# Conservative force in a time-dependent potential

1. Feb 29, 2016

### Happiness

A conservative force is one that is derivable from a potential energy function $V$. If $V$ is time-dependent, is it still possible to have a conservative force or work done such that the work done is only dependent on the initial state $(x_i, y_i, z_i, t_i)$ and final state $(x_f, y_f, z_f, t_f)$ of the particle and independent of the trajectory taken?

2. Feb 29, 2016

### vanhees71

No, because in that case the work-energy theorem turns out to be
$$\frac{m}{2} \vec{v}(t)^2+\text{const}=\int \mathrm{d} t \vec{F}(\vec{x},t) \cdot \vec{v}=-\int \mathrm{d} t \vec{v} \cdot \vec{\nabla} U(\vec{x},t).$$
Now
$$\frac{\mathrm{d}}{\mathrm{d} t} U(\vec{x},t)=\vec{v} \cdot \vec{\nabla} U(\vec{x},t) + \partial_t U(\vec{x},t),$$
i.e., the integrand on the right-hand side of the work-energy theorem is not a total time derivative as in the case, when $U$ is not explicitly time-dependent.

This is a consequence of Noether's theorem: A system is symmetric under time translations if and only if the Lagrangian is not explicitly time dependent modulo a total time derivative of a function $\Omega(q,t)$.

3. Feb 29, 2016

### Happiness

How about we treat time as the fourth spatial dimension? Where $\vec{v}$ becomes a four-dimensional vector. Then
$$\frac{\mathrm{d}}{\mathrm{d} t} U(\vec{x},t)=\vec{v} \cdot \vec{\nabla} U(\vec{x},t)$$
So the integrand on the right-hand side of the work-energy theorem becomes a total time derivative.

What's wrong with this?

4. Feb 29, 2016

### vanhees71

Well, that comes close to relativistic dynamics in terms of a world parameter. That generalized dynamics has to be developed carefully. For a good introduction (although unfortunately full of typos even in the math) see

Barut, Electrodynamics and classical theory of particles and fields, Dover