Are generalised coordinates truly independent?

Question

Say we have a system with two generalised coordinates $x$ and $y$. When we solve the equations of motion we find $x=x(t)$ and $y=y(t)$. I can invert one of these solutions to find $t=t(y)$ and therefore get $x=x(t(y))$ which therefore gives me $x(y)$. Does the equation of motion impose a constraint? Are generalised coordinates independent?

Answer 1

You are losing information by doing that transformation. In particular, you will only know about the orbit of the particle, and will lose all information about the velocity.

In general, a generalized coordinate transform from a set of $x^{a}$ to a set of $y^{a}$ will only be valid if, for the matrix $M_{ab} = \frac{\partial y^{a}}{\partial x^{b}}$, you have ${\rm det}\left( M_{ab}\right) \neq 0$

Answer 2

1. When we ask if generalized coordinates $q^j$ are independent, we by definition mean before using any differential$^1$ equations of motion. A differential equation of motion is by definition not considered a constraint.

2. Generalized coordinates could be dependent if we have further constraints implemented via Lagrange multipliers.

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$^1$ By equations of motion, we mean Lagrange equations (as opposed to purely kinematic identities). By differential equations of motion, we mean equations of motion with time derivatives.

Answer 3

You shouldn't try to make $t$ a coordinate; it's a label for the coordinates, over which the coordinate-dependent Lagrangian is integrated to form the action. (The most obvious problem this causes is that the momentum of time is undefined, viz. $\frac{\partial L}{\partial \dot{t}}=\frac{\partial L}{\partial 1}$.) It would be like trying to change the fields in a theory so one is replaced with the spacetime coordinates that label the fields (note these are integrated over to obtain a field theory's action, which is any fields are analogous to generalised coordinates).