Divergence of a variable vector

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SUMMARY

The discussion centers on the divergence of the vector field defined as v = (a·r)r, where r = xi + yj + zk and a is a constant vector. The user initially calculated the divergence as ∇·v = 2(a·r) but later realized the correct divergence is ∇·v = 4(a·r) based on the proper application of vector calculus. The confusion arose from the misinterpretation of the scalar multiplication and the divergence operation, which was clarified through the example involving the scalar λ.

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gtfitzpatrick
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Homework Statement



v = (a.r)r where r=xi+yj+zk and a is a constant vector
show \nabla.v = 4(a.r)

I let a= ai+bJ+ck

then (a.r) = ax+by+cz

then this (a.r)r = ax^{2}i+by^{2}j+cz^{2}k

\nabla.v = da1/dx+da2/dz+da3/dz =2ax+2by+2cz

which is equal to 2(a.r)

am i wrong or the book?
 
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gtfitzpatrick said:
then this (a.r)r = ax^{2}i+by^{2}j+cz^{2}k
This is not true. For example, consider the scalar \lambda,

\lambda\left(x\hat{i}+y\hat{j}+z\hat{k}\right) = \left(\lambda x\hat{i}+ \lambda y\hat{j}+\lambda z\hat{k}\right)

And in your case,

\lambda = ax+by+cz
 
got ya, thanks a million
 

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