# 6.5 Divergence and curl  (Page 5/9)

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We can now use what we have learned about curl to show that gravitational fields have no “spin.” Suppose there is an object at the origin with mass ${m}_{1}$ at the origin and an object with mass ${m}_{2}.$ Recall that the gravitational force that object 1 exerts on object 2 is given by field

$\text{F}\left(x,y,z\right)=\text{−}G{m}_{2}{m}_{2}⟨\frac{x}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}},\frac{y}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}},\frac{z}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}}⟩.$

## Determining the spin of a gravitational field

Show that a gravitational field has no spin.

To show that F has no spin, we calculate its curl. Let $P\left(x,y,z\right)=\frac{x}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}},$ $Q\left(x,y,z\right)=\frac{y}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}},$ and $R\left(x,y,z\right)=\frac{z}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}}.$ Then,

$\begin{array}{cc}\hfill \text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}& =\text{−}G{m}_{1}{m}_{2}\left[\left({R}_{y}-{Q}_{z}\right)\text{i}+\left({P}_{z}-{R}_{x}\right)\text{j}+\left({Q}_{x}-{P}_{y}\right)\text{k}\right]\hfill \\ & =\text{−}G{m}_{1}{m}_{2}\left[\begin{array}{c}\left(\frac{-3yz}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5\text{/}2}}-\left(\frac{-3yz}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5\text{/}2}}\right)\right)\text{i}\hfill \\ +\left(\frac{-3xz}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5\text{/}2}}-\left(\frac{-3xz}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5\text{/}2}}\right)\right)\text{j}\hfill \\ +\left(\frac{-3xy}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5\text{/}2}}-\left(\frac{-3xy}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5\text{/}2}}\right)\right)\text{k}\hfill \end{array}\right]\hfill \\ & =0.\hfill \end{array}$

Since the curl of the gravitational field is zero, the field has no spin.

Field $\text{v}\left(x,y\right)=⟨-\frac{y}{{x}^{2}+{y}^{2}},\frac{x}{{x}^{2}+{y}^{2}}⟩$ models the flow of a fluid. Show that if you drop a leaf into this fluid, as the leaf moves over time, the leaf does not rotate.

$\text{curl}\phantom{\rule{0.2em}{0ex}}\text{v}=0$

## Using divergence and curl

Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.

If F is a vector field in ${ℝ}^{3},$ then the curl of F is also a vector field in ${ℝ}^{3}.$ Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field. To give this result a physical interpretation, recall that divergence of a velocity field v at point P measures the tendency of the corresponding fluid to flow out of P . Since $\text{div}\phantom{\rule{0.2em}{0ex}}\text{curl}\phantom{\rule{0.2em}{0ex}}\left(\text{v}\right)=0,$ the net rate of flow in vector field curl( v ) at any point is zero. Taking the curl of vector field F eliminates whatever divergence was present in F .

## Divergence of the curl

Let $\text{F}=⟨P,Q,R⟩$ be a vector field in ${ℝ}^{3}$ such that the component functions all have continuous second-order partial derivatives. Then, $\text{div}\phantom{\rule{0.2em}{0ex}}\text{curl}\phantom{\rule{0.2em}{0ex}}\left(\text{F}\right)=\nabla ·\left(\nabla \phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{F}\right)=0.$

## Proof

By the definitions of divergence and curl, and by Clairaut’s theorem,

$\begin{array}{cc}\hfill \text{div curl}\phantom{\rule{0.2em}{0ex}}\text{F}& =\text{div}\left[\left({R}_{y}-{Q}_{z}\right)\text{i}+\left({P}_{z}-{R}_{x}\right)\text{j}+\left({Q}_{x}-{P}_{y}\right)\text{k}\right]\hfill \\ & ={R}_{yx}-{Q}_{xz}+{P}_{yz}-{R}_{yx}+{Q}_{zx}-{P}_{zy}\hfill \\ & =0.\hfill \end{array}$

## Showing that a vector field is not the curl of another

Show that $\text{F}\left(x,y,z\right)={e}^{x}\text{i}+yz\text{j}+x{z}^{2}\text{k}$ is not the curl of another vector field. That is, show that there is no other vector G with $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{G}=\text{F}.$

Notice that the domain of F is all of ${ℝ}^{3}$ and the second-order partials of F are all continuous. Therefore, we can apply the previous theorem to F .

The divergence of F is ${e}^{x}+z+2xz.$ If F were the curl of vector field G , then $\text{div}\phantom{\rule{0.2em}{0ex}}\text{F}=\text{div curl}\phantom{\rule{0.2em}{0ex}}\text{G}=0.$ But, the divergence of F is not zero, and therefore F is not the curl of any other vector field.

Is it possible for $\text{G}\left(x,y,z\right)=⟨\text{sin}\phantom{\rule{0.1em}{0ex}}x,\text{cos}\phantom{\rule{0.1em}{0ex}}y,\text{sin}\left(xyz\right)⟩$ to be the curl of a vector field?

No

With the next two theorems, we show that if F is a conservative vector field then its curl is zero, and if the domain of F is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.

## Curl of a conservative vector field

If $\text{F}=⟨P,Q,R⟩$ is conservative, then $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}=0.$

## Proof

Since conservative vector fields satisfy the cross-partials property, all the cross-partials of F are equal. Therefore,

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