2

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.

Kyle Kanos
  • 29,127
StarBucK
  • 1,688

2 Answers2

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!

user_phys
  • 359
1

  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.

Qmechanic
  • 220,844