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--- 물리:xy모형 [2023/09/03 19:46] – [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}}\\
@@ Line 242: / Line 261: @@
 \begin{eqnarray}
 \kappa^{2n} \exp\left[ -\beta\sum_{i\neq j} p_i p_j \ln \tau \right] &\approx& \kappa^{2n} \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \left[ \ln (\tau+d\tau) - \frac{d\tau}{\tau} \right] \right\}\\
-&=& \kappa^{2n} \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \ln (\tau+d\tau) \right\} \exp \left( - \beta \sum_{i \neq j} p_i p_j \frac{d\tau}{\tau} \right)
+&=& \kappa^{2n} \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \ln (\tau+d\tau) \right\} \exp \left( \beta \sum_{i \neq j} p_i p_j \frac{d\tau}{\tau} \right)
 \end{eqnarray}
 여기에서 $p_i = -p_j$인 인접한 소용돌이 쌍들이 대부분을 기여하므로 $\sum_{i \neq j} p_i p_j \approx -2n p^2$으로 근사하면, 위 식은
 \begin{eqnarray}
 \kappa^{2n} \exp \left( - 2n \beta p^2 \frac{d\tau}{\tau} \right) \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \ln (\tau+d\tau) \right\}
-&\approx& \kappa^{2n} \left( 1 - 2n \beta p^2 \frac{d\tau}{\tau} \right) \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \ln (\tau+d\tau) \right\}
+&\approx& \kappa^{2n} \left( 1 - 2n \beta p^2 \frac{d\tau}{\tau} \right) \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \ln (\tau+d\tau) \right\}\\
 &\approx& \left[ \kappa \left( 1 - \beta p^2 \frac{d\tau}{\tau} \right) \right]^{2n} \exp \left\{ -\beta\sum_{i\neq j} p_i p_j \ln (\tau+d\tau) \right\}
 \end{eqnarray}
@@ Line 258: / Line 277: @@
 &\approx& \kappa\tau^2 \left[ 1 - (\beta p^2-2) \frac{d\tau}{\tau} \right]
 \end{eqnarray}
+====결과====
+차단 길이 $\tau$를 $\tau+d\tau$로 변경함에 따라 계의 맺음변수들이 다음처럼 변화한다:
+\begin{eqnarray}
+\beta p^2 &\longrightarrow& (\beta p^2)' = \beta p^2 \left[ 1 - (2\pi)^2 (\beta p^2) (\kappa \tau^2)^2 \frac{d\tau}{\tau} \right]\\
+\kappa \tau^2 &\longrightarrow& (\kappa \tau^2)' = \kappa \tau^2 \left[ 1 - (\beta p^2 - 2) \frac{d\tau}{\tau} \right].
+\end{eqnarray}
+만일 $x \equiv \beta p^2 - 2$와 $y \equiv 2\pi \kappa \tau^2$을 정의한다면 아래처럼 쓸 수 있다:
+\begin{eqnarray}
+dx &=& -(x+2)^2 y^2 \frac{d\tau}{\tau}\\
+dy &=& -xy \frac{d\tau}{\tau}.
+\end{eqnarray}
+$\lambda \equiv \ln \tau$로 정의하는 것도 일반적이다:
+\begin{eqnarray}
+\frac{dx}{d\lambda} &=& -(x+2)^2 y^2\\
+\frac{dy}{d\lambda} &=& -xy.
+\end{eqnarray}
+$(x,y)=(0,0)$인 고정점 주변에서는 아래처럼 근사할 수 있고
+\begin{eqnarray}
+\frac{dx}{d\lambda} &=& -4y^2\\
+\frac{dy}{d\lambda} &=& -xy,
+\end{eqnarray}
+$x = 2\pi K -2$로 쓸 수도 있으므로 $K$와 $y$에 대해 정리하면 아래의 꼴로 나타나기도 한다:
+\begin{eqnarray}
+\frac{dK^{-1}}{d\lambda} &=& 2\pi y^2\\
+\frac{dy}{d\lambda} &=& (2-2\pi K) y.
+\end{eqnarray}
 ======참고문헌======
    * J. M. Kosterlitz, //The critical properties of the two-dimensional xy model//, J. Phys. C: Solid State Phys. **7**, 1046 (1074).
    * P. W. Anderson, G. Yuval, and D. R. Hamann, //Exact Results in the Kondo Problem. II. Scaling Theory, Qualitatively Correct Solution, and Some New Results on One-Dimensional Classical Statistical Models//, Phys. Rev. B **1**, 4464 (1970).
    * M. Kardar, //Statistical Physics of Fields//, Cambridge University Press (Cambridge, UK, 2007).