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Ma241 Advanced Mechanics

Lecturer: Dr. Sergey Frolov

Website: Link

1 Lagrange's Equations & Conservation Laws
2 Integrating the Equations of Motion
- 2.1 Motion In One Dimension
- 2.2 Motion In A Central Field
3 Collisions & Scattering
4 Oscillations
5 Rigid Body Motion
6 Hamiltonian Dynamics
7 Special Relativity
8 Continuous Systems and Fields
9 References and Links

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Lagrange's Equations & Conservation Laws

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The Lagrangian

Every mechanical system is completely characterised by a function

$L = L(q, \dot{q}, t)$

of the generalised coordinates and velocities, and maybe of time, known as the Lagrangian. In general, the Lagrangian is given by

$L = T - U$

where $T$ and $U$ are the kinetic and potential energies of the system.

Examples:

Free particle of mass $m$ : $L = \frac{1}{2} m v^2$
Particle in a central field (polar coordinates): $L = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 - U(r)$
Relativistic free particle: $L = - m_0 c^2 \sqrt{1 - \frac{1}{c^2} \dot{x}_i^2}$

Properties: The Lagrangian is additive, may be scaled by a constant and is defined to within an additive total time derivative of any function.

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Principle of Least Action

The action of the system is

$S = \int_{t_1}^{t_2} L(q, \dot{q}, t) dt$

and the Principle of Least Action states that for any system developing between times $t_1$ and $t_2$ the action is a minimum.

We can then derive Lagrange's equations of motion by considering a variation from the actual path from $t_1$ to $t_2$ , $q + \delta q$ , but which has the same values at the endpoints (ie $\delta q(t_1) = \delta q(t_2) = 0$ ) giving Lagrangian

$L(q + \delta q, \dot{q} + \delta \dot{q}, t)$

We then have

$\delta S = \delta \int_{t_1}^{t_2} L\, dt$

and the Principle of Least Action states that this variation must be zero. Now,

$\delta S = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q}\right) dt = 0$

and since

$\delta \dot{q} = \frac{d}{dt} (\delta q)$

we rewrite this as

$\delta S = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \delta q + \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}}\delta q\right) -\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \delta q \right) dt = 0$

giving

$\delta S = \left[\frac{\partial L}{\partial \dot{q}} \delta q \right]_{t_1}^{t_2} + \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}}\right) \delta q \, dt = 0$

but since the variation is zero at the endpoints the first term is zero, and hence the remaining integral must equal zero for all $\delta q$ , in other words the integrand is zero. Hence we have Lagrange's equations of motion:

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

where we now use the index i to indicate that this equation holds for each generalised coordinate (ie each degree of freedom of the system).

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Conservation Laws

Energy: Suppose the Lagrangian does not depend explicitly on time. Then we can write the total time derivative as

$\frac{d L}{dt} = \sum_i \frac{\partial L}{\partial q_i} \frac{\partial q}{\partial t} + \sum_i \frac{\partial L}{\partial \dot{q}_i} \frac{\partial \dot{q}_i} {\partial t}$

or, using Lagrange's equations,

$\frac{d L}{dt} = \sum_i \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}}\right) \dot{q}_i + \sum_i \frac{\partial L}{\partial \dot{q}_i} \ddot{q}_i$

$\Rightarrow \frac{d L}{dt} = \sum_i \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \dot{q}_i \right)$

and hence we have

$\frac{d}{dt} \left( \sum_i \dot{q}_i \frac{\partial L}{\partial \dot{q_i}} - L\right) = 0$

so the quantity

$E \equiv \sum_i \dot{q}_i \frac{\partial L}{\partial \dot{q_i}} - L$

known as the energy, is conserved.

Linear Momentum: Consider an infinitesimal displacement $\vec{\epsilon}$ of the system. By virtue of the homogeneity of space, the Lagrangian is invariant under this displacement, which has the effect of transforming the radius vector of every particle in the system such that

$r \rightarrow \vec{r} + \vec{\epsilon}$

and the change in the Lagrangian is

$\delta L = \sum_a \frac{\partial L}{\partial \vec{r}_a} \delta \vec{r}_a = \vec{\epsilon} \cdot \sum_a \frac{\partial L}{\partial \vec{r}_a}$

summing over the particles in the system. As $\vec{\epsilon}$ is arbitrary, the condition that the Lagrangian remains unchanged is

$\delta L = \sum_a \frac{\partial L}{\partial \vec{r}_a} = 0$

$\Rightarrow \frac{d}{dt} \sum_a \frac{\partial L}{\partial \vec{v}_a} = 0$

using Lagrange's equations. Hence we have the conserved quantity

$\vec{P} \equiv \sum_a \frac{\partial L}{\partial \vec{v}_a} = \sum_a m_a \vec{v}_a$

the linear momentum of the system.

Angular Momentum: Consider an infinitesimal rotation. By virtue of the isotropy of space, the Lagrangian is again invariant. The changes in the positions and velocities are given by

$\delta \vec{r} = \delta \vec{\phi} \times \vec{r}$

$\delta \vec{v} = \delta \vec{\phi} \times \vec{v}$

where the infinitesimal rotation is represented by the vector $\delta \vec{\phi}$ . The change in the Lagrangian is then

$\delta L = \sum_a \left( \frac{\partial L}{\partial \vec{r}_a} \delta \vec{r}_a + \frac{\partial L}{\partial \vec{v}_a} \delta \vec{v}_a \right) = 0$

$\sum_a \left( \dot{\vec{p}}_a \cdot \delta \vec{\phi} \times \vec{r}_a + \vec{p}_a \cdot \delta \vec{\phi} \times \vec{v}_a \right) = 0$

and then using a vector dot and cross product identity,

$\delta \vec{\phi} \sum_a \left( \vec{r}_a \times \dot{\vec{p}}_a + \vec{v}_a \times \vec{p}_a \right) = \delta \vec{\phi} \frac{d}{dt} \sum_a \vec{r}_a \times \vec{p}_a = 0$

so the quantity

$\vec{M} \equiv \sum_a \vec{r}_a \times \vec{p}_a$

known as the angular momentum, is conserved.

Cyclic Coordinates: If the Lagrangian is such that

$\frac{\partial L}{\partial q_i} = 0$

for some coordinate $q_i$ then we say that coordinate is cyclic, and have (from Lagrange's equations) the conserved quantity

$P_i = \frac{\partial L}{\partial \dot{q}_i}$

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Noether's Theorem

Consider an infinitesimal transformation of the generalised coordinates such that

$q^i \rightarrow q^i + \sum_{\lbrace \alpha \rbrace} q^{i\alpha} \epsilon_{\alpha}$

and

$L \rightarrow L + \sum_{\lbrace \alpha \rbrace} L^{\alpha} \epsilon_{\alpha}$

then Noether's Theorem states that the quantities

$J^{\alpha} = \sum_{i=1}^s \frac{\partial L}{\partial \dot{q}^i} q^{i\alpha} - L^{\alpha}$

are conserved.

Proof: Consider $S = \int_{t_1}^{t_2} L\, dt$ , then

$\delta S = \delta \int_{t_1}^{t_2} L\, dt = \int_{t_1}^{t_2} \left(\frac{\partial L}{\partial q^i} \delta q^i + \frac{\partial L}{\partial \dot{q}^i} \delta \dot{q}^i \right) dt$

and rewriting as in our derivation of Lagrange's equations,

$\int_{t_1}^{t_2}\delta L\, dt = \int_{t_1}^{t_2} \left[\left(\frac{\partial L}{\partial q^i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}^i} \right) \delta q^i + \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}^i} \delta q^i\right) \right] dt$

and the first term is zero if the equations of motion are satisfied. Now, from our original transformation we also have that

$\int \delta L \, dt = \int_{t_1}^{t_2} \frac{d}{dt} \left(L^{\alpha} \epsilon_{\alpha} \right) dt$

and therefore we have

$\int_{t_1}^{t_2} dt \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}^i} \delta q^i\right) = \int_{t_1}^{t_2} \frac{d}{dt} \left(L^{\alpha} \epsilon_{\alpha} \right) dt$

hence

$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}^i} \delta q^i - L^{\alpha} \epsilon_{\alpha} \right) = 0$

and as $\delta q^i = q^{i\alpha} \epsilon_{\alpha}$ we have the conserved quantity

$J^{\alpha} = \sum_i\frac{\partial L}{\partial \dot{q}^i} q^{i\alpha} - L^{\alpha}$

Note that in the case that the transformation leaves the Lagrangian invariant then $\delta L = 0$ and so we have that

$0 = \int_{t_1}^{t_2} dt \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}^i} \delta q^i\right)$

and the conserved quantity is

$J^{\alpha} = \frac{\partial L}{\partial \dot{q}^i} q^{i\alpha}$

Further reading: A translation of Emmy Noether's original paper

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Integrating the Equations of Motion

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Motion In One Dimension

From the energy

$E = \frac{1}{2} m \dot{x}^2 + U(x)$

we get the following

$t = \sqrt{\frac{m}{2}} \int \frac{dx}{\sqrt{E - U(x)}} + \mbox{constant}$

If the motion is finite we have that the period is

$T = \sqrt{2 m} \int_{x_1}^{x_2} \frac{dx}{\sqrt{E - U(x)}}$

where the limits of integration are the limits of the motion (or turning points), given by

$U(x) = E$

i.e. the points where the velocity $\dot{x} = 0$ .

