I've been trying to solve the following problem for a long time.

Let's consider a particle of mass $m$ in $\mathbb{R}^3$ with polar coordinates $(r,\theta,\phi)$. The particle moves on the orbit described by the equation $r=2a\sin\theta,\,\,a>0$, with velocity $|\mathbf{v}|=k/\sin^2\theta,\,\,k>0$.

The task is to show that the force is central and to determine such a force.

My idea has been to use the two equations given to prove the two components \begin{equation} F_\theta=m\left( r\,\ddot\theta + 2\dot{r}\,\dot\theta - r\,\dot\varphi^2\sin\theta\cos\theta \right)\\ F_\phi=m\left( r\ddot\varphi\,\sin\theta + 2\dot{r}\,\dot\varphi\,\sin\theta + 2 r\,\dot\theta\,\dot\varphi\,\cos\theta \right) \end{equation} to vanish; unfortunately, I wasn't able to get to any interesting result.

Can anybody give me some help on this?

  • $\begingroup$ I am not sure if these equations of motion make sense, since usually $r$ is positive, however in your equation it can be negative as well, and the velocity will be infinite when $\sin\theta=0$. $\endgroup$ – fibonatic Aug 22 '14 at 10:28
  • $\begingroup$ No, $r$ can't be negative, since $0<\theta<\pi$. Velocity can be infinite, but only in the case $r=0$; I have supposed this to be a singularity for the force $\endgroup$ – popoolmica Aug 22 '14 at 10:31
  • $\begingroup$ You are right, I swapped the meaning of $\theta$ with $\varphi$. $\endgroup$ – fibonatic Aug 22 '14 at 11:59
  • $\begingroup$ Those are spherical coordinates and not polar. Since there is no dependence on $\phi$ I suppose $F_{\phi} = 0$? $\endgroup$ – John Alexiou Aug 22 '14 at 13:19

An orbit in three dimensions is generally specified by giving how three quantities depend on time, or by giving how two coordinates depend on the third one.You have however omittied saying how $\phi$ depends on $\theta$; I will thus simply assume that $\phi$ is a constant, hence $v_\phi = 0$.

In this case the total angular momentum $$ \mathbf L = m\mathbf r\times\mathbf v $$ is given by, in spherical coordinates: $$ \mathbf L = mr \hat e_r\times (v_\theta\hat e_\theta + v_\phi\hat e_\phi) = -mrv_\theta\hat e_\phi $$ If this turns out to be a constant, then we shall know that this is a central force.

However, we do not have $v_\theta$ yet, only the modulus $v$. For our orbit, we already know that $v_\phi = 0$. Also, $$ \dot r = 2a \cos\theta \dot \theta $$ while $$ v_\theta = r\dot\theta = 2 a \sin\theta\dot\theta\;. $$ Putting all of this together, we find that $$ |\mathbf v| = 2 a\dot\theta $$ but we also know that $$ |\mathbf v| = \frac{k}{\sin^2\theta} $$ Hence $$ \dot\theta = \frac{k}{2a}\frac{1}{\sin^2\theta} $$ which allows us to find that $$ v_\theta = r\dot\theta = 2a\sin\theta \frac{k}{2a}\frac{1}{\sin^2\theta} = \frac{k}{\sin\theta} $$ and the only non-zero component of the angular momentum becomes $$ \mathbf L = -mr v_\theta \hat e_\phi = - 2 a k m \hat e_\phi $$which remains constant along the orbit, i.e. as $\theta$ changes. The force is central.

A central force has a potential that depends only on $r$, not on $\theta$ or $\phi$. The total energy, which is conserved, is $$ E = \frac{m}{2}|\mathbf v|^2 + V(r) $$ can now be rewritten as $$ E = \frac{16 a^4 k^2 m}{r^4} + V(r)\;. $$ This quantity must remain constant as $r$ varies along the orbit. The only way to obtain this is to have $$ V(r) = -\frac{16 a^4 k^2 m}{r^4} + V_0 $$ with $V_0$ a constant.

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  • $\begingroup$ Great, many thanks. Probably the exercise is not exactly well stated, since it doesn't say anything about $\phi$. I guess your assumption about it is right. $\endgroup$ – popoolmica Aug 22 '14 at 15:02

