Hamiltonian formalism exercises

Exercise 1 Choose a set of generalized coordinates that totally specify the mechanical state of the following systems: A particle of mass ${m}$ that moves along an ellipse. Let ${x=a\cos\theta}$ and ${y=b\sin\theta}$. Then the generalized coordinate is ${\theta}$. Perhaps you might be surprised to find out that we need only a single coordinate to describe the motion of this particle since its motion is described on a plane which is a two dimensional entity. But the particle motion is restricted to be along the ellipse and that constraint decreases the particle's degrees of freedom to one instead of being two. A cylinder that moves along an inclined plane. If the cylinder rotates we need ${x}$, the distance travelled, and ${\theta}$, the angle of rotation. If the cylinder doesn't rotate we only need ${x}$. The two masses on a double pendulum. The generalized coordinates are ${\theta_1}$ and ${\theta_2}$. Do you see why?

Exercise 2 Derive the transformation equations for the double pendulum. It is ${x_1=l_1\cos\theta_1}$, ${x_2=l_1\cos\theta_1+l_2\cos\theta_2}$, ${y_1=l_1\sin\theta_1}$ and ${y_2=l_1\sin\theta_1+l_2\sin\theta_2}$

Exercise 3 Show that ${\dfrac{\partial \dot{\vec{r}}_\nu}{\partial \dot{q}_\alpha}=\dfrac{\partial\vec{r}_\nu}{\partial q_\alpha}}$. ${\begin{aligned} \vec{r}_\nu&=\vec{r}_\nu(q_1,q_2,\cdots,q_n,t)\Rightarrow \\ \dot{\vec{r}_\nu}&=\dfrac{\partial\vec{r}_\nu }{\partial q_1}\dot{q}_1+\cdots+\dfrac{\partial\vec{r}_\nu }{\partial q_n}\dot{q}_n+\dfrac{\partial\vec{r}_\nu}{\partial t} \Rightarrow \\ \dfrac{\partial \dot{\vec{r}}_\nu}{\partial\dot{q}_\alpha}&=\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha} \end{aligned}}$

Exercise 4 Consider a system of particles that experiences an increment ${dq_j}$ on its generalized coordinates. Derive the following expression ${dW=\displaystyle \sum_\alpha \Phi_\alpha dq_\alpha}$ for the total work done by the force and indicate the physical meaning of ${\Phi_\alpha}$. First note that $\displaystyle d\vec{r}_\nu =\sum_{\alpha=1}^n \dfrac{\partial \vec{r}_\nu}{\partial q_\alpha}dq_\alpha$ For ${dW}$ it is ${\begin{aligned} dW &=\sum_{\nu=1}^N\vec{F}_\nu\cdot d\vec{r}_\nu\\ &=\sum_{\nu=1}^N\left( \sum_{\alpha=1}^n \vec{F}_\nu\cdot\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha} \right) dq_\alpha\\ &= \sum_{\alpha=1}^n \Phi_\alpha dq_\alpha \end{aligned}}$ with ${\displaystyle \Phi_\alpha=\sum_{\nu=1}^N \vec{F}_\nu\cdot\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha}}$ being the generalized force.

Exercise 5 Show that ${\Phi_\alpha=\dfrac{\partial W}{\partial q_\alpha}}$. We have ${\displaystyle dW=\sum_\alpha\frac{\partial W}{\partial q_\alpha}dq_\alpha}$ and ${\displaystyle dW=\sum_\alpha\Phi_\alpha dq_\alpha}$. Hence ${\displaystyle \sum_\alpha\left( \Phi_\alpha- \frac{\partial W}{\partial q_\alpha}\right)dq_\alpha=0}$. Since ${dq_\alpha}$ are linearly independent it is ${\Phi_\alpha=\dfrac{\partial W}{\partial q_\alpha}}$.

Exercise 6 Derive the lagrangian for a simple pendulum and obtain an equation to describe its motion. The generalized coordinate for the simple pendulum is ${\theta}$ and the transformation equations are ${x=l\sin\theta}$ and ${y=-l\cos\theta}$. For the kinetic energy it is ${K=1/2mv^2=1/2m(l\dot{\theta})^2=1/2ml^2\dot{\theta}^2}$. for the potential ${V=mgl(1-\cos\theta)}$. Hence the Lagrangian is ${L=K-V=1/2ml^2\dot{\theta}^2-mgl(1-\cos\theta)}$. ${\dfrac{\partial L}{\partial \theta}=-mgl\sin\theta}$ and ${\dfrac{\partial L}{\partial \dot{\theta}}=ml^2\dot{\theta}}$. Hence the Euler Lagrange equation is ${\begin{aligned} \frac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}-\dfrac{\partial L}{\partial \theta}&=0\Rightarrow\\ ml^2\dot{\theta}+mgl\sin\theta&=0\Rightarrow\\ \ddot{\theta}+g/l\sin\theta=0 \end{aligned}}$

Exercise 7 Two particles of mass ${m}$ are connected with each other and to two points ${A}$ and ${B}$ by springs with constant factor ${k}$. The particles are free to slide along the direction of ${A}$ and ${B}$. Use the Euler-Lagrange equations to derive the equations of motion of the particles. The kinetic energy is ${K=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2}$. The potential energy is ${V=1/2kx^2_1+1/2k(x_2-x_1)^2+1/2kx^2_2}$. Hence the Lagrangian is ${L=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2-1/2kx^2_1-1/2k(x_2-x_1)^2-1/2kx^2_2}$. The partial derivatives of the Lagrangian are: ${\dfrac{\partial L}{\partial x_1}=k(x_2-x_1)}$ ${\dfrac{\partial L}{\partial x_2}=k(x_1-x_2)}$ ${\dfrac{\partial L}{\partial \dot{x_1}}=m\dot{x}_1}$ ${\dfrac{\partial L}{\partial \dot{x_2}}=m\dot{x}_2}$ And the Euler-Lagrange equations are: ${m\ddot{x}_1=k(x_2-x_1)}$ ${m\ddot{x}_2=k(x_1-2x_2)}$

Exercise 8 A particle of mass ${m}$ moves subject to a conservative force field. Use cylindrical coordinates to derive: The Lagrangian. The kinetic energy is ${K=1/2m(1/2m\dot{\rho}^2+\rho^2\dot{\phi}^2+\dot{z^2})}$. The potential is ${V=V(\rho,\phi,z)}$. The equations of motion. ${m(\ddot{\rho}-\rho\dot{\phi}^2)=-\dfrac{\partial v}{\partial \rho}}$ ${m\dfrac{d}{dt}(\rho^2\dot{\phi}=-\dfrac{\partial V}{\partial \phi}}$ ${m\ddot{z}=-\dfrac{\partial V}{\partial z}}$

Exercise 9 A double pendulum oscillates on a vertical plane. Calculate: The Lagrangian. The transformation equations for the coordinates are ${x_1=l_1\cos\theta_1}$ ${y_1=l_1\sin\theta_1}$ ${x_2=l_1\cos\theta_1+l_2\cos\theta_2}$ ${y_2=l_1\sin\theta_1+l_2\sin\theta_2}$ Applying ${\dfrac{d}{dt}}$ to the previous equations ${\dot{x}_1=-l_1\dot{\theta}_1\sin\theta_1}$ ${\dot{y}_1=l_1\dot{\theta}_1\cos\theta_1}$ ${\dot{x}_2=-l_1\dot{\theta}_1\sin\theta_1-l_2\dot{\theta}_2\sin\theta_2}$ ${\dot{y}_2=l_1\dot{\theta}_1\cos\theta_1+l_2\dot{\theta}_2\cos\theta_2}$ Hence the kinetic energy is $\displaystyle K=1/2m_1l^2_1\theta^2_1+1/2m\left[ l^2_1\dot{\theta}^2_1+l^2_2\dot{\theta}^2_2+2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2) \right]$ And the potential is $\begin{aligned}\displaystyle V&=m_1g(l_1+l_2-l_1\cos\theta_1)\\ &+m_2g\left[l_1+l_2-(l_1\cos\theta_1+l_2\cos\theta_2)\right]\end{aligned}$ As always the Lagrangian is ${L=K-V=\cdots}$ The equations of motion. ${\begin{aligned}\dfrac{\partial L}{\partial \theta_1}&=-m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)\\ &- m_1gl_1\sin\theta_1-m_2gl_1\sin\theta_1\end{aligned}}$ ${\dfrac{\partial L}{\partial \dot{\theta_1}}=m_1 l_1^2\dot{\theta}_1^2+m_2l_1^2\dot{\theta}_1+m_2l_1l_2\dot{\theta}_2\cos(\theta_1-\theta_2)}$ ${\dfrac{\partial L}{\partial \theta_2}=m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)-m_2gl_2\sin\theta_2}$ ${\dfrac{\partial L}{\partial \dot{\theta_2}}=m_2l_2^2\dot{\theta}_2^2+m_2l_1l_2\dot{\theta}_1\cos(\theta_1-\theta_2)}$ ${\begin{aligned}-(m_1+m_2)gl\sin\theta_1&=(m_1+m_2)l_1\ddot{\theta}_1+ m_2l_1l_2\ddot{_2}\cos(\theta_1-\theta_2)\\ &+ m_2l_1l_2\dot{\theta}_2\sin(\theta_1-\theta_2)\end{aligned}}$ and ${\begin{aligned}-m_2gl_2\sin\theta_2 &=m_2l_2\ddot{\theta}_2+m_2l_1l_2\ddot{\theta}_1\cos(\theta_1-\theta_2)\\ &+ m_2l_1l_2\dot{\theta}_1^2\sin(\theta_1-\theta_2)\end{aligned}}$ Assume that ${m_1=m_2=m}$ e ${l_1=l_2=l}$ and write the equations of motion. Left as an exercise for the reader. Write the previous equations in the limit of small oscillations. If ${\theta\ll1}$ implies ${\sin \theta\approx\theta}$ and ${\cos \theta\approx1}$. Hence the equations of motion are ${2l\ddot{\theta}_1+l\ddot{\theta}_2=-2g\theta_1}$ ${l\ddot{\theta}_1+l\ddot{\theta}_2=-g\theta_2}$

Exercise 10 A particle moves along the plane ${xy}$ subject to a central force that is a function of the distance between the particle and the origin. Find the hamiltonian of the system. The generalized coordinates are ${r}$ and ${\theta}$. The potential is of the form ${V=V(r)}$. The Lagrangian is ${L=1/2m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)}$. The conjugate momenta are: ${p_r=\dfrac{\partial L}{\partial \dot{r}}=m\dot{r}}$ ${p_\theta=\dfrac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}}$ For the Hamiltonian it is ${\begin{aligned} H&=\displaystyle \sum_{\alpha=n}^np_\alpha\dot{q}_\alpha\\ &=p_r\dot{r}+p_\theta\dot{\theta}-(1/2m(\dot{r^2}+r^2\dot{\theta}^2)-V(r))\\ &=\dfrac{p_r^2}{2m}+\dfrac{p_theta^2}{2mr^2}+V(r) \end{aligned}}$ Write the equations of motion. ${\dot{r}=\dfrac{\partial H}{\partial p_r}=\dfrac{p_r}{m}}$ ${\dot{\theta}=\dfrac{\partial H}{\partial p_\theta}=\dfrac{p_\theta}{mr^2}}$ ${\dot{p}_r=-\dfrac{\partial H}{\partial r}=\dfrac{p_\theta}{mr^3}-V'(r)}$ ${\dot{p}_\theta=-\dfrac{\partial H}{\partial \theta}=0)}$

Exercise 11 A particle describes a one dimensional motion subject to a force $\displaystyle F(x,t)= \frac{k}{x^2}e^{-t/\tau}$ where${k}$ and ${\tau}$ are positive constants. Find the lagrangian and the hamiltonian. Compare the hamiltonian with the total energy and discuss energy conservation for this system. Since ${F(x,t)= \frac{k}{x^2}e^{-t/\tau}}$ it follows ${V=\dfrac{k}{x}e^{-t/\tau}}$. For the kinetic energy it is ${K=1/2m\dot{x}^2}$. Hence the lagrangian is $\displaystyle L=1/2m\dot{x}^2-\dfrac{k}{x}e^{-t/\tau}$ . Now ${p_x=m\dot{x}\Rightarrow\dot{x}=\dfrac{p_x}{m}}$. For the Hamiltonian it is ${H=p_x\dot{x}-L=\dfrac{p_x^2}{2m}+\dfrac{k}{x}e^{-t/\tau}}$. Since ${\dfrac{\partial L}{\partial t}=0}$ the system isn't conservative. Since ${\dfrac{\partial U}{\partial \dot{x}}=0}$ it is ${H=E}$.

Exercise 12 Consider two functions of the generalized coordinates and the generalized momenta, ${g(q_k,p_k)}$ and ${h(q_k,p_k)}$. The Poisson brackets are defined as: $\displaystyle [g,h]=\sum_k \left(\frac{\partial g}{\partial q_k}\frac{\partial h}{\partial p_k}-\frac{\partial g}{\partial p_k}\frac{\partial h}{\partial q_k}\right)$ Show the following properties of the Poisson brackets: ${\dfrac{dg}{dt}=[g,H]+\dfrac{\partial g}{\partial t}}$. Left as an exercise for the reader. ${\dot{q}_j=[q_j,H]}$ e ${\dot{p}_j=[p_j,H]}$. Left as an exercise for the reader. ${[p_k,p_j]=0}$ e ${[q_k,q_j]=0}$. Left as an exercise for the reader. ${[q_k,p_j]=\delta_{ij}}$. Left as an exercise for the reader. If the Poisson brackets between two functions is null the two functions are said to commute. Show that if a function ${f}$ doesn't depend explicitly on time and ${[f,H]=0}$ the function is a constant of movement.