# 1: Mechanics


## Point-kinetics in a fixed coordinate system

### Definitions

The position $$\vec{r}$$, the velocity $$\vec{v}$$ and the acceleration $$\vec{a}$$ are defined by: $$\vec{r}=(x,y,z)$$, $$\vec{v}=(\dot{x},\dot{y},\dot{z})$$, $$\vec{a}=(\ddot{x},\ddot{y},\ddot{z})$$. The following holds:

$s(t)=s_0+\int|\vec{v}(t)|dt~;~~~\vec{r}(t)=\vec{r}_0+\int\vec{v}(t)dt~;~~~\vec{v}(t)=\vec{v}_0+\int\vec{a}(t)dt$

When the acceleration is constant this gives: $$v(t)=v_0+at$$ and $$s(t)=s_0+v_0t+ \frac{1}{2} at^2$$.

For the unit vectors in a direction $$\perp$$ to the orbit $$\vec{e}_{\rm t}$$ and parallel to it $$\vec{e}_{\rm n}$$:

$\vec{e}_{\rm t}=\frac{\vec{v}}{|\vec{v}|}=\frac{d\vec{r}}{ds}~~~\dot{\vec{e}_{\rm t}}= \frac{v}{\rho}\vec{e}_{\rm n}~;~~~\vec{e}_{\rm n}=\frac{\dot{\vec{e}_{\rm t}}}{|\dot{\vec{e}_{\rm t}}|}$

For the curvature $$k$$ and the radius of curvature $$\rho$$: $\vec{k}=\frac{d\vec{e}_{\rm t}}{ds}=\frac{d^2\vec{r}}{ds^2}=\left|\frac{d\varphi}{ds}\right| ~;~~~\rho=\frac{1}{|k|}$

### Polar coordinates

Polar coordinates are defined by: $$x=r\cos(\theta)$$, $$y=r\sin(\theta)$$. So, for the unit coordinate vectors: $$\dot{\vec{e}_{r}}=\dot{\theta}\vec{e}_{\theta}$$, $$\dot{\vec{e}_{\theta}}=-\dot{\theta}\vec{e}_{r}$$

The velocity and the acceleration are derived from: $\vec{r}=r\vec{e}_{r} \; ,\;\; \vec{v}=\dot{r}\vec{e}_{r}+r\dot{\theta}\vec{e}_{\theta} \; ,\;\; \vec{a}=(\ddot{r}-r\dot{\theta}^2)\vec{e}_{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\vec{e}_{\theta}$.

## Relative motion

For the motion of a point D w.r.t. a point Q: $$\displaystyle\vec{r}_{\rm D}=\vec{r}_{\rm Q}+\frac{\vec{\omega}\times\vec{v}_{\rm Q}}{\omega^2}$$ with $$\vec{\rm QD}=\vec{r}_{\rm D}-\vec{r}_{\rm Q}$$ and $$\omega=\dot{\theta}$$.

Further a prime on a symbol $$\alpha=\ddot{\theta}$$ $$'$$ means that the quantity is defined in a moving system of coordinates. In a moving system: $$\vec{v}=\vec{v}_{Q}+\vec{v}\,'+\vec{\omega}\times\vec{r}\,'$$ and $$\vec{a}=\vec{a}_{Q}+\vec{a}\,'+\vec{\alpha}\times\vec{r}\,'+2\vec{\omega}\times\vec{v}\,'+\vec{\omega}\times(\vec{\omega}\times\vec{r}\,')$$ with $$\vec{\omega}\times(\vec{\omega}\times\vec{r}\,')=-\omega^2\vec{r}\,'_n$$

## Point-dynamics in a fixed coordinate system

### Force, (angular)momentum and energy

Newton’s 2nd law connects the force on an object and the resulting acceleration of the object where the momentum is given by $$\vec{p}=m\vec{v}$$:

$\vec{F}(\vec{r},\vec{v},t)=\frac{d\vec{p}}{dt}=\frac{d(m\vec{v}\,)}{dt}=m\frac{d\vec{v}}{dt}+ \vec{v}\,\frac{dm}{dt}\mathop{=}\limits^{m={\rm const}}m\vec{a}$ Newton’s 3rd law is given by: $$\vec{F}_{\rm action}=-\vec{F}_{\rm reaction}$$.