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Motion In A Central Field

The Lagrangian in polar coordinates is:

$L = \frac{1}{2}m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 - U(r)$

and from the cyclic coordinate $\phi$ we have the conserved quantity of angular momentum:

$M \equiv \frac{\partial L}{\partial \dot{\phi}} = m r^2 \dot{\phi}$

while the energy is given by:

$E = \frac{1}{2}m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\phi}^2 + U(r)$

and eliminating $\dot{\phi}$ we arrive at

$t = \int \frac{dr}{\sqrt{\frac{2}{m} \left[E - U(r) \right] - \frac{M^2}{m^2 r^2}}} + \mbox{constant}$

and using $d \phi = \frac{M dt}{mr^2}$ from the angular momentum equation,

$\phi = \int \frac{\frac{M dr}{r^2}}{\sqrt{{2m} \left[E - U(r) \right] - \frac{M^2}{r^2}}} + \mbox{constant}$

These two formulae describe the motion of the particle in the central field.

The turning points of the motion as regards distance from the center occur when $\dot{r} = 0$ , i.e. when

$U_{eff}(r) = E$ where we have defined the effective potential $U_{eff}(r) = U(r) + \frac{M^2}{2mr^2}$

If the range of $r$ is bounded only by $r \geq r_{min}$ then the motion comes from and returns to infinity.

If $r_{min} \leq r \leq r_{max}$ the motion is finite, and lies between the two circles of radii $r_{min}$ and $r_{max}$ . In the time it takes for $r$ to vary from $r_{min}$ to $r_{max}$ and back again the radius vector turns through an angle given by

$\Delta \phi = 2 \int_{r_{min}}^{r_{max}} \frac{\frac{M dr}{r^2}}{\sqrt{{2m} \left[E - U(r) \right] - \frac{M^2}{r^2}}}$

and the condition for the orbit to close is that this angle is a rational fraction of $2 \pi$ , i.e.

$\Delta \phi = \frac{m}{n} 2 \pi \,\,\,\,\,\,\, m,n \in Z$ .

The presence of the centrifugal energy $\frac{M^2}{2mr^2}$ in general prevents the particle from falling to the center of the field. This can only happen if the potential energy tends to $- \infty$ as $r \rightarrow 0$ . Using

$\frac{1}{2} m \dot{r}^2 = E - U(r) - \frac{M^2}{2mr^2} > 0$

$\Rightarrow r^2 U(r) + \frac{M^2}{2m} < E r^2$

we find that $r$ will tend to zero only if

$\lim_{r \rightarrow 0} \left[r^2 U(r) \right] < - \frac{M^2}{2m}$

i.e. the potential of the form $- \frac{\alpha}{r^2}, \alpha > \frac{M^2}{2m}$ or $- \frac{\alpha}{r^n}, n>2$ .

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Collisions & Scattering

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Disintegration of Particles

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Elastic Collisions

Consider the collision of a particle of mass $m_1$ with a particle of mass $m_2 > m_1$ which is initially at rest. We let $\chi$ be the direction through which the motion of $m_1$ is turned in the centre of mass system, and $\theta_1, \theta_2$ be the directions of motion of the particles in the lab frame after the collision. We then have the following relationships:

$\tan \theta_1 = \frac{m_2 \sin \chi}{m_1 + m_2 \cos \chi}$

$\theta_2 = \frac{1}{2} (\pi - \chi)$

$v_1' = \frac{\sqrt{m_1^2 + m_2^2 + 2m_1 m_2 \cos \chi} \,\, v}{m_1 + m_2}$

$v_2' = 2 \frac{m_1 v}{m_1 + m_2} \sin \frac{\chi}{2}$

where $v$ is the initial velocity of $m_1$ and $v_1',v_2'$ are the velocities of the two masses in the lab frame after the collision.

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Scattering in a Central Field

Consider the deflection of a single particle of mass $m$ in a central field $U(r)$ , with the centre of the field being the centre of mass of the system. The path of the particle is symmetric about a line from the centre of the field to the point of closest approach. This line has length $r_0$ and makes an angle of $\phi_0$ with the x-axis. From the symmetry of the situation we have that the scattering angle is

$\chi = |\pi - 2 \phi_0|$

In polar coordinates, we have the conserved quantities of energy and angular momentum:

$E = \frac{1}{2}m \dot{r}^2 + \frac{1}{2}mr^2 \dot{\phi}^2 + U(r) = \frac{1}{2}m v_{\infty}^2$

$J = mr^2 \dot{\phi} = m \rho v_{\infty}$

where $\rho$ is the perpendicular distance from the centre of the field to the initial path of the particle (known as the impact parameter), and $v_{\infty}$ is its initial velocity.

Using these relationships, we can derive the following formula for the angle $\phi_0$ :

$\phi_0 = \int_{r_0}^{\infty} \frac{\frac{ \rho}{r^2} dr }{\sqrt{1 - \frac{\rho^2}{r^2} - \frac{U(r)}{E}}}$

The lower limit $r_0$ can be calculated using the fact that at $r=r_0, \dot{r} = 0$ . This integral gives $\phi_0$ and hence $\chi$ as a function of $\rho$ .

For a beam of particles, the differential scattering cross-section is given by

$d \sigma = \frac{dN}{n}$

where dN is the number of particles scattered per unit time between $\chi$ and $\chi + d \chi$ and n is the number of particles passing in unit time through unit area of the beam cross-section.

We assume that only particles whose initial paths lie between $\rho$ and $\rho + d \rho$ from the centre of the field are scattered between the angles $\chi$ and $\chi + d \chi$ . The number of such particles is equal to n multiplied by the area between two circles with radius $\rho$ and $\rho + d \rho$ , giving

$dN = n 2 \pi \rho \, d \rho$

hence,

$d \sigma = 2 \pi \rho \, d \rho = 2 \pi \rho(\chi) \left|\frac{d \rho (\chi)}{d \chi}\right| d \chi$

We can also express this in terms of the solid angle element

$d \Omega = 2 \pi \sin \chi d \chi$

giving:

$d \sigma = \frac{ \rho(\chi)}{\sin \chi} \bigg|\frac{d \rho (\chi)}{d \chi}\bigg| d \Omega$

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Oscillations

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Oscillations in One Dimension

Oscillatory motion involves the motion of a particle about a stable equilibrium, $U(q_0)$ where the potential energy is a minimum. We can consider the potential at a point $q$ close to $q_0$ and expand the potential in a Taylor series:

$U(q) \approx U(q_0) + U'(q_0) (q - q_0) + \frac{1}{2} U''(q_0) (q - q_0)^2 \dots$

and as the first two terms are constant and zero respectively, we let $x = q - q_0$ and $k = U''(q_0)$ and obtain the potential in the following form:

$U(x) = \frac{1}{2} k x^2$

giving Lagrangian

$L = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} kx^2$

and equations of motion

$m \ddot{x} + k x = 0 \Rightarrow \ddot{x} + \omega^2 x = 0$

where

$\omega = \sqrt{\frac{k}{m}}$

is the frequency of the oscillations. The general solution is

$x = c_1 \cos \omega t + c_2 \sin \omega t = a \cos (\omega t + \alpha)$

where $a = \sqrt{c_1^2 + c_2^2}$ is the amplitude of the oscillations and $\alpha = \tan^{-1} - \frac{c_2}{c_1}$ is the initial phase.

The total energy is

$E = \frac{1}{2} m \omega^2 a^2$

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Forced Oscillations in One Dimension

The Lagrangian is

$L = \frac{1}{2}m \dot{x}^2 - \frac{1}{2} k x^2 - U_e (x,t)$

with the final term due to an external force. We have that

$U_e (x,t) \approx U_e (0,t) + x \frac{\partial U_e}{\partial x} \Big|_{x=0}$

and the first term can be neglected as a total time derivative. We then define

$F(t) = - \frac{\partial U_e}{\partial x} \Big|_{x=0}$

so that the Lagrangian becomes

$L = \frac{1}{2}m \dot{x}^2 - \frac{1}{2} k x^2 + F(t) x$

giving equations of motion

$m \ddot{x} + k x = F(t) \Rightarrow \ddot{x} + \omega^2 x = \frac{1}{m} F(t)$

The general solution is a linear combination of the homogeneous and particular solutions.

