Problem 1 Classical Mechanics

Transcription

1 Problem 1 Classical Mechanics A pendulum of length l and mass m is attached to a mass M, which is free to move along the axis Ox, as described by the figure. The system is under the influence of a uniform gravitational acceleration g along the direction Oy. We are considering only the motion in the plane of the figure so, at any time, the system's configuration is fully specified by (1) the position X of the mass M along Ox and (2) the angle θ that the pendulum makes with the vertical (a) [2 pts.] In the limit X = 0 (this is the case where the mass M moves along Ox with a constant velocity), find the equation of motion for θ. (b) [3 pts.] Find the potential energy V of the system in terms of X, θ and their velocities (c) [5 pts.] Find the kinetic energy T of the system in terms of X, θ and their velocities. (d) [7 pts.] Find the equation of motion for X. You do not need to attempt solving this differential equation. (e) [7 pts] Find the equation of motion for theta. You do not need to solve this differential equation but show that, in the limit X = 0, it corresponds to your answer to (a). (f) [4 pts.] From the equations of motion for X and θ, and in the limit of small amplitude oscillations, find the normal mode frequency in terms of the given quantities.

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4 Problem 2 Astro/Modern Physics White dwarfs and neutron stars are two types of stellar remnants. In both of them, gravity is balanced by degenerate pressure force. In a white dwarf the degenerate pressure is provided by electrons, while in a neutron star it is provided by neutrons. In this problem, you will be guided to find the mass-radius relation for white dwarfs and the ratio of the size of white dwarfs to that of neutron stars. We only care about the dependence on physical variables in this problem, not the exact values of some numerical factors. So you can neglect irrelevant numerical factors or irrelevant physical constants. For example, you can write ρ M/R 3, instead of ρ = M/(4πR 3 /3). In this problem, you can treat white dwarfs and neutron stars as uniform spheres, and electrons and neutrons as non-relativistic particles. Take the mass of a proton, neutron, or nucleon all to be m p, and that of an electron to be m e (m p 1836 m e ). (a) [3 pts.] Consider a white dwarf of mass M and radius R, made of carbon and oxygen (a.k.a. C/O white dwarf). There is one electron for every two nucleons. What is the mean inter-electron separation l? (b) [3 pts.] Treat l as the uncertainty in the electron s position, what would be the momentum p of an electron? (c) [4 pts.] As an order-of-magnitude estimate, what is the pressure provided by the electrons? (d) [5 pts.] From the point view of the gravitational force, express the pressure in the white dwarf in terms of G (gravitational constant), M, and R. (hint: this is an order-of-magnitude estimate, you can use dimensional analysis or derive the pressure by evaluating the gravitational force and the relevant area at some special position in the white dwarf). (e) [5 pts.] By equating the results in (c) and (d), drive the mass-radius relation (e.g., if it is a power law, expressed as R M n, and you need to calculate the value of the index n). (f) [5 pts.] Consider a neutron star of the same mass M as the white dwarf. You can regard it as purely made of neutrons. What is the ratio R WD /R NS of the radius of a white dwarf over that of a neutron star? Substitute the relevant physical constants and give the numerical value.

5 Problem 2 Astro/Morden Physics Solution (a) [3 pts.] Consider a white dwarf of mass M and radius R, made of carbon and oxygen (a.k.a. C/O white dwarf). There is one electron for every two neucleons. What is the mean interelectron separation l? The number of neucleons in the white dwarf is M m p, so the number of electrons is N e = 1 2 The number density of electrons is M m p. n e N e R 3 1 M 1 2m p R3. (11) The inter-electron separation l is determined by n e l 3 = 1, which gives l = n 1/3 e ( ) 1/3 2mp R. (12) M (b) [3 pts.] Treat l as the uncertainty in the electron s position, what would be the momentum p of an electron? From the uncertainty principle, we have pl /2, (13) which gives ( ) M 1/3 p h/l = hn 1/3 h e = 2m p R. (14) (c) [4 pts.] As an order-of-magnitude estimate, what is pressure provided by the electrons? We can adopt the following picture electrons all move in one direction with momentum p, hit an imaginary wall, and bounce back. (Considering the istropic distribution of the moving directions only leads to a correction factor that we do not care here). On an area of da on the wall, over time interval dt, the number of electrons hitting the wall is n e vdtda, where v is the electron velocity. After bouncing back, the momentum of each electron changes by 2m e v. The total momentum change of all the electrons hitting the area da over time interval dt is then n e vdtda 2m e v. The pressure P (force per unit area) is then P = n evdtda 2m e v dtda n e m e v 2 p 2 ( = n e n5/3 e 1 M m e m e 2 m p ) 5/3 1 R 5 1 m e. (15) In the last two steps, we have used the results in (b). (d) [5 pts.] From the point view of the gravitational force, express the pressure P in the white dwarf in terms of G (gravitational constant), M, and R. (hint: this is an order-of-magnitude estimate, you can use, e.g., dimensional analysis or derive the pressure by evaluating the gravitational force and the relevant area at some special position in the white dwarf). 3

