Differences

This shows you the differences between two versions of the page.

--- 물리:xy모형 [2023/09/04 17:39] – [피적분함수의 $\tau$를 $\tau+d\tau$로 변경] admin
+++ 물리:xy모형 [2023/09/04 18:06] – [$i$와 $j$를 제외한 소용돌이들 주변으로의 적분] admin
@@ Line 214: / Line 214: @@
 \end{eqnarray}
 ====$i$와 $j$를 제외한 소용돌이들 주변으로의 적분====
-$$ 2\pi \tau d\tau \int_{\overline{D}(i,j)} d\mathbf{r}_j \left\{ 1 + \beta^2 p^4 \sum_k \frac{\tau^2}{\left| \mathbf{r}_j - \mathbf{r}_k \right|^2} + \beta^2 p^2 \sum_{k \neq l} p_k p_l \frac{\tau^2 (\mathbf{r}_j - \mathbf{r}_k) \cdot (\mathbf{r}_j - \mathbf{r}_l)}{\left| \mathbf{r}_j - \mathbf{r}_k \right|^2 \left| \mathbf{r}_j - \mathbf{r}_l \right|^2} \right\}
+$$ 2\pi \tau d\tau \int_{\overline{D}(i,j)} d\mathbf{r}_j \left\{ 1 + \beta^2 p^4 \sum_k \frac{\tau^2}{\left| \mathbf{r}_j - \mathbf{r}_k \right|^2} + \beta^2 p^2 \sum_{k \neq l} p_k p_l \frac{\tau^2 (\mathbf{r}_j - \mathbf{r}_k) \cdot (\mathbf{r}_j - \mathbf{r}_l)}{\left| \mathbf{r}_j - \mathbf{r}_k \right|^2 \left| \mathbf{r}_j - \mathbf{r}_l \right|^2} \right\}$$
-\approx 2\pi \tau d\tau \left( A - 2\pi \tau^2 \beta^2 p^2 \sum_{k \neq l} p_k p_l \ln \left| \frac{\mathbf{r}_k - \mathbf{r}_l}{\tau} \right| \right)$$
+첫 번째 항의 적분은 계의 전체 면적 $A$를 준다(제외되는 반경 $\tau$는 작으므로 무시):
+$$\int_{\overline{D}(i,j)} d\mathbf{r}_j  \approx A.$$
+두 번째 항의 적분은 계의 반경을 $R$이라 했을 때에 $R$이 매우 크다면 마치 $\mathbf{r}_k$가 원점에 있는 것처럼 다음처럼 구해진다:
+$$\int_{\overline{D}(i,j)} \frac{d\mathbf{r}_j}{\left| \mathbf{r}_j - \mathbf{r}_k \right|^2} \approx 2\pi \ln \frac{R}{\tau}.$$
+세 번째 항의 적분 역시 마찬가지로 $\tau$가 작고 $R$이 큰 극한에서 행한다. 편의상 $\mathbf{r}_k = (\rho,0)$, $\mathbf{r}_l = (-\rho,0)$이라고 한다면 이 적분은
+\begin{eqnarray}
+\int_0^R \int_0^{2\pi} \frac{(r^2-\rho^2)}{(r^2+\rho^2+2\rho r\cos\theta) (r^2+\rho^2-2\rho r\cos\theta)}r d\theta dr
+&=& \pi \ln\left[ \frac{1}{4} \left( 1 + \frac{R^2}{\rho^2} \right) \right]\\
+&\approx& \pi \ln\left( \frac{R^2}{\left| \mathbf{r}_k - \mathbf{r}_l \right|^2} \right)\\
+&=& 2\pi \ln\left( \frac{R}{\left| \mathbf{r}_k - \mathbf{r}_l \right|} \right).
+\end{eqnarray}
+위 결과들을 모두 더하면
+\begin{eqnarray}
+\pi \tau d\tau \left(A + 2\pi \tau^2 \beta^2 p^4 \sum_k \ln \frac{R}{\tau} + 2\pi \tau^2 \beta^2 p^2 \sum_{k \neq l} p_k p_l \ln \frac{R}{\left| \mathbf{r}_k  - \mathbf{r}_l \right|} \right)
+&\approx& 2\pi \tau d\tau \left(A - 2\pi \tau^2 \beta^2 p^2 \sum_{k\neq l} p_k p_l \ln \frac{R}{\tau} + 2\pi \tau^2 \beta^2 p^2 \sum_{k \neq l} p_k p_l \ln \frac{R}{\left| \mathbf{r}_k  - \mathbf{r}_l \right|} \right)\\
+&=& 2\pi \tau d\tau \left( A - 2\pi \tau^2 \beta^2 p^2 \sum_{k \neq l} p_k p_l \ln \left| \frac{\mathbf{r}_k - \mathbf{r}_l}{\tau} \right| \right).
+\end{eqnarray}
+따라서
 \begin{eqnarray}
 Z &=& \sum_{n} \frac{1}{(n!)^2} \kappa^{2n} \int_{D_{2n}} d\mathbf{r}_{2n} \cdots \int_{D_{1}} d\mathbf{r}_{1} e^{-\beta H_{2n}}\\