# What causes a force field to be “nonconservative?”

A conservative force field is one in which all that matters is that a particle goes from point A to point B. The time (or otherwise) path involved makes no difference.

Most force fields in physics are conservative (conservation laws of mass, energy, etc.). But in many other applications, the time paths DO matter, meaning that the force field is not "conservative."

What causes a force field to be "non-conservative?" Could you give some examples (probably outside of physics)?

-

## migrated from math.stackexchange.comJul 9 '12 at 14:51

This question came from our site for people studying math at any level and professionals in related fields.

I am not sure if this qualifies as a force field, but if there is friction in the system, the force is definitely not conservative –  Thomas Rot Jul 9 '12 at 13:24
@ThomasRot: Good point. Almost all physics problems say, "neglect frictional forces." –  Tom Au Jul 9 '12 at 13:41
Non-conservative only matters in macroscopic level, in atom level every type of force is conservative. –  Shuhao Cao Jul 9 '12 at 14:16
@ShuhaoCaoL: Figured that might be the case. Thanks for the confirmation. –  Tom Au Jul 9 '12 at 14:20

A force field $F_i(x)$ is conservative if for every curve $C$ from a point $y_1$ to a point $y_2$, we have $\int\limits_C F_i(x)\mathrm{d}x^i$, so that the energy difference between $y_1$ and $y_2$ is independent of the curve taken from one to the other. Equivalently, the integral around a closed curve must be zero, $\oint\limits_C F_i(x)\mathrm{d}x^i=0$ for every closed curve $C$. Alternatively, we require $\nabla\times F=0$, so that we can write $F=\nabla V$; that is, the curl of the force field is zero so that the force field can be expressed as a divergence. Generalizations of this elementary account to higher dimensions in terms of differential forms are possible.

Although Shuhao Cao's comment that whether a physical theory is macroscopic or microscopic will determine whether the theory is conservative is very often correct, nonetheless phenomenological microscopic theories may find it convenient to include nonconservative force fields. For example, the effect of an externally imposed magnetic field on an otherwise microscopic model may be nonconservative. (see Ron's comment below, which points out that variation of an externally imposed magnetic field over time may be used to give an example of a nonconservative field in 4D. The implication of restriction to 3D that is established by my first paragraph has to be removed.)

-
the effect of an externally imposed magnetic field is never nonconservative, since the vector potential does not affect the energy, only the momentum. The way in which magnetic fields are non-conservative is in time-dependent circumstances, where a changing B makes EMF. –  Ron Maimon Jul 9 '12 at 17:28
@Ron Silly me, is obvious. Will edit. –  Peter Morgan Jul 9 '12 at 17:54

I) In this answer we would like to relax the conventional definition of a conservative force to include e.g. the Lorentz force.

II) The standard definition of a conservative force is given on Wikipedia (October 2013) roughly as follows:

A force field ${\bf F}={\bf F}({\bf r})$ is called a conservative force if it meets any of these three equivalent conditions:

1. The force can be written as the negative gradient of a potential $U=U({\bf r})$: $$\tag{1} {\bf F} ~=~ - {\bf \nabla} U.$$ Equivalently, condition (1) means that the one-form $\phi:={\bf F}\cdot \mathrm{d}{\bf r}$ is exact: $\phi=-\mathrm{d}U$, where the exterior derivative is $\mathrm{d}:=\mathrm{d}{\bf r}\cdot{\bf \nabla}$.

2. The position space is simply connected and the curl of ${\bf F}$ is zero: $$\tag{2} {\bf \nabla} \times {\bf F} ~=~ {\bf 0}.$$ Equivalently, condition (2) means that the one-form $\phi:={\bf F}\cdot \mathrm{d}{\bf r}$ is closed: $\mathrm{d}\phi=0$.

3. There is zero net work $W$ done by the force ${\bf F}$ when moving a particle through a closed curve ${\rm r}: S^1 \to \mathbb{R}^3$ that starts and ends in the same position: $$\tag{3} W ~\equiv~ \oint_{S^1} \!\mathrm{d}s~ {\bf F}({\bf r}(s)) \cdot {\bf r}^{\prime}(s) ~=~ 0.$$

We stress that the parameter $s$ does not have to be actual time $t$. In fact time $t$ doesn't enter conditions (1-3) at all. The curve in condition (3) could be any virtual loop. In particular, the curve and its parametrization $s$ do in principle not have to reflect how an actual point particle would travel along a trajectory in a certain pace determined by some equations of motion, let alone move forward in time.

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

$$\tag{4} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}, \qquad {\bf v}~=~\dot{\bf r},$$

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

$$\tag{5} S_{\rm pot}[{\bf r}]~:=~\int_{t_i}^{t_f} \!\mathrm{d}t~U({\bf r}(t),\dot{\bf r}(t),t),$$

and note that eq. (4) can be rewritten with the help of a functional derivative as

$$\tag{6} F_i(t)~=~-\frac{\delta S_{\rm pot}}{\delta x^i(t)}, \qquad i~\in~\{1,2,3\}.$$

Technically at this point we need to impose pertinent boundary conditions (BC) (e.g. Dirichlet BC) at initial and final time, $t_i$ and $t_f$, respectively, in order for the functional derivative (6) to exists. These BC are implicitly assumed from now on.

