.png)
繼續上次的範圍
這裡考慮\(\gamma_{ij}\)和\(\gamma_{ji}\)的關係 \[\begin{aligned}
\psi_{ij}(t)&=\left \langle x_i(t)x_j(0) \right \rangle\\
&=\epsilon_i\epsilon_j\left \langle x_i(-t)x_j(0) \right \rangle\\
&=\epsilon_i\epsilon_j\left \langle x_ix_j(t) \right \rangle
\end{aligned}\] 取時間導數 \[\begin{aligned}
\left \langle \dot{x}_i(t)x_j(0) \right \rangle = \epsilon_i\epsilon_j
\left \langle x_i(0)\dot{x}_j(t) \right \rangle
\end{aligned}\] 那麼
\[\begin{aligned}
-\gamma_{ik}\beta_{kl}\left \langle x_l(t)x_j(0) \right \rangle +\left
\langle y_i(t)x_j(0) \right \rangle =
-\gamma_{jk}\beta_{kl}\epsilon_i\epsilon_j\left \langle
x_i(0)x_l(t) \right \rangle +\epsilon_i\epsilon_j\left \langle
x_i(0)y_j(t) \right \rangle
\end{aligned}\]
現在拿\(t\to0^+\)並注意\(\left \langle y_i(t)x_j(0) \right
\rangle\)和\(\epsilon_i\epsilon_j\left
\langle x_i(0)y_j(t) \right \rangle\)在\(t>0\)時都不見了 \[\begin{aligned}
-\gamma_{ik}\beta_{kl}\left \langle x_l(0)x_j(0) \right
\rangle=-\gamma_{jk}\beta_{kl}\epsilon_i\epsilon_j\left \langle
x_i(0)x_l(0) \right \rangle
\end{aligned}\] 最後得到得關係就是Onsager’s的倒數關係 \[\begin{aligned}
\gamma_{ij}=\epsilon_i\epsilon_j\gamma_{ji}
\end{aligned}\]
擴散流體中的線性動力學(Linearized dynamics in a diffusive fluid)
設系統平衡時的熵為\(S(x_i)\)並且此時的\(x_i=0\)。而平衡狀態的熵是最大的定為\(S_{max}\)，換句話說如果不是平衡時的熵就會比平衡時的小，可以寫為
\[\begin{aligned}
S=S_{max}-\frac{1}{2}\beta_{ij}x_ix_j
\end{aligned}\] 而其中的\(x_i\)是，我猜是用類似之前漲落時類似的\(x\)的設定方式 設系統平衡時的熵為\(S(x_i)\)並且此時的\(x_i=0\)。而平衡狀態的熵是最大的定為\(S_{max}\)，換句話說如果不是平衡時的熵就會比平衡時的小，可以寫為
\[\begin{aligned}
\dot{x}_i=\gamma_{ik}\frac{\partial S}{\partial x_k}+f_i
\end{aligned}\] 跟之前一樣的結論，所以就是 \[\begin{aligned}
\gamma_{ik} = \epsilon_i\epsilon_k \gamma_{ki}
\end{aligned}\] 擴散流體中的熱力學它不會流動，所以\(v=0\) \[\begin{aligned}
dU=TdS-pdV+\mu dN
\end{aligned}\] 因為微積分的另一項是\(0\)所以也可以得到 \[\begin{aligned}
0=SdT-Vdp+Nd\mu
\end{aligned}\] 將剛剛的式子除與體積 \[\begin{aligned}
du=Tds+\mu dn
\end{aligned}\] 移項一下 \[\begin{aligned}
ds = \frac{1}{T}du+\left ( -\frac{\mu}{T} \right )dn
\end{aligned}\] 可以得到類似連續方程的東西 \[\bar{u_0}=
\begin{cases}
\partial_tn+\overset{\rightharpoonup}\nabla\cdot\overset{\rightharpoonup}j_n=0\\
\\
\partial_tu+\overset{\rightharpoonup}\nabla\cdot\overset{\rightharpoonup}j_u=0
\end{cases}\] 其中\(\overset{\rightharpoonup}j_n\)和\(\overset{\rightharpoonup}j_u\)必須是\(\overset{\rightharpoonup}\nabla\left \langle
\frac{1}{T} \right \rangle\)和\(\overset{\rightharpoonup}\nabla\left \langle
-\frac{\mu}{T} \right \rangle\)的函數 \[\begin{aligned}
\partial_tS&=\frac{1}{T}\partial_tu-\frac{\mu}{T}\partial_tn\\
&=-\frac{1}{T}\overset{\rightharpoonup}\nabla\cdot\overset{\rightharpoonup}j_u-\left
( -\frac{\mu}{T} \right
)\overset{\rightharpoonup}\nabla\cdot\overset{\rightharpoonup}j_n\\
&=-\overset{\rightharpoonup}\nabla\cdot\left (
\frac{1}{T}\overset{\rightharpoonup}j_u-\frac{\mu}{T}\overset{\rightharpoonup}j_n \right
)+\overset{\rightharpoonup}j_u\cdot\overset{\rightharpoonup}\nabla\left
( \frac{1}{T} \right
)+\overset{\rightharpoonup}j_n\cdot\overset{\rightharpoonup}\nabla\left
( -\frac{\mu}{T} \right )
\end{aligned}\] 移項一下 \[\begin{aligned}
\partial_tS+\overset{\rightharpoonup}\nabla\cdot \underbrace {\left
(\frac{1}{T}\overset{\rightharpoonup}j_u-\frac{\mu}{T}\overset{\rightharpoonup}j_n \right
) }_{flux of entropy}=\underbrace {
\overset{\rightharpoonup}j_u\cdot\overset{\rightharpoonup}\nabla\left
( \frac{1}{T} \right
)+\overset{\rightharpoonup}j_n\cdot\overset{\rightharpoonup}\nabla\left
( -\frac{\mu}{T} \right )}_{entropy procluction rate}
\end{aligned}\] 本構關係(constitutive relations) \[\begin{aligned}
\overset{\rightharpoonup}j_u& =
L_{uu}\overset{\rightharpoonup}\nabla\left ( \frac{1}{T} \right
)+L_{un}\overset{\rightharpoonup}\nabla\left ( -\frac{\mu}{T} \right
)\\
\overset{\rightharpoonup}j_n&=L_{nu}\overset{\rightharpoonup}\nabla\left
( \frac{1}{T} \right )+L_{nn}\overset{\rightharpoonup}\nabla\left
( -\frac{\mu}{T} \right )
\end{aligned}\] 可以發現\(L_{un}=L_{nu}\)相反是相等的，和熵的關係是
\[\begin{aligned}
L_{\alpha\beta}\left ( \overset{\rightharpoonup}\nabla f_\alpha \right
) \cdot \left ( \overset{\rightharpoonup}\nabla f_\beta \right ) \geq 0
\end{aligned}\] 當\(\mu\)是常數時\(\overset{\rightharpoonup}\nabla
T\not=0\)代表擴散的粒子流\(L_{nu}\not=0\)
當\(\overset{\rightharpoonup}\nabla
T\not=0\)但\(\overset{\rightharpoonup}\nabla\mu\not=0\)時代表能量通量\(L_{un}\not=0\)