# 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 Mechanics 06

— 7. Newton and Euler-Lagrange Equations —

We’ve seen in the previous examples that solving a problema while using the lagrangian formalism would lead us to the same equations of Newton’s formalism.

In this section we’ll show that both formulations are indeed equivalent for conservative systems.

We have ${\dfrac{\partial L}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=0 }$ for ${i=1,2,3}$. The previous equation can be written in the following form:

$\displaystyle \dfrac{\partial (K-U)}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial (K-U)}{\partial \dot{x}_i}=0$

Since our analysis doesn’t depend on the type of coordinates one uses we’ll choose to use rectangular coordinates. Hence it is ${K=K(\dot{x}_i}$ and ${U=U(x)}$. It is ${\dfrac{\partial T}{\partial x_i}=0}$ and ${\dfrac{\partial U}{\partial \dot{x}_i}}$. Hence ${-\dfrac{\partial U}{\partial \dot{x}_i}=\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x}_i}}$. Since for a conservative system it is ${-\dfrac{\partial U}{\partial \dot{x}_i}=F_i}$ it follows that ${F_i=\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x}_i}}$ {\begin{aligned} \dfrac{d}{dt}\dfrac{\partial T}{\partial \dot{x}_i} &= \dfrac{d}{dt}\dfrac{\partial \sum_j(1/2m\dot{x}_j^2)}{\partial \dot{x}_i} \\ &= \dfrac{d}{dt}(m\dot{x}_i) \\ &= \dot{P}_i \end{aligned}} Finally it is ${F_i=\dot{P}_i}$ which is Newton’s Second Axiom. Since the dynamics of a particle are a result of this axiom the dynamics of a given particle have to be the same on both formulations of mechanics.

— 8. Symmetry considerations —

As you certainly noticed in the previous examples the absence of generalized coordinate on the lagrangian of the system implied the conservation of a momentum (angular or linear). These coordinates that don’t appear on the lagrangian are called cyclic coordinates in the literature.

Obviously that the presence or absence of cyclical coordinates on a lagrangian depend on the choice of coordinate that one makes. But the fact that a moment is conserved cannot depend on the choice of the set of coordinates one makes.

Since the right choice of coordinates is linked to the symmetry that the system exhibits one can conclude that symmetry and conserved quantities are intrinsically connected.

In this section we’ll understand why symmetry considerations are so important in contemporary Physics and what is the relationship between symmetry and conservation. If a system exhibits some kind of continuous symmetry this symmetry will always manifest in the form some conserved quantity. The mathematical proof of this theorem (and its multiple generalizations) is Noether’s theorem and I won’t provide a proof of it here. Instead we’ll look into the consequences of three types of continuous symmetry and I’ll provide you links for you to study Noether’s theorem:

— 8.1. Continuous symmetry for time translations —

As we saw in Newtonian Mechanics 01 a frame is said to be inertial if time is homogeneous. When one says that time is homogeneous one is saying that one can perform a continuous time translation (formally one says ${t \rightarrow t+\delta t}$) and the characteristics of the mechanical system won’t change.

Let ${L}$ denote the lagrangian of an isolated system. Since the system is isolated its physical characteristics must remain unchanged for all times. This is equivalent to saying that the lagrangian can’t depend on time (${\dfrac{\partial L}{\partial t}=0}$.)

Hence the total derivative is just

$\displaystyle \frac{dL}{dt}= \sum_j \frac{\partial L}{\partial q_j}\dot{q}_j+ \sum_j \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j$

Using Euler-Lagrange equations 13 for generalized coordinates it is

{\begin{aligned} \frac{dL}{dt} &= \sum_j \dot{q}_j\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}+ \sum_j \frac{\partial L}{\partial \dot{q}_j}\ddot{q}_j \Rightarrow \\ &\Rightarrow \frac{dL}{dt}-\sum_j\frac{d}{dt}\left( \dot{q}_j\frac{\partial L}{\partial \dot{q}_j} \right)= 0 \\ &\Rightarrow \frac{d}{dt} \left( L-\sum_j\dot{q}_j\frac{\partial L}{\partial \dot{q}_j}=0\right) \end{aligned}}

In conclusion it is

$\displaystyle L-\sum_j\dot{q}_j\dfrac{\partial L}{\partial \dot{q}_j}=-H \ \ \ \ \ (14)$

where ${-H}$ (the ${-}$ sign will be apparent later) is some constant.

Suppose that ${U=U(x_{\alpha,i})}$ and ${x_{\alpha,i}=x_{\alpha,i}(q_j)}$. Then it is ${U=U(q_j)}$ and ${\dfrac{\partial U}{\partial \dot{q}_j}=0}$. Hence ${\dfrac{\partial L}{\partial \dot{q}_j}=\dfrac{\partial (K-U)}{\partial \dot{q}_j}=\dfrac{\partial K}{\partial \dot{q}_j}}$

Then we can write equation 14 as ${\displaystyle (K-U)-\sum_j\dot{q}_j\dfrac{\partial K}{\partial \dot{q}_j}=-H}$. From this it follows ${K+U=H}$.

