The second part explores the usage of Lagrange mechanics in problems involving constraints and symmetries.

6.1: Contract Forces

6.11: Contact forces as Constraints

Contract forces arises by the virtue of physical contact between two objects. In origin, contact forces are electromagnetic. As two objects touch, the atoms push against one another through electromagnetic forces. Contact forces are not fundamental: they are the sum of microscopic interactions.

Consider a block on a horizontal ground. The net effect of the contact force is to constrain its vertical motion: the block can move sidewards and not up and down. More of such contract forces translate into constraints on the degree of freedom.

Consider a mechanical system parameterized by $N$ coordinates $q_k, k=1,2,\cdots, N$. However, due to some constraint forces, there are $P$ algebraic relations amongst these coordinates, given by:

where $l=1,\cdots, P$. This means that there are $N_P$ generalized coordinates or degrees of freedom.

The notation here can be confusing. $C$ represents constraints, and $C_l$ is the $l-th$ constraint in the system, representing equations that restrict the motion of the system. The constraints are usually expressed as functions of the generalized coordinates $q_k$ and time $t$, as shown by the above equation.

For example, in the problem of a block on the horizontal ground, let the ground be at $z=0$. Then, we can define $C=z$, so that the algebraic relation $C=0$ implies $z=0$. We started from three coordinates($N=3$) $x, y$ and $z$ and specified the constraint $z=0$ ($P=1$), where $z$ is the vertical position. Then, there are $N=P=2$ degrees of freedom. The Lagrange is given by:

6.12: A New Lagrangian

Let’s consider a new Lagrange defined as:

where $L$ is the original Lagrangian, The $P$ parameters labeled $\lambda_l$ are called Lagrange multipliers, and $C_l$ are constraints equations. In the new Lagrangian, we introduced $P$ additional degrees of freedom labeled $\lambda_l$ with $l=1,\cdots, P$.

We now assume that the constraint equations are not satisfied a priori. As a result, there are $N+P$ degrees of freedom and equations of motion. Using Euler’s equation, equations of motion for $L'$ are (we apply Euler’s equation to the newly defined Lagrange $L'$, not the one we used in the example for the block. Note that there are two equations of motion for $q_ks$ and $\lambda_ls$, where $q_k$ is the generalized coordinates for the original Lagrange $L=T-U$ and $\lambda_l$ is the generalized coordinates for the newly defined Lagrange.) :

Because $L$ is independent of $\lambda_l$ and $\dot \lambda_l$, we know that ${\partial L}/{\partial \dot\lambda_l}=0$. The equation becomes:

The $P$ parameters labeled $\lambda_l$ are called Lagrange multipliers. The equation for motion for $q_ks$ is given by:

In terms of the original Lagrange $L=T-U$, the equation for motion for $L'$ becomes:

For example, with the block on a horizontal floor, $L'$ is equal to:

Equations of motions are:

In this case $\lambda_1$ is the normal force.

Meanings of constraints $C$ and Lagrange multiplers $\lambda_l$ can be abstract. The term $\lambda_l$ represents a Lagrange multiplier associated with the $l$th constraint $C_l$. The nature of Lagrange multipliers are scalar functions that adjust to enforce the constraints in the system. For example, in the problem of a block on a horizontal ground, $C_{l=1}$ is the constraint that $z=0$ (meaning that the block cannot move vertically and can only slide ways) and the multiplier $\lambda_{l=1}$ is the normal force enforcing the constraint. Thus, physically, the multiplier can be interpreted as a force.

6.13: Constraints and Lagrange Multiplers

The addition term involving $\lambda$ is defined as the generalized constraint force $F_k$ that enforce the constraints onto the $q_k$ dynamics.

