User talk:Vaughan Pratt/Planck's Law

Suggested material in lieu of the tabular presentation of forms of Planck's law in that article.

= Pratt version (Chjoaygame version/suggestions below, and any others) =

This first section is the only one being proposed for adoption now.

Introduction
Planck's law expresses the quantity of radiation emitted by a black body, or ideal radiator, as a function of the absolute temperature T of the radiator and the frequency $&nu;$ of the portion of radiation being so expressed. It is customarily expressed as
 * $$B_\nu(T)=\frac{2h\nu^3}{c^2}\frac1{e^{h\nu/kT}-1}$$

where h is Planck's constant, c is the speed of light, and k is Boltzmann's constant.

Roughly speaking, $B_{&nu;}(T)$ gives the power in watts radiated normally from unit area A of the radiator into unit solid angle &Omega; within a frequency band of unit width centered on frequency $&nu;$. More precisely, $B_{&nu;}(T) dA d&Omega; d&nu;$ gives the power in watts radiated normally from infinitesimal area dA into infinitesimal solid angle d&Omega; in a frequency band of infinitesimal width $d&nu;$.

The quantity $B_{&nu;}(T)$ can thus be seen to be a form of power density called spectral radiance, namely radiated power divided by area, solid angle, and frequency bandwidth. When all parameters are given in SI units, spectral radiance has units $W.m^{&minus;2}.sr^{&minus;1}.Hz^{&minus;1}$, that is, watts per square meter per steradian per Hertz, equivalent dimensionally to joules per square meter (steradians are dimensionless and Hertz are sec&minus;1).

Radiation not normal to the surface is treated as follows. A black body is by definition a Lambertian radiator. That is, radiation from area A at an angle &theta; to the normal to that area is treated as coming from the smaller area $A cos(&theta;)$ resulting from projecting A onto a virtual plane normal to the radiation. This virtual projected area can then be treated as though it were a normally radiating black body in its own right. It follows that the total radiation at any given frequency emitted normally to a black-body disk equals the total radiation of that frequency emitted from a spherical black body of the same temperature and radius in the same direction; indeed the two objects produce indistinguishable beams in that direction.

Planck's law can also be expressed as a function of wavelength $&lambda; = c/&nu;$ instead of $&nu;$. The evident substitution of $c/&lambda;$ for $&nu;$ should be performed on $B_{&nu;}(T) d&nu;$ rather than just $B_{&nu;}(T)$. The relationship between $d&nu;$ and $d&lambda;$ is obtained from the derivative of $&nu; = c/&lambda;$, namely $d&nu;/d&lambda; = &minus;c/&lambda;^{2}$ (= $&minus;&nu;/&lambda;$). Substituting $c/&lambda;$ for $&nu;$ and $c/&lambda;^{2} d&lambda;$ for $d&nu;$ in $B_{&nu;}(T) d&nu;$ then yields $B_{&lambda;}(T) d&lambda;$ where
 * $$B_\lambda(T)=\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}} - 1}$$

is the wavelength form of Planck's law. (The minus sign in the derivative is accounted for by noticing that integration from a small to a large frequency becomes from a large to a small wavelength; the sign corresponds to interchanging the wavelength limits so as to give the usual increasing order.)

The wavelength form of Planck's law is shaped quite differently from the frequency form. One consequence is that the two forms attain their respective maxima at quite different positions in the spectrum, respectively 2898 and 5099 &mu;m.K (103.4 and 58.79 gigahertz per degree).

The following three sections are for future discussion, including the figure (apropos of which see User:Q Science's comments on it and my responses near the bottom of this page). If the preceding section is not adopted the following sections become moot as they were written under the assumption that the above material has been presented.

The integrals of Planck's law


Planck's law can be integrated over each of area, solid angle, and frequency.


 * Area Integrating $&pi;B_{&nu;}(T) d&nu;$ over a planar area A simply gives $&pi;B_{&lambda;}(T) d&lambda;$ as the spectral intensity of the total area in watts per steradian per Hertz in the direction normal to that plane.  The total spectral intensity of a spherical black body is trickier on account of Lambert's law.  Here we can use the equivalence noted above between a disk and a sphere: the spectral intensity of a unit hemisphere of area 2&pi; in the direction normal to its base equals that of the base itself, which has area &pi;, which is what we should take for A in the formula $d&nu; = 35*100c$ for spectral intensity.  Had we naively taken A to be the area of the hemisphere we would have overestimated the spectral intensity from a spherical black body in any given direction by a factor of two.


