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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
Show that a gravitational field has no spin.
To show that F has no spin, we calculate its curl. Let $P(x,y,z)=\frac{x}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}},$ $Q(x,y,z)=\frac{y}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}},$ and $R(x,y,z)=\frac{z}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3\text{/}2}}.$ Then,
Since the curl of the gravitational field is zero, the field has no spin.
Field $\text{v}(x,y)=\u27e8-\frac{y}{{x}^{2}+{y}^{2}},\frac{x}{{x}^{2}+{y}^{2}}\u27e9$ 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.
$\text{curl}\phantom{\rule{0.2em}{0ex}}\text{v}=0$
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 ${\mathbb{R}}^{3},$ then the curl of F is also a vector field in ${\mathbb{R}}^{3}.$ 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 .
Let $\text{F}=\u27e8P,Q,R\u27e9$ be a vector field in ${\mathbb{R}}^{3}$ such that the component functions all have continuous second-order partial derivatives. Then, $\text{div}\phantom{\rule{0.2em}{0ex}}\text{curl}\phantom{\rule{0.2em}{0ex}}\left(\text{F}\right)=\nabla \xb7\left(\nabla \phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{F}\right)=0.$
By the definitions of divergence and curl, and by Clairaut’s theorem,
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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 ${\mathbb{R}}^{3}$ 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 $\text{G}(x,y,z)=\u27e8\text{sin}\phantom{\rule{0.1em}{0ex}}x,\text{cos}\phantom{\rule{0.1em}{0ex}}y,\text{sin}\left(xyz\right)\u27e9$ 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.
If $\text{F}=\u27e8P,Q,R\u27e9$ is conservative, then $\text{curl}\phantom{\rule{0.2em}{0ex}}\text{F}=0.$
Since conservative vector fields satisfy the cross-partials property, all the cross-partials of F are equal. Therefore,
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