Jump to content

Photon radiant emmittance and energy density derivation...


Recommended Posts

Posted (edited)

Photon mass:

[math]\boxed{m_{\gamma} = 0}[/math]

 

Photon species total effective degeneracy number:

[math]\boxed{N_{\gamma} = 2}[/math]

 

Photon radiation energy radiant emmittance Bose-Einstein distribution integration via substitution: (ref. 1)

[math]j^{*} = \sigma_{\gamma} T_{\gamma}^4 = \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \cos \phi \sin \phi \; d \phi \; d \theta \; \frac{N_{\gamma} E_{\gamma}^4}{c^2 (2 \pi \hbar)^3} \int_{0}^\infty \frac{E_t (\omega)^3}{e^{\frac{E_t (\omega)}{E_1 (T_{\gamma})}} - 1} d \omega = \frac{\pi N_{\gamma} (k_B T_{\gamma})^4}{c^2 (2 \pi \hbar)^3} \left( \frac{\pi^4}{15} \right) = \frac{N_{\gamma} \pi^2 (k_B T_{\gamma})^4}{120 c^2 \hbar^3}[/math]

 

Radiant emmittance Stefan-Boltzmann constant: (ref. 2)

[math]\boxed{\sigma_{\gamma} = \frac{N_{\gamma} \pi^2 k_B^4}{120 c^2 \hbar^3}}[/math]

 

Radiant emmittance Stefan-Boltzmann law: (ref. 2)

[math]\boxed{j^{*} = \frac{N_{\gamma} \pi^2 (k_B T_{\gamma})^4}{120 c^2 \hbar^3}}[/math]

 

Photon radiation energy density Bose-Einstein distribution integration via substitution: (ref. 3, pg. 43, eq. 204-206)

[math]\epsilon_{\gamma} = \alpha_{\gamma} T_{\gamma}^4 = \int_{0}^{2 \pi} \int_{0}^{\pi} \sin \theta \; d \theta \; d \phi \; \frac{N_{\gamma} E_{\gamma}^4}{(2 \pi \hbar c)^3} \int_{0}^\infty \frac{E_t (\omega)^3}{e^{\frac{E_t (\omega)}{E_1 (T_{\gamma})}} - 1} d \omega = \frac{4 \pi N_{\gamma} (k_B T_{\gamma})^4}{(2 \pi \hbar c)^3} \left( \frac{\pi^4}{15} \right) = \frac{N_{\gamma} \pi^2 (k_B T_{\gamma})^4}{30 (\hbar c)^3}[/math]

 

Photon radiation constant:

[math]\boxed{\alpha_{\gamma} = \frac{N_{\gamma} \pi^2 k_B^4}{30 (\hbar c)^3}}[/math]

 

Photon radiation energy density:

[math]\boxed{\epsilon_{\gamma} = \frac{N_{\gamma} \pi^2 (k_B T_{\gamma})^4}{30 (\hbar c)^3}}[/math]

 

Any discussions and/or peer reviews about this specific topic thread?

 

Reference:
Wikipedia - Stefan–Boltzmann law - derivation from Planck's law: (ref. 1)
https://en.wikipedia.org/wiki/Stefan–Boltzmann_law#Derivation_from_Planck.27s_law

Wikipedia - Stefan-Boltzmann law: (ref. 2)
https://en.wikipedia.org/wiki/Stefan–Boltzmann_law

PHYS: 652 Cosmic Inventory I: Radiation: (ref. 3)
http://www.nicadd.niu.edu/~bterzic/PHYS652/Lecture_09.pdf

 

 

Edited by Orion1
source code correction
Posted

Your answer differs from that of the derivation on Wikipedia.

For discussion, I imagine the citations given would be a good place to start. Given that the Stefan-Boltzmann law is >100 years old, probably not a lot of recent papers on it.