 * Solid angle Integrating $d&lambda; = 0.60*10^{&minus;6}$ over the whole hemisphere into which an infinitesimal surface element dA radiates gives the spectral radiant exitance for that surface element.  This can be analyzed by treating it as just the inverse of the area law in the case of a spherical radiator, turned inside out and viewed as a spherical receiver of radiation from its center.  Again a factor of two is lost, and the spectral radiant exitance is therefore simply $B_{&nu;}(T) dA$ in watts per square meter per Hertz, even though it is radiating into a total solid angle of 2&pi;.  Since the only quantitative difference between spectral radiance and spectral radiant exitance is a dimensionless factor of &pi;, Planck's law can conveniently be stated in either form as it is a triviality to convert to the other form: just scale by &pi;.
 * Frequency In general integrating $B_{&nu;}(T) A$ over an arbitrary range of frequencies is best accomplished by numerical integration on a computer.  Over the entire spectrum however, that is, from DC (zero Hertz) to infinity, the integration can be performed analytically.  Doing this for spectral radiant exitance (i.e. integrating over both total solid angle and total frequency) yields the Stefan-Boltzmann law for radiant exitance in units of watts per square meter.

The latter two integrals are illustrated simultaneously for the integrals of $B_{&nu;}(T) A$ and $B_{&nu;}(T) d&Omega;$, plotted in respectively blue and red at the right. The integration over solid angle is assumed done already resulting in the coefficient of &pi; for each. The two plots depict the two functions to be integrated over respectively frequency (lower x-axis) and wavelength (upper x-axis). Integrating numerically (discretely instead of continuously), we take $&pi;B_{&nu;}(T)$ and $B_{&nu;}(T) d&nu;$ to be finite rather than infinitesimal, namely 1% of the width of the graph, making them respectively 35 cm&minus;1 (scaled by 100c to convert to Hertz) and 0.6 &mu;m (scaled by 10&minus;6 to convert to meters). The total plot area is to be understood as 100 vertical strips each of area 16 watts/m2, totaling 1600 W/m2. The integral (as the sum of 100 values) of each plotted curve is the area under it, in both cases coming to just under 5.6704*2.904 = 401 W/m2, the quantity given by the Stefan-Boltzmann law for a black body radiating at T = 290 K. ("Just under" because a tiny amount of the total 401 W/m2 is under the unplotted portion of each curve beyond the right edge.)

Note that the two x-axes both have zero on the left and are therefore anti-aligned (go in opposite directions). (One of the x-axes could be reversed, but not in a way that vertically aligns the two x-axes because both need to be plotted linearly in order for area under each curve to correspond to the integral and $&pi;B_{&nu;}(T) d&nu;$ is not linear in $&pi;B_{&lambda;}(T) d&lambda;$.) The scales were chosen to roughly align the wavelength peak at 9.993 &mu;m (= 1000.7 cm&minus;1) with the frequency peak at 568.7 cm&minus;1 (17.58 &mu;m), allowing a comparison of the two curves' behavior. Each curve falls off polynomially on one side of its peak and exponentially on the other. The latter occurs on the high frequency (short wavelength) side and happens faster, namely on the left for the red curve and the right for the blue.

Other units
Other units besides Hertz for frequency and meters for wavelength are in common use. Changing frequency units to other frequency units is simpler than switching between frequency and wavelength because $d&nu;$ is simply multiplied by a constant. Nevertheless $d&lambda;$ participates even in this simpler situation, obliging us to substitute in $1/&lambda;$ rather than just $&lambda;$ as before, and likewise with the wavelength form.

