2024-02-23 Central forces cont.

Central forces and the Hamiltonian

$$H=T+u$$ $$H=\sum _i{\dot{q}}_i\frac{\partial L}{\partial {\dot{q}}_i}-L$$ $$L=T-u$$ $$\frac{\partial L}{\partial {\dot{q}}_i}=P_i$$ $$L=\frac{1}{2}m{\dot{x}}^2-u$$ $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}=P$$ $$\dot{x}=\frac{P}{m}$$ $$H=\dot{x}m\dot{x}=\frac{1}{2}m{\dot{x}}^2+u$$ $$=\frac{1}{2}m{\dot{x}}^2+u$$ $$=\frac{P^2}{2m}+u$$

Hamiltonian form:

$$H(q,P,t)$$ $$\frac{\partial H}{\partial q_i}=-P_i$$ $$\frac{dH}{d{P}_i}=\dot{q}$$ $$L=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}-u(|\vec{r}|)=\frac{1}{2}\mu {\dot{\vec{r}}}^2-u(|\vec{r})$$ $$\frac{dL}{dt}=0$$

Energy is conserved, $E\equiv \text{Constant}$

Spherical symmetry of the problem implies that angular momentum $\vec{L}$ is conserved

$$\vec{L}=\text{constant}$$

Therefore $\vec{r}$, $\vec{p}$ are in the same plane

Central Forces → motion in a plane

Using Polar Coordinates example

$$x=r\cos \theta ,\dot{x}=-r\dot{\theta }\sin \theta +\dot{r}\cos \theta$$ $$y=r\sin \theta ,\dot{y}=r\dot{\theta }\sin \theta +\dot{r}\sin \theta$$ $${\dot{\vec{r}}}^2=({\dot{x}}^2+{\dot{y}}^2)={\dot{r}}^2+r^2+r^2{\dot{\theta }}^2$$ $$L=\frac{1}{2}\mu ({\dot{r}}^2+r^2{\dot{\theta }}^2)-u(r)$$

Do Euler-Lagrange Equation

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r}=0$$ $$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}=\mu \ddot{r}$$ $$\frac{\partial L}{\partial r}=\mu r{\dot{\theta }}^2+\frac{\partial u}{\partial r}$$ $$\mu \ddot{r}-\mu r{\dot{\theta }}^2+\frac{\partial u}{\partial r}=0\text{\hspace{1em}(1)}$$ $$\frac{d}{dt}(\mu r^2\dot{\theta })=0\text{\hspace{1em}(2)}$$

Combining these two

$$2\mu r\dot{r}\dot{\theta }+\mu r^2\ddot{\theta }=0$$

This yields us 2 coupled differential equations for $r(t)$ and $\theta (t)$. We want to divorce these equations

$$\theta (t)$$ $$\frac{d}{dt}(\mu r^2\dot{\theta })=0$$ $$\therefore \mu r^2\dot{\theta }=\text{const}$$ $$\frac{\partial L}{\partial \theta }=0=\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}=\frac{d}{dt}{P}_{\theta }\to \frac{\partial L}{\partial \dot{\theta }}=P_{\theta }$$ $$\dot{\theta }=\frac{l}{\mu r^2}\text{\hspace{1em}(3)}$$ $$\frac{d\theta }{dt}=\frac{l}{\mu r^2}⟹\theta ={\theta }_0+\int \frac{l}{\mu r(t{)}^2}dt$$

Referencing equation 1 above, we can plug equation 3 into that

$$\mu \ddot{r}-\frac{l^2}{\mu r^3}+\frac{\partial u}{\partial r}=0$$ $$\mu \ddot{r}=\frac{l^2}{\mu r^3}-\frac{\partial u}{\partial r}$$

Multiply by 1 ($\frac{\dot{r}}{\dot{r}}$)

$$u\dot{r}\ddot{r}=\left(\frac{l^2}{\mu l^3}-\frac{\partial u}{\partial r}\right)\dot{r}\text{\hspace{1em}(4)}$$

