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I have a set of parametric equations in spherical coordinates that supposedly form circle trajectories. See below:

r=C1

theta=C2*Sin[beta]*Sin[phi[t]]

phi=(C2*Sin[beta]*(Cos[theta[t]]/Sin[theta[t]])*Cos[phi[t]])+(C2*Cos[beta])

C1 and C2 are constants and beta is some angle, say 15 degrees, or (15/180)*Pi radians.

These are circle trajectories on the surface of a sphere, hence the constant r-component.

My question is this: How do I solve these trajectories and plot them on the surface of a sphere. This is what I have done already:

Step 1:

Solve the coupled differential equation with NDSolve. See below:

Answer = NDSolve[
   {theta'[t] == C2*Sin[beta]*Sin[phi[t]],
    phi'[t] == (C2*Sin[beta]*Cos[theta[t]]*
        Cos[phi[t]]/Sin[theta[t]]) + (C2*Cos[beta]),
    theta[tMin] == StartingTheta,
    phi[tMin] == StartingPhi},
   {theta, phi},
   {t, tMin, tMax}];

where I have defined StartingTheta=(45/180)*Pi and StartingPhi=(180/180)*Pi. tMin is 0 and tMax is, say, a 1000.

This will form ONE circle trajectory on the surface of the sphere. Changing StartingTheta will give another circle trajectory and so forth.

Step 2: Create the Sphere. I did that - see below:

sphere = ParametricPlot3D[
   {Cos[v]*Cos[u],
    Sin[v]*Cos[u],
    Sin[u]},
   {v, 0, 2*Pi},
   {u, -Pi/2, Pi/2}];

Step 3:

Evaluate. This is where I am struggling. See below what I have done so far:

Trajectory = ParametricPlot3D[
   Evaluate[
    {????????????????}
     /. Answer],
   {t, tMin, tMax}
   ]

At the place where I inserted all the question marks is the problem, I am not sure what form I should be evaluating. I tried a few obvious expressions but I keep getting straight lines.

The next step I suppose is quite easy:

Show[sphere,Trajectory]

If anybody out there is able to help me with what I should be evaluating in order to plot these circle trajectories on the surface of the sphere, it would be greatly appreciated.

Thanx!

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  • $\begingroup$ Welcome to Mathematica.SE. Could you try formatting your question? The information for this can be found above the top right corner of the edit box. $\endgroup$ – Sjoerd C. de Vries Sep 30 '15 at 12:39
  • $\begingroup$ Thanx for the help! $\endgroup$ – Maxwell Sep 30 '15 at 13:05
  • $\begingroup$ I think there is a mixup in the second two statements. You define theta and use phi[t] in the definition. Then you define phi and use theta[t] in the definition. I am pretty sure you didn't mean that. Take another look and see if you can alter the definitions. Most of the time when we make function definitions we use something like theta[t_]:=C2*Sin[t]... $\endgroup$ – Jack LaVigne Sep 30 '15 at 22:31
  • $\begingroup$ @JackLaVigne, I think these definitions are missing a derivative - they are the same that are solved in the differential equation. $\endgroup$ – gpap Oct 1 '15 at 10:14
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I didn't have time to look at this properly as your functions don't work straight out of the box because they require additional definitions. Also, I think there is a derivative missing in your original funciton definitions.

Anyway, I will use the parameters from Jack LaVigne's answer (but I have modified the range of t that only needs to span one period):

C2 = 1.;
beta = 15. Degree;
StartingTheta = 45. Degree;
StartingPhi = 180. Degree;
tMin = 0;
tMax = 2 π.;
Answer = NDSolve[
   {theta'[t] == C2*Sin[beta]*Sin[phi[t]],
    phi'[t] == (C2*Sin[beta]*Cos[theta[t]]*
        Cos[phi[t]]/Sin[theta[t]]) + (C2*Cos[beta]),
    theta[tMin] == StartingTheta,
    phi[tMin] == StartingPhi},
   {theta, phi},
   {t, tMin, tMax}];

Because of the parametric dependence on t, in retrospect, mesh functions is not the optimal way to go (at least as far as I can tell) so it's best to just plot the path together with the sphere. First define the point as a function of t:

ans[t_] = {Cos[v]*Cos[u], Sin[v]*Cos[u], Sin[u]} /. 
    u -> Answer[[1, 1, 2]][t] /. v -> Answer[[1, 2, 2]][t];

and then overlap it to the sphere for a full period in t:

Show[
 ParametricPlot3D[{Cos[v]*Cos[u], Sin[v]*Cos[u], Sin[u]}, {u, 0, 
   2*Pi}, {v, -Pi/2, Pi/2}, PlotStyle -> None
  ],
 Graphics3D[
  {Red,
   Thickness[0.01], Line[Table[ans[t], {t, 0, 2 π, π/64}]]}
  ]
 ]

enter image description here

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  • $\begingroup$ Thanx for the help, much appreciated! I tried several different techniques of my own including the ideas above - I will post my final answer/code once I have smoothed it over. Hopefully someone in the future could gain something from it :) Agian, thanx for the help! $\endgroup$ – Maxwell Oct 1 '15 at 15:40
  • $\begingroup$ @gpap Got my vote. $\endgroup$ – Jack LaVigne Oct 3 '15 at 0:05
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I am quite certain that I don't understand your function but may be able to help with the format for plotting with MeshFunction.

C2 = 1.;
beta = 15. Degree;
StartingTheta = 45. Degree;
StartingPhi = 180. Degree;
tMin = 0;
tMax = 1000.;

Using these parameters

Answer = NDSolve[
   {theta'[t] == C2*Sin[beta]*Sin[phi[t]],
    phi'[t] == (C2*Sin[beta]*Cos[theta[t]]*
        Cos[phi[t]]/Sin[theta[t]]) + (C2*Cos[beta]),
    theta[tMin] == StartingTheta,
    phi[tMin] == StartingPhi},
   {theta, phi},
   {t, tMin, tMax}]

results in two InterpolatingFunctions for theta (Answer[[1,1,2]]) and phi (Answer[[1,2,2]]).

The theta function is cyclic

Plot[Answer[[1, 1, 2]][v], {v, 0, 6 π}, PlotLabel -> "theta"]

Mathematica graphics

The phi function starts low and continues to grow although it does have some periodicity.

Plot[Answer[[1, 2, 2]][u], {u, 0, 6 π}, PlotLabel -> "phi"]

Mathematica graphics

Below is the format using these two functions as MeshFunction. theta is being applied to the parameter u and phi is applied to v.

ParametricPlot3D[{Cos[v]*Cos[u], Sin[v]*Cos[u], Sin[u]}, {u, 0, 
  2*Pi}, {v, -Pi/2, Pi/2},
 Mesh -> {16, 8},
 MeshFunctions -> {
   Answer[[1, 1, 2]][#4] &,
   Answer[[1, 2, 2]][#5] &
   },
 MeshStyle -> {Red, Blue}
 ]

Mathematica graphics

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  • $\begingroup$ Hey, this is not quite right - the parameter dependence is on t, not θ and φ which is what you have used. But seeing as he needs this for the range of the parameter, it doesn't really make sense as for ever t you'd only get a point on the sphere using mesh functions. $\endgroup$ – gpap Oct 1 '15 at 10:12
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Thanx for the answers I received concerning this question. @gpap - your answer was/is superb! Below is another take on it:

I used all the coding in the original question above, but this time I inserted the standard Cartesian-Spherical transformations where I had all the question marks.

Trajectory = ParametricPlot3D[
   Evaluate[
    {1*Sin[theta[t]]*Cos[phi[t]],
      1*Sin[theta[t]]*Sin[phi[t]],
      1*Cos[theta[t]]}
     /. Answer],
   {t, tMin, tMax}, PlotStyle ->Green];

This formes ONE trajectory around the surface of the sphere. Changing the StartingTheta will generate another unique trajectory. The beta angle involved is the off-set angle form the pole of the sphere. If beta was equal to zero, all these trajectories would have been concentric around the pole of the sphere. For good measure I added a new axis to show all the trajectories are concentric around this new axis.

Sphere with trajectories on surface

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  • $\begingroup$ Hi, it's fine to un-accept an answer if it doesn't tick the right boxes but I was under the impression mine did. There was no requirement to change the angle in the original question by the way. $\endgroup$ – gpap Oct 6 '15 at 10:10
  • $\begingroup$ Hi @gpap, sorry I did not realize by ticking my answer I was 'un-ticking' yours, I thought there was space for two answers - just done differently. Mistake corrected - you are ticked again, my apologies. As long as the goal was met, I don't care much who gets the brownie points. Concerning the beta, I just added that to give some context to whoever finds these answers useful someday in the future. $\endgroup$ – Maxwell Oct 6 '15 at 11:37
  • 1
    $\begingroup$ Hey, no worries. I didn't mean to press you into changing your choice: it's also perfectly acceptable to answer your own question. It is however good practice to not rush to accept so that the best answer is selected and this kind of switcharoo is avoided $\endgroup$ – gpap Oct 6 '15 at 17:27

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