# Rotation operator for a point in a coordinate system linearly derived from Cartesian coordinates

For some experimental and practical reason, I have created a new coordinate system in the form

$$x^\prime_i=T_{ij}x_j$$

where $T_{ij}$ isn't a square matrix. $x_i$ is standard Cartesian coordinates, and $x^\prime_j$ is a point in the new system. I have to mention that the new system's axes are not linearly independent. So the last relation can be written as

$$\left(\matrix{x_0^\prime\\x_1^\prime\\x_2^\prime\\x_3^\prime}\right)=\left( \matrix{T_{11} & T_{12} & T_{13} \\ T_{21} & T_{22} & T_{23} \\ T_{31} & T_{32} & T_{33} \\ T_{41} & T_{42} & T_{43} } \right)\cdot \left(\matrix{x\\y\\z}\right)$$

The matrix $T_{ij}$ is well defined.

What I need is a rotation operator that will transform a point in the primed system, as the standard rotation operator does. So say I have the standard rotation matrix in Cartesian coordinates around the z-axis:

$$R_{ij}= \left( \matrix{\cos{\theta}&-\sin{\theta}&0\\ \sin{\theta}&\cos{\theta}&0\\0&0&1} \right)$$

So to rotate a point in Cartesian coordinates, we use the standard operator formula:

$$P^\prime_i=R_{ij}P_j$$

where $P_j$ is the point before rotation, and $P^\prime_i$ is the point after rotation.

How can I write this rotation formula for a point in the new coordinates system that uses 4 points? How will the rotation matrix look like? I expect a rotation matrix that is $4\times4$, but I don't know how to derive it. Please help in that.

• My feeling it that your problem is underconstrained, because in the ' system you have a degree of freedom more. Commented Apr 18, 2014 at 17:06
• By the way, your notation is confusing with the primes, that are use to indicate rotation and 4d-system. Commented Apr 18, 2014 at 17:07
• @Bernhard I'm not certain, but I suspect that there isn't a new degree of freedom on the primed system because the axes aren't linearly independent. Given three of the coordinates, the last would be uniquely defined. Commented Apr 18, 2014 at 17:18
• @Bernhard Never mind, I'm wrong. Commented Apr 18, 2014 at 17:23
• @Draksis I've written the proof in my answer (not sure if it is proof from a mathematicians point of view, but for physicist it is hopefully enough ;)) Commented Apr 18, 2014 at 17:25

To be consistent with notation, I use the $x'$ for the transformation to the new system and $\tilde{x}$ for the rotation. Thus, as you defined

$$x'_i=T_{ij}x_j,$$ $$\tilde{x}_i=R_{ij}x_j.$$

We know that

$$\tilde{x}'_i=T_{ij}\tilde{x}_j=T_{ij}R_{jk}x_k.\tag{1}$$

You are looking for the transformation matrix $Q_{ij}$, such that

$$\tilde{x}'_i=Q_{ij}x'_j,$$

or

$$\tilde{x}'_i=Q_{ij}x'_j=Q_{ij}T_{jk}x_k\tag{2}$$

Naively, one could now write from (1) and (2)

$$T R = Q T ,$$ $$Q=T R T^{-1}.$$

However, $T$ is not a square matrix, and does not invertable. In other words, such a matrix $Q$ can not be determined uniquely.

Or, looking at it as $$T_{ij}R_{jk} = Q_{pq}T_{qr}$$

You know that these are twelve equations, because both procut matrices are $4\times3$. $Q_{ij}$ is a $4\times4$ matrix, with, thus, $16$ unknowns. In other words, there are infinitely many possibilities. Unless, of course, you add constraints.