2D simulation - Functions

First some basic definitions:

Next, here is a list of all the vector fields in this simulation:

• 1/r single line: This field is the electric field around an infinitely long line of charge. It is inversely proportional to the distance from the line.
  \begin{align*} &\mathbf F=-\frac{\mathbf u_r}{r}=-\frac{\mathbf r}{r^2}=\left(-\frac{x}{x^2+y^2},-\frac{y}{x^2+y^2}\right)\quad(r\neq0)\\ & U= -\log r\quad(r\neq0)\\ &\nabla \cdot \mathbf F=0\\ & \nabla \times \mathbf F=0 \end{align*}
  This is a two-dimensional cross section of a three-dimensional field. In three dimensions, the divergence of this field is nonzero; but in this cross section, the divergence is zero everywhere except the origin. (It is important to realize that the two-dimensional divergence of a field can be different than its three-dimensional divergence.)

• 1/r double lines: This is the field around two infinitely long conductors. The distance between them is adjustable.
• 1/r^2 single: This field is associated with gravity and electrostatic attraction. The gravitational field around a planet and the electric field around a single point charge are similar to this field. The field points towards the origin and is inversely proportional to the square of the distance from the origin.
  \begin{align*} &\mathbf F=-\frac{\mathbf u_r}{r^2}=-\frac{\mathbf r}{r^3}=\left(-\frac{x}{(x^2+y^2)^{3/2}},-\frac{y}{(x^2+y^2)^{3/2}}\right)\quad(r\neq0)\\ & U= \frac{1}{r}\quad(r\neq0)\\ &\nabla \cdot \mathbf F=\frac{1}{r^3}\\ & \nabla \times \mathbf F=0 \end{align*}
  This is a two-dimensional cross section of a three-dimensional field. In three dimensions, the divergence of this field is zero except at the origin; but in this cross section, the divergence is positive everywhere (except at the origin, where it is negative).
• 1/r^2 double: This field is associated with gravity and electrostatic attraction. The gravitational field around two planets and the electric field around two negative point charges are similar to this field. The separation of the two centers is adjustable.
• 1/r rotational: In this field, the particle is being pulled in a circle around the center. The speed of its motion is inversely proportional to its distance from the line. This is the magnetic field of an infinitely long current-carrying wire.
  The divergence of this field is zero everywhere as we would expect for a magnetic field. Surprisingly, the curl is also zero everywhere except at the origin, even though the particles are clearly moving in a circle. To see why, set the particle movement to "Curl Detectors". You can see that even though the particles are moving in a circle, the particles themselves are not turning. This is because the force on the far side of the particles (the side away from the center) is slightly weaker than the force on the near side; this is just enough to cause the particles to turn slightly in the opposite direction of the force. This counterbalances the rotation caused by the circular motion of the particles. So, the particles remain pointed in the same direction as they go around in a circle.
  Since the curl is nonzero at the origin, we cannot come up with a potential function for this field.
  \begin{align*} &\mathbf F=-\frac{\mathbf u_{\theta}}{r}=\frac{1}{r^2}(-y,x)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)\quad(r\neq0)\\ &\nabla \cdot \mathbf F=0\\ & \nabla \times \mathbf F=0 \end{align*}
• 1/r rotational potential: This field is an attempt to represent the inverse rotational field using a potential. Because of the discontinuity at the center, this is the best we can do. We come close to the inverse rotational field except that there is a line at the top which the particles cannot cross.
  \begin{align*} & U=\theta\quad(\theta\neq0)\\ &\mathbf F=-\frac{\mathbf u_{\theta}}{r}=\frac{1}{r^2}(-y,x)=\left(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right)\quad(r\neq0)\\ &\nabla \cdot \mathbf F=0\\ & \nabla \times \mathbf F=0 \end{align*}
• 1/r rotational double: Here we have two centers of rotation, with an adjustable distance between them.
• 1/r rotational double + ext: Same as above, but a constant external field is added in. The strength and angle of the external field is adjustable.
• 1/r rotational dipole: Here we have two centers of rotation, with rotation in opposite directions, with an adjustable distance between them.
• 1/r rotational dipole + ext: Same as above, but a constant external field is added in. The strength and angle of the external field is adjustable. (In fluid dynamics this field is called the Lamb dipole).
• one direction: This is a very simple field where the field vectors are pointing in the same direction everywhere with the same magnitude. \begin{align*} &\mathbf F=a\,\mathbf i=(a,0)\\ & U=a\,x\\ &\nabla \cdot \mathbf F=0\\ & \nabla \times \mathbf F=0 \end{align*}
• 1/r^2 sphere: This field is associated with gravitation. Consider the gravitational field around a planet. Outside the planet, the field acts as if the planet were a point mass; the field is inversely proportional to the square of the distance from the center of the planet. Inside the planet, the field is linear to the distance from the center of the planet.
  Outside the planet:
  \begin{align*} &\mathbf F=-\frac{\mathbf r}{r^3}=\left(-\frac{x}{(x^2+y^2)^{3/2}},-\frac{y}{(x^2+y^2)^{3/2}}\right)\quad(r\neq0)\\ & U=\frac{1}{r}\quad(r\neq0)\\ &\nabla \cdot \mathbf F=\frac{1}{r^3}\\ & \nabla \times \mathbf F=0 \end{align*}
  Inside the planet (radius = a):
