Solutions

Exercise list

Lecture 8

Exercise 8.1

Exercise 8.2

Lecture 10

Exercise 10.1

Exercise 10.2

Exercise 10.3

Lecture 11

Exercise 11.1

Exercise 11.2

Exercise 11.3

Exercise 11.4

Exercise 11.5

Appendix 1

Exercise A1.1

Lecture 8

Exercise 8.1

Considering the harmonic oscillator:

\mathcal L = \frac{m}{2}\dot x^2 - \frac{k}{2} x^2

Performing the substitution q = (k \, m)^{{}^1{\mskip -5mu/\mskip -3mu}_4}\,x:

\begin{array}{lll} q = (k \, m)^{\frac{1}{4}}\,x & \Rightarrow & x^2 = \frac{1}{\sqrt{m \, k}} \, q^2\\ \dot q = (k \, m)^{\frac{1}{4}}\,\dot x & \Rightarrow & \dot x = \frac{1}{\sqrt{m \, k}} \, \dot q^2 \end{array}

The Lagrangian becomes:

\mathcal L = \frac{m}{2\sqrt{m \, k}}\dot q^2 - \frac{k}{2\sqrt{m \, k}} q^2 = \frac{1}{2}\sqrt{\frac{k}{m}}\dot q^2 - \frac{1}{2}\sqrt{\frac{m}{k}} q^2

Therefore, written as function of \omega = \sqrt{\frac{k}{m}} the Lagrangian is:

\mathcal L = \frac{1}{2\omega}\dot q^2 -\frac{\omega^2}{2} q^2

Exercise 8.2

Written as function of \omega = \sqrt{\frac{k}{m}} the Lagrangian is:

\mathcal L = \frac{1}{2\omega}\dot q^2 -\frac{\omega^2}{2} q^2

\dot q can be derived as function of q and p:

\begin{aligned} & p = \dfrac{\dot q}{\omega} \quad \Rightarrow \quad \dot q = \omega\,p \end{aligned}

The Hamiltonian can be then written:

\mathcal H = \omega p^2 -\frac{1}{2\omega}\omega^2 p^2 + \frac{\omega^2}{2} q^2 = \frac{\omega}{2}\left( p^2 + q^2 \right)

This quantity, the energy E, is conserved, and a point which starts with a given value of energy H(p,q) = c it remains on that energy.

Lecture 10

Exercise 10.1

From the definition of the Poisson bracket:

\{F,G\} \equiv \sum_i \left( \frac{\partial F}{\partial q_i} \frac{\partial G}{\partial p_i} - \frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q_i} \right)

For a function F(q,p):

\{F(q,p),p_i\} = \sum_j \left( \frac{\partial F(q,p)}{\partial q_j} \frac{\partial p_i}{\partial p_j} - \frac{\partial F(q,p)}{\partial p_j}\frac{\partial p_i}{\partial q_j} \right) = \frac{\partial F(q,p)}{\partial q_i}

since in the sum, all terms \frac{\partial p_i}{\partial q_j} vanish (i.e., each right product), and also all terms \frac{\partial p_i}{\partial p_j} with i \neq j in the left product. Thus, only the terms with j=i remains from the sum, where \frac{\partial p_i}{\partial p_i} = 1.

Exercise 10.2

Using just the definition of the Poisson brackets, computing \dot q:

\begin{aligned} \dot q & = \left\{ q, \mathcal H \right\} = \left\{ q, \frac{1}{2m} p^2 + V(q) \right\} = \left\{ q, \frac{1}{2m} p^2 \right\} + \left\{ q, V(q) \right\} \\ & = \frac{1}{2m} \left\{ q, p^2 \right\} + 0 = \frac{1}{2m} \left[ p\left\{ q, p \right\} + p\left\{ q, p \right\} \right] = \frac{p}{m} \end{aligned}

Computing \dot p:

\begin{aligned} \dot p & = \left\{ p, \mathcal H \right\} = \left\{ p, \frac{1}{2m} p^2 + V(q) \right\} = \frac{1}{2m} \left\{ p, p^2 \right\} + \left\{ p, V(q) \right\}\\ & = 0 - \left\{ V(q), p \right\} = -\frac{\mathrm dV}{\mathrm dq} \end{aligned}

