Chapter 2 Review of Classical Mechanics

2.1 The Principle of Least Action and Lagrangian Mechanics

Exercise 2.1.1 Consider the following system, called a harmonic oscillator. The block has a mass $m$ and lies on a frictionless surface. The spring has a force constant $k$. Write the Lagrangian and get the equation of motion.

Solution. The kinetic energy and potential energy are

Then the Lagrangian is

We can compute

Therefore, the Euler-Lagrange equation is

The equation of motion is

Exercise 2.1.2 Do the same for the coupled-mass problem discussed at the end of Section 1.8. Compare the equations of motion with Eqs. (1.8.24) and (1.8.25).

Solution. The kinetic energy and potential energy of the system are

Then the Lagrangian is

• The Euler-Lagrange equation of $1$:

We get equation of motion

• The Euler-Lagrange equation of $2$:

We get equation of motion

(2.1) and (2.2) are the same as Eqs. (1.8.24) and (1.8.25).

Exercise 2.1.3 A particle of mass $m$ moves in three dimensions under a potential $V(r,\theta,\phi)=V(r)$. Write its $\mathscr{L}$ and find the equations of motions.

Solution. The kinetic energy and potential energy are

Then the Lagrangian is

• Euler-Lagrange equation of $r$:

The equation of motion is

• Euler-Lagrange equation of $\theta$:

The equation of motion is

• Euler-Lagrange equation of $\phi$:

The equation of motion is

where $l$ is a contant.

2.2 The Electromagnetic Lagrangian

2.3 The Two Body Problem

Exercise 2.3.1 Derive Eq. (2.3.6) from (2.3.5) by changing variables.

Solution. Since

The Lagrangian becomes

2.4 How Smart Is a Particle?

2.5 The Hamiltonian Formalism

Exercise 2.5.1 Show that if $T=\sum\limits_i \sum\limits_j T_{i j}(q) \dot{q}_i \dot{q}_j$, where $\dot{q}$'s are generalized velocities, $\sum\limits_i p_i \dot{q}_i=2 T$.

Solution.

Therefore,

Exercise 2.5.2 Using the conservation of energy, show that the trajectories in phase space for the oscillator are ellipses of the form $(x / a)^2+(p / b)^2=1$, where $a^2=2 E / k$ and $b^2=$ $2 m E$.

Solution. The Lagrangian is

So the momentum is

Hamiltonian is

Since $\mathscr{L}$ is not an explicit function of $t$, $\mathscr{H}$ is conservative. Set $\mathscr{H}=E$, where $E$ is a constant, we have

If we denote $a^{2}=2E/k$ and $b^{2}=2mE$, we have

Exercise 2.5.3 Solve Exercise 2.1.2 using the Hamiltonian formalism.

Solution. Start from Lagrangian of the system

Then the momenta are

Then the Hamiltonian of the system is

• Hamilton's canonical equations of $1$:

From the first equation, we know $p_{1}=m\dot{x}_{1}$. Take the time derivative on the both side, we get $\dot{p}_{1}=m\ddot{x}_{1}$. Substitute it into the second equation, we get

• Hamilton's canonical equations of $2$:

From the first equation, we know $p_{2}=m\dot{x}_{2}$. Take the time derivative on the both side, we get $\dot{p}_{2}=m\ddot{x}_{2}$. Substitute it into the second equation, we get

Exercise 2.5.4 Show that $\mathscr{H}$ corresponding to $\mathscr{L}$ in Eq. (2.3.6) is $\mathscr{H}=\left|\mathbf{p}_{\mathrm{CM}}\right|^2 / 2 M+|\mathbf{p}|^2 /$ $2 \mu+V(\mathbf{r})$, where $M$ is the total mass, $\mu$ is the reduced mass, $\mathbf{p}_{\mathrm{CM}}$ and $\mathbf{p}$ are the momenta conjugate to $\mathbf{r}_{\mathrm{CM}}$ and $\mathbf{r}$, respectively.

