2 Classical mechanics

Classical mechanics deals with mass points, i.e. masses without spatial extension, which move at speeds $\ll c$ (transition to relativity theory) and are not too light (transition to quantum mechanics).

2.1 Newtonian mechanics

Classical mechanics is based on Newton's axioms [18]. The most important axiom is Newton's Second Axiom:

The impulse is defined by

\[ \begin{align} \mathbf{p} \coloneqq m\frac{d\mathbf{r}}{dt}, \end{align} \]

where $\mathbf{r} = \mathbf{r}\left(t\right)$ is the trajectory of the mass point. The First Axiom is a special case of the Second and is therefore insignificant. Newton's Third Axiom is:

To complete the theory, two additions to Newton's axioms are introduced. First, one determines that if $N \geq 1$ forces $\mathbf{F}_j$ act on a mass point, in Eq. (2.1) to use the sum of these forces is:

\[ \begin{align} m\frac{d^2\mathbf{r}}{dt^2} = \sum_{j = 1}^{N}\mathbf{F}_j \end{align} \]

Furthermore, it is assumed that the force $\mathbf{F}_1$ is parallel to the connecting line $\mathbf{r}_1 - \mathbf{r}_2$:

\[ \begin{align} \mathbf{F}_1 \parallel \mathbf{r}_1 - \mathbf{r}_2 \end{align} \]

Let two point masses $m_1, m_2$ be given at the locations $\mathbf{r}_1, \mathbf{r}_2$. One defines the center of gravity $\mathbf{R}$ by

\[ \begin{align} \mathbf{R} \coloneqq \frac{m_1\mathbf{r}_1 + m_2\mathbf{r}_2}{M} \end{align} \]

with the total mass

\[ \begin{align} M \coloneqq m_1 + m_2. \end{align} \]

Newton's Second Axiom applies to the acceleration of the center of gravity

\[ \begin{align} \frac{d^2\mathbf{R}}{dt^2} = \frac{m_1\frac{d^2\mathbf{r}_1}{dt^2} + m_2\frac{d^2\mathbf{r}_2}{dt^2}}{M} = \frac{1}{M}\left(\mathbf{F}_1 + \mathbf{F}_2\right), \end{align} \]

where $\mathbf{F}_1$ represents the total force acting on $m_1$ and $\mathbf{F}_2$ represents the total force acting on $m_2$. These can be divided into an internal and an external force:

\[ \begin{align} \mathbf{F}_1 &= \mathbf{F}_1^{\text{(int)}} + \mathbf{F}_1^{\text{(ext)}}, & {} & \mathbf{F}_2 = \mathbf{F}_2^{\text{(int)}} + \mathbf{F}_2^{\text{(ext)}} \end{align} \]

The internal force arises from interactions within the system. Due to Newton's third axiom,

\[ \begin{align} \mathbf{F}_1^{\text{(int)}} = -\mathbf{F}_2^{\text{(int)}}. \end{align} \]

So you get

\[ \begin{align} \frac{d^2\mathbf{R}}{dt^2} = \frac{1}{M}\left(\mathbf{F}_1^{\text{(ext)}} + \mathbf{F}_2^{\text{(ext)}}\right). \end{align} \]

So a kind of Newton's second axiom applies

\[ \begin{align} M\frac{d^2\mathbf{R}}{dt^2} = \sum_{i}^{}\mathbf{F}_i^{\text{(ext)}} \end{align} \]

for the movement of the center of gravity coordinate. The relative coordinate is carried out in the same way

\[ \begin{align} \mathbf{r} \coloneqq \mathbf{r}_2 - \mathbf{r}_1 \end{align} \]

a. This satisfies the equation of motion

\[ \begin{align} \frac{d^2\mathbf{r}}{dt^2} = \frac{d^2\mathbf{r}_2}{dt^2} - \frac{d^2\mathbf{r}_1}{dt^2} = \frac{\mathbf{F}_1}{m_1} - \frac{\mathbf{F}_2}{m_2}\Leftrightarrow \frac{m_1m_2}{m_1 + m_2}\frac{d^2\mathbf{r}}{dt^2} = \frac{m_2}{m_1 + m_2}\mathbf{F}_1 - \frac{m_1}{m_1 + m_2}\mathbf{F}_2. \end{align} \]

If you now only assume internal WW, you get

\[ \begin{align} \mu\frac{d^2\mathbf{r}}{dt^2} = \mathbf{F}_1^{\text{(int)}}. \end{align} \]

This was the reduced mass

\[ \begin{align} \mu \coloneqq \frac{m_1m_2}{M} \end{align} \]

introduced. With vanishing external forces, a mass $\mu$ effectively moves in the force field of the pairwise WW, its position vector is $\mathbf{r}$. Some of the results also hold for $N\in \mathbb{N}$ with $N\geq 1$ particles. First you can get a total mass

\[ \begin{align} M \coloneqq \sum_{i = 1}^{N}m_i \end{align} \]

and a center of gravity coordinate

\[ \begin{align} \mathbf{R} \coloneqq \frac{\sum_{i = 1}^{N}m_i\mathbf{r}_i}{M} \end{align} \]

define. You derive this twice in time and get

\[ \begin{align} M\frac{d^2}{dt^2}\mathbf{R} = \sum_{i = 1}^{N}m_i\frac{d^2}{dt^2}\mathbf{r}_i = \sum_{i = 1}^{N}\mathbf{F}_i. \end{align} \]