6 With dimensional analysis, the gravitational force GM 2 /R 2 and the area R 2, so the pressure is P GM2 R 4. (16) Or we can evaluating the gravitational force a layer at radius R/2 feels, and we would derive the similar results. (e) [5 pts.] By equating the results in (c) and (d), derive the mass-radius relation (e.g., if it is a power law, expressed as R M n, and you need to calculate the value of the index n). Equating the results in (c) and (d), we have ( 1 M 2 m p ) 5/3 1 R 5 1 m e GM2 R 4, (17) and the mass-radius relation is R M 1/3. (18) (f) [5 pts.] Consider a neutron star of the same mass M as the white dwarf. You can regard it as purely made of neutrons. What is the ratio R WD /R NS of the radii of the white dwarf over that of a neutron star? Substitute the relevant physical constants and give the numerical value. From the result in (e), the radius of a white dwarf can be expressed as ( 1 M R WD 2 m p ) 5/3 1 1 m e GM2. (19) For a neutronstar, we need to replace m e with m p in theabove expression. Also the facor 1/2, which comes from the number of electrons being half that of neucleons, should be replaced by 1. We have ( ) M 5/3 1 1 R NS m p GM2. (20) The ratio at the same mass M is then R WD R NS = m p ( ) 1 5/3 m p 578. (21) 2 m e The result is pretty accurate white dwarfs are about the size of the earth (radius 6400km) and neutrons stars typically have radius of 10km, so the size ratio is about

7 Problem 3 Thermodynamics (a) [2 pts.] Write down the First Law of Thermodynamics in differential form; i.e. in the form du =, where U is the internal energy. For a gaseous system, write down ds, the differential of entropy S, in terms of other thermodynamic variables and differentials, and indicate how ds fits into the First Law. 3 The internal energy of a monoatomic gas (such as helium) is given by the formula U = knt, and its 2 equation of state by PV = knt, where k is the Boltzmann constant, N is the number of atoms, P is the pressure, V is the volume, and T is the temperature. (b) [6 pts.] The temperature of the gas changes from T 0 to T 1, while the number of atoms N, and volume V 0, remain fixed. Find the change of the entropy S in this process. (c) [6 pts.] The volume of the gas changes from V0 to V 1, while the number of atoms N and temperature T 0 remain fixed. Find the change of the entropy S in this process. (d) [6 pts.] The gas expands form its initial state given by the parameters (N, V 1, T 1 ) to its final state given by (N, V 2, T 2 ), along a linear path in the V vs. T graph shown. Find the change of the entropy in this S process. V V2, T2 V1, T1 T c) [5 pts.] Write down the Third Law of Thermodynamics. (This law states how entropy or heat capacity behaves when a system approaches zero temperature). The entropy of an ideal gas is given by the famous Sackur-Tetrode equation 3 V 4π mu 2 5 S = Nk ln + N 2 3Nh 2, where h is the Plank constant. Does this equation agree with the third law of thermodynamics? (You must give a physical explanation to your yes/no answer in order to receive any credit)

