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.

  • $\begingroup$ What exactly do you want to know about the functional derivatives and/or two-cycle integral? You have all the makings of a very good question here except that you don't actually ask the question. ;-) $\endgroup$ – David Z Nov 12 '13 at 17:58
  • $\begingroup$ Well, yes, you end with a question mark so it's technically a question, but asking "where can I read about X?" in a specific context like this one is just a copout for asking X directly. For example, you could ask "What does a functional derivative mean?" and it would be a much better question. Or "What is a two-cycle integral?" Or some such thing. $\endgroup$ – David Z Nov 12 '13 at 18:17
  • $\begingroup$ Right, but questions on this site are supposed to be specific. Ask one thing, get one thing answered. Then if you want to know more, you post a followup question, or use what you were given in the original answer to find more information. Here's another way to look at it: answers will always provide references when there is interesting further reading. So by asking for references explicitly, you're eliminating a large class of potential answers, namely those which directly explain the thing you want to know without referring you to an external resource. $\endgroup$ – David Z Nov 12 '13 at 18:33

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.

  • $\begingroup$ Thx v much. I don't grok it yet, but you've laid out the generalization clearly. I suspect a typo just after (A'): should it be "at any other path r_1 : ..." ? $\endgroup$ – Art Brown Nov 24 '13 at 22:23
  • $\begingroup$ @ArtBrown: Thanks, I corrected the typo in an update. $\endgroup$ – Qmechanic Nov 24 '13 at 22:53

I suspect that this is mostly a case of fancy notation being confusing.

Independent of the physical interpretation, the surface integral simply is an integral over a sphere in 3-space (which has to be paramatrized by two variables, called s and t here).

Furthermore, the definition of the functional derivative (as customarily used in physics) can be found in a reference from the answer you mentioned yourself, as well as on its wikipedia page. As far as the connection between the two goes, I wouldn't know. I hope this can get you started though

  • $\begingroup$ Thank you for the link. My apologies, I've decided to split the question up into 3 parts, so now a portion of your answer looks like a non sequitor... $\endgroup$ – Art Brown Nov 12 '13 at 19:50
  • $\begingroup$ Are you referring to the part about the integral? $\endgroup$ – Danu Nov 12 '13 at 20:20
  • $\begingroup$ Actually, the part about the functional derivative. I'm going to save that part for a separate question. $\endgroup$ – Art Brown Nov 12 '13 at 20:26

It took me an embarrassingly long time to understand Qmechanic's construction, so I thought I would add this elaboration, as an aid to anyone as perplexed as I was.

I think this integral (3') is best understood not as a surface integral but as a line integral, just not of your garden-variety gradient of a scalar function ($-\boldsymbol{\nabla} U$ , like 3). Instead, it's the line integral of a functional derivative, thus accounting for the second integration. To wit:

One starts with a functional $S_{pot}(\boldsymbol{r})$, depending on a path function $\boldsymbol{r}(t)$ and its time derivative $\boldsymbol{\dot{r}}(t)$, with boundary conditions established at an initial time $t_i$ and final time $t_f$: $$ S_{pot}(\boldsymbol{r}) = \int_{t_i}^{t_f} L[\boldsymbol{r}(t),\boldsymbol{\dot{r}}(t)] dt $$

Question: given the value of this functional $S_{pot}(\boldsymbol{r_0})$ for a "reference" path $\boldsymbol{r_0}(t)$, how does one calculate its value $S_{pot}(\boldsymbol{r_1})$ for a new path $\boldsymbol{r_1}(t)$?

Well, the functional derivative tells us how this functional changes in response to a small change in the path $\boldsymbol{r}(t) \rightarrow \boldsymbol{r}(t) + \boldsymbol{ \delta r}(t) $:

$$ \delta S_{pot} = \int_{t_i}^{t_f} \frac{\delta S_{pot}}{\boldsymbol{\delta r}} \cdot \boldsymbol{\delta r} $$

where $$ \frac{\delta S_{pot}}{\boldsymbol{\delta r}} = \frac{\partial L}{\partial \boldsymbol{r}} - \frac{d}{dt} \frac{\partial L}{\partial \boldsymbol{\dot{r}}} = - \boldsymbol{F}(\boldsymbol{r},\boldsymbol{\dot{r}})$$

So, one can now construct a sequence of intermediate paths (a continuous contour of paths in the limit) that smoothly deforms the reference path to the new path, introducing a new parameter $s$ to denote position along the contour. Using the same identifier $\boldsymbol{r}$ for the entire contour as for its elements (here's where $\boldsymbol{r}$ acquires a second argument), the contour is denoted:

$$ \boldsymbol{r}(t,s) , \text{ where } \boldsymbol{r}(t,0)=\boldsymbol{r_0}(t) \text{ and } \boldsymbol{r}(t,1)=\boldsymbol{r_1}(t)$$

Along this contour (as opposed to along the path from $t_i$ to $t_f$), $\boldsymbol{\delta r}(t,s)= \boldsymbol{r}'(t,s) \, ds$ , where the "$'$" denotes differentiation with respect to $s$ (to distinguish it from differentiation wrt $t$). It follows that:

$$ S(\boldsymbol{r_1}) = S(\boldsymbol{r_0}) + \int_0^1 ds \int_{t_i}^{t_f} dt \frac{\delta S_{pot}}{\boldsymbol{\delta r}} \cdot \boldsymbol{r}'(t,s) = S(\boldsymbol{r_0}) - \int_0^1 ds \int_{t_i}^{t_f} dt \, \boldsymbol{F}(\boldsymbol{r}(t,s), \boldsymbol{\dot{r}}(t,s)) \cdot \boldsymbol{r}'(t,s)$$

For this result to be contour-independent, a round-trip around any closed $s$-contour must integrate to $0$:

$$ \oint ds \int_{t_i}^{t_f} dt \, \boldsymbol{F}(\boldsymbol{r}(t,s), \boldsymbol{\dot{r}}(t,s)) \cdot \boldsymbol{r}'(t,s) = 0 $$


Again, there's nothing here that isn't already in Qmechanic's answer.

Apparently Volterra originated these ideas. You can find his work online in Cornell's Historical Math Monographs, particularly Chapter II of his 1913 Leçons sur les fonctions de lignes. The presentation is straightforward, clear, and ... in French.


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