It applies again

\[ \begin{align} \mathbf{F}_i = \mathbf{F}_i^{\text{(int)}} + \mathbf{F}_i^{\text{(ext)}} = \sum_{\substack{j = 1,\\j\not = i}}^{N}\mathbf{F}_i^{(j)} + \mathbf{F}_i^{\text{(ext)}}. \end{align} \]

Here $\mathbf{F}_i^{(j)}$ is the force that the $jth particle exerts on the $ith particle. Thus you get

\[ \begin{align} \sum_{i = 1}^{N}\mathbf{F}_i = \sum_{\substack{i, j = 1,\\i\not = j}}^{N}\mathbf{F}_i^{(j)} + \sum_{i = 1}^{N}\mathbf{F}_i^{\text{(ext)}} = \sum_{\substack{i, j = 1,\\i>j}}^{N}\mathbf{F}_i^{(j)} + \mathbf{F}_j^{(i)} + \sum_{i = 1}^{N}\mathbf{F}_i^{\text{(ext)}} = \sum_{\substack{i, j = 1,\\i>j}}^{N}\mathbf{F}_i^{(j)} - \mathbf{F}_i^{(j)} + \sum_{i = 1}^{N}\mathbf{F}_i^{\text{(ext)}} = \sum_{i = 1}^{N}\mathbf{F}_i^{\text{(ext)}}. \end{align} \]

This became Newton's third axiom

\[ \begin{align} \mathbf{F}_i^{(j)} = -\mathbf{F}_j^{(i)} \end{align} \]

used. This means: The internal WWs have no influence on the trajectory of the center of gravity. If no external forces act, the total momentum is constant; this is called law of conservation of momentum.

There is a formulation of Newtonian mechanics for rotations (rotational movements). The angular momentum $\mathbf{L}$ of a particle is defined by

\[ \begin{align} \mathbf{L} \coloneqq \mathbf{r}\times\mathbf{p}.\tag{2.23}\label{eq:def_angular_momentum} \end{align} \]

The torque $\mathbf{F}$ is defined as

\[ \begin{align} \mathbf{D} \coloneqq \mathbf{r}\times\mathbf{F}. \end{align} \]

By deriving Eq. (2.23) follows

\[ \begin{align} \frac{d}{dt}\mathbf{L} = \frac{d\mathbf{r}}{dt}\times\mathbf{p} + \mathbf{r}\times\frac{d\mathbf{p}}{dt} = \mathbf{r}\times\mathbf{F} = \mathbf{D}. \end{align} \]

This is also called Newton's Second Axiom of Rotation. For $N\in\mathbb{N}$ with $N\geq 1$ particles one obtains a total angular momentum of

\[ \begin{align} \mathbf{L} = \sum_{i = 1}^{N}\mathbf{L}_i = \sum_{i = 1}^{m}\mathbf{r}_i\times\mathbf{p}_i. \end{align} \]

This applies to this one

\[ \begin{align} \frac{d\mathbf{L}}{dt} &= \sum_{i = 1}^{N}\mathbf{r}_i\times\left(\mathbf{F}_i^{\text{(int)}} + \mathbf{F}_i^{\text{(ext)}}\right) = \sum_{\substack{i, j = 1,\\i\not = j}}^{N}\mathbf{r}_i\times\mathbf{F}_i^{(j)} + \sum_{i = 1}^{N}\mathbf{r}_i\times\mathbf{F}_i^{\text{(ext)}}\nonumber\\ &= \sum_{\substack{i, j = 1,\\i>j}}^{N}\mathbf{r}_i\times\mathbf{F}_i^{(j)} + \mathbf{r}_j\times\mathbf{F}_j^{(i)} + \sum_{i = 1}^{N}\mathbf{r}_i\times\mathbf{F}_i^{\text{(ext)}} = \sum_{\substack{i, j = 1,\\i>j}}^{N}\mathbf{F}_i^{(j)}\times\left(\mathbf{r}_i - \mathbf{r}_j\right) + \sum_{i = 1}^{N}\mathbf{r}_i\times\mathbf{F}_i^{\text{(ext)}}\nonumber\\ &= \sum_{i = 1}^{N}\mathbf{r}_i\times\mathbf{F}_i^{\text{(ext)}}. \end{align} \]

Newton's third axiom was used again as well as the additional statement that the force between two particles acts along the line connecting them. If no external forces act, the angular momentum is conserved; this is called law of conservation of angular momentum.

The concept of energy should be introduced. A force field is a function $\mathbf{F} = \mathbf{F}\left(\mathbf{r}\right)$, which assigns to each point $\mathbf{r}$ a force $\mathbf{F}\left(\mathbf{r}\right)$ acting there. A mass point moves along the trajectory $\mathbf{r} = \mathbf{r}\left(t\right)$. Then you define the work $W$ that the force field does at the mass point in the time interval $\left[t_1, t_2\right]$ by

\[ \begin{align} W \coloneqq \int_{t_1}^{t_2}\mathbf{F}\left(\mathbf{r}\left(t\right)\right)\cdot\frac{d\mathbf{r}}{dt}dt. \end{align} \]

Using Newton's Second Axiom and partial integration follows

\[ \begin{align} W = \int_{t_1}^{t_2}m\frac{d^2\mathbf{r}}{dt^2}\cdot\frac{d\mathbf{r}}{dt}dt &= \left[m\frac{d\mathbf{r}}{dt}\cdot\frac{d\mathbf{r}}{dt}\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}m\frac{d\mathbf{r}}{dt}\cdot\frac{d^2\mathbf{r}}{dt^2}dt\nonumber\\ \Leftrightarrow W &= \frac{1}{2}m\mathbf{v}\left(t_2\right)^2 - \frac{1}{2}m\mathbf{v}\left(t_1\right)^2. \end{align} \]

