A On the Planck Blackbody Function

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Charles Link

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The graph of the Planck blackbody function has an interesting feature:## \\ ## ## \rho_o=\frac{\int\limits_{0}^{\lambda_{max}} L_{BB}(\lambda,T) \, d \lambda}{\int\limits_{0}^{+\infty} L_{BB}(\lambda, T) \, d \lambda} \approx .2500 ##,
where ## \lambda_{max} ##, in an exact derivation of Wien's law, obeys ## e^{-x}=1-\frac{x}{5} ##, where ## x \neq 0 ##, and ## x=\frac{hc}{\lambda_{max} k_B T} \approx 4.96 ##. From this we can very accurately compute ## \lambda_{max}## as a function of temperature with ## \lambda_{max}T=(\frac{hc}{k_B})(\frac{1}{X_o}) ## where ##X_o ## can be computed to any desired accuracy. The value of ## \rho_o ## can be shown to be independent of temperature ##T ##. ## \\ ## The value of ## \rho_o ## is so close to .2500, that in 1992 it took some calculus expansions long with some very high precision computing to show that the value of this fraction is not exactly equal to ## \frac{1}{4} ##, but instead approximately ## .2501 ##. (As I recall it came out to be ## \rho_o=.2500545...##). Today's very powerful computers could readily get a very accurate numerical number for this ratio to show that it is indeed ## \rho_o=.2500545... ##. Back in 1992, a couple of us made it a group effort to determine if ## \rho_o=\frac{1}{4} ##, or if ## \rho_o \approx .2500 ## , but not exact. After much effort, we concluded ## \rho_o=.2500545... ##.
 
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Charles Link

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@vanhees71 @mfb You might find this of interest.:smile::smile: ## \\ ## Hi @Greg Bernhardt Above is one more for Physics Forums. This one could almost be an Insights article, but this posting here should suffice.:smile::smile:
 
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Charles Link

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If I did some quick transformations correctly ## \rho_o=\frac{\int\limits_{X_o}^{+\infty} \frac{x^3 \,dx}{e^x-1}}{\int\limits_{0}^{+\infty} \frac{x^3 \, dx}{e^x-1}} ## where ## X_o ## is the constant of post 2 above. ## \\ ## Edit: @jedishrfu @fresh_42 To see if I did this correctly, can you evaluate this ratio for me using ## X_o=4.96 ##? Try to get 3 decimal place accuracy. Thanks. :smile::smile: For very high accuracy, you might even try using ## X_o ## as the root not equal to zero of ## e^{-x}=1-\frac{x}{5} ##. A rectangular type integration with about 100,000 intervals from ## 0 ## to ## 100 ## should suffice. Ignore ## x=0 ## in the denominator. No need to integrate that point. For increased accuracy, tighten the resolution around ## x=4.96 ##. ## \\ ## From ## 0 ## to ## .00001 ##, a Taylor expansion of ## e^x \approx 1+x +(\frac{x^2}{2}+...) ## allows for an almost exact ## \int\limits_{0}^{+.00001} x^2 \, dx=\frac{1}{3}(.00001)^3 ##. ## \\ ## For ## x>100 ##, I think it should be possible to show negligible contribution. ## \\ ## Tightening up the resolution round ## x=4.96 ## should allow one to get this ratio of integrals easily accurate to 10 decimal places or more. ## \\ ## I welcome your inputs. I currently don't have access to an EXCEL spreadsheet, but it really requires the high precision of Mathcad and/or its successors. @Dale might you give this a try? ## \\ ## Meanwhile the denominator integral actually has a closed form: ## I_D=\frac{\pi^4}{15} =6.49393940227... ##. See: https://math.stackexchange.com/questions/99843/contour-integral-for-x3-ex-1
 
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As the integral is dominated by the region near X0 you might not get the ratio with three significant figures if you give X0 with just two decimal places.

WolframAlpha calculates 1.62824 for the integral using 4.96 and 1.62394 using 4.965.

We can ask it to solve the original equation for x, the result is X0=4.96511. Using that the numerator is 1.62384.

Starting the integral at 0 we get 6.49394, the ratio is 0.2500546. We have enough significant figures to conclude that the ratio is not 0.25.

WolframAlpha links:
Find X0
Numerator
Denominator

Note that I stopped the integral at 100 because it didn't work well with infinity.
 

Charles Link

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@mfb Many thanks !! :smile::smile::smile::smile:
 

Charles Link

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@mfb See also my final line I just added to post 3. Suggestion is to integrate numerator from ## 0 ## to ## X_0 ## and subtract from the closed form result of ## 0 ## to ## +\infty ##.
 

