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Or Even a derivation for the continuity equation

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- Thread starter islamphysics
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- #1

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Or Even a derivation for the continuity equation

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HallsofIvy

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In cylindrical coordinates, [itex]r= (x^2+ y^2)^{1/2}[/itex] and [itex]\theta= arctan(y/x)[/itex].

[tex]\frac{\partial r}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{\sqrt{x^2+ y^2}}= \frac{x}{r}[/tex]

[tex]= \frac{r cos(\theta)}{r}= cos(\theta)[/tex]

[tex]\frac{\partial r}{\partial y}= sin(\theta)[/tex]

[tex]\frac{\partial \theta}{\partial x}= \frac{1}{1+ (y/x)^2}(-y/x^2)=\frac{-y}{x^2+ y^2}[/tex]

[tex]= \frac{-r sin(\theta)}{r^2}= -\frac{sin(\theta)}{r}[/tex]

[tex]\frac{\partial \theta}{\partial y}= \frac{1}{1+ (y/x)^2}(1/x)= \frac{x}{x^2+ y^2}[/tex]

[tex]= \frac{r cos(\theta)}{r^2}= \frac{cos(\theta)}{r}[/tex]

Using the chain rule

[tex]\frac{\partial \phi}{\partial x}= \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial \phi}{\partial \theta}\frac{\partial \theta}{\partial x}[/tex]

You should be able to finish that, and then the second derivatives yourself. It's tedious but requires no deep mathematics.

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