Gradient（梯度） of a scalar field_gradient of-优快云博客

Definition

The gradient of a scalar field $\phi (x,y,z)$ is defined by
$\phi=\nabla \phi=\boldsymbol{i} \frac{\partial \phi}{\partial x}+\boldsymbol{j} \frac{\partial \phi}{\partial y}+\boldsymbol{k} \frac{\partial \phi}{\partial z}.$
Clearly, $\nabla \phi$ is a vector field whose $x$ -, $y$ - and $z$ - components are the first partial derivatives of $\phi (x,y,z)$ with respect to $x$ , $y$ and $z$ respectively. Also note that the vector field $\nabla \phi$ should not be confused with the vector operator $\phi \nabla$ , which has components $(\phi\partial/\partial x, \phi\partial/\partial y, \phi\partial/\partial z)$ .

Examples

Example 1

Find the gradient of the scalar field $\phi=xy^{2}z^{3}$ .

Solution

From the definition, the gradient of $\phi$ is given by
$\nabla \phi=\boldsymbol{i} \frac{\partial \phi}{\partial x}+\boldsymbol{j} \frac{\partial \phi}{\partial y}+\boldsymbol{k} \frac{\partial \phi}{\partial z}\\ =y^{2}z^{3}\boldsymbol{i}+2xyz^{3}\boldsymbol{j}+3xy^2z^2\boldsymbol{k}.$

Geometrical properties

The gradient of a scalar field $\phi$ has some interesting geometrical properties.
Let us first consider the problem of calculating the rate of change of $\phi$ in some particular direction. For an infinitesimal vector displacement $d\boldsymbol{r}$ , forming its scalar product with $\nabla \phi$ we obtain
$\nabla \phi \cdot d\boldsymbol{r}=(\boldsymbol{i} \frac{\partial \phi}{\partial x}+\boldsymbol{j} \frac{\partial \phi}{\partial y}+\boldsymbol{k} \frac{\partial \phi}{\partial z}) \cdot (\boldsymbol{i}dx+\boldsymbol{j}dy+\boldsymbol{k}dz)\\ =\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz\\ =d\phi,$
which is the infinitesimal change in $\phi$ in going from position $\boldsymbol{r}$ to $\boldsymbol{r}+d\boldsymbol{r}$ . In particular, if $\boldsymbol{r}$ depends on some parameter $u$ such that $\boldsymbol{r}(u)$ defines a space curve then the total derivative of $\phi$ with respect to $u$ along the curve is simply
$\frac{d\phi}{du}=\nabla \phi \cdot \frac{d\boldsymbol{r}}{du}.$

In the particular case where the parameter $u$ is the arc length s along the curve, the total derivative of $\phi$ with respect to s along the curve is given by
$\frac{d\phi}{ds}=\nabla \phi \cdot \hat{\boldsymbol{t}},$
where $\hat{\boldsymbol{t}}$ is the unit tangent to the curve at the given point.
In general, the rate of change of $\phi$ with respect to the distance s in a particular direction $\boldsymbol{a}$ is given by
$\frac{d\phi}{ds}=\nabla \phi \cdot \hat{\boldsymbol{a}}$
and is called the directional derivative. Since $\hat{\boldsymbol{a}}$ is a unit vector we have
$\frac{d\phi}{ds}=\lvert \nabla \phi \rvert cos\theta$
where $\theta$ is the angle between $\hat{\boldsymbol{a}}$ and $\nabla \phi$ as shown in figure below. Clearly $\nabla \phi$ lies in the direction of the fastest increase in $\phi$ , and $\lvert \nabla \phi \rvert$ is the largest possible value of $d\phi/ds$ . Similarly, the largest rate of decrease of $\phi$ is $d\phi/ds=-\lvert \nabla \phi \rvert$ in the direction of $-\nabla \phi$ .
在这里插入图片描述