We want to relate the generalized constraint force with the actual force. For every object $i$ located at position $r_i$, denote the total constraint force acting on it by $F_i$. We also know the relations $r_i(q_k,t)$ that connect the position of every object to the generalized coordinates $q_k$. The work done or energy associated with the constraint forces $F^C_i$ and be expressed as:

Differentiating $r_i(q_k,t)$, we get $dr_i(q_k,t)/dq_k=\partial r_i(q_k,t) /\partial q_k$. Solving for $dr_i(q_k,t)$, we get:

Then, work done is equal to

Rearranging the equation for $F_k$, we get:

6.14: The Second Approach

In chapter four Lagrangian Mechanics , we define action as the integral of the Lagrangian with respect to time. In the process of extremizing the action, we would get (mathematics is not shown here. In order to derive the integral, integral by parts of multi-variable functions is required.):

for some functions $a_{lk}$ of $q_k$ and $t$. If the constraints on the generalized coordinates $q_k$ can be written in the form:

where $a_{lk}$ and $a_{lt}$ are arbitrary functions of $q_k$ and $t$, we can write the following set of $N+P$ equations of motion:

Summary for Finding the Multiplier (Constraint Force)

1. Define the constraints by writing down a constraint equation for each constraint.

2. Define a set of generalized coordinates without applying the constraints yet.

3. Write down the Lagrangian (again, with no constraints applied).

4. Apply the modified Euler-Lagrange equations with constraints and Lagrange multipliers.

   $$a_{lk}\delta q_k+a_{lt}\delta t=0$$

   $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}=\lambda a_{lk}=F_k$$

5. You should now have the equations of motion for each coordinate with Lagrange multipliers.

6. Solve for the Lagrange multipliers, which will give you the constraint forces.

Example 6.1: Rolling Down the Plane

Consider a hoop of radius R and mass M rolling down an inclined plane. The hoop rolls down an inclined plane without slipping. Find the Lagrangian and constraint force.

• Solution

  Steps

  1. Identifying constraints

  2. specifies position/velocity using Cartesian/polar/spherical/cylindrical coordinates

  3. find the Lagrangian $L=T-U$

  4. Use the general form of solution to find coefficients

  5. Find the equations of motion for each generalized coordinate.

  6. After finding general equations of motion for each generalized coordinate, substitute identified constraints and coefficients we found form step 4 to find the Lagrange multiplier.

  7. find the solution

  STEP 1:

  Before solving the problem, we first define the coordinate system. The coordinate system is shifted as shown in the figure. The hoop is described by three variables: the center of mass in two dimensions $r$ and a rotational angle $\theta$. The position $r$ of center of mass is specified as:

  $$r=x\hat{x}+y\hat{y}$$

  From the coordinate we define, we observe two constraints: (1) The vertical position of the center of mass is fixed, so $dy=0$ and $y=R$. (2) The second constraint is that the hoop rolls without slipping, giving us the constraint of:

  $$dx=Rd\theta$$

  The two constraints suggest $3-2=1$ degree of freedom.

  STEP 3:

  Kinetic energy $T$ of the hoop involves linear and rotational kinetic energies, where rotational kinetic energy is equal to $1/2I\omega^2=1/2MR^2\dot\theta^2$. Then, $T$ is equal to:

  $$T=\frac{1}{2}M(\dot x^2+\dot y^2)+\frac{1}{2}MR^2\dot\theta^2$$

  The potential energy of the hoop is equal to:

  $$U=-Mgx\sin\varphi$$

  where $\varphi$ is the angle tilted by the inclined plane. The Lagrangian $L$ is equal to:

  $$L=T-U=\frac{1}{2}M(\dot x^2+\dot y^2)+\frac{1}{2}MR^2\dot\theta^2+Mgx\sin\varphi$$

