Change of coordinates in a potential energy field

  • #1
kal8578
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0
TL;DR Summary
When a particle subjected to a conservative force has a constraint between multiple coordinates, one can change variables in the potential energy. But doing so seems to change the gradient, and thus the force, and I do not understand why.
Hello,

I am having some confusions in what should be basic pointwise Newtonian mechanics, and would like to get some help with that. It is all about changing coordinates in potential energies.

Let us start by considering a point particule in a 2d world with an axis x (left-right) and an axis z (up-down) with constant gravitational acceleration g (like on Earth). Said particle, of mass m, lies on a ground of shape z = f(x).

Capture d’écran 2024-08-28 à 09.02.47.png


In what follows, I will further refine the analysis to z = αx2.

What I think to be true (tell me if it's not !)

While the potential energy of the particle is V(x) = mg αx2, which is quadratic - like 1/2kx2 - its motion is not that of an harmonic oscillator. The reason behind that is that the kinetic energy reads
T = 1/2m(\dot{x}2+\dot{z}2) = 1/2m(1+4α2.x2.)\dot{x}2.

So that conservation of energy yields
1/2m(1+4α2.)\dot{x}2 + mg α2 = cst
Which is not of the form required for an harmonic oscillator, namely $... \dot{x}α2 + ... xα2 = cst$.

To get an harmonic oscillator, the choice of f(x) should be different : essentially, one should introduce s the curvilinear abscissa along the z=f(x) curve, so that T = m \dot{s}α2/2, and choose f so that f(s) is proportional s2.

My problem

Why can I just not write (as the usual ``think of potential energies as landscapes on which a ball can roll'' way of thinking goes):
V(x) = α m g x 2

And use from there Newton's law of motion to derive, with \vec{r} the position of the mass m:
m \ddot{\vec{r}} = - \vec{\nabla} V = - 2 α m g x ex

which projected on the x-axis, yields
m \ddot{x} = - 2 α m g x

which is an harmonic oscillator equation ?

In general,
- Why can't I change coordinates in potential energies?
- How to know which choices of coordinates lead to correct results?

Thanks !

PS : Sorry for the typesetting I don't know how to make LaTeX work on this website
 
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  • #2
kal8578 said:
Why can I just not write (as the usual ``think of potential energies as landscapes on which a ball can roll'' way of thinking goes):
V(x) = α m g x 2
Because the potential is not the only source of forces on the mass. There is also the constraining force provided to keep the mass on the parabola.
kal8578 said:
In general,

- Why can't I change coordinates in potential energies?
You can. It is perfectly fine to do that. However, if the motion is harmonic in some variable ##s##, it obviously will not be harmonic in a variable that is not directly proportional to ##s##.

kal8578 said:
- How to know which choices of coordinates lead to correct results?
The choice that is consistent with the assumptions of the problem.

kal8578 said:
PS : Sorry for the typesetting I don't know how to make LaTeX work on this website
There is a link to https://www.physicsforums.com/help/latexhelp/ right below the post editor pane.
1724843274738.png
 

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