Vector Force Fields and Their Physical Interpretations The vector force field F=(yi,-xj) has a curl of -2. The acceleration of a particle in space is given by:
ax=y/m
ay=-x/m
This vector field has a divergence of 0. Will particles in this vector FORCE field move in circles or will they spiral outwards in a crazy fashion? My conundrum is this, the acceleration of the particles is never towards the center, therefor there is no centripetal acceleration, so how can these particles move in discreet circles?
Does this reasoning make sense or am I doing a nested substitution:
r=Sqrt[x^2+y^2]
r=Sqrt[((1/2)ax*t^2)^2 + ((1/2)ay*t^2)^2]
r=(1/2)Sqrt[t^4 (ax^2 + ay^2)]
r=(1/2)Sqrt[t^4 a^2]
r=(1/2)at^2

 A: From some basic facts of systems of linear differential equations, it follows that the particles will spiral like your simulation showed.  Indeed, solutions of the differential equation have terms involving things like $e^{t/\sqrt{2}} \sin(t/\sqrt{2})$.
The system of differential equations you have can be made a first order system by introducing new variables $w$ and $z$ with $x' = w$ and $y' = z$.  Then $w' = y$ and $z' = -x$ (here I'm taking $m=1$).  We can then write the system in matrix form as
$$
\left(\begin{array}{c} w \\ x \\ y \\ z \end{array}\right)' = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 \end{array}\right)\left(\begin{array}{c} w \\ x \\ y \\ z \end{array}\right).
$$
The type of solutions are then determined by the eigenvalues of the $4 \times 4$ matrix.  If an eigenvalue if $a + ib$, then a solution will be given by $v e^{a+ib} = v e^a + v(\cos b + i \sin b)$ where $v$ is the corresponding eigenvector.  In your system, the eigenvalues are $\frac{1}{\sqrt 2}(\pm 1 \pm i)$, so we get spiral solutions.
Its pretty cool that you made a flash simulation and I wish I could give you an answer without invoking ODE theory, but I can't really think of one.
A: Err, right after I said I coulnd't think of a more elementary way I did.  Suppose a solution is a circle, then we'd have $x^2 + y^2 = const$.  Differentiating both sides gives $2x x' + 2y y' = 0$.  Dividing by 2 and differentiating again gives $x'^2 + xy + y'^2 - xy = 0$.  So $x'^2 + y'^2 = 0$.  This means $x' = y' = 0$.  So the solutions are constant, but only $(x,y) = (0,0)$ is a constant solution to the differential equation.
