# Why can't we ascribe a (possibly velocity dependent) potential to a dissipative force?

Sorry if this is a silly question but I cant get my head around it.

-
Your example is not a good one because if you google "velocity dependent potential" you will find articles that propose them, example fisica.cab.cnea.gov.ar/colisiones/staff/barra/research/… . – anna v Feb 13 '12 at 7:11
@Anna so velocity dependent potentials are actually used to describe conservative forces, is what you meant, right? I will edit my question. – yayu Feb 13 '12 at 7:13
Have a look at this presentation google.gr/… . – anna v Feb 13 '12 at 8:05
Related mathoverflow question: mathoverflow.net/q/146042/13917 – Qmechanic Dec 2 '13 at 14:25

## 4 Answers

To be concrete, let us here assume that the dissipative force is a friction force

$$\tag{1} {\bf F}~=~-k {\bf v}$$

proportional to the velocity ${\bf v}=\dot{\bf r}$ of the point particle.

Recall that a velocity dependent potential $U=U({\bf r},{\bf v},t)$ of a force ${\bf F}$ by definition satisfies

$$\tag{2} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$

cf. Ref. 1. Next define the potential part of the action as

$$\tag{3} S_p~:=~\int \!dt~U,$$

and note that eq. (2) can be rewritten with the help of a functional derivative as

$$\tag{4} F_i(t)~\stackrel{(2)+(3)}{=}~ -\frac{\delta S_p}{\delta x^i(t)}, \qquad i~\in~\{1,2,3\}.$$

Since functional derivatives commute

$$\tag{5} \frac{\delta}{\delta x^i(t)} \frac{\delta S_p}{\delta x^j(t^{\prime})} ~=~\frac{\delta}{\delta x^j(t^{\prime})} \frac{\delta S_p}{\delta x^i(t)},$$

we derive the following consistency condition (6) for a force with a velocity dependent potential

$$\tag{6} \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} ~\stackrel{(4)+(5)}{=}~[(i,t) \longleftrightarrow (j,t^{\prime})].$$

Eq. (6) is a functional analog of a Maxwell relation. However the friction force (1) satisfies precisely the opposite symmetry

$$\tag{7} \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} ~\stackrel{(1)+(8)}{=}~ -k~ \delta_{ij} \frac{d}{dt}\delta(t-t^{\prime}) ~=~ k~ \delta_{ij} \frac{d}{dt^{\prime}}\delta(t-t^{\prime}) ~=~-[(i,t) \longleftrightarrow (j,t^{\prime})].\qquad$$

In eq. (7) was used that

$$\tag{8} \frac{\delta x^i(t)}{\delta x^j(t^{\prime})} ~=~\delta_j^i ~\delta(t-t^{\prime}).$$

Comparing eqs. (6) and (7), we conclude that the friction force (1) can not have a velocity dependent potential (2).

References:

1. H. Goldstein, Classical Mechanics, Chapter 1.
-
I +1'd your answer because it gives a precise reason that frictional force cannot be derived from a potential. Perhaps you would want to better your answer by pointing out that a velocity-dependent force can lead to a conservative potential if the force does not have a component along the direction of velocity. For example, magnetic force depends on the velocity of a charged particle but leads to a potential because the force is perpendicular to the velocity and hence has no component along the velocity. – Chin Yeh Oct 27 '13 at 18:31
Velocity dependent potential for the Lorentz force is discussed in e.g. this Phys.SE answer. – Qmechanic Oct 27 '13 at 19:08
@Qmechanic I'm sorry if its obvious, but I don't see how you derived the expressions in square brackets $[(i,t)\rightarrow (j,t')]$ and $-[(i,t)\rightarrow (j,t')]$. As understand it, this means that frictional forces change sign upon taking the second functional derivative but velocity dependent ones do not? Is there an analogue to conservative forces too? Cheers :) – AngusTheMan Dec 28 '15 at 12:10
@AngusTheMan : I updated the answer. For a discussion of conservative forces, see e.g. my Phys.SE answer here. – Qmechanic Dec 28 '15 at 12:23
@Qmechanic Thank you very much for the update! Doesn't this analysis assume however that the dissipative force does admit a variational principle? – AngusTheMan Dec 28 '15 at 14:11

Because the basic feature of a potential is that it is path-independent. It is a property of a point in phase-space, not of the system's history.

Think of it this way: if you take your system to a little trip in phase space, and come back to your starting point, the potential cannot change in the process (as it is a function of your position in phase-space). But if there's dissipation, you lost energy in the process.

-

Dissipative forces are non conservative. A conservative force is one in which the work done by the force on a body is independent of the path taken. For example, we can move a ball one meter up in multiple ways. We can just move it up, or we can move it to two meters and then let it fall. The net energy supplied to the system by you is the same, it is $mgh$. Now, lets look at processes where the ball comes back to where it is. You can move it to a height of one meter, and let it fall, but you won't be supplying any net energy. Whatever energy you supply will be released during the fall of the ball.

On the other hand, friction/drag/etc are nonconservative. Take a block on a rough surface. Lets say that the kinetic friction force has constant magnitude $f$. Now, move the block $x$ forward, and take it back. You will do work $2fx$ against friction (So friction does work $-2fx$). Even though there was no net change of position, there was work done. Now, work done=change in PE. But, potential at a point must be constant, so change in PE=0! So, potential is not definable.

This happens to most forces which depend upon the velocity of the particle. For example, the magnetic force$^{*}$ ($q\vec{v}\times\vec{B}$), kinetic friction force($-\mu_kN\hat{v}$), etc. It also happens in any case where the field lines of a force form loops (induced electric field lines, for example).

All this can be mathematically encoded as this: If you have a force vector field $\mathbb{\vec{F}}$ (A vector field is a vector that is a function of $(x,y,z)$), then for the field to be conservative, $\nabla\times\mathbb{\vec{F}}=0$

Summing up, we can only define potential for a force which does the same work to get from point A to point B no matter what the path is.

$*$The magnetic force is not exactly nonconservative. It does not do work (it is always perpendicular to the displacement), so we can't really discuss conservativeness.

-

One way to interpret this question is, "what makes a force conservative?" The answer is that conservative forces excite no internal degrees of freedom - there is no transfer of energy to internal energy (no heat flow). When friction is present, then accounting for the energy budget in the system becomes more complicated than the usual interplay between kinetic and potential energy because the internal energy budget becomes important.

-