laplace equation in cylindrical coordinate Watch

cosmos-physics
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#1
hi!
can anyone help me to derive the equation???
can u send a pic from your book???
thanks....
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RDKGames
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(Original post by cosmos-physics)
hi!
can anyone help me to derive the equation???
can u send a pic from your book???
thanks....
Well Laplace's eq. in Cartesian coordinates is \nabla^2 \phi = \phi_{xx} + \phi_{yy} + \phi_{zz}= 0

Switching from Cartesian to Cylindrical we have:

x = r \cos \theta
y = r \sin \theta
z = z

So the operator changes

\dfrac{\partial}{\partial x} = \dfrac{\partial r}{\partial x} \cdot \dfrac{\partial}{\partial r} + \dfrac{\partial \theta}{\partial x} \cdot \dfrac{\partial}{\partial \theta} = r_x \dfrac{\partial}{\partial r} + \theta_x \dfrac{\partial}{\partial \theta}

Hence \dfrac{\partial^2}{\partial x^2} = \left( r_x \dfrac{\partial}{\partial r} + \theta_x \dfrac{\partial}{\partial \theta} \right)^2

Since r = \sqrt{x^2 + y^2} we have r_x = \dfrac{x}{\sqrt{x^2 + y^2}} = \cos \theta

and since \theta = \arctan \dfrac{y}{x} we have that \theta_x = \dfrac{1}{1+\frac{y^2}{x^2}} \cdot \left( -\dfrac{y}{x^2} \right) = -\dfrac{y}{x^2+y^2} =-\dfrac{\sin \theta}{r}

So \dfrac{\partial}{\partial x} = \cos \theta \dfrac{\partial}{\partial r} -\dfrac{\sin \theta}{r} \dfrac{\partial}{\partial \theta}


Move on from there.
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