2024-02-06 Hamilton and Conservation

Continued from 2024-02-02 General Coordinates Continued

We went briefly over two methods to solve the ball rolling down a slope problem. Here is the first method, Substitution

$$L=\frac{1}{2}M{\dot{x}}^2+\frac{1}{2}I\frac{{\dot{x}}^2}{R^2}-Mg(l-x)\sin \varphi$$ $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}=0$$ $$\frac{\partial L}{\partial \dot{x}}=M\dot{x}+\frac{I\dot{x}}{R^2}\text{\hspace{1em}(1)}$$ $$\frac{d}{dt}(\text{Equation 1})=M\ddot{x}+\frac{I}{R^2}\ddot{x}$$ $$\frac{\partial L}{\partial x}=Mg\sin \varphi$$ $$\left(M+\frac{I}{R^2}\right)\ddot{x}-Mg\sin \varphi =0$$ $$\ddot{x}=\frac{Mg\sin \varphi }{M+\frac{I}{R^2}}$$

Method 2, using Lagrange Multipliers

$$\tilde{L}=L+\lambda f=\frac{1}{2}{\dot{x}}^{2}M+\frac{1}{2}I{\dot{\theta }}^2-Mg(l-x)\sin \varphi +\lambda (x-R\theta )$$

Resulting equations: For x

$$\frac{d}{dt}\frac{\partial \tilde{L}}{\partial \dot{x}}-\frac{\partial \tilde{L}}{\partial x}=0$$ $$\frac{\partial \tilde{L}}{\partial \dot{x}}=M\dot{x}$$ $$\frac{d}{dt}\frac{\partial \tilde{L}}{\partial \dot{x}}=M\ddot{x}$$ $$\frac{\partial \tilde{L}}{\partial x}=Mg\sin \varphi +\lambda$$ $$M\ddot{x}-Mg\sin \varphi -\lambda =0\text{\hspace{1em}(2)}$$

For $\theta$

$$\frac{d}{dt}\frac{\partial \tilde{L}}{\partial \dot{\theta }}-\frac{\partial \tilde{L}}{\partial \theta }=0$$ $$\frac{d}{dt}\frac{\partial \tilde{L}}{\partial \dot{\theta }}=I\ddot{\theta }$$ $$\frac{\partial \tilde{L}}{\partial \theta }=-R\lambda$$ $$I\ddot{\theta }+\lambda R=0$$

Use Acceleration relationships

$$\ddot{\theta }=\frac{\ddot{x}}{R}$$ $$I\frac{\ddot{x}}{R}+\lambda R=0\to \lambda =-\frac{I\ddot{x}}{R^2}$$

Plug into equation 2

$$M\ddot{x}-Mg\sin \varphi +\frac{I\ddot{x}}{R^2}=0$$ $$\ddot{x}\left(M+\frac{I}{R^2}\right)=Mg\sin \varphi$$ $$\ddot{x}=\frac{Mg\sin \varphi }{Mr+\frac{I}{R^2}}$$

This matches the last equation! they both give the right answer. Lets solve for $\lambda$ now

$$\lambda =\frac{-Mg\sin \varphi I}{\left(M+\frac{I}{R^2}\right)R^2}=\frac{-Mg\sin \varphi I}{M{R}^2+I}$$

What is $\lambda$? After some dimensional analysis, we can determine that it is a force. In fact, you could call it the forces of constraint (singular). It turns out that in this equation, it is the force of friction on the ball.

For a disk $I=\frac{1}{2}M{R}^2$

$$\ddot{x}=\frac{Mg\sin \varphi }{M+\frac{1}{2}M}=\frac{2g\sin \varphi }{3}$$ $$\lambda =\frac{-Mg\sin \varphi \left(\frac{1}{2}M{R}^2\right)}{M{R}^2+\frac{1}{2}M{R}^2}=-\frac{1}{3}Mg\sin \varphi$$

$$\sum F=m\ddot{x}$$ $$mg\sin \varphi -F=M\ddot{x}$$ $$mg\sin \theta -\frac{I\ddot{x}}{R^2}=m\ddot{x}$$

$$T=I\vec{\alpha }=\vec{r}\times \vec{F}⟹I\alpha =Rf$$ $$\frac{I\ddot{x}}{R}=Rf$$ $$f=\frac{I\ddot{x}}{R^2}$$

NB: there will be problems on the exam that require classical Newtonian mechanics

Conservation laws in Lagrangian Motion

Conservation of energy

If the following is true, energy is conserved 1:

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

Time is homogeneous, L is invariant under translation in time.

• No explicit time-dependent forces of constraint 2: Kinetic energy is only a function of generalized velocity

$$T({\dot{q}}_j)\propto {\dot{q}}_j{\dot{q}}_k$$

3: Velocity and time are independent

$$u=u(q_j)$$

Generalized Lagrangian is

$$L=L(q_j,{\dot{q}}_j)=T({\dot{q}}_j)-u(q_j)$$ $$\frac{dL}{dt}(q_j,{\dot{q}}_j,t)=\sum _j\left[\frac{\partial L}{\partial q_j}{\dot{q}}_j+\frac{\partial L}{\partial {\dot{q}}_j}{\ddot{q}}_j\right]+\cancel{\frac{\partial L}{\partial t}}=0$$

Remember:

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0\to \frac{\partial L}{\partial q}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$$ $$\frac{dL}{dt}=\sum _j\left[\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_j}{\dot{q}}_j+\frac{\partial L}{\partial {\dot{q}}_j}{\ddot{q}}_j\right]$$ $$\frac{dl}{dt}=\sum _j\cdots =0$$

so the sum has to be a constant

$$\frac{d}{dt}\left[L-\sum _j{\dot{q}}_j\frac{dL}{d{\dot{q}}_j}\right]=0$$

we can call this constant $-H$

$$-H=L-\sum _j{\dot{q}}_j\frac{\partial L}{\partial {\dot{q}}_j}$$

for H is constant if $\frac{\partial L}{\partial t}=0$

$$H=\sum _j{\dot{q}}_j\frac{\partial L}{\partial {\dot{q}}_j-L}=E$$

Show H = E Remember $T({\dot{q}}_j)\propto {\dot{q}}_j{\dot{q}}_k$

$$\frac{\partial L}{\partial {\dot{q}}_j}=\frac{\partial }{\partial {\dot{q}}_j}(T-u)=\frac{\partial T}{\partial {\dot{q}}_j}$$

if $u=u(q_j)$

$$H=\sum {\dot{q}}_j\frac{\partial T}{\partial {\dot{q}}_j}-L=\sum _j{\dot{q}}_j\frac{\partial T}{\partial {\dot{q}}_j}-(T-u)$$ $$T=\sum _{j_{1}k}{a}_{j,k}{\dot{q}}_j{\dot{q}}_k$$ $$\frac{\partial T}{\partial {\dot{q}}_l}=\sum _k{q}_{l,k}{\dot{q}}_k+\sum _j{a}_{jl}{\dot{q}}_j$$ $$\sum _l{\dot{q}}_l\frac{\partial L}{\partial {\dot{q}}_l}=\sum _{l,k}{a}_{l,k}{\dot{q}}_l{\dot{q}}_k+\sum _{l,j}{a}_{j,l}{\dot{q}}_j{\dot{q}}_l=2T$$

So

$$H=\sum _j{\dot{q}}_j\frac{\partial L}{\partial {\dot{q}}_j}-L=2T-(T-u)$$ $$=2T-T+u=T+u=E_{\text{total}}$$

This is the Hamiltonian, the same one from Quantum Physics

$$L=T-u=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)-mgy$$ $$x=l\sin \theta ⟹\dot{x}=l\dot{\theta }\cos \theta$$ $$y=-l\cos \theta ⟹\dot{y}=l\dot{\theta }\sin \theta$$ $$H=\sum _j{\dot{q}}_j\frac{\partial L}{\partial {\dot{q}}_j}⟹H=\dot{\theta }\frac{\partial L}{\partial \dot{\theta }}-L$$

However, lets look at an accelerated pendulum-

$$L=T-u=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)-mgy$$ $$x=l\sin \theta +\frac{1}{2}a{t}^2$$ $$\dot{x}=at+l\dot{\theta }\cos \theta$$ $$y=-l\cos \theta$$ $$\dot{y}=l\dot{\theta }\sin \theta$$ $$x^2=a^2{t}^2+l^2{\dot{\theta }}^2{\cos }^2(\theta )+2atl\dot{\theta }\cos \theta$$

where $\frac{\partial L}{\partial t}\ne 0$

$$L=\frac{1}{2}m[a^2{t}^2+l^2{\dot{\theta }}^2+2atl{\dot{\theta }}^2\cos \theta ]+mgl\cos \theta$$

its not conserved. Makes sense because the system is accelerating

Conservation of momentum

1: Space is homogeneous, the Lagrangian is invariant under space translation in the direction of $q_j$

$$\frac{\partial L}{\partial q_j}=0\text{\hspace{1em}(3)}$$ $$P_j=\frac{\partial L}{\partial {\dot{q}}_j}$$

This is our generalized momentum, conserved if (3) is true

In Cartesian coordinates:

$$\frac{\partial L}{\partial x}=0,P_i=\frac{\partial L}{\partial {\dot{x}}_i}\text{ is conserved}$$

Consider Euler-Lagrange Equation again

$$\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_j}-\frac{\partial L}{\partial q_j}=0$$ $$\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_j}=\frac{\partial L}{\partial q_j}=0?$$ $$\frac{\partial L}{\partial q_j}={\dot{P}}_j$$

Force is the negative gradient of potential

If we look at the LHS of the above equation,

$$\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_j}=0$$

means that

$$\frac{\partial L}{\partial {\dot{q}}_j}=P_j=\text{constant }$$

Newton

$$\frac{d\vec{P}}{dt}=\vec{F}$$

Example: 3d projectile

$$\vec{F}=\left(\begin{array}{c}0\\ 0\\ -mg\end{array}\right)$$ $$L=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)-mgz$$ $$\frac{\partial L}{\partial x}=0\to \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=0$$ $$\frac{\partial L}{\partial \dot{x}}=P_x=\text{constant}$$ $$\frac{\partial L}{\partial y}=0⟹\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}=0$$ $$\frac{\partial L}{\partial \dot{y}}=P_y=\text{constant}$$ $$\frac{\partial L}{\partial z}=-mg$$ $$P_z\ne \text{constant}$$ $$\frac{\partial L}{\partial \dot{x}}=m\dot{x}$$ $$\frac{\partial L}{\partial \dot{y}}=m\dot{y}$$

Another example: pendulum Skipping setup, go straight to Lagrangian

$$L=\frac{1}{2}m{l}^2{\dot{\theta }}^2+mgl\cos \theta$$ $$\frac{\partial L}{\partial \theta }=-mgl\sin \theta \ne 0$$ $$P_{\theta }\ne \text{constant}$$ $$P_0=\frac{\partial L}{\partial \dot{\theta }}=m{l}^2\dot{\theta }$$

Angular Momentum