Special Case: We consider sinusoidal driving forces, that is,

$F(t) = f \cos (\gamma t + \beta)$

and guess a particular solution

$x_1 = b \cos (\gamma t + \beta)$

which gives

$b = \frac{f}{m (\omega^2 - \gamma^2)}$

and the general solution is then

$x(t) = a \cos (\omega t + \alpha) + \frac{f}{m (\omega^2 - \gamma^2)} \cos (\gamma t + \beta)$

Resonance occurs at $\omega = \gamma$ . We analyse this case by subtracting a second particular solution, such that we have:

$x(t) = a \cos (\omega t + \alpha) + \frac{f}{m (\omega^2 - \gamma^2)} \Big[ \cos (\gamma t + \beta) - \cos (\omega t + \beta) \Big]$

We then take the limit as $\omega \rightarrow \gamma$ and use L'hopital's rule to find:

$x(t) a \cos (\omega t + \alpha) = \frac{f}{2 m \omega} t \sin (\omega t + \beta)$

General Case: We now let the driving force be arbitrary. Let

$\xi = \dot{x} + i \omega x$

then

$\dot{\xi} - i \omega \xi = \frac{1}{m} F(t)$

and we guess a solution

$\xi = A(t) e^{i \omega t}$

$\Rightarrow \dot{\xi} = \dot{A} e^{i \omega t} + i \omega A(t) e^{i \omega t} = \dot{A} e^{i \omega t} + i \omega \xi$

so we have

$\dot{A} e^{i \omega t} = \frac{1}{m} F(t) \Rightarrow \dot{A} = \frac{1}{m}e^{- i \omega t} F(t)$

giving

$A(t) = \frac{1}{m} \int_0^t F( \tau) e^{- i \omega \tau} d \tau + A_0$

and so

$\xi(t) = e^{i \omega t} \left[\frac{1}{m} \int_0^t F( \tau) e^{- i \omega \tau} d \tau + \xi_0 \right]$

and the general solution is found from

$x(t) = \frac{1}{\omega} \mbox{Im}(\xi(t))$

We could also have chosen the lower limit of integration as being $t_0 = - \infty$ as then the constant term $\xi(t_0)$ is zero.

A system performing forced oscillations gains energy. To find the energy gained by the system, consider that

$| \xi(\infty) |^2 = \frac{1}{m^2} \left| \int_{-\infty}^{\infty} F(t) e^{- i \omega t} dt \right|^2$

using $\xi(t) \overline{\xi(t)} = \xi(t)^2$ . Now,

$E = \frac{1}{2} m ( \dot{x}^2 + \omega^2 x^2 ) = \frac{1}{2} m | \xi |^2$

so the energy aquired is

$E = \frac{1}{2m} \left| \int_{-\infty}^{\infty} F(t) e^{- i \omega t} dt \right|^2$

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Oscillations of a System With Many Degrees of Freedom

Free Oscillations: The potential of the system is now $U(q_i)$ where i runs from 1 to s, a minimum for $q_i = q_{i0}$ . We define new coordinates $x = q_i - q_{i0}$ giving

$U(q_i) = U(q_{i0}) + x_i \frac{\partial U}{\partial q_i} + \frac{1}{2} x_i x_j \frac{\partial^2 U}{\partial q_i \partial q_j} \dots$

and again we have that the first two terms may be ignored as they are constant and zero, respectively. We then have

$k_{ij} = \frac{\partial^2 U}{\partial q_i \partial q_j}$

$m_{ij} = g_{ij}(q_{i0})$

with both of these matrices being symmetric and positive definite, and the Lagrangian is

$L = \frac{1}{2} m_{ij} \dot{x_i} \dot{x}_j - \frac{1}{2} k_{ij} x_i x_j$

giving equations of motion

$m_{ij} \ddot{x_j} + k_{ij} x_j = 0$

summing over j. We guess that solutions will be of the form

$x_j = A_j e^{i \omega t}$

leading to

$( k_{ij} - \omega^2 m_{ij} ) A_j = 0 \,\,\, \mbox{(*)}$

and for non-trivial solutions we must have that

$\mbox{det}( k_{ij} - \omega^2 m_{ij} ) = 0$

which gives a polynomial of degree s in $\omega^2$ , with positive real roots $\omega_{\alpha}^2$ . Having found these roots, we substitute each root individually into (*) to find the $A_j$ corresponding to each root. The general solution is then a superposition of these:

$x_j = \mbox{Re} \left[\, \sum_{\alpha=1}^s A_{j \alpha} e^{i \omega_{\alpha} t}\, \right]$

The $A_{j\alpha}$ are proportional to the minors of the determinant, $\Delta_{j \alpha}$ . We can then write the general solution as

$x_j = \mbox{Re} \left[\, \sum_{\alpha=1}^s \Delta_{j \alpha} C{ \alpha} e^{i \omega_{\alpha} t}\, \right]$

$x_j = \sum_{\alpha} \Delta_{j \alpha} \theta_{\alpha}$

where

$\theta_{\alpha} = \mbox{Re} \left[\,C_{\alpha} e^{i \omega_{\alpha}t} \,\right]$

the normal coordinates of the system, with arbitrary amplitudes and phases but definite frequencies. These satisfy

$\ddot{\theta}_{\alpha} + \omega_{\alpha}^2 \theta_{\alpha} = 0$

and we can express the Lagrangian of the system as

$L = \frac{1}{2} \sum_{\alpha = 1}^s m_{\alpha} \left( \dot{\theta}_{\alpha}^2 - \omega_{\alpha}^2 \theta_{\alpha}\right)$

It is convenient to then define further coordinates

$Q_{\alpha} = \sqrt{m_{\alpha}} \theta_{\alpha}$

such that

$L = \frac{1}{2} \sum_{\alpha = 1}{s} \left( \dot{Q}_{\alpha}^2 - \omega_{\alpha}^2 Q_{\alpha}\right)$

Forced Oscillations: If an external force is present the Lagrangian is

$L = \mathcal{L_0} + \sum_j F_j(t) x_j$

and upon transforming to normal coordinates

$L = \frac{1}{2} \sum_{\alpha = 1}^s \left( \dot{Q}_{\alpha}^2 - \omega_{\alpha}^2 Q_{\alpha}\right) + \sum_{\alpha} f_{\alpha} (t) Q_{\alpha}$

where $f_{\alpha}(t) = \sum_j \frac{F_j(t)}{\sqrt{m_{\alpha}}} \Delta_{j \alpha}$ and the equations of motion are:

$\ddot{Q}_{\alpha} + \omega_{\alpha}^2 Q_{\alpha} = f_{\alpha} (t)$

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Damped Oscillations

One-Dimensional Case: The equation of motion of a particle oscillating while experiencing a damping or frictional force is:

$m \ddot{x} + kx = - \alpha \dot{x}$

where $\alpha$ is a positive constant. Note that this cannot be derived from any Lagrangian. Dividing across by the mass as before and putting $2 \lambda = \frac{\alpha}{m}$ we obtain:

$\ddot{x} + 2 \lambda \dot{x} + \omega_0^2 x = 0$

We guess a solution

$x = C e^{rt} \Rightarrow r^2 + 2 \lambda r + \omega_0^2 = 0$

$\Rightarrow r_{\pm} = -\lambda \pm \sqrt{\lambda^2 - \omega_0^2}$

and the general solution is:

$x(t) = C_+ e^{r_+ t} + C_- e^{r_- t}$

There are three cases to consider:

$\lambda < \omega_0$ then we put

$r = - \lambda \pm i \omega$

where $\omega = \sqrt{\omega_0^2 - \lambda^2}$ and then

$x(t) = a e^{-\lambda t} \cos (\omega t + \phi)$ (light damping)

$\lambda = \omega$ then $r = \lambda$ and the general solution is

$x(t) = (C_1 + C_2 t) e^{-\lambda t}$ (critical damping)

$\lambda > \omega$ then r is negative, and the general solution is

$x(t) = C_+ e^{\left(-\lambda + \sqrt{\lambda^2 - \omega_0^2}\right) t} + C_- e^{\left(-\lambda - \sqrt{\lambda^2 - \omega_0^2}\right) t}$ (heavy damping)

Many-Dimensional Case: The equations of motion are:

$m_{ij} \ddot{x}_{j} + k_{ij} x_j = - \alpha_{ij} \dot{x}_j$

ad our ansatz is

$x_j = A_j e^{rt} \Rightarrow (m_{ij} r^2 + \alpha_{ij} r + k_{ij}) A_j = 0$

and for non-trivial solutions we must have

$\mbox{det}(m_{ij} r^2 + \alpha_{ij} r + k_{ij}) = 0$

giving 2s roots for r, which must have its real part negative.

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Forced Damped Oscillations

We consider only the special case where the driving force is sinusoidal. The equation of motion (in one dimension) is:

$\ddot{x} + 2 \lambda \dot{x} + \omega_0^2 x = \frac{f}{m} e^{i \gamma t}$

where we have written the force in complex form. Then our ansatz is

$x = B e^{i \gamma t}$

giving

$B = \frac{f}{m ( \omega_0^2 - \gamma^2 + 2 i \lambda \gamma)} = b e^{i \delta}$

and hence

$b = \frac{f}{m \sqrt{( \omega_0^2 - \gamma^2 )^2 + 4 \lambda^2 \gamma^2}}$

$\tan \delta = \frac{2 \lambda \gamma}{\gamma^2 - \omega_0^2}$

Our solution x is the real part of $B e^{i \gamma t}$ , so

$x = b \cos ( \gamma t + \delta )$

The maximum value of b occurs when $\gamma = \sqrt{\omega_0^2 - 2 \lambda^2}$ (resonance).

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Rigid Body Motion

[edit]

Angular Velocity

A rigid body may be defined as a system of particles such that the distances between the particles do not vary. To describe the motion of such a body we use two sets of axes: a fixed, inertial set XYZ and a moving set which is rigidly fixed in the body and moves with it.

