# Why so many arguments for the transformation equations of generalized coordinates?

For a system of $N$ particles with $k$ holonomic constraints, their Cartesian coordinates are expressed in terms of generalized coordinates as $$\mathbf{r}_1 = \mathbf{r}_1(q_1, q_2,..., q_{3N-k}, t)$$ $$...$$ $$\mathbf{r}_N = \mathbf{r}_N(q_1, q_2,..., q_{3N-k}, t)$$

Each particle in space can be uniquely identified by 3 independent variables, so why aren't the above of the form $$\mathbf{r}_i = \mathbf{r}(q_{i1}, q_{i2}, q_{i3})?$$

Note there is only one transformation $\mathbf{r}$ for all $\mathbf{r}_i$, a function of only three generalised coordinates and independent of $t$.

The generalized coordinates of a system of $N$ particles apply to the system as a whole, not the individual particles, and accordingly they can (and often do) combine the coordinates of multiple particles.

One common example is that of two-body orbital motion: one generalized coordinate is the position of the center of mass of the system,

$$\mathbf{q}_1 = \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2}{m_1 + m_2}$$

and the other is the displacement vector between the two bodies,

$$\mathbf{q}_2 = \mathbf{r}_2 - \mathbf{r}_1$$

Each of the generalized coordinates depends on physical coordinates from both objects, and if you invert the transformation you'll find that the physical coordinates of each object depend on both generalized coordinates. So you can't just express $\mathbf{r}_i$ in terms of $q_{i(1,2,3)}$ alone.

• So it is not necessary the generalized coordinates refer to the positions of particles right? In your example we could also have the generalized coordinates as the $r, \theta, \phi$ for each particle, whereas $r_{cm}$ and $r_{rel}$ clearly shows that this isn't always the case. Commented Feb 20, 2022 at 13:56
• @AntoniosSarikas I'm not sure I understand your question. I guess you could consider the spherical coordinates of each particle as a set of generalized coordinates, but that wouldn't be an example of a generalized coordinate that combines the physical coordinates of two or more different particles. (So that example would not have been a demonstration of the point I wanted to make in this answer.) Commented Feb 20, 2022 at 22:14
• I wanted to point out that is not necessary the generalized coordinates refer to particle's position, as your example showed. And that is why we must take into account all the generalized coordinates. Commented Feb 21, 2022 at 13:48
• @AntoniosSarikas Hmm okay... well, I suppose all I can say is that I disagree it would be beneficial to add that to the answer. Commented Feb 21, 2022 at 20:32

The $k$ holonomic constraints are used to eliminate $k$ $q$s, so reducing their number from $3N$ to $3N-k$. This then introduces the dependence of some of the transformation equations on t and other $q$s.

You have k holonomic constraints of the form $$\mathbf{f}_1(q_1, q_2,..., q_{3N},t) = 0$$ $$...$$ $$\mathbf{f}_k(q_1, q_2,...,q_{3N}, t) = 0$$ $3N$ q coordinates for the $N$ particles $$(q_1,q_2,q_3), (q_4,q_5,q_6),..., (q_{3N-2},q_{3N-1},q_{3N})$$ $3N$ transformation equations relating cartesian to generalised coordinates $$\mathbf{r}_1 = \mathbf{r}_1(q_1,q_2,q_3)$$ $$...$$ $$\mathbf{r}_N = \mathbf{r}_N(q_{3N-2},q_{3N-1},q_{3N})$$ Using the first constraint to eliminate $q_1$ gives $\mathbf{r}_1= \mathbf{g}_1(q_2,q_3,..,q_{3N},t)$ which is of the same form as one of the transformation equations you quoted. Which $q$ can be eliminated depends upon the constraints of the problem and so in general, the transformation equations are of the form $$\mathbf{r}_i= \mathbf{r}_i(q_1,q_2,..,q_{3N-k},t)$$