  Substituting constraints of $\dot y=0$ and $dx=R\theta$, the equation becomes:

  $$L=M\dot x^2+Mgx\sin\varphi$$

  Using Euler’s equation, the equation of motion is:

  $$0=\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot x}=Mg\sin\varphi-2M\ddot x$$

  giving us:

  $$\ddot x=\frac{1}{2}g\sin\varphi$$

  STEP 4: Finding Coefficients

  We are interested in knowing the frictional force, so we will apply the general form of solution we derived previously in the section. We have shown that the solution has a general form of:

  $$\begin{equation} a_{lk}\delta q_k+a_{lt}\delta t=0 \end{equation}$$

  Rearranging the constraint equation for $dx$, we get $0=Rd\theta-dx$, matching the form of the general solution. Representing the constraint in the form $a_{l\theta}\delta\theta+a_{1x}\delta x+a_{lt}\delta t=0$ and comparing the two equations, we get:

  $$a_{1\theta}=R,\quad a_{1x}=-1, \quad a_{1t}=0$$

  STEP 5-6:

  We know that:

  $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}=\lambda a_{lk}=F_k$$

  where $q_k$ is the generalized coordinate. In this problem, $q_k$ are identified as $x$ and $\varphi$.

  $$M\ddot x-Mg\sin\varphi=-\lambda_1=F_x$$

  $$MR^2\ddot\theta=\lambda_1R=F_\theta$$

  STEP 7:

  We know have a system of differential equations

  $$\begin{cases} M\ddot x-Mg\sin\varphi=-\lambda_1=F_x \\ MR^2\ddot\theta=\lambda_1R=F_\theta\\ d\dot x=R\dot\theta \\ \ddot x=\frac{1}{2}g\sin\varphi \end{cases}$$

  giving us:

  $$\frac{1}{2}Mg\sin\varphi-Mg\sin\varphi=-\lambda_1$$

  Then, $\lambda_1$ is equal to:

  $$\begin{cases} \lambda_1=\frac{1}{2}Mg\sin\varphi \\ \ddot x=\frac{1}{2}g\sin\varphi \end{cases}$$

  The force $F_\theta$ is equal to:

  $$F_\theta=\lambda_1R=\frac{1}{2}MgR\sin\varphi$$

Example 6.2: Stacking Barrels

Consider the problem of two cylindrical barrels, one on top of the other, as shown in Figure 6.3. The bottom barrel is fixed in position and orientation, but the top one, of mass m, is free to move. It starts rolling down from its initial position at the top, rolling without slipping due to friction between the barrels. The problem is to find the point along the lower barrel where the top barrel loses contact with it. That is, we need to find the moment when the normal force acting on the top barrel vanishes.

• Solution

  Steps

  1. Identifying constraints

  2. specifies position/velocity using Cartesian/polar/spherical/cylindrical coordinates

  3. Find the Lagrangian $L=T-U$

  4. Use the general form of solution to find coefficients

  5. Find the equations of motion for each generalized coordinate.

  6. After finding general equations of motion for each generalized coordinate, substitute identified constraints and coefficients we found form step 4 to find the Lagrange multiplier.

  7. find the solution

  STEP 1: Identifying Constraints

  Before solving the problem, we need to identifies constraints first. We have three variables in the problem, $r, \theta$ and a rotational angle $\varphi$. Using polar coordinate, the first constraint we identify is $r=R+a$. Another constraint is

  $$ad\varphi=Rd\theta$$

  Because radius $R$ and $a$ are fixed, we know that $r=R+a$ is fixed. Thus, there are three constraints in the total for this problem.

  STEP 2: Position and Velocity

  Using spherical coordinates, the center’s potion of the top barrel is specified using $\vec{s}=(r\hat{r}, r\theta\hat\theta)$ (the is also the barrel’s position of the center mass, assuming a uniform mass distribution). Differentiating $\vec{s}$, velocity of barrel is $v=(\dot r\hat{r}, r\dot\theta\hat\theta)$, where $r=R+a$.