Define the kinetic energy $E_{\text{kin}}$ by

follows

\[ \begin{align} W = \Delta E_{\text{kin}}. \end{align} \]

So the work done by the field becomes kinetic energy. Furthermore, a force field is called conservative, if a potential $U = U\left(\mathbf{r}\right)$ exists with

\[ \begin{align} \mathbf{F} = -\nabla U. \end{align} \]

In this case, according to Eq. (B.47)

\[ \begin{align} \nabla\times\mathbf{F} = \mathbf{0}, \end{align} \]

Thus, using Stokes' theorem, Eq. (15.26)

\[ \begin{align} \int_{\partial A}\mathbf{F}\cdot d\mathbf{s} = 0 \end{align} \]

for each surface $A$, so that the work done is independent of the path. Define the total energy of the particle by

\[ \begin{align} E \coloneqq E_{\text{kin}} + U, \end{align} \]

so this size is obtained:

\[ \begin{align} \frac{dE}{dt} &= \frac{dE_{\text{kin}}}{dt} + \frac{dU}{dt} = \frac{1}{2}m2\frac{d\mathbf{r}}{dt}\cdot\frac{d^2\mathbf{r}}{dt^2} + \nabla U\cdot\frac{d\mathbf{r}}{dt}\nonumber\\ &= \left(m\frac{d^2\mathbf{r}}{dt^2} + \nabla U\right)\cdot\frac{d\mathbf{r}}{dt} = \left(\mathbf{F} + \nabla U\right)\cdot\frac{d\mathbf{r}}{dt} = \left(-\nabla U + \nabla U\right)\cdot\frac{d\mathbf{r}}{dt} = 0 \end{align} \]

The energy of a particle moving under the influence of a conservative force is therefore constant.

There are two other formulations of classical mechanics, the Lagrange formalism and the Hamilton formalism. Both can be derived from Newton's axioms.

2.2 Lagrange formalism

Newtonian mechanics is very convenient for freely moving mass points (atoms in a gas, planets in a solar system) or rigid bodies (ships, aircraft). In principle, such systems may move in any direction. In many situations, however, motion is restricted, for example for a marble in a track, a billiard ball on a table, or a train on rails. Such systems are subject to constraints. If the system consists of exactly one particle with trajectory $\mathbf{r} = \mathbf{r}\left(t\right)$, a constraint can be written in the form

\[ \begin{align} g\left(\mathbf{r}, t\right) = 0\tag{2.38}\label{eq:def_holonome_zwangsbedingung} \end{align} \]

where $g$ is a scalar field. The condition $g=0$ may be interpreted as a surface. This surface may itself depend explicitly on time, as in the case of a vertically accelerated billiard table. Equation (2.38) is called a holonomic constraint; all others are called nonholonomic. For example, this is the case if $g$ depends on $\frac{d\mathbf{r}}{dt}$. Explicitly time-dependent constraints are called rheonomic, whereas time-independent constraints are called scleronomic. In general, for $N$ particles one can formulate $R$ constraints,

\[ \begin{align} g_\alpha\left(\mathbf{r}_1, \dotsc, \mathbf{r}_N, t\right) = 0 \end{align} \]

with $1\leq\alpha\leq R$ and $R\leq 3N-1$. If $R=3N$, no motion would be possible, because each constraint eliminates one degree of freedom.

2.2.1 Lagrange equations of the first kind

Satisfaction of the constraints is enforced by so-called constraint forces. These are perpendicular to the surface defined by the constraint, which is easy to visualize for motion on a table. Friction acts tangentially to the table, but it does not enforce the constraint; therefore it is not a constraint force. For a constraint force $\mathbf{Z}$ one can write

\[ \begin{align} \mathbf{Z}\left(\mathbf{r}, t\right) = \lambda\left(t\right)\nabla g\left(\mathbf{r}, t\right). \end{align} \]

The multiplier $\lambda$ is time-dependent because the constraint force depends on the time-dependent trajectory. The equation of motion is then

\[ \begin{align} m\frac{d^2}{dt^2}\mathbf{r} = \mathbf{F} + \lambda\left(t\right)\nabla g\left(\mathbf{r}, t\right), \end{align} \]

where $\mathbf{F}$ is the sum of all non-constraint forces (gravity, friction, etc.). For two constraints, this generalizes to

\[ \begin{align} m\frac{d^2}{dt^2}\mathbf{r} = \mathbf{F} + \sum_{\alpha = 1}^{2}\lambda_\alpha\left(t\right)\nabla g_\alpha\left(\mathbf{r}, t\right). \end{align} \]

This is plausible: $g_1=0$ and $g_2=0$ define two surfaces on which the trajectory lies. $\nabla g_1$ and $\nabla g_2$ are perpendicular to their respective surfaces, hence also to the trajectory, and they are linearly independent. Therefore, $\sum_{\alpha=1}^{2}\lambda_\alpha\left(t\right)\nabla g_\alpha\left(\mathbf{r}, t\right)$ is a general ansatz for a force perpendicular to the trajectory, i.e. a general constraint force. Together with $g_1\left(\mathbf{r}, t\right)=g_2\left(\mathbf{r}, t\right)=0$, this yields five equations for the five unknown functions $x\left(t\right),y\left(t\right),z\left(t\right),\lambda_1\left(t\right),\lambda_2\left(t\right)$. The system is therefore, in principle, solvable. In general, consider $N$ particles with Cartesian coordinates $x_n$ ($1\le n\le 3N$), where $x_1,x_2,x_3$ are the coordinates of the first particle, $x_4,x_5,x_6$ those of the second, and so on. For $R\le 3N-1$ holonomic constraints, the equations of motion are

Here, $m_1=m_2=m_3$ is the mass of the first particle, and so forth. Together with the constraints, this yields $3N+R$ equations for the $3N+R$ unknown functions $x_n\left(t\right),\lambda_\alpha\left(t\right)$. These equations are called the Lagrange equations of the first kind.

