Take the 2-minute tour ×

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 $t$.

share|improve this question
add comment

2 Answers

up vote 15 down vote accepted

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.

share|improve this answer
add comment

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 $$\mathbf{r}_i= \mathbf{r}_i(q_1,q_2,..,q_{3N-k},t)$$

share|improve this answer
add comment

Your Answer