Consider an arbitrary infinitesimal displacement of a point $P$ in a rigid body, and let the origin of the moving frame of axes be at the centre of mass of the body. In the inertial frame,

$d \vec{r^{\prime}} = d\vec{R} + d \vec{\phi} \times \vec{r}$

where $\vec{r^{\prime}}$ and $\vec{r}$ are the radius vectors of the point $P$ with respect to the origin of the XYZ frame and the moving frame respectively, $\vec{R}$ is the radius vector of the centre of mass of the body, and $d\vec{\phi}$ describes the rotation of the centre of the mass.

Dividing by $dt$ we obtain

$\vec{v} = \vec{V} + \vec{\Omega} \times \vec{r}$

where $\vec{V}$ is the centre of mass velocity and $\vec{\Omega}$ is the angular velocity.

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The Inertia Tensor

The kinetic energy of a rigid body is given by

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

summing over all particles in the body. Subbing in for $v$ we get

$T = \sum \frac{1}{2} m (\vec{V} + \vec{\Omega} \times \vec{r} )^2 = \sum \frac{1}{2} m V^2 + \sum m \vec{V} \cdot (\vec{\Omega} \times \vec{r}) + \sum \frac{1}{2} m (\vec{\Omega} \times \vec{r} )^2$

$\Rightarrow \frac{1}{2} V^2 \sum m + \vec{V} \times \vec{\Omega} \cdot \sum m \vec{r} + \frac{1}{2} \sum m \Big[ \Omega^2 r^2 - (\vec{\Omega} \cdot \vec{r} )^2 \Big]$

The first term is simply the kinetic energy due to the translational motion of the centre of mass, while the second term is zero as our origin is the centre of mass. We thus have

$T = T_{trans} + T_{rot}$

where, in tensor form,

$T_{rot} = \frac{1}{2} \sum m ( \Omega_i^2 x_i^2 - \Omega_i x_i \Omega_k x_k)$

$\Rightarrow T_{rot} = \frac{1}{2} \sum m ( \Omega_i \Omega_k \delta_{ik} x_l^2 - \Omega_i \Omega_k x_i x_k)$

$\Rightarrow T_{rot} = \frac{1}{2} \Omega_i \Omega_k \sum m ( \delta_{ik} x_l^2 - x_i x_k)$

so then

$T = \frac{1}{2} \mu V^2 + \frac{1}{2} I_{ik} \Omega_i \Omega_k$

where $\mu$ is the total mass of the body and

$I_{ik} = \sum m ( \delta_{ik} x_l^2 - x_i x_k)$

is the inertia tensor, which we can write explicitly as:

$\begin{pmatrix}\sum m (y^2 + z^2) & - \sum m x y & - \sum m x z \\ -\sum m yx & \sum m (x^2 + z^2)& - \sum m y z \\ - \sum m z x & - \sum m z y & \sum m (x^2 + y^2) \\ \end{pmatrix}$

The diagonal components $I_{xx}, I_{yy}, I_{zz}$ are called the moments of inertia about the corresponding axes.

If the body is continuous, then the sum becomes an integral:

$I_{ik} = \int \rho ( x_l^2 \delta_{ik} - x_i x_k) dV$

We can always reduce the inertia tensor to diagonal form by choosing an appropriate set of axes, known as the principal axes of inertia, $I_1, I_2, I_3$ , in which case

$T_{rot} = \frac{1}{2} (I_1 \Omega_1^2 + I_2 \Omega_2^2 + I_3 \Omega_3^2)$

Note that none of the principal moments of inertia can exceed the sum of the other two.

A body with $I_1 \neq I_2 \neq I_3$ is called an asymmetric top.

A body with $I_1 = I_2 \neq I_3$ is called a symmetric top.

A body with $I_1 = I_2 = I_3$ is called a spherical top.

If a body is symmetrical then the position of the centre of mass and directions of the principal axes must have the same symmetry as the body.

We can also calculate the tensor $I_{ik}^' = \sum m ( \delta_{ik} x_l^{2'} - x_i^' x_k^')$ about a different origin $O'$ . Then, if $x_i = x_i' + a$ ,

$I_{ik}' = I_{ik} + \mu ( a^2 \delta_{ik} - a_i a_k )$

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Angular Momentum of a Rigid Body

We have

$\vec{M} = \sum m \vec{r} \times \vec{v} = \sum m \vec{r} \times (\vec{V} + \vec{\Omega} \times \vec{r})$

summing over all the particles in the body. Letting our origin be the centre of mass of the body, then $\vec{V} = 0$ , and so

$\vec{M} = \sum m \vec{r} \times (\vec{\Omega} \times \vec{r})$

$\vec{M} = \sum m \Big(r^2 \vec{\Omega} - \vec{r} (\vec{r} \cdot \vec{\Omega} ) \Big)$

and in component form

$M_i = \sum m ( x_k^2 \Omega_i - x_i x_k \Omega_k$

$\Rightarrow M_i = \Omega_k \sum m ( x_j^2 \delta_{ik} - x_i x_k )$

$\Rightarrow M_i = I_{ik} \Omega_k$

$\Rightarrow \vec{M} = I \vec{\Omega}$

If $x_1, x_2, x_3$ are the principal axes of inertia, then

$M_1 = I_1 \Omega_1, M_2 = I_2 \Omega_2, M_3 = I_3 \Omega_3$

For a spherical top, $I_1 = I_2 = I_3 = I$ so $\vec{M} = I \vec{\Omega}$ , pointing in the same direction.

For a symmetric top, $I_1 = I_2 \neq I_3$ . In the case of free rotations, the angular momentum is constant. We let $x_2$ be perpendicular to the plane of $\vec{M}$ and $x_3$ , giving

$M_2 = 0 \Rightarrow \Omega_2 = 0$

$\Omega_1 = \frac{M_1}{I_1} = \frac{M \sin \theta}{I_1}$

$\Omega_3 = \frac{M_3}{I_3} = \frac{M \cos \theta}{I_3}$

where $\theta$ is the angle between the $x_3$ axis and $\vec{M}$ .

We can decompose $\vec{\Omega} = \vec{\Omega}_{Pr} + \vec{\Omega}_3'$ , where $\vec{\Omega}_{Pr}$ is the precessional rotation about $\vec{M}$ , and $\vec{\Omega}_3'$ is along $x_3$ . We have that

$\Omega_1 = \Omega_{pr} \sin \theta$

$\Rightarrow \Omega_{pr} = \frac{M}{I_1}$

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Equations of Motion of a Rigid Body

For each particle we have

$\dot{\vec{p}} = \dot{\vec{f}}$

where $\vec{p}$ is each particle's momentum, and $\vec{f}$ is the total force on each particle. Summing, we obtain

$\dot{\vec{P}} = \vec{F}$

which includes only external forces.

Let $U$ be the potential of a rigid body in an external field. Then, when the body undergoes a translation $\delta \vec{R}$ the radius vector of each point in the body translates as $\vec{r} \rightarrow \vec{r} + \delta \vec{R}$ . Hence,

$\delta U = \sum \frac{\partial U}{\partial \vec{r}} \cdot \delta \vec{r} = \sum \frac{\partial U}{\partial \vec{r}} \cdot \delta \vec{R} = \delta \vec{R} \cdot \sum \frac{\partial U}{\partial \vec{r}} = - \delta \vec{R} \cdot \sum \vec{f} = - \vec{F} \cdot \delta \vec{R}$

$\Rightarrow \vec{F} = - \frac{\partial U}{\partial \vec{R}}$

We now consider the total time derivative of the angular momentum:

$\dot{\vec{M}} = \frac{d}{dt} \left(\sum \vec{r} \times \vec{p}\right)$

$\Rightarrow \dot{\vec{M}} = \sum \dot{\vec{r}} \times \vec{p} + \sum \vec{r} \times \dot{\vec{p}}$

Now, as $\vec{v} = \vec{V} + \dot{\vec{r}}$ and $\vec{V} = 0$ in the centre of mass frame we have that $\vec{v} = \frac{\vec{p}}{m} = \dot{\vec{r}}$ , hence the first term is zero. Therefore,

$\dot{\vec{M}} = \vec{K} = \sum \vec{r} \times \vec{f}$

where $\vec{K}$ is called the torque. The torque depends on the choice of origin: consider moving origin a distance $\vec{a}$ , then $\vec{r}^{\prime} = \vec{r} - \vec{a}$ giving

$\vec{K} = \vec{K}^{\prime} + \vec{a} \times \vec{F}$

If the total force is zero, the torque is independent of choice of origin and the body is known as a couple.

Using a similar argument to the above, we have that

$\vec{K} = - \frac{\partial U}{\partial \vec{\phi}}$

The above equations of motion could also have been derived using the Lagrangian

$L = \frac{1}{2} \mu V^2 + \frac{1}{2} I_{ik} \Omega_i \Omega_k - U (\vec{R}, \vec{\phi})$ .

Consider the case where force is perpendicular to the torque, then we can find an $\vec{a}$ such that $\vec{K}^{\prime} = 0$ and so $\vec{K} = \vec{a} \times \vec{F}$ . Note that this choice of $\vec{a}$ is not unique as we can always transform as $\vec{a} \rightarrow \vec{a} + \vec{a}_p$ where $\vec{a}_p$ is parallel to the torque and so contributes nothing in the cross product. This gives a line along which the force acts.