Taking the derivative with respect of time of the first equation and using the second it is possible to derive Newton’s equation of motion:

\begin{aligned} & \ddot q = \dfrac{\dot p}{m} = -\frac{1}{m} \frac{\mathrm dV}{\mathrm dq} \\ & m \, \ddot q = -\frac{\mathrm dV}{\mathrm dq} = F_q \\ & F_q = m \, a_q \end{aligned}

Exercise 10.3

Angular momentum is the quantity which is conserved when there are rotation symmetry.

\begin{aligned} & \delta x = - y\,\varepsilon \\ & \delta y = x\,\varepsilon \\ & \delta z = 0 \end{aligned}

The last equation is simply stating that there is no change in z for a rotation in the plane. Looking at the momentum \mathbf p it possible to write a similar set for each component:

\begin{aligned} & \delta p_x = - p_y\,\varepsilon \\ & \delta p_y = p_x\,\varepsilon \\ & \delta p_z = 0 \end{aligned}

If there is a symmetry there is a conserved quantity:

\mathcal Q = \sum_i p_i \,f_i(\mathbf q) = x\,p_y - y\,p_y = L_z

This is the angular momentum on the z direction, perpendicular to the xy plane. The other components in the three dimensional space of \mathbf L can be derived in a similar way, and will have indexes in a cyclic fashion x \rightarrow y \rightarrow z:

\begin{aligned} & L_z = x\,p_y - y\, p_x \\ & L_x = y\,p_z - z\, p_y \\ & L_y = z\,p_x - x\, p_z \end{aligned}

Using the cross product, it can be written in a short form as:

\mathbf L = \mathbf r \times \mathbf p

Using the Poisson brackets and take them between x, y, z and L_z:

\begin{aligned} \left\{ x, L_z \right\} = & \left\{ x, x\,p_y - y\, p_x \right\} = \left\{ x, - y\, p_x \right\} = -y \left\{ x, p_x \right\} = -y \\ \left\{ y, L_z \right\} = & \left\{ y, x\,p_y - y\, p_x \right\} = \left\{ y, x \, p_y \right\} = x \left\{ y, p_y \right\} = x \\ \left\{ z, L_z \right\} = & \left\{ z, x\,p_y - y\, p_x \right\} = 0 \end{aligned}

These quantity are the infinitesimal change \delta divided by the fix quantity \varepsilon.

Calculating the Poisson bracket with respect to momentum:

\begin{aligned} \left\{ p_x, L_z \right\} = & \left\{ p_x, x\,p_y - y\, p_x \right\} = \left\{ p_x, x\, p_y \right\} = p_y \left\{ p_x, y \right\} = -p_y \\ \left\{ p_y, L_z \right\} = & \left\{ p_y, x\,p_y - y\, p_x \right\} = \left\{ p_y, -y\, p_x \right\} = -p_x \left\{ p_y, x \right\} = p_x \\ \left\{ p_z, L_z \right\} = & \left\{ p_z, x\,p_y - y\, p_x \right\} = 0 \end{aligned}

This has been calculated for rotation invariance; it is possible to do a similar analysis for translation invariance, and the quantity which is conserved in that case is the momentum.

\delta q = \varepsilon \quad f = 1 \quad \Rightarrow \quad \mathcal Q = p

Calculating the Poisson bracket between different component of the angular momentum:

\begin{aligned} & \left\{ L_z, L_x \right\} = \left\{ x\,p_y - y\,p_x, y\,p_z - z\,p_y \right\} = z\,p_x - x\,p_z = L_y \\ & \left\{ L_x, L_y \right\} = \left\{ y\,p_z - z\,p_y, z\,p_x - x\,p_z \right\} = x\,p_y - z\,p_x = L_z \\ & \left\{ L_y, L_z \right\} = \left\{ z\,p_x - x\,p_z, x\,p_y - y\,p_x \right\} = y\,p_x - x\,p_y = L_x \end{aligned}

Lecture 11

Exercise 11.1

The cross product of two vectors \mathbf{V} and \mathbf{A} (\mathbf{V} \times \mathbf{A}), is a vector which the i^{th} component can be expressed using the Levi-Civita symbol \epsilon_{ijk} and the components of the vectors \mathbf{V} and \mathbf{A}. The Levi-Civita symbol is defined as 1 for even permutations of (i, j, k), -1 for odd permutations, and 0 if any two indices are equal.

