So, the property is not applicable to this problem.

[Nat3]

Homework Statement

$$\vec r = <x, y, z>, r = \left | \vec r \right |$$

The problem is, verify the identity:

$$\nabla \cdot (r\vec r) = 4r$$

Homework Equations

My book has the following property:
$$(c\vec a)\cdot \vec b = c(\vec a \cdot \vec b)$$

The Attempt at a Solution

I tried using the above property to rewrite the problem as:
$$r(\nabla \cdot\vec r)$$

But that gives 3r and the correct answer is 4r.

Why does the property not work? How should I correctly solve the problem?

Thanks for your advice.

 

[Dick]

You need to use the product rule.
$$r \vec r$$ is <sqrt(x^2+y^2+z^2)x,sqrt(x^2+y^2+z^2)y,sqrt(x^2+y^2+z^2)z>. You need to use the product rule to evaluate the divergence. Equivalently div(r vec r)=grad(r).vec r+r*div(vec r). Sorry about my bad TeX skills. But you are missing the first term.

 

[I like Serena]

Have you tried writing the equation out in cartesian coordinates, do the deravative, and work back to polar coordinates again?

 

[Nat3]

But why doesn't the property from my textbook work? I understand that the way Dick did it, distributing the r inside the vector, I would need to use the product rule when differentiating, but according to the property I listed I shouldn't have to do that.

I'm really confused!

 

[I like Serena]

But why doesn't the property from my textbook work? I understand that the way Dick did it, distributing the r inside the vector, I would need to use the product rule when differentiating, but according to the property I listed I shouldn't have to do that.

I'm really confused!

I'm afraid that the Del operator and indeed derivation in general do not simply comply with your property.
The Del operator may be written as a vector, and in a number of aspects it behaves like one, but in this case it doesn't.

More specifically, you cannot bring a factor outside of a derivation, if that factor is relevant to derivation (that is, not a constant).