Exploring Black Body Energy Spectrums: hv vs kbT

T)dn).15. Differentiate with respect to β: <E> = \frac{∫nhvexp(-nhv/kbT)dn}{∫exp(-nhv/kbT)dn}.16. Simplify: <E> = \frac{∫nhvexp(-nhv/kbT)dn}{Z} = \frac{∫nhvexp(-nhv/kbT)dn}{∑exp(-nhv/kbT)}.17. Use the definition of the partition function: Z = ∑exp(-nhv/kbT) = ∫exp(-nhv/kbT)dn.18. Substitute this into the equation
  • #1
HendryJacoby
1
0
The question is as follows:

The black body energy spectrum is [tex]\rho[/tex](T,v)dv=[tex]\frac{8\piv2<E(v)>}{c3}[/tex]dv where v is the frequency of the EM wave and <E(v)> is the average energy at v. Assuming the energy of a EM wave of v can only take multiples of hv, the from the Boltzman probability P(E)=exp(-E/kbT/[tex]\sumexp(-E/kbT[/tex]), calculate the average energy <E(v)>=[tex]\sumEP(E)[/tex] and [tex]\rho[/tex](T,v). Discuss the result in the limits of hv<<kbT and hv>>kbT, and compare the results with the Rayleigh-Jean Law and Wien's result.

I am fairly sure that answering this question intails basically deriving [tex]\rho[/tex](T,v) which involves putting the probabilty series within the average energy series. Then taking the simplified expression for average energy and inserting it into the black body spectrum equation. I know from online research that [tex]\rho[/tex](T,v)=[tex]\frac{2hv3}{c2}[/tex][tex]\frac{1}{e\frac{hv}{kbT}-1}[/tex].

Basically I get to the point where <E>=-[tex]\partiallnZ[/tex]/[tex]\partial[/tex][tex]\beta[/tex] where z=[tex]\sum[/tex]exp(-nhv[tex]\beta[/tex]) and [tex]\beta[/tex]=1/kbT.

I can see that if I am able to continue the derivation then the exp term can allow me to answer the questions about limit behavior. My problem is with finishing the derivation. Can anyone help?
 
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  • #2


Thank you for your question. You are correct in your approach to derive the black body energy spectrum. Here is a step-by-step guide to completing the derivation:

1. Start with the Boltzmann probability distribution: P(E) = exp(-E/kbT)/∑exp(-E/kbT), where E is the energy of the EM wave and kb is the Boltzmann constant.

2. Use the relation E = nhv, where n is the number of photons and v is the frequency of the EM wave.

3. Rewrite the sum as a series: ∑exp(-E/kbT) = ∑exp(-nhv/kbT).

4. Substitute this into the probability distribution: P(E) = exp(-E/kbT)/∑exp(-nhv/kbT).

5. Use the definition of the partition function Z: Z = ∑exp(-E/kbT) = ∑exp(-nhv/kbT).

6. Rewriting the sum in terms of Z: P(E) = exp(-E/kbT)/Z.

7. Take the natural log of both sides: lnP(E) = -E/kbT - lnZ.

8. Differentiate with respect to β = 1/kbT: ∂lnP(E)/∂β = -E/kbT² - ∂lnZ/∂β.

9. Using the definition of β and rearranging: ∂lnP(E)/∂β = -E/kbT² + <E>/kbT².

10. Solving for <E>: <E> = -∂lnZ/∂β.

11. Substitute back into the black body energy spectrum equation: \rho(T,v) = \frac{8\piv²<E(v)>}{c³}dv.

12. Substitute the expression for <E> into the above equation: \rho(T,v) = \frac{8\piv²(-∂lnZ/∂β)}{c³}dv.

13. Use the definition of Z: Z = ∑exp(-nhv/kbT) = ∫exp(-nhv/kbT)dn.

14. Substitute this into the equation for <E>: <E> = -∂lnZ/∂β = -∂/∂β(ln∫exp(-
 

1. What is black body energy spectrum?

Black body energy spectrum refers to the distribution of energy emitted by a black body at different wavelengths. This spectrum is dependent on the temperature of the black body and follows a specific curve known as the Planck curve.

2. What is the relationship between hv and kbT in the black body energy spectrum?

Hv and kbT are both variables used to describe the distribution of energy in the black body energy spectrum. Hv represents the energy of a single photon at a specific wavelength, while kbT represents the thermal energy of the black body at a given temperature. The ratio of hv to kbT determines the shape and intensity of the spectrum.

3. How does the temperature of a black body affect its energy spectrum?

The temperature of a black body directly affects the shape and intensity of its energy spectrum. As the temperature increases, the peak of the spectrum shifts to shorter wavelengths and the overall intensity increases. At extremely high temperatures, the spectrum will approach a continuous distribution.

4. What is the significance of the black body energy spectrum in astrophysics?

The black body energy spectrum is significant in astrophysics because it can provide information about the temperature and composition of celestial bodies. By analyzing the spectrum of light emitted by a star or planet, scientists can determine its temperature and the elements present in its atmosphere.

5. How does the black body energy spectrum relate to the laws of thermodynamics?

The black body energy spectrum is related to the laws of thermodynamics because it follows the principles of energy conservation and entropy. According to the second law of thermodynamics, the total energy emitted by a black body must increase with temperature, and the third law states that at absolute zero temperature, the spectrum will approach zero intensity at all wavelengths.

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