Central Forces

Alexei Gilchrist

2 Central Forces

A single object or particle moving under the influence of a central force.

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1 Co-ordinate system

In three dimensions, three numbers need to be specified to pin down the position of an object. With a central force, the potential can only depend on \(r\), the radial distance to the origin, so it makes sense to have this as one of the co-ordinates. The other two are usually given as angles. With a bit of cunning we can immediately reduce the dimensionally of the problem.

It should be pretty clear that regardless of what the direction of the initial velocity of the object, the dynamics are going to be confined to the plane that contains the velocity vector and the position vector. The forces will only act in the direction of the position vector so there is nothing that can take the object out of this plane.

Within this plane the position of the object can be specified by the distance to the origin and an angle \((r, \theta)\), and the velocity will be \((\dot{r}, r\dot{\theta})\). This velocity is in an instantaneous co-ordinate system that moves with the object. Alternatively, we can fix some axes and define

\[\begin{align*}x &= r \cos\theta\\ y &= r \sin\theta\end{align*}\]

In which case the velocity will be

\[\begin{align}\dot{x} &= \dot{r} \cos\theta - r\sin\theta \dot{\theta}\\ \dot{y} &= \dot{r} \sin\theta + r\cos\theta \dot{\theta} \nonumber\end{align}\]

2 Equations of motion

To find the Lagrangian we need the kinetic and potential energies. The potential energy is easy, it’s just some function of the radial distance: \(U(r)\). The kinetic energy is only a little more tricky, but we already have the objects velocity in two coordinate systems above. In the instantaneous co-ordinate system it’s straight forward:

\[\begin{equation*}T(r,\theta) = \frac{1}{2}m (\dot{r}^2+r^2\dot{\theta}^2)\end{equation*}\]

Similarly in the fixed coordinate system, we can write down the kinetic energy then convert using \eqref{eq:velocities}:

\[\begin{equation*}T(x,y) = \frac{1}{2}m (\dot{x}^2+\dot{y}^2) = \frac{1}{2}m \left[ (\dot{r} \cos\theta - r\sin\theta \dot{\theta})^2 + (\dot{r} \sin\theta + r\cos\theta \dot{\theta})^2 \right] = \frac{1}{2}m (\dot{r}^2+r^2\dot{\theta}^2)\end{equation*}\]

Hence the Lagrangian for the system is

\[\begin{equation}\mathcal{L}(r,\theta) = \frac{1}{2}m (\dot{r}^2+r^2\dot{\theta}^2) - U(r)\end{equation}\]

The Euler-Lagrange equation for the generalised coordinate \(\theta\) will yield

\[\begin{gather}\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{\theta}} = \frac{\partial\mathcal{L}}{\partial \theta}\nonumber \\ \frac{d}{dt}\left(mr^2\dot{\theta}\right) = 0 \label{eq:angular-momentum}\end{gather}\]

This implies that the expression \(mr^2\dot{\theta}\) doesn’t change in time, it has some value set by the initial conditions and it doesn’t change from that value at any point in the dynamics. We may as well label that constant

\[\begin{equation}mr^2\dot{\theta}=L\end{equation}\]

since, as you may have recognised already, it’s the angular momentum. What we have discovered is that for central forces the total angular momentum is conserved.

The Euler-Lagrange equation for the generalised coordinate \(r\) yields

\[\begin{gather} \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{r}} = \frac{\partial\mathcal{L}}{\partial r}\nonumber \\ m\ddot{r} = mr\dot{\theta}^2- \frac{\partial U}{\partial r} \nonumber\\ m\ddot{r} = \frac{L^2}{mr^3}- \frac{\partial U}{\partial r} \label{eq:r-eq}\end{gather}\]

where in the last step we used \eqref{eq:L} to replace \(\dot{\theta}\).

3 Effective Potential

We can rewrite the expression \eqref{eq:r-eq} so that the differentiation with respect to \(r\) acts on the whole left side:

\[\begin{equation*}m\ddot{r} = -\frac{\partial}{\partial r}\left(U(r)+\frac{L^2}{2mr^2}\right).\end{equation*}\]

Now the equations of motion look just like Newton’s law written in terms of an effective potential \(U_\mathrm{eff}(r)\): \(F=-\partial U_\mathrm{eff}(r)/\partial r=m a\), Where

\[\begin{equation}U_\mathrm{eff}(r) = U(r)+\frac{L^2}{2mr^2}.\end{equation}\]

Notice that as \(r\rightarrow 0\), the second term in \eqref{eq:Ueff} goes to infinity unless \(L=0\). So if there in any initial angular momentum in the system, there will be a minimum distance \(r_\mathrm{min}\) that the object can attain with a given energy, regardless of what the actual potential is.

4 Example Potentials

We will look at the solutions for motions in different potentials elsewhere, but to finish off this node lets examine several common potentials and infer general properties of the dynamics in each case.

4.1 Gravity \(U(r)\sim 1/r\)

\label{sub:gravity}

The gravitational potential energy (with zero potential at \(r\rightarrow \infty\) ) is

\[\begin{equation*}U(r) = -G\frac{Mm}{r} = -\alpha \frac{m}{r}\end{equation*}\]

where \(G=\) is the gravitation constant. So in total the effective potential is

\[\begin{equation*}U_\mathrm{eff}(r) = -\alpha \frac{m}{r} +\frac{L^2}{2mr^2}.\end{equation*}\]

By playing with the parameters in the simulation you should be able to find values where the objects motion is unbounded and values where it’s contained to lie somewhere in a circular or ring region. For the right parameters you can constrain the allowed region enough that you can determine the shape of the orbit.

4.2 Spring \(U(r)\sim r^2\)

\label{sub:spring}

In contrast to the gravitational potential, the spring-force potential goes as \(r^2\), specifically,

\[\begin{equation*}U(r) = \frac{1}{2}k r^2 = \alpha r^2\end{equation*}\]

where \(k\) is the spring constant. Again, the effective potential has an additional term due to the angular momentum

\[\begin{equation*}U_\mathrm{eff}(r) = \alpha r^2 +\frac{L^2}{2mr^2}.\end{equation*}\]

For this system there are only bounded motions. Of course if you stretch the spring enough it would break, but this is beyond what is described by the simple potential we are using!