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.

  • $\begingroup$ 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 $\endgroup$ Commented Mar 8, 2018 at 17:41
  • $\begingroup$ 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. $\endgroup$ Commented Mar 8, 2018 at 17:42

2 Answers 2


Even though it may be possible to have a 'bijective motion' for a specific trajectory, it is not possible to find a function which does this for all trajectories.

A potential (for the force field) has the property that the work done by the force along any path is given by the difference in values of the potential at the starting point and end point.

I hope this helps!

  • $\begingroup$ I do not see how this answers the question. How does this resolve the doubts about Bijective motion and Existence of the primitive? $\endgroup$ Commented Mar 8, 2018 at 17:35
  • $\begingroup$ @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. $\endgroup$
    – StarBucK
    Commented Mar 8, 2018 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.

  2. See e.g this Phys.SE post for a proof that a class of dissipative forces doesn't have (possibly velocity-dependent) potentials.


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