# Stokes' Theorem

• Apr 4th 2011, 05:52 AM
craig
Stokes' Theorem
Show, using Stokes' Theorm, that

$\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = 0$

Firstly Stokes gives us:

$\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = \int_S \nabla \times [\phi \nabla \phi] \cdot \mathbf{dS}$

Expanding $\displaystyle \nabla \times [\phi (\nabla \phi)]$ gives:

$\displaystyle \phi \nabla \times \nabla \phi + \nabla \phi \times \nabla \phi$

The second part of this is zero for any vector, so this leaves the integral as:

$\displaystyle \int_S \phi \nabla \times \nabla \phi \cdot \mathbf{dS}$

Anyone any ideas?

• Apr 4th 2011, 06:06 AM
TheEmptySet
Quote:

Originally Posted by craig
Show, using Stokes' Theorm, that

$\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = 0$

Firstly Stokes gives us:

$\displaystyle \oint_C \phi \nabla \phi \cdot \mathbf{dl} = \int_S \nabla \times [\phi \nabla \phi] \cdot \mathbf{dS}$

Expanding $\displaystyle \nabla \times [\phi (\nabla \phi)]$ gives:

$\displaystyle \phi \nabla \times \nabla \phi + \nabla \phi \times \nabla \phi$

The second part of this is zero for any vector, so this leaves the integral as:

$\displaystyle \int_S \phi \nabla \times \nabla \phi \cdot \mathbf{dS}$

Anyone any ideas?

Remember that $\displaystyle \phi(x,y,z)$ is a scalar function!

$\displaystyle \displaystyle \nabla \phi =\frac{\partial \phi}{\partial x}\mathbf{i}+\frac{\partial \phi}{\partial y}\mathbf{j}+\frac{\partial \phi}{\partial z}\mathbf{k}$

$\displaystyle \displaystyle \phi \nabla \times \nabla \phi=\phi \left(\nabla \times \nabla \phi \right)$

Expand out what is inside the parenthesis and you will get zero (As long as $\displaystyle \phi$ is twice continuously differentiable)

Remember 2nd continuous partial derivatives commute Clairaut's theorem

Symmetry of second derivatives - Wikipedia, the free encyclopedia
• Apr 4th 2011, 08:05 AM
craig
Quote:

Originally Posted by TheEmptySet
Remember that $\displaystyle \phi(x,y,z)$ is a scalar function!

$\displaystyle \displaystyle \phi \nabla \times \nabla \phi=\phi \left(\nabla \times \nabla \phi \right)$

Ahh of course! I was thinking of it as a vector function.

If we take $\displaystyle \displaystyle \left(\nabla \times \nabla \phi \right)$, we get,

$\displaystyle \displaystyle \left(\frac{\partial}{\partial y} \frac{\partial \phi}{\partial z} - \frac{\partial}{\partial z} \frac{\partial \phi}{\partial y},... \right) = \left(0,0,0 \right)$

Hence the integral is equal to zero!

Many thanks for that :)