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Context
This is an interim problem related to a Green's function solution for a boundary-value problem in the cylindrical coordinate system.
Question
How do I convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinate system?
My worked solution is given in the answer below.
$$(x-x')^2 + (y-y')^2 + (z-z')^2 =r^2 + {r'}^2 -2\,r\,{r}' \,\cos\left(\theta - \theta'\right) + (z-z')^2$$
My attempt
I know that to tranform from Cartesian to cylindrical requires the following three equations \begin{align} r &= \sqrt{x^2 + y^2} \\ \theta &= \arctan{\left(\frac{y}{x}\right)} \\ z &= z \quad. \end{align}
Directly, the question reduces to how to covert the following to cylindrical system. $$(x-x')^2 + (y-y')^2.$$ Its natural to begin with $r$ and $r'$. \begin{align} (x-x')^2 + (y-y')^2 &= x^2 - 2\,x\,x' +{x'}^2 + y^2 - 2\,y\,{y'} +{y'}^2 \\ &= x^2 + y^2 +{x'}^2 +{y'}^2 - 2\,x\,x' - 2\,y\,{y'} \\ &= r^2 +{r'}^2 - 2\,x\,x' - 2\,y\,{y'} \end{align} Now the problem is to convert $ - 2\,x\,x' - 2\,y\,{y'} $ to cylindrical coordinate system.
The next part of the answer required trial and error, and some good fortune. \begin{align*} r\,r' \,\cos\left(\theta - \theta'\right) &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{y}{x}} - \arctan{\frac{y'}{x'}}\right) \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{\frac{y}{x} - \frac{y'}{x'}}{1+ \frac{y}{x} + \frac{y'}{x'}}} \right) \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{\frac{y}{x} - \frac{y'}{x'}}{1+ \frac{y\,y'}{x\,x'} }} \right) \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{ y\, x' - y'\,x }{x\,x'+ y\,y' }} \right) \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{1}{\sqrt{1+\left(\frac{ y\, x' - y'\,x }{x\,x'+ y\,y' }\right)^2}} \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+ y\,y' }{\sqrt{(x\,x'+ y\,y' )^2+\left( y\, x' - y'\,x \right)^2}} \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+ y\,y' }{\sqrt{\left(x\,x'\right)^2+ 2\,x\,x' \, y\,y' + \left( y\,y' \right)^2+\left( y\, x' \right)^2 -2\, y\, x'\,y'\,x + \left( y'\,x \right)^2}} \\ &= \sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+ y\,y' }{\sqrt{\left(x\,x'\right)^2+ \left( y\,y' \right)^2+\left( y\, x' \right)^2 + \left( y'\,x \right)^2}} \\ &= x\,x'+ y\,y' \end{align*} At this point the nut is cracked, and I rebuild the solution as follows: \begin{align} - 2\,x\,x' - 2\,y\,{y'} &= -2\,r\,r' \,\cos\left(\theta - \theta'\right) \\ r^2 + {r'}^2 - 2\,x\,x' - 2\,y\,{y'} &=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right) \\ (x-x')^2 + (y-y')^2 &=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right) \\ (x-x')^2 + (y-y')^2 + (z-z')^2 &=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right) + (z-z')^2 \end{align}
