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Advanced Classical Mechanics/Central Forces

Central forces are forces between two particles that depend only on the distance between the particles and point from one particle to another. Central forces are conservative (why?). Furthermore, if we look at a central force between two particles we can separate the motion of the center of mass from the relative motion of the particles reducing the complexity of the problem.

Separation of Motion
We have two particles of masses $$m_1$$ and $$m_2$$ located at positions $$\vec r_1$$ and $$\vec r_2$$. Furthermore, let's define $$\vec r=\vec r_1 - \vec r_2$$. Let's assume that the potential depends only on the relative position of the two particles. We can write out the Lagrangian as

$$ L = T - V = \frac{1}{2} m_1 \dot \vec r_1^2 + \frac{1}{2} m_2 \dot \vec r_2^2 - V \left ( \vec r \right ) + m_1 \vec g \cdot \vec r_1 + m_2 \vec g \cdot \vec r_2 $$

where $$\vec g$$ is the direction of the acceleration due to a uniform gravitational field.

Let's define $$\vec R$$ to be the position of the center of mass

$$ \vec R=\frac{m_1 \vec r_1 + m_2 \vec r_2}{m_1+m_2} $$

so

$$ \vec r_1 = \vec R + \frac{m_2 \vec r}{m_1 + m_2} $$ and $$ \vec r_2 = \vec R - \frac{m_1 \vec r}{m_1 + m_2}. $$

Lagrangian
Let's write out the Lagrangian using the new coordinates

$$ L = \frac{1}{2} m_1 \left ( \dot \vec R^2 + 2 \dot \vec R \cdot \dot \vec r \frac{m_2}{m_1+m_2} + \dot \vec r^2 \frac{m_2^2}{\left (m_1+m_2\right )^2} \right ) + $$

$$ \frac{1}{2} m_2 \left ( \dot \vec R^2 - 2 \dot \vec R \cdot \dot \vec r \frac{m_1}{m_1+m_2} + \dot \vec r^2 \frac{m_1^2}{\left (m_1+m_2\right )^2} \right ) - $$

$$ V \left (\vec r \right ) + m_1 \vec g \cdot \left ( \vec R + \frac{m_2 \vec r}{m_1 + m_2} \right ) + m_2 \vec g \cdot \left ( \vec R - \frac{m_1 \vec r}{m_1 + m_2} \right ) $$

Notice that the middle terms in the first two sets of parentheses and the final terms in the final set cancel each other out and we are left with

$$ L = \frac{1}{2} M \dot \vec R^2 + \frac{1}{2} \mu \dot \vec r^2 - V \left (\vec r \right ) + M \vec g \cdot \vec R $$

where $$M=m_1+m_2$$ is the total mass and the reduced mass is $$ \mu = \frac{m_1 m_2}{m_1 + m_2} $$

So we find that for two bodies the dynamics reduces to the motion of the center of mass (possibly in a uniform gravitational field) and the motion of single particle about the center of mass.

Angular Momentum
Let's calculate the total angular momentum of the system. We know from the results on many-particle systems that the angular momentum of the system will separate into the angular momentum about the center of mass and that of the center of mass. Furthermore, if $$V\left ( \vec r \right ) = V \left ( \left | \vec r \right | \right )$$, the internal forces do not contribute to the change in the angular momentum. We have

$$ \vec L = m_1 \vec r_1 \times \vec v_1 + m_2 \vec r_2 \times v_2 $$

$$ = m_1 \left ( \vec R + \frac{m_2 \vec r}{m_1 + m_2} \right ) \times \left ( \vec V + \frac{m_2 \vec v}{m_1 + m_2}\right ) + m_2 \left (  \vec R - \frac{m_1 \vec r}{m_1 + m_2} \right ) \times \left ( \vec V - \frac{m_1 \vec v}{m_1 + m_2}\right ) $$

where $$\vec v=\vec v_1 - \vec v_2$$ and $$\vec V = \left ( m_1 \vec v_1 + m_2 \vec v_2 \right ) / M$$.

$$ \vec L = M \vec R \times \vec V + \left [ m_1 \frac{m_2^2}{\left ( m_1 + m_2 \right )^2} + m_2 \frac{m_1^2}{\left ( m_1 + m_2 \right)^2} \right ] \left ( \vec r \times \vec v \right ) = M \vec R \times \vec V + \mu \vec r \times \vec v. $$

We know from the results on many-particle systems in a uniform gravitational field that the internal angular momentum ($$\mu \vec r \times \vec v$$) of the two-body system is conserved. We could have seen this from separation of the internal and external degrees of freedom in the Lagrangian.

General Results
We can treat the central force between two particles as a force between a fixed center that we take to be the origin and a single particle. We have

$$ L = \frac{1}{2} \mu \dot \vec r^2 - V \left (\left | \vec r \right |\right ), $$

Taking the particular case of the results from the first section where the potential does not depend on direction, we know that the angular momentum of the particle is conserved. This means that the motion will be restricted to a plane that we will take to be the equatorial plane ($$\theta=\pi/2$$) without a loss of generality. Let's write the Lagrangian in spherical coordinates. First let's write the position of the particle in terms of the spherical coordinates. We have

$$ \vec r = \left [ \begin{matrix} r \sin \theta \cos \phi \\ r \sin \theta \sin \phi \\ r \cos \theta \end{matrix} \right ] $$

and

$$ \dot \vec r = \left [ \begin{matrix} \dot r \sin \theta \cos \phi + r \dot \theta \cos \theta \cos \phi - r \dot \phi \sin \theta \sin \phi \\ \dot r \sin \theta \sin \phi + r \dot \theta \cos \theta \sin \phi + r \dot \phi \sin \theta \cos \phi \\ \dot r \cos \theta - r \dot \theta \sin \theta \end{matrix} \right ] $$

so

$$ \dot \vec r^2 = \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta. $$

Lagrangian
Let's write out the Lagrangian in terms of the spherical coordinates

$$ L = \frac{1}{2} \mu \left ( \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta\right ) - V(r). $$

Conserved Quantities
First let's notice that $$\partial L/\partial \phi = 0$$ so $$p_\phi$$ is conserved. Also $$\partial L/\partial t=0$$ so $$H$$ is conserved as well.

Let's calculate the conserved momentum,

$$ p_\phi = \frac{\partial L}{\partial \dot \phi} = \mu r^2 \dot \phi\sin^2 \theta $$

which has units of angular momentum. It is simply the angular momentum about the $$z-$$axis.

Equations of Motion
We can also use Lagrange's equation to get the equations of motion

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot \theta} - \frac{\partial L}{\partial \theta} = 0, $$

$$ \frac{d}{dt} \left ( \mu r^2 \dot \theta \right ) = 2 \mu r \dot r \dot \theta + \mu r^2 \ddot \theta = \mu r^2 \dot \phi^2 \sin \theta \cos \theta. $$

If $$\dot \theta=0$$ and $$\theta=0,\pi/2$$ or $$\pi$$ initial, then $$\ddot \theta=0$$ and $$\theta$$ stays constant. We take $$\theta=\pi/2$$ and $$\dot \theta=0$$ initially, so the particle remains in the equatorial plane as we expected from the conservation of angular momentum.

Now for the radial motion,

$$ \frac{d}{dt} \frac{\partial L}{\partial \dot r} - \frac{\partial L}{\partial r} = 0, $$

$$ \frac{d}{dt} \left ( \mu \dot r \right ) - r \mu \dot \theta^2 - r \mu \dot \phi^2 \sin^2 \theta + \frac{\partial V}{\partial r} = 0. $$

Using the initial conditions we have

$$ \ddot r = -\frac{1}{\mu} \frac{\partial V}{\partial r} + r \dot \phi^2 = -\frac{1}{\mu} \frac{\partial V}{\partial r} + \frac{p_\phi^2}{\mu^2} \frac{1}{r^3}. $$

where the dependence of the acceleration on the angle $$\phi$$ has been included through the conserved quantity $$p_\phi$$, separating the radial equation from the angular equation. The first term is the force between the two bodies and the second term is a 'fictitous' force because of the spherical coordinates. We will solve this differential equation for two special cases, but we can get a good idea of the nature of the motion by looking at the conserved quantities.

Bounded Motion
First, we have

$$ p_\phi = \mu r^2 \dot \phi = \mu dA/dt $$

so the path subtends equal areas in equal times. Given a solution for the radial motion we can determine the angular motion

$$ \dot \phi = \frac{p_\phi}{\mu} \frac{1}{r^2}. $$

Second we have the Hamiltonian,

$$ H = \dot r \frac{\partial L}{\partial \dot r} + \dot \theta \frac{\partial L}{\partial \dot \theta} + \dot \phi \frac{\partial L}{\partial \dot \phi} - L = \frac{1}{2} \mu \left ( \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta\right ) + V(r) $$

Because the Hamiltonian is conserved, let's take its initial value to be

$$ E = \frac{1}{2} \mu \dot r^2 + \frac{1}{2} \frac{p_\phi^2}{\mu} \frac{1}{r^2} + V(r). $$

We can solve this for $$\dot r^2$$ to get

$$ \dot r^2 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{2 V(r)}{\mu} $$

which is very similar to the one-dimensional case but with an added term for the centrifugal force. Unless $$V \sim r^n$$ as $$r\rightarrow 0$$ with $$n\leq -2$$ and $$V$$ is negative, motion of the the particle is bounded if $$p_\phi \neq 0$$. This point of closest approach is called the pericenter. Depending on the potential and the initial conditions, the motion may be bounded at large $$r$$ as well. In particular if the potential is attractive and positive there is always a point of maximum distance. The point of maximum radius is called the apocenter.

If the motion is bounded at both large and small radius, we can calculate the time for the particle to go from the minimum to maximum radius -- this is one half of the period of the complete radial motion as we did for the one-dimensional pendulum. The angular motion as we shall find can have a different (and not usually comeasureable) period. We have

$$ \frac{d r}{d t} = \sqrt { \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{2V(r)}{\mu} } $$

and rearranging we have

$$ P = 2 \int_{r_{\rm min}}^{r_{\rm max}} \frac{dr}{ \sqrt { \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{2 V(r)}{\mu} } } $$

Central Force Solution
For a spherical harmonic oscillator, $$ V = k r^2/2. $$

Bounds of Motion
Let's first calculate the bounds of the motion we have

$$ 0 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{k r^2}{\mu} $$

which yields a quadratic equation for $$r^2$$,

$$ r^4 - \frac{2 E}{k} r^2 + \frac{p_\phi^2}{\mu k} = 0 $$

which yields the solutions

$$ r_{\pm}^2 = \frac{E}{k} \pm \sqrt{\frac{E^2}{k^2} - \frac{p_\phi^2}{\mu k}} $$

Radial Motion
We will first use the conservation of energy to calculate the radial motion, we have

$$ t_1 - t_0 = \int_{r_0}^{r_1} \frac{dr}{\sqrt{\frac{2E}{\mu}-\frac{k}{\mu} r^2 - \frac{p_\phi^2}{r^2 \mu^2} }} = \sqrt{\frac{\mu}{k}} \int_{r_0}^{r_1} \frac{r dr}{\sqrt{\left (r_+^2-r^2\right )\left (r^2-r_-^2\right ) }} $$

$$ = \frac{1}{2} \sqrt{\frac{\mu}{k}} \int_{r_0^2}^{r_1^2} \frac{d \left ( r^2 \right )}{\sqrt{\left (r_+^2-r^2\right )\left (r^2-r_-^2\right ) }}. $$

To perform this integral let's make the substitution

$$ s=\frac{2 r^2 - \left (r_+^2 + r_-^2 \right )}{r_+^2 - r_-^2} $$

to give

$$ t_1 - t_0 = \frac{1}{2} \sqrt{\frac{\mu}{k}} \int_{r=r_0}^{r=r_1} \frac{d s}{\sqrt{1-s^2}} = \left. \frac{1}{2} \sqrt{\frac{\mu}{k}} \arcsin s \right |_{r=r_0}^{r=r_1} $$

so

$$ r^2 = \frac{r_+^2 + r_-^2}{2} + \frac{r_+^2-r_-^2}{2} \sin \left [ 2 \sqrt{\frac{k}{\mu}} \left ( t - t_0 \right ) + \psi_0 \right ] $$

$$ = \frac{E}{k} + \sqrt{\frac{E^2}{k^2} - \frac{p_\phi^2}{\mu k}} \sin \left [ 2 \sqrt{\frac{k}{\mu}} \left ( t - t_0 \right ) + \psi_0 \right ] $$

with

$$ \psi_0 = \arcsin \left [ \frac{2 r_0^2 - \left (r_+^2 + r_-^2 \right )}{r_+^2 - r_-^2} \right ] = \arcsin \left [ \frac{r_0^2 - \frac{E}{k}}{\sqrt{\frac{E^2}{k^2} - \frac{p_\phi^2}{\mu k}}} \right ] $$

Angular Motion
What remains is to solve for the motion in the $$\phi-$$direction. Using the conservation of angular momentum we have

$$ \frac{\partial \phi}{\partial t} = \frac{p_\phi}{\mu} \frac{1}{r^2} = \frac{p_\phi}{\mu} \frac{1}{\frac{r_+^2 + r_-^2}{2} + \frac{r_+^2-r_-^2}{2} \sin \left [ 2 \sqrt{\frac{k}{\mu}} \left ( t - t_0 \right ) + \psi_0 \right ]}. $$

We can integrate this directly to yield

$$ \phi = \frac{p_\phi}{\sqrt{k\mu} r_+ r_-} \arctan \left \{ \frac{1}{2 r_+ r_-} \left [ \left (r_+^2 + r_-^2 \right ) \tan \left ( \sqrt{\frac{k}{\mu}} \left ( t-t_0 \right ) + \psi_0 \right ) + r_+^2 -r_-^2 \right ] \right \}. $$

Notice that the angular frequency is $$\sqrt{k/m}$$, one-half of the radial frequency.