The i^{th} component of the cross product is defined as:

(\mathbf{V} \times \mathbf{A})_i = \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{ijk} V_j A_k

For the first component:

(\mathbf{V} \times \mathbf{A})_1 = \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{1jk} V_j A_k

Expanding the sums, we specifically get terms involving \epsilon_{123}, \epsilon_{132}, …, leading to:

(\mathbf{V} \times \mathbf{A})_1 = \epsilon_{123} V_2 A_3 + \epsilon_{132} V_3 A_2

Given \epsilon_{123} = 1 (cyclic permutation) and \epsilon_{132} = -1 (anti-cyclic permutation), it simplifies to:

(\mathbf{V} \times \mathbf{A})_1 = V_2 A_3 - V_3 A_2

This process can be analogously applied for i=2 and i=3, leading to the full cross product vector.

The expression for the second component (\mathbf{V} \times \mathbf{A})_2 can be derived similarly:

(\mathbf{V} \times \mathbf{A})_2 = \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{2jk} V_j A_k

Expanding the sums and considering the non-zero terms involving \epsilon_{213}, \epsilon_{231}, …:

(\mathbf{V} \times \mathbf{A})_2 = \epsilon_{213} V_1 A_3 + \epsilon_{231} V_3 A_1

Given \epsilon_{213} = -1 (anti-cyclic permutation) and \epsilon_{231} = 1 (cyclic permutation), it simplifies to:

(\mathbf{V} \times \mathbf{A})_2 = -V_1 A_3 + V_3 A_1

Applying the same approach for the third component (\mathbf{V} \times \mathbf{A})_3:

(\mathbf{V} \times \mathbf{A})_3 = \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{3jk} V_j A_k

Considering the relevant terms with non-zero \epsilon values:

(\mathbf{V} \times \mathbf{A})_3 = \epsilon_{312} V_1 A_2 + \epsilon_{321} V_2 A_1

Given \epsilon_{312} = 1 (cyclic permutation) and \epsilon_{321} = -1 (anti-cyclic permutation), this simplifies to:

(\mathbf{V} \times \mathbf{A})_3 = V_1 A_2 - V_2 A_1

Hence, the cross product \mathbf{V} \times \mathbf{A} in component form is:

\mathbf{V} \times \mathbf{A} = \left( V_2 A_3 - V_3 A_2, V_3 A_1 - V_1 A_3, V_1 A_2 - V_2 A_1 \right)

The second part of the exercise is to prove:

V_i A_j - V_j A_i = \sum_k \epsilon_{ijk} (\mathbf{V} \times \mathbf{A})_k

Substituting (\mathbf{V} \times \mathbf{A})_k with its definition in terms of \epsilon_{kmn} and vector components:

V_i A_j - V_j A_i = \sum_k \epsilon_{ijk} \left( \sum_{m=1}^{3} \sum_{n=1}^{3} \epsilon_{kmn} V_m A_n \right)

Simplifying using the property of the Levi-Civita symbol:

V_i A_j - V_j A_i = \sum_{m=1}^{3} \sum_{n=1}^{3} \left( \sum_k \epsilon_{ijk} \epsilon_{kmn} \right) V_m A_n

Applying the Levi-Civita contraction identity:

V_i A_j - V_j A_i = \sum_{m=1}^{3} \sum_{n=1}^{3} (\delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}) V_m A_n

Recognizing that only terms where m=i and n=j (and vice versa) contribute:

V_i A_j - V_j A_i = V_i A_j - V_j A_i

Exercise 11.2

Let V(x, y, z) be a scalar field. The gradient of V, denoted as \nabla V, in Cartesian coordinates is given by:

\nabla V = \frac{\partial V}{\partial x} \mathbf{i} + \frac{\partial V}{\partial y} \mathbf{j} + \frac{\partial V}{\partial z} \mathbf{k}

The curl of a vector field \mathbf{F} = M\mathbf{i} + N\mathbf{j} + P\mathbf{k} is defined as:

\nabla \times \mathbf{F} = \left( \frac{\partial P}{\partial y} - \frac{\partial N}{\partial z} \right) \mathbf{i} + \left( \frac{\partial M}{\partial z} - \frac{\partial P}{\partial x} \right) \mathbf{j} + \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \mathbf{k}

Applying the curl operator to \nabla V:

\nabla \times (\nabla V) = \left( \frac{\partial}{\partial y}\left(\frac{\partial V}{\partial z}\right) - \frac{\partial}{\partial z}\left(\frac{\partial V}{\partial y}\right) \right) \mathbf{i} + \left( \frac{\partial}{\partial z}\left(\frac{\partial V}{\partial x}\right) - \frac{\partial}{\partial x}\left(\frac{\partial V}{\partial z}\right) \right) \mathbf{j} + \left( \frac{\partial}{\partial x}\left(\frac{\partial V}{\partial y}\right) - \frac{\partial}{\partial y}\left(\frac{\partial V}{\partial x}\right) \right) \mathbf{k}

Given that the order of partial differentiation is interchangeable for continuous and twice differentiable functions (Schwarz’s theorem), we have:

\frac{\partial^2 V}{\partial y \partial z} = \frac{\partial^2 V}{\partial z \partial y}, \quad \frac{\partial^2 V}{\partial z \partial x} = \frac{\partial^2 V}{\partial x \partial z}, \quad \frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x}

Therefore, each term in the expression for \nabla \times (\nabla V) cancels out, leading to:

\nabla \times (\nabla V) = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} = \mathbf{0}

This proves that the curl of the gradient of any scalar field V(x, y, z) is always zero.

Exercise 11.3

Considering \mathbf A:

\begin{aligned} & A_x = 0 \\ & A_y = b\,x \\ & A_z = 0 \end{aligned}

Taking the curl, the only derivative is \frac{\partial A_y}{\partial x} = b, and taking the curl is:

B_z = \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = b - (0) = b

Considering \mathbf A{'} instead:

\begin{aligned} & A_x{'} = -b\,y \\ & A_y{'} = 0 \\ & A_z{'} = 0 \end{aligned}

Taking again the curl:

B_z = \frac{\partial A_y{'}}{\partial x} - \frac{\partial A_x{'}}{\partial y} = (0) - -b = b So these gives the same uniform magnetic field oriented in the z direction and therefore they differ by a gradient:

\mathbf A = \mathbf A{'} + \nabla V

Therefore:

\nabla V = \mathbf A - \mathbf A{'} since \frac{\partial V}{\partial x} = - by, \frac{\partial V}{\partial y} = -bx and \frac{\partial V}{\partial z} = 0, the potential function which satisfies these equation and has this gradient is:

V(x,y,z) = -bxy

Exercise 11.4

The Newton-Lorentz equation of motion are:

m\mathbf{a} = \frac{e}{c}(\mathbf{v} \times\mathbf{B})

Considering a Lagrangian for a particle moving in a constant magnetic field:

\mathcal L = \frac{m}{2}\left( \dot x^2 + \dot y^2 + \dot z^2 \right) +\frac{e}{c}\left( \dot x A_x + \dot y A_y + \dot z A_z \right)

From this Lagrangian, it is possible to compute the Hamiltonian:

\mathcal H = \sum_i \left(p_i\dot q_i\right) - \mathcal L

In general, the Hamiltonian is a function or p and q and therefore it is necessary to solve the equation for the canonical momentum to retrieve \dot x and use it in the Hamiltonian:

\dot x_i = \frac{1}{m}\left(p_i - \frac{e}{c}A_i \right)

and using the canonical momentum Lagrangian in case of magnetic field:

\begin{aligned} \mathcal H & = \sum_i \left(p_i\dot x_i\right) - \sum_i \left[\frac{m}{2}\left( \dot x_i \right)^2 + \frac{e}{c} \dot x_i A_i \right] \\ & = \sum_i \left[\left(p_i\dot x_i\right) - \frac{m}{2}\left( \dot x_i \right)^2 - \frac{e}{c} \dot x_i A_i \right] \\ & = \sum_i \left[p_i \left( \frac{p_i}{m} - \frac{e}{m\,c}A_i \right) - \frac{m}{2}\left( \frac{p_i}{m} - \frac{e}{m\,c}A_i \right)^2 - \frac{e}{c} \left(\frac{p_i}{m} - \frac{e}{m\,c}A_i \right) A_i \right] \\ & = \sum_i \left[p_i \left( \frac{p_i}{m} - \frac{e}{m\,c}A_i \right) - \frac{m}{2}\left( \frac{p_i}{m} - \frac{e}{m\,c}A_i \right)^2 - \frac{e}{c} \left(\frac{p_i}{m} - \frac{e}{m\,c}A_i \right) A_i \right] \\ & = \sum_i \left[\frac{p_i^2}{m} - \frac{e\,p_i}{m\,c}A_i - \frac{p_i^2}{2\,m} + \frac{e\,p_i}{c\,m}A_i - \frac{e^2}{2\,m\,c^2}A_i^2 - \frac{e}{m\,c}p_iA_i + \frac{e^2}{m\,c^2}A_i^2 \right] \\ & = \sum_i \left[\frac{p_i^2}{2\,m} - \frac{e}{m\,c}p_iA_i + \frac{e^2}{2\,m\,c^2}A_i^2 \right] \\ & = \sum_i \left[\frac{1}{2\,m} \left(p_i^2 - \frac{2\,e}{c}p_iA_i + \frac{e^2}{c^2}A_i^2 \right) \right] \\ & = \sum_i \left[\frac{1}{2\,m} \left(p_i- \frac{e}{c}A_i \right)^2 \right] \end{aligned}

Incidentally, the Hamiltonian is just the kinetic energy:

\mathcal H = \sum_i \left[\frac{1}{2\,m} \left(p_i- \frac{e}{c}A_i \right)^2 \right] = \sum_i \frac{1}{2}m\ \dot x_i^2 = \frac{1}{2}m\,v^2

From the Hamiltonian it is possible to compute the equations of motion:

\begin{array}{c} \dfrac{\partial H}{\partial p_i} = \dot q_i \\[10pt] \dfrac{\partial H}{\partial q_i} = - \dot p_i \end{array}

Substituting:

\begin{aligned} \dot x_i & = \dfrac{\partial H}{\partial p_i} = \frac{1}{m}\left( p_i - \frac{e}{c}A_i \right) \\ \dot p_i & = - \dfrac{\partial H}{\partial x_i} = - \frac{1}{m} \sum_j \left( p_j - \frac{e}{c}A_j \right) - \frac{e}{c}\frac{ \partial A_j}{\partial x_i} \\ & = \frac{e}{c} \sum_j \left( p_j - \frac{e}{c}A_j \right) \frac{ \partial A_j}{\partial x_i} = \frac{e}{c} \sum_j \frac{ \partial A_j}{\partial x_i} \dot x_j \end{aligned}

The first equation results in this way because \mathbf A(\mathbf x) is not function of \mathbf p and for the i component it just depends from p_i; for the second the definition of \dot x_i.

It is now possible to write the equation of motion:

\begin{aligned} m\,a_x & = m\,\ddot x_i = \dot p_i -\frac{e}{c}\dot A_i = \dot p_i - \sum_j \frac{e}{c}\frac{\partial A_i}{\partial x_j}\dot x_j = \frac{e}{c} \sum_j \frac{ \partial A_j}{\partial x_i} \dot x_j - \sum_j \frac{e}{c}\frac{\partial A_i}{\partial x_j}\dot x_j \\ & = \frac{e}{c} \sum_{j\neq i} \left( \frac{ \partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j} \right) \dot x_j = \frac{e}{c} \left( \dot {\mathbf x} \times \mathbf B \right) \end{aligned}

This results coincides with the Newton-Lorentz equations of motion.