2.2.2 Lagrange equations of the second kind

If one is not interested in the explicit constraint forces, they can be eliminated. We do this next.

As noted above, each constraint $g_\alpha$ removes one degree of freedom. Hence the number of degrees of freedom is

\[ \begin{align} f = 3N - R \end{align} \]

This is the number of quantities required to specify the positions of all $N$ particles. These quantities are called generalized coordinates $q_1,\dotsc,q_f$. They must satisfy two conditions:

1. They must determine the Cartesian coordinates, i.e. $x_n=x_n\left(q_1,\dotsc,q_f,t\right)$ for all $1\le n\le 3N$.

2. They must incorporate the constraints: $g_\alpha\left(q_1,\dotsc,q_f,t\right)=0$ for $1\le\alpha\le R$ and all admissible $q_i$.

The second condition can be expressed formally as

\[ \begin{align} \frac{dg_\alpha}{dq_k} = \sum_{n = 1}^{3N}\frac{\partial g_\alpha}{\partial x_n}\frac{\partial x_n}{\partial q_k} = 0. \end{align} \]

Multiplying Eq. (2.43) by $\frac{\partial x_n}{\partial q_k}$ yields

\[ \begin{align} m_n\frac{d^2x_n}{dt^2}\frac{\partial x_n}{\partial q_k} = F_n\frac{\partial x_n}{\partial q_k} + \sum_{\alpha = 1}^{R}\lambda_\alpha\left(t\right)\frac{\partial g_\alpha\left(x_1, \dotsc, x_{3N}, t\right)}{\partial x_n}\frac{\partial x_n}{\partial q_k}. \end{align} \]

Summing over $n$ gives

\[ \begin{align} \sum_{n = 1}^{3N}m_n\frac{d^2x_n}{dt^2}\frac{\partial x_n}{\partial q_k} = \sum_{n = 1}^{3N}F_n\frac{\partial x_n}{\partial q_k}.\tag{2.47}\label{eq:lagrange_deriv_3} \end{align} \]

This equation holds for all $1\le k\le f$. Constraint forces no longer appear. Introduce the notation

\[ \begin{align} x \coloneqq \left(x_1, \dotsc, x_n\right), & {} & \newdot{x} \coloneqq \left(\newdot{x}_1, \dotsc, \newdot{x}_n\right)\\ q \coloneqq \left(q_1, \dotsc, q_n\right), & {} & \newdot{q} \coloneqq \left(\newdot{q}_1, \dotsc, \newdot{q}_n\right) \end{align} \]

The quantities $\newdot{q}_i$ are called generalized velocities. Now take the total time derivative of the transformation

\[ \begin{align} x_n = x_n\left(q, t\right) \end{align} \]

to obtain

\[ \begin{align} \newdot{x}_n = \frac{d}{dt}x_n\left(q, t\right) = \sum_{k = 1}^{f}\frac{\partial x_n}{\partial q_k}\newdot{q}_k + \frac{\partial x_n}{\partial t} = \newdot{x}_n\left(q, \newdot{q}, t\right)\tag{2.51}\label{eq:transformation_lagrange_geschwindigkeiten} \end{align} \]

Hence

\[ \begin{align} \frac{\partial\newdot{x}_n}{\partial\newdot{q}_k} = \frac{\partial x_n}{\partial q_k}. \end{align} \]

For the kinetic energy $T$,

\[ \begin{align} T = T\left(\newdot{x}\right) = \sum_{n = 1}^{3N}\frac{1}{2}m_n\newdot{x}_n^2.\tag{2.53}\label{eq:newton_kinetische_energie} \end{align} \]

insert Eq. (2.51):

\[ \begin{align} T = T\left(q, \newdot{q}, t\right) = \sum_{i, k = 1}^{f}m_{i, k}\left(q, t\right)\newdot{q}_i\newdot{q}_k + \sum_{k = 1}^{f}b_k\left(q, t\right)\newdot{q}_k + c\left(q, t\right) \end{align} \]

From Eq. (2.53),

\[ \begin{align} \frac{\partial T\left(q, \newdot{q}, t\right)}{\partial q_k} = \sum_{n = 1}^{3N}m_n\newdot{x}_n\frac{\partial\newdot{x}_n}{\partial q_k}.\tag{2.55}\label{eq:lagrange_deriv_2} \end{align} \]

and also

\[ \begin{align} \frac{\partial T\left(q, \newdot{q}, t\right)}{\partial\newdot{q}_k} = \sum_{n = 1}^{3N}m_n\newdot{x}_n\frac{\partial\newdot{x}_n}{\partial\newdot{q}_k} = \sum_{n = 1}^{3N}m_n\newdot{x}_n\frac{\partial x_n}{\partial q_k}. \end{align} \]

Taking the total time derivative gives

\[ \begin{align} \frac{d}{dt}\frac{\partial T}{\partial\newdot{q}_k} = \sum_{n = 1}^{3N}m_n\frac{d^2x_n}{dt^2}\frac{\partial x_n}{\partial q_k} + \sum_{n = 1}^{3N}m_n\newdot{x}_n\frac{\partial\newdot{x}_n}{\partial q_k}.\tag{2.57}\label{eq:lagrange_deriv_1} \end{align} \]