If we substitute the values of $$r_+$$ and $$r_-$$ we get

$$ \phi = \arctan \left \{ \sqrt{\frac{\mu E^2}{p_\phi^2 k}} \left [  \tan \left ( \sqrt{\frac{k}{\mu}} \left ( t-t_0 \right ) + \psi_0 \right ) + \sqrt{ 1 - \frac{p_\phi^2 k}{\mu E^2}} \right ] \right \}. $$

It is possible to write $$r$$ as a function of $$\phi$$ and find that it is the equation for an ellipse centered at the origin. This is left as an exercise for the reader.

The Shape of the Orbit
Let's step back and try to determine $$r(\phi)$$ instead of $$r(t)$$. First we define $$u=1/r^2$$ so we have

$$ \frac{du}{d\phi} = -\frac{2}{r^3} \frac{dr}{d\phi} $$

and using the chain rule we get

$$ \dot r = \frac{dr}{d\phi} \dot \phi = -\frac{r^3}{2} \dot \phi \frac{du}{d\phi} = - \frac{r}{2} \frac{p_\phi}{\mu} \frac{du}{d\phi}. $$

We can substitute this into the energy equation to get

$$ \dot r^2 = \frac{r^2}{4} \frac{p_\phi^2}{\mu^2} \left ( \frac{du}{d\phi} \right )^2 = \frac{1}{4 u} \frac{p_\phi^2}{\mu^2} \left ( \frac{du}{d\phi} \right )^2 =\frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} u - \frac{k}{\mu u} $$

and rearranging

$$ \left ( \frac{du}{d\phi} \right )^2 + 4 u^2 - \frac{8 E \mu}{p_\phi^2} u = -\frac{4 k \mu}{p_\phi^2}. $$

We add a constant to both sides of the equation to complete the square

$$ \left [ \frac{d}{d\phi} \left ( u - \frac{E \mu}{p_\phi^2} \right ) \right ]^2 + 4 \left ( u - \frac{ E \mu}{p_\phi^2} \right )^2  = -\frac{4 k \mu}{p_\phi^2} + 4 \left ( \frac{E \mu}{p_\phi^2} \right )^2 $$

We find that the derivative of a function squared plus four times the function itself yields a constant, so the function must be a sine or cosine. We have $$ u - \frac{E \mu}{p_\phi^2} = \left [ \left ( \frac{E \mu}{p_\phi^2} \right )^2  - \frac{ k \mu}{p_\phi^2} \right ]^{1/2} \cos  2 \left  ( \phi - \phi_0 \right ). $$

We can write this in terms of the radius

$$ r^2 =\frac{p_\phi^2}{E\mu} \frac{1}{ \left ( 1 - \frac{k p_\phi^2}{E^2 \mu} \right )^{1/2} \cos  2 \left  ( \phi - \phi_0 \right ) + 1}. $$

This is the equation for an ellipse centered on the origin. The constant in front of the cosine is related to the shape of the ellipse

$$ \left ( 1 - \frac{k p_\phi^2}{E^2 \mu} \right )^{1/2} = \frac{e^2}{2-e^2} $$

where $$e^2=1-b^2/a^2$$ is the square of the eccentricity of the ellipse ($$a$$ and $$b$$ are the semimajor and semiminor axes). If and only if the spring constant is negative (repulsive spring), then the eccentricity can be greater than one (up to $$\sqrt{2}$$ and the curve is a hyperbola centered on the origin. An interesting case is $$k=0$$ so $$e=1$$.  This gives a straight line passing a distance $$p_\phi/\sqrt(2E\mu)$$ from the origin.

Cartesian Solution
We can solve the spherical harmonic oscillator in Cartesian coordinates as well. We have for the Lagrangian

$$ L = \frac{1}{2} \mu \left ( \dot x^2 + \dot y^2 + \dot z^2 \right ) - \frac{1}{2} k \left ( x^2 + y^2 + z^2 \right ) $$

which yields the following equation of motion for the $$x-$$coordinate

$$ \mu \ddot x + k x = 0 $$

and similarly for the other directions. We find the following solution for the motion in the $$x-$$direction

$$ x(t) = A \cos \left ( \omega t + a \right ) $$

with $$\omega^2=k/\mu$$ and similarly for the other coordinates. Since the value of $$\omega$$ is the same for all three directions we find that the figure is an ellipse. We can also calculate the value of $$r^2$$

$$ r^2 = A^2 \cos^2 \left ( \omega t + a \right ) + B^2 \cos^2 \left ( \omega t + b \right ) $$

where for simplicity and without loss of generality we have restricted the motion to the $$x-y-$$plane. We have

$$ r^2 = \frac{A^2}{2} \left [ 1 + \cos \left (2 \omega t + a \right ) \right ] + \frac{B^2}{2} \left [ 1 + \cos \left (2 \omega t + b \right ) \right ] $$

$$ = \frac{A^2+B^2}{2} + \frac{A^2}{2} \cos \left (2 \omega t + a \right ) + \frac{B^2}{2} \cos \left (2 \omega t + b \right ) $$

We can rewrite these two cosines as a single cosine,

$$ r^2 = \frac{A^2+B^2}{2} + \frac{1}{2} \sqrt{ A^4 + B^2 + 2 A^2 B^2 \cos \left (a - b\right ) } \cos \left [ 2\omega t + \psi_0 \right ] $$

where

$$ \psi_0 = \arccos \frac{A^2 \cos a + B^2 \cos b}{\sqrt{ A^4 + B^2 + 2 A^2 B^2 \cos \left (a - b\right ) }} $$

which has the same form as the result for the spherical analysis.

Gravity
For gravity we have

$$ V = - \frac{G m_1 m_2}{r} = - \frac{G \mu M}{r} = $$

yielding the following Lagrangian,

$$ L = \frac{1}{2} \mu \left ( \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta\right ) + \frac{G \mu M}{r}. $$

Bounds of the Motion
Let's first calculate the bounds of the motion we have

$$ 0 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} + \frac{2 G M}{r} $$

which yields a quadratic equation for $$r$$

$$ r^2 + \frac{G M \mu}{E} r - \frac{p_\phi^2}{E \mu} = 0 $$

which yields the solutions

$$ r_{\pm} = -\frac{G M\mu}{2 E} \pm \sqrt{ \left (\frac{G M \mu}{2 E}\right)^2 + \frac{p_\phi^2}{E \mu}}. $$

For the case of an attractive force $$G>0$$ we find that if $$E>0$$, only $$r_+$$ is greater than zero and this represents the point of closest approach. On the other hand if $$E<0$$ we have $$r_-< r < r_+$$. For a repulsive force $$G<0$$ and again only $$r_+$$ is greater than zero if $$E>0$$. If both $$G$$ and $$E$$ are negative, we cannot find a positive value of $$r$$ with $$\dot r^2>0$$.

Radial Motion
We can also solve for the radial velocity

$$ \dot r^2 =\frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} + \frac{2 G M}{r} = \frac{2 E}{\mu r^2} \left ( r^2 + \frac{G M \mu}{E} r - \frac{p_\phi^2}{E \mu} \right ) = \frac{2 E}{\mu r^2} \left ( r - r_+ \right ) \left ( r - r_- \right ). $$

Yielding the following equation for the time evolution

$$ t_1 - t_0 = \sqrt {\frac{\mu}{2E} } \int_{r_0}^{r_1} \frac{r}{\sqrt{\left ( r - r_+ \right ) \left ( r - r_- \right )} }d r $$

and let's make the substitution

$$ s=\frac{2 r - \left (r_+ + r_- \right )}{r_+ - r_-} $$

to give the following integral

$$ t_1 - t_0 = \sqrt {-\frac{\mu}{2E} } \frac{r_+-r_-}{2} \int_{r=r_0}^{r=r_1} \frac{s}{\sqrt{1-s^2}} + \frac{r_+ + r_-}{\sqrt{1-s^2}} d s $$

where we have assumed that $$E<0$$. This gives

$$ t_1 - t_0 = \left. \sqrt {-\frac{\mu}{2E} } \frac{r_+-r_-}{2} \left [ \left ( r_+ + r_- \right ) \arcsin s - \sqrt{1-s^2} \right ]  \right |_{r=r_0}^{r=r_1} $$

and for $$E>0$$ we have

$$ t_1 - t_0 = \left. \sqrt {\frac{\mu}{2E} } \frac{r_+-r_-}{2} \left [ \left ( r_+ + r_- \right ) {\rm arcsinh} s - \sqrt{s^2-1} \right ]  \right |_{r=r_0}^{r=r_1} $$

Unfortunately we cannot invert this to find the radius as a function of time, but we can determine, the time to go from the minimum to the maximum radius in the $$E<0$$ case

$$ P = \pi \sqrt{-\frac{\mu}{2E}} \left ( r_+^2 - r_-^2 \right ) = \pi \sqrt{-\frac{\mu}{2E}} \frac{G M \mu}{-E} \sqrt{\left(\frac{G M \mu}{E}\right)^2 + \frac{4 p_\phi^2}{E \mu}} $$

The Shape of the Orbit
Let's step back and try to determine $$r(\phi)$$ instead of $$r(t)$$. First we define $$u=1/r$$ so we have

$$ \frac{du}{d\phi} = -\frac{1}{r^2} \frac{dr}{d\phi} $$

and using the chain rule we get

$$ \dot r = \frac{dr}{d\phi} \dot \phi = -r^2 \dot \theta \frac{du}{d\theta} = - \frac{p_\phi}{\mu} \frac{du}{d\phi}. $$

We can substitute this into the energy equation to get

$$ \dot r^2 = \frac{p_\phi^2}{\mu^2} \left ( \frac{du}{d\phi} \right )^2 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} u^2 + 2 G M u $$

and rearranging

$$ \left ( \frac{du}{d\phi} \right )^2 + u^2 - \frac{2 G M \mu^2 u}{p_\phi^2} = \frac{2E\mu}{p_\phi^2} $$

Let's define the length $$l=p_\phi^2/(G M \mu^2)$$ and simplify the equation

$$ \left ( \frac{du}{d\phi} \right )^2 + u^2 - \frac{2}{l} u = \frac{2E}{G M \mu l}. $$

Let's add $$1/l^2$$ to both sides of the equation and get

$$ \left [ \frac{d}{d\phi} \left( u - \frac{1}{l} \right ) \right ]^2 + \left ( u - \frac{1}{l} \right )^2 = \frac{2E}{G M \mu l} + \frac{1}{l^2}. $$

We have a function squared plus its first derivative squared being a constant. The function is clearly a trigonometric function we have

$$ u = \frac{1}{r} = \frac{e}{l} \cos \left ( \phi- \phi_0 \right ) + \frac{1}{l} $$

where

$$ e^2 = \frac{2E l}{G M \mu} + 1. $$

Rearranging the equation gives

$$ r \left [ e \cos \left ( \phi - \phi_0 \right ) + 1 \right ] = l $$

which is the polar equation for an ellipse with its focus at the origin for $$01$$ the curve is a hyperbola. $$e=1$$ is a parabola and $$e=0$$ is a circle. (Contrast this with the spherical harmonic oscillator where the center of the ellipse was at the origin).

If we have a repulsive force (e.g. the force between two positive charges) $$e \geq 1$$ because $$E \geq 0$$. The resulting curve is a hyperbola.

The figure shows the various conic sections as a function of the value $$e$$, the eccentricity. The blue parabola has $$e=1$$, the aqua ellipse has $$e=0.5$$ and the green circle has $$e=0$$. The red hyperbola has $$e=2$$. All of the curves have a focus at the origin and $$l=1$$. The semi-latus rectum is the official name for the quantity $$l$$ which sets the size of the curve. In all cases the curve intersects the $$y-$$axis at $$y=\pm 1$$. In a coordinate free way, the semi-latus rectum is the distance between the curve and the focus in the direction perpendicular to the point of closest approach.

What is the meaning of the two branches of the hyperbola? The left branch is for an attactrive force and the right branch is for a repulsive force.


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Advanced Classical Mechanics/Conservative Systems

We have found that when the forces can be described by a potential energy function it is straightforward to use the Lagrangian techniques to find the equations of motion. Given a set of force, how can we determine the potential function? How can we determine whether there is a potential function?

In conservative forces always mechanical energy is conserved ie, Kinetic energy + Potential energy remains constant at every point.For example, spring force gravitational force and electro static force.