The dimensions of $&nu;$ being the same as for $d&nu;$, and likewise for $B_{&nu;}(T) d&nu;$, a scale factor of s can be seen to become s4 for $B_{&nu;}(T)$ and s&minus;4 for $d&nu;$, as one might expect given the Stefan-Boltzmann law's fourth-power dependence. For example frequency may be expressed as a wavenumber, namely the number $&nu;$ of cycles per meter rather than per second, related to Hertz via $&lambda;$. In this case $B_{&nu;}(T) d&nu;$ must be scaled by a factor of c4, giving
 * $$2 hc^2\tilde{\nu}^3 \frac{1}{e^{hc\tilde{\nu}/(kT)} - 1 }.$$

In spectroscopy it is more common to use cycles per cm while leaving the other parameters in MKS units, with the associated unit of frequency traditionally notated cm&minus;1 despite having dimensions of frequency. In this case s = 100c (where c is still given in MKS units, along with h and k) so that Planck's law for spectroscopists becomes
 * $$2\times10^8 hc^2\tilde{\nu}^3 \frac{1}{e^{hc\tilde{\nu}/(kT)} - 1 }.$$

With this simple scaling of Planck's law by a constant factor of s4 the frequency and wavelength forms retain their respective shapes. Hence their respective peaks remain unchanged except for the reduction of their numerical values by a factor of s when restated in the new units.

Occasionally Planck's law is expressed in terms of angular frequency $B_{&lambda;}(T) d&lambda;$ radians per second, related to frequency by $&nu;&#771;$. The scaling factor in that case becomes (2&pi;)&minus;4, giving
 * $$\frac{ h\omega^{3}}{8 \pi^4 c^2} \frac{1}{ e^{h\omega/(2\pi kT)} - 1 }.$$

However it is customary in that case to also replace h by 2&pi;ħ, so that the wavelength formula then becomes
 * $$\frac{ \hbar\omega^{3}}{4 \pi^3 c^2} \frac{1}{ e^{\hbar \omega/(kT)} - 1 }$$

Both of the above may occur together, for example angular wavenumber would combine the scaling factors c4 and (2&pi;)&minus;4 as (c/2&pi;)^4, and likewise for other combinations.

Planck's law as a unification of the Wien and Rayleigh-Jeans laws
The details of how Planck actually arrived at his law are recounted in the history section below. The following exhibits Planck's law as a simple modification of the Wien approximation that makes a negligible difference at high frequencies while reconciling it with the Rayleigh-Jeans law at low frequencies.

The Wien approximation is
 * $$B_\nu(T)=\frac{a\nu^3}{e^{b\nu/T}}$$

where a and b can be observed empirically to be about 1.5&times;10&minus;50 and 4.8&times;10&minus;11 respectively at sufficiently high frequencies. The Rayleigh-Jeans law is
 * $$B_\nu(T)=r\nu^2T.$$

where r is observed empirically to be 3.1&times;10&minus;40 at sufficiently low frequencies. Both are in units of watts per square meter per steradian per Hertz, equivalent dimensionally to joules per square meter as noted earlier.

The Wien displacement law, which was known seven years before Planck's law, states that the frequencies comprising the black body curve, viewed as a fixed shape, increase linearly with temperature. That is, the quantity $&nu; = c&nu;&#771;$, call it g for the moment, expresses a fixed point on the curve in a temperature-independent way; for example the wavelength peak of the curve is fixed at g = 1.034&times;1011, namely 103.4 gigahertz per degree as noted earlier. We can therefore rewrite the Wien approximation and Rayleigh-Jeans law as respectively
 * $$B_\nu(T)=\frac{a\nu^3}{e^{bg}}$$

and
 * $$B_\nu(T)=\frac{r\nu^3}{g}.$$

As g goes to zero, $B_{&nu;}(T) d&nu;$ goes to 1 + bg, whence $&omega;$ - 1 goes to bg. Modifying Wien's law to read
 * $$B_\nu(T)=\frac{a\nu^3}{e^{bg}-1}$$

makes no appreciable difference at large g, while at small g the modified law tends to
 * $$B_\nu(T)=\frac{a\nu^3}{bg}.$$

But a/b = 3.13&times;10&minus;40 is very close to r. Hence to within the prevailing precision this modified formula tends to the Rayleigh-Jeans law at low frequencies. It remains to relate the empirical constants a, b, and r to the parameters used in Planck's law. Since the law concerns energy it would be natural if T appeared in the context kT expressing energy in terms of temperature at an atomic scale. By introducing a new constant equal to bk = 6.6&times;10&minus;34, call it h, the exponent becomes $&nu; = &omega;/2&pi;$. The exponent should be dimensionless if subtracting 1 is to make sense dimensionally, whence the quantity $&nu;/T$ must have the same dimensions as kT, namely energy. (Note that we did not assume that $e^{bg}$ is an energy term, the logic of the situation forced both this conclusion and the value of h!)