For the power $$P$$: $$P=\dot{W}=\vec{F}\cdot\vec{v}$$. For the total energy $$W$$, the kinetic energy $$T$$ and the potential energy $$U$$: $$W=T+U~;~~~\dot{T}=-\dot{U}$$ with $$T= \frac{1}{2} mv^2$$.

The kick $$\vec{S}$$ is given by: $$\displaystyle\vec{S}=\Delta\vec{p}=\int\vec{F}dt$$

The work $$A$$, delivered by a force, is $$\displaystyle A=\int\limits_1^2\vec{F}\cdot d\vec{s}=\int\limits_1^2F\cos(\alpha)ds$$

The torque $$\vec{\tau}$$ is related to the angular momentum $$\vec{L}$$: $$\vec{\tau}=\dot{\vec{L}}=\vec{r}\times\vec{F}$$; and
$$\vec{L}=\vec{r}\times\vec{p}=m\vec{v}\times\vec{r}$$, $$|\vec{L}|=mr^2\omega$$. The following equation is valid:

$\tau=-\frac{\partial U}{\partial \theta}$

Hence, the conditions for a mechanical equilibrium are: $$\sum\vec{F}_i=0$$ and $$\sum\vec{\tau}_i=0$$.

The force of friction is usually proportional to the force perpendicular to the surface, except when the motion starts, when a threshold has to be overcome: $$F_{\rm fric}=f\cdot F_{\rm norm}\cdot\vec{e}_{\rm t}$$.

### Conservative force fields

A conservative force can be written as the gradient of a potential: $$\vec{F}_{\rm cons}=-\vec{\nabla}U$$. From this follows that $$\nabla\times\vec{F}=\vec{0}$$. For such a force field also:

$\oint\vec{F}\cdot d\vec{s}=0~\Rightarrow~U=U_0-\int\limits_{r_0}^{r_1}\vec{F}\cdot d\vec{s}$

So the work delivered by a conservative force field depends not on the trajectory covered but only on the starting and ending points of the motion.

### Gravitation

The Newtonian law of gravitation is (in GRT one also uses $$\kappa$$ instead of $$G$$):

$\vec{F}_{\rm g}=-G\frac{m_1 m_2}{r^2}\vec{e}_{r}$

The gravitational potential is then given by $$V=-Gm/r$$. From Gauss' law it then follows: $$\nabla^2 V=4\pi G\varrho$$.

### Orbital equations

If $$V=V(r)$$ one can derive from the equations of Lagrange for $$\phi$$ the conservation of angular momentum:

$\frac{\partial {\cal L}}{\partial \phi}=\frac{\partial V}{\partial \phi}=0\Rightarrow\frac{d}{dt}(mr^2\phi)=0\Rightarrow L_z=mr^2\phi=\mbox{constant}$

For the radial position as a function of time it can be found that:

$\left(\frac{dr}{dt}\right)^2=\frac{2(W-V)}{m}-\frac{L^2}{m^2r^2}$

The angular equation is then:

$\phi-\phi_0=\int\limits_0^r\left[\frac{mr^2}{L}\sqrt{\frac{2(W-V)}{m}-\frac{L^2}{m^2r^2}}~\right]^{-1}dr \stackrel{r^{-2}{\rm field}}{=} \arccos\left(1+\frac{\frac{1}{r}-\frac{1}{r_0}}{\frac{1}{r_0}+km/L_z^2}\right)$

If $$F=F(r)$$: $$L=$$ constant, if $$F$$ is conservative: $$W=$$ constant, if $$\vec{F}\perp\vec{v}$$ then $$\Delta T=0$$ and $$U=0$$.

#### Kepler’s orbital equations

In a force field $$F=kr^{-2}$$, the orbits are conic sections with the origin of the force in one of the foci (Kepler’s 1st law). The equation of the orbit is:

$r(\theta)=\frac{\ell}{1+\varepsilon\cos(\theta-\theta_0)}~,~~\mbox{or:~~} x^2+y^2=(\ell-\varepsilon x)^2$

with

$\ell=\frac{L^2}{G\mu^2M_{\rm tot}}~;~~~\varepsilon^2=1+\frac{2WL^2}{G^2\mu^3M^2_{\rm tot}}=1-\frac{\ell}{a} ~;~~~a=\frac{\ell}{1-\varepsilon^2}=\frac{k}{2W}$

$$a$$ is half the length of the long axis of the elliptical orbit in case the orbit is closed. Half the length of the short axis is $$b=\sqrt{a\ell}$$. $$\varepsilon$$ is the excentricity of the orbit. Orbits with an equal $$\varepsilon$$ are of equal shape. Now, five types of orbits are possible:

1. $$k<0$$ and $$\varepsilon=0$$: a circle.
2. $$k<0$$ and $$0<\varepsilon<1$$: an ellipse.
3. $$k<0$$ and $$\varepsilon=1$$: a parabola.
4. $$k<0$$ and $$\varepsilon>1$$: a hyperbola, curved towards the centre of force.
5. $$k>0$$ and $$\varepsilon>1$$: a hyperbola, curved away from the centre of force.