We dismiss the possibility that one would like to call a force with explicit time dependence for a conservative force. Let us therefore drop explicit time dependence from now on. However, see this Phys.SE post.

IV) Seen in the light that velocity dependent potentials (4) are extremely useful in Lagrangian formulations, it is tempting to generalize the notion of a conservative force in the following non-standard way:

A velocity dependent force field ${\bf F}={\bf F}({\bf r},{\bf v})$ is called a conservative force if it meets any of these three equivalent conditions:

1. The force can be written as the negative functional gradient of a potential action $S_{\rm pot}[{\bf r}]=\int_{t_i}^{t_f} \!\mathrm{d}t~U({\bf r}(t),\dot{\bf r}(t))$: $$\tag{1'} {\bf F} ~=~ -\frac{\delta S_{\rm pot}}{\delta {\bf r}} ~\equiv~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}} .$$ Equivalently, condition (1') means that the one-form $\Phi:=\int_{t_i}^{t_f}\!\mathrm{d}t~ F_i(t)\mathrm{d}x^i(t)$ is exact in path space: $\Phi=-\mathrm{d}S_{\rm pot}$, where the exterior derivative is $\mathrm{d}:=\int_{t_i}^{t_f}\!\mathrm{d}t~ \mathrm{d}x^i(t)\frac{\delta}{\delta x^i(t)}$.

2. The position space is simply connected and the force ${\bf F}$ satisfies a closedness condition wrt. to functional derivatives $$\tag{2'} \frac{\delta F_i(t)}{\delta x^j(t^{\prime})} ~=~[(i,t) \longleftrightarrow (j,t^{\prime})].$$ Equivalently, condition (2') means that the one-form $\Phi:=\int_{t_i}^{t_f}\!\mathrm{d}t~ F_i(t)\mathrm{d}x^i(t)$ is closed in path space: $\mathrm{d}\Phi=0$. The equivalent Helmholtz conditions [2] wrt. to partial and total derivatives read $$\frac{\partial F_i}{\partial x^j} -\frac{1}{2}\frac{d}{dt}\frac{\partial F_i}{\partial v^j} ~=~[i \longleftrightarrow j], \qquad \frac{\partial F_i}{\partial v^j}~=~-[i \longleftrightarrow j].$$

3. The following integral (3') over a two-cycle ${\rm r}: S^2 \to \mathbb{R}^3$ vanishes always: $$\tag{3'} \oint_{S^2}\!\mathrm{d}t \wedge \mathrm{d}s~ {\bf F}({\bf r}(t,s),\dot{\bf r}(t,s)) \cdot {\bf r}^{\prime}(t,s) ~=~ 0.$$

Here a dot and a prime mean differentiation wrt. $t$ and $s$, respectively.

With this definition (1'-3') of a conservative force, then e.g. the Lorentz force and the Coriolis force become conservative forces, while the friction force ${\bf F}=-k {\bf v}$ will stay a non-conservative force, cf. this and this Phys.SE answers.

It should be said that there are straightforward generalizations of conditions (1'-3'):

1. Firstly one may allow the force ${\bf F}={\bf F}({\bf r}, {\bf v}, {\bf a}, {\bf j},\ldots)$ to depend on acceleration, jerk, etc.

2. Secondly, one can generalize to generalized positions $q^i$, generalized velocities $\dot{q}^i$, and generalized forces $Q_i$, etc.

Finally, let us mention that this construction (1'-3') is in spirit related to the inverse problem for Lagrangian mechanics.

References:

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

2. H. Helmholtz, Ueber die physikalische Bedeutung des Prinzips der kleinsten Wirkung, J. für die reine u. angewandte Math. 100 (1887) 137.

-
I imagine conditions (1'-3') are well-known to experts. Unfortunately, I don't know any references. –  Qmechanic Oct 28 '13 at 10:58
@ArtBrown : Thanks for the feedback. Concerning terminology, I found what you said in a footnote in Section 1.5 in the 2nd (but not the 3rd!) edition of Goldstein. There is one (relatively obvious) generalization that I didn't mention: This concerning generalized position coordinates $q^j$ instead of the particle position ${\bf r}$. I think that the adjective generalized should be used for this generalization. –  Qmechanic Oct 29 '13 at 19:42
(cont.) In other words, $U({\bf r},{\bf v},t)$ and $U(q,\dot{q},t)$ should be called velocity dependent potential and generalized velocity dependent potential, respectively. Similarly, ${\bf F}({\bf r},{\bf v})$ and $Q_j(q,\dot{q},t)$ should be called velocity dependent force and generalized velocity dependent force, respectively. Of course in practice, you might want to shorten that! –  Qmechanic Oct 29 '13 at 19:43
OK, a test of whether I understand what's going on: in (3'), I think the integration over t should be from t_i to t_f, so technically the integral is not over the sphere S^2. Es verdad? Thx. –  Art Brown Dec 1 '13 at 5:27