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| <math> | | <math> |
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− | 《数学》
| + | p_i = \frac{\Omega_B(E - E_i)}{\Omega_{(S,B)}(E)}. |
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| + | </math> |
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| + | Assuming that the heat bath's internal energy is much larger than the energy of S (E ≫ E<sub>i</sub>), we can Taylor-expand <math>\Omega_B</math> to first order in E<sub>i</sub> and use the thermodynamic relation <math>\partial S_B/\partial E = 1/T</math>, where here <math>S_B</math>, <math>T</math> are the entropy and temperature of the bath respectively: |
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− | p_i = \frac{\Omega_B(E - E_i)}{\Omega_{(S,B)}(E)}. | + | 假设热水池的内能远大于热水池的内能''S'' (''E'' ≫ ''E<sub>i</sub>'') ,我们可以对E<sub>i</sub> 进行一阶泰勒展开 <math>\Omega_B</math> ,并利用热力学关系式 <math>\partial S_B/\partial E = 1/T</math>,这里<math>S_B</math>, <math>T</math> 分别是热水池的熵和温度: |
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− | P _ i = frac { Omega _ b (e-e _ i)}{ Omega _ {(s,b)}(e)}.
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− | :<math>
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| + | <math> |
| + | k \ln p_i &= k \ln \Omega_B(E - E_i) - k \ln \Omega_{(S,B)}(E) \\[5pt] |
| </math> | | </math> |
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− | p_i = \frac{\Omega_B(E - E_i)}{\Omega_{(S,B)}(E)}.
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− | Assuming that the heat bath's internal energy is much larger than the energy of S (E ≫ E<sub>i</sub>), we can Taylor-expand <math>\Omega_B</math> to first order in E<sub>i</sub> and use the thermodynamic relation <math>\partial S_B/\partial E = 1/T</math>, where here <math>S_B</math>, <math>T</math> are the entropy and temperature of the bath respectively:
| + | <math> |
| + | &\approx -\frac{\partial\big(k \ln \Omega_B(E)\big)}{\partial E} E_i + k \ln\Omega_B(E) - k \ln \Omega_{(S,B)}(E) |
| + | </math> |
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− | 假设热水池的内能远大于热水池的内能''S'' (''E'' ≫ ''E<sub>i</sub>'') ,我们可以对E<sub>i</sub> 进行一阶泰勒展开 <math>\Omega_B</math> ,并利用热力学关系式 <math>\partial S_B/\partial E = 1/T</math>,这里<math>S_B</math>, <math>T</math> 分别是热水池的熵和温度:
| + | <math> |
| + | &\approx -\frac{\partial\big(k \ln \Omega_B(E)\big)}{\partial E} E_i + k \ln\Omega_B(E) - k \ln \Omega_{(S,B)}(E) |
| + | </math> |
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| <math> | | <math> |
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| <math> | | <math> |
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− | Thus
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| p_i \propto e^{-E_i/(kT)} = e^{-\beta E_i}. | | p_i \propto e^{-E_i/(kT)} = e^{-\beta E_i}. |
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− | :<math>
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| </math> | | </math> |
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| Z = \sum_i e^{-\beta E_i} = \frac{\Omega_{(S,B)}(E)}{\Omega_B(E)}. | | Z = \sum_i e^{-\beta E_i} = \frac{\Omega_{(S,B)}(E)}{\Omega_B(E)}. |
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− | :<math>
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| </math> | | </math> |
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| <math>\langle E \rangle = \sum_s E_s P_s = \frac{1}{Z} \sum_s E_s | | <math>\langle E \rangle = \sum_s E_s P_s = \frac{1}{Z} \sum_s E_s |
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− | [数学]长角 e rangle = sum _ s e _ s p _ s = frac {1}{ z } sum _ s e _ s
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− | | + | e^{- \beta E_s} = - \frac{1}{Z} \frac{\partial}{\partial \beta} |
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− | e^{- \beta E_s} = - \frac{1}{Z} \frac{\partial}{\partial \beta} | + | Z(\beta, E_1, E_2, \cdots) = - \frac{\partial \ln Z}{\partial \beta} |
− | | + | |
− | E ^ {-beta e _ s } =-frac {1}{ z } frac { partial beta }
| + | </math> |
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− | : <math>\langle E \rangle = \sum_s E_s P_s = \frac{1}{Z} \sum_s E_s
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− | Z(\beta, E_1, E_2, \cdots) = - \frac{\partial \ln Z}{\partial \beta} | |
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− | Z (beta,e_1,e_2,cdots) =-frac { partial ln z }{ partial beta }
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− | e^{- \beta E_s} = - \frac{1}{Z} \frac{\partial}{\partial \beta}
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− | </math> | |
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| or, equivalently, | | or, equivalently, |
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| 或者,等价地说, | | 或者,等价地说, |
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− | <math>\langle E\rangle = k_B T^2 \frac{\partial \ln Z}{\partial T}.</math>
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| : <math>\langle E\rangle = k_B T^2 \frac{\partial \ln Z}{\partial T}.</math> | | : <math>\langle E\rangle = k_B T^2 \frac{\partial \ln Z}{\partial T}.</math> |
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| 如果微态能量依赖于参数 λ 的方式 | | 如果微态能量依赖于参数 λ 的方式 |
− | <math>E_s = E_s^{(0)} + \lambda A_s \qquad \mbox{for all}\; s </math>
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| : <math>E_s = E_s^{(0)} + \lambda A_s \qquad \mbox{for all}\; s </math> | | : <math>E_s = E_s^{(0)} + \lambda A_s \qquad \mbox{for all}\; s </math> |
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| 那么 A 的期望值就是 | | 那么 A 的期望值就是 |
− | <math>\langle A\rangle = \sum_s A_s P_s = -\frac{1}{\beta}
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− | 1. a rangle = sum _ s a _ s p _ s =-frac {1}{ beta }
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− | \frac{\partial}{\partial\lambda} \ln Z(\beta,\lambda).</math>
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| : <math>\langle A\rangle = \sum_s A_s P_s = -\frac{1}{\beta} | | : <math>\langle A\rangle = \sum_s A_s P_s = -\frac{1}{\beta} |
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| 这为我们提供了一种计算许多微观量的期望值的方法。我们将这个量人为地加到微态能量上(或者用量子力学的语言,加到哈密顿量上) ,计算出新的配分函数和期望值,然后在最终的表达式中将 λ 设置为零。这类似于量子场论路径积分表述中使用的源场方法。 | | 这为我们提供了一种计算许多微观量的期望值的方法。我们将这个量人为地加到微态能量上(或者用量子力学的语言,加到哈密顿量上) ,计算出新的配分函数和期望值,然后在最终的表达式中将 λ 设置为零。这类似于量子场论路径积分表述中使用的源场方法。 |
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| === Relation to thermodynamic variables 与热力学变量的关联 === | | === Relation to thermodynamic variables 与热力学变量的关联 === |
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