# Transformation equations from Generalized coordinates to rectangular

I've a proper set of generalized coordinates {$$q_j$$} ,$$j=1...n$$ for a system. This set determines the configuration of the system, also these can be used to determine the rectangular coordinates of all the particles of the system.

A transformation that gives us the rectangular coordinates from these generalized ones should only depend on {$$q_j$$} but Thornton Marion say about the coordinate transformation

$$x_{ i}=x_{i}\left(q_{1}, q_{2}, \ldots, q_{n}, t\right)$$.

Why do we need an additional variable of time to determine the rectangular coordinates $$x_i$$ if {$$q_j$$} are enough?

• Well, add to the generalized and to the rectilinear coordinates time $t$ as an extra coordinate. And introduce another parametrizing parameter, say $s = t$. Then you are in the setting you want, just one dimension higher. Jul 17, 2021 at 1:08

Here is an example:

Let $$O\,\vec{e}_x\,\vec{e}_y\,\vec{e}_z$$ be an inertial coordinate system placed in a gravitational field of constant magnitude and direction $$-g\,\vec{e}_z$$.

Assume you have a bead of mass $$M$$ restricted to movie only along the vertical circular ring of radius $$r$$ on the picture under the constant gravitational acceleration $$-g\,\vec{e}_z$$. Assume that the circular ring spins with constant angular velocity $$\omega \, \vec{e}_z$$ along the coordinate axis $$O\,\vec{e}_z$$. Then the angle between the axis $$O \,\vec{e}_x$$ and the vector $$\vec{OP}$$ changes with time an for each moment of time $$t$$ it is $$\omega\, t$$. The position of the bead on the ring is described by the angle $$\phi$$ which is the angle between the coordinate axis $$O\,\vec{e}_z$$ and the radius-vector $$\vec{OM}$$. Thus, $$\phi$$ is your generalized coordinate. Then, given the angle $$\phi$$ and a specific moment of time, the position of the bead in the 3D space with respect to the inertial coordinate system is given by the (time-dependent) transformation

\begin{align} &x = r\sin(\phi)\cos(\omega\,t)\\ &y = r\sin(\phi)\sin(\omega\,t)\\ &x = r\cos(\phi) \end{align}

Now, the Lagrangian of this bead in the inertial coordinate frame is the standard kinetic minus potential energy Lagrangian

$$L \, =\, \frac{M}{2}\left( \, \Big(\frac{dx}{dt}\Big)^2 + \Big(\frac{dy}{dt}\Big)^2 + \Big(\frac{dz}{dt}\Big)^2\,\right) \, -\, M\,g\,z$$

But since the fact that the bead is restricted to the ring, we can express the cartesian inertial coordinates in terms of the generalized coordinate $$\phi$$, taking into account the predictable time-dependence of the ring's rotation

$$\frac{M}{2}\left( \, \Big(\frac{d}{dt}\big(\,r\sin(\phi)\cos(\omega\,t)\,\big)\,\Big)^2 + \Big(\frac{d}{dt}\big(\,r\sin(\phi)\sin(\omega\,t)\,\big)\,\Big)^2 + \Big(\frac{d}{dt}\big(\,r\cos(\phi)\,\big)\,\Big)^2\,\right) \, -\, M\,g\,r\,\cos(\phi)$$

After performing all the differentiations and trigonometric simplifications (and if I have calculated it correctly), the final Lagrangian becomes

$$L \, =\, \frac{Mr^2}{2} \Big(\frac{d\phi}{dt}\Big)^2 \, +\, \frac{Mr^2\omega^2}{2} \sin^2(\phi) \,-\, Mgr \cos(\phi)$$

And the corresponding Euler-Lagrange equation is $$\frac{d}{dt} \left(\,\frac{\partial L}{\partial \dot{\phi}}\,\right) \, =\, \frac{\partial L}{\partial {\phi}}$$ $$\frac{d^2\phi}{dt^2} \, =\, \omega^2 \sin(\phi) \cos(\phi) \,+\, \frac{g}{r} \sin(\phi)$$

• Thanks for the effort :) Jul 22, 2021 at 7:59

The constraints holding the system together can be explicitly time dependent. If that is the case the Lagrangian will depend explicitly on time. Remember that the case in which $$L(q,\dot q,t)$$ does not depend explicitly on time is the case where energy is conserved. This should make it clear that the absence of $$t$$ (i,e $$L=L(q,\dot q)$$) is a special case --- even though in most exmples in testbooks it is the only case considered..

• Could you please give me an example? Jul 14, 2021 at 4:55
• A rigid pendulum with a time dependent length $l(t)$: $L[\theta, \dot \theta, t]= ml^2(t)\dot \theta^2 /2- l(t)mg\cos\theta$ Jul 14, 2021 at 11:51

Why do we need an additional variable of time to determine the rectangular coordinates $$x_i$$ if {$$q_j$$} are enough

There are many scenarios where you don't need an explicit time-dependency. But there are also scenarios where an explicit time-dependency makes perfect sense.

Consider for example the transformation from cylindrical coordinates ($$r, \phi$$) on a carousel (rotating with angular velocity $$\omega$$) to cartesian coordinates ($$x,y$$) on the ground:

\begin{align} x&=r\cos(\phi+\omega t) \\ y&=r\sin(\phi+\omega t) \end{align}

This $$t$$-dependent transformation comes in handy, when you want to describe a body's motion on the rotating carousel using the generalized coordinates ($$r,\phi$$).

• Why isn't $x=r cos(\phi)$ ? Jul 14, 2021 at 4:47
• Because I chose the generalized coordinates to be a rotating frame. When a body is at rest on the carousel, then it is moving relative to the ground Jul 14, 2021 at 6:26
• Thank you very much. Jul 22, 2021 at 7:59
• @ThomasFritsch Are both descriptions equivalent? I mean we could get the $x$ coordinate based on one of the following transformations: \begin{align} x&=r\cos(\phi+\omega t) \\ x&=r\sin(\phi') \end{align} where $\phi'$ has now "absorbed" the motion of the carousel. Is there any benefit from writing the transformation equations as: $$x = x(q_1, \ldots, q_n, t)$$ instead of $$x = x(q_1, \ldots, q_n)$$? Jul 11, 2022 at 16:11

Within Newtonian mechanics, the logic is usually the opposite: We are given $$N$$ point-particle positions in rectangular 3-space $$\mathbb{R}^3$$. They satisfies $$3N-n$$ holonomic constraints, so that we can (at least locally) define $$n$$ generalized coordinates.

References:

1. H. Goldstein, Classical Mechanics; Chapter 1.