In a uniform field of force described by $\vec{f}_{\alpha} = e_{\alpha} \vec{E}$ then $\vec{F} = \vec{E} \sum_{\alpha} e_{\alpha}$ and the total torque is given by

$\vec{K} = \vec{r}_0 \times \vec{F}$

where $\vec{r}_0 = \frac{\sum_{\alpha} e_{\alpha} \vec{r}}{\sum_{\alpha} e_{\alpha}}$

e.g. in a uniform gravitational field, where $\vec{r}_0$ is the centre of mass position vector.

[edit]

Rotations and Euler Angles

Euler angles

A rotation in 2-dimensions is given by the following matrix:

$\begin{pmatrix}\cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{pmatrix}$

We describe the rotation in 3-dimensions of the moving axes $x_1 x_2 x_3$ by:

i) first rotating the axes through an angle $\phi$ in the XY plane.

ii) rotating through an angle $\theta$ about the line ON, which is the intersection of the $x_1 x_2$ plane with the XY plane, known as the line of nodes.

iii) rotating through an angle $\psi$ around the $x_3$ axis.

$\theta, \phi$ and $\psi$ are the Euler angles. $\theta \in [0, \pi]$ , $\phi, \psi \in [0, 2 \pi]$ . This gives a general rotation matrix:

$\begin{pmatrix}\cos \phi & \sin \phi & 0 \\ - \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & - \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} \cos \psi & \sin \psi & 0 \\ - \sin \psi & \cos \psi & 0 \\ 0 & 0 & 1\end{pmatrix}$

We can now find the components of $\vec{\Omega}$ along $x_1, x_2, x_3$ in terms of the Euler angles. Decomposing the rates of change of each angle onto each axis, we find:

$\Omega_1 = \dot{\phi} \sin \theta \sin \psi + \dot{\theta} \cos \psi$

$\Omega_2 = \dot{\phi} \sin \theta \cos \psi - \dot{\theta} \sin \psi$

$\Omega_3 = \dot{\phi} \cos \theta + \dot{\psi}$

If $x_1, x_2, x_3$ are the principal axes, these expressions allow us to find the rotational kinetic energy.

For a symmetrical top, $I_1 = I_2 \neq I_3$ , and

$T_{rot} = \frac{1}{2} I_1 ( \dot{\phi}^2 \sin^2 \theta + \dot{\theta}^2 ) + \frac{1}{2} I_3 (\dot{\phi} \cos \theta + \dot{\psi})^2$

We let $\vec{M}$ be along the Z axis and let the $x_1$ axis coincide with the line of nodes at the instant considered, so that $\psi = 0$ . Hence,

$M_1 = I_1 \Omega_1 = I_1 \dot{\theta} = 0$

as $x_1$ is perpendicular to the Z axis, meaning that $\theta$ is constant, and

$M_2 = I_1 \Omega_2 = I_1 \dot{\phi} \sin \theta = M \sin \theta$

giving the precessional velocity

$\dot{\phi} = \frac{M}{I_1}$

and

$M_3 = I_3 \Omega_3 = I_3 ( \dot{\phi} \cos \theta + \dot{\psi}) = M \cos \theta$

giving the angular velocity about the $x_3$ axis,

$\Omega_3 = \frac{M}{I_3} \cos \theta$

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Euler's Equations

Let $\vec{A}$ be a vector, then its total rate of change can be written

$\frac{d \vec{A}}{dt} = \frac{d' \vec{A}}{dt} + \vec{\Omega} \times \vec{A}$

where the first term is the rate of change of the vector with respect to the moving axes, and the second is its rotational rate of change. We hence have that

$\frac{d' \vec{P}}{dt} + \vec{\Omega} \times \vec{P} = \vec{F}$

$\frac{d' \vec{M}}{dt} + \vec{\Omega} \times \vec{M} = \vec{K}$

In component form, $\left( \frac{d' \vec{P}}{dt}\right)_i = \frac{dP_i}{dt}$ , and letting $\vec{P} = \mu \vec{V}$ we get

$\mu \left( \frac{d V_1}{dt} + \Omega_2 V_3 - \Omega_3 V_2\right) = F_1$

$\mu \left( \frac{d V_2}{dt} + \Omega_3 V_1 - \Omega_1 V_3\right) = F_2$

$\mu \left( \frac{d V_3}{dt} + \Omega_1 V_2 - \Omega_2 V_1\right) = F_3$

or in tensor notation

$\mu \left(\frac{d V_i}{dt} + \varepsilon_{ijk}\Omega_j \Omega_k \right) = F_i$

Letting $M_i = I_i \Omega_i$ ,

$I_1 \frac{d \Omega_1}{dt} + (I_3 - I_2) \Omega_2 \Omega_3 = K_1$

$I_2 \frac{d \Omega_2}{dt} + (I_1 - I_3) \Omega_3 \Omega_1 = K_2$

$I_3 \frac{d \Omega_3}{dt} + (I_2 - I_1) \Omega_1 \Omega_2 = K_3$

These six equations are known as Euler's equations.

[edit]

Hamiltonian Dynamics

[edit]

Hamilton's Equations of Motion

Consider a general Lagrangian, $L(q^i, \dot{q}^i,t)$ . We can write its differential as

$d L = \frac{\partial L}{\partial q^i} dq^i + \frac{\partial L}{\partial \dot{q}^i}d \dot{q}^i$

or in terms of the generalised momentum $p_i \equiv \frac{\partial L}{\partial q^i}$

$d L = \dot{p}_i dq^i + p_i d \dot{q}^i$

$\Rightarrow d L = d(p_i \dot{q}^i)- \dot{q}^i dp_i + \dot{p}_i dq^i$

$\Rightarrow d(p_i\dot{q}^i - L) = \dot{q}^i dp_i - \dot{p}_i dq^i$

where the quantity $p_i\dot{q}^i - L \equiv H$ is the Hamiltonian of the system, a function of $p_i$ and $q^i$ . We then have Hamilton's equations of motion

$\dot{p}_i = - \frac{\partial H}{\partial q^i}$

$\dot{q}^i = \frac{\partial H}{\partial p_i}$

a set of $2s$ first-order differential equations for the quantities $p_i,q^i$

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Legendre's Transformation

The transformation $q^i, \dot{q}^i \Rightarrow q^i, p_i$ is an example of Legendre's transformation, which is a mathematical method for switching from one set of independent coordinates to another. Given a function $f(x,y)$ such that

$df = u dx + v dy$

such that $\frac{\partial f}{\partial x} = u$ we can transform to a function $g(u,y)$ by rewriting the differential as

$d\underbrace{(f - xy)}_{g(u,y)} = - xdu + vdy$

As a more involved example, we can apply Legendre's transformation to the general Lagrangian

$L = \frac{1}{2} g_{kl}(q) \dot{q}^k \dot{q}^l - U(q)$

where $g_{kl}$ is the metric of the configuration space or manifold. We have

$p_i = \frac{\partial L}{\partial \dot{q}^i} = g_{ij} \dot{q}^j$

and

$\dot{p}_i = \frac{\partial L}{\partial q^i} = \frac{1}{2} \frac{\partial g_{kl}}{\partial q^i} \dot{q}^k \dot{q}^l - \frac{\partial U}{\partial q^i}$

We now introduce a new tensor $g^{ij}$ such that

$g^{ij} g_{jk} = \delta_k^i$

Now, $p_i = g_{ij} \dot{q}^j$

$\Rightarrow g^{ki} p_i = g^{ki} g_{ij} \dot{q}^j = \delta_j^k \dot{q}^j = \dot{q}^k$

and hence we have $2s$ first-order equations

$\dot{q}^i = g^{ij} p_j$

$\dot{p}_i = \frac{1}{2} \frac{\partial g_{kl}}{\partial q^i} g^{kj} p_j g^{ln} p_n - \frac{\partial U}{\partial q^i}$

Now we can rewrite the second equation using the product rule as

$\frac{\partial g_{kl}}{\partial q^i} \dot{q}^k \dot{q}^l = \left( \frac{\partial}{\partial q^i} \underbrace{(g_{kl} g^{kj})}_{=1} - g_{kl} \frac{\partial g^{kj}}{\partial q^i}\right) g^{ln} p_j p_n$

$\qquad = - g_{kl} g^{ln} \frac{\partial g^{kj}}{\partial q^i} p_j p_n = - \frac{\partial g^{nj}}{\partial q^i} p_n p_j$

as $g_{kl} g^{ln} = \delta_k^n$ . Hence we have

$\dot{q}^i = g^{ij} p_j = \frac{\partial}{\partial p_i} \left( \frac{1}{2} g^{kl} p_k p_l + U\right)$

$\dot{p}_i = - \frac{\partial}{\partial q^i} \left( \frac{1}{2} g^{kl} p_k p_l + U\right)$

where

$H(p,q) = \frac{1}{2} g^{kl} p_k p_l + U(q)$

is the Hamiltonian of the system.

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The Routhian

It is sometimes convenient to replace only some of the generalised velocities in momentum. This leads to a function known as the Routhian, which is a Lagrangian with respect to some of the coordinates, and a Hamiltonian with respect to others.