  STEP 3: Find the Lagrangian

  Because the lower barrel is stationary, only the upper barrel contributes to the system’s kinetic energy. Since the object rolls without slipping, we need to consider both translational and rotational kinetic energies, where $K_{rot}=1/2\omega^2=1/2ma^2\dot\varphi^2$. Then, the system’s kinetic energy of the system is equal to:

  $$T=K_{trans}+K_{rot}=\frac{1}{2}m(\dot r^2+r^2\dot\theta^2)+\frac{1}{2}ma^2\varphi^2$$

  Potential energy of the system is:

  $$U=mgr\cos\theta$$

  The Lagrangian is equal to:

  $$L=\frac{1}{2}m(\dot r^2+r^2\dot\theta^2)+\frac{1}{2}ma^2\varphi^2-mgr\cos\theta$$

  Substituting the constraint $ad\varphi=Rd\theta$, we get:

  $$L=\frac{1}{2}m(\dot r^2+r^2\dot\theta^2)+\frac{1}{2}mR^2\theta^2-mgr\cos\theta$$

  Note that we do NOT substitute the constraint $0=\dot r$ here when writing the Lagrangian. This is because $\dot r$ is part of the generalized coordinates, which will be differentiated later to find equations of motion. If we substitute $\dot r=0$, we will get a wrong answer. This might sound confusing and abstract. An analogy to understand this is that we want to find the slope of $f(x)=ax^2+bx+c(x)$ at $x=2$, where $c(x)=x-2$ and $c(x=2)=0$. In order to find $f’(2)$, we will substitute $c(x)=x-2$ into $f(x)$ instead of $c(2)=0$, giving us $f(x)=ax^2+bx+x-2$. Differentiating $f(x)$, we get $f’(2)=2ax+b+1=4a+2x+1$, whereas $f’(2)=2ax+b=4a+b=4a+b$ if $f(x)=ax^2+bx$.

  STEP 4: Find coefficients

  From the Lagrangian, we can see that there are two generalized coordinates $r$ and $\theta$. Thus, the general form of solution is equal to $a_{lr}\delta{r}+a_{l\theta}\delta \theta+a_{lt}\delta t=0$. Comparing the equation with the constraint $0=r-R-a$, we get:

  $$a_{1r}=1\quad, a_{1t}=a_{1\theta}=0$$

  Note that we use the constraint $r=R+a$ instead of $ad\varphi=Rd\theta$ because both $r$ and $\theta$, the generalized coordinates, are present, whereas the latter one includes unknown variable.

  STEP 5: Find equations of motion

  Using Euler’s equation, equations of motion for $r$ and $\theta$ are:

  $$0=\frac{\partial L}{\partial r}-\frac{d}{dt}\frac{\partial L}{\partial \dot r}=a_{1r}\lambda_1=mr\dot\theta^2-mg\cos\theta-\frac{d}{dt}(m\dot r)mr\dot\theta^2=mr^2\dot\theta^2-mg\cos\theta-m\ddot r$$

  $$0=\frac{\partial L}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial \dot \theta}=a_{1\theta}\lambda_1=mgr\sin\theta+mR^2\theta-\frac{d}{dt}(mr^2\dot\theta)=mgr\sin\theta+mR^2\theta-mr^2\ddot\theta$$

  STEP 6: Finding the Lagrange multiplier $\lambda_l$

  Because $a_{1r}=1$ and $a_{1\theta}=0$, we get:

  $$\lambda_1=F_r=mr^2\dot\theta^2-mg\cos\theta-m\ddot r$$

  Substitute the constraint $\dot r=0$, $\lambda_1$ is equal to:

  $$F_r=mr^2\dot\theta^2-mg\cos\theta =N$$

  where $N$ is normal force. Energy is:

  $$E=T+U=\frac{1}{2}m(R+a)^2\dot\theta^2+\frac{1}{2}mR^2\dot\theta^2-mg(R+a)\cos\theta$$

