On the Planck Blackbody Function

In summary: The blackbody on the slide is placed so the emission peak is at the center of the screen. The radiated energy is then calculated as a function of wavelength. To our surprise, we found that the radiated energy was about 25% to the left of the peak on the graph! ## \\ ## This discrepancy between theory and observation was not resolved until Wien's law was found to be valid and e^{-x}=1-\frac{x}{5}.
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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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@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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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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  • #4
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.
 
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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 ##.
 
  • #8
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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mfb said:
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.
 
  • #10
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.
 
  • #11
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:
 
  • #12
Using Mathematica, we can really go big on the digits. I get
$$
X_0 \approx 4.96511423174427630369875913132289394405558499
$$
and then
$$
\rho_0 \approx 0.25005454682271047905965004646011027863057
$$
 
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  • #13
erbahar said:
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 ratio ## \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.
 
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  • #14
DrClaude said:
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:
 
  • #15
Charles Link said:
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...)
 
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:oldsmile:
 
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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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1. What is the Planck Blackbody Function?

The Planck Blackbody Function, also known as the Planck radiation law, is a mathematical function that describes the spectral density of electromagnetic radiation emitted by a blackbody at a given temperature. It was developed by physicist Max Planck in 1900 and is a fundamental concept in understanding the properties of light and heat.

2. How is the Planck Blackbody Function used in science?

The Planck Blackbody Function is used in a variety of fields, including astrophysics, cosmology, and thermodynamics. It is used to calculate the amount and distribution of radiation emitted by objects at different temperatures, such as stars and planets. It also helps scientists understand the properties of light and its interactions with matter.

3. What is the significance of the Planck Blackbody Function?

The Planck Blackbody Function is significant because it provided a solution to the "ultraviolet catastrophe" problem, which was a discrepancy between classical physics and experimental observations of blackbody radiation. It also laid the foundation for the development of quantum mechanics and our understanding of the behavior of subatomic particles.

4. How does the Planck Blackbody Function relate to the Planck constant?

The Planck constant, denoted by the symbol h, is a fundamental physical constant that relates the energy of a photon to its frequency. It is a crucial component of the Planck Blackbody Function, as it is used to calculate the intensity of radiation emitted by a blackbody at a given temperature. In fact, the Planck constant was first introduced by Max Planck in his derivation of the Planck Blackbody Function.

5. Can the Planck Blackbody Function be applied to real-world objects?

Yes, the Planck Blackbody Function can be applied to real-world objects that emit thermal radiation, such as stars, planets, and even everyday objects like a lightbulb. However, it is important to note that the Planck Blackbody Function is a theoretical model and may not perfectly describe the behavior of all objects. In some cases, other factors such as absorption and reflection may need to be considered in addition to the Planck Blackbody Function.

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