Solution. Start from Lagrangian

where total mass $M=m_{1}+m_{2}$ and reduced mass $\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}$. Then the momenta satisfy

Therefore the Hamiltonian is

2.6 The Electromagnetic Force in the Hamiltonian Scheme

2.7 Cyclic Coordinates, Poisson Brackets, and Canonical Transformations

Exercise 2.7.1 Show that

Note the similarity between the above and Eqs. (1.5.10) and (1.5.11) for commutators.

Solution.

Exercise 2.7.2 (i) Verify Eqs. (2.7.4) and (2.7.5). (ii) Consider a problem in two dimensions given by $\mathscr{H}=p_x^2+p_y^2+a x^2+b y^2$. Argue that if $a=b,\left\{l_z, \mathscr{H}\right\}$ must vanish. Verify by explicit computation.

Solution.

(i)

and

(ii) The Hamiltonian given is $\mathscr{H}=p_{x}^{2}+p_{y}^{2}+ax^{2}+by^{2}$. If $a=b$, $\mathscr{H}$ has a symmetry under simultaneous rotations in the $x-y$ and $p_{x}-p_{y}$ planes, under which $l_{z}$ (the generator) is conserved. Therefore, $\{l_{z},\mathscr{H}\}=0$. We check this as follows:

But

So

which vanishes if $a=b$.

Exercise 2.7.3 Fill in the missing steps leading to Eq. (2.7.18) starting from Eq. (2.7.14).

Solution. Consider the following transformations:

If this transformation is canonical, then the variables $\bar{q}_{i}$ and $\bar{p}_{i}$ satisfy Hamilton's equation:

If we write Hamiltonian $\mathscr{H}$ as a function of new variables, we can get partial derivatives

The time derivative of any function $\omega$ can be written as a Poisson bracket with Hamiltonian $\mathscr{H}$:

Therefore, for transformed velocities, we have

In order to satisfy Hamilton's equation, we must have

We could do the same calculation for the time derivative of transform momentum

In order to satisfy Hamilton's equation, we must have

Thus, we can conclude that in order for the transformation to be canonical, the conditions are

Exercise 2.7.4 Verify that the change to a rotated frame

is a canonical transformation.

Solution. To show this is a canonical transformation, we must evaluate the Poisson brackets. Before computing Poisson brackets, we can first compute non-vanishing derivatives

where $q_{1}=x$, $q_{2}=y$ and $p_{1}=p_{x}$, $p_{2}=p_{y}$.

since neither coordinate depends on any momentum. Similarly,

since Poisson bracket contains derivatives of $\bar{p}_{i}$ with respect to $q_{i}$ and these are all zero.

The remaining Poisson brackets are of the form $\{\bar{q}_{i},\bar{p}_{j}\}$.

Similarly,

Therefore, the change of rotated frame is a canonical transformation.

Exercise 2.7.5 Show that the polar variables

are canonical. ( $\hat{e}_\rho$ is the unit vector in the radial direction.)

Solution. The non-vanishing derivatives are

Now, let's evaluate Poisson brackets

since coordinates don't depend on the momenta.

The remaining Poisson brackets are of the form $\{\bar{q}_{i},\bar{p}_{j}\}$.

Thus all the Poisson brackets are correct, so the transformation is canonical.

Exercise 2.7.6 Verify that the change from the variables $\mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1, \mathbf{p}_2$ to $\mathbf{r}_{\mathrm{CM}}, \mathbf{p}_{\mathrm{CM}}, \mathbf{r}$, and $\mathbf{p}$ is a canonical transformation. (See Exercise 2.5.4).

Solution. The transformation from the coordinates $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ of the masses $m_{1}$ and $m_{2}$ to relative position $\mathbf{r}$ and the position of the center of mass $\mathbf{r}_{CM}$ are

where $M:=m_{1}+m_{2}$ is the total mass. The conjugate momenta is the original system are

The conjugate momenta transform according to

where $\mu:=\frac{m_{1}m_{2}}{M}$ is the reduced mass.