Some Examples
Let's do an example,

$$ F_x = - \frac{k x}{\left ( x^2 + y^2 \right )^{3/2}}, $$ $$ F_y = - \frac{k y}{\left ( x^2 + y^2 \right )^{3/2}}. $$

How much work does the force do moving from $$(x_0,y_0)$$ to $$(x,y)$$?

$$ W = \int_{(x_0,y_0)}^{(x,y)} \left( F_x dx + F_y dy \right ) = \int_{(x_0,y_0)}^{(x,y)} \left( -\frac{k x}{\left ( x^2 + y^2 \right )^{3/2}} dx - \frac{k y}{\left ( x^2 + y^2 \right )^{3/2}} dy \right ) $$

If we look at the two terms, let's try to integrate each one with respect to $$x$$ and $$y$$.

$$ F_x = \frac{\partial}{\partial x} \left [ \frac{k}{\left (x^2 + y^2 \right)^{1/2}} \right ] $$ and $$ F_y = \frac{\partial}{\partial y} \left [ \frac{k}{\left (x^2 + y^2 \right)^{1/2}} \right ]. $$

They are both derivatives of the same function

$$ \frac{k}{\left (x^2 + y^2 \right)^{1/2}}. $$

Let's calculate the total derivative of the function

$$ d \left [ \frac{k}{\left (x^2 + y^2 \right)^{1/2}} \right ] = -\frac{k x}{\left ( x^2 + y^2 \right )^{3/2}} dx - \frac{k y}{\left ( x^2 + y^2 \right )^{3/2}} dy $$

and substitute it back into the integral

$$ W = \int_{(x_0,y_0)}^{(x,y)} \left( F_x dx + F_y dy \right ) = \int_{(x_0,y_0)}^{(x,y)} d \left [ \frac{k}{\left (x^2 + y^2 \right)^{1/2}} \right ] = \frac{k}{\left (x^2 + y^2 \right)^{1/2}} - \frac{k}{\left (x_0^2 + y_0^2 \right)^{1/2}} $$

The integral does not depend on path. We could write

$$ F_x = -\frac{\partial V}{\partial x} $$ and $$ F_y = -\frac{\partial V}{\partial y} $$ with $$ V = -\frac{k}{\left (x^2 + y^2 \right)^{1/2}}. $$

What about $$ F_x = 3 B x^2 y^2 $$ and $$ F_y = 2 B x^3 y $$?

What about $$ F_x = a x y $$ and $$ F_y = b x y $$?

 * If the forces are such that the work done when the system is moved from one configuration to another only depends on the initial and final   configurations, then the force is conservative.

Dry Friction
The force of friction between two dry surfaces is proportional to the normal force between the surfaces in a direction that opposes the relative motion between the surfaces. For an object resting on a surface we have $$F=\mu m g$$ that we can resolve into components as

$$ F_x = -\mu m g \frac{dx}{ds} $$

and

$$ F_y = -\mu m g \frac{dy}{ds}. $$

Let's write out the work done by the force,

$$ W = \int \left ( F_x dx + F_y dy \right ) = -\mu m g \int \left ( \frac{dx}{ds} dx + \frac{dy}{ds} dy \right ) $$

$$ = -\mu m g  \int \left ( \frac{dx}{ds} \frac{dx}{ds} ds + \frac{dy}{ds} \frac{dy}{ds} \right ) = -\mu m g \int ds = -\mu m g s $$

where $$s$$ is the length of the path. In particular, the work done by the frictional path does not necessarily vanish over a closed path.

The Potential Energy
In general,

$$ V = - \int_{x_i^0,y_i^0,z_i^0}^{x_i,y_i,z_i} \sum_{i=1}^p \left ( F_{x_i} d x_i + F_{y_i} d y_i +F_{z_i} d z_i \right ) $$

where $$i$$ counts over the various particles. Specifically, $$F_{x_i}$$ is the force on particle $$i$$ in the $$x-$$direction.

For this to be true,

$$ F_{x_i} = -\frac{\partial V}{\partial x_i}, $$ $$ F_{y_i} = -\frac{\partial V}{\partial y_i}, $$ $$ F_{z_i} = -\frac{\partial V}{\partial z_i}. $$

So let's calculate for example

$$ \frac{\partial F_{x_3}}{\partial y_4} = - \frac{\partial^2 V}{\partial y_4 \partial x_3} = \frac{\partial F_{y_3}}{\partial x_3}. $$

If the forces are conservative, then the partial derivatives of the forces are equal to each other. This is a quick way to check whether the forces are indeed conservative!

We can make some observativions from this result. First, if the force varies only along the direction that it points, then the force is conservative. An important example of this is the central force. Second, if and only if the curl of the force is zero, the force is conservative.

Why calculate V?
It takes quite a bit of effort to calculate the potential energy. Why don't we just use the forces directly?

 * 1) You can use the transformation rules to find the generalized forces $$F_{q_r} = \sum_i F_{x_i} \frac{\partial x_i}{\partial q_r}$$ but this is really ugly and could be even more effort than finding the potential energy.
 * 2) Calculate $$V$$ in whatever coordinates are convenient and then transform the coordinates so $$F_{q_r} = -\frac{\partial V}{\partial q_r}$$.
 * 3) Sometimes you can get $$V$$ by inspection.
 * 4) The generalized forces for $$V$$ and $$V+C$$ are the same so you can drop any constant term.  For example, we can write $$V=mgr\left (1- \cos\theta\right), mg \left (l - r\cos\theta\right ), -mgr \cos \theta$$ for the potential energy of a pendulum.

Do conservative systems conserve energy?
Conservative systems are those which you can write the force as the gradient for a potential function. We have seen forces where this is not the case: dry friction and a particle in magnetic field. In the first case we have a non-conservative force that does not conserve energy. In the second case, we have a non-conservative force that does conserve energy.

It is straightforward to come up with an example for a conservative force that does not conserve energy. Specifically we could have $$ \vec F = -\vec \nabla V \left (\vec r, t\right ). $$ Because the Lagrangian now depends on time explicitly, the Hamiltonian is no longer conserved.

Jiggly Pendulum

Let's write down the Cartesian coordinates of the bob of mass $$m$$,

$$ \vec x = \left [ l \sin\theta,  z(t)-l \cos\theta   \right ]. $$

Taking the time derivative of the position gives the velocity

$$ \vec v = \dot \vec x = \left [ l \dot \theta \cos\theta,  \dot z(t)+l \dot \theta \sin\theta   \right ]. $$

Using these results we can calculate the kinetic energy of the bob

$$ T = \frac{1}{2} m v^2 = \frac{m}{2} \left [ l^2 \dot \theta^2 \cos^2 \theta + \left ( \dot z(t) + l \dot \theta \sin\theta \right )^2 \right ] $$

$$ = \frac{m}{2} \left ( l^2 \dot \theta^2 + \dot z(t)^2 + 2 l \dot\theta \dot z(t) \sin \theta \right ) $$

and the potential energy

$$ V = m g \left ( z(t) - l \cos \theta \right ). $$

For this system we have $$\partial L/\partial t \neq 0$$ so $$d H/d t \neq 0$$; however, we can define a potential energy so the system is conservative.

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text:
Advanced Classical Mechanics/Constraints and Lagrange's Equations

There is more to classical mechanics than $$F=ma$$. Often the motion of a system is constrained in some way.

What are constraints?

 * The particles could be restricted to travel along a curve or surface. Specifically one could have some function of the coordinates of each particle and time vanish. These restrictions are either kinematical or geometrical in nature.

$$ f\left (\vec r_1,\vec r_2, \vec r_3, \cdots, t \right ) = 0 $$

This is called a holonomic constraint. For example we could have

$$ \left ( \vec r_i - \vec r_j \right )^2 - c_{ij}^2 = 0 $$

which expresses that the distances between two particles that make up a rigid body are fixed.

$$r^2 - a^2 \geq 0$$ for a particle travelling outside the surface of a sphere or constraints that depend on velocities as well,
 * There are non-holonomic constraints.  For example, one could have

$$ f\left (\vec r_1,\vec r_2, \vec r_3, \cdots, \vec v_1,\vec v_2, \vec v_3, \cdots , t \right ) = 0. $$

A familar example of the latter is a ball rolling on a surface.

We will be dealing exclusively with holonomic constraints in this course.

What are the consequences?

 * The coordinates are no longer independent. They are related through the equations of constraint.
 * The force of constraint are not given so they must be determined from the solution (if you actually want them at all).
 * If the constraints are holonomic, the equations of constraint can be used to eliminate some of the coordinates to get a set of generalized independent coordinates.

These generalized coordinates usually will not fall into pairs or triples that transform as vectors, e.g. the natural coordinates for motion restricted to a sphere are spherical coordinates ($$\theta,\phi$$).

D'Alembert's Principle
D'Alembert's principle relies on the concept of virtual displacements. The idea is that you can imagine freezing the system in time and jiggling each of the particles in a way consistent with the various constraints at that particular time and determine the work (virtual work) needed to perform these virtual displacements.

We will denote the virtual displacement of a particle as $$\delta \vec r_i$$. The virtual displacement must be consistent with the constraints.

For each particle we have

$$ \vec F_i = \dot \vec p_i ~\rm{so}~ \vec F - \dot \vec p_i= 0 $$

and summing over the particles we have

$$ \sum_i \left ( \vec F_i - \dot \vec p_i \right ) \cdot \delta \vec r_i = 0. $$

Let's divide the force on each particle into applied forces and constraints

$$ \vec F_i = \vec F_i^{(a)} + \vec f_i $$

so we have

$$ \sum_i \left ( \vec F_i - \dot \vec p_i \right ) \cdot \delta \vec r_i = \sum_i \left ( \vec F_i^{(a)} - \dot \vec p_i \right ) \cdot \delta \vec r_i + \sum_i \vec f_i \cdot \delta \vec r_i= 0. $$

If we assume that the forces of constraint do no virtual work, then the last term vanishes. This is a reasonable assumption because the force of constraint to restrict a particle to a surface is normal to that surface but a displacement consistent with the constraints is tangent to the surface so the dot product will vanish and the force of constraint performs no virtual work. If the constraints are a function of time, the forces of constraint can perform work on the particles but the whole idea of D'Alembert's principle is that we have frozen time.

This leaves us with

$$ \sum_i \left ( \vec F_i^{(a)} - \dot \vec p_i \right ) \cdot \delta \vec r_i = 0, $$

D'Alembert's principle. The forces of constraints are gone! However, the coordinates are not independent, so the equation is only a statement about the sum of the various forces, momenta and virtual displacements.

Generalized Coordinates
We can try to find a set of independent coordinates given the constraints. Let's write

$$ \vec r_i = \vec r_i \left ( q_1, q_2, q_3, \cdots, t \right ) $$

and so

$$ \vec v_i = \frac{d \vec r_i}{d t} = \sum_k \frac{\partial \vec r_i}{\partial q_k} \dot q_k + \frac{\partial \vec r_i}{\partial t} $$

and a virtual displacement is

$$ \delta \vec r_i = \sum_j \frac{\partial \vec r_i}{\partial q_j} \delta q_j. $$

Notice that there is no $$dt$$ in the virtual displacement because time is held fixed. Now let's look at the first term in D'Alembert's equation

$$ \sum_i \vec F_i \cdot \delta \vec r_i = \sum_i \vec F_i \cdot  \left ( \sum_j \frac{\partial \vec r_i}{\partial q_j} \delta q_j \right ) =  \sum_j \left ( \sum_i \vec F_i \cdot  \frac{\partial \vec r_i}{\partial q_j} \right )  \delta q_j = \sum_j Q_j \delta q_j $$

where $$Q_j$$ is called a generalized force.

The second term takes a bit more work. We have

$$ \sum_i \dot \vec p_i \cdot \delta \vec r_i = \sum_i m_i \ddot \vec r_i  \cdot \delta \vec r_i = \sum_i m_i \ddot \vec r_i \left ( \sum_j \frac{\partial \vec r_i}{\partial q_j} \delta q_j \right ). $$

Let's focus on a particular one of the generalized coordinates j. We have

$$ \sum_i m_i \ddot \vec r_i \cdot \frac{\partial \vec r_i}{\partial q_j} = \sum_i \left [ \frac{d}{dt} \left ( m_i \dot \vec r_i \cdot \frac{\partial \vec r_i}{\partial q_j} \right ) - m_i \dot \vec r_i \cdot \frac{d}{dt} \left ( \frac{\partial \vec r_i}{\partial q_j} \right ) \right ] $$

$$= \sum_i \left [ \frac{d}{dt} \left ( m_i \vec v_i \cdot \frac{\partial \vec r_i}{\partial q_j} \right ) - m_i \dot \vec r_i \cdot \frac{\partial \vec v_i}{\partial q_j}  \right ] $$

We can use the fact that

$$ \frac{\partial \vec v_i}{\partial \dot q_j} = \frac{\partial \vec r_i}{\partial q_k} $$

from the defintion of $$\vec v_i$$ to get

$$= \sum_i \left [ \frac{d}{dt} \left ( m_i \vec v_i \cdot \frac{\partial \vec v_i}{\partial \dot q_j} \right ) - m_i \dot \vec r_i \cdot \frac{\partial \vec v_i}{\partial q_j}  \right ] $$

$$= \frac{d}{dt} \left [ \frac{\partial}{\partial \dot q_j} \sum_i \left ( \frac{1}{2} m_i v_i^2 \right ) \right ] - \frac{\partial}{\partial q_j} \left [ \sum_i \left ( \frac{1}{2} m_i v_i^2 \right ) \right ]. $$

Notice that the quantity in the innermost parenthesis is just the total kinetic energy of the system $$T$$ so we have

$$ \sum_j \left \{ \left [ \frac{d}{dt} \left ( \frac{\partial T}{\partial \dot q_j} \right ) - \frac{\partial T}{\partial q_j} \right ] - Q_j \right \} \delta q_j = 0. $$

Lagrange's Equations
If the constraints are holonomic, we can pick the $$q_j$$ to be independent so the various $$\delta q_j$$ are completely arbitrary and the quantity in braces must vanish to yield

$$ \left [ \frac{d}{dt} \left ( \frac{\partial T}{\partial \dot q_j} \right ) - \frac{\partial T}{\partial q_j} \right ] = Q_j. $$

These expressions are sometimes called Lagrange's equations, but the term Lagrange's equations is often reserved for the case of a conservative force. In this case we have

$$ \vec F_i = - \vec \nabla_i V $$ so

$$ Q_j = \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j} = - \sum_i \vec \nabla_i V\cdot \frac{\partial \vec r_i}{\partial q_j} = - \frac{\partial V}{\partial q_j} $$

In this case we can rearrange the equation to give

$$ \left [ \frac{d}{dt} \left ( \frac{\partial T}{\partial \dot q_j} \right ) - \frac{\partial (T-V)}{\partial q_j} \right ] = 0. $$

Furthermore, if the forces do not depend explicitly on the velocities we can define, the Lagrangian to be $$ L = T - V $$

and write Lagrange's equations in their traditional form

$$ \left [ \frac{d}{dt} \left ( \frac{\partial L}{\partial \dot q_j} \right ) - \frac{\partial L}{\partial q_j} \right ] = 0. $$

Conserved Quantities
For each coordinate we can define a conjugate momentum to be

$$ p_i = \frac{\partial L}{\partial {\dot q}_i}. $$

This is called the generalized momentum conjugate to the coordinate $$q_i$$. Let's look at Lagrange's equations with this definition

$$ \frac{d p_i}{dt} = \frac{\partial L}{\partial q_i} $$

so if the Lagrangian does not depend on a particular coordinate $$q_i$$, then the momentum conjugate to the coordinate does not change with time; it is conserved. This conserved momentum is called a first integral. The coordinate that doesn't affect the Lagrangian is called a cyclic coordinate.