Exercise 11.5

Considering a uniform magnetic field which lies in the z direction with a magnitude b; is possible to chose either A_x = -b\,y or A_y = b\,x, both are leading to the same equations of motions.

Starting from the case:

A_x = 0; \quad A_y = b\,x \quad A_z = 0

The Hamiltonian is:

\mathcal H = \sum_i \left[\frac{1}{2\,m} \left(p_i- \frac{e}{c}A_i \right)^2 \right] =\frac{1}{2\,m} \left[ p_x^2 + \left(p_y - \frac{e}{c}b\,x \right)^2 + p_z^2 \right]

As pointed out previously, the energy is conserved; since y and z are cyclic coordinates, p_y and p_z will be conserved. Since A_z = 0, p_z = m\,v_z and therefore the velocity in the z direction will be constant; however, p_y = c is not equivalent to the conservation of the velocity in the y direction, as the canonical momentum conjugated to y is p_y = m\,v_y + \frac{e}{c}b\,x.

\frac{\mathrm d}{\mathrm dt}\left( m\,v_y + \frac{e}{c}b\,x \right) = m\,a_y + \frac{e}{c}b\,v_x = 0

Therefore:

a_y = - \frac{e}{c} \frac{b}{m} v_x

Since the Hamiltonian has an explicit dependence on x, it is not possible to use a symmetry to compute p_x and therefore it is necessary to use Hamilton equations; it is however possible to change the gauge and use a different Hamiltonian in which x is now a cyclic coordinate. Considering the case:

A_x = -b\,y; \quad A_y = 0 \quad A_z = 0

The Hamiltonian is now:

\mathcal H = \sum_i \left[\frac{1}{2\,m} \left(p_i- \frac{e}{c}A_i \right)^2 \right] =\frac{1}{2\,m} \left[ \left(p_x + \frac{e}{c}b\,y \right)^2 + p_y^2 + p_z^2 \right]

Now the momentum conjugated to x is conserved:

\frac{\mathrm d}{\mathrm dt}\left( m\,v_x - \frac{e}{c}b\,y \right) = m\,a_x - \frac{e}{c}b\,v_y = 0

Therefore:

a_x = \frac{e}{c} \frac{b}{m} v_y

These are the Newton-Lorentz equations in the special case of uniform magnetic field.

Since there is no acceleration in the z direction, it is possible to assume that the velocity is zero and just consider the motion in the xy plane; the equations of motion are the one just derived:

\begin{aligned} & m\,a_x = \frac{e}{c} b\,v_y = \frac{e}{c} (\mathbf v \times \mathbf B)_x \\ & m\,a_y = -\frac{e}{c}b\, v_y = \frac{e}{c} (\mathbf v \times \mathbf B)_y \\ &m\,\mathbf a = \frac{e}{c} (\mathbf v \times \mathbf B) \end{aligned}

The acceleration is perpendicular to the velocity, but the magnitude of the velocity is constant, and also the magnitude of the field \mathbf B is constant, therefore the magnitude of the cross product is also constant; therefore it moves on a circle:

m\frac{v^2}{r} = \frac{e}{c}\,v\,b

This give the radius of the orbit:

r = \frac{m\,c}{e\,b}|v|

The position of the center is not explicit, so there is an infinite number of circular orbits, each for every point, with the radius that increase for higher velocities.

Appendix 1

Exercise A1.1

From Lecture 2 the components of the velocity and the acceleration are:

\begin{aligned} & x(t) = r\,\cos(\omega\,t) \\ & y(t) = r\,\sin(\omega\,t) \\ & a_x(t) = -r\,\omega^2 \cos(\omega\,t) \\ & a_y(t) = -r\, \omega^2 \sin(\omega\,t) \end{aligned}

Therefore magnitude of the acceleration is:

a = \sqrt{a_x^2 + a_y^2} = \sqrt{r^2 \omega^4 \cos(\omega\,t)^2 + r^2 \omega^4 \sin(\omega\,t)^2} = \omega^2r

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