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:

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

This leads to:

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

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.