  Using the initial condition $\theta(0)=0$, we get $\dot\theta(0)=0$. Then, energy $E=-mg(R+a)$. This implies that:

  $$mg(R+a)=\frac{1}{2}m(R+a)^2\dot\theta^2+\frac{1}{2}mR^2\dot\theta^2-mg(R+a)\cos\theta$$

  $$g(R+a)=\frac{1}{2}(R+a)^2\dot\theta^2+\frac{1}{2}R^2\dot\theta^2-g(R+a)\cos\theta$$

  $$\dot\theta^2=\frac{2g(R+a)}{2R^2+a^2+2Ra}(1-\cos\theta)$$

  Normal force $N=\lambda_1=mr^2\dot\theta^2-mg\cos\theta$. T At the moment when barrels lose contact, $N=0$, giving us:

  $$0=mr^2\dot\theta^2-mg\cos\theta$$

  Solving for $\dot\theta$, we get:

  $$\dot\theta_c^2=\frac{g\cos\theta_c}{m(R+a)}$$

  where $\theta_c$ is the critical angle at which the condition $N=0$ is satisfied and barrels loses contact.

Example 6.3:

A classic problem is that of a bob pendulum consisting of a point mass $m$ at the end of a rope of length l swinging in a plane (see Figure 6.4). We would like to determine the tension in the rope as a function of the angle θ. We start with two variables

• Solution

  Step 1: Identifying Constraint(s)

  The first constraint we identified is that the string has a fixed length, reducing the degree of freedom of $2-1=1$. Mathematically, the constraint can be expressed as:

  $$r=l\rightarrow d r=\delta r0$$

  For a pendulum, we usually use polar coordinates $r$ $r$$\theta$, where position $\vec{s}$ is given by $\vec{s}=(r, r\theta)$ and velocity $\vec{r}=(\dot{r}, r\dot\theta)$.

  Step 2: The Lagrangian

  The Lagrangian $L$ is equal to:

  $$L=\frac{1}{2}m(\dot{r}^2+r^2\dot\theta^2)-mgr\cos\theta$$

  Equations of motion are:

  $$mr\dot\theta^2-mg\cos\theta+m\ddot r=\lambda$$

  $$-mgr\sin\theta+m\ddot\theta=0$$

  Substituting the constraints of $r=l$ and $\dot r=\ddot r =0$, we get:

  $$ml\dot\theta^2-mg\cos\theta=\lambda$$

6.3: Cyclic Coordinates and Generalized Momenta

In chapter 4 The Variational Principle, we defined a cyclic coordinate $q_k$ as a generalized coordinate $q_k$ is missing but the corresponding velocity $\dot q_k$ is presents in the Lagrangian. The generalized momentum is defined as:

and is conserved.

The question is: why is a particular coordinate cyclic, from a physical point of view?

In order to answer the question, consider the example of a moving object with mass $m$ under a uniform gravitational field, where gravity acts in $z$ direction. The Lagrangian $L=T-U$ is equal to:

where $x$ and $y$ are cyclic coordinates. The corresponding generalized momenta $p_x$ and $p_y$ are conserved:

The answer for the conservation is quit clear. The physical environment is invariant under displacements in $x$ and $y$ directions, but not under $z$, the vertical direction. A change in $z$ means that $m$ gets closer or farther from the ground, but a change in $x$ and $y$ make no difference. We say that there is a symmetry under horizontal displacements, but not under vertical displacement.

If a generalized coordinate $q_i$ is missing from the Lagrangian, it means that there is a symmetry under changes in that coordinate: the physical environment, whether a potential energy or a constraint, is independent of that coordinate. If the environment possesses a symmetry, the corresponding generalized momentum will be conserved.

Note: The concept of “symmetry” here might sound abstract and confusing. It is better illustrated in example 6.5.

Example 6.5: A star orbiting s spheroidal galaxy

A particular galaxy consists of an enormous sphere of stars somewhat squashed along one axis, so it becomes spheroidal in shape, as shown in the figure. A star at the outer fringes of the galaxy experiences the general gravitational pull of the galaxy. Are there any conserved quantities in the motion of this star?

• Solution

  There is no given quantities, such as mass and radius, so we cannot solve the problem mathematically using Lagrange mechanics (The statement can be proved mathematically, but the derivation could be cumbersome. We will use symmetry to solve it in a smarter way. Check the last part for the mathematical derivation).

