# Change of coordinates in a potential energy ?

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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 $z = f(x)$.

In what follows, I will further refine the analysis to $z = \alpha x^2$.

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

While the potential energy of the particle is $V(x) = mg \alpha x^2$, which is quadratic - like $\frac{1}{2}kx^2$ - its motion is not that of an harmonic oscillator. The reason behind that is that the kinetic energy reads

T = \frac{1}{2}m(\dot{x}^2+\dot{z}^2) = \frac{1}{2}m(1+4\alpha^2x^2)\dot{x}^2

So that conservation of energy yields

\frac{1}{2}m(1+4\alpha^2x^2)\dot{x}^2 + mg \alpha x^2 = cst

Which is not of the form required for an harmonic oscillator, namely $\dots \dot{x}^2 + \dots x^2 = cst$.

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 $T = m \dot{s}^2/2$, and choose $f$ so that $f(s) \propto s^2$.

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) = \alpha 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 \alpha m g x \vec{e}_x

which projected on the $x$-axis, yields

m \ddot{x} = - 2 \alpha 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 !

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