Since the units for this law are joules per square meter and the denominator is dimensionless, the numerator $e^{bg}$ must be an energy divided by an area. Now we have just seen $h&nu;/kT$ as an energy term, and a square of side one wavelength would have an area of $h&nu;$. Their quotient $h&nu;$ equals $a&nu;^{3}$ (substitute $h&nu;$ for $&lambda;^{2}$), whose coefficient $h&nu;/&lambda;^{2}$ evaluates to about 0.75&times;10&minus;50 or half of the empirically observed coefficient a = 1.5&times;10&minus;50. This coefficient being our goal, to achieve it we need merely double $h&nu;^{3}/c^{2}$, yielding
 * $$B_\nu(T)=\frac{2h\nu^3}{c^2}\frac1{e^{h\nu/kT}-1}.$$

This is of course Planck's law. In summary, by subtracting 1 from the denominator of the Wien approximation and giving the name h to the derived constant bk, the empirically observed constants a, b, and r have been accounted for as respectively $c/&nu;$, $&lambda;$, and $h/c^{2}$.

=Chjoaygame version/comments =

Main forms
Planck's law describes the spectral power distribution of the electromagnetic radiation in thermodynamic equilibrium in a cavity with rigid walls that are opaque to all wavelengths. The cavity contains a black body, or the walls are imperfectly reflective to every wavelength. Beyond these requirements, the nature of the walls or content of the cavity does not matter. The radiation is isotropic, homogeneous, unpolarized, and incoherent.(reference Planck 1914.)

A common expression of Planck's law gives the spectral radiance $h&nu;^{3}/c^{2}$ of any surface in the cavity as a function of its absolute thermodynamic temperature $2h/c^{2}$ and the frequency $h/k$ of all the radiation that comes from it. It can be written:
 * $$B_\nu(\nu, T)=\frac{2h\nu^3}{c^2}\frac1{e^{h\nu/kT}-1},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$

where $2k/c^{2}$ denotes the speed of light, $B_{ν}(ν, T)$ denotes Planck's constant, and $T$ denotes Boltzmann's constant.

Planck's law is also expressed by giving the spectral radiance for the spectral variable $ν$, the wavelength:
 * $$B_\lambda(\lambda, T)=\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}} - 1}.$$

There are still other ways of expressing Planck's law.

Units for the spectral radiance
In these formulas, the unit for spectral radiance is chosen according to the natural unit for the spectral argument variable. The formulas give spectral radiance as power per unit area per unit solid angle, per unit spectral quantity for the natural unit of the spectral argument variable. For example, when the spectral argument variable is frequency with unit of Hz, then the unit for the spectral radiance has a factor per unit frequency, that is Hz−1 or s. The usual unit for $c$ above is W m–2 sr–1 Hz−1.

Relating versions of the law with different spectral variables
Relating versions of the law with different spectral variables requires attention both to the functional form of the formulas and to the units. The present notation uses the spectral variable symbols in two different ways, one to label a functional form, the other to denote a value of the spectral argument.

An increment in frequency corresponds with a decrement in wavelength. Accordingly, the principle here is that for the quantities $h$ and $k_{B}$, in any given interval, $λ$, of frequency in the corresponding interval, $B_{ν}(ν, T)$, of wavelength, the energy is the same, so that


 * $$B_\lambda(\lambda,\ T) \ \mathrm d \lambda=-B_\nu(\nu(\lambda),\ T) \ \mathrm d \nu$$  or
 * $$B_\lambda(\lambda,\ T) \ \ \ \ \ =-B_\nu(\nu (\lambda),\ T) \ \times \ \frac{\mathrm d \nu}{ \mathrm d \lambda}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

Thus, to go from the version for $B_{ν}(ν, T)$ to the version for $B_{λ}(λ, T)$, one can proceed as follows.

