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- Homework Statement
- Find the Entropy of spin-1/2 Paramagnetic gas

- Relevant Equations
- ## S = k_B \ln \Omega(U) ##

## S = \Sigma_r P_r \ln P_r ##.

As we know, dipole can be only arranged either parallel or anti-parallel with respect to applied magnetic field ## \vec{H} ## if we are to use quantum mechanical description, then parallel magnetic dipoles will have energy ## \mu H ## and anti-parallel magnetic dipoles have energy ## -\mu H##.

Let the number of magnetic dipoles with ## m_s = + \frac{1}{2} ## spin are ## n_{\uparrow} ## and number of magnetic dipoles with ## m_s = - \frac{1}{2} ## spin are ## n_{\downarrow} ##. Total number of dipoles are ## N = n_{\uparrow} + n_{\downarrow} ##. So, ## n_{\downarrow} = N - n_{\uparrow} ##. And, total internal energy ## U = n_{\uparrow} \mu H - n_{\downarrow} \mu H = ( 2 n_{\uparrow} - N ) \mu H \Longrightarrow n_{\uparrow} = \dfrac{1}{2} \left( \dfrac{U}{\mu H} + N \right) ##.

Now, total number of microstates having internal energy U will be $$\Omega (U) = \dfrac{N!}{n_{\uparrow} ! ( N - n_{\uparrow} ) !}$$. Using ## S = k_B ln \Omega(U) ##, we get ## S = k_B [ \ln N! - \ln n_{\uparrow} ! - \ln (N - n_{\uparrow}) ! ] ##. Using

Using expression ## S = N k_B \Sigma_r P_r ln P_r ##, we have ## S = N k_B [ p \ln (p) + (1-p) \ln (1-p) ] ##. If we take thermodynamic probability ## p = \dfrac{e^{\frac{- \mu H}{k_B T}}}{e^{\frac{- \mu H}{k_B T}} + e^{\frac{\mu H}{k_B T}}} ##. Upon simplifying this, I do not seem to get above expression. Answer comes in terms of hyperbolic functions.

Now, my confusion is - which expression is correct one - method 1 or method 2 ? Answer is given - $$S = N k_B \left[\left( \dfrac{-N \mu H+ U }{2 N \mu H} \right) \ln \left( \dfrac{N \mu H+U }{2 \mu H} \right) - \left( \dfrac{N \mu H+ U }{2 N \mu H} \right) \ln \left( \dfrac{N \mu H- U }{2 N \mu H}\right) \right]$$. . Now, I am not getting this expression from above ones. This is nearer from first method, but it does not seem obtainable from first method as well. I think Answer can be incorrect.

I am confused about which formula to use. Of course, only one answer is there. But I require clarification about which step I can proceed with under what Conditions.I am beginner in this subject. So I want clarification like, in what limit does this stirling approximation apply ? When temperature tend toward infinity or something else ? I have seen expression ## S = N k_B \ln 2 ## for infinite temperature limit. and ##S = 0## for ## T \rightarrow 0 ##.

Kindly share any relevant link or resource which address this issue. I am looking into Statistical Mechanics book by F.Rief, it has section in 6th chapter for using approximations for canonical ensembles. It describes that calculation of number of microstates ## \Omega(E) ## in range ## E < E_r < E + \delta E## can be difficult to calculate. Is this related to this? I don't think but still mentioning in case I am missing out something.

Thanks.

**Method 1:**Let the number of magnetic dipoles with ## m_s = + \frac{1}{2} ## spin are ## n_{\uparrow} ## and number of magnetic dipoles with ## m_s = - \frac{1}{2} ## spin are ## n_{\downarrow} ##. Total number of dipoles are ## N = n_{\uparrow} + n_{\downarrow} ##. So, ## n_{\downarrow} = N - n_{\uparrow} ##. And, total internal energy ## U = n_{\uparrow} \mu H - n_{\downarrow} \mu H = ( 2 n_{\uparrow} - N ) \mu H \Longrightarrow n_{\uparrow} = \dfrac{1}{2} \left( \dfrac{U}{\mu H} + N \right) ##.

Now, total number of microstates having internal energy U will be $$\Omega (U) = \dfrac{N!}{n_{\uparrow} ! ( N - n_{\uparrow} ) !}$$. Using ## S = k_B ln \Omega(U) ##, we get ## S = k_B [ \ln N! - \ln n_{\uparrow} ! - \ln (N - n_{\uparrow}) ! ] ##. Using

*stirling approximation*, $$S = k_B [ N ln N - N - (n_{\uparrow} \ln n_{\uparrow} - n_{\uparrow}) - ( (N - n_{\uparrow}) \ln (N - n_{\uparrow}) - (N - n_{\uparrow}) ) ] $$. Upon simplifying this we get $$S = k_B \left[ N \ln N - \left( \dfrac{N \mu H- U }{2 \mu H} \right) \ln \left( \dfrac{N \mu H- U }{2 \mu H} \right) - \left( \dfrac{N \mu H+ U }{2 \mu H} \right) \ln \left( \dfrac{N \mu H+ U }{2 \mu H}\right) \right]$$.**Method 2:**Using expression ## S = N k_B \Sigma_r P_r ln P_r ##, we have ## S = N k_B [ p \ln (p) + (1-p) \ln (1-p) ] ##. If we take thermodynamic probability ## p = \dfrac{e^{\frac{- \mu H}{k_B T}}}{e^{\frac{- \mu H}{k_B T}} + e^{\frac{\mu H}{k_B T}}} ##. Upon simplifying this, I do not seem to get above expression. Answer comes in terms of hyperbolic functions.

Now, my confusion is - which expression is correct one - method 1 or method 2 ? Answer is given - $$S = N k_B \left[\left( \dfrac{-N \mu H+ U }{2 N \mu H} \right) \ln \left( \dfrac{N \mu H+U }{2 \mu H} \right) - \left( \dfrac{N \mu H+ U }{2 N \mu H} \right) \ln \left( \dfrac{N \mu H- U }{2 N \mu H}\right) \right]$$. . Now, I am not getting this expression from above ones. This is nearer from first method, but it does not seem obtainable from first method as well. I think Answer can be incorrect.

I am confused about which formula to use. Of course, only one answer is there. But I require clarification about which step I can proceed with under what Conditions.I am beginner in this subject. So I want clarification like, in what limit does this stirling approximation apply ? When temperature tend toward infinity or something else ? I have seen expression ## S = N k_B \ln 2 ## for infinite temperature limit. and ##S = 0## for ## T \rightarrow 0 ##.

Kindly share any relevant link or resource which address this issue. I am looking into Statistical Mechanics book by F.Rief, it has section in 6th chapter for using approximations for canonical ensembles. It describes that calculation of number of microstates ## \Omega(E) ## in range ## E < E_r < E + \delta E## can be difficult to calculate. Is this related to this? I don't think but still mentioning in case I am missing out something.

Thanks.

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