Analyzing LHS

$$\mu \dot{r}\ddot{r}=\frac{d}{dt}\left(\frac{1}{2}\dot{\mu }{r}^2\right)$$

Analyzing RHS

$$\frac{dg(r)}{dt}=\frac{\partial g}{\partial r}\frac{\partial r}{\partial t}$$

Using these two, we can look at the equation 4 from above and use the chain rule example from above

$$\left(\frac{l^2}{\mu r^3}-\frac{\partial u}{\partial r}\right)\dot{r}=-\frac{d}{dt}\left(\frac{1}{2}\frac{l^2}{\mu r^2}+u\right)$$

\mu \dot{r}\ddot{r} = \left( \frac{l^{2}}{\mu r^{3}} - \frac{ \partial u }{ \partial r } \right) \dot{r}

\frac{d}{dt}\left( \frac{1}{2}\mu \dot{r}^{2} \right) = -\frac{d}{dt} \left( \frac{1}{2} \frac{l^{2}}{\mu r^{2}}+u \right)

\frac{d}{dt} \left( \frac{1}{2} \mu \dot{r}^{2} + \frac{1}{2} \frac{l^{2}}{\mu r^{2}}+u \right) =0

$E=$ constant Compare to original E-L equation

E=T+ u = \frac{1}{2}\mu(\dot{r^{2} + r^{2} \dot{\theta}^{2}})+u(r) = \frac{1}{2} \mu \dot{r}^{2} + \frac{1}{2} \frac{l^{2}}{\mu^{2} r^{2} }+ u(r)

Solve for $\dot{r}$

\dot{r} = \pm \sqrt{ \frac{2}{\mu} \left( E-u-\frac{l^{2}}{2\mu r^{2}} \right) } = \frac{dr}{dt}

t = \int\limits_{r_{0}}^{r} \frac{1}{\pm \sqrt{ E-u-\frac{l^{2}}{2\mu r^{2}} }} , dr

Rearrange and integrate -> $r(t)$ Can find $\theta(r)$

\dot{\theta}=\frac{ d \theta }{ d t } = \frac{l}{\mu r^{2}}

\frac{d\theta}{dr} = \frac{ d \theta }{ d t } \frac{ d t }{ d r } = \frac{\dot{\theta}}{\dot{r}}

d\theta = \frac{\dot{\theta}}{\dot{r}}dr

\theta = \int  \frac{l}{\mu r^{2}}\left[ \frac{2}{\mu}\left( E-u-\frac{l^{2}}{2\mu r^{2}} \right) \right]^{-1/2} \, dr

This will again depend on the potential for $u\propto r ^{n+1}$, $F(r) \propto r ^{n}$ for $n=1,2,3$ Solutions will be oscillating (sin & cos) Orbits: ($r(\theta)$) Go back $r(t)$

\mu \ddot{r} - \mu r \dot{\theta}^{2} + \frac{ \partial u }{ \partial r } =0

u \ddot{r} - \frac{l^{2}}{\mu r^{3}} = - \frac{ \partial u }{ \partial r } = F(r)

\frac{\mu d^{2}}{dt^{2}} r - \frac{l^{2}}{\mu r^{3}} = F(r)

$$Asidetime:$$

\frac{d\theta}{dt}=\frac{l}{\mu r^{2}}

\mu r^{2}d\theta = ldt

$$\ast \ast or\ast \ast$$

\frac{d}{dt} = \frac{l}{\mu r^{2}} \frac{d}{d\theta}

\frac{d^{2}}{dt^{2}} = \frac{l}{\mu r^{2}}\frac{d}{dt} \left( \frac{l}{\mu r^{2}} \frac{d}{d\theta} \right)

$$Puttingthisbackin:$$

\mu \frac{l}{ur^{2}} \frac{d}{d\theta}\left( \frac{l}{\mu r^{2}} \frac{ d r }{ d \theta } \right) - \frac{l^{2}}{\mu r^{3} } = F(r)

If we know the orbit $r(\theta)$ then we can express the force or vice versa: Solving for $r(\theta)$

