# Why can't $U=b\mathbf{v} \cdot \mathbf{r}$ be considered a potential for the drag force ${\bf F}=-b {\bf v}$?

Consider the function $$U(\mathbf{r},\dot{\mathbf{r}})=b\mathbf{r} \cdot \mathbf{v}=bx\dot{x}+by\dot{y}+bz\dot{z}$$, where $$b$$ is the drag coefficient. Why can't this be considered as a potential energy for the drag force? It can easily seen that $$\frac{\partial U}{\partial x}=b\dot{x}$$, for the $$x$$ coordinate and similarly for the $$y$$ and $$z$$ coordinates. Collecting in vector form, and using the well-known relation $$\mathbf{F}=-\nabla U$$, one gets $$\mathbf{F}_{\mathrm{drag}}=-\nabla U=-b\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}.$$ But according to Wikipedia, a force is conservative if it can be written as the negative gradient of a potential energy, so this would make drag conservative, and it says nothing about velocity dependence, so that doesn't matter, but it is usually regarded as a non-conservative force. Then, why can't drag be regarded as a conservative force?

• Wikipedia article that you mention states that "A conservative force is dependent only on the position of the object." – atarasenko Aug 14 '19 at 22:34

1. A velocity-dependent potential $$U=U({\bf r},{\bf v},t)$$ of a force $${\bf F}$$ satisfies by definition $${\bf F}~=~\color{red}{\frac{d}{dt} \frac{\partial U}{\partial {\bf v}}} - \frac{\partial U}{\partial {\bf r}} \tag{1}$$ rather than $${\bf F}~=~ - \frac{\partial U}{\partial {\bf r}},\qquad\qquad (\leftarrow\text{Wrong!} ) \tag{2}$$ cf. Ref. 1. The idea being that in order for a notion of potential $$U$$ to be useful, it should yield an action formulation. The extra red term in eq. (1) can be viewed as a consequence of Euler-Lagrange (EL) equations.
2. A proof that the drag force $${\bf F}~=~-b {\bf v} \tag{3}$$ does not have a velocity-dependent potential $$U=U({\bf r},{\bf v},t)$$ is given in this Phys.SE post.