In the last term we used the identity

\[ \begin{align} \frac{d}{dt}\frac{\partial x_n}{\partial q_k} = \sum_{l = 1}^{f}\frac{\partial^2x_n}{\partial q_l\partial q_k}\newdot{q}_l + \frac{\partial^2 x_n}{\partial t\partial q_k} = \frac{\partial}{\partial q_k}\left(\sum_{l = 1}^{f}\frac{\partial x_n}{\partial q_l}\newdot{ q_l} + \frac{\partial x_n}{\partial t}\right) = \frac{\partial}{\partial q_k}\frac{dx_n}{dt} \end{align} \]

Now define the generalized forces:

\[ \begin{align} Q_k \coloneqq \sum_{n = 1}^{3N}F_n\frac{\partial x_n}{\partial q_k} \end{align} \]

Insert this definition, together with Eqs. (2.57) and (2.55), into Eq. (2.47):

\[ \begin{align} \sum_{n = 1}^{3N}m_n\frac{d^2x_n}{dt^2}\frac{\partial x_n}{\partial q_k} = \sum_{n = 1}^{3N}F_n\frac{\partial x_n}{\partial q_k} & {} & \Leftrightarrow \sum_{n = 1}^{3N}m_n\frac{d^2x_n}{dt^2}\frac{\partial x_n}{\partial q_k} = Q_k\nonumber\\ \Leftrightarrow\frac{d}{dt}\frac{\partial T}{\partial\newdot{q}_k} - \sum_{n = 1}^{3N}m_n\newdot{x}_n\frac{\partial\newdot{x}_n}{\partial q_k} = Q_k & {} & \Leftrightarrow \frac{d}{dt}\frac{\partial T}{\partial\newdot{q}_k} - \frac{\partial T}{\partial q_k} = Q_k \end{align} \]

Assume the forces have the form

\[ \begin{align} F_n = -\frac{\partial U\left(x, \newdot{x}, t\right)}{\partial x_n} + \frac{d}{dt}\frac{\partial U\left(x, \newdot{x}, t\right)}{\partial\newdot{x}_n}\tag{2.61}\label{eq:lagrange_kraefte} \end{align} \]

with a potential $U=U\left(x,\newdot{x},t\right)$. For velocity-independent conservative forces, this reduces to the familiar Newtonian form; the second term is included to account for velocity dependence. Since $\left(x,\newdot{x}\right)$ can be mapped bijectively to $\left(q,\newdot{q}\right)$, one may write

\[ \begin{align} U = U\left(q, \newdot{q}, t\right). \end{align} \]

Therefore,

\[ \begin{align} Q_k = \sum_{n = 1}^{3N}F_n\frac{\partial x_n}{\partial q_k} = -\sum_{n = 1}^{3N}\frac{\partial U}{\partial x_n}\frac{\partial x_n}{\partial q_k} + \frac{d}{dt}\sum_{n = 1}^{3N}\frac{\partial U}{\partial\newdot{x}_n}\frac{\partial x_n}{\partial q_k} = -\frac{\partial U}{\partial q_k} + \frac{d}{dt}\frac{\partial U}{\partial\newdot{q}_k}. \end{align} \]

Hence,

\[ \begin{align} \frac{d}{dt}\frac{\partial\left(T - U\right)}{\partial\newdot{q}_k} = \frac{\partial\left(T - U\right)}{\partial q_k}. \end{align} \]

Define the Lagrangian $L$ by

\[ \begin{align} L\left(q, \newdot{q}, t\right) \coloneqq T\left(q, \newdot{q}, t\right) - U\left(q, \newdot{q}, t\right). \end{align} \]

Then

for $1\leq k\leq f$. These are the Lagrange equations of the second kind.

2.3 Hamilton formalism

First define the canonical impulses by

\[ \begin{align} p_i \coloneqq\frac{\partial L}{\partial\newdot{q}_i} \end{align} \]

for $1\leq i\leq f$. One now eliminates the generalized velocities $\newdot{q}$:

\[ \begin{align} p_i = \frac{\partial L\left(q, \newdot{q}, t\right)}{\partial\newdot{q}_i}\to\newdot{q}_j = \newdot{q}_j\left(q, p, t\right) \end{align} \]

The Hamilton function $H$ is defined by

\[ \begin{align} H\left(q, p, t\right) \coloneqq \sum_{i = 1}^{f}\newdot{q}_i\left(q, p, t\right)p_i - L\left(q, \newdot{q}\left(q, p, t\right), t\right). \end{align} \]

The Hamilton's equations or also canonical equations follow by partial differentiation:

The space of $\left(q, p\right)$ is called phase space.

2.3.1 Poisson bracket

Any two quantities $F$, $K$ can only depend on $\left(q, p, t\right)$. One defines the Poisson bracket $\left\lbrace F, K\right\rbrace$ of $F$ and $K$ by

\[ \begin{align} \left\lbrace F, K\right\rbrace \coloneqq \sum_{i = 1}^f\left(\frac{\partial F}{\partial q_i}\frac{\partial K}{\partial p_i} - \frac{\partial F}{\partial p_i}\frac{\partial K}{\partial q_i}\right).\tag{2.73}\label{eq:poisson_bracket_def} \end{align} \]