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Poisson Brackets

Consider some function $f = f(p,q,t)$ , then

$\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial q^k} \frac{d q^k}{d t} + \frac{\partial f}{\partial p_k} \frac{d p_k}{dt}$

summing over $k$ . Then, using Hamilton's equations of motion,

$\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial q^k} \frac{\partial H}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial H}{\partial q^k}$

$\Rightarrow \frac{df}{dt} = \frac{\partial f}{\partial t} + \lbrace H, f\rbrace$

where

$\lbrace H, f \rbrace = \frac{\partial H}{\partial p_k}\frac{\partial f}{\partial q^k} - \frac{\partial H}{\partial q^k}\frac{\partial f}{\partial p_k}$

called the Poisson bracket of $H$ and $f$ . We have that

$\dot{q}^i = \lbrace H, q^i \rbrace$

$\dot{p}_i = \lbrace H, p_i \rbrace$

If $f$ is an integral of the motion, then

$\lbrace H,f \rbrace = 0$

Thus for two arbitrary functions $f,g$ of $p,q,t$ we have

$\lbrace f,g \rbrace = \sum_{k=1}^{s} \left( \frac{\partial f}{\partial p_k}\frac{\partial g}{\partial q^k} - \frac{\partial f}{\partial q^k}\frac{\partial g}{\partial p_k}\right)$

Properties of the Poisson Bracket:

$\lbrace p_i, q^j\rbrace - \delta_i^j$ , $\lbrace p_i, p_j \rbrace = 0$ and $\lbrace q^i, q^j \rbrace = 0$
$\lbrace f, q^k\rbrace = \frac{\partial f}{\partial p_k}$ and $\lbrace f, p_k\rbrace = -\frac{\partial f}{\partial q^k}$
$\lbrace f,g\rbrace = - \lbrace g,f \rbrace$
$\lbrace f,c\rbrace = 0 \, \forall c$ not depending on $p,q$
$\lbrace f_1 + f_2,g \rbrace = \lbrace f_1,g \rbrace + \lbrace f_2,g \rbrace$
$\lbrace f_1 f_2,g \rbrace = f_2 \lbrace f_1 ,g \rbrace + f_1 \lbrace f_2,g \rbrace$
$\frac{\partial}{\partial t} \lbrace f,g\rbrace = \lbrace \frac{\partial f }{\partial t},g\rbrace + \lbrace f,\frac{\partial g}{\partial t}\rbrace$
Jacobi's identity: $\lbrace f,\lbrace g,h\rbrace \rbrace + \lbrace g,\lbrace h,f\rbrace \rbrace + \lbrace h,\lbrace f,g\rbrace \rbrace =0$
Poisson's theorem: If $f,g$ integrals of the motion then $\lbrace f,g \rbrace$ is an integral of the motion.

To prove Poisson's theorem, note that we have

$\frac{df}{dt} = \frac{\partial f}{\partial t} + \lbrace H,f\rbrace = 0$

$\Rightarrow \frac{\partial f}{\partial t} = - \lbrace H,f\rbrace$

and similarly,

$\frac{\partial g}{\partial t} = - \lbrace H,g\rbrace$

so then

$\frac{d}{dt} \lbrace f,g \rbrace = \frac{\partial}{\partial t} \lbrace f,g \rbrace + \lbrace H , \lbrace f,g \rbrace \rbrace$

$\Rightarrow \frac{d}{dt} \lbrace f,g \rbrace = \left \lbrace \frac{\partial f}{\partial t} ,g \right \rbrace + \left \lbrace f ,\frac{\partial g}{\partial t}\right \rbrace + \lbrace H , \lbrace f,g \rbrace \rbrace$

$\Rightarrow \frac{d}{dt} \lbrace f,g \rbrace = - \lbrace \lbrace H,f \rbrace, g \rbrace + \lbrace f, - \lbrace H,g \rbrace \rbrace + \lbrace H , \lbrace f,g \rbrace \rbrace$

$\Rightarrow \frac{d}{dt} \lbrace f,g \rbrace = \lbrace g, \lbrace H,f \rbrace \rbrace + \lbrace f, \lbrace g,H \rbrace \rbrace + \lbrace H , \lbrace f,g \rbrace \rbrace = 0$

using $\lbrace f,g \rbrace = - \lbrace g,f \rbrace$ and Jacobi's identity.

For the coordinates $(q^i, p_i)$ we may introduce new notation

$x^M \,\,\,\, M= 1 \dots 2s$

such that

$x^i = q^i \,\,\,\, i= 1 \dots s$

$x^{s+i} = p_i \,\,\,\, i = 1 \dots s$

so we have

$\lbrace x^M, x^N \rbrace = \omega^{MN}$

the Poisson bivector, where

$\omega^{MN} = \begin{cases} - \delta^{ij} \,\,\,\, M=i, N = j+s \\ \delta^{ij} \,\,\,\, M=i+s, N = j\end{cases}$

$\omega^{MN} = \begin{pmatrix} 0 & -\mathbb{I}_s \\ \mathbb{I}_s & 0 \end{pmatrix}$

where $\mathbb{I}_s$ is the identity $s \times s$ matrix. We then have that

$\lbrace f, g \rbrace = \frac{\partial f}{\partial x^M} \omega^{MN} \frac{\partial g}{\partial x^N}$

and the equations of motion are given by

$\dot{x}^M = \lbrace H, x^M \rbrace$

All properties of the Poisson bracket are satisfied by any constant skew-symmetric matrix $\omega^{MN}$ . If $\omega^{MN} = \omega^{MN}(x)$ then we require that Jacobi's identity be satisfied. Hence if

$\lbrace x^M, x^N \rbrace = \omega^{MN}(x)$

$\lbrace f(x), g(x) \rbrace = \frac{\partial f}{\partial x^M} \omega^{MN}(x) \frac{\partial g}{\partial x^N}$

$\lbrace f,\lbrace g,h\rbrace \rbrace + \lbrace g,\lbrace h,f\rbrace \rbrace + \lbrace h,\lbrace f,g\rbrace \rbrace =0$

then we have the most general Poisson bracket. Any manifold $M$ parametrised by $x^M$ equipped with a Poisson structure is said to be a Poisson manifold.

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Canonical Transformations

In the Lagrangian formalism we may always transform from one set of generalised coordinates $q^i$ to another, $Q^i = Q^i(q,t)$ . Such a transformation is called a point transformation. In the Hamiltonian formalism however, the generalised momenta $p_i$ are on an equal footing with the generalised coordinates, thus we may make transformations to new coordinates and momenta of the form

$Q^i = Q^i(p,q,t), P_i =P_i (p,q,t)$

We wish to find such transformations such that the equations of motion of the new coordinates are of the same form as the old, that is,

$\dot{Q}^i = \frac{\partial H^{\prime}}{\partial P_i}$

$\dot{P}_i = - \frac{\partial H^{\prime}}{\partial Q^i}$

Now, we can consider deriving Hamilton's equations of motion from an action principle, where we have

$\delta S = \delta \! \int(p_i \dot{q}^i - H) dt = \delta \! \int (p_i dq^i - H dt)$

For the new coordinates,

$\delta S^{\prime} = \delta \! \int (P_i dQ^i - H^{\prime} dt)$

and the two integrands are equivalent if they differ only by the differential of some function $S_1$ of the coordinates, momenta and time, that is

$p_i dq^i - H dt = P_i dQ^i - H^{\prime} dt - dS_1$

such $S = S^{\prime} - S_1(t_1) + S_2(t_2)$ , where the last two terms on the right are constant and so have no effect on the variational principle. A transformation satisfying this condition is said to be canonical.

So we have

$dS_1 = - p_i dq^i + P_i dQ^i - (H^{\prime} - H)dt$

which implies we have a function $S_1 = S_1 (q,Q,t)$ such that

$p_i = - \frac{\partial S_1}{\partial q^i}$

$P_i = \frac{\partial S_1}{\partial Q^i}$

$H^{\prime} - H = - \frac{\partial S_1}{\partial t}$

$S_1 (q,Q,t)$ is called the generating function of the canonical transformation.