The function ${H}$ is called the Hamiltonian and its definition is given by equation 14.

Furthermore one can identify the Hamiltonian with the total energy of the system if the following conditions are met:

• The equations of coordinate transformations are time independent which implies that the kinetic energy is a quadratic homogeneous function of ${\dot{q}_j}$
• The potential energy is velocity independent so that the terms ${\dfrac{\partial U}{\partial \dot{q}_j}}$ can be eliminated.

— 8.2. Continuous symmetry for space translations —

As we saw in Newtonian Mechanics 01 a frame is said to be inertial if space is homogeneous. When one says that space is homogeneous one is saying that the lagrangian is invariant under space translations. Formally one says that ${\delta L=0}$ for ${\vec{r}_\alpha \rightarrow \vec{r}_\alpha+\delta\vec{r}}$.

Without loss of generality let us consider just one particle. Now ${L=L(x_i),\dot{x_i}}$ and ${\displaystyle \delta \vec{r} = \sum_i\delta x_i \vec{e}_i}$. In this case the variation in ${L}$ due to ${\delta \vec{r}}$ is

$\displaystyle \delta L = \sum_i \frac{\partial L}{\partial x_i}\delta x_i + \sum_i\frac{\partial L}{\partial \dot{x}_i}\delta \dot{x}_i=0$

Now ${\delta x_i=\delta\dfrac{dx_i}{dt}=\frac{d}{dt}\delta x_i=0}$ so the expression for the variation in the lagrangian becomes

$\displaystyle \delta L = \sum_i \frac{\partial L}{\partial x_i}\delta x_i=0$

For the previous expression to be identically ${0}$ one has to have ${\dfrac{\partial L}{\partial x_i}=0}$ since the ${\delta x_i}$ are arbitrary variations.

According to Euler-Lagrange equations 13 one also has ${\dfrac{\partial L}{\partial \dot{x}_i}=\mathrm{const}}$.

{\begin{aligned} \frac{\partial (K-U)}{\partial \dot{x}_i} &= \frac{\partial K}{\partial\dot{x}_i}\\ &= \frac{\partial}{\partial \dot{x}_i}\left( 1/2m\sum_j\dot{x}_j^2 \right) \\ &= m\dot{x}_i \\ &= P_i \end{aligned}}

Hence the homogeneity of space to translations implies the conservation of linear momentum in an isolated system.

— 8.3. Continuous symmetry for space rotations —

As we saw in Newtonian Mechanics 01 a frame is said to be inertial if space is isotropic. When one says that space is isotropic one is saying that the lagrangian is invariant under space rotations. Formally one says that ${\delta L=0}$ for ${\vec{r}_\alpha \rightarrow \vec{r}_\alpha+\delta\vec{r}}$ where ${\delta\vec{r}=\delta \vec{\theta} \times \vec{r}}$.

First we have (we’re considering one particle) ${\delta\vec{\dot{r}}=\delta \vec{\theta} \times \vec{\dot{r}}}$

$\displaystyle \delta L = \sum_i \frac{\partial L}{\partial x_i}\delta x_i + \sum_i\frac{\partial L}{\partial \dot{x}_i}\delta \dot{x}_i=0$

From ${p_i=\dfrac{\partial L}{\partial \dot{x}_i}}$ and ${\dot{p}_i=\dfrac{\partial L}{\partial x_i}}$ it follows

{\begin{aligned} \delta L &= \sum_i\dot{p}_i\delta x_i+ \sum_i p_i\delta\dot{x}_i\\ &= \dot{\vec{p}}\cdot\delta\vec{r}+ \vec{p}\cdot\delta\dot{\vec{r}} \\ &= \dot{\vec{p}}\cdot(\delta \vec{\theta} \times \vec{r})+ \vec{p}\cdot(\delta \vec{\theta} \times \dot{\vec{r}}) \\ &= \delta\vec{\theta}\cdot(\vec{r}\times\dot{\vec{p}}) + \delta\vec{\theta}\cdot(\dot{\vec{r}}\times\vec{p})\\ &= \delta\vec{\theta}\cdot (\vec{r}\times\dot{\vec{p}} + \dot{\vec{r}}\times\vec{p}) \end{aligned}}

Since ${\delta\vec{\theta}\cdot (\vec{r}\times\dot{\vec{p}} + \dot{\vec{r}}\times\vec{p})=\delta\vec{\theta}\cdot\dfrac{d}{dt}(\vec{r}\times\vec{p})}$ and ${\delta L=0}$, it follows ${\delta\vec{\theta}\cdot\dfrac{d}{dt}(\vec{r}\times\vec{p})=0}$.