Other combinations are not possible: the total energy in a repulsive force field is always positive so $$\varepsilon>1$$.

If the surface between the orbit covered between $$t_1$$ and $$t_2$$ and the focus C around which the planet moves is $$A(t_1,t_2)$$, Kepler’s 2nd law is

$A(t_1,t_2)=\frac{L_{\rm C}}{2m}(t_2-t_1)$

Kepler’s 3rd law is, with $$T$$ the period and $$M_{\rm tot}$$ the total mass of the system is:

$\frac{T^2}{a^3}=\frac{4\pi^2}{GM_{\rm tot}}$

### The virial theorem

The virial theorem for one particle is:

$\left\langle m\vec{v}\cdot\vec{r} \right\rangle=0\Rightarrow\left\langle T \right\rangle=-\frac{1}{2}\left\langle \vec{F}\cdot\vec{r} \right\rangle=\frac{1}{2}\left\langle r\frac{dU}{dr} \right\rangle=\frac{1}{2} n\left\langle U \right\rangle\mbox{ if } U=-\frac{k}{r^n}$

The virial theorem for a collection of particles is:

$\left\langle T \right\rangle=-\frac{1}{2}\left\langle \sum\limits_{\rm particles}\vec{F}_i\cdot\vec{r}_i+ \sum\limits_{\rm pairs}\vec{F}_{ij}\cdot\vec{r}_{ij} \right\rangle$

These propositions can also be written as: $$2E_{\rm kin}+E_{\rm pot}=0$$.

## Point dynamics in a moving coordinate system

### Fictitious forces

The total force in a moving coordinate system can be found by subtracting the fictitious forces from the forces working in the reference frame: $$\vec{F}\,'=\vec{F}-\vec{F}_{\rm app}$$. The different fictictous forces are:

1. Transformation of the origin: $$F_{\rm or}=-m\vec{a}_a$$
2. Rotation: $$\vec{F}_{\alpha}=-m\vec{\alpha}\times\vec{r}\,'$$
3. Coriolis force: $$F_{\rm cor}=-2m\vec{\omega}\times\vec{v}$$
4. Centrifugal force: $$\vec{F}_{\rm cf}=m\omega^2\vec{r}_n\,'=-\vec{F}_{\rm cp}$$ ; $$\displaystyle\vec{F}_{\rm cp}=-\frac{mv^2}{r}\vec{e}_{r}$$

### Tensor notation

Transformation of the Newtonian equations of motion to $$x^\alpha=x^\alpha(x)$$ gives:

$\frac{dx^\alpha}{dt}=\frac{\partial x^\alpha}{\partial \bar{x}^\beta}\frac{d\bar{x}^\beta}{dt};$

The chain rule gives:

$\frac{d}{dt}\frac{dx^\alpha}{dt}=\frac{d^2x^\alpha}{dt^2}=\frac{d}{dt} \left(\frac{\partial x^\alpha}{\partial \bar{x}^\beta}\frac{d\bar{x}^\beta}{dt}\right)= \frac{\partial x^\alpha}{\partial \bar{x}^\beta}\frac{d^2\bar{x}^\beta}{dt^2}+ \frac{d\bar{x}^\beta}{dt}\frac{d}{dt}\left(\frac{\partial x^\alpha}{\partial \bar{x}^\beta}\right)$

so:

$\frac{d}{dt}\frac{\partial x^\alpha}{\partial \bar{x}^\beta}=\frac{\partial }{\partial \bar{x}^\gamma}\frac{\partial x^\alpha}{\partial \bar{x}^\beta}\frac{d\bar{x}^\gamma}{dt}= \frac{\partial^2x^\alpha}{\partial\bar{x}^\beta\partial\bar{x}^\gamma}\frac{d\bar{x}^\gamma}{dt}$

