\displaystyle \lambda B_{\lambda }(T)}$ are equal. They have the same units, as you may check for yourself. I pointed this out in 2011 but it was promptly deleted, this article being a war zone at the time and as such immune to anything resembling rational discussion.
I'd been contributing to this article since 2008, but after a month of insane edit wars in 2011, with you and Headbomb being the main obstacles to progress, I gave this article up as a lost cause.
I therefore have no intention of resuming editing it as long as this situation continues. In the meantime the article has accumulated an absurd quantity of rubbish and is in urgent need of being trimmed down to something more sane. Under the circumstances there is no point trying to fix this.
Competently written articles don't suffer this fate, witness Planck's_law#Percentiles, Boolean algebra, and the lead of Hyperbola, each of which I wrote single-handedly many years ago and which have survived largely untouched since then because there's very little urgently needing fixing. Come to think of it, the other 30-odd articles I've created have done about as well, with the only one I had to seriously defend being Proof (truth), and that only because a high-energy editor was unable to accept that "proof" had any meaning outside mathematics and had to be shouted down by a bunch of non-mathematicians for whom "proof" meant more than just what logic textbooks defined it to be. Vaughan Pratt (talk) 12:44, 12 November 2014 (UTC)

Let me address your argument in more detail. You seem to be arguing that the peaks are at different points in the EM spectrum on the ground that "the quantities are not just numbers. They have very different units". However changing the units from MKS to CGS would result in different numbers with different units without however moving the peak. Hence a difference of units and values can't be the reason the peak moved.

The peak moved as a result of a nonlinear distortion of the x-axis. For example if you plot ${\displaystyle B_{\nu }(T)}$ on log-linear paper the peak moves to a different frequency in the EM spectrum, despite the fact that neither the numbers nor the units changed. Not surprisingly this also happens with ${\displaystyle B_{\lambda }(T)}$: plotting it on log-linear paper moves the peak to a different wavelength.

What is surprising, at least at first glance, is that the two peaks move to the same position in the EM spectrum, one given by frequency and the other by the corresponding wavelength.

This is true despite the fact that ${\displaystyle B_{\nu }(T)}$ and ${\displaystyle B_{\lambda }(T)}$ have different values with different units.

The common peak of these different curves, when each is plotted on log-linear paper, is the wavelength-frequency-neutral peak.

This is not so surprising once you notice that log(x) and log(1/x) differ only by a constant factor, namely −1. The plot of spectral intensity by log(frequency) is merely the mirror image of that by log(wavelength), a linear distortion.

It should also be pointed out that ${\displaystyle B_{\nu }(T)}$ and ${\displaystyle B_{\lambda }(T)}$ are really the same distribution in the sense that for any two points in the EM spectrum, whether represented as frequencies ${\displaystyle \nu _{1}}$ and ${\displaystyle \nu _{2}}$ or wavelengths ${\displaystyle \lambda _{1}=c/\nu _{1}}$ and ${\displaystyle \lambda _{2}=c/\nu _{2}}$, the areas under the two curves between those two points, defined as the respective integrals ${\displaystyle \int _{\nu _{1}}^{\nu _{2}}B_{\nu }(T)d\nu }$ and ${\displaystyle \int _{\lambda _{2}}^{\lambda _{1}}B_{\lambda }(T)d\lambda }$, are equal.

Given this equality, one might well conclude that the point in the EM spectrum where black body radiation at temperature T is most intense must be at the peak of the distribution. As explained above, this line of reasoning is fallacious because a nonlinear distortion of the x-axis, in this case reciprocation, can move the peak, though a linear one cannot. But two nonlinear distortions, namely log in each case, can bring the two peaks into coincidence. Vaughan Pratt (talk) 19:17, 12 November 2014 (UTC)

Vaughan Pratt, what you said above:
"For example if you plot ${\displaystyle B_{\nu }(T)}$ on log-linear paper the peak moves to a different frequency in the EM spectrum, despite the fact that neither the numbers nor the units changed."
is not correct. To change the position of the peak you have to change the function being plotted. I think you're confusing energy emitted per unit area per unit solid angle per unit spectral measurement with emitted per unit area per unit solid angle per fractional change in spectral measurement. The peak in the latter function is at the same point whether the spectral measurement is wavelength or frequency, because the fractional change in wavelength is the same as the fractional change in frequency. Djr32 (talk) 00:13, 13 November 2014 (UTC)

Mea culpa, good catch. What I wrote near the top of this section, "the more relevant quantities ${\displaystyle \nu B_{\nu }(T)}$ and ${\displaystyle \lambda B_{\lambda }(T)}$ are equal", is correct, and is how ${\displaystyle B_{\nu }(T)}$ and ${\displaystyle B_{\lambda }(T)}$ must be scaled vertically in order to compensate for the horizontal compression of each curve at respectively high frequencies and long wavelengths when plotted logarithmically, if the integral is to be faithfully represented by the area under the curve. In a senior moment I pictured the compensation happening automatically when transferring to log-linear paper but of course it doesn't. The wavelength-frequency-neutral peak is the peak of ${\displaystyle \nu B_{\nu }(T)}$, and hence of ${\displaystyle \lambda B_{\lambda }(T)}$ since they're equal, with the common function being ${\displaystyle 2hr/E}$ where ${\displaystyle r=(\nu /\lambda )^{2}}$ and ${\displaystyle E=e^{h\nu /kT}-1=e^{hc/\lambda kT}-1}$ (bearing in mind that ${\displaystyle \lambda \nu =c}$ when asking what it is a function of). Thanks for catching that! Vaughan Pratt (talk) 04:10, 13 November 2014 (UTC)