\frac{1}{r^{2}} \frac{ d r }{ d \theta } = - \frac{ d }{ d \theta } \frac{1}{r}

\frac{l^{2}}{\mu}\left[ \frac{1}{r^{2}} \frac{d}{d\theta}\left( -\frac{d}{d\theta} \left( \frac{1}{r} \right) \right)-\frac{1}{r^{3}} \right] = F(r)

\frac{l^{2}}{\mu r^{2}} \left[ \frac{d^{2}}{d\theta^{2}} \left( \frac{1}{r} \right) +\frac{1}{r} \right] = -F(r)

$$Centrifugalenergyandtheeffectivepotential$$

E-u - \frac{l^{2}}{2\mu r^{2}}

Define potential energy $u_{e} = \frac{l^{2}}{2\mu r^{2}}$

F_{e} = -\frac{ \partial u }{ \partial r } = \frac{l^{2}}{\mu r^{3}}=\mu r\theta^{2}

$$Effectivepotential$$

v=u + \frac{l^{2}}{2\mu r^{2}}

F(r) = -\frac{k}{r^{2}}

V(r) = -\frac{k}{r} + \frac{l^{2}}{2\mu r^{2}}

## Dynamics of systems of particles Motion of the system = Motion of the center of mass of the system + the motion of particles relative to the center of mass For $n$ particles of mass $m_{i}$ at positions $\vec{r}_{i}$

\vec{R}{cm} = \frac{\sum^{n}{l=1} m_{i}\vec{r}{i}}{\sum^{n}{i=1} m_{i}}

=\frac{1}{M_{tot}} \sum^{N}{i=1} m{i}\vec{r}_{i}

$$Ifinsteadwehaveacontinuousdistribution$$

\vec{R}_{cm} = \frac{\int \vec{r}dm}{\int , dm }

\rho = \frac{m}{v}

dm = \rho dV

=\frac{\int \vec{r}\rho , dv }{\int \rho , dv }

$$Foravariablemassdensity$$

\rho=\rho(r)

$$Totallinearmomentumofasystem$$

\vec{P}{tot} = \sum^{N}{i=1} \vec{P}{i} = \sum^{N}{i=1} m_{i} \vec{v}_{i}

N particles of $\vec{P}_{i}$ momentum Conservation of linear momentum

\frac{d\vec{P}_\text{tot}}{dt}=0

if $\vec{F}_\text{tot}=0$ and internal forces obey newtons 3rd law i.e $F_{ij}=-F_{ji}$ Velocity of center of mass

\vec{V}{com} = \frac{d\vec{R}{com}}{dt} = \frac{1}{M_{tot}} \sum^{N}{i=1} m{i} \frac{d\vec{r}{i}}{dt}= \frac{1}{m{tot}} \sum^{N}{i=1} m{i}\vec{v}_{i}

=\frac{\vec{P}{tot}}{M{tot}}

\vec{P}\text{tot} = M\text{tot} \vec{V}_{com}

$$Takingtimederivativesofbothsidestogiveexternalforcesandaccelerationvianewtonssecondlaw$$

\vec{P}{tot} = M{tot}\vec{V}_{com}

\frac{ d \vec{P}{tot} }{ d t } = \vec{F}\text{ext} = \frac{d}{dt} (M_\text{tot}\vec{V}\text{com}) = M\text{tot} \frac{d\vec{V}_\text{com}}{dt}

\frac{d\vec{V}\text{cm}}{dt}=\vec{A}\text{com}

$$CenterofmassaccelerationExternalforcesactonthecenterofmassIfthereisrotationofthesystemthenwemustconsiderthetotalangularmomentumofthesystem$$

\vec{L}\text{tot} = \vec{R}\text{com} \times \vec{P}\text{tot} + \sum^{N}{i=1} \vec{r}{i}’ \times \vec{P}‘{i}

The first term $\vec{R}_\text{com}\times \vec{P}_\text{tot}$ is angular momentum of the center of mass about the origin The second term $\sum^{N}_{i=1} \vec{r}'_{i} \times \vec{P}_{i}'$ is the angular momentum of particles about the center of mass