Bose-Einstein numerical integration

In summary, the conversation discusses the integration of total energy density over all photon energies between two different temperature values. It is suggested to use the equation ##N/V = C T^3## and integrate ##T^3## with the proper constant outside the integral. The procedure for numerical integration is the same for both energy density and intensity. The constant for integration over wavelength is ##8 \pi h c##, while for integration over temperature it is ##-8 \pi (kT)^4 (hc)^{-3}##. The use of wavelength or frequency in Planck's law depends on the context and can be converted using the factor ##C/4##.
  • #1
TeslaPow
40
1
Want to integrate the total energy density over all photon energies between two
temperature values from 500K to 5800K, but not sure how to proceed.

Here is some examples to help:

bose einstein photon.jpg
Bose-Einstein Density.jpg
 

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  • #2
TeslaPow said:
Want to integrate the total energy density over all photon energies between two
temperature values from 500K to 5800K, but not sure how to proceed.
I don't see how it would make sense physically to integrate over temperature.

In any case, you have ##N/V = C T^3## [see the last line of the solution of 40. (a)], which shouldn't be too hard to integrate :smile:
 
  • #3
DrClaude said:
I don't see how it would make sense physically to integrate over temperature.

In any case, you have ##N/V = C T^3## [see the last line of the solution of 40. (a)], which shouldn't be too hard to integrate :smile:

By C you mean: 8 *pi * k / (hc)^3 ?
 
  • #4
TeslaPow said:
By C you mean: 8 *pi * k / (hc)^3 ?
That should be ##k^3## and you forgot the ##\Gamma(3) \zeta(3) \approx 2.40## factor.
 
  • #5
DrClaude said:
That should be ##k^3## and you forgot the ##\Gamma(3) \zeta(3) \approx 2.40## factor.

So I also have to multiply the constant outside the integral with 2.40? What value will I use for T in the same equation on the left side if I integrate between 500K and 5500K ?
 
  • #7
TeslaPow said:
So I also have to multiply the constant outside the integral with 2.40? What value will I use for T in the same equation on the left side if I integrate between 500K and 5500K ?
I don't understand. You said you
TeslaPow said:
Want to integrate the total energy density over all photon energies between two
temperature values from 500K to 5800K, but not sure how to proceed.
which I take to mean
$$
\int_{500\ \mathrm{K}}^{5800\ \mathrm{K}} \frac{N}{V} dT
$$
So you simply need to integrate ##T^3## with the proper constant outside the integral.

TeslaPow said:
When looking at Stefan-Boltzmann law, and how the procedure is done there,
As I said above, I do not understand what an integral over temperature physically means here. Usually one is interested in the value at a given ##T## or in the change with respect to ##T##.
TeslaPow said:
is this the equation to use if I want to find the integral between 500K and 5500K?
Not if you are interested in the energy density.

Maybe it would help to explain what physical situation you are considering?
 
  • #8
I thought maybe that an integration was necessary on the energy density, but it seems that the Wien displacement law is used to find the peak curve and then you use Stefan Boltzmann law to integrate between wavelengths within that peak. Stefan law is the Planck radiation formula multiplied by C/4. Does this seem reasonable?
 
Last edited:
  • #9
TeslaPow said:
I thought maybe that an integration was necessary on the energy density, but it seems that the Wien displacement law is used to find the peak curve and then you use Stefan Boltzmann law to integrate between wavelengths within that peak. Stefan law is the Planck radiation formula multiplied by C/4. Does this seem reasonable?
Everything comes from Planck's law, which gives the energy density of a photon gas per wavelength/frequency. Wien's displacement law is simply the maximum of that distribution. Stefan's law gives, which is indeed Planck's law multiplied by ##C/4## if you use the correct ##C##, gives you the power per unit area emitted by a blackbody.

Since you didn't say what you are trying to calculate, I cannot say more about what it is you should be considering.
 
  • #10
Is the procedure for the numerical integration for Planck's radiation law the same for the energy density
as it is for the intensity? How is the value calculated here in b) ?
Uten navn.jpg

stefan.jpg
 
  • #11
TeslaPow said:
Is the procedure for the numerical integration for Planck's radiation law the same for the energy density
as it is for the intensity?
Yes.

TeslaPow said:
How is the value calculated here in b) ?
There is no integration here, you simply plug in the values of λ and T.
 