I don't agree that ${\displaystyle \nu B_{\nu }(T)}$ and ${\displaystyle \lambda B_{\lambda }(T)}$ are the more relevant quantities, ${\displaystyle B_{\nu }(T)}$ is the fundamental quantity as the modes are quantised linearly in ${\displaystyle \nu }$. The area under the curve being proportional to the integral is a property of linear/linear graphs, not logarithmic axes, and if you're going to use a logarithmic axis I think it's a mistake to scale the B axis in some strange way to recover this property. I don't think there is a problem here: there are two (maybe even three) quantities of interest, they have peaks in different places, and the article correctly reflects this. Djr32 (talk) 18:49, 15 November 2014 (UTC)
Sorry, I'm not following. Are you saying that when the x-axis (the spectrum) is scaled logarithmically one should plot the same B-values as when the x-axis is scaled linearly? Vaughan Pratt (talk) 00:15, 23 November 2014 (UTC)
Assuming that's what you're saying, it would indeed simplify things to take ${\displaystyle B_{\nu }(T)}$ as the fundamental quantity, and to plot it independently of how the x-axis is scaled, whether proportional to the logarithm of frequency, its reciprocal, or any other scaling. Vaughan Pratt (talk) 20:05, 25 November 2014 (UTC)
To clarify, I'd be opposed to any such simplification, whether plotting spectral intensity as a function of either the reciprocal of frequency or the logarithm thereof. You appear to be agreeing in the case of reciprocal but disagreeing in the case of logarithm. Vaughan Pratt (talk) 18:10, 5 December 2014 (UTC)

## undid good faith edit; reason

The main thing about Kirchhoff's law of thermal radiation is not that equality. It is that thermal equilibrium has underlying it the universal spectral distribution that is eventually expressed by Planck's law, but was not explicitly known to Kirchhoff, nor to anyone till Planck discovered it. Based on this, and derivative from it, one can define absorption and emission coefficients that have that equality under conditions of thermal equilibrium.Chjoaygame (talk) 11:37, 19 June 2015 (UTC)

## Formulas mistake

Both formulas for the spectral radiance have mistake and even the transition from the one to the other is wrong. look at Rybicki, G. and Lightman, A. P. "The Planck Spectrum." Radiative Processeys in Astrophysics. New York: Wiley-Interscience, pp. 3-4 and 20-23, 1979. for the right function — Preceding unsigned comment added by Pantelis156 (talkcontribs) 8:06, 21 August 2015 (UTC)

Editor Pantelis156 seems to find some discrepancy between the article's and Rybicki & Lightman's formulas for spectral radiance (which, by the way on their page 3 they call specific intensity or brightness). Their formulas are on their page 22, equations (1.51) and (1.52). By my reading there is no discrepancy, and so no need for me to copy their formulas into this talk page. Perhaps Editor Pantelis156 is confusing radiance with energy density? I don't think I have missed something there.Chjoaygame (talk) 12:43, 21 August 2015 (UTC)

Pantelis156 (talk) 14:34, 21 August 2015 (UTC)You are correct,I confused it with energy densityPantelis156 (talk) 14:34, 21 August 2015 (UTC)

Well it's not so much a mistake, as it is incomplete. The formula ${\displaystyle B_{\lambda }(\lambda ,T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}}$ has a numerator of 1 in the second fraction. What this means is the result of the equation is measured in units of "intensity per meter of spectrum". If you want the much more convenient units of "intensity per nanometer of spectrum", that numerator must be changed to 0.000000001. Or else, the result will need to be divided by 1000000000. This is not mentioned in the article, but should be. Benhut1 (talk) 06:40, 23 November 2016 (UTC)

And you'd have a different factor if you want the answer in ergs/(s·cm2) per angstrom of spectrum. Unit conversions don't belong in this article. 11:25, 23 November 2016 (UTC)

## Which definition of Planck's constant is being used here?

Nowhere in this article does it say what definition of Planck's constant is being used. There are 2 possible ones that I can see from looking at the Wikipedia page for that constant. These are "6.626070040(81)×10−34" and "2π". And it's not just this one constant. This entire article has the problem that none of the constants refered to have the their values stated anywhere in this article. Please improve this article by including all the values for the constants that are used. Benhut1 (talk) 06:10, 23 November 2016 (UTC)

2π is not Planck's constant. I don't know where you saw that, but I suggest you recheck your sources. And if you want to know the value of Planck's constant, look up Planck's constant. That's why we link to it. We also don't mention the value of the speed of light or Boltzman's constant for the same reasons. 11:21, 23 November 2016 (UTC)