Now we calculate the Poisson brackets to check whether it is a canonical transformation.

Note that the new coordinates depend only on the old coordinates, and conversely, the new momenta depend only on the old momenta. Also notice that $r_{i}$ depends only on the $i$ components of $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$, and $p_{j}$ depends only on the $j$ components of $\mathbf{p}_{1}$ and $\mathbf{p}_{2}$.

Since the Poisson brackets $\{\bar{q}_{i},\bar{q}_{j}\}$ and $\{\bar{p}_{i},\bar{p}_{j}\}$ all invoke taking derivatives of coordinates with respect to momenta or momenta with respect to coordinates, we have

where $i$ and $j$ takes on the values $x$, $y$ and $z$. Then what we left to check are $\{\bar{q}_{i},\bar{p}_{j}\}$. There are three cases $\{r_{i},p_{j}\}$, $\{r_{CMi},p_{CMj}\}$, $\{r_{CMi},p_{j}\}$ or $\{r_{i},p_{CMj}\}$.

(1) $\{r_{i},p_{j}\}$

• $i=j$

where $q_{\alpha}$ and $p_{\alpha}$ sum over all $6$ components of the original position vectors $\{r_{1x},r_{1y},r_{1z},r_{2x},r_{2y},r_{2z}\}$ that we denote as $\{r_{1i},r_{2i}\}$ and momentum vectors $\{p_{1x},p_{1y},p_{1z},p_{2x},p_{2y},p_{2z}\}$ that we denote as $\{p_{1i},p_{2i}\}$, respectively.

• $i\neq j$

(2) $\{r_{CMi},p_{CMj}\}$

• $i=j$

• $i\neq j$

(3) $\{r_{CMi},p_{j}\}$ or $\{r_{i},p_{CMj}\}$

• $i=j$

• $i\neq j$

Thus all the Poisson brackets are correct, so the transformation is canonical.

Exercise 2.7.7 Verify that

is a canonical transformation.

Solution. The partial derivatives are

The only remaining term to verify is

Thus the transformation is canonical.

Exercise 2.7.8 We would like to derive here Eq. (2.7.9), which gives the transformation of the momenta under a coordinate transformation in configuration space:

(1) Argue that if we invert the above equation to get $q=q(\bar{q})$, we can derive the following counterpart of Eq. (2.7.7):

(2) Show from the above that

(3) Now calculate

Use the chain rule and the fact that $q=q(\bar{q})$ and not $q(\bar{q}, \dot{\bar{q}})$ to derive Eq. (2.7.9).

(4) Verify, by calculating the Poisson braket in Eq. (2.7.18), that the point transformation is canonical.

Solution. (1) Since $q_{i}=q_{i}(\bar{q}_{1},\ldots,\bar{q}_{n})$,

(2) Since the velocities $\dot{\bar{q}}_{j}$ are independent variables, if we hold the coordinates $\bar{q}$ constant, we will have

(3) We can use the Lagrangian to see how the momenta $p_{i}$ transform under the coordinate change. The definition of the canonical momentum is

If we write the Lagrangian in terms of the new coordinates and velocities $\mathscr{L}=\mathscr{L}(\bar{q},\dot{\bar{q}})$, then the momenta in the new coordinate system are

At this point, it's worth noting that although $\mathscr{L}(\bar{q},\dot{\bar{q}})$ and $\mathscr{L}(q,\dot{q})$ are different functions, they have the same value at each point in the configuration space. That is, if we choose some point that has the coordinates $(q,\dot{q})$ in the $q$ system and coordinates $(\bar{q},\dot{\bar{q}})$ in the $\bar{q}$ system, then, numerically at that one point, we must have $\mathscr{L}(\bar{q},\dot{\bar{q}})=\mathscr{L}(q,\dot{q})$. Because of this, we can write

That is, if we are keeping $\bar{q}$ constant, the derivative of $\mathscr{L}$ with respect to $\dot{\bar{q}}_{i}$ must be the same (numerically) no matter what coordinates we are using to write $\mathscr{L}$. Therefore, we can use the latter form and then use the chain rule to write out the derivative:

Because the coordinates $q$ don't depend on the velocities $\dot{\bar{q}}$, the first term on the RHS is zero. We can use (2.3) in the second term, and we have

where we used the definition of canonical momentum at the last equality. We have derived Eq. (2.7.9).