Since ${\delta\vec{\theta}}$ is an arbitrary vector it follows ${\dfrac{d}{dt}(\vec{r}\times\vec{p})=0}$. Hence ${\vec{r}\times\vec{p}}$ is constant.

In conclusion one can say that space isotropy implies the conservation of angular momentum. Another important result is that whenever a mechanical system has a symmetry axis the angular momentum about that axis is a conserved quantity.

— 9. Hamiltonian Dynamics —

As was seen previously if the potential energy of a system doesn’t depend on velocity then ${p_i=\dfrac{\partial L}{\partial \dot{x}_i}}$. Consequently one can introduce the following definition:

 Definition 8 In a system of generalized coordinates ${q_j}$ the generalized momentum is given by the following expression $\displaystyle p_j=\frac{\partial L}{\partial \dot{q}_j} \ \ \ \ \ (15)$

As a consequence of the previous definition it is ${\dot{p}_j=\frac{\partial L}{\partial q_j}}$.

And the Hamiltonian can be written as a Legendre Transformation of the Lagrangian

$\displaystyle H=\sum_j p_j\dot{q}_j-L \ \ \ \ \ (16)$

Since ${\dot{q}_j=\dot{q}_j(q_k,p_k,t)}$ equation 16 can be written

$\displaystyle H(q_k,p_k,t)=\sum_j p_j\dot{q}_j-L(q_k,\dot{q}_k,t) \ \ \ \ \ (17)$

So we have ${H=H(q_k,p_k,t)}$ and ${L=L(q_k,\dot{q}_k,t)}$. Hence the differential f ${H}$ is

$\displaystyle dH=\sum_k\left( \frac{\partial H}{\partial q_k}dq_k+\frac{\partial H}{\partial p_k}dp_k \right) + \frac{\partial H}{\partial t}dt \ \ \ \ \ (18)$

Calculating ${\dfrac{\partial H}{\partial q_k}}$ and ${\frac{\partial H}{\partial p_k}}$ via 16 and substituting into 18 it is

$\displaystyle dH=\sum_k (\dot{q}_kdp_k-\dot{p}_kdq_k)-\frac{\partial L}{\partial t}dt \ \ \ \ \ (19)$

Identifying the coefficients of ${dq_k}$, ${dt_k}$ and ${dt}$ it follows:

$\displaystyle \dot{q}_k=\frac{\partial H}{\partial p_k} \ \ \ \ \ (20)$

and

$\displaystyle -\dot{p}=\frac{\partial H}{\partial q_k} \ \ \ \ \ (21)$

Which are called the canonical equations of motion. When one uses these equations to study the time evolution of a physical system one is using Hamiltonian Dynamics.

One has ${-\dfrac{\partial L}{\partial t}=\dfrac{\partial H}{\partial t}}$. Furthermore one also have ${\dfrac{dH}{dt}=\dfrac{\partial H}{\partial t}}$ which implies that if the Hamiltonian function doesn’t depend explicitly on ${t}$ then ${H}$ is a conserved quantity.

 Example 7 A particle of mass ${m}$ moves on the surface of a cylinder subject to a force which points to the center of the cylinder (the origin of our frame) and is proportional to the distance between the particle and the origin. Since ${\vec{F}=-k\vec{r}}$ it follows that ${U=1/2kr^2=1/2k(R^2+z^2)}$ For the velocity it is ${v^2=\dot{r}^2+r^2\dot{\theta}^2+\dot{z}^2}$. Since ${r=R}$ is a constant in our example ${K=1/2m(R^2\dot{\theta}^2+\dot{z}^2)}$ Hence the Lagrangian is ${L=1/2m(R^2\dot{\theta}^2+\dot{z}^2)-1/2k(R^2+z^2)}$. The generalized coordinates are ${\theta}$ and ${z}$ and the generalized momenta are $\displaystyle p_\theta=\frac{\partial L}{\partial \dot{\theta}}=mR^2\dot{\theta}$ and $\displaystyle p_z=\frac{\partial L}{\partial \dot{z}}=m\dot{z}$ Since this a conservative system and and transformation equations between coordinates systems don’t depend on time, ${H}$ is the total energy of the system and it is a function of ${\theta}$, ${p_\theta}$, ${z}$ and ${p_z}$. But ${\theta}$ doesn’t appear in the Lagrangian (thus it is a cyclic coordinate). $\displaystyle H(z,p_\theta,p_z)=K+U= \frac{p_\theta^2}{2mR^2}+\frac{p_z^2}{2m} +1/2kz^22$ for the equations of motion it is $\displaystyle \dot{p}_\theta=-\frac{\partial H}{\partial \theta}=0$ $\displaystyle \dot{p}_z=-\frac{\partial H}{\partial z}=-kz$ $\displaystyle \dot{\theta}=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{mR^2}$ $\displaystyle \dot{z}=\frac{\partial H}{\partial p_z}=\frac{p_z}{m}$ From the previous relationships one sees that the angular momentum about the ${z}$ axis is constant ${p_\theta=mR^2\dot{\theta}}$. Which is equivalent to saying that the ${z}$ axis is a axis of symmetry of the system. We also have ${m\ddot{z}=-kz\Rightarrow m \ddot{z}+kz=0\Rightarrow \ddot{z}+k/mz=0\Rightarrow\ddot{z}+\omega_0^2}$ with ${\omega_0^2=k/m}$. Which means that the particle describes an harmonic motion along the ${z}$ axis. To finalize our treatment of classical dynamics let us just make a quick summary of Lagrangian and Hamiltonian dynamics. The generalized coordinates and the respective generalized momenta are said to be canonical coordinates. Coordinates that don’t appear explicitly on the expressions ok ${K}$ and ${U}$ are called cyclical coordinates. A coordinate that is cyclical in ${H}$ also is cyclical in ${L}$. A generalized coordinate and and it’s corresponding generalized momentum are said to be canonical coordinates.

