# 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 ${L=K-V=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2-1/2kx^2_1-1/2k(x_2-x_1)^2-1/2kx^2_2}$. 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 $\displaystyle V=m_1g(l_1+l_2-l_1\cos\theta_1)+m_2g\left[l_1+l_2-(l_1\cos\theta_1+l_2\cos\theta_2)\right]$ As always the Lagrangian is ${L=K-V=\cdots}$ The equations of motion. ${\dfrac{\partial L}{\partial \theta_1}=-m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)-m_1gl_1\sin\theta_1-m_2gl_1\sin\theta_1}$ ${\dfrac{\partial L}{\partial \dot{\theta_1}}=m_1 l_1^2\dot{\theta}_1^2+m_2l_1^2\dot{\theta}_1+m_2l_1l_2\dot{\theta}_2\cos(\theta_1-\theta_2)}$ ${\dfrac{\partial L}{\partial \theta_2}=m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)-m_2gl_2\sin\theta_2}$ ${\dfrac{\partial L}{\partial \dot{\theta_2}}=m_2l_2^2\dot{\theta}_2^2+m_2l_1l_2\dot{\theta}_1\cos(\theta_1-\theta_2)}$ ${(m_1+m_2)l_1\ddot{\theta}_1+m_2l_1l_2\ddot{_2}\cos(\theta_1-\theta_2)+m_2l_1l_2\dot{\theta}_2\sin(\theta_1-\theta_2)=-(m_1+m_2)gl\sin\theta_1}$ and ${m_2l_2\ddot{\theta}_2+m_2l_1l_2\ddot{\theta}_1\cos(\theta_1-\theta_2)+m_2l_1l_2\dot{\theta}_1^2\sin(\theta_1-\theta_2)=-m_2gl_2\sin\theta_2}$ Assume that ${m_1=m_2=m}$ e ${l_1=l_2=l}$ and write the equations of motion. Is 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.

# Newtonian formalism exercises

 Exercise 1 A particle of mass ${m}$ moves in the plane ${xy}$. Its position vector is ${\vec{r}=a\cos \omega t \vec{i}+b\sin \omega t \vec{j}}$ with ${a,b,\omega}$ positive constants and ${a>b}$. Show that: The particle’s trajectory is an ellipse. For ${x=a\cos\omega\ t}$ and ${y=b\sin\omega\ t}$ it is $\displaystyle \left( \frac{x}{a} \right)^2+\left( \frac{y}{b} \right)^2=\cos^2\omega t+\sin^2\omega t=1$ Which is the equation of an ellipse The particle is subject to a central force oriented to the origin It is {\begin{aligned} \vec{F} &= m\dfrac{d\vec{v}}{dt}\\ &= m\dfrac{d^2\vec{r}}{dt^2}\\ &= m\dfrac{d^2}{dt^2}a\cos \omega t \vec{i}+b\sin \omega t \vec{j}\\ &=-m\omega^2\vec{r} \end{aligned}} Which is a central force oriented to the origin (note the minus sign).

 Exercise 2 A particle of constant mass ${m}$ is subject to to a force ${F}$. Suppose that for ${t_1}$ e ${t_2}$ the velocities are ${\vec{v}_1}$ and ${\vec{v}_2}$, respectively. Show that the work the force does on the particle equals its change in kinetic energy. {\begin{aligned} W &= \int_{t_1}^{t_2}\vec{F}\cdot d\vec{r}\\ &= \int_{t_1}^{t_2}\vec{F}\cdot\dfrac{d\vec{r}}{dt}dt \\ &= \int_{t_1}^{t_2}\vec{F}\cdot\vec{v}dt\\ &= \int_{t_1}^{t_2}m\dfrac{d\vec{v}}{dt}\cdot\vec{v}dt\\ &=m\int_{t_1}^{t_2}\vec{v}\cdot d\vec{v}\\ &=\dfrac{1}{2} m\int_{t_1}^{t_2}d(\vec{v}\cdot\vec{v})\\ &=1/2m(v_2^2-v_1^2) \end{aligned}}

