Question

Vector Calculus
Prove : (a is constant vector.)
\[\nabla \times [\dfrac{a\times \bar r}{r^n}] = \dfrac{2-n}{r^n}a + \dfrac{n(a.\bar r)\bar r}{r^{n+2}}\]

So i used (any other method?) the identity

\(\nabla \times \left( \mathbf{v \times F} \right) = \left[ \left( \mathbf{ \nabla \cdot F } \right) + \mathbf{F \cdot \nabla} \right] \mathbf{v}- \left[ \left( \mathbf{ \nabla \cdot v } \right) + \mathbf{v \cdot \nabla} \right] \mathbf{F}\)

Answer 1

plugged in

\(\nabla .a = 0 \\ \nabla .\bar r =3\)

Answer 2

in

\(\nabla \times \left( \mathbf{a \times \bar r } \right) = \left[ \left( \mathbf{ \nabla \cdot \bar r } \right) + \mathbf{\bar r \cdot \nabla} \right] \mathbf{a}- \left[ \left( \mathbf{ \nabla \cdot a } \right) + \mathbf{a \cdot \nabla} \right] \mathbf{\bar r}\)

Answer 3

what next ?

Answer 4

also if you see, i just removed r^n outside...is that legal ?

Answer 5

where did u get div a =0 and div r'=3 from

Answer 6

oh, is that incorrect ?

Answer 7

no, jw

Answer 8

\(\operatorname{div}\,\mathbf{F} = \nabla\cdot\mathbf{F}
=\frac{\partial U}{\partial x}
+\frac{\partial V}{\partial y}
+\frac{\partial W}{\partial z
}\)

Answer 9

ohh ok i see equating from that identity

Answer 10

what does
\(\bar r . \nabla\)
even mean ?? and how di i calculate it ?

Answer 11

lol i was looking at that.. i was thought when it goes that way.. its just undefined

Answer 12

it has to be a scalar, thats what i am sure of

Answer 13

taught* >_> and umm, what did u get for the expression for F

Answer 14

F here is r = xi +yj +zk