[대학원-전자기학]Vector Calculus Problem 1

Problem 1: Prove \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

Answer:
\[\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x_1-x_2}{|\vec{\eta}^3|} & \frac{y_1-y_2}{|\vec{\eta}^3|} & \frac{z_1-z_2}{|\vec{\eta}^3|} \end{vmatrix}\] Where \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\vec{\eta} = \vec{r}-\vec{r}\prime\], \[|\vec{\eta}|=\sqrt{(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\] \[\begin{aligned} \nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3} &= (\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\eta}^3|}))\vec{x} \nonumber \\ & -(\frac{\partial}{\partial x}(\frac{z_1-z_2}{|\vec{\eta}^3|}) - \frac{\partial}{\partial z}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{y} \nonumber \\ & +(\frac{\partial}{\partial x}(\frac{y_1-y_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial y}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{z} \nonumber\end{aligned}\]

Since \[\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(z_1 - z_2)(y_1 - y_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

\[\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(y_1 - y_2)(z_1 - z_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

They cancel out each others. Same procedure occur for the other two components. Hence, \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

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