latex编写程序,一步步推导证明:\nabla \times \left ( \nabla \Phi \right ) =0
时间: 2023-11-25 10:05:17 浏览: 35
根据矢量分析理论,我们可以得到以下结论:
$$\nabla \times \left ( \nabla \Phi \right ) =0$$
其中,$\Phi$是标量场,$\nabla$是梯度算子,$\nabla \times$是旋度算子。
证明过程如下:
$$\nabla \times \left ( \nabla \Phi \right ) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial \Phi}{\partial x} & \frac{\partial \Phi}{\partial y} & \frac{\partial \Phi}{\partial z} \end{vmatrix}$$
$$=\left ( \frac{\partial^2 \Phi}{\partial y \partial z} - \frac{\partial^2 \Phi}{\partial z \partial y} \right ) \hat{i} + \left ( \frac{\partial^2 \Phi}{\partial z \partial x} - \frac{\partial^2 \Phi}{\partial x \partial z} \right ) \hat{j} + \left ( \frac{\partial^2 \Phi}{\partial x \partial y} - \frac{\partial^2 \Phi}{\partial y \partial x} \right ) \hat{k}$$
由于混合偏导数的交换次序不影响结果,所以有:
$$\frac{\partial^2 \Phi}{\partial y \partial z} = \frac{\partial^2 \Phi}{\partial z \partial y}, \frac{\partial^2 \Phi}{\partial z \partial x} = \frac{\partial^2 \Phi}{\partial x \partial z}, \frac{\partial^2 \Phi}{\partial x \partial y} = \frac{\partial^2 \Phi}{\partial y \partial x}$$
因此,$\nabla \times \left ( \nabla \Phi \right ) =0$。