# Which inertial observer should I choose when using $\Delta U = -\Delta K$?

I just read and undertood that

• For any conservative force $F$ (which is not necessarily constant), we can define a potential energy change $\Delta U$ of an object on which the force exerts as $\Delta U=-W$ where $W$ is the work done by the force on the object.

• For any force $F$ (which is not necessarily conservative), we can define a kinetic energy change $\Delta K$ of an object on which the force $F$ acts as $\Delta K =W$ where $W$ is the work done by the force on the object.

The non-linearity of $K$ makes $\Delta K$ unequal to any non-inertial observer. For example, according to a stationary observer, an object moving to the right at speed from $v_i$ to $v_f$ has $\Delta K =1/2 m (v_f^2 - v_i^2)$. But according to another observer moving at speed $V$ to the left relative to the stationary observer, the change of kinetic energy becomes $\Delta K = 1/2m \left[(V+v_f)^2 - (V+v_i)^2\right]$.

Let's now assume the force acting on the object $m$ is conservative so we can define $\Delta U$.

• According to the stationary observer,

\begin{align*} \Delta U &= - W\\ &= - \Delta K\\ &= - 1/2 m (v_f^2 - v_i^2) \end{align*}

• According to the other observer moving to the left,

\begin{align*} \Delta U &= - W\\ &= - \Delta K\\ &= - 1/2m \left[(V+v_f)^2 - (V+v_i)^2\right] \\ &= -1/2 m (v_f^2-v_i^2 + 2Vv_f - 2Vv_i) \end{align*}

## Question

Which $\Delta U$ is the correct one?

• If momentum is conserved for a system of particles, then energy is conserved in all frames of reference. If you simply posit a field of force acting on a single particle, then momentum is not conserved, and conservation of energy is not frame-independent. – Ben Crowell Oct 20 '14 at 23:44

The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, that is, the work done by it on a closed path is zero, or alternativelly, between two points is not path dependent. Realize that this is related to the referential, since a closed path in one frame wil not in general be closed on another. So if this relation holds for the stationary observer, it can not hold in your example for the moving observer.
Edit: For example, consider the force acting along the $x$ axis given by $\vec{F}(x)=-kx\hat{\imath}$, that is one that will lead to an simple harmonic motion. This force is clearly conservative, and we can associate the potential $U(x)=\dfrac{1}{2}kx^2$ to it. Consider now a frame that is moving to the right with velocity $v$. To this frame the force is now given by $\vec{F}(x')=-k(x'+vt)\hat{\imath}$, since we have the relation $x=x'+vt$. But since this force now depends also on the time, it will not performe a null work over a closed path in the moving reference frame. Take for example the path $x'(t)=t(1-t)$, for $0\leq t\leq1$. This path is closed in the moving frame, but is not in the rest one, since $x(0)=0$ and $x(1)=v$, which gives a net work done by the force of $W=-\dfrac{1}{2}kv^2$. To calculate the work done by the force in the moving frame we do $$W'=\int_C F(x'(t),t)dx'(t)=-\int_0^1k(t(1-t)+vt)(1-2t)dt=\frac{kv}{6},$$ which is not null, so the force is not conservative in this frame.