In general the Lagrangian will depend on the coordinates, velocities and time; what happens if $$\partial L/\partial t$$ vanishes? Is there a conserved quantity similar to the momenta?

Let calculate the total derivative of the Lagrangian with respect to time,

$$ \frac{d L}{d t} = \sum_i \left [ \frac{\partial L}{\partial q_i} {\dot q}_i + \frac{\partial L}{\partial {\dot q}_i} {\ddot q}_i \right ] + \frac{\partial L}{\partial t}. $$

Now let's use the definition of the momenta and the Lagrange's equation to simplify things a bit,

$$ \frac{d L}{d t} = \sum_i \left [ \frac{d p_i}{d t} {\dot q}_i + p_i {\ddot q}_i \right ] + \frac{\partial L}{\partial t}  = \sum_i \frac{d ( p_i {\dot q}_i )}{d t}  + \frac{\partial L}{\partial t} $$

and rearranging

$$ \frac{\partial L}{\partial t} = \frac{d}{dt} \left ( L - \sum_i p_i {\dot q}_i \right ).$$

So if

$$ \partial L/\partial t = 0 $$

then the Hamiltonian,

$$ H = L - \sum_i p_i {\dot q}_i, $$

is conserved.

Lagrangians with Non-Conservative Forces
From the analysis so far it would appear that one can only construct a Lagrangian when the forces that act on a particle are conservative (they can be derived from a potential $$V$$). It is indeed the case that truly dissipative forces such as friction cannot be directly included in a Lagrangian formulation, but forces that can be written in the form $$d/dt (\partial L/\partial {\dot q})$$ may be included in the Lagrangian. Although this seems very restrictive, an important force of this class is the magnetic force on a charged particle.

$$ \vec F = q \left( \vec E + \vec v \times \vec B \right ). $$

In electrostatics the electric field is simply the gradient of the electrostatic potential, but for more general fields we have

$$ \vec E = -\vec \nabla \phi - \frac{\partial \vec A}{\partial t} $$

where $$\vec A$$ is the vector potential. The magnetic field may be written as

$$ \vec B = \vec \nabla \times \vec A. $$

Let's consider the function

$$ V = q \phi \left (\vec r, t\right )- q {\dot \vec r} \cdot \vec A \left (\vec r, t \right ) $$

and calculate

$$ \frac{d}{dt} \left ( \frac{\partial V}{\partial {\dot x}} \right ) - \frac{\partial V}{\partial x} = - q \frac{d A_x}{d t} - q \frac{\partial \phi}{\partial x} + q \left ( \dot x \frac{\partial A_x}{\partial x} + \dot y \frac{\partial A_y}{\partial x} + \dot z \frac{\partial A_z}{\partial x} \right ). $$

The total time derivative of the vector potential generates several terms (due to the chain rule) to yield

$$ \frac{d}{dt} \left ( \frac{\partial V}{\partial {\dot x}} \right ) - \frac{\partial V}{\partial x} = - q \left ( \frac{\partial A_x}{\partial t} + \dot x \frac{\partial A_x}{\partial x} + \dot y \frac{\partial A_x}{\partial y} + \dot z \frac{\partial A_x}{\partial z} \right ) - q \frac{\partial \phi}{\partial x} + q \left ( \dot x \frac{\partial A_x}{\partial x} + \dot y \frac{\partial A_y}{\partial x} + \dot z \frac{\partial A_z}{\partial x} \right ). $$

Notice that the terms proportional to $$\dot x$$ cancel leaving

$$ \frac{d}{dt} \left ( \frac{\partial V}{\partial {\dot x}} \right ) - \frac{\partial V}{\partial x} = - q \left ( \frac{\partial \phi}{\partial x} + \frac{\partial A_x}{\partial t} \right ) + q \left [ \dot y \left ( \frac{\partial A_x}{\partial y} - \frac{\partial A_y}{\partial x} \right ) + \dot z \left ( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right )   \right ]. $$

The first term is simply the charge of the particle times the x-component of the electric field. The second term is the charge of the particles times the x-component of $$\vec v \times \vec B$$, so the following Lagrangian will yield the equations of motion

$$ L = \frac{1}{2} m {\dot r}^2 + q \vec \dot r \cdot \vec A \left (\vec r, t\right ) - q \phi \left ( \vec r, t \right ). $$

If the vector potential and the scalar potential do not depend on time, then $$\partial L/\partial t = 0$$ and

$$ H = \frac{\partial L}{\partial \dot \vec r} \cdot \dot \vec r - L = \frac{1}{2} m {\dot r}^2 + q \phi \left ( \vec r, t \right ) $$

is conserved. The force is not conservative but the system is.


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Advanced Classical Mechanics/Continuum Mechanics

We can extend the formalism of Lagrangian mechanics to cover the motion of continuous materials: membranes, strings and fields.

Deriving the Lagrange Equations of a Continuum
We saw for a string that the potential and kinetic energy can be written as a integral over the length of the string.

$$ L = \int_0^L {\mathcal L} \left (\phi, \frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial t} \right ) dx $$ where $${\mathcal L}$$ is called the Lagrangian density.

The Action
Let's write out the action

$$ S = \int dx dt {\mathcal L} \left (\phi, \frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial t} \right ). $$

We would like to find the function $$\phi(x,t)$$ that minimizes the action, so we let $$\phi=\phi_0 + \eta$$ and calculate $$ \delta S = \int dx dt \left [ \left .\frac{\partial {\mathcal L}}{\partial \phi} \right |_{\phi_0} \eta + \left .\frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial x}\right )} \right |_{\phi_0} \frac{\partial \eta}{\partial x} + \left. \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial t}\right )} \right |_{\phi_0} \frac{\partial \eta}{\partial t} \right ]. $$

Let integrate the last two items by parts and drop the boundary term because we assume that the value of $$\eta$$ vanishes on the boundary, we get

$$ \delta S = \int dx dt \eta \left [ \left .\frac{\partial {\mathcal L}}{\partial \phi} \right |_{\phi_0} - \frac{d}{dx} \left .\frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial x}\right )} \right |_{\phi_0} - \frac{d}{dt} \left. \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial t}\right )} \right |_{\phi_0} \right ]. $$

Lagrange's Equations
Because the function $$\eta$$ is completely arbitrary then for the value of $$\delta S$$ to vanish, the integrand itself must vanish and we obtain the Lagrange equation for a field in one spatial dimension:

$$ \frac{d}{dx} \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial x}\right )} + \frac{d}{dt} \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial t}\right )} - \frac{\partial {\mathcal L}}{\partial \phi} = 0 $$

that the actual motion must satify.

Each new dimension that the field lives in adds a new term of the form

$$ \frac{d}{dz} \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial \phi}{\partial z}\right )} $$

to the equations of motion.

Electrostatics
Let's do an example of an electric field with a bunch of charges around. The energy density of an electric field with charges around is

$$ ??? \quad \left({\mathcal \phi}\right )^2 + \rho \phi $$

We would like to minimize the integral of $${\mathcal E}$$ over a volume so it is natural to take $$??? {\mathcal L} = $$. We have

$$ {\mathcal L} = \frac{\epsilon_0}{2} \left [ \left ( \frac{\partial \phi}{\partial x} \right )^2 + \left ( \frac{\partial \phi}{\partial y} \right)^2 \right ] + \rho \phi $$

and Lagrange's equations are

$$ ??? \quad \frac{d}{dx} \frac{\partial {\mathcal L} }{\partial \left ( \frac{\partial \phi}{} \right) } + \frac{d}{dy} \frac{\partial {\mathcal L}} {\partial \left ( \frac{\partial}{} \right)} $$

A String
As a problem toward the beginning of the term, you found that the potential energy of a string could be given by

$$ V = \int \frac{1}{2} \lambda \left ( \frac{\partial z}{\partial x} \right )^2 dx $$

where $$\lambda$$ is the string tension.

We can also write the kinetic energy of the string as an integral

$$ T = \int \frac{1}{2} \left ( \frac{\partial z}{\partial t} \right )^2 d m = \int \frac{1}{2} \left ( \frac{\partial z}{\partial t} \right )^2 \mu d x $$

where $$\mu=M/L$$, the mass per unit length of the string. We can combine these two expressions to get the Langrangian density

$$ {\mathcal L} = \frac{1}{2} \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left ( \frac{\partial z}{\partial x} \right )^2 \right ]. $$

Now let's apply Lagrange's equations to this Lagrangian density,

$$ \frac{d}{dx} \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial z}{\partial x}\right )} + \frac{d}{dt} \frac{\partial {\mathcal L}}{\partial \left ( \frac{\partial z}{\partial t}\right )} + - \frac{\partial {\mathcal L}}{\partial z} = 0 $$

$$ \mu \frac{d}{dt} \frac{\partial z}{\partial t} - \lambda \frac{d}{dx} \frac{\partial z}{\partial x} = 0 $$

$$ \frac{\partial^2 z}{\partial t^2} - \frac{\lambda}{\mu} \frac{\partial^2 z}{\partial x^2} = 0 $$

We can find a solution to this equation of the form, $$ z(x,t) = f(x-vt)\left. \right. $$ where $$v=\pm \sqrt{\lambda/\mu}$$. Let's substitute this into the equation to get

$$ v^2 f - \frac{\lambda}{\mu} f = \frac{\lambda}{\mu} f - \frac{\lambda}{\mu} f = 0 $$

This solution is a wave travelling in the positive or negative x-direction. The function f can take any form whatsoever and it always travels at the same speed. This type of wave is called a 'simple wave'.

The Discrete Fourier Transform
When a string vibrates it causes the air around it to vibrate as well; these vibrations travel through the air as waves that we can detect with the acoustic spectrometers attached to the sides of our heads. Our ears are sensitive to the pitch or frequency of the sound that the string produces so it is natural to look at the waves along a string as a sum of waves of a particular frequency, so we have $$ z(x,t) = g(x) \cos \left ( \omega t + \phi \right )\left. \right. $$ where φ is the phase of the wave.

Let's substitute this into the equation of motion to get

$$ -\omega^2 g(x) \cos \left ( \omega t + \phi \right )- \frac{\lambda}{\mu} g'' \left ( \cos\omega t + \phi \right ) = 0 $$ so $$ g'' = -\frac{\omega^2 \mu}{\lambda} g $$ and $$ g = A \sin \left ( \omega \sqrt{\frac{\mu}{\lambda}} x \right ) + B \cos \left ( \omega \sqrt{\frac{\mu}{\lambda}} x \right ) $$

Now we must apply the boundary conditions of the string. Specifically, we have $$z(0,t)=z(L,t)=0$$ so

$$ g = A \sin \left ( \frac{n\pi}{L} x \right ) $$

where n is an integer so we have a formula for the frequency of the wave in terms of the form of the oscillation along the string

$$ \omega = n \frac{\pi}{L} \sqrt{\frac{\lambda}{\mu}} $$

The general solution will allow a sum of different frequencies so

$$ z(x,t) = \sum_n A_n \sin \left ( \frac{n\pi}{L} x \right ) \cos \left ( n \frac{\pi}{L} \sqrt{\frac{\lambda}{\mu}} + \phi_n \right ) $$

Let's suppose that we pluck the string so that a t=0, the string is not straight but it is not moving either, so we have $$\phi_n=0$$ in the formula above, and we get

$$ z(x,0) = \sum_n A_n \sin \left ( \frac{n\pi}{L} x \right ). $$

We would like to find the coefficients $$A_n$$ that tell us what pitches the string will emit when it is released.

