How do we justify that potential energy in a spring is Galilean invariant (to the extent that Newtonian mechanics holds)? A spring can store elastic potential energy by elastically deforming and moving its atoms out of their minima potentials. The atoms themselves can be modeled as balls connected by Hooke-like springs which are themselves supposed to model (an approximation of) the electromagnetic field. From this I surmise that the elastic potential energy of a compressed spring can be ultimately reduced down to electromagnetic energy.
Now my question is, is there a good argument that demonstrates that this is Galilean invariant as far as Newtonian mechanics holds?
This is difficult to think through because the EM field is inherently a relativistic object, and as far as I know the energy of the field is only defined in terms of its energy density. Note that for example kinetic energy is not Galilean invariant, so the question of whether potential energy is Galilean invariant is not obvious to me.
The biggest issue here is that at the atomic level, if we switch to a relatively moving reference frame, then we are no longer considering electrostatics, so it's not clear how to proceed in this scenario.
 A: 
From this I surmise that the elastic potential energy of a compressed spring can be ultimately reduced down to electromagnetic energy.

Well, yes, but you have to use the quantum mechanical rules of electromagnetism (QED) rather than classical electromagnetism (Maxwell’s equations). However, near the equilibrium point any potential can be approximated to second order by a quadratic potential. This gives rise to a force of the same form as Hooke’s law. With this justifiable simplification, the analysis is straightforward:
Suppose two particles of mass $m$ under a Hooke’s law (approximate) force with a stiffness $k$ start at their equilibrium separation $d$ each with velocity $v_0$ directly away from the other. Solving Newton’s 2nd law gives us $$x(t)=\pm \frac{d}{2} \pm \frac{v_0}{\omega} \sin (\omega t)$$ where $\omega^2=2k/m$. From this, $v(t)=\pm v_0 \cos(\omega t)$, and so the KE is $KE= m v_0^2 \cos(\omega t)^2$ which starts at $m v_0^2$ and goes to 0 at $t=\pi/(2\omega)$ which implies that the elastic potential energy is $m v_0^2$ at that time.
Now, if we perform a Galilean transformation $x\rightarrow x + u t$ and solve again we get $$x(t)=\pm \frac{d}{2} \pm \frac{v_0}{\omega} \sin (\omega t) + u t$$ so $v(t)=u \pm v_0 \cos(\omega t)$ and the KE is $KE=mu^2+m v_0^2 \cos(\omega t)^2$ which starts at $mu^2 + m v_0^2$ and goes to $mu^2$ at $t=\pi/(2\omega)$ which implies that the elastic potential energy is $m v_0^2$ at that time.
So the elastic potential energy is the same in all frames even though the KE is not.
A: The potential energy of a spring is given by $U=\frac{k \Delta x^2}{2}$.  Consider the Galilean transformations applicable in 1 dimension:

*

*Uniform motion, with velocity $\vec{v}$, is defined by $(\vec{x},t) \mapsto (\vec{x}+\vec{v}t,t)$.  Therefore $$\Delta x'=x'-x_0'=x+vt-x_0-vt=x-x_0=\Delta x, $$ i.e. the potential energy is invariant.


*Translation, defined by $(\vec{x},t) \mapsto (\vec{x}+\vec{a}, t+s)$.  Therefore $$\Delta x'=x'-x_0'=x+a-x_0-a=x-x_0=\Delta x$$ i.e. the potential energy is invariant.
Thus, the potential energy of a spring is invariant under Galilean transformations.
I have no idea and I doubt anyone has an answer at the atomic level.
