# Plot particle motion in potential

Consider the anisotropic harmonic potential in two dimensions $(q_1,q_2)$ given by

$$V(q_1,q_2) = \frac{m}{2} \, q_1^2 + \frac{k}{2} \, q_2^2,$$

or V = m/2 q1^2 + k/2 q2^2; in Mathematica.

The Newtonian e.o.m.s of a particle moving through this potential are

$$\ddot{q}_1 = -q_1, \qquad \ddot{q}_2 = -\omega^2 \, q_2,$$

where $\omega = \sqrt{k/m}$ is the angular frequency of oscillations in the $q_2$-direction. Given the initial conditions $q_i(0) = q_{i,0}$ and $p_i(0) = p_{i,0}$, the e.o.m.s are solved by

\begin{aligned} q_1(t) &= q_{1,i} \cos(t) + \frac{p_{1,i}}{m} \, \sin(t),\\ q_2(t) &= q_{2,i} \cos(\omega t) + \frac{p_{2,i}}{m \omega} \, \sin(\omega t), \end{aligned} \qquad \text{with p_i = m \dot{q}_i.}

In Mathematica:

DSolve[{q1''[t] == -q1[t], q2''[t] == -ω^2 q2[t], q1 == q10,
q1' == p10/m, q2 == q20, q2' == p20/m}, {q1[t], q2[t]}, t]
//FullSimplify

{{q1[t] -> q10 Cos[t] + (p10 Sin[t])/m, q2[t] -> q20 Cos[t ω] + (p20 Sin[t ω])/(m ω)}}


A 3d plot of the potential looks like this.

Plot3D[V /. {m -> 1, k -> 3}, {q1, -5, 5}, {q2, -5, 5},RegionFunction -> Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 12 /. {m -> 1, k -> 3}]] What I would like to do now is draw the particle as a ball moving through this potential with a fixed energy. Any help would be much appreciated.

Some remarks about the physics behind this simulation: The particle always reaches a certain height both in $q_1$- and $q_2$-direction before rolling back down again and up the other side. This is because the Hamiltonian $H = \frac{p_1^2}{2 m} + \frac{p_2^2}{2 m} + V(q_1,q_2)$ does not couple the degrees of freedom in $q_1$- and $q_2$-direction. Therefore, the total energies $E_1$ and $E_2$ available in dimensions $q_1$ and $q_2$ are conserved separately.

For long times $t \to \infty$ and irrational angular frequency $\omega \notin \mathbb{Q}$ (which ensures that the trajectory never closes, thus making the system ergodic), the particle's trajectory should therefore trace out a rectangle $R$ whose length and width are determined by $E_1$ and $E_2$.

Update: With anderstood's and BlacKow's help, I was able to piece together this solution that does exactly what I want.

V = m/2 q1^2 + k/2 q2^2; \[Omega] = Sqrt[k/m];
sol = {q1[t], q2[t], m/2 q1[t]^2 + k/2 q2[t]^2} /.
DSolve[{q1''[t] == -q1[t], q2''[t] == -\[Omega]^2 q2[t],
q1 == q10, q1' == p10/m, q2 == q20,
q2' == p20/m}, {q1[t], q2[t]}, t] // FullSimplify

Manipulate[Block[{m = 1, k = 5, q10 = 5, p10 = 1, q20 = 2, p20 = 1},
surf = Plot3D[V, {q1, -7, 7}, {q2, -5, 5},
RegionFunction -> Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 25], PlotStyle -> Opacity[0.5]];
Show[surf, Graphics3D@{Blue, Ball[sol /. {t -> tf}, 0.4]},
ParametricPlot3D[sol, {t, 0, tf}]]], {tf, 0.1, 100}] • @corey979 Sorry, I'm having trouble posting this question. I keep getting the error Your post appears to contain code that is not properly formatted as code. Please indent all code by 4 spaces using the code toolbar button or the CTRL+K keyboard shortcut. For more editing help, click the [?] toolbar icon. Looking on Meta, it appears this issue has come up before. Dec 18 '16 at 14:20
• Just copy and paste you code from the notebook, select it and hit the {} button. Dec 18 '16 at 14:23
• Exactly what I did. This isn't my first question but I never encountered this problem before. Dec 18 '16 at 14:24
• Can you post the whole question and ignore the message? Or it won't let you post, except for this fragment? Dec 18 '16 at 15:13
• It seems you were able to post it. What changed? Dec 18 '16 at 20:49

First, store the 3D trajectory (in the space $(q_1,q_2,V)$):

sol = {q1[t], q2[t], m/2 q1[t]^2 + k/2 q2[t]^2} /.
DSolve[{q1''[t] == -q1[t], q2''[t] == -\[Omega]^2 q2[t],
q1 == q10, q1' == p10/m, q2 == q20,
q2' == p20/m}, {q1[t], q2[t]}, t] // FullSimplify


Plot the potential surface:

surf = Plot3D[
V /. {m -> 1, k -> 3, \[Omega] -> Sqrt[k/m]}, {q1, -5, 5}, {q2, -5,
5}, RegionFunction ->
Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 12 /. {m -> 1, k -> 3}]]


Then select some initial conditions and plot the trajectory using ParametricPlot3D. Show it together with the surface using... Show:

p10 = 3; p20 = 1; q20 = -1.5; q10 = 2;
traj = ParametricPlot3D[
sol /. \[Omega] -> Sqrt[k/m] /. {m -> 1, k -> 3}, {t, 0, 50},
PlotStyle -> Red];
Show[pot, traj] • Looks good! One question: Is there a way to avoid repeating the replacement /. {m -> 1, k -> 3} everywhere m or k come up? For example, could they be assigned specific values all throughout a cell? Dec 19 '16 at 14:10
• @Casimir I you want to keep these values throughout the notebook, just add at the beginning: m=1; k=3;. And you should also avoid using "redundant" variables, i.e. $\omega$, which depends on k and m (I guess...). Dec 19 '16 at 15:20
• @Casimir Then you can use Block[{k=1,m=3}, ...]. Dec 19 '16 at 15:23
• @anderstood equipotential surface is a surface where potential energy is the same, so it's defined by $V(q) = const$; it will be an ellipse in $(q_1,q_2)$ coordinates. And here $V$ is not a total energy but, potential energy only. Dec 19 '16 at 18:56
• @BlacKow You're right, edited. Dec 19 '16 at 19:10

Using @anderstood answer we can play with graphics to make the surface transparent and plot ball movement:

V[q1_, q2_] := m/2 q1 q1 + k/2 q2 q2
surf = Plot3D[
V[q1, q2] /. {m -> 1, k -> 3, \[Omega] -> Sqrt[k/m]}, {q1, -5,
5}, {q2, -5, 5},
RegionFunction ->
Function[{q1, q2}, m/2 q1^2 + k/2 q2^2 <= 12 /. {m -> 1, k -> 3}],
Mesh -> None,
ColorFunction ->
Function[{z}, Opacity[0.4, #] &@ColorData["TemperatureMap"][z]]]

traj[t_] :=
Evaluate[Flatten@sol /. \[Omega] -> Sqrt[k/m] /. {m -> 1,
k -> 3} /. {p10 -> 3, p20 -> 1, q20 -> -1.5, q10 -> 2}]
frames = Table[
Show[surf, Graphics3D@{Red, Ball[traj[t], 0.2]}], {t, 0, 10,
0.1}];
Export["Documents/animBall.gif", frames] 