Magnitude of a vector, cubed, in Einstein Summation Notation
Evaluate $\nabla \cdot (r^3v)$. The answer will be in terms of r.
Where v represents the position vector and r represents the scalar magnitude of the position vector.
I started by writing this in the notation,
$d\over(dx_i)$$((r_jr_j)^3v)_i$
I'm not exactly sure on what "evaluating" will be in terms of doing it in summation notation? Since the instructions specify to evaluate, I'm assuming there's a tensor in there somewhere. Also, I am having trouble in expressing $r^3$ in summation notation. I do know that the magnitude of a vector is just $r \cdot r$, but how do i incorporate the cubed?
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$\begingroup$The magnitude of the vector $v$ is $r = \sqrt{v\cdot v} = \sqrt{v_1^2 + \dots + v_n^2}$.
The $i$th component of $r^3 v$ is then $(r^3 v)_i = \left(\sqrt{v_1^2 + \dots + v_n^2}\right)^3 v_i = \left(v_1^2 + \dots + v_n^2\right)^{3/2} v_i$.
Because $v$ is the 'position' vector, $v = (x_1,\dots,x_n)$, i.e. $(r^3v)_i = \left(x_1^2 + \dots + x_n^2\right)^{3/2} x_i$.
Now just differentiate with respect to $x_i$: $$ \frac{\partial}{\partial x_i} \left(x_1^2 + \dots + x_n^2\right)^{3/2} x_i = (3/2) \left(x_1^2 + \dots + x_n^2\right)^{1/2}(2 x_i)(x_i) + \left(x_1^2 + \dots + x_n^2\right)^{3/2}. $$ Noting that $\left(x_1^2 + \dots + x_n^2\right)^{1/2} = r$, we have $$ \frac{\partial}{\partial x_i} (r^3v)_i = 3 r x_i^2 + r^3 $$ (alternatively, $3rv_i^2 + r^3$). Finally, note that this answer is in Einstein notation: $$ \nabla\cdot(r^3v) = \frac{\partial}{\partial x_i} (r^3v)_i = 3 r x_i^2 + r^3, $$ but if you sum over the components, you get $$ \nabla\cdot(r^3v) = 3 r (x_1^2 + \dots + x_n^2) + r^3 = 3r^3 + r^3 = 4r^3. $$
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