  \begin{align*} &\mathbf F=-\frac{\mathbf r}{a^3}=\left(-\frac{x}{a^3},-\frac{y}{a^3}\right)\quad(a\neq0)\\ & U=\frac{3}{2a}-\frac{r^2}{2a^3}\quad(a\neq0)\\ &\nabla \cdot \mathbf F=-\frac{2}{a^3}\\ & \nabla \times \mathbf F=0 \end{align*}
• const radial: In this field, the particle is being pulled towards the center with a constant force no matter where it is.
  \begin{align*} &\mathbf F=a\,\mathbf u_r\\ & U=-a\,r\\ &\nabla \cdot \mathbf F=-\frac{a}{r}\quad(r\neq0)\\ & \nabla \times \mathbf F=0 \end{align*}
• linear radial: In this field, the particle is being pulled towards the center with a force proportional to its distance from the center, similar to the force on an object attached to a spring. This force looks more interesting when Display: Particles (Force) is used instead of a velocity field.
  \begin{align*} &\mathbf F=-k\,r \,\mathbf u_r\\ & U=-\frac12 k\, r^2\\ &\nabla \cdot \mathbf F=-2kr\\ & \nabla \times \mathbf F=0 \end{align*}
• inverse to y axis: There is a force towards the y axis inversely proportional to the distance.
  \begin{align*} &\mathbf F=-\frac{1}{x}\mathbf i\quad(x\neq0)\\ & U=\frac{1}{2x^2}\quad(x\neq0)\\ &\nabla \cdot \mathbf F=\frac{1}{x^2}\\ & \nabla \times \mathbf F=0 \end{align*}
• 1/r^2 rotational: In this field, the particle is being pulled in a circle around the center. The speed of its motion is inversely proportional to the square of the distance from the center.
  The curl of this field is nonzero everywhere as we might expect for a field that causes particles to rotate. However, note that the curl is negative everywhere (except at the center); this means that the curl vector is in the negative z direction, which implies clockwise movement. This is surprising, since the particles are clearly moving counter-clockwise. To see why this is, set the particle movement to "Curl Detectors". You can see that even though the particles are moving counter-clockwise, the particles themselves are turning very slowly in a clockwise direction. This is because the force on the far side of the particles (the side away from the center) is weaker than the force on the near side; this is enough to cause the particles to turn slightly in the opposite direction of the force. This more than makes up for the counter-clockwise rotation caused by the motion of the particles.
  \begin{align*} &\mathbf F=-\frac{\mathbf u_{\theta}}{r^2}=\frac{1}{r^3}(-y,x)=\left(-\frac{y}{(x^2+y^2)^{3/2}},\frac{x}{(x^2+y^2)^{3/2}}\right)\\ & \nabla \times \mathbf F=-\frac{\mathbf k}{r^3}\quad (r\neq 0)\\ &\nabla \cdot \mathbf F=0 \end{align*}
• pendulum potential: Consider a pendulum which is free to move in both the $x$ and $y$ direction. We can describe its position using the angle it makes with the positive $x$ axis (theta) and the angle it makes with the $xy$ plane (phi). Its potential energy is determined by its height, which is equal to $h-l \cos(\text{theta}) \cos(\text{phi})$ where $l$ is the length of the pendulum and $h$ is the height of the support it is attached to.
  If we replace theta with $x$ and phi with $y$, then we can graph the potential energy, and we can treat the force on the pendulum as a vector field, and see how particles act when placed in the vector field.
  In the simulation, the angles range from $-\pi$ to $\pi$ ($-180^{\circ}$ to $180^{\circ}$). The center of the graph is when both angles are zero and the pendulum is at rest. If you set the particle movement to "Force", you can see the particles swing back and forth like a pendulum would. You can see this better if you set the number of particles to a small number.
  If you set the floor colors to "Potential", you can see a white line around the center red area; this is the point where one or both of the angles is at 90 degrees; in this case the pendulum is at the same height as its support and it doesn't matter what the other angle is. In the green areas, the pendulum is actually above its support.
  Sometimes the particles manage to make it over the top of the hills on either side; this corresponds to a pendulum swinging up and over the top.
  \begin{align*} & U=\cos x\cos y\\ &\mathbf F=(-\sin x\cos y,-\cos x \sin y)\\ &\nabla \cdot \mathbf F=-2\cos x \cos y\\ & \nabla \times \mathbf F=0 \end{align*}
• user-defined potential: this allows you to enter an arbitrary potential function into a text box. The field vectors will be calculated from it. A sample potential is x*x-y*y, which is a saddle potential. Here are some other sample expressions, to give you an idea of the input format:
  x+y
  -x^2+5.2*y
  2*sin(x)+3*y
  1000/(r^2)-5*y
  abs(x)
  exp(2*log(x+10))+sqrt(y+10)
  
  
  After entering a new potential, press return so that the change takes effect. If nothing happens, then the field may be too weak; either increase the field strength, or multiply the field by a large constant. If particles jump around randomly with no apparent pattern, the field may be too strong.
  Each coordinate ranges from -10 to 10, so the edges of the view are at (±10,±10). The center of the view is the origin.
• user-defined field: this allows you to enter an arbitrary field into text boxes (one for each coordinate). If you try to enter some of the predefined fields (like (x,x2)) keep in mind that the fields will not appear exactly the same, because many of the predefined fields have a constant scaling factor that isn't reflected in the name of the field.

References

[1] Marsden, J. E.; Tromba. A. J. (2012). Vector Calculus. 6th ed. USA: W. H. Freeman and Company.

[2] Trench, W. F. (2013). Elementary Differential Equations with Boundary Value Problems. Books and Monographs.