# Energy non-conservation for time-dependent potentials

Written in a book I read that the "total energy is not preserved when the potential depends explicitly on time", i.e. $U=U(x,t)$. Is there any proof or explanation for this?

• – Qmechanic Nov 16 '11 at 18:17

It is easy to understand on a "ball & wall" problem. If you throw a ball in the wall, the ball total energy is conserved during reflection: $E=\frac{mv^2}{2}+U(x)$. The potential energy $U(x)$ is explicitly time-independent here.
If you keep the ball at rest but hit it with a wall, the ball energy changes. Now the wall is moving and its position explicitly depends on time. It transfers some energy to the ball. This is the case of $U=U(x-Vt)$ .
Yes, generically it will then not be conserved. On the other hand, if there is no explicit time dependence, then time translations $t\to t+a$ will be a symmetry of the action. The corresponding Noether charge is the total energy $E$, and it will be conserved due to Noether's first Theorem.
$$U(t) = - F(t)x$$