Let's take the integral of both sides times $$\sin\left ( \frac{m\pi}{L} x \right )$$ to get

$$ \int_0^L z(x,0) \sin\left ( \frac{m\pi}{L} x \right ) dx = \int_0^L \sum_n A_n \sin \left ( \frac{n\pi}{L} x \right ) \sin\left ( \frac{m\pi}{L} x \right ) dx. $$

$$ = \int_0^L \sum_n \frac{A_n}{2} \left [ \cos \left ( \frac{(n+m)\pi}{L} x \right ) + \cos\left ( \frac{(n-m)\pi}{L} x \right ) \right ] dx = \int_0^L \frac{A_m}{2} d x = \frac{A_m L}{2}. $$

Rearranging we have

$$ A_m = \frac{2}{L} \int_0^L z(x,0) \sin\left ( \frac{m\pi}{L} x \right ) dx $$

which defines the discrete Fourier transform.

A Membrane
As a problem toward the beginning of the term, you found that the potential energy of a string could be given by

$$ V = \int \frac{1}{2} \lambda \left [ \left ( \frac{\partial z}{\partial x} \right  )^2 + \left ( \frac{\partial z}{\partial y} \right  )^2\right ] dx dy $$

where $$\lambda$$ is the surface tension.

We can also write the kinetic energy of the membrane as an integral

$$ T = \int \frac{1}{2} \left ( \frac{\partial z}{\partial t} \right )^2 d m = \int \frac{1}{2} \left ( \frac{\partial z}{\partial t} \right )^2 \mu d x dy $$

where $$\mu=M/A$$, the mass per unit area of the membrane. We can combine these two expressions to get the Langrangian density

$$ {\mathcal L} = \frac{1}{2} \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left [ \left ( \frac{\partial z}{\partial x} \right )^2 + \left ( \frac{\partial z}{\partial y} \right  )^2 \right ] \right ] = \frac{1}{2} \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left ( \nabla z \right )^2 \right ] $$

If we apply Lagrange's equations we obtain,

$$ \mu \frac{d}{dt} \frac{\partial z}{\partial t} - \lambda \frac{d}{dx} \frac{\partial z}{\partial x} - - \lambda \frac{d}{dy} \frac{\partial z}{\partial y}= 0 $$

$$ \mu \frac{d}{dt} \frac{\partial z}{\partial t} - \lambda \nabla^2 z = 0 $$

Different Coordinates
The value of the action does not depend on the coordinates that you use, but as we shall see the Lagrangian density does. First let's write the action using Cartesian coordinates

$$ S = \int dx dy {\mathcal L} = \int dx dy = \frac{1}{2} \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left ( \nabla z \right )^2 \right ]. $$

Sometimes it is useful to use polar coordinates for a circular drum perhaps. Here the action is

$$ S = \int dr d\theta {\mathcal L} = \int dr d\theta r \frac{1}{2} \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left ( \nabla z \right )^2 \right ]. $$

Notice that an r appears in front of the "Cartesian" Lagrangian density. This is just so the units work out, so we find that

$$ {\mathcal L}_{\rm polar} = {\mathcal L}_{\rm Cartesian} \times r $$

In both coordinate systems the action has the same units but the Lagrangian density does not. We also need to rewrite the spatial derivatives in polar coordinates to get

$$ {\mathcal L}_{\rm polar} = \frac{1}{2} r \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left ( \nabla z \right )^2 \right ] = \frac{1}{2} r \left [ \mu \left ( \frac{\partial z}{\partial t} \right )^2 - \lambda \left [ \left ( \frac{\partial z}{\partial r} \right )^2 + \left ( \frac{1}{r} \frac{\partial z}{\partial \theta} \right )^2\right ] \right ] $$

If we apply the Lagrange equations we find that the Lagrange density depends explicitly on the coordinate r. We obtain,

$$ r \mu \frac{d}{dt} \frac{\partial z}{\partial t} - \lambda \frac{d}{dr} \left ( r \frac{\partial z}{\partial r} \right ) - \lambda \frac{1}{r} \frac{d}{d\theta} \frac{\partial z}{\partial \theta}= 0 $$

and divide both sides by r to get

$$ \mu \frac{\partial^2 z}{\partial t^2} - \lambda \left [ \frac{1}{r} \frac{d}{dr} \left ( r \frac{\partial z}{\partial r} \right ) + \frac{1}{r^2} \frac{\partial^2 z}{\partial \theta^2} \right ] = 0 $$

or using the definition of the Laplacian in cylindrical coordinates we get

$$ \mu \frac{\partial^2 z}{\partial t^2} - \lambda \nabla^2 z = 0 $$

Solutions to the Two-Dimensional Wave Equation
When analyzing a drum or something with boundary conditions, it is natural to look for solutions to these equations with the form $$ z(x,y,t) = X(x) Y(y) g(t) $$ in Cartesian coordinates, $$ z(r,\theta,t) = R(r) f(\theta) g(t) $$ in cylindrical coordinates, or $$ z(\theta,\phi,t) = Y(\theta,\phi) g(t). $$ Generally you can pick the function g to vary harmonically in time and the solutions to the spatial dependence are known as harmonics. In Cartesian coordinates we get sines and cosines (the harmonic functions). In cylidrical coordinates, we get sines and cosines in the $$\theta$$-direction and Bessel functions in the radial direction. If your drum is the surface of a sphere, the solutions are called spherical harmonics: sines and cosines in the $$\phi-$$direction and polynomials of $$\cos\theta$$ in the other direction.

A Slung Slinky
As a final example we will study a slinky hanging from a horizontal surface and the waves that can propagate along it. Let's first draw a picture to show what we are talking about

Let's use the variable $$\phi_i$$ to denote the angle that each pendulum makes with the vertical. Also let's restrict ourselves to the small-angle approximation. We can write the kinetic energy of each mass as

$$ T_i = \frac{1}{2} m r^2 {\dot \phi_i}^2 $$

There are two sources of potential energy: gravity and the springs. The expression for gravity is straightforward we have

$$ V_{{\rm gravity},i} = - m g r \left ( 1 - \cos \phi_i \right ) \approx \frac{1}{2} m g r \phi_i^2. $$

The potential energy of the springs is a bit more complicated. First, we have to come up with a convention so that we don't double count the springs. Let's identify a spring with the mass to its left so we have

$$ V_{{\rm spring},i} = \frac{1}{2} k \left \{ r \left[ \left ( \frac{b}{r} + \sin\phi_{i+1} - \sin\phi_i \right )^2 + \left (\cos \phi_{i+1}-\cos\phi_i\right)^2 \right ]^{1/2} - l \right \}^2 $$

Now let's take the small-angle approximation,

$$ V_{{\rm spring},i} \approx \frac{1}{2} k \left ( b + r \phi_{i+1} - r \phi_i - l \right )^2 = \frac{1}{2} k (b-l)^2 + k r (b-l) \left (\phi_{i+1}-\phi_{i}\right ) + \frac{1}{2} k  r^2 \left (\phi_{i+1}-\phi_{i}\right )^2. $$

We can simplify this further. The first term is a constant; let's drop it. The second term sums to zero when one combines the contribution from every spring. The final term is

$$ V_{{\rm spring},i} \approx \frac{1}{2} k  r^2 \left (\phi_{i+1}-\phi_{i}\right )^2 =   \frac{1}{2} k  r^2 l^2 \left (\frac{\partial \phi}{\partial x}\right )^2 $$

where we have replaced $$\phi_i$$ with $$\phi(x_i)$$.

The Lagrange density is given by

$$ {\mathcal L} = \frac{T_i - V_i}{l} = \frac{1}{2} \frac{m}{l} r^2 {\dot \phi}^2 - \frac{1}{2} \frac{m}{l} g r \phi^2 - \frac{1}{2} k  r^2 l \phi'^2 $$

The Klein-Gordon Equation
Let's apply Lagrange's equations to the Lagrange density to get the following equation of motion

$$ \mu {\ddot \phi} - \lambda \phi'' + \mu \frac{g}{r} \phi = 0 $$

where $$\mu=m r^2/l$$ and $$\lambda=k r^2 l$$. This equation of motion is very similar to that of the stretched string but there is an extra term. Let's divide both sides by $$\mu$$ to get

$$ {\ddot \phi} - \frac{\lambda}{\mu} \phi'' + \omega_0^2 \phi = 0 $$

where $$\omega_0^2=g/r$$ is the natural frequency of the hanging masses. This is known as the Klein-Gordon equation, and it describes the evolution of massive scalar field. If we let the spring constant $$k$$ go to zero, then $$\lambda$$ goes to zero as well and we have the simple equation for a pendulum.

$$ {\ddot \phi} + \omega_0^2 \phi = 0 $$

To get an appreciation for the solutions, let's substitute a harmoncally varying wave

$$ \phi = A(x) \cos \omega t $$

to get

$$ -\omega^2 A - \frac{\lambda}{\mu} A'' + \omega_0^2 A = 0 $$ and $$ A'' = - \frac{\mu}{\lambda} \left ( \omega^2 - \omega_0^2 \right ) A $$

If the frequency of the wave $$\omega$$ is greater than the natural frequency of the pendulum, the wave varies sinusoidally in space as well as time (it propagates along the slinky). On the other hand, low-frequency waves are damped, so the slung slinky acts as a high-pass filter.

The form of the equation above encourages us to use solutions of the form

$$ \phi = \cos \left ( k x - \omega t \right ) $$

with the dispersion relation

$$ k^2 = \frac{\mu}{\lambda} \left ( \omega^2 - \omega_0^2 \right ). $$

Phase Velocity
Again we see that when $$\omega<\omega_0$$ things get a bit funny. We can define the phase velocity of the wave

$$ v_{\rm phase} = \frac{\omega}{k} = \sqrt{\frac{\lambda}{\mu}} \left ( 1 - \frac{\omega_0^2}{\omega^2} \right )^{-1/2}. $$

This measures the speed at which the crest of a wave of a particular frequency passes by.

Group Velocity
Let's say that we want to send a signal down the slinky. How long fast will the signal travel? Specifically, we would like to displace the end of the slinky back and forth and watch the compression travel along the spring.

We could naturally think of a signal as the sum of two waves of a single frequency.

$$ A (x, t) = \cos \left [ \left ( k + \Delta k \right ) x - \left ( \omega + \Delta \omega \right ) t \right ] + \cos \left [ \left ( k - \Delta k \right ) x - \left ( \omega - \Delta \omega \right ) t \right ]. $$

Using some trignometric identites we see that this is just an amplitude-modulated wave

$$ A(x, t) = 2 \cos (k x - \omega t ) \cos ( \Delta k x - \Delta \omega t ). $$

The amplitude modulation is a wave that travels on top of the carrier wave at a speed of $$ \Delta \omega/\Delta k \approx d \omega/d k $$ for narrow bandwidth.

The group velocity measures the speed of a "wave packet" or how long a signal takes to travel along the slinky

$$ v_{\rm group} = \frac{d \omega}{d k} = \frac{\lambda}{\mu} \frac{k}{\omega} = \frac{\lambda}{\mu} \frac{1}{v_{\rm phase}} = \sqrt{\frac{\lambda}{\mu}} \left ( 1 - \frac{\omega_0^2}{\omega^2} \right )^{1/2}. $$

If we think about the wave described above in more detail, we have a rapid back-and-forth oscillation that travels at the phase velocity and on top of it a variation in the amplitude of the rapid oscillation. If we ignored the frequency of the pendulums (taking $$g\rightarrow 0$$ or $$r\rightarrow \infty$$), these two variations would travel at the speed $$\sqrt{\lambda/\mu}$$ and one would simply see a pattern travelling at a constant speed along the slinky.

Hanging the slinky introduces dispersion, this means that the pattern generally changes as it travels along the slinky.

The Sine-Gordon or Pendulum Equation
The slung Slinky is related to another important equation of physics. Let's take the exact expression for the gravitational potential energy

$$ V_{{\rm gravity},i} = - m g r \left ( 1 - \cos \phi_i \right ) $$

The new Lagrange density is given by

$$ {\mathcal L} = \frac{T_i - V_i}{l} = \frac{1}{2} \frac{m}{l} r^2 {\dot \phi}^2 - \frac{m}{l} g r \left ( 1 - \cos\phi \right ) - \frac{1}{2} k  r^2 l \phi'^2 $$

and the equation of motion is

$$ \mu {\ddot \phi} - \lambda \phi'' + \mu \frac{g}{r} \sin \phi = 0. $$

This is known as the sine-Gordon equation (those physicists are really good punsters). This equation is no longer linear in the variable $$\phi$$ so it is much more difficult to solve than the Klein-Gordon equation. Specifically, we expect that the velocity of the wave will depend not only on the frequency but also the amplitude.

We can define our scale of time ($$\sqrt{r/g}$$) and distance ($$\sqrt{\mu r/(g\lambda)}$$) so that the equation takes the simpler form

$$ {\ddot \phi} - \phi'' +  \sin \phi = 0. $$

Let's look for solutions that have a fixed pattern travelling at a fixed velocity. This was not possible (except for a monochromatic wave) for the Klein-Gordon equation because of the dispersion. Here we have dispersion and non-linearity. Let's use

$$ \phi = A(x - v t) $$

to get

$$ v^2 A - A + \sin A = 0 $$

so

$$ A'' = \frac{\sin A}{1-v^2} $$

Here the value of velocity depends on the amplitude of the wave, $$v=2$$ for an amplitude of 0.01, $$v=1.91$$ for an amplitude of 1 and 1.64 for an amplitude of 2.

