VC3.26

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Here r=xi+yj+zk,r^2=|r|^2=x^2+y^2+z^2\, --(1)

Since a is a constant vector,\frac{\partial a}{\partial x}=0\, --(2)

From (1),\frac{\partial r}{\partial x}=i,\frac{\partial r}{\partial x}=\frac{x}{r}\, --(3)

Now,\mathrm{curl}(\frac{a\times r}{r^3})=\sum {i\times\frac{\partial}{\partial x}[\frac{a\times r}{r^3}]}\, by definition of curl --(4)

But \frac{\partial}{\partial x}[\frac{a\times r}{r^3}]=\frac{\partial}{\partial x}(r^{-3}(a\times r)\,

=-3r^{-4}\frac{\partial r}{\partial x}(a\times r)+r^{-3}\frac{\partial}{\partial x}(a\times r)\,

=-\frac{3r^{-4}x}{r}(a\times r)+r^{-3}[\frac{\partial a}{\partial x}\times r+a\times\frac{\partial r}{\partial x}]\, by (3)

=-3r^{-5}x(a\times r)+r^{-3}(a\times i)\, by (2),(3)

Therefore,i\times\frac{\partial}{\partial x}[\frac{a\times r}{r^3}]\,

=i\times[-3r^{-5}x(a\times r)+r^{-3}(a\times i)]\,

=-3r^{-5}xi\times(a\times r)+r^{-3}i\times(a\times i)\,

=-3r^{-5}x[(i\cdot r)a-(i\cdot a)r]+r^{-3}[(i\times i)a-(i\cdot a)i]\, --(5)

Let a=a_1i+a_2j+a_3k\, so that i\cdot a=a_1\, etc. --(6)

Also i\cdot r=i\cdot(xi+yj+zk)=x\, --(7)

Using (6) and (7),(5)reduces to

i\times\frac{\partial}{\partial x}[\frac{a\times r}{r^3}]\,

=3r^{-5}x[xa-a_1r]+r^{-3}(a-a_1i)\,

=-\frac{3x^2}{r^5}a+\frac{3a_1x}{r^5}r+\frac{1}{r^3}a-\frac{1}{r^3}a_1i\,

=\sum i\times\frac{\partial}{\partial x}[\frac{a\times r}{r^3}]\,

=\sum[-\frac{3x^2}{r^5}a+\frac{3a_1x}{r^5}r+\frac{1}{r^3}a-\frac{a_1}{r^3}i]\,

=-\frac{3}{r^5}[\sum x^2]a+\frac{3}{r^5}(\sum a_1 x)r+\frac{1}{r^3}(\sum 1)a-\frac{1}{r^3}\sum a_1 i\, --(8)

But \sum x^2=x^2+y^2+z^2=r^2,\sum a_1x=a_1x+a_2y+a_3z=r\cdot a\,

and \sum 1=3,\sum a_1i+a_2j+a_3k=a\,

Therefore \sum i\times\frac{\partial}{\partial x}[\frac{a\times r}{r^3}]\,

=-\frac{3}{r^5}r^2 a+\frac{3}{r^5}(r\cdot a)r+\frac{3}{r^3}a-\frac{1}{r^3}a\, by (8)

or \mathrm{curl}(\frac{a\times r}{r^3})=-\frac{a}{r^{-3}}+\frac{3}{r^5}(a\cdot r)r\, by (4)

Hence the required is proved.

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