# Newtonian Mechanics 05

— 1. Variational Calculus —

 Definition 1 A functional is a mapping from vector spaces to into real numbers.

Let ${\displaystyle J = \int _{x_1}^{x^2} f\{y(x),y\prime (x),x\}dx}$. Suppose that ${x_1}$ and ${x_2}$ are constants, the functional form of ${f}$ is known.

According to definition 1 ${J}$ is a functional and the goal of the Calculus of Variations is to determine ${y(x)}$ such that the value of ${J}$ is an extremum.

Let ${y=y(\alpha, x)}$ be a parametric representation of ${y}$ such that ${y(0,x)=y(x)}$ is the function that makes ${J}$ an extremum.

We can write ${y(\alpha, x)=y(0,x)+ \alpha\eta(x}$, where ${\eta (x)}$ is a function of ${x}$ of the class ${C^1}$ (that means that ${\eta}$ is a continuous function whose derivative is also continuous) with ${\eta (x_1)=\eta (x_2)=0}$.

Now ${J}$ is of the form ${\displaystyle J(\alpha) = \int _{x_1}^{x^2} f\{y(\alpha, x),y\prime (\alpha, x),x\}dx}$

Therefore the condition for ${J}$ to be an extremum is

$\displaystyle \frac{dJ}{d\alpha}(\alpha=0)=0$

 Example 1 Let ${y(x)=x}$. Take ${y(\alpha, x)= x+ \alpha\sin x}$ as a parametric representation of ${y}$. Let ${f=\left(dy/dx\right)^2}$, ${x_1=0}$ and ${x_2=2\pi}$. Given the previous parametric equation find ${\alpha}$ such that ${J}$ is a minimum. Now ${\eta (0)=\eta (2\pi)=0}$ and ${dy/dx=1+\alpha\cos x}$. Hence ${\displaystyle J(\alpha)= \int_0^{2\pi}(1+2\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}$. By the previous expression ${J(\alpha)}$ it is trivial to see that the minimum value is reached when ${\alpha=0}$
 Exercise 1 Given the points ${(x_1,y_1)=(0,0)}$ and ${(x_2,y_2)=(1,0)}$, calculate the equation of the curve that minimizes the distance between the points Now ${y(\alpha, x)=y(0,x)+\alpha \eta (x) = 0+\alpha(x^2-x)}$. It is ${\eta (x) = x^2-x}$, ${ds=\displaystyle \sqrt{dx^2+dy^2}=\sqrt{1+(dy/dx)^2}dx}$ And it is ${s= \displaystyle \int _0^1 \sqrt{1+(dy/dx)^2}}$ with ${dy/dx=\alpha (2x-1)}$. The rest is left as an exercise for the reader.

— 2. Euler Equations —

In the following section we’ll analyze the condition for ${J}$ to be an extremum:

{\begin{aligned} \frac{\partial J}{\partial \alpha} &= \frac{\partial}{\partial \alpha} \int _{x_1}^{x_2}f(y,y\prime,x)dx \\ &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+ \frac{\partial f}{\partial y\prime}\frac{\partial y\prime}{\partial \alpha}\right) dx \end{aligned}}

Since it is ${\partial y /\partial \alpha = \eta (x)}$ and ${\partial y\prime /\partial \alpha = d\eta/dx}$ it follows

$\displaystyle \frac{\partial J}{\partial \alpha}= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\eta (x)+ \frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}\right) dx$

Now ${\displaystyle \int _{x_1}^{x_2}\frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}dx=\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}- \int _{x_1}^{x_2}\frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right)\eta (x) dx}$.

For the first term it is ${\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}=0}$ since ${\eta (x_1)=\eta (x_2)=0}$ by hypothesis.

