# How to understand dynamics $\dot x_i=\partial_jA_{ij}$ from skew-symmetric potential $A$?

We speak of a dynamical system with a potential if there is a scalar possibly depending on coordinate such that the vector field is exactly the (negative) gradient of the potential. That means each point moves along the direction with the steepest gradient descent. Also we know that the direction of gradient is perpendicular to level set. All of these are about understanding of dynamics in terms of potential.

But how about vector potential? Given a vector potential $A$, what can we say about the system $\dot x=\nabla\times A$? I have no idea what geometric intuitions can be drawn from $A$. Even I don't know why we call it a potential if there's nothing similar to scalar potential besides something to do with $\nabla$.

Similar questions I have for $n$ dimension where $A$ is a second-order skew-symmetric tensor and the system is $\dot x_i=\partial_jA_{ij}$.

Thank you for any of your ideas.

• The magnetic vector potential is usually called a "potential" because it is related to the magnetic field by $\mathbf B = \nabla\times\mathbf A$ which is analogous to $\mathbf F = -\nabla V$ in the sense that one obtains one vector field by applying a differential operator to another field (the corresponding potential). – joshphysics Oct 29 '13 at 16:59
• Also, the dynamical system of classical mechanics is (taking $m=1$) $$\ddot{x}=-\vec{\nabla}\phi(x)$$ So maybe you want to ask which are the solutions and the geometric interpretations of the dynamical system $$\ddot{x}=\vec{\nabla}\times \vec{A}(c)$$ which is the one we can construct with a vector potential. Why are you interested in just one time derivative of $x$? – Federico Carta Oct 29 '13 at 17:09
• taking the gradient of a scalar, or the curl of a vectorfield is basically the same operation. they can be unified by considering the so called exterior derivative $d$ acting on 0-forms<->scalars and 1-forms<->vectors respectively. the result is a form of degree (n+1), so translating $B=\nabla\times A$ to $B=dA$ where A is a 1-form means B has to be a 2-form. Turns out 2-forms are in three dimensions isomorphic to 1-forms and thus can be (almost) identified with three-component objects - here: the vector $\vec{B}$ Almost because the vector is an axial one, not a proper polar one! – Nephente Oct 29 '13 at 18:58
• You might find this useful: arxiv.org/abs/physics/9803023 If potential energy is the timelike component of a four-vector, then there must be a corresponding spacelike part which would logically be called the potential momentum. The potential four-momentum consisting of the potential momentum and the potential energy taken together is just the gauge field of the associated force times the charge associated with that force. The canonical momentum is the sum of the ordinary and potential momenta. – Alfred Centauri Oct 29 '13 at 20:36
• @joshphysics That's what exactly I've thought, nothing more? – Shuchang Oct 30 '13 at 2:44

OP considers an equations of motion of the form

$$\tag{1}\dot{\bf x}~=~{\bf B}({\bf x}),$$

where the vector field ${\bf B}$ is of the form$^1$

$$\tag{2} {\bf B}~=~{\bf \nabla}\times {\bf A}.$$

In other words, ${\bf B}$ is divergence-free

$$\tag{3} {\bf \nabla}\cdot {\bf B}~=~0.$$

Eq.(3) is locally eqivalent to eq. (2), cf. Poincare's Lemma. Let

$$\tag{4} \Omega~:=~\mathrm{d}x^1 \wedge \mathrm{d}x^2\wedge \mathrm{d}x^3$$

be a volume form in the configuration space. The velocity vector field (1) preserves this volume form

$$\tag{5} {\cal L}_{\bf B} \Omega~=~0,$$

which can be viewed as a Liouville's theorem for this system.

--

$^1$A vector field ${\bf A}$ is dual to a two-forms $\frac{1}{2} A_{ij}\mathrm{d}x^i \wedge \mathrm{d}x^j$ in 3 dimensions, as nephente mentions in a comment.

• Oh. So much appreciation for answering me after so long a time. I'm actually concerned about any n dimension. Is it true similarly n-volume is conserved? – Shuchang Apr 6 '14 at 21:43
• The volume form $\Omega$ and the divergence-free vector field ${\bf B}$ generalize to any dimension $n$. – Qmechanic Apr 6 '14 at 21:51