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:


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:


Thus, they say, one can find the law of transformation of the energy between the two frames by making the substitution:


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.

  • $\begingroup$ They are assuming that the potential function is a scalar. $\endgroup$ Jan 14 '17 at 15:40
  • $\begingroup$ I don't understand what difference this makes. I also assumed that $U(\mathbf{r})$ is a scalar-valued function... $\endgroup$
    – Lachy
    Jan 14 '17 at 15:49
  • 1
    $\begingroup$ A scalar is the name often given to a quantity that doesn't change under change of (inertial) reference frame (usually under Lorentz transformation, I never heard it used in classical mechanics but that's what is probably the meaning Lewis intended) $\endgroup$ Jan 14 '17 at 15:49
  • $\begingroup$ If the potential depends on the difference of the position of the particles, or on the difference between some fixed point and the particle positions (i.e a spring with one point fixed), the extra term $\mathbf{V}t$ cancels out when computing the differences. I think the reason is this, namely that potential energy is referred to some "background" whose transformation cancels the new term. $\endgroup$
    – user139175
    Jan 14 '17 at 16:11
  • $\begingroup$ @yoric Yes, I think this is it. The fact they don't specifically state that the system is closed made me think that the potential can't be assumed to depend only on the difference between particle positions. But as you say, the potential will still depend on the position of a particle with respect to an origin that is itself moved by the transformation. Is that right? $\endgroup$
    – Lachy
    Jan 14 '17 at 17:12

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 $U'(h') = mg(h - vt + vt) = U(h)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.