Hence

{\begin{aligned} \frac{\partial J}{\partial \alpha} &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}- \frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right) \frac{\partial y}{\partial \alpha}\right)dx \\ &= \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime} \right)\eta (x) dx \end{aligned}}

Remembering that ${\partial J / \partial\alpha(\alpha=0)=0}$ and taking into account the fact that ${\eta (x)}$ is an arbitrary function one can conclude that

$\displaystyle \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime}=0$

The previous equation is known as the Euler’s Equation

 Example 2 As an example let us try to derive the equations of motion of a particle that moves in a constant force field starting its movement from the state of rest. The particles starts from point ${x_1, y_1}$ and goes to point ${x_2, y_2}$. From the enunciate it follows ${K+U=c}$. Let us take our original point as being our reference point for the potential. Then it is ${k+U=0}$. As always it is ${k=1/2mv^2}$. For the potential it is ${U=-Fx=-mgx}$. From the previous equations it follows that ${v=\sqrt{2gx}}$. From the definition of velocity it follows that $\displaystyle t=\int _{x_1,y_1}^{x_2,y_2} \frac{ds}{v}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{1+y\prime^2}}{\sqrt{2gx}}dx$ Let ${f=\sqrt{\frac{1+y\prime^2}{x}}}$ since ${(2g)^{-1/2}}$ is only a constant factor and can be omitted from our analysis. Given the functional form of ${f}$ it is ${df/dy=0}$ and Euler’s Equation just is: $\displaystyle \frac{d}{dx}\frac{\partial f}{\partial y\prime}=0$ From the previous relationship it is $\displaystyle \frac{\partial f}{\partial y\prime}=(2a)^{-1/2}=\mathrm{const}$ Hence it is {\begin{aligned} \frac{y\prime^2}{x(1+y\prime^2)} &= \frac{1}{2a} \Rightarrow\\ y &= \int \frac{x}{\sqrt{2ax-x^2}}dx \end{aligned}} Making the change of variables ${x=a(1-\cos \theta)}$ it follows ${dx=a\sin \theta d\theta}$. Hence the expression for ${y}$ is ${y=\int a(1-\cos \theta)d\theta\Rightarrow y=a(\theta-\sin \theta)+A}$. Since our particle starts from the origin it is ${A=0}$. Thus the solution to our initial problem is {\begin{aligned} x &= a(1-\cos \theta) \\ y &= a(\theta-\sin \theta) \end{aligned}} Which are the parametric equations of a cycloid.

To close our thoughts on the Euler equation let us say that there also is a second form for the Euler equation. The second form is

$\displaystyle f-y\prime\frac{\partial f}{\partial y\prime}= \mathrm{const}$

and is used in the cases where ${f}$ doesn’t depend explicitly on ${x}$.

— 3. Euler Equation for ${n}$ variables —

Let ${f}$ be of the form ${f=f\{ y_1(x),y\prime _1(x),y_2(x),y\prime _2(x),\cdots,y_n(x),y\prime _n(x), x \}}$.

Now we have ${y_i(\alpha, x)= y_i(0,x)+\alpha \eta (x)}$ and ${\displaystyle \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime} \right)\eta _i (x) dx}$ for each of the values of ${i}$. Since ${\eta _i(x)}$ are independent functions it follows that for ${\alpha=0}$

$\displaystyle \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime}=0$

That is to say we have ${n}$ independent Euler equations.

— 4. Hamilton’s Principle —

Minimum principles have a long history in the history of Physics:

In modern Physics one uses a more general extremum principle and the focus of this section will be to state this principle and flesh out its consequences.

 Definition 2 The lagrangian (sometimes called the lagrangian function), ${L}$, of a particle is the difference between its kinetic and potential energies. $\displaystyle L=K-U \ \ \ \ \ (1)$
 Definition 3 The action, ${S}$, of a particle’s movement (be it a real or virtual one) is: $\displaystyle \int_{t_1}^{t_2}(K-U)dt \ \ \ \ \ (2)$
 Axiom 1 Given a collection of paths that a particle can take between points ${x_1}$ and ${x_2}$ in the the time interval ${\Delta t= t_2-t_1}$ the actual path that the particle takes is the one that makes the action stationary $\displaystyle \delta S=\delta \int_{t_1}^{t_2}(K-U)dt=0 \ \ \ \ \ (3)$

For rectangular coordinates it is ${T=T(x_i)}$, ${U=U(x_i)}$, so ${L=T-U=L(x_i,\dot{x}_i)}$ (where ${\dot{x}_i=\dfrac{dx_i}{dt}}$ is called Newton’s notation).