$\frac{d^2x^\alpha}{dt^2}=\frac{\partial x^\alpha}{\partial \bar{x}^\beta}\frac{d^2\bar{x}^\beta}{dt^2}+ \frac{\partial^2x^\alpha}{\partial\bar{x}^\beta\partial\bar{x}^\gamma}\frac{d\bar{x}^\gamma}{dt} \left(\frac{d\bar{x}^\beta}{dt}\right)$

Hence the Newtonian equation of motion

$m\frac{d^2x^\alpha}{dt^2}=F^\alpha$

will be transformed into:

$m\left\{\frac{d^2x^\alpha}{dt^2}+\Gamma_{\beta\gamma}^\alpha \frac{dx^\beta}{dt}\frac{dx^\gamma}{dt}\right\}=F^\alpha$

The apparent forces are projected from the origin to the side affected by $$\displaystyle\Gamma_{\beta\gamma}^\alpha\frac{dx^\beta}{dt}\frac{dx^\gamma}{dt}$$.

## Dynamics of masspoint collections

### The centre of mass

The velocity w.r.t. the centre of mass $$\vec{R}$$ is given by $$\vec{v}-\dot{\vec{R}}$$. The coordinates of the centre of mass are given by:

$\vec{r}_{\rm m}=\frac{\sum m_i\vec{r}_i}{\sum m_i}$

In a 2-particle system, the coordinates of the centre of mass are given by:

$\vec{R}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$

With $$\vec{r}=\vec{r}_1-\vec{r}_2$$, the kinetic energy becomes: $$frac{1}{2} T=M_{\rm tot}\dot{R}^2+ frac{1}{2} \mu\dot{r}^2$$, with the reduced mass $$\mu$$ given by:

$\displaystyle\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$

The motion of the centre of mass and relative to it can be separated:

$\dot{\vec{L}}_{\rm outside}=\vec{\tau}_{\rm outside}~;~~~ \dot{\vec{L}}_{\rm inside}=\vec{\tau}_{\rm inside}$ $\vec{p}=m\vec{v}_{\rm m}~;~~~\vec{F}_{\rm ext}=m\vec{a}_{\rm m}~;~~~\vec{F}_{12}=\mu\vec{u}$

### Collisions

With collisions, where B are the coordinates of the collision and C an arbitrary other position: $$\vec{p}=m\vec{v}_{\rm m}$$ is constant, and $$T= \frac{1}{2} m\vec{v}_{\rm m}^{\,2}$$ is constant. The changes in the relative velocities can be derived from: $$\vec{S}=\Delta\vec{p}=\mu(\vec{v}_{\rm aft}-\vec{v}_{\rm before})$$. Further  $$\Delta\vec{L}_{\rm C}=\vec{\rm CB}\times\vec{S}$$, $$\vec{p}~\parallel~\vec{S}=$$constant and $$\vec{L}$$ w.r.t. B is constant.

## Dynamics of rigid bodies

### Moment of Inertia

The angular momentum in a moving coordinate system is given by:

$\vec{L}'=I\vec{\omega}+\vec{L}_n'$

where $$I$$ is the moment of inertia with respect to a central axis, which is given by:

$I=\sum\limits_i m_i\vec{r}_i~^2~;~~~T'=W_{\rm rot}=\frac{1}{2} \omega I_{ij}\vec{e}_i\vec{e}_j=\frac{1}{2} I\omega^2$

or, in the continuous case:

$I=\frac{m}{V}\int{r'}^2_ndV=\int{r'}^2_ndm$

Further:

$L_i=I^{ij}\omega_j~;~~~I_{ii}=I_i~;~~~I_{ij}=I_{ji}=-\sum\limits_km_kx_i'x_j'$

Steiner’s theorem is: $$I_{\rm w.r.t. D}=I_{\rm w.r.t. C}+m(DM)^2$$ if axis C $$\parallel$$ axis D.