  The problem can be solved easily by identifying symmetries. Recall that if an environment possess symmetry, the corresponding generalized momentum $p_i$ is conserved. Imaging moving translationally, we observe that any translational movement will make one nearer or farther from the galaxy, so no translational symmetries are presents. Thus, no linear momentum is conserved.

  Consider rotating about the axis. The shape of the galaxy remains unchanged. Thus, the angular momentum $p_{\varphi}$ is conserved, where $\varphi$ is the azmuthal angle.

Example 6.6: A Charged Particle Outside a Charged Rod

An infinite straight dielectric rod is oriented in the z direction, and given a uniform electric charge per unit length. A point charge is free to move outside it, as shown in Figure 6.7. Are there any conserved quantities for the motion of the particle?

• Solution

  When finding symmetries, we consider translational and rotational movements. When moving alone $x$ and $y$ directions, $p_x$ and $p_y$ are not conserved. When rotating around the $z$ axis, a symmetry exists. Thus, $p_z$ and angular momentum $L$ are conserved.

6.4: A Less Straightforward Example

Consider a system consisted two masses $m_1$ and $m_2$. Let the distances be $q_1$ and $q_2$ as shown in the figure.

The Lagrangian $L=T-U=\frac{1}{2}m_1\dot q_1^2+\frac{1}{2}m_2\dot q_2^2-U(q_1-q_2)$. The action $S$ is equal to:q

Consider the simple transformation

where $C$ is some arbitrary constant. This gives us:

so terms for kinetic energy remain unchanged after the transformation. In addition, we know that:

The action is equal to:

In the next section, we will dip deeper into infinitesimal transformations.

6.5: Infinitesimal Transformations

6.51: Direct Transformations

A direct transformation deforms the generalized coordinates of a system as in:

where $\Delta$ is used to label a direct transformation. Note that $\Delta q_k(t, q)$ can possibly be a function of time, where in section 6.4 we discussed the special case when $\Delta q_k(t, q)$ is a constant.

6.52: Indirect Transformations

An indirect transformation affects the generalized coordinates indirectly, through the transformation of the time coordinate:

Note that again that the shift in time is assumed to be small, and the shift can be a function f of time. The small shifts in time bring small shifts in generalized coordinates and affect them indirectly, giving us:

where we use Taylor expansion in $\delta t$ to linear order only. Rearranging the equation, we get

6.53: Combined Transformations

We want consider a transformation that includes both direct and indirect transformations. We write:

Thus, for an indirect transformation

where we need to provide a set of functions $\delta t(t, q)$ and $\delta q_k(t, q)$ to specify a particular transformation and determine $\Delta q_k(t, q)$.

The question seems to be complicated. In order to understand the formula, we consider physical meanings of the expression. The term $\delta q_k(t,q)$ represents the infinitesimal variation in the generalized coordinates $q_k$. The second term $\Delta q_l(t,q)$ accounts for how $q_k$ changes indirectly because time $t$ is changed. The last term represents the change in the generalized coordinate $q_k$ due to its time derivative $\dot q_k$, capturing the direct effect of the time shift on $q_k$.

Example 6.7: Translations

Consider a single particle in three dimensions, described by the three Cartesian coordinates $r^x = x, r^y = y$, and $r^z = z$. We also have the time coordinate $r^t = c t$. An infinitesimal constant spatial translation can be realized by

• Solution

  To specify a particular transformation, we need to provide the sets of function $\delta q_k(t, q)$ and $\delta t(t, q)$ to determine $\Delta q(t, q)$. Because we want an infinitesimal constant spatial translation, we know that the time coordinate remains unchanged, giving us:

  $$\delta r^i(t,x)=\delta\epsilon^i, \quad \delta t(t,q)=0$$

  By equation (7) for an indirect transformation, we know that $\Delta r(t, x)$\ is equal to:

  $$\Delta r^i(t, x)=\delta \epsilon^i$$

  Note that the notation here can be misleading. $\delta \epsilon^i$ doe NOT represent exponentiation. A constant spatial translation in three dimensional by shifting in $x, y$ and $z$ directions. The shift is represented using $\Delta r=(\Delta x, \Delta y, \Delta z)=(r^{x_f}-r^{x_i}, r^{y_f}-r^{y_i}, r^{z_f}-r^{z_i})=\delta\epsilon^i$, where $i=x, y,z$ and $\epsilon^i$ are three small constants for three shifts in $x, y$ and $z$ directions. A translation in space is then defined by:

  $$\{\delta t(t, x)=0, \delta r(t, x)=\delta\epsilon^i\}$$

  Similarly, for a constant infinitesimal shift in time, the spatial coordinate remains unchanged, giving us:

  $$\delta r(t,x)=0,\quad \delta(t, x)=\delta\epsilon$$

  Then,

  $$\Delta r^i(t, x)=-\dot r^i\delta t(t, q)=-\dot r^i\delta\epsilon$$

  The translation is defined by:

  $$\{\delta r^i(t, x)=0, \delta t(t, x)=\delta\epsilon\}$$

  Note: It might be confusing at first to understand how we lead to the conclusions that $\delta r(t,x)=\delta\epsilon^i$ and $\delta t(t, x)=\delta\epsilon$.

  A constant translation in time means that every time coordinate $t$ is uniformly shifted by a constant small amount $\delta\epsilon$. Mathematically speaking, since $t’=t+\delta\epsilon$, where $\epsilon$ is a constant and the symbol $\delta$ represents a change in time. This gives :

  $$t'-t=\delta t=\delta\epsilon$$

  Similarly, for a shift in space coordinate, we know that at time $t$, coordinates are:

  $$r(t)=(r^x, r^x, r^z)=(x,y,z)$$

  At time $t’$, each coordinates is shifted by amounts of $\delta\epsilon^x, \delta\epsilon^y,$ and $\delta\epsilon^z$. Then:

  $$r(t')=(x+\delta\epsilon^x, y+\delta\epsilon^y, z+\delta\epsilon^z)$$

  Thus, $\delta r^i(t,x)=\delta\epsilon^i$.

Example 6.8: Rotations

To describe constant rotations, we consider for simplicity a particle moving in two dimensions. We use the coordinates $r^x = x$ and $r^y = y$.

• Solution

  Again, for a constant spatial transformation, the time coordinate $t$ remains unchaged, giving us:

  $$\delta t(t,x)=0, \quad \delta r^i(t,x)=\delta\epsilon^i$$

  where $i=x, y$ and $\delta\epsilon^i$ represents two constants. A rotation in matrix form is given by:

  $$\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta, & \cos\theta \end{bmatrix}$$

  The rotation in coordinates is given by:

  $$\begin{bmatrix} r^{x'} \\ r^{y'} \end{bmatrix}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta, & \cos\theta \end{bmatrix}\begin{bmatrix}r^{x} \\ r^y \end{bmatrix}$$

  Using small angle approximation, we know that $\cos\theta=1$ and $\sin\theta=\theta$. The rotation is then equal to:

  $$\begin{bmatrix} r^{x'} \\ r^{y'} \end{bmatrix}=\begin{bmatrix} 1 & \theta \\ -\theta, & 1\end{bmatrix}\begin{bmatrix}r^{x} \\ r^y \end{bmatrix}$$

Example 6.9 Lorentz Transformation

6.6: Symmetry

We define a symmetry as a transformation that leaves the action unchanged in form, where action $S$ is defined as:

where $L$ is the Lagrangian. Using the general/combined transformation given by $\Delta q_k(t, q)$ and $\delta t(t, q$), it follows that:

where we use the Leibniz product’s rule $\delta (ab)=(\delta a)b+a(\delta b)$, since $\delta$ is an infinitesimal change. The first term the $\delta S$ is the change in the measure of the integrand:

The second term has two parts:

After some derivations (the note here is not complete. The complete derivation is in the textbook. Need to com back after studying mathematics from chapter two.), $\delta S$ is equal to: