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Math Help - Scalar and vector fields

  1. #1
    Senior Member chella182's Avatar
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    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.
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  2. #2
    Senior Member
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    Vector identities

    Rewrite your expression in component form:
    <br />
\nabla \times (f \vec v)  = \epsilon_{ijk} \partial_j (f \vec v_k)<br />
    where <br />
\epsilon_{ijk} <br />
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):
    <br />
\partial_j (f \vec v_k) = (\partial_j f) \vec v_k + f (\partial_j \vec v_k)<br />

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

    <br />
= \epsilon_{ijk} ( (\partial_j f) \vec v_k + f (\partial_j \vec v_k)<br />
)
    <br />
= \epsilon_{ijk} (\partial_j f) \vec v_k + \epsilon_{ijk} f (\partial_j \vec v_k)<br />

    <br />
= \epsilon_{ijk} (\partial_j f) \vec v_k + f \epsilon_{ijk}  (\partial_j \vec v_k)<br />

    <br />
= (\nabla f) \times \vec v + f \nabla \times \vec v
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  3. #3
    Senior Member chella182's Avatar
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    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
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