# Divergence of a vector field

by 1MileCrash
Tags: divergence, field, vector
 P: 1,177 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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P: 24,981
 Quote by 1MileCrash 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.
P: 1,177
 Quote by Dick I think you are misunderstanding the definition of divergence. ∇.F doesn't mean grad(F).F. Look it up.
I will, thanks!

P: 1,177

## 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?
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P: 24,981
 Quote by 1MileCrash 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.
 P: 1,177 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 $\nabla f$, since f is a scalar being multiplied by some vector, del, the result is a vector, which is the gradient of f. Cool. :)
 Quote by 1MileCrash 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 $\nabla f$, since f is a scalar being multiplied by some vector, del, the result is a vector, which is the gradient of f. Cool. :)