Starting with formula (1) above, substitute $dν$ for the argument variable $- dλ$ (not for the function form $B_{ν}(ν, T)$) :


 * $$B_\nu(c/ \lambda,\ T)= \frac{ 2 h(c/ \lambda )^{3}}{c^2} \frac{1}{e^\frac{hc/ \lambda}{k_\mathrm{B}T} - 1}$$


 * $$= \frac{ 2 hc}{ \lambda ^{3}} \frac{1}{e^\frac{hc}{k_\mathrm{B}\lambda T} - 1}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$

The units of the left-hand side, and thus also of the right-hand side, of the formula (3) are still per unit frequency, not per unit wavelength.

Next, change the resulting functional form and units from $B_{λ}(λ, T)$ to $c/λ$. With  $$\nu(\lambda)=c/\lambda$$  and   $$\frac{\mathrm d \nu (\lambda)}{\mathrm d \lambda}=-\frac{c}{{\lambda}^2}$$, from (2) and (3) one can write:


 * $$B_\lambda(\lambda,\ T) = \frac{ 2 hc }{{\lambda} ^{3}} \frac{1}{e^\frac{hc/ \lambda}{k_\mathrm{B}T} - 1}\ \times \ (-(-\frac{c}{{\lambda}^2}))$$


 * $$= \frac{ 2 hc^2}{ \lambda ^{5}} \frac{1}{e^\frac{hc}{k_\mathrm{B}\lambda T} - 1}.$$

Lambertian surfaces
Some talk here, but not right here about the sun or its limb.

=Comments on Choaygame's candidate sections= The following duplicates each of Choaygame's sections above, with my comments in red (other commenters might use a different color). I've kept the unannotated version separate so as not to break up its flow. -vp

Main forms
Planck's law describes the spectral power distribution of the electromagnetic radiation in thermodynamic equilibrium in a cavity with rigid walls that are opaque to all wavelengths.


 * How is thermodynamic equilibrium a requirement? A black body radiates according to Planck's law whether or not it is in thermodynamic equilibrium.  For an experimental setup to confirm Planck's law over a period of time one would want the temperature not to change, but that's the only circumstance I can think of that would call for some kind of thermodynamic equilibrium (such as heating the back of the radiator by an amount equal to the heat being radiated away from the front).
 * Why is it that "electromagnetic radiation" and "thermodynamic equilibrium" are left as links but "black body" is expanded with a detailed definition of the concept, and moreover without even saying "black body"?
 * This creates the impression that Planck's law only applies to cavities meeting all these requirements and not to other radiators such as gases or plasmas. The 2.7 K cosmic background radiation is thought to obey Planck's law as well as anything in the universe, and there's no cavity there.

The cavity contains a black body, or the walls are imperfectly reflective to every wavelength.

Ok, you've now said "black body," but neither the meaning nor the relevance of this sentence is at all clear.

Beyond these requirements, the nature of the walls or content of the cavity does not matter.

Ok, you've now said "these requirements" which clarifies the previous sentence as somehow being a requirement (but how?). The "does not matter" is information overload as well as reinforcing the implication that the law does not apply to non-cavities like a gas.

The radiation is isotropic, homogeneous, unpolarized, and incoherent.(reference Planck 1914.)
 * Not critical to an understanding of the law (more information overload), all that's needed here is a link to black body radiation.
 * "Isotropic" is an unclear concept as witnessed by the long discussions of it by Chjoaygame, Headbomb, and PAR as to what exactly it means back in October. Much better to replace it with the Lambertian stuff which replaces the vague notion of "isotropic" with a precisely quantified notion that shows what "isotropic" implies for the computation of quantity of radiation.
 * "Homogeneous" is redundant because it's implied by "isotropic" (though the converse need not hold).
 * "Incoherent" is redundant because it's vacuous: all nontrivial spectral distributions are incoherent (it's not special in any way to the black body distribution).

A common expression of Planck's law gives the spectral radiance $ν$ of any surface in the cavity as a function of its absolute thermodynamic temperature $B_{ν}$ and the frequency $B_{ν}$ of all the radiation that comes from it. It can be written:
 * $$B_\nu(\nu, T)=\frac{2h\nu^3}{c^2}\frac1{e^{h\nu/kT}-1},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$

where $B_{λ}$ denotes the speed of light, $B_{ν}(ν, T)$ denotes Planck's constant, and $T$ denotes Boltzmann's constant.

Great, finally a statement of Planck's law.
 * "Any surface in the cavity" further reinforces the very limited applicability of the law implied earlier.
 * "of all the radiation that comes from it." Not true, it only gives the portion of the radiation at frequency $ν$.  (Yes we know what you meant but don't assume the reader does.)

Planck's law is also expressed by giving the spectral radiance for the spectral variable $c$, the wavelength:
 * $$B_\lambda(\lambda, T)=\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}} - 1}.$$

There are still other ways of expressing Planck's law.

This confuses "form" and "expression." The frequency and wavelength forms differ in that when plotted their respective x-axes are not linearly related, which makes it important to distinguish them. The other expressions aren't different forms in that sense but simply different scalings of the same form resulting from different units, which for any practical purpose is a triviality. You seem to think that this is not the case for e.g. "spectral energy per unit volume" but that's not so, it can be plotted in either of those two forms and there is no special third form for it.

(There is the logarithmic form, which I and User:Q Science would like to see included as a third form intermediate between the above two, but three is the limit as far as the literature is concerned. Moreover all three forms can be found in Goudy and Young as well as in Greenhouse effect.  I included this form in the article in October but Headbomb deleted it and after much arguing with Headbomb about a variety of things I decided there were more productive things I could be doing for the next six months than invest the considerable effort needed to make a case against Headbomb.)

Units for the spectral radiance
In these formulas, the unit for spectral radiance is chosen according to the natural unit for the spectral argument variable. The formulas give spectral radiance as power per unit area per unit solid angle, per unit spectral quantity for the natural unit of the spectral argument variable. For example, when the spectral argument variable is frequency with unit of Hz, then the unit for the spectral radiance has a factor per unit frequency, that is Hz−1 or s. The usual unit for $h$ above is W m–2 sr–1 Hz−1.

Relating versions of the law with different spectral variables
Relating versions of the law with different spectral variables requires attention both to the functional form of the formulas and to the units. The present notation uses the spectral variable symbols in two different ways, one to label a functional form, the other to denote a value of the spectral argument.

An increment in frequency corresponds with a decrement in wavelength. Accordingly, the principle here is that for the quantities $k_{B}$ and $&nu;$, in any given interval, $λ$, of frequency in the corresponding interval, $B_{ν}(ν, T)$, of wavelength, the energy is the same, so that


 * $$B_\lambda(\lambda,\ T) \ \mathrm d \lambda=-B_\nu(\nu(\lambda),\ T) \ \mathrm d \nu$$  or
 * $$B_\lambda(\lambda,\ T) \ \ \ \ \ =-B_\nu(\nu (\lambda),\ T) \ \times \ \frac{\mathrm d \nu}{ \mathrm d \lambda}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

Thus, to go from the version for $B_{ν}(ν, T)$ to the version for $B_{λ}(λ, T)$, one can proceed as follows.

Starting with formula (1) above, substitute $dν$ for the argument variable $- dλ$ (not for the function form $B_{ν}(ν, T)$) :


 * $$B_\nu(c/ \lambda,\ T)= \frac{ 2 h(c/ \lambda )^{3}}{c^2} \frac{1}{e^\frac{hc/ \lambda}{k_\mathrm{B}T} - 1}$$


 * $$= \frac{ 2 hc}{ \lambda ^{3}} \frac{1}{e^\frac{hc}{k_\mathrm{B}\lambda T} - 1}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$

The units of the left-hand side, and thus also of the right-hand side, of the formula (3) are still per unit frequency, not per unit wavelength.

Next, change the resulting functional form and units from $B_{λ}(λ, T)$ to $c/λ$. With  $$\nu(\lambda)=c/\lambda$$  and   $$\frac{\mathrm d \nu (\lambda)}{\mathrm d \lambda}=-\frac{c}{{\lambda}^2}$$, from (2) and (3) one can write:


 * $$B_\lambda(\lambda,\ T) = \frac{ 2 hc }{{\lambda} ^{3}} \frac{1}{e^\frac{hc/ \lambda}{k_\mathrm{B}T} - 1}\ \times \ (-(-\frac{c}{{\lambda}^2}))$$


 * $$= \frac{ 2 hc^2}{ \lambda ^{5}} \frac{1}{e^\frac{hc}{k_\mathrm{B}\lambda T} - 1}.$$

This derivation is certainly needed, but note that it is considerably longer (nine sentences) than my version of the same derivation (four sentences), which (omitting the side discussion of the minus sign since you don't go into it yourself) reads as follows.