The following are immediate consequences:

\[ \begin{align} \left\lbrace F + G, K\right\rbrace &= \left\lbrace F, K\right\rbrace + \left\lbrace G, K\right\rbrace,\tag{2.74}\label{eq:poisson_bracket_prop_1}\\ \left\lbrace F, K\right\rbrace &= -\left\lbrace K, F\right\rbrace,\tag{2.75}\label{eq:poisson_bracket_prop_2}\\ \left\lbrace F, F\right\rbrace &= 0.\tag{2.76}\label{eq:poisson_bracket_prop_3} \end{align} \]

Apply in the Hamilton formalism

\[ \begin{align} \frac{\partial q_i}{\partial q_j} = \delta_{i, j}, & {} & \frac{\partial q_i}{\partial p_j} = 0, & {} & \frac{\partial q_i}{\partial t} = 0,\\ \frac{\partial p_i}{\partial q_j} = 0, & {} & \frac{\partial p_i}{\partial p_j} = \delta_{i, j}, & {} & \frac{\partial p_i}{\partial t} = 0. \end{align} \]

Furthermore are

\[ \begin{align} \left\lbrace F, q_j\right\rbrace = -\frac{\partial F}{\partial p_j}, & {} & \left\lbrace F, p_j\right\rbrace = \frac{\partial F}{\partial q_j}. \end{align} \]

This follows further

\[ \begin{align} \left\lbrace q_i, q_j\right\rbrace = 0, & {} & \left\lbrace p_i, p_j\right\rbrace = 0, & {} & \left\lbrace p_i, q_j\right\rbrace = -\delta_{i, j}. \end{align} \]

For the total time derivative of $F$ we now get

\[ \begin{align} \frac{dF}{dt} = \sum_{i = 1}^f\frac{\partial F}{\partial q_i}\newdot{q}_i + \sum_{i = 1}^f\frac{\partial F}{\partial p_i}\newdot{p}_i + \frac{\partial F}{\partial t} = \sum_{i = 1}^f\frac{\partial F}{\partial q_i}\frac{\partial H}{\partial p_i} - \sum_{i = 1}^f\frac{\partial F}{\partial p_i}\frac{\partial H}{\partial q_i} + \frac{\partial F}{\partial t} \end{align} \]

This implies

\[ \begin{align} \frac{dH}{dt} = \frac{\partial H}{\partial t}. \end{align} \]

The Hamilton function is constant if and only if it does not explicitly depend on time. From Eq. (2.82) follows the equations (2.70) - (2.71) in the form

\[ \begin{align} \newdot{p}_i = \left\lbrace p_i, H\right\rbrace, & {} & \newdot{q}_i = \left\lbrace q_i, H\right\rbrace. \end{align} \]

2.4 Harmonic oscillator

The so-called harmonic oscillator is used as an example for the application of the theory of classical mechanics. This is about the evolution of a generalized coordinate $x = x\left(t\right)$ according to the differential equation

\[ \begin{align} \frac{d^2x}{dt^2} = -kx.\tag{2.85}\label{eq:harm_osz_base} \end{align} \]

If $x$ is the Cartesian coordinate of a mass point with mass $m$ and $k$ is the quotient of the spring constant and mass, then it is an undamped spring, but there are also other examples.

2.4.1 Undampened fall

If you put in Eq. (2.85) a solution $x = x_0e^{-i\omega_0t}$ follows

$\omega$ is called the eigenfrequency.

For the kinetic energy $K$ as a function of time applies

\[ \begin{align} K = \frac{1}{2}m\left(\frac{dx}{dt}\right)^2 = \frac{1}{2}mx_0^2\omega_0^2\sin\left(\omega_0t\right)^2, \end{align} \]

here $m$ is the mass of the oscillating mass point. For the potential energy applies

\[ \begin{align} U = \frac{1}{2}mkx^2 = \frac{1}{2}mkx_0^2\cos\left(\omega_0t\right)^2. \end{align} \]

This therefore applies to the total energy $E$

\[ \begin{align} E &= K + U = \frac{1}{2}mx_0^2\omega_0^2\sin\left(\omega_0t\right)^2 + \frac{1}{2}mkx_0^2\cos\left(\omega_0t\right)^2\nonumber\\ &= \frac{1}{2}mkx_0^2\left[\sin\left(\omega_0t\right)^2 + \cos\left(\omega_0t\right)^2\right] = \frac{1}{2}mkx_0^2 = \frac{1}{2}m\omega_0^2x_0^2.\tag{2.89}\label{eq:e_tot_harm_osc} \end{align} \]

Furthermore, on average half of the energy is stored in kinetic and potential energy:

\[ \begin{align} \newoverline{K} &= \frac{1}{4}mx_0^2\omega_0^2 =\frac{1}{4}mx_0^2k = \frac{E}{2},\nonumber\\ \newoverline{U} &= \frac{1}{4}mkx_0^2 = \frac{1}{4}m\omega_0^2x_0^2 = \frac{E}{2}. \end{align} \]

2.4.2 Dampened case

Now you dampen the oscillator linearly, modifying Eq. (2.85) accordingly

\[ \begin{align} x'' = -kx - 2dx'\tag{2.91}\label{eq:harm_osz_dampened} \end{align} \]

with a damping $d \geq 0$. If you insert a solution of the form $x = x_0e^{-i\omega t}$ here, you get

\[ \begin{align} -\omega^2 = -k + 2id\omega \Leftrightarrow \omega^2 + 2id\omega - k = 0. \end{align} \]

From this it follows using the pq formula Eq. (A.5)

\[ \begin{align} \omega = -id \pm \sqrt{-d^2 + k}. \end{align} \]

In the case $d = 0$, Eq. (2.86).

