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Suppose force on a particle depends on both position of the particle and time.

In this case, how to find potential energy of the particle?

To be specific, force is $$ F(r,t) = (k/r^2) \exp(-\alpha t);$$ where $k$ and $\alpha$ are constants. $r$ is position of the particle from force centre and $t$ is time.

I have read that - Potential energy can not be defined for non-conservative forces. Force which depends on time is non-conservative force. Is this correct ?

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  1. A velocity-dependent generalized potential $U=U({\bf r},{\bf v},t)$ satisfies by definition $$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.\tag{1} $$

  2. If there is no velocity-dependence but possible explicit time dependence, this simplifies to the well-known gradient form $$ {\bf F}~=~ - \frac{\partial U}{\partial {\bf r}},\tag{2} $$ which can be readily applied to OP's example, essentially because $t$ and ${\bf r}$ don't talk to each other in eq. (2).

  3. Whether the definition of a conservative force allows for explicitly time-dependence is ultimately a matter of convention. On one hand, the notion of potential energy still exists in the presence of explicit time dependence. Note in particular that the definition could involve virtual work along virtual paths where time $t$ is fixed/frozen. On the other hand, no sane person would probably call a physical system with explicit time dependence for 'conservative'. See also this related Phys.SE post.

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  • $\begingroup$ I read that potential energy can not be defined for non-conservative forces. Force which depends on time, is non-conservative. Is this correct? $\endgroup$ – atom Dec 9 '19 at 10:00
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    $\begingroup$ Read where? Which page? $\endgroup$ – Qmechanic Dec 9 '19 at 10:01
  • $\begingroup$ I don't remember it. Also, position $r$ is function of time $t$ : $r(t)$. How can you say t and r don't talk each other ? $\endgroup$ – atom Dec 9 '19 at 10:24
  • $\begingroup$ @atom In $U=U(r,\dots)$ the $r$ is a spatial coordinate for all points in space. How this function varies in space determines the force field. This force field doesn't depend on any trajectory if the potential energy is not velocity dependent. This is what Qmechanic means. i.e. $$\mathbf F(\mathbf r,t)=-\nabla U(\mathbf r,t)$$ and then if you want to know the force your object experiences along some trajectory $\mathbf x(t)$, then you plug into your force field function to get $\mathbf F(\mathbf x(t),t)$ $\endgroup$ – Aaron Stevens Dec 9 '19 at 11:01
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    $\begingroup$ It is certainly true that potential energy cannot be defined for non-conservative forces. But the force function you give is conservative at every time $t$, and so you can define a potential energy. It has the slightly unusual property of being explicitly time-dependent though. $\endgroup$ – Clara Diaz Sanchez Dec 9 '19 at 11:11

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