The function ${L}$ can be identified with the function ${f}$ that we saw on Newtonian Mechanics 04 if one makes the obvious analogies

• ${x \rightarrow t}$
• ${y_i(x) \rightarrow x_i(t)}$
• ${y\prime_i(x) \rightarrow x\prime_i(t)}$
• ${f(y_i(x),y\prime_i (x),x) \rightarrow L(x_i,\dot{x}_i,t)}$

In this case the Euler equations are called the Euler-Lagrange equations and it is

$\displaystyle \frac{\partial L}{\partial x_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_i}=0$

 Example 3 Let us study the harmonic oscillator under Langrangian formalism $\displaystyle L=T-U=1/2m\dot{x}^2-1/2kx^2 \ \ \ \ \ (4)$ First it is ${\dfrac{\partial L}{\partial x_i}=-kx}$. Then we have ${\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=\dfrac{d}{dt}m\dot{x}=m\ddot{x}}$. Hence it is ${\dfrac{\partial L}{\partial x_i}-\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{x}_i}=0 \Rightarrow m\ddot{x}+kx=0 \Rightarrow m\ddot{x}=-kx}$ which is just the harmonic oscillator dynamic equation that we already know.
 Example 4 Consider a planar pendulumplanar pendulum write its Lagrangian and derive its equation of motion. The Lagrangian for the planar pendulum is $\displaystyle L=1/2ml^2\dot{\theta}^2-mgl(1-\cos \theta) \ \ \ \ \ (5)$ If we consider ${\theta}$ to be a rectangular coordinate (which it isn’t!) it follows that the equation of motion is: $\displaystyle \ddot{\theta}+g/l\sin \theta=0$ This is precisely the equation of motion of a planar pendulum and this result is apparently unexpected since we only analyzed the Lagrangian for rectangular coordinates.

— 5. Generalized coordinates —

Consider a mechanical system constituted by ${n}$ particles. In this case one would need ${3n}$ quantities to describe the position of all particles (since we have 3 degrees of freedom). In the case of having any kind of restraints on the motion of the particles the number of quantities needed to describe the motion of particle is less than ${3n}$. Suppose that one has ${m}$ restrictions than the degrees of freedom are ${3n-m}$.

Let ${s=3n-m}$. These ${s}$ coordinates don’t need to be rectangular, polar, cylindrical nor spherical. These coordinates can be of any kind provided that they completely specify the mechanical state of the system.

 Definition 4 The set of ${s}$ coordinates that totally specify the mechanical state of ${n}$ particles is defined to be the set of generalized coordinates. The generalized coordinates are represented by $\displaystyle q_1,q_2,\cdots,q_s$

Since we defined the generalized coordinates of a system of particles one can also define its set of generalized velocities.

 Definition 5 The set of ${s}$ velocities of a system of ${n}$ particles described by a set of generalized coordiinattes is defined to be the set of generalized velocities. The generalized velocities are represented by $\displaystyle \dot{q_1},\dot{q_2},\cdots,\dot{q_s}$

Let ${\alpha}$ denote the particle, ${\alpha=1,2,\cdots,n}$, ${i}$ represent the degrees of freedom ${i}$, ${i=1,2,3}$ and ${j}$ the number generalized coordinates ${j=1,2,\cdots,s}$.

$\displaystyle x_{\alpha,i}=x_{\alpha,i}(q_1,q_2,\cdots,q_s,t)=x_{\alpha,i}(q_j,t) \ \ \ \ \ (6)$

For the generalized velocities it is

$\displaystyle \dot{x}_{\alpha,i}=\dot{x}_{\alpha,i}(q_j,t) \ \ \ \ \ (7)$

The inverse transformations are

$\displaystyle q_j=q_j(x_{\alpha,i},t) \ \ \ \ \ (8)$

and

$\displaystyle \dot{q_j}=\dot{q}_j(x_{\alpha,i},t) \ \ \ \ \ (9)$

Finally let us note that we also need ${m=3n-s}$ equations of constraint

$\displaystyle f_k=f_k(x_{\alpha,i},t) \ \ \ \ \ (10)$

with ${k=1,2,\cdots,m}$.

 Example 5 Consider a point particle that moves along the surface of a semi-sphere of radius ${R}$ whose center is the origin of the coordinate system. The relevant equations are ${x^2+y^2+z^2-R^2\geq 0}$ and ${z\geq 0}$. Let ${q_1=x/R}$, ${q_2=y/R}$ and ${q_3=z/R}$ be our generalized coordinates. Furthermore we have the condition ${q_1^2+q_2^2+q_3^2=1}$ as a constraint equation. Hence ${q_3=\sqrt{1-(q_1^2+q_2^2)}}$
 Definition 6 Configuration space is the vector space defined by the generalized coordinates

The time evolution of a mechanical system can be represented as a curve in the configuration space.

— 6. Euler-Lagrange Equations in generalized coordinates —

Since ${K}$ and ${U}$ are scalar functions ${L}$ is also a scalar function. Therefore ${L}$ is an invariant for coordinate transformations.

Hence it is

$\displaystyle L=K(\dot{x}_{\alpha,i})- U(x_{\alpha,i})=T(q_j,\dot{q}_j,t)-U(q_j,t) \ \ \ \ \ (11)$

and ${L=L(q_j,\dot{q}_j,t)}$.

