# Transformation matrix of parabolic coordinates

I am working on the exercises in "Spacetime and Geometry" from Sean Carroll. In exercise 4a) of section 3 we should find the transformation Matrix $$A$$ and its inverse, that transforms the parabolodidal coordinates $$(u,v,\phi)$$ to the Cartesian coordinates. It is given, that \begin{align} x = uv\cos{\phi}, \qquad y = uv\sin{\phi}, \qquad z = \frac{1}{2}(u^2-v^2) \end{align} First of all, I am not very experienced in transformation matrices. My first observation was, that this coordinate transformation is not linear, so why does the transformation matrix even exist? I found solutions online, where the matrix is calculated by \begin{align} A = \left( \begin{array}{rrr} \partial_ux & \partial_vx & \partial_\phi x \\ \partial_uy & \partial_vy & \partial_\phi y \\ \partial_uz & \partial_vz & \partial_\phi z \end{array}\right) = \left( \begin{array}{rrr} vcos(\phi) & ucos(\phi) & -uvsin(\phi) \\ vsin(\phi) & usin(\phi) & uvcos(\phi) \\ u & -v & 0 \end{array}\right) \end{align}

It makes sense to me to calculate the matrix this way, since $$x'^\mu=\frac{\partial x^\mu}{\partial x'^\nu}x^\nu$$. My question is, shouldn't this matrix $$A$$ satisfy \begin{align} x'=Ax \end{align} where $$x'=(u,v,\phi)^T,\quad x = (x,y,z)^T$$

• \begin{align*} &\begin{bmatrix} dx \\ dy \\ dz \\ \end{bmatrix}=\mathbf{A}\, \begin{bmatrix} du \\ dv \\ d\theta \\ \end{bmatrix}\quad &\begin{bmatrix} du \\ dv \\ d\theta \\ \end{bmatrix}=\mathbf{A}^{-1}\, \begin{bmatrix} dx \\ dy \\ d\theta \\ \end{bmatrix} \end{align*}
– Eli
Nov 17, 2022 at 15:08

The correct version of $$x'^\mu=\frac{\partial x^\mu}{\partial x'^\nu}x^\nu$$ should be $$dx'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}dx^\nu$$ Notice that the prime is in the numerator. In 1D this becomes the chain rule:$$dy=\frac{dy}{dx}dx$$, which is a way to check this equation. This explains why this transformation is possible: only for infinitesimal transformations around a point can we approximate the transformation as a linear map. You could check this equation numerically by calculating $$dx'^\mu$$ using two ways. First by using the Jacobian you gave (the transformation matrix A) and secondly by calculating $$dx'^\mu\approx x'^\mu(x^\nu+dx^\nu)-x'^\mu(x^\nu)$$, both using small values for $$dx^\nu$$.
\begin{align*} &\text{with }\\ &\mathbf{R}= \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \begin{bmatrix} u\,v\,\cos(\phi) \\ u\,v\,\sin(\phi) \\ \frac{1}{2}(u^2-v^2) \\ \end{bmatrix}\quad, \mathbf{q}=\begin{bmatrix} u \\ v \\ \phi \\ \end{bmatrix} \Rightarrow\\\\ &A_{ij}=\frac{\partial R_i}{\partial q_j}= \left[ \begin {array}{ccc} v\cos \left( \phi \right) &u\cos \left( \phi \right) &-uv\sin \left( \phi \right) \\ v\sin \left( \phi \right) &u\sin \left( \phi \right) &uv\cos \left( \phi \right) \\ u&-v&0\end {array} \right]\\ \end{align*} from here the metric $$~\mathbf{G}$$ \begin{align*} &\mathbf{G}=\mathbf{A}^T\,\mathbf{A}= \left[ \begin {array}{ccc} {u}^{2}+{v}^{2}&0&0\\ 0& {u}^{2}+{v}^{2}&0\\ 0&0&{u}^{2}{v}^{2}\end {array} \right] \quad\Rightarrow\quad\text{line element}\\ &ds^2=du^2(u^2+v^2)+dv^2(u^2+v^2)+d\phi^2\,u^2\,v^2 \end{align*} the rotation matrix (transformation matrix) $$~\mathbf{S}~$$ between $$~[\,u~,v~,\phi]~$$ and $$~[\,x~,y~,z]~$$ system is:
\begin{align*} &\mathbf{S}=\begin{bmatrix} \hat{e}_u & \hat{e}_v & \hat{e}_\phi \\ \end{bmatrix}\quad\text{where }\quad, \hat{e}_u=\frac{\partial \mathbf{R}}{\partial u}\quad, \hat{e}_v=\frac{\partial \mathbf{R}}{\partial v}\quad, \hat{e}_\phi=\frac{\partial \mathbf{R}}{\partial \phi} \\\\ &\mathbf{S}= \left[ \begin {array}{ccc} {\frac {v\cos \left( \phi \right) }{\sqrt {{u}^{2}+{v}^{2}}}}&{\frac {u\cos \left( \phi \right) }{\sqrt {{u}^{2} +{v}^{2}}}}&-\sin \left( \phi \right) \\ {\frac {v \sin \left( \phi \right) }{\sqrt {{u}^{2}+{v}^{2}}}}&{\frac {u\sin \left( \phi \right) }{\sqrt {{u}^{2}+{v}^{2}}}}&\cos \left( \phi \right) \\ {\frac {u}{\sqrt {{u}^{2}+{v}^{2}}}}&-{ \frac {v}{\sqrt {{u}^{2}+{v}^{2}}}}&0\end {array} \right]\quad,\mathbf{S}^T\,\mathbf{S}=\mathbf{I}_3\\ \end{align*}