# Are generalised coordinates truly independent?

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?

• This answer of mine addresses a related confusion. Upshot: You need to distinguish between the coordinates $(x,y)$ of the state space and the specific choice of a path $(x(t),y(t))$. – ACuriousMind Jul 17 '17 at 9:44

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 ${\rm det}\left( M_{ab}\right) \neq 0$

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.

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).