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9 Problem 4 Electricity and Magnetism Consider a disk of radius b and thickness s made of a material with a resistivity ρ. The disk rotates with the fixed angular frequency ω around an ideally conductive axle of radius a, which extends through the resistive disk. The rim of the disk is made of an ideally conductive material and is of negligible thickness. Brush contacts A and B are sliding on the rim of the axle and rim as shown in the figure. They are connected to an ideal solenoid (not shown on the figure) of length l with n coils per unit length and whose axis coincides with the axis of the disk. The solenoid also has radius b, such that the entire disk fits exactly inside the solenoid. s (a) [4 pts.] Find the inductance L of the solenoid in terms of the given quantities. (b) [6 pts.] Find the resistance R of the disk as measured between points A and B. (c) [2 pts.] Suppose the magnetic field produced by the solenoid is in the upward direction in the figure. Do you expect a current induced by the motion of the resistive material in the magnetic field to flow through the disk from A to B or from B to A? In the rest of the problem, we assume the brush contacts are connected to the solenoid so the magnetic field it produces is pointing upward on the figure. (d) [6 pts.] Express the electromotive force E between A and B resulting from the motion of the disk in the magnetic field in terms of the current I flowing through the solenoid. (e) [7 pts.] The system appears as a circuit with a motion induced electromotive force Ɛ, a resistance R, and an inductance L. Write the differential equation for the current and establish a condition for the magnetic field to be spontaneously maintained as long as the disk keeps rotating at the fixed angular frequency ω.

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12 Problem 5 Quantum Mechanics A particle of mass m is located in a 1-dimensional potential given by V(x) = 1 2 mω2 x 2, where ω is a positive constant with unit of inverse seconds (s 1). (a) [5 pts. ] Write down the time-independent Schrödinger equation for this system in terms of the wave function φ(x), its derivatives with respect to x, the given constants m, ω and Planck s constant ħ. We approximate the wave function of the ground state by the trial function A φ(x) = b 2 + x 2 where A is the normalization constant, and b is a free parameter that roughly corresponds to the width scale of the wave function. (b) [4 pts] Normalize the trial wave function φ(x), i.e. find A in terms of b. (c) [6 pts] Find the expectation values, x, x 2, p, and p 2 for the normalized trial wave function φ(x) obtained from part (b), in terms of b, m, ω, and ħ. Here x and p denote the position and momentum operators, respectively. (d) [2 pts] Find the expectation value E of the Hamiltonian (energy) operator for the normalized trial wave function φ(x) obtained from part (b), in terms of b, m, ω, and ħ. (e) [ 4 pts. ] Find the value of b, in terms of m, ω, and ħ, that minimizes E for the normalized trial wave function φ(x) obtained from part (b), and hence find the approximation for the ground state energy for this system. (f) [4 pts] Calculate the product σ x σ p for the optimized ground state trial wave function φ(x) obtained from part (e). The uncertainty σ O of an observable O is defined as its root-mean-square expectation value, where σ 2 O O 2 O 2. Show that the Heisenberg Uncertainty Relation is satisfied: σ x σ p ħ 2. You might find the following definite integrals useful: 1 (b 2 + x 2 ) 2 dd 1 (b 2 + x 2 ) 3 dd 1 (b 2 + x 2 ) 4 dd = π 2b 3, x 2 = 3π 8b 5, x 2 = 5π 16b 7, x 2 (b 2 + x 2 ) 2 dd = π 2b, x 4 (b 2 + x 2 ) 2 dd = (b 2 + x 2 ) 3 dd (b 2 + x 2 ) 4 dd = π 8b 3, x 4 (b 2 + x 2 ) 3 dd = π 16b 5, x 4 (b 2 + x 2 ) 4 dd = 3π 8b = π 16b 3