2.4.2.1 Swing case

In the case

\[ \begin{align} -d^2 + k > 0 \Leftrightarrow d^2 < k \end{align} \]

applies

\[ \begin{align} x\left(t\right) = \underbrace{x_0}_{\text{Anfangsamplitude}}\cdot\underbrace{\exp\left(-dt\right)}_{\text{Einhüllende}}\cdot\underbrace{\exp\left(\mp it\sqrt{-d^2 + k}\right)}_{\text{Oszillation}} \end{align} \]

or

\[ \begin{align} x\left(t\right) = x_0\exp\left(-dt\right)\exp\left(-it\sqrt{-d^2 + k}\right) + x_1\exp\left(-dt\right)\exp\left(it\sqrt{-d^2 + k}\right). \end{align} \]

$x_0$ and $x_1$ result from the initial conditions. In this case, the damping has two consequences:

1. The natural frequency is now

   \[ \begin{align} \omega = \pm\sqrt{k - d^2}.\tag{2.97}\label{eq:eigenfrequency_damp} \end{align} \]

2. The initial amplitude decreases exponentially over time with the decay time $\tau = \frac{1}{d}$.

This case is called swing case.

2.4.2.2 Creep case

In the case

\[ \begin{align} -d^2 + k < 0 \Leftrightarrow d^2 > k \end{align} \]

applies

\[ \begin{align} \omega = -id \pm \sqrt{-d^2 + k} = -id \pm \sqrt{\left(-1\right)\cdot\left(d^2 - k\right)} = -id \pm \sqrt{-1}\sqrt{\underbrace{d^2 - k}_{> 0}} = -id \pm i\sqrt{d^2 - k} = i\left(\underbrace{-d \pm \sqrt{d^2 - k}}_{< 0}\right). \end{align} \]

From this it follows

\[ \begin{align} x\left(t\right) = x_0\exp\left[-i^2\left(-d \pm \sqrt{d^2 - k}\right)t\right] = x_0\exp\left[\left(-d \pm \sqrt{d^2 - k}\right)t\right]. \end{align} \]

or

\[ \begin{align} x\left(t\right) = x_0\exp\left[\left(-d + \sqrt{d^2 - k}\right)t\right] + x_1\exp\left[\left(-d - \sqrt{d^2 - k}\right)t\right]. \end{align} \]

So there is no longer any oscillation. This case is called creep case.

2.4.2.3 Aperiodic limit case

In borderline cases

\[ \begin{align} -d^2 + k = 0 \Leftrightarrow d^2 = k\tag{2.102}\label{eq:cond_ap} \end{align} \]

the two linearly independent solutions, which were found in the oscillating case and in the aperiodic limit case, shrink to one. Since it is based on a second-order differential equation, there must be another solution in this case. This is where the approach is made

\[ \begin{align} x\left(t\right) = x_1t\exp\left(-dt\right).\tag{2.103}\label{eq:harm_osz_ap_ansatz} \end{align} \]

This implies

\[ \begin{align} x'\left(t\right) &= x_1\exp\left(-dt\right) - dx_1t\exp\left(-dt\right),\\ x''\left(t\right) &= -dx_1\exp\left(-dt\right) - dx_1\exp\left(-dt\right) + d^2x_1t\exp\left(-dt\right) = -2dx_1\exp\left(-dt\right) + d^2x_1t\exp\left(-dt\right). \end{align} \]

From this it follows by inserting into Eq. (2.91)

\[ \begin{align} -2dx_1\exp\left(-dt\right) + d^2x_1t\exp\left(-dt\right) &= -kx_1t\exp\left(-dt\right) - x_12d\exp\left(-dt\right) + 2d^2x_1t\exp\left(-dt\right)\nonumber\\ \Leftrightarrow-2dx_1 + d^2x_1t &= -kx_1t - 2x_1d + 2d^2x_1t\nonumber\\ \Leftrightarrow-2d + d^2t &= -kt - 2d + 2d^2t\nonumber\\ \Leftrightarrow d^2t &= -kt + 2d^2t\nonumber\\ \Leftrightarrow d^2 &= -k + 2d^2\nonumber\\ \Leftrightarrow-d^2 &= -k\nonumber. \end{align} \]

This is according to Eq. (2.102) is met. Eq. (2.103) solves Equation (2.91). So the general solution in this case is:

\[ \begin{align} x\left(t\right) = x_0\exp\left(-dt\right) + x_1t\exp\left(-dt\right) = \left(x_0 + x_1t\right)\exp\left(-dt\right). \end{align} \]

This case is called aperiodic limit. The four cases of the harmonic oscillator are shown in Fig. 2.1.

2.4.3 Powered case

Now you drive the system with the circular frequency $\omega$, modifying Eq. (2.91) too

\[ \begin{align} x'' + kx + 2dx' = y_0\exp\left(-i\omega t\right) \end{align} \]

with a real excitation amplitude $y_0 > 0$. If you use the approach $x = x_0e^{-i\omega t}$ again, you get

\[ \begin{align} -\omega^2x_0 + kx_0 - 2id\omega x_0 &= y_0\nonumber\\ \Leftrightarrow -\omega^2 + k - 2id\omega &= \frac{y_0}{x_0}\\ \Leftrightarrow \omega^2 - k + 2id\omega &= -\frac{y_0}{x_0}\\ \Leftrightarrow \omega^2 + 2id\omega + \left(\frac{y_0}{x_0} - k\right) &= 0.\tag{2.110}\label{eq_harm_osz_force_deriv_0} \end{align} \]

