Consider a particle moving along some trajectory in the $x$-$y$ plane, in a viscous medium.

Then its equation of motion is given by:

$$\mathbf{F}_d = - b \mathbf{v} .$$

it's well-known from the Gradient theorem(fundamental theorem of line integral) that if there exists a scalar-valued function $\varphi$ that satisfy:

$\mathbf{F}_d=\nabla\varphi$,then this implies $\mathbf{F}_d$ is conservative.

I wanna show through a proof by contradiction that $\varphi$ does not exist for $\mathbf{F}_d$.

Let(for the sake of Reductio ad absurdum) $\mathbf{F}_d=\nabla\varphi$.

Consider an arbitrary curve which is parameterized by the position vector $\mathbf{r}(t)=<x(t),y(t)>$.

consider that our particle is moving on that curve.

Therefore $\mathbf{F}_d$ by definition is given by:


And let $\nabla\varphi$ be given by : $<\dfrac{\partial \varphi }{\partial x},\dfrac{\partial \varphi }{\partial y}>$ From our hypothesis $\mathbf{F}_d=\nabla\varphi$ We have that:

$$\varphi(x,y)=\int -b\dfrac{dx(t)}{dt} dx = \int -b\dfrac{dy(t)}{dt} dy. $$

Assuming that our functions are well-behaved then we get:

$$\varphi=-b( x\dfrac{dx}{dt} + y\dfrac{dy}{dt} ).$$

So Although I expected that I'd arrive at some contradiction I did not, In a sense I proved that $\mathbf{F}_d$ is conservative (Although it's not!)

So what possibly I did wrong?


2 Answers 2


The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity ${\bf v}$. The formula reads (see e.g. Ref. 1)

$$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$

rather than just $$\tag{2} {\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. \qquad\qquad(\leftarrow\text{Wrong}!)$$

Velocity dependence of conservative forces and potentials is also discussed in my Phys.SE answer here. To see why the friction force $$\tag{3} {\bf F}~=~-b {\bf v}$$ has no velocity-dependent potential $U$, see e.g. this Phys.SE post and this MO.SE post.


  1. H. Goldstein, Classical Mechanics; eq. (1.58).

To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero, $$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0. $$ But if you insert the actual values from your calculations, $$\oint \vec{\nabla} \phi \cdot \vec{dl} = -b \int \vec{r}'(t) \cdot \vec{r}'(t) dt = -b\int\left| \vec{r}'(t) \right|^2 dt \neq 0. $$


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