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Line 16: |
| and <math> K = J/k_B T </math> | | and <math> K = J/k_B T </math> |
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| <math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N}\sum_{S_{N+1}} e^{K S_N S_{N+1} } | | <math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N}\sum_{S_{N+1}} e^{K S_N S_{N+1} } |
| </math> | | </math> |
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| Performing the sum of the possible values of <math> S_{N+1} </math> we get: | | Performing the sum of the possible values of <math> S_{N+1} </math> we get: |
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| <math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K S_N ) \right] | | <math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K S_N ) \right] |
| </math> | | </math> |
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| Taking into account that <math> \cosh(K) = \cosh(-K) </math> | | Taking into account that <math> \cosh(K) = \cosh(-K) </math> |
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| <math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{kS_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K ) \right] | | <math> Q_{N+1} = \sum_{S_1} \sum_{S_2} e^{K S_1S_2} \sum_{S_3} e^{K S_2 S_3} \cdots \sum_{S_N} e^{K S_{N-1} S_N} \left[ 2 \cosh ( K ) \right] |
| </math> | | </math> |
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Revision as of 12:46, 23 February 2007
Model:
Consider a system with
spins in a row.
The energy of the system will be given by
,
where each variable
can be either -1 or +1.
The partition function of the system will be:
,
where
represents the possible configuration of the N spins of the system,
and
Performing the sum of the possible values of
we get:
Taking into account that
Therefore:
The Helmholtz free energy in the thermodynamic limit will be