We can also rewrite our differential expression as

$d \underbrace{(S_1 - P_iQ^i)}_{=S_2(q,P,t)} = - p_i dq^i - Q^i dP_i - (H^{\prime} - H)dt$

hence

$p_i = - \frac{\partial S_2}{\partial q^i}$

$Q^i = - \frac{\partial S_2}{\partial P_i}$

Similarly we can obtain $S_3(Q,p,t)$ and $S_4(p,P,t)$ such that

$q^i = \frac{\partial S_3}{\partial p_i}$

$P_i = \frac{\partial S_3}{\partial Q^i}$

and

$q^i = \frac{\partial S_4}{\partial p_i}$

$Q^i = - \frac{\partial S_4}{\partial P_i}$

In all cases we have

$H^{\prime} - H = - \frac{\partial S_i}{\partial t}$

The Poisson bracket is invariant under a canonical transformation,

$\lbrace f,g \rbrace_{p,q} = \lbrace f,g \rbrace_{P,Q}$

$\lbrace f,g \rbrace_{x} = \lbrace f,g \rbrace_{X}$

Now, we have $\lbrace X^M, X^N \rbrace = \omega^{MN}$ (true as the transformed coordinates satisfy Hamilton's equations of motion). So

$\left \lbrace f(X(x)), g(X(x))\right \rbrace_x = \frac{\partial f}{\partial x^k} \omega^{kl} \frac{\partial g}{\partial x^l} = \frac{\partial f}{\partial X^M} \frac{\partial X^M}{\partial x^k}\omega^{kl} \frac{\partial X^N}{\partial x^l} \frac{\partial g}{\partial X^N} = \frac{\partial f}{\partial X^M} \lbrace X^M, X^N \rbrace_x \frac{\partial g}{\partial X^N}$

hence if the transformation is canonical we get

$\left \lbrace f, g\right \rbrace_x = \frac{\partial f}{\partial X^M} \omega^{MN} \frac{\partial g}{\partial X^N} = \lbrace f,g \rbrace_X$

Thus we have the following relationships which can be used to check whether a transformation is canonical:

$\lbrace Q^i, Q^j\rbrace_{p,q} = 0$

$\lbrace P_i,P_j\rbrace_{p,q} = 0$

$\lbrace P_i, Q^j\rbrace =\delta_i^j$

Given some function $V(p,q,t)$ we can generate a canonical transformation using the formulae

$P_i = p_i + \lbrace V,p_i\rbrace + \frac{1}{2!} \lbrace V, \lbrace V, p_i\rbrace \rbrace + \frac{1}{3!} \lbrace V, \lbrace V, \lbrace V,p_i \rbrace \rbrace \rbrace + \dots$

$\Rightarrow P_i = p_i + \sum_{n=1}^{\infty} \frac{1}{n!} \underbrace{\lbrace V \lbrace V \dots \lbrace V}_{n}, p_i \rbrace \dots \rbrace$

and

$Q^i = q^i + \sum_{n=1}^{\infty} \frac{1}{n!} \underbrace{\lbrace V \lbrace V \dots \lbrace V}_{n}, q^i \rbrace \dots \rbrace$

Any function $f(p,q)$ transforms as

$f(p,q) \rightarrow \tilde{f}(p,q) = f + \sum_{n=1}^{\infty} \frac{1}{n!} \underbrace{\lbrace V \lbrace V \dots \lbrace V}_{n}, f \rbrace \dots \rbrace$

which we can notate as

$\tilde{f} = e^{\lbrace V,}\cdot f$

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Action as a Generating Function

We may consider the action as a function of the initial and final positions and times,

$S(q^{(1)},q^{(2)},t_1,t_2) = \int_{t_1}^{t_2} L (q(t),\dot{q}(t),t) dt$

Let $q^{(1)},t_1,t_2$ be fixed so that $S = S(q^{(2)})$ , then

$\delta S(q^{(2)}) = \int_{t_1}^{t_2} \left( \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}}\delta \dot{q}\right) dt = 0$

$\Rightarrow \delta S(q^{(2)}) = \int_{t_1}^{t_2} \left( \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}}\delta q\right) + \frac{\partial L}{\partial q} \delta q -\frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \delta q \right) dt = 0$

where the latter terms clearly vanish as the path is a trajectory, hence

$\delta S(q^{(2)}) = \frac{\partial L}{\partial \dot{q}}(t_2) \delta q^{(2)}$

$\delta S(q^{(2)}) = p^{(2)} \delta q^{(2)}$

$\frac{\partial S(q^{(2)})}{\partial q_i^{(2)}} = p_i^{(2)}$

Now let $q^{(1)},t_1$ be fixed so that $S = S(q^{(2)},t_1)$ , then

$\frac{d}{dt_2} S = \frac{d}{dt_2} \int_{t_1}^{t_2} L dt = L (q^{(2)},\dot{q}^{(2)},t)$

and also

$\frac{dS}{dt_2} = \frac{\partial S}{\partial t_2} + \frac{\partial S}{\partial q^{(2)}} \dot{q}^{(2)} = \frac{\partial S}{\partial t_2} + p^{(2)} \dot{q}^{(2)}$

$\Rightarrow \frac{\partial S}{\partial t_2} = L^{(2)} - p^{(2)} \dot{q}^{(2)} = - H (p^{(2)},q^{(2)},t_2)$

Similarly, if we let $q^{(2)},t_2$ be fixed we find that

$d S = - p_i^{(1)} d q_i^{(1)} + H (p^{(1)},q^{(1)},t_1)$

so in general we have

$S = S(q^{(1)},q^{(2)},t_1,t_2)$

and

$dS = p_i^{(2)} d q_i^{(2)} - p_i^{(1)} d q_i^{(1)} - H^{(2)} dt_2 + H^{(1)} dt_1$

corresponding to a generating function of the type $S_1(q,Q,t)$ where we have

$q_i \equiv q_i^{(1)} \,\,\,\, p_i \equiv p_i^{(1)} \,\,\,\, H^{\prime} \equiv H^{(2)}$

$Q_i \equiv q_i^{(2)} \,\,\,\, P_i \equiv p_i^{(2)} \,\,\,\, H^{\prime} \equiv H^{(1)}$

and

$dt_2 = dt_1 = dt$

Hence we see that the dynamics of the system correspond to a canonical transformations (from the coordinates and momenta at the time $t_1$ to the coordinates and momenta at the time $t_2$ ).

[edit]

Liouville's Theorem

Volume in phase space is preserved.

[edit]

The Hamilton-Jacobi Equation

[edit]

Hamilton-Jacobi Equation

Recall the action

$S = S(q^{(1)},q^{(2)},t_1,t_2)$

and

$dS = p_i^{(2)} d q_i^{(2)} - p_i^{(1)} d q_i^{(1)} - H^{(2)} dt_2 + H^{(1)} dt_1$

and now fix $t_1, q^{(1)},p^{(1)}$ such that $t_1 \equiv t_0 \Rightarrow dt_1 = 0$ , $q^{(1)} \equiv q_0$ and $p^{(1)} \equiv p_0$ , and let $q^{(2)}, p^{(2)}, t_2 \equiv q,p,t$ . Then consider our new action

$\tilde{S} (q,q_0,t,t_0)$

with

$d\tilde{S} = p dq - p_0 dq_0 - H(q,p,t) dt$

$\Rightarrow d(\tilde{S} + p_0 d q_0) = p dq + q_0 dp_0 - H dt$

$\Rightarrow dS(q,p_0,t)= p dq + q_0 dp_0 - H dt$

giving

$p = \frac{\partial S}{\partial q} \,\,,\,\, q_0 = \frac{\partial S}{\partial p_0} \,\,,\,\, H(q,p,t) = - \frac{\partial S}{\partial t}$

the last of which means our new Hamiltonian $H^{\prime}$ is identically zero; which means the new coordinates and momenta are therefore constant. It also gives us the Hamilton-Jacobi equation:

$\frac{\partial S}{\partial t} + H \left( q_1 \dots q_s, \frac{\partial S}{\partial q_1} \dots \frac{\partial S}{\partial q_s}, t\right) = 0$

So effectively what we are doing are transforming via the generating function $S$ from some coordinates $p,q$ to a set of constant coordinates $p_0,q_0$ , and then solving the Hamilton-Jacobi equations for $S$ , which allows us to calculate $p,q$ as functions of $p_0,q_0$ and time, thus solving the system.

Now assume we have a solution to the Hamilton-Jacobi equation, $S(q_1 \dots q_s, \alpha_1 \dots \alpha_s, t) + A$ where the $\alpha_i$ are $s$ independent constants, and $A$ is an arbitrary constant which will vanish as the Hamilton-Jacobi equation contains only derivatives of $S$ . We now identify $P_i = \alpha_i$ , (i.e. we canonically transform to the new constant momenta), so we have

$p_i = \frac{\partial S (q, \alpha,t)}{\partial q_i}$

allowing us to evaluate the $\alpha_i$ in terms of $q$ and $p$ , and

$Q_i = \beta_i = \frac{\partial S (q, \alpha,t)}{\partial \alpha_i}$

where the $\beta_i$ are our new constant coordinates and may be evaluated using initial conditions. Thus we can solve for

$q_i = q_i(\alpha, \beta,t)$

$p_i = p_i(\alpha, \beta,t)$

giving us the motion of the system.

If the Hamiltonian is independent of time, $H=H(q,p)$ then

$S(q, \alpha,t) =S_0 (q,\alpha) - E(\alpha) t$

and so $H\left(q, \frac{\partial S_0}{\partial q}\right) = E(\alpha)$ and $\frac{d}{dt} S_0 = \frac{\partial S_0}{\partial q} \dot{q} = p \dot{q} \Rightarrow S_0 = \int^t p_i \dot{q}_i dt$ , called the abbreviated action.

