How does effective potential transform under coordinate transformation?

Question

Let us say we have an equation of motion of the following form,

$$\ddot{x}=g\tag{1}$$

For this system an effective potential can be defined as,

$$\ddot{x}=-\dfrac{d}{dx}U_\text{eff}$$

$$U_\text{eff}=-gx\tag{2}$$

Now let us do a coordinate transformation of the simplest sort, i.e.

$$y=x/2$$

Under this $(1)$ becomes

$$2\ddot{y}=g \implies \ddot{y}=g/2$$

$$\ddot{y}=-\dfrac{d}{dy}U_\text{eff}^*$$

$$\implies U_\text{eff}^*=-\dfrac{g}{2}y$$

However, putting the transformation in $(2)$ directly,

$$\tilde U_\text{eff}=-2gy$$

I expected that under coordinate transformations,

$$U_\text{eff}^*=\tilde U_\text{eff}$$

But that is not the case as shown. Why am I wrong?

Answer 1

You can see the transformation as a change of unit of measure. Let me change the constant in the transformation to make a more physically understandable example.

If $x$ is measured in meters and $y$ in centimeters, you have the transformation $y = 100 \ x$. However, the value of $g$ should also change otherwise $100 \ \ddot y = g$ would compare cm/s$^2$ to the left and m/s$^2$ to the right.

Hope this helps.

PS: of course, this has nothing to do with the potential energy itself. Also, the equation between 1 and 2 is wrong because the mass is missing, but that doesn't influence the question nor the answer.