VC3.40

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We have the identity,\nabla \times (f\times g)=f(\nabla \cdot g)-g(\nabla \cdot f)+(g\cdot \nabla )f-(f\cdot \nabla )g\, --(1)

By putting f=F and g=r in (1),we have

\nabla \times (F\times r)=F(\nabla \cdot r)-r(\nabla \cdot F)+(r\cdot \nabla )F-(F\cdot \nabla )r\, --(2)

Now,\nabla \cdot r=\nabla \cdot (xi+yj+zk)={\frac  {\partial }{\partial x}}(x)+{\frac  {\partial }{\partial y}}(y)+{\frac  {\partial }{\partial z}}(z)=3\, --(3)

Let F=F_{1}i+F_{2}j+F_{3}k\, --(4)

Therefore,F\cdot \nabla =(F_{1}i+F_{2}j+F_{3}k)\cdot (i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}})=F_{1}{\frac  {\partial }{\partial x}}+F_{2}{\frac  {\partial }{\partial y}}+{\frac  {\partial }{\partial z}}\, --(5)

and so (F\cdot \nabla )r=(F_{1}{\frac  {\partial }{\partial x}}+F_{2}{\frac  {\partial }{\partial y}}+{\frac  {\partial }{\partial z}})(xi+yj+zk)\,

=F_{1}i+F_{2}j+F_{3}k\,,by (4) --(6)

Using (3) and (6),(2) becomes \nabla \times (F\times r)=3F-r(\nabla \cdot f)+(r\cdot \nabla )F-F=2F-(\nabla \cdot F)r+(r\cdot \nabla )F\,

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