I have been trying to convert the Del operator from Cartesian to Cylindrical coords since like 5 days. but still i can't see why my way doesn't work. It worked for the 3D heat equation and 3D wave equation but for vector quantities no :( ...
This is the way i followed
[tex] \nabla P = \frac{\partial P}{\partial x} i + \frac{\partial P}{\partial y} j + \frac{\partial P}{\partial z} k [/tex]
Since there is no significant change in z vector, i will work on converting the i and j terms. therefore,
i know that,
[tex] \frac{\partial P}{\partial x} = \frac{\partial P}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial P}{\partial \theta } \frac{\partial \theta}{\partial x} [/tex]
[tex] \frac{\partial P}{\partial y} = \frac{\partial P}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial P}{\partial \theta } \frac{\partial \theta}{\partial y} [/tex]
Taking the Cartesian - polar derivatives, therefore,
[tex] \frac{\partial P}{\partial x} = \frac{\partial P}{\partial r} \frac{1}{cos ~ \theta} - \frac{\partial P}{\partial \theta } \frac{\partial \theta}{rsin ~ \theta} [/tex]
[tex] \frac{\partial P}{\partial y} = \frac{\partial P}{\partial r} \frac{1}{sin ~ \theta} + \frac{\partial P}{\partial \theta } \frac{\partial \theta}{rcos ~ \theta} [/tex]
Now, it can be shown that the vectors can be transformed to
[tex] i = cos ~ \theta ~ e_r - sin ~ \theta ~ e_\theta [/tex]
[tex] j = sin ~ \theta ~ e_r + cos ~ \theta ~ e_\theta [/tex]
Finallly substituting all of these to get stuck as follows,
[tex] \nabla P = \frac{\partial P}{\partial r}e_r - \frac{\partial P}{\partial \theta} \frac{cos ~ \theta}{r ~ sin ~ \theta}e_r - \frac{\partial P}{\partial r} \frac{sin ~ \theta}{cos ~ \theta}e_\theta + \frac{\partial P}{\partial \theta} \frac{1}{r}e_\theta + \frac{\partial P}{\partial r}e_r + \frac{\partial P}{\partial \theta} \frac{sin ~ \theta}{r ~ cos ~ \theta}e_r + \frac{\partial P}{\partial r} \frac{cos ~ \theta}{sin ~ \theta}e_\theta + \frac{\partial P}{\partial \theta} \frac{1}{r}e_\theta + \frac{\partial P}{ \partial z}e_z \ [/tex]
I thought that it would cancel out cleanly but didn't. Any ideas ?
This is the way i followed
[tex] \nabla P = \frac{\partial P}{\partial x} i + \frac{\partial P}{\partial y} j + \frac{\partial P}{\partial z} k [/tex]
Since there is no significant change in z vector, i will work on converting the i and j terms. therefore,
i know that,
[tex] \frac{\partial P}{\partial x} = \frac{\partial P}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial P}{\partial \theta } \frac{\partial \theta}{\partial x} [/tex]
[tex] \frac{\partial P}{\partial y} = \frac{\partial P}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial P}{\partial \theta } \frac{\partial \theta}{\partial y} [/tex]
Taking the Cartesian - polar derivatives, therefore,
[tex] \frac{\partial P}{\partial x} = \frac{\partial P}{\partial r} \frac{1}{cos ~ \theta} - \frac{\partial P}{\partial \theta } \frac{\partial \theta}{rsin ~ \theta} [/tex]
[tex] \frac{\partial P}{\partial y} = \frac{\partial P}{\partial r} \frac{1}{sin ~ \theta} + \frac{\partial P}{\partial \theta } \frac{\partial \theta}{rcos ~ \theta} [/tex]
Now, it can be shown that the vectors can be transformed to
[tex] i = cos ~ \theta ~ e_r - sin ~ \theta ~ e_\theta [/tex]
[tex] j = sin ~ \theta ~ e_r + cos ~ \theta ~ e_\theta [/tex]
Finallly substituting all of these to get stuck as follows,
[tex] \nabla P = \frac{\partial P}{\partial r}e_r - \frac{\partial P}{\partial \theta} \frac{cos ~ \theta}{r ~ sin ~ \theta}e_r - \frac{\partial P}{\partial r} \frac{sin ~ \theta}{cos ~ \theta}e_\theta + \frac{\partial P}{\partial \theta} \frac{1}{r}e_\theta + \frac{\partial P}{\partial r}e_r + \frac{\partial P}{\partial \theta} \frac{sin ~ \theta}{r ~ cos ~ \theta}e_r + \frac{\partial P}{\partial r} \frac{cos ~ \theta}{sin ~ \theta}e_\theta + \frac{\partial P}{\partial \theta} \frac{1}{r}e_\theta + \frac{\partial P}{ \partial z}e_z \ [/tex]
I thought that it would cancel out cleanly but didn't. Any ideas ?
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