 * Planck's law can also be expressed as a function of wavelength $ν$ instead of $B_{ν}$. The evident substitution of $B_{ν}$ for $B_{λ}$ should be performed on $&lambda; = c/&nu;$ rather than just $&nu;$. The relationship between $c/&lambda;$ and $&nu;$ is obtained from the derivative of $B_{&nu;}(T) d&nu;$, namely $B_{&nu;}(T)$ (= $d&nu;$).  Substituting $d&lambda;$ for $&nu; = c/&lambda;$ and $d&nu;/d&lambda; = &minus;c/&lambda;^{2}$ for $&minus;&nu;/&lambda;$ in $c/&lambda;$ then yields $&nu;$ where
 * $$B_\lambda(T)=\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}} - 1}$$
 * is the wavelength form of Planck's law.

Does your considerably longer derivation meet a need not met in mine?

Lambertian surfaces
Some talk here, but not right here about the sun or its limb.

Why a whole section on material already covered at Lambert's cosine law and black body radiation? A single five-sentence paragraph gives a sufficient account for the purposes of applications to Planck's law, namely:


 * Radiation not normal to the surface is treated as follows. A black body is by definition a Lambertian radiator. That is, radiation from area A at an angle θ to the normal to that area is treated as coming from the smaller area A cos(θ) resulting from projecting A onto a virtual plane normal to the radiation. This virtual projected area can then be treated as though it were a normally radiating black body in its own right. It follows that the total radiation at any given frequency emitted normally to a black-body disk equals the total radiation of that frequency emitted from a spherical black body of the same temperature and radius in the same direction; indeed the two objects produce indistinguishable beams in that direction.

Moreover Planck's law is useless in any practical application without knowing how it is distributed over solid angle, which is what this describes. As such it should be one of the very first things that is said about Planck's law if it is to be at all useful to anyone. That's why I put it in the first section of my version. right after saying that $c/&lambda;^{2} d&lambda;$ gives the radiation in watts from unit area into unit solid angle over unit bandwidth.

Incidentally the "roughly speaking" there refers to the fact that Planck's law applied directly to unit solid angle is a few percent factor of 3 (!) high. This is fixed by using infinitesimal solid angle $d&nu;$ instead at some cost in understandability for readers who don't routinely use calculus. No comparable inaccuracy results from using 1 Hz in place of $B_{&nu;}(T) d&nu;$, nor any inaccuracy at all for using 1 square meter in place of dA provided the square meter is planar, but for uniformity it seems preferable to pass from (1,1,1) to $B_{&lambda;}(T) d&lambda;$ at the same time.

= Discussion with User:Q Science about the plot in my 2nd section =

Blackbody Curve Plots
I think the first section should have a plot of the 2 versions of the curve with the wavenumber and wavelength correctly aligned, ie, the same x-axis. The y-axis should also be on a common scale, not percent of total. Currently, there are no plots of this type anywhere on wikipedia. As a result, the readers will never get a feeling for how radically different the functions really are.

The scaling in the existing image (above) is ok for discussing the area under the curve, but hides the significant difference in curve shapes. Q Science (talk) 17:46, 9 May 2012


 * Easy enough to do, though zero can't be on either scale (since 1/0 is infinity) making it impractical to show as much of the curve as when the scales are both increasing. Moreover one of the two forms will have to be plotted backwards, so it will appear as the mirror image of how that form is normally plotted, with zero on the right instead of on the left as customary.  Thirdly the two scales can't both be linear, which does hide the shape of whichever curve you plot non-linearly (the shapes are correctly shown in my version, meaning as they each appear in textbooks).  These may help explain why no plots of this type can be found on Wikipedia.  The way I plotted the two forms replaces those three drawbacks with one drawback: the scales are skew-aligned.  Vertical alignment creates all three of the problems I pointed out.  --Vaughan Pratt (talk) 19:17, 9 May 2012 (UTC)
 * Incidentally the frequency-wavelength-neutral curve $B_{&nu;}(T)$ avoids all this by being a single curve that
 * has its peak in between the wavelength and frequency peaks,
 * permits plots spanning many octaves of spectrum without unduly stretching or compressing one end,
 * yields the frequency and wavelength forms simply by dividing by respectively $d&Omega;$ and $d&nu;$, and
 * has a frequency-wavelength-symmetric expression that does not mention c, namely
 * $$2h\left(\frac{\nu}{\lambda}\right)^2/(e^{h\nu/kT}-1).$$

It is possible to keep the wavenumber on the bottom and plot the wavelength function verses wavenumber. That places both plots on the same axis and in the same orientation. For the y-axis, you could scale the wavelength version such that the area of the 2 curves is the same. Doing it this way shows that the peaks are in different locations and that the wavelength height is about 20% below the wavenumber height.