Solitons
The solutions above are strictly periodic, but we can also find waves that travel at velocities less than one that are not periodic (i.e. they are solitary waves or solitons).

The wider waves have amplitudes closer to $$2\pi$$ and velocities closer to zero. The speed of the solitons are 0.1, 0.5, 0.8 and 0.9 from widest to narrowest and the amplitudes range from 6.17 down to 6.11.

The differential equation actually has a simple solution as well. It is called the "kink" solution

$$ \phi = 4 \tan^{-1} \left [ \exp \left ( \frac{x-vt}{\sqrt{1-v^2}} \right ) \right ] $$

The speed at which the kink travels depends on the maximum slope of the kink.

The steepest slope travels at $$v=0.9$$; then we have 0.8 and zero (a frozen kink).


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Advanced Classical Mechanics/Dissipative Forces

Dissipative forces are forces of such nature that energy is lost from a system when motion takes place. Of course energy is in general conserved but it is lost from the degrees of freedom of interest into heat (the random motion of internal degrees of freedom) or radiation (the motion of new particles created by the motion -- light usually).

The force can often be represented by $$ \left |\vec F \right | = a\left ( \vec r, t \right ) \left | \vec v \right |^n. $$ Depending on the value of the index $$n$$ we have different types of dissipative forces.

Frictional Forces (Dry Friction)
The force of kinetic friction is supposed to be proportional to the normal force and independent of area of contact or speed. It is simply a property of the materials in contact. Specifically, we have $$ \left| \vec F \right | = \mu \left | F_{\rm normal} \right | $$ so $$n=0$$.

The frictional force opposes the relative motion of two surfaces.

Viscous Forces
Here the frictional force increases as the first power of the relative speed between the surfaces and opposes the relative motion. Viscous friction is important for wet surfaces at small relative velocties.

High-Velocity Friction
At higher velocities, the force of friction increases as a higher power of the relative velocity. For example, $$n=2$$ gives a good approximation to the dissipative force experiences by objects travelling through fluids at high Reynolds number $$Re = (vL)/\nu$$ where $$\nu$$ is the viscousity of the fluid.

Determining the Generalized Force
The Lagrangian method requires us to find the generalized forces corresponding to the different coordinates that we have used to characterize the degrees of freedom of the motion. There are two general ways to determine the generalized forces.

 * 1) Calculate them in Cartesian coordinates and convert to the generalized coordinates
 * 2) Use the power function

Partial derivatives
For any force that we know in Cartesian (or any other set of coordinates for that matter), we can find the generalized force by using the definition of the generalized coordinates in terms of the coordinates in which the force is known. The definition of the generalized force is $$ Q_j = \sum_i \vec F_i \cdot \frac{\partial \vec r_i}{\partial q_j}. $$

To do this you have to first determine the dissipative force in Cartesian coordinates and the full transformation between the Cartesian coordinates and the generalized coordinates -- it can be pretty painful for even simple problems.

Damped Pendulum
Let's look at a pendulum of length $$l$$. We will use the angle $$\theta$$ and the vertical to describe the single degree of freedom. The position of the bob is given by

$$ x=l \sin \theta, y=-l \cos \theta\, $$

and the velocities are

$$ \dot x= l \cos \theta \dot \theta, \dot y = l \sin \theta \dot \theta \,. $$

The viscous force opposes the direction of motion and it is proportional to the velocity so we have

$$ F_x= - a l \cos \theta \dot \theta, F_y = - a l \sin \theta \dot \theta \,. $$

We also need the partial derivatives for the transformation

$$ \frac{\partial x}{\partial \theta}= l \cos \theta, \frac{\partial y}{\partial \theta} = l \sin \theta \, $$

to calculate the generalized force

$$ F_\theta = F_x \frac{\partial x}{\partial \theta} + F_y \frac{\partial y}{\partial \theta} = - a l^2 \cos^2 \theta \dot \theta - a l^2 \sin^2 \theta \dot \theta = - a l^2 \dot \theta $$

Power function
The great advantage of using the potential instead of the generalized forces directly was that the definition of the partial derivative took care of all of the heavy lifting involved in going from one system of coordinates to another. It turns out that one can use an analogous quantity called the power function to do the same thing (sometimes this is called the dissipation function). Like for the potential the power function for various simple forms of forces can simply be written down and used.

In analogy with the potential we define the power function such that force on particle $$i$$ in the $$x$$-direction is

$$ F_{i,x} = \frac{\partial P}{\partial {\dot x}_i} $$

and similarly for the other directions and particles. Of course, not all forces can be written in this way but many dissipative forces can. Let's write out the generalized force using the standard formula

$$ Q_j = \sum_i \vec F_i \frac{\partial \vec r_i}{\partial q_j} = \sum_i \vec F_i \frac{\partial v_i}{\partial \dot q_j} = \sum_i \frac{\partial P}{\partial \vec v_i} \frac{\partial v_i}{\partial \dot q_j} = \frac{\partial P}{\partial \dot q_j} $$

where we used the equality $$ \partial \vec r_i/\partial q_j = \partial \vec v_i/\partial \dot q_j $$ in the first step, definition of the power function in the second step, and the definition of the partial derivative in the final step.

Damped Pendulum
Let's look at the pendulum again. The viscous force is

$$ \vec F = -a \vec v $$

so the power function is

$$ P = -\frac{1}{2} a v^2 = -\frac{1}{2} a \left ( l \dot \theta \right )^2 $$

and

$$ F_\theta = \frac{\partial P}{\partial \dot \theta} = -a l^2 \dot \theta. $$

Let's combine the results for the power function with the Lagrangian for the pendulum, we have

$$ L = T - V = \frac{1}{2} m l^2 \dot \theta^2 + g m l \cos \theta $$

and Lagrange's equations including the power function are

$$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot \theta}\right) - \frac{\partial L}{\partial \theta} = Q_{{\rm NC},\theta} = \frac{\partial P}{\partial \dot \theta} $$

which yields

$$ m l^2 \ddot \theta + g m l \sin \theta = -a l^2 \dot \theta. $$

If we take the small angle limit of this equation we get

$$ \ddot \theta = - \frac{g}{l} \theta - \frac{a}{m} \dot \theta, $$

the equation for a damped linear oscillator.

In this example, we have included all of the forces that we could have included in the Lagrangian, leaving only the dissipative force to be included in the power function. Of course, one could have included non-dissipative forces in the power function as well, but one must be careful not to include the same force twice, so it is generally a good idea to include as much as possible in the Lagrangian, leaving only the forces that cannot be included in the Lagrangian for the power function. This has the added advantage that one can still look at the Lagrangian for first integrals if the power function does not depend on a particular coordinate.

More Generally
If the dissipative force is given by

$$ \left | \vec F \right | = a \left | \vec v \right |^n $$ or more precisely $$ \vec F = -a \left | \vec v \right |^{n-1} \vec v $$

then the power function is given by

$$ P = -\frac{1}{n+1} a \left | \vec v \right |^{n+1}. $$

More generally if the dissipative force points in the direction of the relative velocity it can also be written as a power function. If

$$ \vec F_i = \phi_i \left ( \vec r_i, \left | \vec v_i \right |, t \right ) \frac{\vec v_i}{\left | \vec v_i \right |} $$

then

$$ P = \sum_i \int \phi_i d \left | \vec v_i \right | $$

A Useful Example with Dry Friction
It turns out that the simplest looking type of friction ('dry friction') actually can yield some useful surprises. Let's calculate the frictional force between a package and a moving conveyor belt.

Let's assume that the belt is moving along at a velocity $$v_y$$ in the $$y-$$direction and calculate the frictional forces that resist moving the package across the belt in the $$x$$ and $$y$$ directions.

Let's write the power function for the dry friction between the belt and the package,

$$ P = -\mu m g v $$

where $$\mu$$ is the coefficient of kinetic friction between the belt and the package, $$m$$ is the mass of the package and the relative speed between the package and the belt is given by

$$ v = \sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }. $$

We can calculate the frictional force resisting motion in the two directions

$$ F_x = \frac{\partial P}{\partial \dot x} = -\mu m g \frac{\dot x}{\sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }}, F_y = \frac{\partial P}{\partial \dot y} = -\mu m g \frac{\dot y - v_y}{\sqrt{ \dot x^2 + \left ( \dot y - v_y \right )^2 }}. $$

Let's imagine that $$v_y$$ is much larger $$\dot x$$ and $$\dot y$$ then we have

$$ F_x \approx -\mu m g \frac{\dot x}{v_y}, F_y \approx \mu m g. $$

The $$y-$$ component indicates a force that tries to drag the package along the belt. It is approximately equal to the kinetic friction that you are used to. The $$x-$$ component is much smaller by a factor of $$\dot x/v_y$$ which could be large. You can try this at the supermarket by blocking a big box of corn flakes on its side as the belt is moving. Even though the friction between the belt and the box is large (you can feel this force with the hand that impedes the motion of the box), the force to move the box across the belt is quite small.

This effect is used for more important purposes when removing a cork from a bottle. It is much easier to removing the cork while twisting it round than to pull it out directly.


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Advanced Classical Mechanics/Energy and Angular Momentum

Let's start with some basic definitions. First m is the inertial mass of a particle -- we will consider only particles with constant masses (although the physics of rockets provides an important counter-example). A particle for us is something that can be described completely by its position $$\vec{r}$$ and mass (it doesn't have any important internal degrees of freedom).

Momentum and Force
We can define the velocity and momentum of the particle,

$$ \vec{v} = \frac{d \vec {r}}{dt} $$ and $$ \vec{p} = m \vec{v} $$.

Because the acceleration is the rate of change of velocity we have

$$ \vec{F} = m \vec{a} = m \frac{d \vec{v}}{dt} = \frac{d\vec{p}}{dt}, $$

so if the force vanishes, then the momentum is constant in time or conserved.

Angular Momentum and Torque
Let's also define the angular momentum and the torque,

$$ \vec{L} = \vec{r} \times \vec{p} $$ and $$ \vec{N} = \vec{r} \times \vec{F} = \vec{r} \times \frac{d \vec{p}}{d t}. $$

Let's calculate the rate of change of the angular momentum

$$ \frac{d}{dt} \vec{L} = \frac{d \vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{dt } = \vec{N} $$

so if the torque vanishes, the angular momentum is conserved. Specifically if the force is always directed along the position vector, $$\vec r$$, the torque vanishes and the angular momentum is conserved. This is called a central force.

Work and Energy
We can define the work done on a particle while moving it from position #1 to position #2 to be

$$ W_{12} = \int_1^2 \vec{F} \cdot d \vec{s}. $$

Again let's assume that the mass of the particle is constant so

$$ W_{12} = \int_1^2 \vec{F} \cdot d \vec{s} = m \int_1^2 \frac{d \vec{v}}{dt} \cdot \vec{v} d t = \frac{m}{2} \int_1^2 \frac{d}{dt} \left ( v^2 \right ) dt = \frac{m}{2} \left ( v_2^2 - v_1^2 \right ) = T_2 - T_1 $$

where we have defined the kinetic energy,

$$T=\frac{1}{2} m v^2$$.

If the integral $$W_{12}$$ depends only on the locations of the points 1 and 2 and not the path, then the force or system is conservative. In particular,

$$ \oint \vec{F} \cdot d \vec{s} = 0 $$ for such a system.

This means that the force is the gradient of some other function

$$ \vec{F} = -\vec{\nabla} V(r) $$

call the potential energy $$V$$. In this case we find that

$$ W_{12} = V_1 - V_2 = T_2 - T_1 $$

so

$$ T_1 + V_1 = T_2 + V_2 $$

This is called the conservation of mechanical energy.


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Advanced Classical Mechanics/Introduction

Previous physics courses have presented many applications of Newton's second law $$F=ma$$ where the force, $$F$$, is given or is a combination of constraining forces and given ones. The downside was that each phenomenon has its own set of forces to remember and different constraints required different formulations.

Physics is about unifying principles.
 * Can we unify these diverse forces into a single principle?
 * The answer is yes, yes. We can find two equivalent principles (ways of looking at classical mechanics).
 * But we get more than that. These principles with only slight modification apply to quantum mechanics, electrodynamics and general relativity - in other words to all of physics.

The first is The Principle of Least Action that we will cover near the beginning of the course. This principle is often the easiest route to finding and solving the equations of motion of a system with constraints. We can formulate electrodynamics and general relativity with the principle, and it also provides a powerful (but not so easy) route to quantum mechanics.

The second is Hamilton's Equations. This formulation provides the easiest route to quantum mechanics and also gives us the powerful tool of Phase Space.

To get to the principle of least action we shall start with $$F=ma$$ (as in Schaum, Goldstein and Kibble). We could have given the principle as an axiom (as in Landau and Lifshiftz) and found $$F=ma$$ as a consequence. Since you are already familiar with $$F=ma$$ the first path is probably more comfortable.


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Advanced Classical Mechanics/Linear Motion

We will examine the dynamics of a particle that is free to travel along only one dimension. It could be restricted to travel along a line or in a circle for example. The results for one-dimensional motion of a single particle provide an foundation to understand motion in more than one dimension and the dynamics of many-particle systems.

