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I derived the equations of motion for a particle constrained on the surface of a sphere Parametrizing the trajectory as a function of time through the usual $\theta$ and $\phi$ angles, these equations read:

$$ \ddot{\theta} = \dot{\phi}^2 \sin \theta \cos \theta $$

$$ \ddot{\phi} = - 2 \dot{\phi} \dot{\theta} \frac{1}{\tan \theta} $$

I've obtained them starting from the Lagrangian of the system and using the Euler-Lagrange equations.

My question is simple: is there a way (a clever substitution, maybe), to go on and solve the differential equations? I would be interested even in a simpler, partially integrated solution. Or is a numerical solution the only way?

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    $\begingroup$ What Lagrangian did you use for the system? $\endgroup$ – DelCrosB Sep 26 '16 at 16:17
  • $\begingroup$ @DelCrosB: The Lagrangian in eq. (1) of my Phys.SE answer here. $\endgroup$ – Qmechanic Sep 26 '16 at 19:00
  • $\begingroup$ @Qmechanic: That's what I would use as well. But I don't see how the EL equations in the OP come from there. Isn't $\phi$ a cyclic coordinate, giving $\ddot{\phi}=0$ as it should be? $\endgroup$ – DelCrosB Sep 26 '16 at 19:28
  • $\begingroup$ 1) This is probably better suited to Math Stack Exchange. 2) Questions like this are required to show effort and to identify a specific conceptual issue that you want to ask about. Asking "how do I solve this?" is not supported on this site. See the help center for more information. $\endgroup$ – DanielSank Sep 27 '16 at 7:10
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Note that you can rewrite your second equation as $$ \frac{\ddot{\phi}}{\dot{\phi}} = -2\cot{(\theta)}\dot{\theta} $$ Each side is an exact differential in one variable, so we can integrate, and Wolfram|Alpha gives $$ \ln{(\dot{\phi})}=-2\ln{(\sin{(\theta)})}+C $$ for some integration constant $C$. We can exponentiate to get $$ \dot{\phi}=\frac{B}{\sin{(\theta)}^2} $$

Substituting this into the first equation yields $$ \ddot{\theta}=B^2\frac{\cos{(\theta)}}{\sin{(\theta)}^3} $$ This, too, can be integrated via the "energy trick": multiply by $ \theta $, then integrate. The LHS integrates by parts to $\dot{\theta}^2$ but the RHS looks sufficiently complicated I don't want to type it out on my phone.

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In case the particle is not subject to any external forces except those maintaining the constraint, there is no need to write and solve the equations of motion in particular system of coordinates. The particle will move with constant speed around some great circle on the sphere. Which circle it will be and the speed of motion are determined by the initial position and velocity of the particle.

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I checked, and these equations of motions correspond to motion in of an otherwise free Lagrangian. Something that will make your life easier in solving this is to recognize that angular momentum is conserved here. Because velocity is guaranteed to be perpendicular to the radius: $$\begin{align} |\mathbf{L}| &= mvr \\ & = mr \left(\dot{\theta} + \dot{\phi}\sin\theta\right) \\ |\dot{\mathbf{L}}| & = mr\left(\ddot{\theta} + \ddot{\phi}\sin\theta + \dot{\phi}\dot{\theta}\cos\theta\right) = 0.\end{align}$$ You should be able to combine the equations of motion you have to show that last line. You also have that $|L| = I\omega$ and, because $I$ is fixed, that means that the rate of change of some angular variable is constant. The final hint I give is that you should look at how rotations can be defined as rotations around an axis by an angle.

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Considering you are aware of conservation of total angular momentum in a sphere (if not, I will prove it below), from the lagrangian I think you are using you get:

$$\mathcal{L}=\dfrac{1}{2}R^2\left(\dot\theta^2+\sin^2\theta\,\dot\phi^2\right)$$

$$l_\theta=\dfrac{\partial\mathcal{L}}{\partial\dot\theta}=\dot\theta$$

$$l_\phi=\dfrac{\partial\mathcal{L}}{\partial\dot\phi}=\sin^2\theta\,\dot\phi=const\quad \left(\text{since }\dfrac{\partial\mathcal{L}}{\partial\phi}=0\right)$$

for $r=R\,$ fixed for a sphere of radius $R$. You can see $l_\theta$ and $l_\phi$ are the conjugated momenta associated to $\theta$ and $\phi$, respectively.

The total angular momentum of the system $L$ obeys the following:

$$L^2=mR^2\left(\dot\theta^2+\sin^2\theta\,\dot\phi^2\right)$$

Defining $l^2=\dfrac{L^2}{mR^2}$ and using what we found above: $\,\dot\theta^2=l_\theta^2\quad\text{and}\quad\,\dot\phi^2=\dfrac{l_\phi^2}{\sin^4\theta}$.

Thus:

$l^2=l_\theta^2+\dfrac{l_\phi^2}{\sin^2\theta}$

We would like to show that this total angular momentum is conserved as well. Noting that differentiating respect to a parameter $\lambda$ we get this is conserved for the curve parametrized by $\lambda$:

$$\dfrac{d\,l^2}{d\lambda}=\dfrac{d}{d\lambda}\,\dot\theta^2+\dfrac{d}{d\lambda}\left(\dfrac{l_\phi^2}{\sin^2\theta}\right)=2\left(\ddot\theta-\dot\phi^2\sin\theta\cos\theta\right)\dot\theta=0$$

because the result involves the equation of motion for $\theta$ you already computed, when is equal to zero.

Furthermore,

$$\dot\theta=\sqrt{l^2-\dfrac{l_\phi^2}{\sin^2\theta}}$$

$$\dot\phi=\dfrac{l_\phi}{\sin^2\theta}$$

and from here you can as well try to integrate both equations separately. My recommendation would be trying to find $\phi=\phi(\theta)$, so for instance you can make:

$$\dot\phi=\dfrac{d\phi}{d\theta}\dot\theta$$

What's more:

$$\dfrac{d\phi}{d\theta}\sqrt{l^2-\dfrac{l_\phi^2}{\sin^2\theta}}=\dfrac{l_\phi}{\sin^2\theta}$$

$$\Rightarrow\quad\dfrac{d\phi}{d\theta}=\dfrac{l_\phi}{l}\dfrac{1}{\sin\theta\sqrt{\sin^2\theta-\left(\frac{l_\phi}{l}\right)^2}}$$

Finally, integrating respect to $\theta$ leads to :

$$\phi(\theta)=\phi_0+\arctan\left(\dfrac{\frac{l_\phi}{l}\cos\theta}{\sqrt{\sin^2\theta-\left(\frac{l_\phi}{l}\right)^2}}\right)$$

You can use any plotter you know for seeing how this can give you portions of arc of a sphere (parallels and meridians, e.g.) for a Parametric 3D Plot, setting $(r=R,\theta\in[0,\pi],\phi=\phi(\theta))$. You can get, for example, the equator for $l_\phi=0$.

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Solving using spherical coordinates is somehow messy in evaluating integral. You could try Cartesian coordinates.

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    $\begingroup$ Hi, welcome to Physics SE! Answers usually have a bit more detail, and this may be more suitable as a comment. Try gaining the reputation needed to comment everywhere by asking questions and writing answers! $\endgroup$ – user191954 Jun 29 '18 at 16:34

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