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Conservative force in a time-dependent potential

  1. Feb 29, 2016 #1
    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?
     
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  3. Feb 29, 2016 #2

    vanhees71

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    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)##.
     
  4. Feb 29, 2016 #3
    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?
     
  5. Feb 29, 2016 #4

    vanhees71

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    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
     
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