证明标量函数的梯度的旋度为0
时间: 2023-06-20 10:10:26 浏览: 207
设 $f(x,y,z)$ 为标量函数,其梯度为 $\nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}$。则其旋度为:
$$
\begin{aligned}
\nabla \times \nabla f &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{vmatrix} \\
&= \left(\frac{\partial^2 f}{\partial y \partial z} - \frac{\partial^2 f}{\partial z \partial y}\right)\mathbf{i} + \left(\frac{\partial^2 f}{\partial z \partial x} - \frac{\partial^2 f}{\partial x \partial z}\right)\mathbf{j} + \left(\frac{\partial^2 f}{\partial x \partial y} - \frac{\partial^2 f}{\partial y \partial x}\right)\mathbf{k} \\
&= 0 \cdot \mathbf{i} + 0 \cdot \mathbf{j} + 0 \cdot \mathbf{k} \\
&= \mathbf{0}
\end{aligned}
$$
因此,标量函数的梯度的旋度为0。