(4) Point transformation is given by

In this case, the coordinate transformation to $\bar{q}$ is completely arbitrary, but the momentum transformation must follow the formula given. The derivatives $\frac{\partial q_i}{\partial \bar{q}_j}$ in the formula for $\bar{p}_i$ are taken at constant $\bar{q}$. Since the coordinate formulas depend only on the old coordinates, and the momentum formulas depend only on the old momenta, the Poisson brackets satisfy

For the mixed brackets, we have

Thus the point transformation is a canonical transformation.

Exercise 2.7.9 Verify Eq. (2.7.19) by direct computation. Use the chain rule to go from $q, p$ derivatives to $\bar{q}, \bar{p}$ derivatives. Collect terms that represent Poisson braket of the latter. Solution. The Poisson bracket of two functions is defined as

Calculating the Poisson bracket requires knowing $\omega$ and $\sigma$ as functions of the coordinates $q_i$ and momenta $p_i$ in the particular coordinate system we're using.

The simplest way of finding out is to write the canonical transformation as

We can then write the Poisson bracket in the new coordinates as

Assuming the transformation is invertible, we can use the chain rule to calculate the derivatives with respect to the barred coordinates. This gives the following (Here we use Einstein summation convention):

For a canonical transformation, the Poisson brackets in the last equation satisfy

Applying these conditions to the above, we find

Thus the Poisson bracket is invariant under a canonical transformation.

2.8 Symmetries and Their Consequences

Exercise 2.8.1 Show that $p=p_1+p_2$, the total momentum, is the generator of infinitesimal translations for a two-particle system.

Solution. Since $g=p_{1}+p_{2}$, it generates the infinitesimal transformations

So to order $\varepsilon$, these give the canonical transformations $x_{i}\to \bar{x}_{i}(x_{j},p_{j})$ and $p_{i}\to \bar{p}_{i}(x_{j},p_{j})$ with

which is precisely a spatial transformation of the whole system by an amount $\varepsilon$.

Exercise 2.8.2 Verify that the infinitesimal transformation generated by any dynamical variable $g$ is a canonical transformation. (Hint: Work, as usual, to first order in $\varepsilon$.)

Solution. If the coordinates and momenta after the infinitesimal transformation generated by dynamical variable $g$ becomes

Then the Poisson brackets between new coordinate and momentum is

Therefore, the infinitesimal transformation generated by any dynamical variable $g$ is a canonical transformation.

Exercise 2.8.3 Consider

whose invariance under the rotation of the coordinates and momenta leads to the conservation of $l_z$. But $\mathscr{H}$ is also invariant under the rotation of \textit{just the coordinates}. Verify that this is a \textit{noncanonical} transformation. Convince yourself that in this case it is not possible to write $\delta \mathscr{H}$ as $\varepsilon\{\mathscr{H}, g\}$ for any $g$, i.e., that no conservation law follows.

Solution. Rotation of just the coordinates:

Then the Poisson brackets are

Thus, rotation of just the coordinates is not a canonical transformation.

If $\delta \mathscr{H}=\varepsilon\{\mathscr{H},g\}=\varepsilon\left(\frac{\partial \mathscr{H}}{\partial x}\frac{\partial g}{\partial p_{x}}-\frac{\partial \mathscr{H}}{\partial p_{x}}\frac{\partial g}{\partial x}+\frac{\partial \mathscr{H}}{\partial y}\frac{\partial g}{\partial p_{y}}-\frac{\partial \mathscr{H}}{\partial p_{y}}\frac{\partial g}{\partial y}\right)$, we have

which means that

According to last exercise, this is a canonical transformation. Therefore, there doesn't exists any $g$, such that $\delta \mathscr{H}=\varepsilon\{\mathscr{H},g\}$.