Posted (edited)

Photon species total effective degeneracy number:
[math]\boxed{N_{\gamma} = 2}[/math]

Planck's law: (ref. 1)
[math]\boxed{I_{\gamma}(\nu,T_{\gamma}) = \frac{N_{\gamma} h \nu^3}{c^2 (e^{\frac{E_t}{E_{\gamma}}} - 1)}}[/math]

Radiant emmittance integration via substitution: (ref. 2)
[math]j^* = \int d\Omega \int_0^\infty I_{\gamma}(\nu,T_{\gamma}) \; d\nu[/math]

[math]\int d\Omega = \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \cos \phi \sin \phi \; d \phi \; d\theta[/math]

[math]j^* = \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \cos \phi \; \sin \phi \; d \phi \; d\theta \int_0^\infty I_{\gamma}(\nu,T_{\gamma}) \; d\nu[/math]

[math]j^* = \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \cos \phi \; \sin \phi \; d \phi \; d\theta \; \frac{N_{\gamma} h}{c^2} \int_0^\infty \frac{\nu^3}{e^{\frac{E_t}{E_{\gamma}}} - 1} \; d\nu[/math]

Differential calculus theorem:
[math]\boxed{\frac{\int_a^b f(u)^n \; du}{\int_a^b f(v)^n \; dv} = \left( \frac{du}{dv} \right)^{n+1}}[/math]

[math]\int_0^\infty \frac{\nu^3}{e^{\frac{E_t}{E_{\gamma}}} - 1} \; d\nu = \left( \frac{d\nu}{du} \right)^4 \int_0^\infty \frac{u^3}{e^u - 1} \; du[/math]

[math]j^* = \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \cos \phi \; \sin \phi \; d \phi \; d\theta \; \frac{N_{\gamma} h}{c^2} \left( \frac{d\nu}{du} \right)^4 \int_0^\infty \frac{u^3}{e^u - 1} \; du[/math]

[math]\frac{d\nu}{du} = \frac{E_{\gamma}}{h}[/math]

[math]j^* = \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \cos \phi \; \sin \phi \; d \phi \; d\theta \; \frac{N_{\gamma} h}{c^2} \left( \frac{E_{\gamma}}{h} \right)^4 \int_{0}^\infty \frac{E_t (\nu)^3}{e^{\frac{E_t (\nu)}{E_{\gamma} (T_{\gamma})}} - 1} d \nu[/math]

[math]j^* = \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \cos \phi \; \sin \phi \; d \phi \; d\theta \; \frac{N_{\gamma} E_{\gamma}^4}{c^2 h^3} \int_{0}^\infty \frac{E_t (\nu)^3}{e^{\frac{E_t(\nu)}{E_{\gamma} (T_{\gamma})}} - 1} d \nu[/math]

[math]j^{*} = \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \cos \phi \sin \phi \; d \phi \; d \theta \; \frac{N_{\gamma} E_{\gamma}^4}{c^2 (2 \pi \hbar)^3} \int_{0}^\infty \frac{E_t (\omega)^3}{e^{\frac{E_t (\omega)}{E_{\gamma} (T_{\gamma})}} - 1} d \omega = \frac{N_{\gamma} \pi^2 (k_B T_{\gamma})^4}{120 c^2 \hbar^3}[/math]

Radiant emmittance Stefan-Boltzmann law: (ref. 3)
[math]\boxed{j^{*} = \frac{N_{\gamma} \pi^2 (k_B T_{\gamma})^4}{120 c^2 \hbar^3}}[/math]

Is there a formal name or formal method name for this differential calculus theorem?
[math]\boxed{\frac{\int_a^b f(u)^n \; du}{\int_a^b f(v)^n \; dv} = \left( \frac{du}{dv} \right)^{n+1}}[/math]

Any discussions and/or peer reviews about this specific topic thread?

Reference:
Wikipedia - Planck's law: (ref. 1)
https://en.wikipedia.org/wiki/Planck's_law

Wikipedia - Stefan–Boltzmann law - derivation from Planck's law: (ref. 2)
https://en.wikipedia.org/wiki/Stefan–Boltzmann_law#Derivation_from_Planck.27s_law

Wikipedia - Stefan-Boltzmann law: (ref. 3)
https://en.wikipedia.org/wiki/Stefan–Boltzmann_law

Edited by Orion1
source code correction
Posted
1 hour ago, Orion1 said:

Any discussions and/or peer reviews about this specific topic thread?

Re-posting, asking the same question and ignoring a response is not a tactic I would advise.

Posted
1 hour ago, Orion1 said:

j=Nγπ2(kBTγ)4120c23

Is there a formal name or formal method name for this differential calculus theorem?
baf(u)ndubaf(v)ndv=(dudv)n+1

 

looks like a differentation by parts 

 

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.