General Results
Before looking at some specific examples let's look at a rather general case. A particle moves along one dimension subject to a force that only depends on position along that dimension. We have

$$ m \ddot x = F(x). $$

Let's define the kinetic energy of the particle to be

$$T=\frac{1}{2}m \dot x^2.$$

The rate of change of the kinetic energy is

$$ \dot T = m \dot x \ddot x = \dot x F(x) $$

and let's calculate the total change in the kinetic energy from one time to another

$$ T_2 - T_1 = \int_{t_1}^{t_2} \dot T d t = \int_{t_1}^{t_2} \frac{d x}{d t} F(x) d t = \int_{x_1}^{x_2} F(x) d x = V_1 - V_2 $$

where we have defined the potential energy

$$ V(x) = -\int_{x_0}^x F(x) dx. $$

If we rearrange things we get

$$ T_2 + V_2 = T_1 + V_1 $$

so $$E=T+V$$ is a constant over the motion. This is the conservation of energy. We can invert the integral to calculate the force in terms of the potential energy

$$ F(x) = -\frac{dV}{dx} $$

or we can calculate the velocity of the particle as a function of position using the fact that energy is conserved $$ E = m \dot x^2/2 + V(x) $$ so

$$ \frac{1}{2} m \dot x^2 = E - V(x). $$

In particular the motion is restricted to the region where $$V(x) \leq E$$.

The Pendulum
Let's consider a small ball of mass $$m$$ suspended from a pivot by a light rod of length $$l$$. The ball is restricted to move in a vertical circle. It is useful to write the position of the ball in terms of the angle $$\theta$$ that the rod makes with the vertical direction.

The position of the bob is

$$ x = l \sin \theta,  y = -l \cos \theta $$

using Cartesian coordinates. The force is $$F_y = -m g$$. Let's construct the potential energy

$$ V = -\int_0^y F_y d y' = \int_0^y m g d y' = m g y = -mg l \cos \theta $$

and the kinetic energy

$$ T = \frac{1}{2} m v^2 = \frac{1}{2} m \left ( \dot x^2 + \dot y^2 \right ) = \frac{1}{2} m \left ( l^2 \cos^2 \theta \dot \theta^2 + l^2 \sin^2 \theta \dot \theta^2 \right ) = \frac{1}{2} m l^2 \dot \theta^2. $$

The Period

Let's say the bob is released at an angle $$\theta_0$$, how long does in take to reach the bottom, $$\theta=0$$. This is one quarter of the period of the pendulum. The total energy of the bob is

$$ E = T + V = - m g l \cos \theta_0 = \frac{1}{2} m l^2 \dot \theta^2 - m g l \cos \theta. $$

We can solve this equation for $$\dot \theta$$ to get

$$ \dot\theta = \frac{d \theta}{d t} = \left [ \frac{2 g}{l} \left ( \cos \theta - \cos \theta_0\right ) \right]^{1/2} $$

We can solve this differential equation by the separation of variables

$$ -\sqrt{\frac{l}{2g}} \int_{\theta_0}^0 \frac{1}{\sqrt{\cos \theta - \cos\theta_0} } d \theta = \int_0^{P/4} d t = \frac{P}{4} $$

where $$P$$ is the period of the pendulum. The integral is not elementary but we can do it numerically or using special functions. However, if $$\theta$$ is small we can use $$\cos\theta \approx 1-\theta^2/2$$ to get another integral

$$ P \approx 4 \sqrt{\frac{l}{g}} \int_0^{\theta_0} \frac{1}{\sqrt{\theta_0^2 - \theta^2}} d \theta = 4 \sqrt{\frac{l}{g}} \int_0^1 \frac{1}{\sqrt{1 - u^2}} d u $$

$$ \approx \left. -4 \sqrt{\frac{l}{g}} \arccos \theta \right |_0^1=2\pi \sqrt{\frac{l}{g}} $$.

We can also calculate the period as a function of amplitude numerically or using special functions

Bounded and Unbounded Motion
The motion of the pendulum has two different characters depending on the initial energy of the bob. In the previous example we considered a situation where the bob starts without any motion, so the initial energy is purely potential. On the other hand, the energy of the bob could be more arbitrary. Let's write

$$ E = \frac{1}{2} m \dot \theta^2 - m g l \cos \theta. $$

Remember that the total energy of the particle is defined up to an additive constant. The cosine ranges from -1 to 1, so if

$$ E < m g l $$

the motion of the bob is periodic because at some value of $$\theta$$ the velocity of the bob vanishes and then changes sign. Specifically,

$$ \theta_{max} = \arccos \left ( \frac{E}{m g l} \right ). $$

On the other hand if $$E>mgl$$ the equation has no real solution. In this case the velocity of the bob doesn't change sign and the bob goes round and round, slowing as it goes over the top.

Motion near Equilibrium
The motion of a particle can only be in equilibrium if the total forces on it vanish. If the force only depends on the position of the particle (as it does for a conservative force), then the slope of potential energy must vanish at the equilibrium point because $$ F(x) = - V'(x). $$

For simplicity (and without loss of generality) we can take the point of equilibrium to be $$x=0$$ and take the potential energy to be zero at this point. We can expand the potential energy as a Taylor series about any point. About the point of equilibrium we have

$$ V(x) = V(0) + x V'(0) + \frac{1}{2} x^2 V''(0) + \cdots. $$

For the equilibrium point the first two terms vanish so we have

$$ V(x) \approx \frac{1}{2} k x^2~{\rm where}~k=V''(0). $$

The motion of a particle near almost any point of equilibrium is described by this potential energy, so it appears often in all branches of physics and it is useful to analyze in some detail. If the constant $$k$$ is positive, the potential energy curve is concave up. The motion of a particle of any energy is periodic between the points where $$E=V(x)$$,

$$ x = \pm \sqrt{\frac{2E}{k}}. $$

On the other hand, if $$k$$ is negative, the particle approaches the origin and slows down as it gets closer. If $$E<0$$, the particle turns around at

$$ x = \pm \sqrt{-\frac{2E}{k}}. $$

If the energy of the particle is positive it surmounts the barrier and continues travelling on the other side.

The force corresponding to the potential energy function is $$F=-kx$$, yielding an equation of motion

$$ m \ddot x + k x = 0. $$

This is a linear differential equation for $$x(t)$$ with constant coefficients. Such equations can generally be solved by linear combinations of exponential functions. Let's try the substitution

$$ x(t) = A e^{a t} + B e^{b t} $$

to get

$$ m A a^2 e^{a t} + k A e^{a t} + m B b^2 e^{b t} + k B e^{b t} = 0 $$

which we can solve for any values of $$A$$ and $$B$$ as long as

$$ a = \sqrt{-\frac{k}{m}} $$ and $$ b = -\sqrt{-\frac{k}{m}}. $$

If $$k$$ is negative, it is handy to use the related hyperbolic functions to look at the motion

$$ x(t) = C \cosh p t + D \sinh p t\, $$

where $$p=\sqrt{-k/m}$$. From the discussion earlier if the energy of the particle is less than zero, then the motion is restricted to one side of the barrier and the variable $$x$$ always has the same sign. Looking at the solution, this means that $$ $$ On the other hand, if the energy is positive we have $$|D|>|C|$$.
 * C|>|D|.

If $$k$$ is positive, the constants $$a$$ and $$b$$ are imaginary. In this case we can use

$$ x(t) = C \cos \omega t + D \sin \omega t\, $$

where $$\omega=\sqrt{k/m}$$.

Alternative Solution
We can also solve for the motion using the same techniques as we used for the pendulum. For conservative linear motion this is always possible, but for motion in several dimensions, one can use this technique only if there are additional conserved quantities. Using the conservation of energy has the advantage that it doesn't appear that we are guessing for the solution. The energy of the particle is conserved

$$ E = T + V = \frac{1}{2} m {\dot x}^2 + \frac{1}{2} k x^2. $$

We can solve this equation for the velocity of the particle as a function of position

$$ {\dot x} = \frac{d x}{dt} = \pm \sqrt{ \frac{2E}{m} - \frac{k}{m} x^2 }. $$

We can solve this differential equation by the separation of variables

$$ \int_{t_0}^{t_1} dt = \pm \int_{x_0}^{x_1} \frac{1}{\sqrt{\frac{2E}{m} - \frac{k}{m} x^2 }} dx = \pm \sqrt{\frac{m}{2E}} \int_{x_0}^{x_1} \frac{1}{\sqrt{1 - \frac{k}{2E} x^2 }} dx $$

We have four possibilities depending on the signs of $$E$$ and $$k$$. If $$E$$ and $$k$$ are both positive the integral is

$$ t_1 - t_0 = \pm \sqrt{\frac{m}{k}} \left [ \arcsin \left (\sqrt{\frac{k}{2E}} x_1 \right ) -  \arcsin \left (\sqrt{\frac{k}{2E}} x_0  \right ) \right ] $$

and if we invert this formula we get

$$ x_1 = \sqrt{\frac{2E}{k}} \sin \left [ \sqrt{\frac{k}{m}} \left ( t_1 - t_0 \right ) + \arcsin \left (\sqrt{\frac{k}{2E}} x_0  \right )\right ] $$

where we have dropped the $$\pm$$ sign for clarity.

If $$E$$ is negative and $$k$$ is positive, there is no solution. If $$k$$ is negative, the solutions are unbounded. For positive $$E$$ we have

$$ x_1 = \sqrt{-\frac{2E}{k}} \sinh \left [ \sqrt{-\frac{k}{m}} \left ( t_1 - t_0 \right )  + {\rm arcsinh} \left (\sqrt{-\frac{k}{2E}} x_0  \right )\right ] $$

and for negative $$E$$ we have

$$ x_1 =  \sqrt{\frac{2E}{k}} \cosh \left [ \sqrt{-\frac{k}{m}} \left ( t_1 - t_0 \right ) +  {\rm arccosh} \left (\sqrt{\frac{k}{2E}} x_0  \right )\right ]. $$

Complex Representation
We found earlier that the general solution for small oscillations was

$$ x = A e^{pt} + B e^{-pt}. $$

When $$k$$ is positive, the constant $$p$$ is purely imaginary. It is customary to define $$p^2=-\omega^2=-k/m$$, yielding the periodic solution

$$ x = A e^{i \omega t} + B e^{-i \omega t}. $$

This solution is of course a complex number. However, both the real and imaginary parts of this solution are also solutions of the differential equation, or one could let $$A={\bar B}$$ yielding the real solution,

$$ x = (c - i d) (\cos \omega t + i \sin \omega t) + (c + i d) (\cos \omega t - i \sin \omega t) = 2 c \cos \omega t + 2 d \sin \omega t $$

that we obtained earlier. In general we considering periodic phenomena, we will use solutions of the form $$x(t)=A e^{i\omega t}$$ with the proviso that since the physical quantity $$x$$ that we will take the real part of the solution.

Damped Linear Oscillator
We found that the motion of a particle near a position of stable equilibrium under a conservative force approximates that of a simple harmonic oscillator. There may be some energy loss as well, so there may be a force that depends on the velocity. If we take $$x$$ and $$\dot x$$ as small quantities, a good approximation to the total force is

$$ F = -k x - \lambda \dot x $$

yielding an equation of motion

$$ m \ddot x + \lambda \dot x + k x = 0. $$

Using the earlier results we will try a solution of the form $$x=A e^{pt}$$ which yields an equation

$$ A e^{pt} \left ( mp^2 + \lambda p + k \right ) = 0. $$

The roots of this equation are

$$ p = -\frac{\lambda}{2m} \pm \sqrt { \left ( \frac{\lambda}{2m} \right )^2 - \frac{k}{m} }. $$

There are three regimes of the damping.

Large Damping
If

$$ \left ( \frac{\lambda}{2m} \right )^2 > \frac{k}{m}, $$

then both roots for $$p$$ are real and negative. Specifically we have

$$ \gamma_{\pm} = \frac{\lambda}{2m} \pm \sqrt { \left ( \frac{\lambda}{2m} \right )^2 - \frac{k}{m} }. $$

and the solution

$$ x = A e^{-\gamma_+ t} + B e^{-\gamma_- t}. $$

At late times, the second term dominates and the damping time is approximately $$1/\gamma_-$$.

Small Damping
If

$$ \left ( \frac{\lambda}{2m} \right )^2 < \frac{k}{m}, $$

then both roots for $$p$$ are complex and the general solution is

$$ x = {\rm Re} \left ( A e^{i\omega t - \gamma t} \right ) = a e^{-\gamma t} \cos \left ( \omega t - \theta \right ) $$

where $$\gamma=\lambda/(2m)$$,

$$ \omega = \sqrt{ \omega_0^2 - \gamma^2 } $$

and $$\omega_0^2 = k/m$$.