Charles Link

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A historical note: The question came up,for a couple of us in 1992, what percentage of radiated energy lies to the left of the peak on the intensity vs. wavelength curve for a blackbody type source? ## \\ ## We consulted what is known in the IR (infrared) industry as a blackbody slide rule. The result of the mark on the slide rule showed it was so close to the .25 point for ## 0 ## to ## \lambda_{max} ## that the possibility existed that the number might be exactly .25. ## \\ ## We did arrive at the result that ## \rho_o=.2500545...##, but our approach was much more complex mathematically than the route that @mfb took above. ## \\ ## In any case, the result that ## \rho_o \approx \frac{1}{4} ##, along with Wien's law, ## \lambda_{max}T=2898 \, \mu \, K ##, can be useful in estimating the efficiency in the visible region of the spectrum of various incadescent sources. The temperature of the filament of an incadescent bulb is typically ## T \approx \, ## 2500-2800 K. Efficiencies in the visible are typically in the 10% range for incadescent sources. ## .40 \, \mu < \lambda_{visible}<.75 \, \mu ##.
 
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Charles Link

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As the integral is dominated by the region near X0 you might not get the ratio with three significant figures if you give X0 with just two decimal places.

WolframAlpha calculates 1.62824 for the integral using 4.96 and 1.62394 using 4.965.

We can ask it to solve the original equation for x, the result is X0=4.96511. Using that the numerator is 1.62384.

Starting the integral at 0 we get 6.49394, the ratio is 0.2500546. We have enough significant figures to conclude that the ratio is not 0.25.

WolframAlpha links:
Find X0
Numerator
Denominator

Note that I stopped the integral at 100 because it didn't work well with infinity.
@mfb The WolframAlpha links don't seem to be working anymore. Perhaps you can reset them and/or permanently download them.
 

Charles Link

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Scratch the last comment=the "links are working again. Adding a 5 in the next digit of ##X_o ## changed the 888 considerably, so, yes, I would agree with @mfb in his above assessment.
 
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That was interesting, thanks!
This semester I'm giving modern physics course, did the BB radiation a few weeks ago. Never thought about the ratios before though... Will mention it this week then, thanks again! :smile:
 

DrClaude

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Using Mathematica, we can really go big on the digits. I get
$$
X_0 \approx 4.96511423174427630369875913132289394405558499
$$
and then
$$
\rho_0 \approx 0.25005454682271047905965004646011027863057
$$
 

Charles Link

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That was interesting, thanks!
This semester I'm giving modern physics course, did the BB radiation a few weeks ago. Never thought about the ratios before though... Will mention it this week then, thanks again! :smile:
By using this ration ## \rho_o \approx \frac{1}{4} ##, and Stefan-Boltzmann ## \frac{\sigma T^4}{\pi}=\int\limits_{0}^{+\infty}L_{BB}(\lambda,T) \, d \lambda ##, (where Stefan's constant ## \sigma=\frac{\pi^2 k_B^4}{60 \hbar^3c^2} =5.6697 ## E-8 Watts/## (m^2 K^4)##), and Wien's law ## \lambda_{max}T=2898 \, \mu \, K ##, it really tells you the most important features of the Planck Plackbody function. ## \\ ## Meanwhile radiated power per unit area (radiant emittance) for a blackbody is ## M=\sigma T^4 ##, and brightness (radiance) ## L=\frac{M}{\pi} ##, so that irradiance ## E ## (watts/m^2) at a distance ## s ## on axis from a small blackbody source of area ## A_{BB} ## is given by ## E=\frac{LA_{BB}}{s^2} ## , and integrated over hemisphere with a ## \cos(\theta) ## fall-off , we get power ## P=M A_{BB}=L_{BB}A_{BB} \pi ##. ## \\ ## The exact expression for spectral radiance (brightness) is ## L_{BB}(\lambda,T)=\frac{2hc^2}{\lambda^5 (e^{hc/(\lambda k_B T)}-1) }##. ## \\ ## @Greg Bernhardt You just might want to feature this one. It is rather educational.
 

Charles Link

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Using Mathematica, we can really go big on the digits. I get
$$
X_0 \approx 4.96511423174427630369875913132289394405558499
$$
and then
$$
\rho_0 \approx 0.25005454682271047905965004646011027863057
$$
Mathematica must use a lot of decimal places for the exponential constant ## e ## in order to get these integrals so precise and accurate. :smile::smile::smile:
 

DrClaude

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Mathematica must use a lot of decimal places for the exponential constant ## e ## in order to get these integrals so precise and accurate. :smile::smile::smile:
As many as you want :wink:

(I don't like the new smilies...)
 

BvU

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:oldsmile:
 

BvU

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Thanks Charles, but I really don't deserve a like for that ....
If only I could cash in all or even just a few of those likes :partytime: I'd be a gold member by now !
 
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