Black Body Entropy: Solving the Puzzle of q and N

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SUMMARY

The discussion focuses on the interpretation of the variable q in the context of black body entropy and Einstein solids. The expression for entropy, S = k ln((q + N - 1)! / (q! (N - 1)!)), applies to both Einstein solids and Planck's blackbody radiation resonators, with q representing the total number of energy quanta in a system of N oscillators. It is established that q must be dimensionless, leading to the conclusion that q cannot denote the energy hν, but rather can be expressed as q = E_tot / hv, where E_tot is the total energy. This clarification is crucial for understanding the relationship between energy units and entropy in these systems.

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LCSphysicist
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Homework Statement
I am having trouble to understand how the entropy of a black body was derived .
Relevant Equations
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1607312031349.png

I thought that would be something like, using similar counts from Einstein solid, ##S = kln(\frac{(q+N-1)!}{q!(N-1)!})##
Where q is ##hv##
v is frequency

But the results are not similar, so i am little stuck
 
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Herculi said:
I thought that would be something like, using similar counts from Einstein solid, S=kln((q+N−1)!q!(N−1)!)
Yes. With a proper interpretation of ##q##, this expression for ##S## will apply to both the Einstein solid and Planck's blackbody-radiation resonators (oscillators).

Where q is hv
v is frequency
From the above expression for ##S##, ##q## must be dimensionless since ##N## and 1 are dimensionless. So, ##q## can't denote the energy ##h \nu##. What does ##q## actually represent for the Einstein solid? What does ##q## represent for the blackbody radiation system?
 
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TSny said:
Yes. With a proper interpretation of ##q##, this expression for ##S## will apply to both the Einstein solid and Planck's blackbody-radiation resonators (oscillators).

From the above expression for ##S##, ##q## must be dimensionless since ##N## and 1 are dimensionless. So, ##q## can't denote the energy ##h \nu##. What does ##q## actually represent for the Einstein solid? What does ##q## represent for the blackbody radiation system?
In Einstein solid, it is in fact the energy unit, as far as i know. I never noticed that you pointed. Thinking better, maybe q can be equal to the total energy E divided by the energy unit hv, that is, ##q = E_{tot}/hv##?
 
Herculi said:
In Einstein solid, it is in fact the energy unit, as far as i know. I never noticed that you pointed. Thinking better, maybe q can be equal to the total energy E divided by the energy unit hv, that is, ##q = E_{tot}/hv##?
Yes, ##q## is the total number of energy quanta in the system of N oscillators.
 
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