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

Question

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

Answer 1

To be concrete, let us here assume that the dissipative force $$ {\bf F}~=~-f(v^2)~ {\bf v} \tag{1} $$ has a direction opposite of the velocity ${\bf v}=\dot{\bf r}$ of the point particle. Here $f=f(v^2)$ is a function that may depend on the speed square $v^2\equiv {\bf v}^2$. Drag is of this form (1). Linear friction/drag corresponds to a constant $f$-function. The $f$-function is a square root for quadratic drag. The $f$-function is a reciprocal square root for kinetic friction.

Recall that a velocity-dependent potential $U=U({\bf r},{\bf v},t)$ of a force ${\bf F}$ by definition satisfies $$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}, \tag{2} $$ cf. Ref. 1. For a discussion of the notion of conservative forces, see e.g. my Phys.SE answer here.

Theorem: The dissipative force (1) can not have a velocity-dependent potential$^1$ (2).

Proof of theorem:

1. Define the potential part of the action as $$ S_p~:=~\int \!dt~U,\tag{3}$$ and note that eq. (2) can be rewritten with the help of a functional derivative as $$ F_i(t)~\stackrel{(2)+(3)}{=}~ -\frac{\delta S_p}{\delta x^i(t)}, \qquad i~\in~\{1,\ldots,n\}, \tag{4} $$ where $n$ is the number of spatial dimensions.

2. Since functional derivatives commute $$ \frac{\delta}{\delta x^i(t)} \frac{\delta S_p}{\delta x^j(t^{\prime})} ~=~\frac{\delta}{\delta x^j(t^{\prime})} \frac{\delta S_p}{\delta x^i(t)},\tag{5}$$ we derive the following consistency condition (6) for a force with a velocity-dependent potential $$ \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} ~\stackrel{(4)+(5)}{=}~[(i,t) \longleftrightarrow (j,t^{\prime})].\tag{6} $$ Eq. (6) is a functional analog of a Maxwell relation, and equivalent to the Helmholtz conditions, cf. this Phys.SE post.

3. The functional derivative of the dissipative force (1) reads $$ \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} ~\stackrel{(1)+(8)+(9)}{=}~ -T_{ij}(t) \frac{d}{dt}\delta(t\!-\!t^{\prime}), \tag{7} $$ where we have defined $$ T_{ij}~:=~f(v^2) ~\delta_{ij} + 2 f^{\prime}(v^2)~v^iv^j\tag{8}.$$ In eq. (7) was used that $$ \frac{\delta x^i(t)}{\delta x^j(t^{\prime})} ~=~\delta_j^i ~\delta(t\!-\!t^{\prime})\tag{9}.$$

4. The functional Maxwell relation (6) becomes $$\begin{align} 0~\stackrel{(6)}{=}~&\frac{\delta F_j(t^{\prime})}{\delta x^i(t)}-\frac{\delta F_i(t)}{\delta x^j(t^{\prime})}\cr ~\stackrel{(7)}{=}~& -T_{ij}(t^{\prime}) \frac{d}{dt^{\prime}}\delta(t\!-\!t^{\prime})+ T_{ij}(t)\frac{d}{dt}\delta(t\!-\!t^{\prime}) \cr ~=~&\left\{T_{ij}(t^{\prime}) + T_{ij}(t)\right\}\frac{d}{dt}\delta(t\!-\!t^{\prime})\cr ~=~& 2T_{ij}(t)\frac{d}{dt}\delta(t\!-\!t^{\prime})+\delta(t\!-\!t^{\prime})\frac{dT_{ij}(t)}{dt} .\end{align} \tag{10} $$

5. Eq. (10) implies that $$ T_{ij}~\stackrel{(10)}{=}~0.\tag{11} $$ This in turn implies that $$ 0~\stackrel{(10)}{=}~T_{ii}~\stackrel{(8)}{=}~n f(v^2) + 2v^2 f^{\prime}(v^2), \tag{12} $$ and $$ 0~\stackrel{(10)}{=}~v^i T_{ij} v^j~\stackrel{(8)}{=}~v^2 f(v^2) + 2v^4 f^{\prime}(v^2). \tag{13} $$ Eqs. (12) & (13) show that $$ (n-1)v^2f(v^2)~\stackrel{(12)+(13)}{=}~0\tag{14} $$ so that the function $$ f(v^2)~\stackrel{(14)}{=}~0\tag{15} $$ vanishes for $n>1$. $\Box$

6. Special case of one spatial dimension $n=1$: Eqs. (12) & (13) also restrict the $n=1$ case, but it is instructive to carefully redo the analysis for the $n=1$ case alone. We now change the notation so that $v$ denotes a 1-vector velocity rather than the speed. Then the functional derivative of the dissipative force $F(v)$ reads $$\frac{\delta F(v(t))}{\delta x(t^{\prime})} ~=~ F^{\prime}(v(t)) \frac{d}{dt}\delta(t\!-\!t^{\prime}). \tag{16} $$ The functional Maxwell relation (6) becomes $$\begin{align} 0~\stackrel{(6)}{=}~&\frac{\delta F(v(t))}{\delta x(t^{\prime})}-\frac{\delta F(v(t^{\prime}))}{\delta x(t)}\cr ~\stackrel{(16)}{=}~& F^{\prime}(v(t)) \frac{d}{dt}\delta(t\!-\!t^{\prime})- F^{\prime}(v(t^{\prime}))\frac{d}{dt^{\prime}}\delta(t\!-\!t^{\prime}) \cr ~=~&\left\{F^{\prime}(v(t)) + F^{\prime}(v(t^{\prime}))\right\}\frac{d}{dt}\delta(t\!-\!t^{\prime})\cr ~=~& 2F^{\prime}(v(t))\frac{d}{dt}\delta(t\!-\!t^{\prime})+\delta(t\!-\!t^{\prime})\frac{dF^{\prime}(v(t))}{dt} .\end{align} \tag{17} $$ Eq. (17) implies that $$F^{\prime}(v)~=~0,\tag{18}$$ i.e. the force $F(v)$ is independent of $v$. This is not a dissipative force of the form (1). $\Box$

References:

1. H. Goldstein, Classical Mechanics, Chapter 1.

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$^1$ Although the dissipative force (1) does not have a velocity dependent potential (2), it does have a Rayleigh's dissipation function ${\cal F}(v^2)$. It is the antiderivative of $\frac{1}{2}f(v^2)$, i.e. $$f(v^2)~ {\bf v}~\stackrel{(1)}{=}~-{\bf F}~=~\frac{\partial{\cal F}}{\partial {\bf v}}~=~2{\cal F}^{\prime}(v^2)~{\bf v}\tag{19}.$$

Answer 2

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.

Answer 3

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.

Answer 4

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.