% Bernd Berg Oct 24 2018.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2018 
Solutions} \smallskip

\centerline{\bf Set 10} 

\begin{description} 
\item {\bf 23. Partial Derivatives.}
\end{description}

\noindent (a) From $x'=x^2\,y^2$ and $y'=x^2/y^2$ we get
$$ x^4=x'\,y'~~{\rm and}~~y^4=\frac{x'}{y'}~~\Rightarrow~~x=\left(x'
\,y'\right)^{1/4}~~{\rm and}~~y=\left(\frac{x'}{y'}\right)^{1/4}\ .$$
(b) \begin{eqnarray} \nonumber
  \frac{\partial~}{\partial x'}\,\frac{\partial x'}{\partial x} 
  &=& \frac{\partial~}{\partial x'}\,2x\,y^2 = 2\,
  \frac{\partial~}{\partial x'}\,(x'\,y')^{1/4}\,\left(\frac{x'}
  {y'}\right)^{1/2} = 2\,\frac{\partial~}{\partial x'}\,
  \left(\frac{x'^3}{y'}\right)^{1/4} \\ \nonumber 
  &=& \frac{1}{2}\,\left(\frac{x'^3}{y'}\right)^{-3/4}\,3\,
  \frac{x'^2}{y'} = \frac{3}{2}\,\left(\frac{x^6\,y^6\,y^2}{x^2}\,
  \right)^{-3/4}\,\frac{x^4\,y^4\,y^2}{x^2} \\ \nonumber 
  &=& \frac{3}{2}\,\left(x\,y^2\right)^{-3}\,x^2\,y^6 = \frac{3}{2x}\ .
  \\ \nonumber
  \frac{\partial~}{\partial x}\,\frac{\partial x'}{\partial x'} 
  &=& \frac{\partial~}{\partial x}\,1 = 0\,.
\end{eqnarray}
(c) \begin{eqnarray} \nonumber
  \frac{\partial^2 x'}{\partial x\,\partial x}\,\frac{\partial x}
  {\partial x'} &=& 2y^2\,\left(x'y'\right)^{-3/4}\,\frac{y'}{4} 
  = \frac{y^2}{2}\,\left(x^4\right)^{-3/4}\,\frac{x^2}{y^2} = 
  \frac{1}{2x} \\ \nonumber
  \frac{\partial^2 x'}{\partial x\,\partial y}\,\frac{\partial y}
  {\partial x'} &=& 4\,x\,y\,\left(\frac{x'}{y'}\right)^{-3/4}\,
  \left(\frac{1'}{4y'}\right) = x\,y\,\left(y^4\right)^{-3/4}\,
  \frac{y^2}{x^2} = \frac{1}{x}\,.
\end{eqnarray}
So the sum
$$ \frac{\partial^2 x'}{\partial x\,\partial x}\,\frac{\partial x}
  {\partial x'} + \frac{\partial^2 x'}{\partial x\,\partial y}\,
  \frac{\partial y} {\partial x'} = \frac{3}{2x} $$
agrees with the first result of~(b). \medskip

\noindent (d) The operators
$$ \frac{\partial~}{\partial x}~~{\rm and}~~
   \frac{\partial~}{\partial y} $$ 
commute because $x$ and $y$ are independent variables. To the contrary
$x$ and $x'$ are not independet. So the order of the differentiations
$$ \frac{\partial~}{\partial x}~~{\rm and}~~
   \frac{\partial~}{\partial x'}$$
matters. \vfil \eject

\noindent (d) We have
$$ p^{1'}_{1i}\,p^i_{1'} = \frac{\partial^2 x^{1'}}{\partial x^1\,
   \partial x^1}\,\frac{\partial x^1}{\partial x^{1'}} + 
   \frac{\partial^2 x^{1'}}{\partial x^1\,\partial x^2}\,
  \frac{\partial x^2} {\partial x^{1'}} = \frac{3}{2x^1}\, $$
where the result follows from the previous calculation.

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