- Summary
- It is concluded in MB Statistics that ##\beta = -\frac{1}{k_BT}## But this is based on Boltzmann Statistics when degeneracy of quantum states is not taken into account. If it is taken into account, the equation to solve for ##\beta## gives a different value. Is this correct or am I misunderstanding something?

Hello,

The relationship between entropy ##S##, the total number of particles ##N##, the total energy ##U(β)##, the partition function ##Z(β## and a yet to be defined constant ##β## is:

$$S(\beta)=k_BN \cdot \ln(Z(\beta)) - \beta k_B \cdot U(\beta)$$

Which leads to:

$$\frac{dS}{d\beta} = -k_B\beta \cdot \frac{dU}{d\beta}$$

And since ##dS = \frac{dU}{T}##, this means that ##\beta = -\frac{1}{k_BT}##. This is the known derived value for ##\beta## for the Maxwell-Boltzmann Distribution. Source starting from sheet 40

However, this derivation does not take the degeneracy of quantum states into account. If it does then ##S(β)## would have an extra parameter in its formula. If the number of quantum states of an energy level is ##gj##, then this wold be:

$$S = k_BN \cdot \ln(Z(\beta)) - k_B\beta\cdot U(\beta) + k_B\cdot \sum^n_{j=1}\bigg[\ln(g_j)\cdot \frac{N}{Z(\beta)} \cdot e^{\beta E_j}\bigg]$$

And since ##\ln(g_j) = \ln\big(\frac{N}{Z(\beta)}e^{\beta E_j}\big)- \beta E_j - \frac{N}{Z(\beta)}##, this would eventually give me:

$$\frac{dS}{d\beta} = k_B\cdot\left( - \beta\cdot\frac{dU}{d\beta} + \frac{U(\beta)\cdot N}{Z(\beta)} - U(\beta) \right)$$

Derivation under Part II (I asked about the derivation there and someone helped me with it)

But using again ##dS = \frac{dU}{T}##, this relationship doest

Is it permitted for ##β## to have a different value than ##-\frac{1}{k_BT}## when quantum states is taken into account or am I misunderstanding something here?

The relationship between entropy ##S##, the total number of particles ##N##, the total energy ##U(β)##, the partition function ##Z(β## and a yet to be defined constant ##β## is:

$$S(\beta)=k_BN \cdot \ln(Z(\beta)) - \beta k_B \cdot U(\beta)$$

Which leads to:

$$\frac{dS}{d\beta} = -k_B\beta \cdot \frac{dU}{d\beta}$$

And since ##dS = \frac{dU}{T}##, this means that ##\beta = -\frac{1}{k_BT}##. This is the known derived value for ##\beta## for the Maxwell-Boltzmann Distribution. Source starting from sheet 40

However, this derivation does not take the degeneracy of quantum states into account. If it does then ##S(β)## would have an extra parameter in its formula. If the number of quantum states of an energy level is ##gj##, then this wold be:

$$S = k_BN \cdot \ln(Z(\beta)) - k_B\beta\cdot U(\beta) + k_B\cdot \sum^n_{j=1}\bigg[\ln(g_j)\cdot \frac{N}{Z(\beta)} \cdot e^{\beta E_j}\bigg]$$

And since ##\ln(g_j) = \ln\big(\frac{N}{Z(\beta)}e^{\beta E_j}\big)- \beta E_j - \frac{N}{Z(\beta)}##, this would eventually give me:

$$\frac{dS}{d\beta} = k_B\cdot\left( - \beta\cdot\frac{dU}{d\beta} + \frac{U(\beta)\cdot N}{Z(\beta)} - U(\beta) \right)$$

Derivation under Part II (I asked about the derivation there and someone helped me with it)

But using again ##dS = \frac{dU}{T}##, this relationship doest

*not*give ##\beta = -\frac{1}{k_BT}##.Is it permitted for ##β## to have a different value than ##-\frac{1}{k_BT}## when quantum states is taken into account or am I misunderstanding something here?