13 Solution to Problem 5 Quantum Mechanics A particle of mass m is located in a 1-dimensional potential given by V(x) = 1 2 mω2 x 2, where ω is a positive constant with unit of inverse seconds (s 1). (a) [5 pts. ] Write down the time-independent Schrödinger equation for this system in terms of the wave function φ(x), its derivatives with respect to x, the given constants m, ω and Planck s constant ħ. Solution: Time independent Schrödinger equation for a particle of mass m in a 1d potential V(x) is given in general by ħ2 φ(x) + V(x)φ(x) = Eφ(x) 2m dd2 So with V(x) = 1 2 mω2 x 2 we then have ħ2 d 2 d 2 2m dd 2 φ(x) mω2 x 2 φ(x) = Eφ(x) We approximate the wave function of the ground state by the trial function A φ(x) = b 2 + x 2 (b) [4 pts] Normalize the trial wave function φ(x), i.e. find A in terms of b. Assuming A to be a real value, then 2 φ(x) A φ(x)dd = b 2 + x 2 dd = A 2 And so A = 2b3 π, 1 (b 2 + x 2 ) 2 dd φ(x) = 2b3 π 1 b 2 + x 2 = A 2 π 2b 3 = 1 (c) [6 pts] Find the expectation values, x, x 2, p, and p 2 for the normalized trial wave function φ(x) obtained from part (b), in terms of b, m, ω, and ħ. Here x and p denote the position and momentum operators, respectively. x = φ(x) xx(x)dd = 2b3 π x (b 2 + x 2 ) 2 dd = 0 The integral for x vanishes because the integrand is an odd function integrated over to. x 2 = φ(x) x 2 φ(x)dd = 2b3 π x 2 (b 2 + x 2 ) 2 dd = 2b3 π π 2b = b2 The momentum operator acts on φ(x) by pp(x) = iħ d dd φ(x) = iħa d dd 1 x b 2 + x2 = 2iħA (b 2 + x 2 ) 2 = 2iħA x (b 2 + x 2 ) 2 And so its expectation value is then p = φ(x) pp(x)dd = 2iħ 2b3 π x (b 2 + x 2 ) 3 dd = 0 Where the integral again vanishes because the integrand is an odd function. The momentum squared operator then gives

14 p 2 φ(x) = p[pp(x)] = iħ d dd [pp(x)] = iħa d dd 2iħA x (b 2 + x 2 ) 2 = 2ħ2 A 2 d dd x (b 2 + x 2 ) 2 = = 2ħ 2 A 2 1 (b 2 + x 2 ) 2 4x 2 (b 2 + x 2 ) 3 And so p 2 = φ(x) p 2 φ(x)dd = 2ħ 2 2b3 π 1 (b 2 + x 2 ) 3 dd x 2 4 (b 2 + x 2 ) 4 dd = 4ħ2 b 3 3π π 8b 5 4 π 16b 5 = 4ħ2 b 3 3π π 8b 5 2π ħ2 8b5 = 2b 2 (d) [2 pts] Find the expectation value E of the Hamiltonian (energy) operator for the normalized trial wave function φ(x) obtained from part (b), in terms of b, m, ω, and ħ. E = H = 1 2m p mω2 x 2 = 1 2m p mω2 x 2 = ħ2 4mb mω2 b 2 (e) [ 4 pts. ] Find the value of b, in terms of m, ω, and ħ, that minimizes E for the normalized trial wave function φ(x) obtained from part (b), and hence find the approximation for the ground state energy for this system. To find the minimum E, we take its derivative with respect to b and set it to zero: d E = 2 ħ2 dd 4mb 3 + mω2 b = 0 mω 2 b ħ2 2mb 3 = 0 ħ 2 ħ b4 = 2m 2 b = ω2 2mω And so the approximate ground state energy is: E 0 = ħ2 4mb mω2 b 2 = ħ2 2mω + 1 ħ 4m ħ 2 mω2 2mω = 1 2 ħω (f) [4 pts] Calculate the product σ x σ p for the optimized ground state trial wave function φ(x) obtained from part (e). The uncertainty of an observable O is defined as its root-mean-square expectation value: σ 2 O O 2 O 2. Show that the Heisenberg Uncertainty Relation is satisfied: σ x σ p ħ 2. We have x 2 = b 2, and x = 0, so And we have p 2 = ħ2 2b 2, and p = 0, so And σ x 2 = b 2 0 = b 2 σ x = b σ 2 x = ħ2 2b 2 0 = ħ 2 2b 2 σ x = ħ 2b σ x σ p = ħ 2b b = ħ 2 > ħ 2