The amplitude is independent of time, therefore $\omega$ is real. For $x_0$ you now write down in polar coordinates

\[ \begin{align} x_0 = \left|x_0\right|\exp\left(i\phi\right), \end{align} \]

the real number $\phi$ is called the phase. If it is positive, the oscillator runs ahead of the excitation; if it is negative, the oscillator runs behind the excitation. Putting this into Eq. (2.110), you get

\[ \begin{align} \omega^2 + 2id\omega + \left(\frac{y_0}{\left|x_0\right|}e^{-i\phi} - k\right) &= \omega^2 + 2id\omega + \left(\frac{y_0}{\left|x_0\right|}\cos\left(\phi\right) - \frac{y_0}{\left|x_0\right|}i\sin\left(\phi\right) - k\right) = 0.\tag{2.112}\label{eq_harm_osz_force_deriv_1} \end{align} \]

$\omega$ is not an unknown here, it is given by the excitation. The unknowns are the reaction of the system $\left|x_0\right|, \phi$. To determine this, write down the real and imaginary parts of Equation. (2.112):

\[ \begin{align} \omega^2 + \left(\frac{y_0}{\left|x_0\right|}\cos\left(\phi\right) - k\right) &= 0,\tag{2.113}\label{eq_harm_osz_force_deriv_2}\\ 2d\omega - \frac{y_0}{\left|x_0\right|}\sin\left(\phi\right) &= 0 \Rightarrow 2d\omega = \frac{y_0}{\left|x_0\right|}\sin\left(\phi\right) \Rightarrow \left|x_0\right| = \frac{y_0}{2d\omega}\sin\left(\phi\right).\tag{2.114}\label{eq_harm_osz_force_deriv_3} \end{align} \]

If you put Eq. (2.114) in Eq. (2.113), you get

\[ \begin{align} \omega^2 + \left(\frac{y_0}{\frac{y_0}{2d\omega}\sin\left(\phi\right)}\cos\left(\phi\right) - k\right) &= 0\nonumber\\ \Leftrightarrow\omega^2 + \left(\frac{2d\omega}{\tan\left(\phi\right)} - k\right) &= 0\nonumber\\ \Leftrightarrow\omega^2\tan\left(\phi\right) + 2d\omega - k\tan\left(\phi\right) &= 0\nonumber\\ \Leftrightarrow\tan\left(\phi\right)\left(k - \omega^2\right) &= 2d\omega\nonumber\\ \Leftrightarrow\tan\left(\phi\right) &= \frac{2d\omega}{k - \omega^2}. \end{align} \]

The function

is called phase spectrum. Because $-\pi/2 < \phi < \pi/2$ holds

\[ \begin{align} \sin = \tan\cdot\cos = \tan\sqrt{1 - \sin^2} \Rightarrow \sin^2 = \tan^2 - \tan^2\sin^2 \Rightarrow \sin^2\left(1 + \tan^2\right) = \tan^2 \Rightarrow \sin = \frac{\tan}{\sqrt{1 + \tan^2}}. \end{align} \]

If you now set Eq. (2.116) in Eq. (2.114), you get

\[ \begin{align} \left|x_0\right| = \frac{y_0}{2d\omega}\sin\left(\phi\right) = \frac{y_0}{2d\omega}\frac{\tan\left(\phi\right)}{\sqrt{1 + \tan\left(\phi\right)^2}} = \frac{y_0}{2d\omega}\frac{\frac{2d\omega}{k - \omega^2}}{\sqrt{1 + \frac{4d^2\omega^2}{\left(k - \omega^2\right)^2}}} = \frac{y_0}{\left(k - \omega^2\right)\sqrt{1 + \frac{4d^2\omega^2}{\left(k - \omega^2\right)^2}}}. \end{align} \]

From this it follows

This function is called amplitude spectrum. One defines the auxiliary function $h$ by

\[ \begin{align} h\left(\omega\right) \coloneqq \left(k - \omega^2\right)^2 + 4d^2\omega^2 \Rightarrow h'\left(\omega\right) = -4\omega\left(k - \omega^2\right) + 8d^2\omega = -4\omega k + 4\omega^3 + 8d^2\omega. \end{align} \]

Setting the derivative to zero results in

\[ \begin{align} -4\omega k + 4\omega^3 + 8d^2\omega &\hastobe 0 \Rightarrow -4k + 4\omega^2 + 8d^2 = 0 \Rightarrow -k + \omega^2 + 2d^2 = 0\nonumber\\ \Rightarrow \omega^2 &= k - 2d^2. \end{align} \]

Here we limit ourselves to this case

\[ \begin{align} k - 2d^2 > 0 \Leftrightarrow d < \sqrt{\frac{k}{2}}. \end{align} \]

The frequency

is called resonance frequency. For the second derivative of $h$ one obtains

\[ \begin{align} h''\left(\omega\right) = -4k + 12\omega^2 + 8d^2. \end{align} \]

From this it follows

\[ \begin{align} h''\left(\omega_\text{res}\right) = -4k + 12\left(k - 2d^2\right) + 8d^2 = -4k + 12k - 24d^2 + 8d^2 = 8k - 16d^2 = 8\left(k - 2d^2\right) > 0. \end{align} \]

$\omega_\text{res}$ ist also die Frequenz maximaler Amplitude. Man beachte den Unterschied zwischen Eigenfrequenz Eq. (2.97) und Resonanzfrequenz Eq. (2.123). Das Resonanzverhalten ist in Abb. 2.2 dargestellt.

The system considered so far in this section has exactly one degree of freedom, namely the generalized deflection $x$. Conceptually, however, many of the findings can be generalized to systems with many degrees of freedom. Such complex systems (houses, circuits, the climate) also have their own modes.