Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Related: – 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 $U=U(x-Vt)$ .

share|cite|improve this answer

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.

share|cite|improve this answer

The Noether argument is true, but it is also simple to see by thinking about it--- if you start with a free particle at position -50, turn on a spring potential kx^2/2 for a short time, it will start moving towards the center. Then you turn off the spring, and the particle is moving. In general, any force you apply to the particle will have a description in terms of a time dependent potential of the form

$$ U(t) = - F(t)x$$

So whatever force you apply to a particle, you have a time dependent potential. The notion of potential is then only really mathematically useful when there are further conditions, most importantly that it is time independent.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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