For a two-dimensional cartesian coordinate system $(x,y)$ describing two particles of unit mass, one frequently encounters a transformation into center-of-mass and relative coordinates $(R,r)$ defined by

$$ \begin{pmatrix}R\\r\end{pmatrix} = \begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} $$

How can this concept reasonably be extended to higher dimensions?

Here are my thoughts. In principle, the problem can be approached quite directly as the following: Find a unitary transformation matrix $V$ with a row of ones on top, $$ V = \begin{pmatrix}1&1&\cdots&1&1\\\\ & &U\\ &\end{pmatrix} $$ This can be accomplished, for example, by starting from the identity matrix, plugging in the row of ones on top, and applying Gram-Schmidt orthonormalization to obtain the matrix U. This, however, is not going to produce a "balanced" set of coordinates (whatever that means). Thus, I'm looking for approaches where $U$ is reasonably chosen, say as compact or sparse as possible, or with other suitable properties. This is also where I'd be interested what references in the literature did.


1 Answer 1


How can this concept reasonably be extended to higher dimensions?

first create a vector $\vec{x}$ with n components , for example $n=5$

$$\vec{x}= \left[ \begin {array}{c} x_{{1}}\\ x_{{2}} \\ x_{{3}}\\ x_{{4}} \\ x_{{5}}\end {array} \right] $$

then create the vector $\vec{r}$ with $n-1$ relative equations

$$\vec{r}=\left[ \begin {array}{c} x_{{1}}-x_{{2}}\\ x_{{2}}- x_{{3}}\\ x_{{3}}-x_{{4}}\\ x_{{4} }-x_{{5}}\end {array} \right] $$

the matrix

$U=\frac{\partial \vec{r}}{\partial \vec{x}}\quad,(n-1\times n)$ the Jacobi-Matrix

$$U=\left[ \begin {array}{ccccc} 1&-1&0&0&0\\ 0&1&-1&0&0 \\ 0&0&1&-1&0\\0&0&0&1&-1 \end {array} \right] $$

you can write a small program to build the matrix V

enter image description here

$$V=\left[ \begin {array}{ccccc} 1&1&1&1&1\\ 1&-1&0&0&0 \\ 0&1&-1&0&0\\ 0&0&1&-1&0 \\ 0&0&0&1&-1\end {array} \right] $$

  • $\begingroup$ Actually I considered this choice, but discarded it, as it's not a unitary transformation, and thus won't produce an orthonormal coordinate system. But nevertheless, thanks for your answer! $\endgroup$
    – davidhigh
    Jul 15, 2020 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.