# Understanding the math behind velocity-dependent conservative forces, Part 1

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.

• 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. ;-) – David Z Nov 12 '13 at 17:58
• 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. – David Z Nov 12 '13 at 18:17
• 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. – 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.

• 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 : ..." ? – Art Brown Nov 24 '13 at 22:23
• @ArtBrown: Thanks, I corrected the typo in an update. – 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

• 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... – Art Brown Nov 12 '13 at 19:50
• Are you referring to the part about the integral? – Danu Nov 12 '13 at 20:20
• Actually, the part about the functional derivative. I'm going to save that part for a separate question. – 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 $0$:

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

Voila.