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.