Hence we can write Hamilton’s Principle (Section 4) in the form

$\displaystyle \delta \int_{t_1}^{t_2} L(q_j,\dot{q}_j,t) dt=0 \ \ \ \ \ (12)$

That is

• ${x \rightarrow t}$
• ${y_i(x) \rightarrow q_j(t)}$
• ${y\prime_i(x) \rightarrow q\prime_j(t)}$
• ${f(y_i(x),y\prime_i (x),x) \rightarrow L(q_j,\dot{q}_j,t)}$

are the analogies to be made now.

Finally the Euler-Lagrange Equations are

$\displaystyle \frac{\partial L}{\partial q_j}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}=0 \ \ \ \ \ (13)$

for ${j=1,2,\cdots,s}$

To finalize this section let us note the conditions of validity for the Euler-Lagrange equations:

• The system is conservative.
• The equations of constraint have to be functions between the coordinates of the particles and can also be a function of time.
 Example 6 Consider the motion of a particle of mass ${m}$ along the surface of a half-angle cone under the action of the force of gravity. The equations are ${z=r\cot\alpha}$ and ${v^2=\dot{r}^2\csc^2\alpha+r^2\dot{\theta}^2}$ Now for the potential energy it is ${U=mgz=mgr\cot\alpha}$. Thus the lagrangian is $\displaystyle L=1/2m(\dot{r}^2\csc^2\alpha+r^2\dot{\theta}^2)-mgr\cot\alpha$ Since ${\dfrac{\partial L}{\partial \theta}=0}$ it is ${\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}=0}$. Hence it is ${\dfrac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}=\mathrm{const}}$. The angular momentum about the ${z}$ axis is ${mr^2\dot{\theta}=mr^2\omega}$. Thus ${mr^2\omega=\mathrm{const}}$ expresses the conservation of angular momentum about the axis of symmetry of system. It is left as an exercise for the reader to find the Euler-Lagrange equation for ${r}$.

# Newtonian Mechanics 04

— 1. Variational Calculus —

 Definition 1 A functional is a mapping from vector spaces to into real numbers.

Let ${\displaystyle J = \int _{x_1}^{x^2} f\{y(x),y\prime (x),x\}dx}$. Suppose that ${x_1}$ and ${x_2}$ are constants, the functional form of ${f}$ is known.

According to definition 1 ${J}$ is a functional and the goal of the Calculus of Variations is to determine ${y(x)}$ such that the value of ${J}$ is an extremum.

Let ${y=y(\alpha, x)}$ be a parametric representation of ${y}$ such that ${y(0,x)=y(x)}$ is the function that makes ${J}$ an extremum.

We can write ${y(\alpha, x)=y(0,x)+ \alpha\eta(x}$, where ${\eta (x)}$ is a function of ${x}$ of the class ${C^1}$ (that means that ${\eta}$ is a continuous function whose derivative is also continuous) with ${\eta (x_1)=\eta (x_2)=0}$.

Now ${J}$ is of the form ${\displaystyle J(\alpha) = \int _{x_1}^{x^2} f\{y(\alpha, x),y\prime (\alpha, x),x\}dx}$

Therefore the condition for ${J}$ to be an extremum is

$\displaystyle \frac{dJ}{d\alpha}(\alpha=0)=0$

 Example 1 Let ${y(x)=x}$. Take ${y(\alpha, x)= x+ \alpha\sin x}$ as a parametric representation of ${y}$. Let ${f=\left(dy/dx\right)^2}$, ${x_1=0}$ and ${x_2=2\pi}$. Given the previous parametric equation find ${\alpha}$ such that ${J}$ is a minimum. Now ${\eta (0)=\eta (2\pi)=0}$ and ${dy/dx=1+\alpha\cos x}$. Hence ${\displaystyle J(\alpha)= \int_0^{2\pi}(1+2\alpha\cos x +\alpha^2\cos ^2x)dx=2\pi+\alpha^2\pi}$. By the previous expression ${J(\alpha)}$ it is trivial to see that the minimum value is reached when ${\alpha=0}$
 Exercise 1 Given the points ${(x_1,y_1)=(0,0)}$ and ${(x_2,y_2)=(1,0)}$, calculate the equation of the curve that minimizes the distance between the points. Now ${y(\alpha, x)=y(0,x)+\alpha \eta (x) = 0+\alpha(x^2-x)}$. It is ${\eta (x) = x^2-x}$, ${ds=\displaystyle \sqrt{dx^2+dy^2}=\sqrt{1+(dy/dx)^2}dx}$ And it is ${s= \displaystyle \int _0^1 \sqrt{1+(dy/dx)^2}dx}$ with ${dy/dx=\alpha (2x-1)}$. The rest is left as an exercise for the reader.