Exercise 2.8.4 Consider $\mathscr{H}=\frac{1}{2} p^2+\frac{1}{2} x^2$, which is invariant under infinitesimal rotations in phase space (the $x-p$ plane). Find the generator of this transformation (after verifying that it is canonical). (You could have guessed the answer based on Exercise 2.5.2.).

Solution. Consider a one-dimensitonal system with

and perform a infinitesimal rotation in phase space $x-p$ plane:

This is a canonical transformation since

If $g(x,p)$ is the generator

The solution of these two equations is

where $C$ is a constant of integration. The equality is just the Hamiltonian itself.

In fact, the canonical transformation is just the time evolution with $\theta=t$.

Exercise 2.8.5 Why is it that a noncanonical transformation that leaves $\mathscr{H}$ invariant does not map a solution into another? Or, in view of the discussions on consequence II, why is it that an experiment and its transformed version do not give the same result when the transformation that leaves $\mathscr{H}$ invariant is not canonical? It is best to consider an example. Consider the potential given in Exercise 2.8.3. Suppose I release a particle at $(x=a, y=0)$ with $\left(p_x=b, p_y=0\right)$ and you release one in the transformed state in which $(x=0, y=a)$ and $\left(p_x=b, p_y=0\right)$, i.e., you rotate the coordinates but not the momenta. This is a noncanonical transformation that leaves $\mathscr{H}$ invariant. Convince yourself that at later times the states of the two particles are not related by the same transformation. Try to understand what goes wrong in the general case.

Solution. If the Hamiltonian is invariant under a regular canonical transformation and we can find a generator $g$ such that an infinitesimal version of this transformation is given by

then $g$ is conserved.

If we are dealing with a finite regular canonical transformation where we go from $(q, p) \rightarrow(\bar{q}, \bar{p})$, and the Hamiltonian is invariant under this transformation, then it turns out that if a trajectory $(q(t), p(t))$ satisfies Hamilton's equations of motion:

then the trajectory obtained by transforming every point in the original trajectory $(q(t), p(t))$ to the barred system $(\bar{q}(t), \bar{p}(t))$ is also a solution of Hamilton's equations in the sense that

The proof of this is a bit subtle, but goes as follows. To begin, review the derivation of the conditions for a transformation to be canonical. This derivation applied to a passive transformation, in which the two sets of parameters $(q, p) \rightarrow(\bar{q}, \bar{p})$ refer to the same point in phase space. The transformation we're considering here is an active transformation, in which $(q, p) \rightarrow(\bar{q}, \bar{p})$ actually moves the point in phase space. The original derivation (for passive transformations) relied on the fact that the numerical value of the Hamiltonian is the same in both coordinate systems, since both $(q, p)$ and $(\bar{q}, \bar{p})$ refer to the same point in phase space. However, for our active transformation, we're assuming that the Hamiltonian is invariant under the transformation, that is $H(\bar{q}, \bar{p})=H(q, p)$, where $(q, p)$ and $(\bar{q}, \bar{p})$ now refer to different points in phase space. Since the assumption that the Hamiltonian satisfies $H(\bar{q}, \bar{p})=H(q, p)$ was all that we used in the original derivation, the same derivation works both for passive transformations (always) and for active transformations (if the Hamiltonian is invariant under the active transformation). We therefore end up with the equations

Since the transformation is specified to be canonical, the conditions on the Poisson brackets apply here:

The result is that the transformed trajectory also satisfies Hamilton's equations (2.4) and (2.5).

We can now revisit the 2-d harmonic oscillator to show that a noncanonical transformation violates these results. The Hamiltonian is

and we consider the transformation where we rotate the coordinates but not the momenta. The transformation is

As we've seen, this is a noncanonical transformation. To see what happens, we'll consider the initial conditions

The mass is started off at a point on the $x$ axis with a momentum only in the $x$ direction. In this case, the mass behaves like a one-dimensional harmonic oscillator, moving along the $x$ axis only. To be precise, we can work out Hamilton's equations of motion:

The equations for $y$ and $p_y$ are the same, with $x$ replaced by $y$ everywhere. We can solve these ODEs in the usual way, by differentiating the first one and substituting the second one into the first to get

This has the general solution

We can do the same for $x$ and get

Applying the initial conditions, we get

Plugging these into the equations of motion (2.10) and (2.11) and solving for $B$ and $D$ we get the final solution

Now suppose we start off with $x(0)=0$, $y(0)=a$, $p_x(0)=b$ and $p_y(0)=0$. That is, we have rotated the coordinates through $\frac{\pi}{2}$, but not the momenta. We now begin with the mass on the $y$ axis, but moving in the $x$ direction, so as time progresses, it will have components of momentum in both the $x$ and $y$ directions. Although it's fairly obvious that this motion will not be simply the motion in the first case rotated through $\frac{\pi}{2}$, let's go through the equations. By the same technique as above, we can solve the equations to get

If we look at the system at, say, $t=\frac{\pi}{2 \omega}$, then $\cos \omega t=0$ and $\sin \omega t=1$. The mass that started off on the $x$ axis will be at position $(x, y)=\left(\frac{b}{m \omega}, 0\right)$ and so will the mass that started off on the $y$ axis. Since the two masses are in the same place, obviously one is not the rotated version of the other.

Another, probably easier, way to see this is that since the first mass moves only along the $x$ axis, if the rotated version of the trajectory was also to be a solution, the rotated trajectory would have to lie entirely along the $y$ axis, which is certainly not true for the mass that starts off on the $y$ axis, but with a momentum $p_x \neq 0$.

In the general case, if the transformation is noncanonical, then the Poisson brackets in (2.6) and (2.7) don't satisfy the conditions (2.8) and (2.9), with the result that Hamilton's equations aren't satisfied in the $(\bar{q}, \bar{p})$ coordinates. (There may be a deeper, physical interpretation that I've missed, but from a mathematical point of view, that's what goes wrong.)

Exercise 2.8.6 Show that $\partial S_{\mathrm{cl}} / \partial x_f=p\left(t_f\right)$.

Solution. The situation is as shown in the following diagram:

The two trajectories now take the same time, but in the modified trajectory, the particle moves a distance $\Delta x$ further. Since both paths take the same time, there is no extra contribution $\mathscr{L}\Delta t$. In this case $\eta(t)>0$, since the new (blue) curve $x(t)$ is above the old (red) one $x_{\mathrm{cl}}(t)$. The total variation in the action is now

At $t=t_{f}$, $\eta(t_{f})=\Delta x$, we get

Exercise 2.8.7 Consider the harmonic oscillator, for which the general solution is

Express the energy in terms of $A$ and $B$ and note that it does not depend on time. Now choose $A$ and $B$ such that $x(0)=x_1$ and $x(T)=x_2$. Write down the energy in terms of $x_1, x_2$, and $T$. Show that the action for the trajectory connecting $x_1$ and $x_2$ is

Verify that $\partial S_{\mathrm{cl}} / \partial T=-E$.

Solution. For the case of the one-dimensional harmonic oscillator, we have

The general solution for the position is given by

The total energy is given by

where we just multiplied out the second line, cancelled terms and used $\cos ^2 x+\sin ^2 x=1$.

To get the action, we need the Lagrangian:

The action for a trajectory from $t=0$ to $t=T$ is then

To proceed further, we need to specify $A$ and $B$, since these depend on the boundary conditions (that is, on where we require the mass to be at $t=0$ and $t=T$ ). If we require $x(0)=x_1$ and $x(T)=x_2$, then

Plugging these into (2.12) gives the energy as

Plugging $A$ and $B$ into (2.13), we get:

Taking the derivative, we get

Thus the result is verified for the harmonic oscillator.