Until you pointed it out, I hadn't noticed that you have both values increasing to the right. I don't remember seeing that in other sources. I think your text needs to explain why you have done this since readers may not get it without some help.

I completely support adding a section on the frequency-wavelength-neutral curve. Q Science (talk) 20:17, 9 May 2012 (UTC)


 * Thanks for supporting the FWN curve, Q Science. I do have a couple of questions about these red and purple curves though.
 * Maybe I've misinterpreted your axis labels, but if as indicated on the y-axis your units are 1000 W/m2/cm^{-1}, doesn't that mean that the area of just the bottom square from 500 to 1000 cm^{-1}, which is 500 cm^{-1} wide, will be 1000x500 = 500,000 W/m2? That seems several thousand times too large for that square, and there are 7-8 squares worth of area under the red curve.
 * I would have interpreted that to mean
 * (1 / 1000) x 500 = 0.5 W/m2
 * But the correct value should be should be about 50 W/m2. Good catch. Please suggest a better way to label the axis (other than simply using 0.1, 0.2 ...). Q Science (talk) 06:15, 10 May 2012 (UTC)
 * Ah, I see, by "Power x 1000" you didn't mean to multiply the numbers 1, 2, 3 on the y-axis by 1000 but that the value of the curve when multiplied by 1000 is the number on the y-axis. What's wrong with using 0.1, 0.2, ...?  Alternatively you can write mW instead of W and label with 100, 200, 300, ... --Vaughan Pratt (talk) 19:02, 10 May 2012 (UTC)
 * While the red curve looks fine (modulo my constant-factor question), the purple curve can't possibly work. One can only compare the peaks of the two curves when each is plotted on its own linear scale, as standardly done (I don't know where you're finding your claimed counterexamples).  You've plotted the red curve on a linear scale, which is fine, but in order to correctly show the purple curve you need to plot it on a scale that is linear in wavelength.  If you try to plot a fifth power of lambda on a scale that is linear in nu, the area under it can't be the correct area, which by the definition of integration has to be the fifth power of lambda when plotted on a scale that is linear in lambda, not in nu.
 * It would great if there were a way to plot the red and purple curves on the same graph and have lambda and nu vertically aligned, but so far I haven't found one, and am pretty sure one doesn't exist. --Vaughan Pratt (talk) 02:12, 10 May 2012 (UTC)


 * The relative positions of the peaks are correct. The wavelength (purple) peak is approximately 9615 um which maps to 1040 cm-1, as shown. Q Science (talk) 06:15, 10 May 2012 (UTC)
 * Yes, the purple curve's peak is in the right place because you've plotted it as a fifth power of lambda. But that doesn't fix the problem that it's plotted on an x-axis that is linear in nu and therefore not linear in lambda.  In order for the area under the purple curve to represent the integral, $(dA,d&Omega;,d&nu;)$ needs to be constant along the x-axis, but you have it varying from large (in fact infinite right at the left) to small.  As a result the area under the purple curve to the left of the peak understates the integral and on the right it overstates it.  --Vaughan Pratt (talk) 17:03, 10 May 2012 (UTC)
 * Good points. What makes this interesting is to make the plots at solar frequencies (5777 K). In that case, the wavelength version peaks around 500 nm (green) and the wavenumber version peaks around 870 nm (infrared). As for equal integrals, I will work on that. From the Planck's law article, the 50% point is 711 nm. Q Science (talk) 21:46, 10 May 2012 (UTC)
 * Right. If you'd like the C code that computed the 711 nm, as well as 502 nm and 882 nm for the wavelength and frequency peaks, I'll email it to you (I have your address).  --Vaughan Pratt (talk) 07:11, 11 May 2012 (UTC)