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 $x'=(u,v,\phi)^T,\quad x = (x,y,z)^T$
Best Answer
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 $dx^\nu$.