Example: Harmonic Oscillator

The Hamiltonian is

$H = \frac{1}{2m} (p^2 + m^2 \omega^2 q^2) \equiv E$

and we have

$S = S_0 - Et$

as the Hamiltonian is independent of $t$ . Then,

$\frac{1}{2m} \left[ \left(\frac{\partial S_0}{\partial q}\right)^2 + m^2 \omega^2 q^2\right] = E = \mbox{constant}$

$\frac{\partial S_0}{\partial q} = \sqrt{2mE - m^2 \omega^2 q^2}$

$\Rightarrow S_0 = \int \sqrt{2mE - m^2 \omega^2 q^2} dq$

Now, we have $Q = \beta = \frac{\partial S}{\partial \alpha}$ and we let $\alpha =E$ , then

$\beta = \frac{\partial S}{\partial E} = \frac{\partial S_0}{\partial E} - t$

and we have that

$\frac{\partial S_0}{\partial E} = \int \frac{m}{\sqrt{2mE - m^2 \omega^2 q^2}} dq = \frac{1}{\omega} \arcsin \left(q \sqrt{\frac{m \omega^2}{2E}}\right)$

$\beta + t = \frac{1}{\omega} \arcsin \left(q \sqrt{\frac{m \omega^2}{2E}}\right)$

$\Rightarrow q = \sqrt{\frac{2E}{m \omega^2}} \sin (\omega t + \omega \beta)$

$p =\frac{\partial S}{\partial q} = \frac{\partial S_0}{\partial q} = \sqrt{2mE - m^2 \omega^2 q^2} = \sqrt{2mE} \cos (\omega t + \omega \beta)$

and $\beta$ is determined from the initial conditions, using $\tan \omega \beta = \frac{q_0}{m \omega p_0}$ .

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Separation of Variables

Let us denote the Hamilton-Jacobi equation by

$F\left(q_i, t, \frac{\partial S}{\partial q_i}, \frac{\partial S}{\partial t}\right) = 0$

and suppose that $q_n$ only appears in some combination $\phi\left(\frac{\partial S}{\partial q_n}, q_n\right)$ , so that

$F \left(q_i^{\prime}, t, \frac{\partial S}{\partial q_i^{\prime}}, \frac{\partial S}{\partial t}, \phi\left(\frac{\partial S}{\partial q_n}, q_n\right)\right) = 0$

where $q_i^{\prime} = q_1 \dots q_{n-1}$ . We then look for a solution to the Hamilton-Jacobi equation of the form

$S = S^{\prime} \left(q_i^{\prime},t) + S_n(q_n\right)$

Supposing this has been found, then

$F \left(q_i^{\prime}, t, \frac{\partial S^{\prime}}{\partial q_i^{\prime}}, \frac{\partial S^{\prime}}{\partial t}, \phi \left(\frac{\partial S_n}{\partial q_n}, q_n\right) \right) = 0$

which must hold for any value of $q_n$ , hence it follows that

$\phi \left(\frac{\partial S_n}{\partial q_n}, q_n\right) = \alpha_n = \mbox{constant}$

from which we can solve for $S_n$ . This process can then be repeated for each variable, assuming they each only occur in some combination

$\phi_i\left(\frac{\partial S}{\partial q_i}, q_i\right)$

so that we obtain

$S = \sum_{k=1}^n S_k(q_k, \alpha_1 \dots \alpha_n) - E(\alpha_1 \dots \alpha_n) t$

Example: Cyclic Coordinate

Assume that $q_n$ does not appear in the Hamiltonian, then we have by the above that

$\frac{\partial S}{\partial q_n} = \alpha_n = p_n$

$\Rightarrow S_n = \alpha_n q_n$

Example: Spherical Coordinates

In spherical coordinates we have the Lagragian

$L = \frac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2) - U(r, \theta, \phi)$

giving Hamiltonian

$H = \frac{1}{2m} \left( p_r^2 + \frac{1}{r^2} p_{\theta}^2 + \frac{1}{r^2 \sin^2 \theta} p_{\phi}^2\right) + U(r, \theta, \phi)$

Suppose that the potential energy is of the form

$U(r, \theta, \phi) = U_1(r, \theta) + \frac{c(\phi)}{r^2 \sin^2 \theta}$

then we have

$H =\frac{1}{2m} \left( p_r^2 + \frac{1}{r^2} p_{\theta}^2 \right) + U_1(r, \theta) + \frac{1}{r^2 \sin^2 \theta} \underbrace{\left( \frac{p_{\phi}^2}{2m} + c(\phi)\right)}_{\phi(\frac{\partial S}{\partial \phi},\phi)}$

thus $S = S^{\prime}(r,\theta) + S_{\phi}(\phi)$ and

$\frac{1}{2m} \left(\frac{\partial S_{\phi}}{\partial \phi}\right)^2 + c(\phi) = \alpha_{\phi}$

hence

$S_{\phi} = \int^{\phi} \sqrt{2m(\alpha_{\phi} - c(\phi)} d \phi$

and then

$H = \frac{1}{2m} \left( p_r^2 + \frac{1}{r^2} p_{\theta}^2 \right) + U_1(r, \theta) + \frac{\alpha_{\phi}}{r^2 \sin^2 \theta}$

and if

$U_1 (r, \theta) = a(r) + \frac{b(\theta)}{r^2}$

we can repeat the process to obtain

$S = S_r(r) + S_{\theta} (\theta) + S_{\phi} (\phi) - Et$

where

$S_{\theta} (\theta) = \int^{\theta} \sqrt{2m\left(\alpha_{\theta} - \frac{\alpha_{\phi}}{\sin^2 \theta} - b(\theta)\right)} \, d \theta$

and

$E = H = \frac{p_r^2}{2m} + a(r) + \frac{\alpha_{\theta}}{r^2}$

$S_r = \int^r \sqrt{2m \left(E - \frac{\alpha_{\theta}}{r^2} - a(r)\right)} \, dr$

[edit]

Adiabatic Invariants

Consider a system undergoing finite motion and let one of the parameters $\lambda$ vary slowly (adiabatically) with time, such that

$T \frac{d \lambda}{dt} <<\lambda$

$T \frac{d (\log \lambda)}{dt} << 1$

where $T$ is the period of oscillations. Now, if $\lambda$ were constant the system would be closed and the energy $E$ would be constant, hence the period $T(E)$ would be fixed. As $\lambda$ varies only slowly it follows that the energy will also vary slowly,

$\frac{dE}{dt}$ ~ $\frac{d \lambda}{dt} \Rightarrow \frac{dE}{dt} = f(\lambda) \frac{d \lambda}{dt}$

hence $E$ is in some sense a function of $\lambda$ . This dependency can be expressed as the constancy of some combination of $E$ and $\lambda$ , implying the existence of a function

$I(E,\lambda) = \mbox{constant}$

known as an adiabatic invariant.

We look now to average the rate of change of the energy. Let $H(p,q,\lambda)$ be the Hamiltonian of the system. Then

$\frac{dE}{dt} = \frac{\partial H}{\partial t} = \frac{\partial H}{\partial \lambda}\frac{d \lambda}{dt}$

and

$\overline{\frac{dE}{dt}} = \frac{d \lambda}{dt} \overline{\frac{\partial H}{\partial \lambda}}$

as $\lambda$ varies only slowly we can takes its rate of change out of the averaging, i.e. we average over constant $\lambda$ . Now,

$\overline{\frac{\partial H}{\partial \lambda}} = \frac{1}{T} \int_o^T \frac{\partial H}{\partial \lambda} dt$

and from Hamilton's equations of motion,

$\frac{dq}{dt} = \frac{\partial H}{\partial p} \Rightarrow dt = \frac{dq}{\frac{\partial H}{\partial p}}$

and we can write the period as

$T = \int_0^T dt = \oint \frac{dq}{\frac{\partial H}{\partial p}}$

a loop integral as we compute the integration over the range of variance of $p$ and $q$ . Hence we have

$\overline{\frac{dE}{dt}} = \frac{d \lambda}{dt} \frac{\oint \frac{\partial H}{\partial \lambda} \frac{dq}{\frac{\partial H}{\partial p}}}{\oint \frac{dq}{\frac{\partial H}{\partial p}}}$

Now the integrations are performed over a path of constant $\lambda \Rightarrow$ constant $E$ , and the momentum is then a definite function of $q,E,\lambda$ , so

$p = p (q,E,\lambda)$

$H = H(p(q,E,\lambda),q, \lambda) = E$

$\Rightarrow \frac{dH}{d\lambda} = \frac{\partial H}{\partial \lambda} + \frac{\partial H}{\partial p} \frac{\partial p}{\partial \lambda} = 0$

thus

$- \frac{\partial p}{\partial \lambda} = \frac{\frac{\partial H}{\partial \lambda}}{\frac{\partial H}{\partial p}}$

and so we have

$\overline{\frac{dE}{dt}} = - \frac{d \lambda}{dt} \frac{\oint \frac{\partial p}{\partial \lambda} dq}{\oint \frac{\partial p}{\partial E}dq}$

writing $\frac{1}{\frac{\partial H}{\partial p}} = \frac{1}{\frac{\partial E}{\partial p}} = \frac{\partial p}{\partial E}$ . We then obtain

$\oint \left( \frac{\partial p}{\partial E} \overline{\frac{dE}{dt}} + \frac{\partial p}{\partial \lambda} \frac{d \lambda}{dt}\right) dq = 0$

$\Rightarrow \frac{d}{dt} \oint \overline{(p dq)} = 0$

We then have

$I = \frac{1}{2\pi} \oint p dq$

an adiabatic invariant (the scaling factor is convention).

So we have $I = I(E,\lambda)$ and

$\frac{\partial I}{\partial E} = \frac{1}{2\pi} \oint \frac{\partial p}{\partial E} dq = \frac{1}{2\pi} \oint \frac{dq}{\frac{\partial H}{\partial p}} = \frac{T}{2 \pi}$