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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?

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  • $\begingroup$ 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. $\endgroup$ Jul 17, 2021 at 1:08

4 Answers 4

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Here is an example:

enter image description here

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 $$\frac{d}{dt} \left(\,\frac{\partial L}{\partial \dot{\phi}}\,\right) \, =\, \frac{\partial L}{\partial {\phi}}$$ $$\frac{d^2\phi}{dt^2} \, =\, \omega^2 \sin(\phi) \cos(\phi) \,+\, \frac{g}{r} \sin(\phi)$$

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  • $\begingroup$ Thanks for the effort :) $\endgroup$
    – Kashmiri
    Jul 22, 2021 at 7:59
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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..

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

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  • $\begingroup$ Why isn't $x=r cos(\phi)$ ? $\endgroup$
    – Kashmiri
    Jul 14, 2021 at 4:47
  • $\begingroup$ 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 $\endgroup$ Jul 14, 2021 at 6:26
  • $\begingroup$ Thank you very much. $\endgroup$
    – Kashmiri
    Jul 22, 2021 at 7:59
  • $\begingroup$ @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 $$x = x(q_1, \ldots, q_n)$$? $\endgroup$ Jul 11, 2022 at 16:11
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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.
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