Using conservation of angular momentum you have (note that this relation implies a monotonic relation between angle and time) $$ mr^2\dot{\varphi}=J\Rightarrow\mathrm{d}t=\frac{mr^2}{J}\mathrm{d}\varphi. $$ The energy is conserved and given by, $$ E=\frac{1}{2}m\dot r^2+\frac{J^2}{2mr^2}+V(r). $$ Changing variables from time to angle ($\dot r=r_\varphi \frac{J}{mr^2}$) $$ E=\frac{1}{2}\frac{J^2}{mr^4}r'^2(\varphi)+\frac{J^2}{2mr^2}+V(r). $$ Noting that $\frac{r'(\varphi)}{r^2}=-\frac{\mathrm{d}}{\mathrm{d}\varphi}\frac{1}{r(\varphi)}$, we change to the variable $u(\varphi)=\frac{1}{r(\varphi)}$. The conservation of energy in terms of $u$ reads $$ E=\frac{1}{2}\frac{J^2}{m}u'^2(\varphi)+\frac{J^2}{2m}u^2(\varphi)+V(\frac{1}{u(\varphi)}). $$ That is, give a trajectory $r(\varphi)$ the potential is given by $$ V(\frac{1}{u(\varphi)})=E-\frac{1}{2}\frac{J^2}{m}u'^2(\varphi)-\frac{J^2}{2m}u^2(\varphi). $$

In your case start by assuming its central force and find the potential.

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  • $\begingroup$ I'm not sure I got your answer: I can't assume the force to be central, since it's what I have to prove. $\endgroup$ – popoolmica Aug 22 '14 at 10:33
  • $\begingroup$ You can assume, and if you find a potential under this assumption that it proves your assumption. $\endgroup$ – PhysicistRRE Aug 22 '14 at 10:36
  • $\begingroup$ You can also use the above to show that the force is given by $F(r)=\frac{J^2}{mr^4}(r''(\varphi)-\frac{2}{r}r'^2-r)$. If the RHS is angle independent then it means that the orbit can be satisfied by a central force. In your case the force is $F(r)=-\frac{8a^2}{r}$ $\endgroup$ – PhysicistRRE Aug 22 '14 at 10:39
  • $\begingroup$ @PhysicistRRE - Showing that a central force is consistent with the givens is a sufficient condition. It does not prove the conjecture unless you can also show that it is a necessary condition. $\endgroup$ – David Hammen Aug 22 '14 at 12:24
  • $\begingroup$ @PhysicistRRE - Wouldn't I obtain a potential with a dependence on total energy and total angular momentum? $\endgroup$ – popoolmica Aug 22 '14 at 15:06

Using energy conservation you can find your answer from $$ V = E-\frac{1}{2}mv^2 $$ Express $V$ in terms of $r$ and you should get the form of the potential, assuming that it is a central force. Then you have to show that the assumption is correct.

Is the problem such that you can assume $\varphi =0$?

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Using differential geometry one can show that

$$ \vec{p}(\theta) = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 a \sin \theta \cos \theta \\ 2 a \sin \theta^2 \\ 0 \end{pmatrix} $$

$$ \vec{v}(\theta,\dot\theta) = \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \frac{{\rm d}}{{\rm d}t} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 a \cos(2 \theta) \dot{\theta} \\ 2 a \sin(2 \theta) \dot{\theta} \\ 0 \end{pmatrix} $$

$$ \vec{a}(\theta,\dot\theta,\ddot\theta) = \begin{pmatrix} x'' \\ y'' \\ z'' \end{pmatrix} = \frac{{\rm d}^2}{{\rm d}t^2} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 a \left(\ddot{\theta}\cos(2\theta) - 2 \dot{\theta}^2 \sin(2 \theta)\right)\\ 2 a \left(\ddot{\theta}\sin(2\theta) + 2 \dot{\theta}^2 \cos(2 \theta)\right) \\ 0 \end{pmatrix} $$

and the orbit radius of curvature $\rho(\theta)$

$$ \frac{1}{\rho(\theta)} = \frac{|y' x'' - y'' x'|}{(x'^2+y'^2)^{1.5}} = \frac{8 a^2 \dot{\theta}^3 }{(4 a^2 \dot{\theta}^2)^{1.5}} = \frac{1}{a} $$

So if the curvature is constant, then the orbit is circular.

At any point, if the tangent direction is $\vec{e}$ and the normal direction $\vec{n}$ then the kinematic decomposition is

$$ \begin{aligned} \vec{v} & = v \vec{e} \\ \vec{a} & = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n} \end{aligned}$$

Given that $|\vec{v}| = v = \frac{k}{\sin^2\theta}$ and thus $\dot{v} = - \frac{k \cos(\theta) \dot{\theta}}{\sin^3( \theta)}$

Together you have:

$$ \begin{aligned} \vec{v} & = \frac{k}{\sin^2(\theta)} \vec{e}(\theta) \\ \vec{a} & = - \frac{k \cos(\theta) \dot{\theta}}{\sin^3( \theta)} \vec{e}(\theta) + \frac{k^2}{a \sin^4(\theta)} \vec{n}(\theta) \\ \vec{F} & = m \vec{a} \end{aligned}$$

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