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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

F ( x , y , z ) = G m 2 m 2 x ( x 2 + y 2 + z 2 ) 3 / 2 , y ( x 2 + y 2 + z 2 ) 3 / 2 , z ( x 2 + y 2 + z 2 ) 3 / 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 ( x , y , z ) = x ( x 2 + y 2 + z 2 ) 3 / 2 , Q ( x , y , z ) = y ( x 2 + y 2 + z 2 ) 3 / 2 , and R ( x , y , z ) = z ( x 2 + y 2 + z 2 ) 3 / 2 . Then,

curl F = G m 1 m 2 [ ( R y Q z ) i + ( P z R x ) j + ( Q x P y ) k ] = G m 1 m 2 [ ( −3 y z ( x 2 + y 2 + z 2 ) 5 / 2 ( −3 y z ( x 2 + y 2 + z 2 ) 5 / 2 ) ) i + ( −3 x z ( x 2 + y 2 + z 2 ) 5 / 2 ( −3 x z ( x 2 + y 2 + z 2 ) 5 / 2 ) ) j + ( −3 x y ( x 2 + y 2 + z 2 ) 5 / 2 ( −3 x y ( x 2 + y 2 + z 2 ) 5 / 2 ) ) k ] = 0.

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

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Field v ( x , y ) = y x 2 + y 2 , 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.

curl v = 0

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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 3 , then the curl of F is also a vector field in 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 div curl ( v ) = 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 F = P , Q , R be a vector field in 3 such that the component functions all have continuous second-order partial derivatives. Then, div curl ( F ) = · ( × F ) = 0 .


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

div curl F = div [ ( R y Q z ) i + ( P z R x ) j + ( Q x P y ) k ] = R y x Q x z + P y z R y x + Q z x P z y = 0.

Showing that a vector field is not the curl of another

Show that F ( x , y , z ) = e x i + y z j + x z 2 k is not the curl of another vector field. That is, show that there is no other vector G with curl G = F .

Notice that the domain of F is all of 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 + 2 x z . If F were the curl of vector field G , then div F = div curl G = 0 . But, the divergence of F is not zero, and therefore F is not the curl of any other vector field.

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Is it possible for G ( x , y , z ) = sin x , cos y , sin ( x y z ) to be the curl of a vector field?


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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 F = P , Q , R is conservative, then curl F = 0 .


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

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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