So, I was reading Classical Mechanics by Goldstein where he just introduced scleronomous and rheonomous constraints.

He writes:

Constraints are further classified according to whether the equations of constraint contain the time as an explicit variable (rheonomous) or are not explicitly dependent on time (scleronomous). A bead sliding in a rigid curved wire fixed in space is obviously subject to scleronomous constraint; if the wire is moving in some prescribed fashion, the constraint is rhenomous. Note that, if the wire moves, say, as a reaction to the bead's motion, then the time-dependence of the constraint enters in the equation of the constraint only through the coordinates of the curved wire. The overall constraint is then scleronomous.

I couldn't get why the case where the wire moves "as a reaction to the bead's motion" is scleronomous.

$\bullet$ Isn't the time-dependence of the constraint enough for the equation of constraint to be schelonomous? (Is there any significance of the word explicit in this present context?)

$\bullet$ Why is the above example of bead-wire not schleronomous? When the wire moves in a "prescribed fashion", the case is rheonomous; but when the wire moves, "as a reaction to the bead's motion", the case becomes sceloronomous. Why is it so? How does the time-dependence by entering "in the equation of the constraint only through the coordinates of the curved wire" make the the equation of constraint rheonomous?


When you start with the problem you can be told explicitly that the wire moves in a prescribed manner - rheonomous.
However if the wire moves as the result of some else happening then you know that the wire will move but you do not know, until you have solved the problem, how the movement of the wire will depend on time - scleronomous.

The example which is often used is that of a simple pendulum where with a fixed point of suspension the constraint is scleronomous.
The position of the pivot $(0,0)$ and the position of the bob $(x,y)$ are linked to the length of the string $l$.
$$x^2+y^2 = l^2$$

If the point of suspension is forced to oscillated with simple harmonic motion the constraint is rheonomous.
The position of the point of suspension $(a\sin \omega t,0)$ and the position of the bob $(x,y)$ are linked to the length of the string $l$.
$$(x-a\sin \omega t )^2+y^2 = l^2\;.$$

If the point of suspension $(x'(t),0)$ is free to move along a horizontal line then the motion is scleronomous because $(x-x'(t))^2+y^2 = l^2$ but although you can guess that $x'(t)$ will depend on time $t$ you do not know when starting the problem how $x'(t)$ actually depends on the time.
You have not directly forced $x'(t)$ to change with time in a prescribed/predetermined way.

  • $\begingroup$ Thanks @farcher for the answer. So, it all boils down to what the problem offers me initially, is it so? And if I had known the function $x'(t)$, would it be rhenomous then? I guess that is what it the matter is indeed. $\endgroup$
    – user36790
    Sep 15 '16 at 10:39

In addition to the above answer, it is key to note that when x'(t) is added to the equations that the position of the constraint becomes a coordinate in the system, and you go from having 2 coordinates, x and y, to having 3 coordinates, x , y and x'. This will make the future Euler-Lagrange Equations more difficult to solve, which makes total sense because obviously a mobile pendulum is more complex than a stationary one. Additionally, if x'(t) is prescribed before, then we don't include it in the systems coordinates, but we must take it as a rheonomous constraint then. Think of it as an externally forced system. This is in correspondence with scleronomous constraints preserving the Hamiltonian of the system (in a broad number of cases) and rheonomous constraints not necessarily preserving the Hamiltonion. This makes sense as the Hamiltonion of an eternally forced system can indeed be time dependent.


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