Why does one neglect potential energy when switching between (classical) reference frames? The title of the post is probably a little broad. In particular, I am looking at a proof from "Mechanics" by Landau and Lifshitz. The authors begin in reference frame $K$, in which a system of $n$ particles have the velocities $\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_n$. The energy of the system is given by:
$E=\frac{1}{2}\sum_am_av_a^2+U$,
where the potential energy function, $U$, depends only on the positions of the particles.
They then say that there is a second reference frame $K'$ which moves at a velocity $\mathbf{V}$ with respect to $K$. In this frame the $i^{\text{th}}$ particle has velocity $\mathbf{v}'_i$, where:
$\mathbf{v}_i=\mathbf{v}'_i+\mathbf{V}$.
Thus, they say, one can find the law of transformation of the energy between the two frames by making the substitution:
$E=\frac{1}{2}\sum_am_a(\mathbf{v}'_i+\mathbf{V})^2+U$.
What I find confusing is that they assume that the potential energy function is unchanged by the change of reference frames. I can see that this would be the case if one considers a closed system in which all interactions depend on the distance between the particles. But they do not make the claim that the system is closed. If there is an external potential, then it seems true that:
$U(\mathbf{r}_i)\rightarrow U(\mathbf{r}_i'+\mathbf{V}t)$.
Is this the case? 
Intuitively, it does seem that the potential energy of a classical particle shouldn't depend on where we view it from. I just can't see why this is necessarily true when I perform the transformation. Thanks.
 A: I think your confusion is using the same symbol $U$ for two different functions.
Start by writing$$\begin{align}
E &= \tfrac{1}{2}\sum_a m_a v_a^2 + U(\mathbf r)\\
E' &= \tfrac{1}{2}\sum_a m_a {v'_a}^2 + U'(\mathbf r')
\end{align}$$
The statement in L&L is correct because there are two changes to the potential energy expression which cancel out. First, the function $U(\mathbf r)$ depends on the coordinate system used for $\mathbf r$. Second, the argument $\mathbf r(t)$ for the same particle, in two different coordinate systems also depends on the coordinate system.
The form of the potential function is different in the two frames, but its value is the same for the same physical configuration of particles. 
Consider a simple example, like gravity near the earth's surface. In a coordinate frame fixed to the earth, a particle of mass $m$ at height $h(t)$ has potential energy $U(h) = mgh$. In a coordinate frame moving upwards with constant speed $v$, the potential energy function is now $U'(h') = mg(h' + vt)$, where $h'$ is measured in the moving frame. But the position of the same particle in the moving frame is $h' = h - vt$, so we get
$U'(h') = mg(h - vt + vt) = U(h)$. 
