Change of coordinates in a potential energy ?

Question

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

$$\begin{equation} T = \frac{1}{2}m(\dot{x}^2+\dot{z}^2) = \frac{1}{2}m(1+4\alpha^2x^2)\dot{x}^2 \end{equation}$$
So that conservation of energy yields
$$\begin{equation} \frac{1}{2}m(1+4\alpha^2x^2)\dot{x}^2 + mg \alpha x^2 = cst \end{equation}$$
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):

$$\begin{equation} V(x) = \alpha m g x^2 \end{equation}$$

And use from there Newton's law of motion to derive, with $\vec{r}$ the position of the mass $m$:

$$\begin{equation} m \ddot{\vec{r}} = - \vec{\nabla} V = - 2 \alpha m g x \vec{e}_x \end{equation}$$

which projected on the $x$-axis, yields

$$\begin{equation} m \ddot{x} = - 2 \alpha m g x \end{equation}$$

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 !

Answer 1

The potential energy landscape for a harmonic oscillator $V(x) = kx^2$ is a landscape the "left-right" direction of which is $x$, but the "up-down" direction is $V$, not a further spatial coordinate.

The case at the beginning of your question concerns a situation where the mass is to move in a two-dimensional space, but constrained to the shape $z=f(x)$, which you specialise to $z=\alpha x^2$. This is a physically different situation; as the particle moves, a reduction in its potential energy is converted into kinetic energy of motion along $x$, but, as you state, also along $z$. Therefore this situation cannot be a harmonic oscillator.

Your approach via Newton's law with the force equal to $-\vec\nabla V$ treats the potential, in which you have used the $z=\alpha x^2$ constraint, as if it were the potential of an unconstrained system. This is not possible, $\vec F=-\vec\nabla V$ simply does not hold in such a case. 

Answer 2

You cannot change coordinates in the potential energies because

1. fancy answer, it is because of general covariance

2. simple answer that you might be looking for. the potential energy transforms in the adjoint representation of the gauge group

Comments

In my understanding, general covariance has a lot to do with the change of coordinates. You just have to do it correctly. If you do, you will change coordinates also in a potential.

Furthermore, the question has little to do with gauge theories.

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are you the same Flamma from the other thread? Kripkep, i feel you didn't understand my answer. coordinate changes in general relativity have to do with the hole argument. you cannot possibly understand it without understanding the hole argument. I stand by both points that I made