Divergence of a vector field


by 1MileCrash
Tags: divergence, field, vector
1MileCrash
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#1
Apr23-12, 09:52 PM
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1. The problem statement, all variables and given/known data

F(x,y,z) = (-x+y)i + (y+z)j + (-z+x)k

Find divergence

2. Relevant equations



3. The attempt at a solution

The gradient is
-i + j + -k

Dotting that with F, I get

x - y + y + z + z - x
=
2z

My book lists the answer as -1. What the heck are they talking about? (they did not ask me to evaluate for any point)
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Dick
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Apr23-12, 10:15 PM
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Quote Quote by 1MileCrash View Post
1. The problem statement, all variables and given/known data

F(x,y,z) = (-x+y)i + (y+z)j + (-z+x)k

Find divergence

2. Relevant equations



3. The attempt at a solution

The gradient is
-i + j + -k

Dotting that with F, I get

x - y + y + z + z - x
=
2z

My book lists the answer as -1. What the heck are they talking about? (they did not ask me to evaluate for any point)
I think you are misunderstanding the definition of divergence. ∇.F doesn't mean grad(F).F. Look it up.
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Apr23-12, 10:18 PM
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Quote Quote by Dick View Post
I think you are misunderstanding the definition of divergence. ∇.F doesn't mean grad(F).F. Look it up.
I will, thanks!

1MileCrash
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Apr23-12, 10:19 PM
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Divergence of a vector field


Is it correct to say that it's like taking grad F, then adding up the resulting components for a scalar?
Dick
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Apr23-12, 10:23 PM
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Quote Quote by 1MileCrash View Post
Is it correct to say that it's like taking grad F, then adding up the resulting components for a scalar?
Yes, if F=(Fx,Fy,Fz) then the divergence of F is dFx/dx+dFy/dy+dFz/dz. It's a scalar.
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Apr23-12, 10:36 PM
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I see my misunderstanding now. The del operator is not the gradient of anything in particular. It's just (d/dx)i + (d/dy)j + (d/dz)k. Dot product that with F leads to the correct definition.

This actually clears up a lot of the past notation. Since del is not a gradient of anything in particular, when we say [itex]\nabla f[/itex], since f is a scalar being multiplied by some vector, del, the result is a vector, which is the gradient of f.

Cool. :)
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Apr23-12, 10:39 PM
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Quote Quote by 1MileCrash View Post
I see my misunderstanding now. The del operator is not the gradient of anything in particular. It's just (d/dx)i + (d/dy)j + (d/dz)k. Dot product that with F leads to the correct definition.

This actually clears up a lot of the past notation. Since del is not a gradient of anything in particular, when we say [itex]\nabla f[/itex], since f is a scalar being multiplied by some vector, del, the result is a vector, which is the gradient of f.

Cool. :)
You've got it, I think. You just dotting the grad operator with the vector. The result is a scalar.


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