Object $$I$$ Object $$I$$

Hollow cylinder

$$I=mR^2$$ Massive cylinder $$I= \frac{1}{2} mR^2$$

Disc, axis in plane disc through m

$$I= \frac{1}{4} mR^2$$ Dumbbell $$I= \frac{1}{2} \mu R^2$$

Hollow sphere

$$I=\frac{2}{3}mR^2$$ Massive sphere $$I=\frac{2}{5}mR^2$$

Bar, axis $$\perp$$ through c.o.m.

$$I=\frac{1}{2}ml^2$$ Bar, axis $$\perp$$ through end $$I=\frac{1}{3}ml^2$$

Rectangle, axis $$\perp$$ plane thr. c.o.m.

$$I=\frac{1}{2}m(a^2+b^2)$$ Rectangle, axis $$\parallel b$$ thr. m $$I=ma^2$$

### Principal axes

Each rigid body has (at least) 3 principal axes which stand $$\perp$$ to each other. For a principal axis:

$\frac{\partial I}{\partial \omega_x}=\frac{\partial I}{\partial \omega_y}=\frac{\partial I}{\partial \omega_z}=0~~\mbox{so}~~L'_n=0$

The following holds: $$\dot{\omega}_k=-a_{ijk}\omega_i\omega_j$$ with $$\displaystyle a_{ijk}=\frac{I_i-I_j}{I_k}$$ if $$I_1\leq I_2\leq I_3$$.

### Time dependence

For the torque $$\vec{\tau}$$: $\vec{\tau}\,'=I\ddot{\theta}~;~~~\frac{d''\vec{L}'}{dt}=\vec{\tau}\,'-\vec{\omega}\times\vec{L}'$ The torque $$\vec{T}$$ is defined by: $$\vec{T}=\vec{F}\times\vec{d}$$.

## Variational Calculus, Hamilton and Lagrange mechanics

### Variational Calculus

Starting with:

$\delta\int\limits_a^b{\cal L}(q,\dot{q},t)dt=0 \;\;\textrm{where} \;\; \delta(a)=\delta(b)=0\mbox{~~and~~} \delta\left(\frac{du}{dx}\right)=\frac{d}{dx}(\delta u)$

the equations of Lagrange can be derived:

$\frac{d}{dt}\frac{\partial {\cal L}}{\partial \dot{q}_i}=\frac{\partial {\cal L}}{\partial q_i}$

When there are additional conditions applying to the variational problem $$\delta J(u)=0$$ of the type $$K(u)=$$constant, the new problem becomes: $$\delta J(u)-\lambda\delta K(u)=0$$.

### Hamilton mechanics

The Lagrangian is given by: $${\cal L}=\sum T(\dot{q}_i)-V(q_i)$$. The Hamiltonian is given by: $$H=\sum\dot{q}_ip_i-{\cal L}$$. In two dimensions: $${\cal L}=T-U= frac{1}{2} m(\dot{r}^2+r^2\dot{\phi}^2)-U(r,\phi)$$.

If the coordinates used are canonical the Hamilton equations are the equations of motion for the system:

$\frac{dq_i}{dt}=\frac{\partial H}{\partial p_i}~;~~~\frac{dp_i}{dt}=-\frac{\partial H}{\partial q_i}$

Coordinates are canonical if the following holds: $$\{q_i,q_j\}=0,~\{p_i,p_j\}=0,~\{q_i,p_j\}=\delta_{ij}$$ where $$\{,\}$$ is the Poisson bracket:

$\{A,B\}=\sum\limits_i\left[\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right]$

The Hamiltonian of an harmonic oscillator is given by $$H(x,p)=p^2/2m+ \frac{1}{2} m\omega^2 x^2$$. With new coordinates $$(\theta,I)$$, obtained by the canonical transform $$x=\sqrt{2I/m\omega}\cos(\theta)$$ and $$p=-\sqrt{2Im\omega}\sin(\theta)$$, with inverse $$\theta=\arctan(-p/m\omega x)$$ and $$I=p^2/2m\omega+ \frac{1}{2} m\omega x^2$$ it follows: $$H(\theta,I)=\omega I$$.

The Hamiltonian of a charged particle with charge $$q$$ in an external electromagnetic field is given by:

$H=\frac{1}{2m}\left(\vec{p}-q\vec{A}\,\right)^2+qV$

This Hamiltonian can be derived from the Hamiltonian of a free particle $$H=p^2/2m$$ with the transform $$\vec{p}\rightarrow\vec{p}-q\vec{A}$$ and $$H\rightarrow H-qV$$. This is elegant from a relativistic point of view: it is equivalent to the transformation of the momentum 4-vector $$p^\alpha\rightarrow p^\alpha-qA^\alpha$$. A gauge transform on the potentials $$A^\alpha$$ corresponds with a canonical transform, which make the Hamilton equations the equations of motion for the system.

