Div and Curl of Vector Fields

Question

I am struggling to compute the div and curl of the the vector field $v$. First, $v$ is defined to be $p^{-1} \nabla p$. Here, $p$ is the distance to the $z$ axis. I don't know what the div and curl mean, so I'm not sure how to begin to calculate it. Secondly, $v$ is defined to be $v \hat{\phi}$. Here, $v$ is a scalar field. $ \hat{\phi}$ is direction vector of $\phi$ (rotating around the $x, y$ axes). Thanks so much.

Answer 1

A vector field $v$ on some domain $\Omega\subset{\mathbb R}^3$ can be the mathematical model of (a) a force field, or (b) a flow field. If $v$ is supposed to be a force field then it makes sense to consider ${\rm curl}\> v$, and if $v$ is rather a flow field then it makes sense to consider ${\rm div}\> v$ as physically intuitive notions. While ${\rm curl}\> v$ measures the local nonconservativity of the force field $v$, the divergence measures the source intensity of the flow field $v$. I won't go into more details here. At any rate it is only in very special circumstances that one would compute as well the curl as the divergence of the same field $v$.

In your example we are given $$v(x,y,z)={1\over p(x,y,z)}\nabla p(x,y,z)$$ on the domain $\Omega:={\mathbb R}^3\setminus\{z\hbox{-axis}\}$, where $$p(x,y,z):=\sqrt{x^2+y^2}>0\qquad\bigl((x,y,z)\in\Omega\bigr)\ .$$ It follows that $$v=\nabla(\log p)\ .$$ Since $v$ is the gradient field of a certain function it automatically follows that $${\rm curl}\> v(x,y,z)\equiv(0,0,0)\qquad\bigl((x,y,z)\in\Omega\bigr)\ .$$ Now for the divergence: One has $${\rm div}\>v={\rm div}\bigl(\nabla(\log p)\bigr)=\Delta(\log p)\ ,$$ where $\Delta$ denotes the so-called Laplace operator: $$\Delta u:=u_{xx}+u_{yy}+u_{zz}\ .$$ In the example at hand the function $$u(x,y,z):=\log p={1\over2}\log(x^2+y^2)$$ depends only on $x$ and $y$. It is a fundamental fact of two-dimensional vector analysis (and easily verified doing the computation) that $$\Delta\left({1\over2}\log(x^2+y^2)\right)\equiv0\ .$$ It follows that $${\rm div}\> v(x,y,z)\equiv0\qquad\bigl((x,y,z)\in\Omega\bigr)$$ as well.

Answer 2

If you don't know what "div" and "curl" mean then where did you get this problem?? If you do not know them then you certainly cannot do this! The "div", also called "divergence" of a vector field, v, is the scalar field that, at each point, gives the rate of at which the vectors tend to point away from or toward (if the divergence is negative) the orgin. In terms of an x, y, z coordinate system, it is given by $\nabla\cdot\vec{v}= \nabla\cdot f(x, y, z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$. With p being "the distance to the z-axis", $(x^2+ y^2)^{1/2}$, $$\nabla p= div p= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}+ \frac{\partial (x^2+ y^2)^{1/2}}{\partial y}$$

$$= \frac{x}{(x^2+ y^2)^{1/2}}+ \frac{y}{(x^2+ y^2)^{1/2}}= \frac{x+ y}{\sqrt{x^2+ y^2}}$$ The "curl" of a vector field is another vector field giving the tendency of the original field to "circulate" around the origin. It is a vector pointing in the direction of circulation with length giving the rate of circulation.
In terms of an x, y, z coordinate system, it is given by $\nabla\times\vec{v}=$$ \nabla\times f(x, y, z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}=$$ \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial y}\right)\vec{i}+ \left(\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$.

Often we represent that "$\nabla$" operator as a symbolic "vector operator", $\nabla= \frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial z}\vec{k}$ so that the "div" and "curl" can be thought of as the "dot product" and "cross product" of vectors.