In a notable answer to this question, Qmechanic formulates conditions for "conservative" velocity-dependent forces (e.g. the Lorentz force, but not velocity-proportional friction) that are analogous to those for traditional velocity-independent conservative forces.
To wit, in a simply-connected domain (for the velocity-independent case, anyway), two sets of three equivalent conditions for a force to be conservative are presented:
$$ \begin{array} {cccc}
\text{ } & \text{velocity-independent force } \boldsymbol{F}( \boldsymbol{r}(t)) & |
& \text{velocity-dependent force } \boldsymbol{F}(\boldsymbol{r}(t),\boldsymbol{\dot{r}}(t)) \\
1) & F_i = – \frac{\partial U}{\partial x^i} & | &
F_i = -\frac{\partial U}{\partial x^i} + \frac{d}{dt} \left( \frac{\partial U}{\partial \dot{x}_i} \right) \\
2) & \boldsymbol{\nabla \times F} = 0 & | &
\frac{\delta F_i(t)}{\delta x_j(t')} – \frac{\delta F_j(t')}{\delta x_i(t)} = 0 \\
3) & \oint_{S^1} dt \, \boldsymbol{F}(\boldsymbol{r}(t)) \boldsymbol{\cdot \, \dot{r}} (t) = 0 & | &
\oint_{S^2} dt \wedge ds \, \boldsymbol{F} ( \boldsymbol{r}(t,s),
\boldsymbol{\dot{r}}(t,s)) \boldsymbol{\cdot \, r'}(t,s) = 0
\end {array}
$$
where $\delta$ denotes a functional derivative, the final integral is over any "two-cycle $r: S^2 \rightarrow \mathbb{R}^3 \,$", and "a dot and a prime mean differentiation wrt. $t$ and $s$, respectively".
I have changed the formulation somewhat; I hope I didn't introduce errors.
I get the maths for the velocity-independent force conditions, but, for the velocity-dependent case, I am a bit puzzled by the functional derivatives and totally baffled by the two-cycle integral.
My question:
-
What is this "two-cycle integral", which looks like no surface integral I've ever seen, and how is it evaluated? (and how did $\boldsymbol{r}$ acquire two arguments?).
– How is this functional derivative evaluated?
– Why are the functional derivative and two-sided integral equivalent to each other and to the potential formula for the force?I suspect this is a rather large subject; references would be appreciated.
Best Answer
We assume that OP is mainly concerned with condition (3').
A) Recall first that for forces ${\bf F}={\bf F}({\bf r})$ that do not dependent on velocity, we may use condition (3) to extend a definition of potential energy $U({\bf r}_{0})$ at a fiducial point ${\bf r}_{0}$ to a potential energy
$$\tag{A} U({\bf r}_{1})~:=~U({\bf r}_{0})-\int_{[0,1]} \!\mathrm{d}s~ {\bf F}({\bf r}(s)) \cdot {\bf r}^{\prime}(s) $$
at any other point ${\bf r}_{1}$ along any curve ${\bf r}: [0,1] \to \mathbb{R}^3$ with boundary conditions
$${\bf r}(s\!=\!0)={\bf r}_{0}\qquad\text{and}\qquad{\bf r}(s\!=\!1)={\bf r}_{1}.$$
Condition (3) ensures that definition (A) does not depend on the curve. Infinitesimally, definition (A) leads to
$$ \delta U~=~-{\bf F}\cdot \delta{\bf r}, $$
which implies condition (1).
A') We now repeat this for velocity dependent forces ${\bf F}={\bf F}({\bf r},{\bf v})$. The only difference is that points are replaced by paths that satisfy pertinent boundary conditions (BC). We may use condition (3') to extend a definition of potential action $S_{\rm pot}[{\bf r}_{0}]$ at a fiducial path ${\bf r}_{0}:[t_i,t_f]\to \mathbb{R}^3$ to a potential action
$$\tag{A'} S_{\rm pot}[{\bf r}_{1}]~:=~S_{\rm pot}[{\bf r}_{0}]-\int_{[t_i,t_f]\times [0,1]} \!\mathrm{d}t \wedge \mathrm{d}s~ {\bf F}({\bf r}(t,s),\dot{\bf r}(t,s)) \cdot {\bf r}^{\prime}(t,s) . $$
at any other path ${\bf r}_{1}:[t_i,t_f]\to \mathbb{R}^3$ along any homotopy ${\bf r}:[t_i,t_f]\times [0,1] \to \mathbb{R}^3$ with boundary conditions
$${\bf r}(t,s\!=\!0)={\bf r}_{0}(t)\qquad\text{and}\qquad{\bf r}(t,s\!=\!1)={\bf r}_{1}(t).$$
Condition (3') ensures that definition (A') does not depend on the homotopy. Infinitesimally, definition (A') leads to
$$ \delta S_{\rm pot}~=~-\int_{[t_i,t_f]} \!\mathrm{d}t~{\bf F} \cdot \delta{\bf r}, $$
which implies condition (1') under suitable BC.