### Motion near equilibrium, linearization

For natural systems near equilibrium the following equations are valid:

$\left(\frac{\partial V}{\partial q_i}\right)_0=0~;~~~V(q)=V(0)+V_{ik}q_iq_k\mbox{~~with~~} V_{ik}=\left(\frac{\partial^2V}{\partial q_i\partial q_k}\right)_0$

With $$T= \frac{1}{2} (M_{ik}\dot{q}_i\dot{q}_k)$$ one obtains the set of equations $$M\ddot{q}+Vq=0$$. If $$q_i(t)=a_i\exp(i\omega t)$$ is substituted, this set of equations has solutions if $${\rm det}(V-\omega^2 M)=0$$. This leads to the eigenfrequencies of the problem: $$\displaystyle\omega^2_k=\frac{a_k^{\rm T}Va_k}{a_k^{\rm T}Ma_k}$$. If the equilibrium is stable: $$\forall k$$ that $$\omega^2_k>0$$. The general solution is a superposition of eigenvibrations.

### Phase space, Liouville’s equation

In phase space:

$\nabla=\left(\sum_i\frac{\partial }{\partial q_i},\sum_i\frac{\partial }{\partial p_i}\right)\mbox{~~so~~} \nabla\cdot\vec{v}=\sum_i\left(\frac{\partial }{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial }{\partial p_i}\frac{\partial H}{\partial q_i}\right)$

If the equation of continuity, $$\partial_t\varrho+\nabla\cdot(\varrho\vec{v}\,)=0$$ holds, this can be written as:

$\{\varrho,H\}+\frac{\partial \varrho}{\partial t}=0$

For an arbitrary quantity $$A$$ :

$\frac{dA}{dt}=\{A,H\}+\frac{\partial A}{\partial t}$

Liouville’s theorem can then be written as:

$\frac{d\varrho}{dt}=0~;~~~\mbox{or:~}\int pdq=\mbox{constant}$

### Generating functions

Starting with the coordinate transformation:

$\left\{\begin{array}{l} Q_i=Q_i(q_i,p_i,t)\\ P_i=P_i(q_i,p_i,t) \end{array}\right.$

one can derive the following Hamilton equations with the new Hamiltonian $$K$$:

$\frac{dQ_i}{dt}=\frac{\partial K}{\partial P_i}~;~~~\frac{dP_i}{dt}=-\frac{\partial K}{\partial Q_i}$

Now, a distinction between 4 cases can be made:

1. If $$\displaystyle p_i\dot{q}_i-H=P_iQ_i-K(P_i,Q_i,t)-\frac{dF_1(q_i,Q_i,t)}{dt}$$, the coordinates follow from: $p_i=\frac{\partial F_1}{\partial q_i}~;~~~P_i=-\frac{\partial F_1}{\partial Q_i}~;~~~K=H+\frac{\partial F_1}{\partial t}$
2. If $$\displaystyle p_i\dot{q}_i-H=-\dot{P}_iQ_i-K(P_i,Q_i,t)+\frac{dF_2(q_i,P_i,t)}{dt}$$, the coordinates follow from: $p_i=\frac{\partial F_2}{\partial q_i}~;~~~Q_i=\frac{\partial F_2}{\partial P_i}~;~~~K=H+\frac{\partial F_2}{\partial t}$
3. If $$\displaystyle-\dot{p}_iq_i-H=P_i\dot{Q}_i-K(P_i,Q_i,t)+\frac{dF_3(p_i,Q_i,t)}{dt}$$, the coordinates follow from: $q_i=-\frac{\partial F_3}{\partial p_i}~;~~~P_i=-\frac{\partial F_3}{\partial Q_i}~;~~~K=H+\frac{\partial F_3}{\partial t}$
4. If $$\displaystyle-\dot{p}_iq_i-H=-P_iQ_i-K(P_i,Q_i,t)+\frac{dF_4(p_i,P_i,t)}{dt}$$, the coordinates follow from: $q_i=-\frac{\partial F_4}{\partial p_i}~;~~~Q_i=\frac{\partial F_4}{\partial P_i}~;~~~K=H+\frac{\partial F_4}{\partial t}$

The functions $$F_1$$, $$F_2$$, $$F_3$$ and $$F_4$$ are called generating functions.

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This page titled 1: Mechanics is shared under a CC BY license and was authored, remixed, and/or curated by Johan Wevers.