# Thread: Scalar and vector fields

1. ## Scalar and vector fields

Show that $\nabla\times(f\mathbf{v})=\nabla f\times\mathbf{v}+f\nabla\times\mathbf{v}$

Here $f(x,y,z)$ is a scalar field and $\mathbf{v}(x,y,z)$ is a vector field.

2. ## Vector identities

Rewrite your expression in component form:
$
\nabla \times (f \vec v) = \epsilon_{ijk} \partial_j (f \vec v_k)
$

where $
\epsilon_{ijk}
$
is the fully antisymmetric symbol which is 0 if any indices are the same, 1 if ijk is cyclic to xyz and -1 if ijk is cyclic to xzy.

Concentrate on what the derivative is doing. This is just like d(uv) = (du)v + u(dv):
$
\partial_j (f \vec v_k) = (\partial_j f) \vec v_k + f (\partial_j \vec v_k)
$

So the proof goes:
$
\nabla \times (f \vec v) = \epsilon_{ijk} \partial_j (f \vec v_k)
$

$
= \epsilon_{ijk} ( (\partial_j f) \vec v_k + f (\partial_j \vec v_k)
)$

$
= \epsilon_{ijk} (\partial_j f) \vec v_k + \epsilon_{ijk} f (\partial_j \vec v_k)
$

$
= \epsilon_{ijk} (\partial_j f) \vec v_k + f \epsilon_{ijk} (\partial_j \vec v_k)
$

$
= (\nabla f) \times \vec v + f \nabla \times \vec v$

3. Thanks.

That looks nothing like the solutions I was given though plus, turns out I'd done it right in the first place I just thought I was writing BS, but apparently not