— 2. Euler Equations —

In the following section we’ll analyze the condition for ${J}$ to be an extremum:

{\begin{aligned} \frac{\partial J}{\partial \alpha} &= \frac{\partial}{\partial \alpha} \int _{x_1}^{x_2}f(y,y\prime,x)dx \\ &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+ \frac{\partial f}{\partial y\prime}\frac{\partial y\prime}{\partial \alpha}\right) dx \end{aligned}}

Since it is ${\partial y /\partial \alpha = \eta (x)}$ and ${\partial y\prime /\partial \alpha = d\eta/dx}$ it follows

$\displaystyle \frac{\partial J}{\partial \alpha}= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\eta (x)+ \frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}\right) dx$

Now ${\displaystyle \int _{x_1}^{x_2}\frac{\partial f}{\partial y\prime}\frac{d \eta}{dx}dx=\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}- \int _{x_1}^{x_2}\frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right)\eta (x) dx}$.

For the first term it is ${\frac{\partial f}{\partial y\prime}\eta (x)|_{x_1}^{x_2}=0}$ since ${\eta (x_1)=\eta (x_2)=0}$ by hypothesis.

Hence

{\begin{aligned} \frac{\partial J}{\partial \alpha} &= \int _{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}- \frac{d}{dx}\left( \frac{\partial f}{\partial y\prime} \right) \frac{\partial y}{\partial \alpha}\right)dx \\ &= \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime} \right)\eta (x) dx \end{aligned}}

Remembering that ${\partial J / \partial\alpha(\alpha=0)=0}$ and taking into account the fact that ${\eta (x)}$ is an arbitrary function one can conclude that

$\displaystyle \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y\prime}=0$

The previous equation is known as the Euler’s Equation

 Example 2 As an example let us try to derive the equations of motion of a particle that moves in a constant force field starting its movement from the state of rest. The particles starts from point ${x_1, y_1}$ and goes to point ${x_2, y_2}$.From the enunciate it follows ${K+U=c}$. Let us take our original point as being our reference point for the potential. Then it is ${k+U=0}$. As always it is ${k=1/2mv^2}$. For the potential it is ${U=-Fx=-mgx}$. From the previous equations it follows that ${v=\sqrt{2gx}}$. From the definition of velocity it follows that $\displaystyle t=\int _{x_1,y_1}^{x_2,y_2} \frac{ds}{v}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{dx^2+dy^2}}{\sqrt{2gx}}=\int _{x_1,y_1}^{x_2,y_2}\frac{\sqrt{1+y\prime^2}}{\sqrt{2gx}}dx$ Let ${f=\sqrt{\frac{1+y\prime^2}{x}}}$ since ${(2g)^{-1/2}}$ is only a constant factor and can be omitted from our analysis. Given the functional form of ${f}$ it is ${df/dy=0}$ and Euler’s Equation just is: $\displaystyle \frac{d}{dx}\frac{\partial f}{\partial y\prime}=0$ From the previous relationship it is $\displaystyle \frac{\partial f}{\partial y\prime}=(2a)^{-1/2}=\mathrm{const}$ Hence it is {\begin{aligned} \frac{y\prime^2}{x(1+y\prime^2)} &= \frac{1}{2a} \Rightarrow\\ y &= \int \frac{x}{\sqrt{2ax-x^2}}dx \end{aligned}} Making the change of variables ${x=a(1-\cos \theta)}$ it follows ${dx=a\sin \theta d\theta}$. Hence the expression for ${y}$ is ${y=\int a(1-\cos \theta)d\theta\Rightarrow y=a(\theta-\sin \theta)+A}$. Since our particle starts from the origin it is ${A=0}$. Thus the solution to our initial problem is {\begin{aligned} x &= a(1-\cos \theta) \\ y &= a(\theta-\sin \theta) \end{aligned}} Which are the parametric equations of a cycloid. Cycloid

To close our thoughts on the Euler equation let us say that there also is a second form for the Euler equation. The second form is

$\displaystyle f-y\prime\frac{\partial f}{\partial y\prime}= \mathrm{const}$

and is used in the cases where ${f}$ doesn’t depend explicitly on ${x}$.

— 3. Euler Equation for ${n}$ variables —

Let ${f}$ be of the form ${f=f\{ y_1(x),y\prime _1(x),y_2(x),y\prime _2(x),\cdots,y_n(x),y\prime _n(x), x \}}$.

Now we have ${y_i(\alpha, x)= y_i(0,x)+\alpha \eta (x)}$ and ${\displaystyle \int _{x_1}^{x_2}\left( \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime} \right)\eta _i (x) dx}$ for each of the values of ${i}$. Since ${\eta _i(x)}$ are independent functions it follows that for ${\alpha=0}$

$\displaystyle \frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y _i\prime}=0$

That is to say we have ${n}$ independent Euler equations.