물리:bbgky_계층

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물리:bbgky_계층 [2022/04/21 17:24] – [푸아송 괄호 계산] jiwon물리:bbgky_계층 [2022/04/22 10:18] – [푸아송 괄호 계산] jiwon
Line 57: Line 57:
 마찬가지로 푸아송 괄호를 풀어서 적어보면 마찬가지로 푸아송 괄호를 풀어서 적어보면
 \begin{align*} \begin{align*}
- \int \prod_{i=s+1}^N d^3p_id^3q_i\sum_{j=1}^N\left[\frac{\partial\rho}{\partial\vec q_j}\cdot\frac{\partial H'}{\partial\vec p_j}-\frac{\partial\rho}{\partial\vec p_j}\cdot\frac{\partial H'}{\partial\vec q_j}\right]+ -\int \prod_{i=s+1}^N d^3p_id^3q_i\sum_{j=1}^N\left[\frac{\partial\rho}{\partial\vec q_j}\cdot\frac{\partial H'}{\partial\vec p_j}-\frac{\partial\rho}{\partial\vec p_j}\cdot\frac{\partial H'}{\partial\vec q_j}\right]
 \end{align*} \end{align*}
 이다. $\partial H'/\partial\vec q_j=0$이 되고, $j\le s$일 때는 이다. $\partial H'/\partial\vec q_j=0$이 되고, $j\le s$일 때는
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 이므로, 이를 모두 모아서 적어보면 이므로, 이를 모두 모아서 적어보면
 \begin{align*} \begin{align*}
- &\int \prod_{i=s+1}^N d^3p_id^3q_i\sum_{j=1}^N\left[\frac{\partial\rho}{\partial\vec q_j}\cdot\frac{\partial H'}{\partial\vec p_j}-\frac{\partial\rho}{\partial\vec p_j}\cdot\frac{\partial H'}{\partial\vec q_j}\right]\\+ &-\int \prod_{i=s+1}^N d^3p_id^3q_i\sum_{j=1}^N\left[\frac{\partial\rho}{\partial\vec q_j}\cdot\frac{\partial H'}{\partial\vec p_j}-\frac{\partial\rho}{\partial\vec p_j}\cdot\frac{\partial H'}{\partial\vec q_j}\right]\\
  =&\int \prod_{i=s+1}^N d^3p_id^3q_i\left[\sum_{n=1}^s\frac{\partial\rho}{\partial\vec p_n}\cdot\sum_{j=s+1}^N\frac{\partial V(\vec q_n-\vec q_j)}{\partial\vec q_n} + \sum_{j=s+1}^N\frac{\partial\rho}{\partial\vec p_j}\cdot\sum_{n=1}^s\frac{\partial V(\vec q_j-\vec q_n)}{\partial\vec  q_j}\right]  =&\int \prod_{i=s+1}^N d^3p_id^3q_i\left[\sum_{n=1}^s\frac{\partial\rho}{\partial\vec p_n}\cdot\sum_{j=s+1}^N\frac{\partial V(\vec q_n-\vec q_j)}{\partial\vec q_n} + \sum_{j=s+1}^N\frac{\partial\rho}{\partial\vec p_j}\cdot\sum_{n=1}^s\frac{\partial V(\vec q_j-\vec q_n)}{\partial\vec  q_j}\right]
 \end{align*} \end{align*}
-이 된다.+이 된다. 두 번째 항 중에서 $j$인덱스를 하나 골라 계산해보면 
 +\begin{align*} 
 + &\int d^3p_jd^3q_j\frac{\partial\rho}{\partial\vec p_j}\sum_{n=1}^s\cdot\frac{\partial V(\vec q_j-\vec q_n)}{\partial\vec q_j}\\ 
 + =&\int d^3q_j\left[\rho\sum_{n=1}^s\cdot\frac{\partial V(\vec q_j-\vec q_n)}{\partial\vec q_j}\right]_{\vec p_j\text{ at }\infty}-\int d^3q_jd^3p_j\rho\sum_{n=1}^s\cdot\frac{\partial^2 V(\vec q_j-\vec q_n)}{\partial\vec p_j\partial\vec q_j}\\ 
 + =&0 
 +\end{align*} 
 +가 된다. 그리고 모든 입자는 같다고 생각하기 때문에 첫 항의 $j$에 대한 합은 $s+1$ 번째 입자를 $N-s$번 고려한 것으로 생각할 수 있다. 정리해보면, 
 +\begin{align*} 
 + &(N-s)\int\prod_{i=s+1}^Nd^3q_id^3p_i \sum_{i=1}^s \frac{\partial V(\vec q_n-\vec q_{s+1})}{\partial\vec q_n}\cdot\frac{\partial\rho}{\partial\vec p_n}\\ 
 + =&(N-s)\sum_{n=1}^s\int d^3q_{s+1}d^3p_{s+1}\frac{\partial V(\vec q_n-\vec q_{s+1})}{\partial\vec q_n}\cdot\frac{\partial\rho_{s+1}}{\partial\vec p_n} 
 +\end{align*} 
 +이다. 이제 계산해두었던 푸아송 괄호들을 다 모아보면 
 +$$ 
 + \frac{\partial\rho_s}{\partial t}-\{H_s,\rho_s\}=(N-s)\sum_{n=1}^s\int d^3q_{s+1}d^3p_{s+1}\frac{\partial V(\vec q_n-\vec q_{s+1})}{\partial\vec q_n}\cdot\frac{\partial\rho_{s+1}}{\partial\vec p_n} 
 +$$ 
 +를 얻는다.
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