 Exercise 3 For the conditions of exercise 1 calculate: The particle’s kinetic energy in the positive semi-major axis and on the positive semi-minor axis. First note that ${\vec{v}=-\omega a\sin \omega t \vec{i}+\omega b \cos \omega t \vec{j}}$. Now ${K=1/2m(\omega^2a^2\sin^2\omega t+\omega^2b^2\cos^2\omega t)}$. Let ${A}$ denote the semi-major axis extension and ${B}$ the semi-minor axis extension. Hence ${K_A=1/2m\omega ^2b^2}$ and ${K_B=1/2m\omega ^2a^2}$ The work done by the force field when it moves the particle from the positive semi-major axis to the positive semi-minor axis. {\begin{aligned} W &= \int_A^B\vec{F}\cdot d\vec{r}\\ &= \int_A^B(-m\omega ^2\vec{r})\cdot d\vec{r}\\ &= -m\omega ^2\int_A^B\vec{r}\cdot d\vec{r}\\ &= -1/2m\omega ^2\int_A^B d(\vec{r}\cdot\vec{r})\\ &=1/2m\omega ^2(a^2-b^2) \end{aligned}} The total work done by the force field by moving the particle along the ellipse. ${W=0}$. Do you see why?

 Exercise 4 Show that if a particle is subject to ${\vec{F}}$ and ${\vec{v}}$ is its instantaneous velocity then the instantaneous power is $\displaystyle \mathcal{P}=\vec{F}\cdot\vec{v}$ By definition it is ${dW=\vec{F}\cdot d\vec{r}}$. Hence {\begin{aligned} \mathcal{P}&=\dfrac{dW}{dt}\\ &= \vec{F}\cdot\dfrac{d\vec{r}}{dt}\\ &= \vec{F}\cdot\vec{v} \end{aligned}}

 Exercise 5 Show that the integral ${\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is independent of a particle’s trajectory if and only if ${ \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Let ${\Gamma=P_1AP_2BP_1}$ denote a closed curve and admit that ${\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is path independent. Then {\begin{aligned} \oint \vec{F}\cdot d\vec{r}&=\int_\Gamma\vec{F}\cdot d\vec{r} \\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}} Where the last equality follows from our assumption that ${\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is path independent. Suppose now that ${ \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. {\begin{aligned} \int_\Gamma\vec{F}\cdot d\vec{r}&=\int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}} Where the last equality follows from our ${ \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Hence ${\displaystyle\int_{P_1AP_2}\vec{F}\cdot d\vec{r}=\int_{P_1BP_2}\vec{F}\cdot d\vec{r}}$.

 Exercise 6 A particle of mass ${m}$ moves along ${x}$ subject to to a conservative force field ${V(x)}$. If the particle’s positions are ${x_1}$ and ${x_2}$ on ${t_1}$ and ${t_2}$, respectively, show that, if ${E}$ is the total energy then $\displaystyle t_2-t_1=\sqrt{\frac{m}{2}}\int_{x_1} ^{x_2}\frac{dx}{\sqrt{E-V(x)}}$ Write ${1/2m\left( \dfrac{dx}{dt} \right)^2+V(x)=E}$ solve in order to ${dt}$ and integrate.

 Exercise 7 Consider a particle of mass ${m}$ that moves vertically on a resistive medium where the retarding force is proportional to the particle’s velocity. Consider that particle initially moves in the downward direction with an initial velocity ${v_0}$ from an height ${h}$. Derive the particle’s equation of motion ${z=z(t)}$. It is ${F=m\dfrac{dv}{dt}=-mgg-kmv}$ with ${v<0}$ and ${-kmv>0}$. Then solve in order to ${dv}$ and integrate to find ${v=\dfrac{dz}{dt}=-\dfrac{g}{k}+\dfrac{kv_0+g}{k}e^{-kt}}$ (the terminal velocity ${v_t}$ is ${v_t=\displaystyle \lim_{t \rightarrow +\infty} v=-g/k}$). Solving in order to ${dz}$ and integrating and it is ${z=h-\dfrac{gt}{k}+\dfrac{kv_0+g}{k^2}(1-e^{-kt})}$.

 Exercise 8 A particle is shot vertically on a region where exists a constant gravitational field with a constant initial velocity ${v_0}$. Show that, in the presence of resistive force proportional to the square of the particle’s instantaneous velocity, the velocity of the particle when it returns to its initial position is: $\displaystyle \frac{v_0 v_t}{\sqrt{v_0^2+v_t^2}}$ Where ${v_t}$ denotes the particle’s terminal velocity.