I have been doing some work in curvilinear coordinates and I often encounter orthogonal coordinates (which are obviously super convenient). I am of course familiar with the usual suspects like cylindrical, spherical, and polar coordinates, but all of these coordinates have a particular thing in common that makes them less than ideal; their Jacobian determinant is something other than $1$, or specifically is nonconstant. Equivalently, we can say that the tangent vectors $\frac{\partial{\bf r}}{\partial y^i}$ are orthogonal, but not constant. It got me wondering about orthogonal transformations in which the Jacobian determinant is constant, or the coordinate tangent vectors are constant.
There are some obvious transformations: for $\mathbb{R}^n$, any transformation from ${\bf x}\to{\bf y}$ of the form
$${\bf y}={\bf x}_0+R{\bf x}$$
where $R$ is an orthogonal matrix is a valid coordinate transformation and the new coordinates are orthogonal coordinates. To show this, the basis vectors are
$$\frac{\partial{\bf r}}{\partial y^i}=\sum_{j=1}^n\frac{\partial {\bf r}}{\partial x^j}\frac{\partial x^j}{\partial y^i}=\sum_{j=1}^n\frac{\partial x^j}{\partial y^i}{\bf e}_j$$
where ${\bf e}_j$ is the $j^{th}$ standard basis vector $(0,…,1,…,0)$ with a $1$ in position $j$. The inverse of the transform is
$${\bf x}=R^T{\bf y}-R^T{\bf x}_0\implies\frac{\partial x^j}{\partial y^i}=R_{ij}$$
So
$$\frac{\partial {\bf r}}{\partial y^i}=\sum_{j=1}^nR_{ij}{\bf e}_j=R_i$$
where $R_{ij}$ is row $i$ column $j$ of $R$ and $R_i$ is the entire row $i$ of $R$. Since $R$ is an orthogonal matrix the rows are all orthogonal, these vectors are all orthogonal.
So it's clear that the given transformation creates orthogonal tangent vectors, since it amounts to a displacement and rotation. The question I have is whether this constitutes all such orthogonal coordinates.
The question is: suppose I have a new set of coordinates for $\mathbb{R}^n$ $(u^1,…,u^n)$ with the property that the vectors $\partial{\bf r}/\partial u^i$ are a collection of orthonormal vectors (enforcing normality for simplicity). Does this imply that these coordinates can be computed from a translation and a rotation from the standard coordinates?
I haven't really been able to show anything useful. Any thoughts are appreciated!
Edit: I can at least supply the information that I would know. In general, to have an orthonormal set of coordinates $(u^1,…,u^n)$, we require two things: orthogonality and normality, so
$$\frac{\partial{\bf r}}{\partial u^i}\cdot\frac{\partial{\bf r}}{\partial u^j}=\delta_{ij}$$
so
$$\sum_{k=1}^n\sum_{m=1}^n\frac{\partial x^k}{\partial u^i}\frac{\partial x^m}{\partial u^j}\frac{\partial{\bf r}}{\partial x^k}\cdot\frac{\partial{\bf r}}{\partial x^m}=\delta_{ij}$$
As we have $\partial{\bf r}/\partial x^k={\bf e}_k$, this is
$$\sum_{k=1}^m\frac{\partial x^k}{\partial u^i}\frac{\partial x^k}{\partial u^j}=\delta_{ij}$$
If $J$ is the Jacobian matrix for the transformation in the form
$$J=\left(\begin{array}{ccc}
\frac{\partial x^1}{\partial u^1} & \cdots & \frac{\partial x^n}{\partial u^1} \\
\vdots & \ddots & \vdots \\
\frac{\partial x^1}{\partial u^n} & \cdots & \frac{\partial x^n}{\partial u^n}
\end{array}\right)$$
This then means $JJ^T=I$, which makes sense: if the Jacobian matrix is orthogonal, then the coordinates remain orthogonal. The question then is: does there exist a nonconstant orthogonal Jacobian matrix that describes a valid coordinate system for $\mathbb{R}^n$?
Edit 2: Another thing I notice is that we can explicitly find the inverse of $J$ by noticing
$$\delta^i_j=\frac{\partial u^i}{\partial u^j}=\sum_{k=1}^n\frac{\partial u^i}{\partial x^k}\frac{\partial x^k}{\partial u^j}$$
which tells us that
$$J^{-1}=\left(\begin{array}{ccc}
\frac{\partial u^1}{\partial x^1} & \cdots & \frac{\partial u^n}{\partial x^1} \\
\vdots & \ddots & \vdots \\
\frac{\partial u^1}{\partial x^n} & \cdots & \frac{\partial u^n}{\partial x^n}
\end{array}\right)$$
Since we also know $J^T=J^{-1}$, this means
$$\frac{\partial x^i}{\partial u^j}=\frac{\partial u^j}{\partial x^i}$$
I don't know if this is helpful, but it's definitely an interesting relationship.
Best Answer
If the Jacobian is a rotation (an element of $SO(n)$ or $O(n)$), the pulled back metric is $R^T R=I$. So you have an isometry to the Euclidean space, which if you are mapping from nD to nD, means that it can only be a global rotation and translation.