15 Problem 6 Classical Mechanics A uniform hoop of mass m and radius r is initially at rest at position A on an inclined plane, which is at an angle θ with respect to the horizontal direction (ground). Under the influence of gravitational acceleration g, it starts to roll down the plane without slipping. Finally, it reaches position B. From position A to position B, the center of the hoop moves a distance L. (a) [4 pts.] Find the translational speed v of the hoop, when it is at position B, in terms of the given quantities. (b) [4 pts.] Find the angular velocity ω of the hoop, around its center-of-mass, when it is at position B. (c) [5 pts.] At position B, find the magnitude of the total acceleration a of the center-of-mass of the hoop. (d) [6 pts.] How long does it take the hoop to roll from position A to position B? (e) [6 pts.] If the hoop is replaced with a disk of uniform density, with the same mass m and the same radius r, how long would it take the disk to roll (without slipping) from position A to position B?

16 Problem 6 Elementary Classical Mechanics Solution (a) [4 pts.] Find the translational speed v of the hoop, when it is at position B, in terms of the given quntaties. Since the hoop rolls down without slipping, the translational speed v and angular velocity ω satisfy v = ωr. During the process, the mechanical energy is conserved, with the kinetic energy of the hoop from the translational motion and the rotation. We have mglsinθ = 1 2 mv Iω2, (1) where the momentum of inertia of the hoop I = mr 2. The translational speed v is solved to be v = glsinθ. (2) (b) [4 pts.] Find the angular velocity ω of the hoop, around its center-of-mass, when it is at position B? ω = v glsinθ r =. (3) r (c) [5 pts.] At position B, find the magntiude of the total acceleration a of the center-of-mass of the hoop? At the positon where the center of the hoop moves a distance l, as in (a) we have i.e., mglsinθ = 1 2 mv Iω2 = mv 2, (4) Taking derivative with respect to time t on both sides, we have Since v = dl/dt and a = dv/dt, we then have glsinθ = v 2. (5) gsinθ dl dt = 2vdv dt. (6) a = 1 gsinθ. (7) 2 (d) [6 pts.] How long does it take the hoop to roll from position A to position B? From (c), we see that the center-of-mass of the hoop moves with a uniform acceleration a, starting at rest. We have L = 1 2 at2, (8) which leads to 4L t = gsinθ. (9) 1

17 (e) [6 pts.] If the hoop is replaced with a disk of uniform density, with the same mass m and the same radius r, how long would it take the disk to roll (without slipping) from position A to position B? To figure out the answer, we only need to replace the momentum of intertia of the hoop, I = mr 2, with that of the disk, I = 1 2 mr2, in all the relevant equations for the hoop case. We then have the time for the disk to roll down from A to B, 3L t = gsinθ. (10) 2

18 NOTE: Parts (a) and (b) are unrelated Problem 7 Electricity and Magnetism (a) [15 pts.] Two identical small sphere, each of mass m = 1.0x10 4 kg, hang on two threads as shown in the figure attached. The threads have length l = 0.30 m, and gravity is directed downward as shown in the figure. At first, the two spheres touch each other. When the same charge q is transferred onto each sphere, the separation between them becomes d = 0.06 m. Determine the charge q on each sphere [Numerical answer and proper units required]. (b) [10 pts.] Show that an electric charge cannot remain in stable equilibrium under the action of electrostatic fields alone. (Hint: start with the assumption that a charge is in fact in such a stable equilibrium, and consider the possible existence of a restoring force for an arbitrary small displacement that can lead to such a situation.)

19 Solutions for problems 7 and 8 (Dated: August 28, 2015) Problem 7 a) The situation described in the problem is depicted in Fig. 1. There are three forces acting on each sphere: tension force T, Coulomb repulsion force F C, and the force of gravity, m g. In equilibrium, the three forces add up to zero: T + F C + m g = 0. Projecting the Newton s second law on the axis perpendicular to a thread, one obtains where the last approximation is due to d l. Using the expression for the Coulomb Force, and the numbers given in the problem we find F C = mg tan θ = mg d 2 l 2 (d/2) mg d 2 2l, F C = 1 4πɛ 0 q 2 d 2, q ± C. b) For definiteness, let us assume that a positive charge is in stable equilibrium under the action of electrostatic forces. Let us draw a closed surface around it, such that there are no other charges inside it, see Fig. 2. Given the assumption of stable equilibrium, there should be a restoring force acting on the charge for an infinitesimal displacement in any direction. Therefore, there should be an electric field that points inward at any point of the surface surrounding the charge. Such an electric field has a nonzero flux through the chosen surface: d S E < 0, d S pointing along the outer normal. By Gauss s law, this would require a negative charge inside the surface, which contradicts the assumption that there are no other charges besides the test one inside the surface. Hence the described situation is impossible, and an isolated charge cannot be in equilibrium under the action of only electrostatic fields. Problem 8 a) The stationary Schrödinger equation can be written simply as ˆp 2 2m ψ = Eψ, where ψ is the particle s wave function, and ˆp = i r is the momentum operator. Since motion of the particle is restricted to the azimuthal direction, the only component of ˆp that enters the Schrödinger equation is the one that FIG. 1: Force diagram