  • #12
DrClaude said:
Yes.

So the value for the constant outside the integral in the energy density is
8 * pi * c^(2) * h ?
 
  • #13
It depends what you are integrating. Are you integrating over wavelength, frequency, temperature?
 
  • #14
DrClaude said:
It depends what you are integrating. Are you integrating over wavelength, frequency, temperature?

Wavelength, is it just using the value I wrote one thread above? What will it be for frequency and temperature?
 
  • #15
The energy density per unit wavelength is given in your post #6
$$
\frac{U}{V} = \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}
$$
so if you integrate over λ, the constant is ##8 \pi h c##.

However, if you make the substitution ##x = hc/\lambda kT##, then
$$
\begin{align*}
\int_{\lambda_1}^{\lambda_2} \frac{U}{V} d\lambda &= \int_{\lambda_1}^{\lambda_2} \frac{8 \pi h c}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1} d \lambda
&= -\frac{8 \pi (kT)^4}{(hc)^3} \int_{x_1}^{x_2} \frac{x^3}{e^{x} - 1} d x
\end{align*}
$$
so the constant is ##-8 \pi (kT)^4 (hc)^{-3}##.
 
  • #16
Uten navn1.jpg

So this is used for the integration?
 
  • #17
TeslaPow said:
View attachment 272035
So this is used for the integration?
In the second case I presented, yes. But in your post #10, problem 24(a) is calculated using a direct integration as a function of λ, as in the first case I presented.
 
  • #18
The integration I outlined is used for temperature? What I don't understand quite is why wavelength is used in both energy density and intensity as the factor C/4 is used to convert between these.
 
  • #19
TeslaPow said:
The integration I outlined is used for temperature?
Again, I have never seen Planck's law being integrated over temperature. I do not understand what this would mean physically. Planck's law gives the energy density of a photon gas at equilibrium , hence at a given temperature.

TeslaPow said:
What I don't understand quite is why wavelength is used in both energy density and intensity as the factor C/4 is used to convert between these.
You can write Planck's law as a function of frequency or wavelength. How to go from energy density of a photon gas to emission from a black body is explained in statistical physics textbooks. A couple of online references that might be useful:
https://en.wikipedia.org/wiki/Stefan–Boltzmann_law#Derivation_from_Planck's_law
Lecture 25. Blackbody Radiation (Ch. 7)www.physics.rutgers.edu › ~gersh (slide 13)
 
  • #21
Is this the right equation to use for the answer in 24 b) ?
 
  • #22
TeslaPow said:
Last question, as for 24 b) in #10, the answer for the first intensity should be I(400nm,T) = 335289 W/m^2

https://www.wolframalpha.com/input/?i=solve(x/(4.593*10^(6))=0.073,x)
I don't understand what you are doing here. You seem to be calculating I(400nm,T) / (integral of I over all λ) instead of I(400nm,T)/I(966nm,T).

TeslaPow said:
This is correct.
 
  • #23
Thanks for your guidance and help, learned a lot and really appreciate it. Here is a video I think you will
like,
 

What is Bose-Einstein numerical integration?

Bose-Einstein numerical integration is a computational method used to numerically solve equations that involve Bose-Einstein statistics. It is based on the principles of quantum mechanics and is used to calculate the behavior of particles at very low temperatures.

How does Bose-Einstein numerical integration work?

Bose-Einstein numerical integration involves discretizing the energy levels of a system and using a numerical algorithm to calculate the probability of particles occupying each energy level. This is then used to calculate the total energy and other properties of the system.

What are the advantages of using Bose-Einstein numerical integration?

One advantage of using Bose-Einstein numerical integration is that it allows for the calculation of properties of a system at very low temperatures, where classical statistical methods are not applicable. It is also a more accurate method compared to other numerical techniques.

What are the limitations of Bose-Einstein numerical integration?

Bose-Einstein numerical integration can be computationally intensive and may not be suitable for systems with a large number of particles. It also assumes that the particles in the system are in thermal equilibrium, which may not always be the case.

What are some real-world applications of Bose-Einstein numerical integration?

Bose-Einstein numerical integration is commonly used in the field of condensed matter physics to study the behavior of particles in systems such as superfluids and Bose-Einstein condensates. It also has applications in astrophysics, where it is used to model the behavior of particles in extreme environments such as neutron stars.

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