We can define the quality factor or 'Q' of the oscillator to be

$$ Q = \frac{m\omega_0}{\lambda} = \frac{\omega_0}{2\lambda}. $$

Critical Damping
In the limiting case we have $$\omega_0=\lambda$$ and the two roots for $$p$$ coincide. The resulting solution $$ x = a e^{-\gamma t} $$ cannot be general because a second-order differential equation but have two constants of integration. Here we have

$$ x = \left ( a + b t \right ) e^{-\gamma t} $$

Driven Oscillators
We can imagine some force driving the oscillator. Perhaps someone is pushing on a child on a swing. We get the following equation of motion,

$$ m \ddot x + \lambda \dot x + k x = F(t). $$

This is an inhomogeneous equation of motion. The general solution to such an equation is the solution to the homogeneous equation (which has two constants of integration) plus a particular solution to the equation. To find the particular let's consider a periodic driving force

$$ F(t) = F_1 e^{i \omega_1 t}. $$

Since the equation of motion is linear, the particular solution for the sum of several driving forces is the sum of the solutions for each driving force. Let's look for a solution that varies at the same frequency as the driving force

$$ x(t) = A_1 e^{i \omega_1 t}. $$

Substituting this into the equation of motion yields an equation for the amplitude of the oscillation

$$ A_1 \left ( -m \omega_1^2 + i \lambda \omega_1 + k \right ) = F_1 $$

and solving for the amplitude yields

$$ A_1 = \frac{F_1}{m} \frac{1}{\omega_0^2 - \omega_1^2 + 2 i \gamma \omega_1}. $$

The amplitude is complex and we can divide it into a magnitude and a phase $$a e^{-i \theta_1}$$. Using the fact that $$1/A_1 = 1/a e^{i \theta_1}$$ we have

$$ a = \frac{1}{\sqrt{\left ( \omega_0^2 - \omega_1^2 \right )^2 + 4 \gamma^2 \omega_1^2 }} ~{\rm and}~ \tan \theta_1 = \frac{2\gamma \omega_1}{\omega_0^2 - \omega_1^2}. $$

This yields the real general solution to our equation of the form

$$ x = a_1 \cos \left (\omega_1 t - \theta_1 \right ) + a e^{-\gamma t} \cos \left ( \omega t - \theta \right ). $$

The second term that depends on the initial conditions decreases exponentially. After a long time this transcient can be neglected and the oscillations are solely governed by the driving force.


text:
Advanced Classical Mechanics/Many-Particle Systems

Here we will look a group of particles with forces that exert forces among themselves and have external forces exerted upon them as well. If we focus on particle $$i$$, we have

$$ \dot{\vec {p}}_i = \sum_j \vec{F}_{ji} + F_i^{(e)} $$

where the first term is the force that particle $$j$$ exerts on particle $$i$$ summed over the other particles. The second term denotes the external force on particle $$i$$.

Weak law of action and reaction -- total momentum
Let's assume the weak law of action and reaction which states that $$\vec{F}_{ij}=-\vec{F}_{ji}$$ and sum the change in the momentum of all the particles

$$ \sum_i \dot{\vec {p}}_i = \frac{d^2}{dt^2} \sum_i m_i \vec{r}_i = \sum_i F_i^{(e)} + \sum_{i,j} F_{ji} $$

The second term vanishes because each force between each pair of particles cancels out. If we define the position of the center of mass of the system as

$$ \vec{R} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \frac{1}{M} \sum_i m_i \vec{r}_i $$

we have

$$ M \frac{d^2 \vec{R}}{dt^2} = \sum_i \vec{F}_i^{(e)} = \vec{F}^{(e)}. $$

The total momentum is

$$ \vec{P} = \sum_i m_i \frac{d \vec{r}_i}{dt} = M \frac{d \vec{R}}{dt}, $$

so if $$\vec{F}^{(e)}=0$$ then $$\vec{P}$$ is conserved.

Strong law of action and reaction -- total angular momentum
The total angular momentum of the system is given by

$$ \vec{L} = \sum_i \vec{r}_i \times \vec{p}_i $$

and its rate of change is

$$ \dot{\vec{L}} = \sum_i \left (\dot{\vec{r}}_i \times \vec{p}_i + \vec{r}_i \times \dot{\vec{p}}_i \right ) = \sum_i \vec{r}_i \times \dot{\vec{p}}_i $$

where the first term vanishes because the linear momentum and velocity point in the same direction.

Substituting in the forces yields

$$ \dot{\vec{L}} = \sum_i \vec{r}_i \times \vec{F}_i^{(e)} + \sum_{i,j} \vec{r}_i \times \vec{F}_{ji}. $$

If we look at the final term more closely we have

$$ \sum_{i,j} \vec{r}_i \times \vec{F}_{ji} = \sum_{i<j} \vec{r}_i \times \vec{F}_{ji} + \vec{r}_j \times \vec{F}_{ij} = \sum_{i<j} \left ( \vec{r}_i - \vec{r}_j \right ) \times \vec{F}_{ji}, $$

so if we assume that the force between the particles points in the same direction as the displacement between the particles (strong law of action and reaction), the forces between the particles do not affect the total angular momentum of the system. This type of force is called a central force. In this case we have $$\sum_{i,j} \vec{r}_i \times \vec{F}_{ji} =0$$ so

$$ \dot{\vec{L}} = \sum_i \vec{r}_i \times \vec{F}_i^{(e)} = \vec{N}^{(e)}. $$

A central force satisfies the strong law of action and reaction (not necessarily the weak law). Gravity and electric forces are central and also satisfy the weak law of action and reaction but magnetic forces generally satisfy neither.

Let's take a closer look at the total angular momentum of the system of particles by defining the displacement of a particle relative to the center of mass.

$$ \vec{r}_i = \vec{s}_i + \vec{R} ~\rm{and}~ \vec{v}_i = \vec{u}_i + \vec{V} ~\rm{where}~ \vec{V} = \frac{d \vec{R}}{dt}. $$

The total angular momentum is given by

$$ \vec{L} = \sum_i m_i \vec{s_i} \times \vec{u}_i + \sum_i m_i \vec{s}_i \times \vec{V} + \sum_i m_i \vec{R} \times \vec{v}_i + \sum_i m_i \vec{R} \times \vec{V}. $$

The two middle terms vanish because

$$ \sum_i m_i \vec{s}_i = \sum_i m_i \left ( \vec{r}_i - \vec{R} \right ) = M \left (\vec{R} - \vec{R}\right ) = 0 $$

from the definition of the center of mass. We are left with

$$ \vec{L} = \sum_i m_i \vec{s_i} \times \vec{u}_i + M \vec{R} \times \vec{V}. $$

That is the angular momentum of a system is the angular momentum of the center of mass plus the angular momentum about the center of mass.

Uniform Gravitational Field
A familar possibility for the external forces and torques is a uniform gravitational field. In this case we have

$$ \vec N^{(e)} = \sum_i m_i \vec r_i \times \vec g = M \vec R \times \vec g. $$

Let's calculate the change in the velocity of the center of mass

$$ \dot \vec V = \frac{1}{M} \sum_i m_i \dot \vec v_i = \frac{1}{M} \sum_i m_i \vec g = \vec g $$

and the change in the angular momentum

$$ \dot \vec L = M \vec R \times \vec g = \frac{d}{dt}\left ( \sum_i m_i \vec{s_i} \times \vec{u}_i \right ) + M \left( \vec V \times \vec V + \vec R \times \dot \vec V \right ) = \frac{d}{dt}\left ( \sum_i m_i \vec{s_i} \times \vec{u}_i \right )  + M \vec R \times \vec g $$

so

$$ \frac{d}{dt}\left ( \sum_i m_i \vec{s_i} \times \vec{u}_i \right )  = 0. $$

In an uniform gravitational field, the internal angular momentum of a body is conserved. Think about footballs and frisbees!

Work and Kinetic Energy
Let's examine the total work performed on the system of particles. We have

$$ W_{12} = \sum_i \int_1^2 \vec{F}_i \cdot d \vec{s}_i = \sum_i \int_1^2 \vec{F}^{(e)}_i \cdot d \vec{s}_i + \sum_{i,j} \int_1^2 \vec{F}_{ji} \cdot d \vec{s}_i. $$

We also have

$$ W_{12} = \sum_i \int_1^2 \vec{F}_i \cdot d \vec{s}_i = \sum_i \int_1^2 m_i \dot{\vec{v}}_i \cdot \vec{v}_i dt = \sum_i \int_1^2 d \left ( \frac{1}{2} m_i v_i^2 \right ) $$

so

$$ W_{12} = T_2 - T_1 ~{\rm where}~ T=\frac{1}{2} \sum_i m_i v_i^2. $$

We can also look at the energy using the center of mass to get

$$ T = \frac{1}{2} \sum_i m_i \left ( \vec{u}_i + \vec{V} \right )^2 = \frac{1}{2} \sum m_i \vec{u}_i^2 + \sum m_i u_i \cdot \vec{V} + \frac{1}{2} M V^2. $$

The middle term vanishes for the reasons outlined earlier.

Potential Energy
First let's look at the external forces

$$ \sum_i \int_1^2 \vec{F}^{(e)}_i \cdot d \vec{s}_i = - \sum_i \int_1^2 \vec{\nabla}_i V_i \cdot d \vec{s}_i = - \sum_i \left .V_i \right |_1^2 $$

where $$ \nabla_i \equiv \frac{d}{d \vec{r}_i} $$ if the external forces are conservative. N.B.: in this section, $$V$$ denotes potential energy not the velocity of the center of mass.

To look at the interparticle forces, let's assume that they too are conservative and that the potential only depends on the distance between the particles so

$$ V_{ij} = V_{ij} \left ( \left | \vec{r}_i - \vec r_j \right | \right ). $$

Such a potential has two nice properties. First,

$$ \vec{F}_{ji} = -\vec{\nabla}_i V_{ij} = \vec{\nabla}_j V_{ij} = - \vec{F}_{ij} $$

so it satisfies the weak law of action-reaction. Second we have

$$ \vec{\nabla}_i V_{ij} \left ( \left | \vec r_i - \vec r_j \right | \right ) = \left ( \vec r_i - \vec r_j  \right ) f \left ( \left | \vec r_i - \vec r_j \right | \right ) $$

so it is central and satifies the strong law of action-reaction. If we use this particular type of force we have

$$ \sum_{i,j} \int_1^2 \vec F_{ji} \cdot d \vec r_i = - \sum_{i<j} \int_1^2 \left ( \vec \nabla_i V_{ij} \cdot d \vec r_i + \nabla_j V_{ij} \cdot d \vec r_j \right ) $$

Let's define $$\vec r_{ij} = \vec r_i - \vec r_j$$. We have

$$ \frac{dV}{d \vec r_{ij}} = \frac{d V}{d \vec r_i} = \nabla_i V = -\nabla_j V $$

and

$$d \vec r_{ij} = d \left ( \vec r_i - \vec r_j \right ) = d \vec r_i - d \vec r_j.$$

If we substitute these results into the expression for the work we have

$$ - \sum_{i<j} \int_1^2 \left ( \vec \nabla_i V_{ij} \cdot d \vec r_i + \nabla_j V_{ij} \cdot  d \vec r_j \right ) = -\sum_{i<j} \int_1^2 \left ( \vec \nabla_{ij} V_{ij} \cdot d r_{ij} \right ) = - \left. \sum_{i<j} V_{ij} \right |_1^2 = - \frac{1}{2} \left. \sum_{i\neq j} V_{ij} \right |_1^2 $$

so we can definte the total potential energy to be the sum of the potential energies of each particle due to the external force plus the sum of the potential energies of each pair of particles:

$$ V = \sum_i V_i + \frac{1}{2} \sum_{i\neq j} V_{ij}. $$

The Virial Theorem
Let's look at particular and common form of central force. In particular, let's assume that

$$ V_{ij}(\vec r_i, \vec r_j) = k \left | \vec r_{ij} \right |^n. $$

and that there is no external potential. This means than

$$ \nabla_i V_{ij} = n V_{ij} \frac{\vec r_{ij}}{\left | \vec r_{ij} \right|^2}. $$

We can define

$$ K = \sum_i \frac{1}{2} m_i \vec r_i^2 $$

as a measure of the size of the system. Let's calculate the second derivative of $$K$$ with respect to time

$$ \frac{d^2 K}{d t^2} = \sum_i \left ( m_i \dot \vec r_i^2 + m_i \vec r_i \cdot \ddot \vec r_i  \right ) = 2 T - \sum_i \vec r_i \cdot\sum_j \nabla_i V_{ij}. $$

Again let's group the particles in the final summation in pairs

$$ \frac{d^2 K}{d t^2} = 2 T - \sum_{i<j} \left ( \vec r_i - \vec r_j \right ) \nabla_i V_{ij} = 2 T - \sum_{i<j} \left ( \vec r_i - \vec r_j \right ) n V_{ij} \frac{\vec r_{i} - \vec r_j}{\left | \vec r_i - \vec r_j \right|^2} = 2 T - n V. $$

If we have a bunch of particles in equilibrium so they are not expanded or contracting on average than the left-hand side of the equation above vanishes. The most familar example of this is the force of gravity with $$n=-1$$, so for a self-gravitating system in equilibrium we have $$2 T + V=0$$. This relationship is used to weigh distant galaxies and clusters of galaxies.

If $$n<0$$ the potential energy of the system vanishes as the particles get further apart, so the total energy of the system $$E=T+V$$ is the negative of the amount of energy needed to break up the system and leave the particles stationary. Let's calculate the total energy of a system in equilibrium in terms of its kinetic energy. We have

$$ E = T + V = \left ( 1 + \frac{2}{n} \right ) T $$

We find that only if $$-2<n<0$$ is the energy of the system negative. For $$n=-2$$ the total energy of the equilibrium system is zero, so the equilibrium is marginally stable; even a tiny bit of additional energy is enough to break it apart. For $$n<-2$$ the equilibrium is unstable. The grouping of particles has more energy than if they separate.

Thankfully the force of gravity has $$n=-1$$. If $$n$$ were less than -2, there would be no self-gravitating equilibrium systems such as stars, galaxies etc. In fact theories of gravity in higher dimensions find that $$n=2-D$$ where $$D$$ is the number of spatial dimensions. Apparently $$D=3$$ is kind of special.