# 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.

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My feeling it that your problem is underconstrained, because in the ' system you have a degree of freedom more. –  Bernhard Apr 18 '14 at 17:06
By the way, your notation is confusing with the primes, that are use to indicate rotation and 4d-system. –  Bernhard Apr 18 '14 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. –  Shivam Sarodia Apr 18 '14 at 17:18
@Bernhard Never mind, I'm wrong. –  Shivam Sarodia Apr 18 '14 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 ;)) –  Bernhard Apr 18 '14 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.

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