# Is it possible to always find a potential associated to a force (even when non-conservative)?

I am totally confused by the non-conservative forces.

I know that a non conservative force is either a force for which the work will be path independant (and only depends on the boundaries), or in an equivalent way, a force that depends on a potential.

But for me it is always possible to find such a potential.

I will take the typical example of the friction as the non conservative force.

I have $$\delta W = F dx = -k \frac{dx}{dt} dx$$

I will forget the $-k$ because it is not very important for what I want to say.

Let's consider I have a bijective motion, I can write $x=f(t)$ and $t=f^{-1}(x)$

Thus, $$\frac{dx}{dt}(t)=\frac{dx}{dt}(f^{-1}(x))=g(x)$$

And I always can find a primitive associated to $g(x)$ : $G'(x)=g(x)$.

Thus, I have $$\delta W=g(x)dx=dG$$

Thus the force is conservative. I don't understand.

Here I made two main assumptions :

• Bijective motion
• Existence of the primitive

I think the second assumption is not the problem (I only have to assume that $f$ is continuous).

Maybe the problem is linked to the first assumption, maybe for bijective motion between $x$ and $t$ the forces are always conservatives.

But I think my mistake is somewhere else, but as I really don't find it I am asking for help.

I would like to have an answer linked to the "little maths" I used here using derivatives and primitives.

• I don't think that "bijective motion between $x$ and $t$" implies that the force(s) is conservative. As an example, a mass or particle that goes in straight line motion until it collides and interact instantly with a wall such that it rebounces in a different direction with a smaller speed (or if you dislike instantaneous times, you can imagine the mass or ball rolling/interaction over the wall for a finite amount of time greater than 0 s). In that case the motion is bijective in the sense that for any given $x$, one can retrieve $t$ univocally. Yet the forces that acted (or are acting) on – thermomagnetic condensed boson Mar 8 '18 at 17:41
• the system aren't conservative. Note also that conservative forces doesn't imply a "bijection of motion" either. Take a mass on a spring, with our without friction. In both cases there's, in general, no bijection of motion because given $x$, one cannot find a unique corresponding $t$. In the case of when friction is involved (damped harmonic motion), it is possible to retrieve $t$ for some $x$'s, but not for all $x$'s. – thermomagnetic condensed boson Mar 8 '18 at 17:42

• @no_choice99 Actually it is more that it is not possible to find the same potential for all trajectories. If my force is of the form $F(t)$ and I have $t=f(x)$, then $F(f(x))$ is my force written in function of the position. And thus, this function actually depends on the trajectory because of the function $f$. Thus it is not possible to have the same potential for all trajectories. – StarBucK Mar 8 '18 at 18:49
1. Upshot: For a (possibly velocity-dependent) potential $U=U({\bf r},{\bf v},t)$, the defining force-potential relation $${\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}} \tag{A}$$ has to be satified in each point of the (tangent) configuration space without the use of equations of motion or a specific trajectory.