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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 .

Proof

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?

No

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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 .

Proof

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

Questions & Answers

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ninjadapaul
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Salomon
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ninjadapaul
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ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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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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