20 2 FIG. 2: Point charge in equilibrium. necessary to maintain the equilibrium. Black arrows schematically represent the direction electrostatic fields that would be corresponds to the azimuthal component of 1 r operator in cylindrical coordinates, p φ = e φ i R φ, where φ is the azimuthal angle. The Schödinger equation becomes b) The general solution to Eq. (1) is 2 2 ψ 2mR 2 = Eψ. (1) φ2 ψ(φ) = Ae iλφ + Be iλφ. (2) Since φ and φ + 2π are equivalent, ψ(φ) must be a periodic function of φ with 2π period. This requires λ = n, n being a non-negative integer. (the non-negativeness comes from the condition that one should not double-count the solutions.) Substituting the solutions back into the Schrödinger equation, we immediately obtain E n = 2 n 2 2mR 2. (3) All states for non-zero n are doubly degenerate, corresponding to two linearly independent solutions for ψ, Eq. (2). The lowest energy state with n = 0 in non-degenerate, as there is a single state with zero energy (just a φ-independent constant). This is in accord with the general statement about the ground state of a single-particle system being non-degenerate and not having zeros. c) The simplest way to write ˆL z is to note that ˆL = ˆr ˆp, and that the motion of the particle is constrained to r = R, and the only component of ˆp is p φ. This immediately yields ˆL z = i φ. (4) It is obvious that L z commutes with the Hamiltonian, [ˆL z, H] = 0, because partial derivatives commute. The underlying reason for that is the presence of rotational symmetry around z-axis, which implies conservation of L z. In quantum-mechanical language this conservation law is expressed as [ˆL z, H] = 0. Since ˆL z commutes with the Hamiltonian, the two operators can be diagonalized simultaneously. In fact, the eigenstates for the Hamiltonian in Eq. (2) were chosen such that each lineraly independent one is also an eigenstate of the z-component of the angular momentum: ˆL z e ±inφ = ± ne ±inφ, (5) thus we see that the angular momentum is quantized in units of, L z = n. The double degeneracy of states found in (b) is nothing but the degeneracy between states corresponding to angular momenta equal in magnitude, but opposite in sign (z and z are equivalent, hence the degeneracy. Would be lifted by, say, magnetic field in z-direction). The ground state, corresponding to L z = 0, is non-degenerate for obvious reasons. d) In the presence of magnetic field the Schrödinger equation is modified according to (ˆp ea) 2 2m ψ = Eψ,

21 Solutions for problems 7 and 8 (Dated: August 28, 2015) Problem 7 a) The situation described in the problem is depicted in Fig. 1. There are three forces acting on each sphere: tension force T, Coulomb repulsion force F C, and the force of gravity, m g. In equilibrium, the three forces add up to zero: T + F C + m g = 0. Projecting the Newton s second law on the axis perpendicular to a thread, one obtains where the last approximation is due to d l. Using the expression for the Coulomb Force, and the numbers given in the problem we find F C = mg tan θ = mg d 2 l 2 (d/2) mg d 2 2l, F C = 1 4πɛ 0 q 2 d 2, q ± C. b) For definiteness, let us assume that a positive charge is in stable equilibrium under the action of electrostatic forces. Let us draw a closed surface around it, such that there are no other charges inside it, see Fig. 2. Given the assumption of stable equilibrium, there should be a restoring force acting on the charge for an infinitesimal displacement in any direction. Therefore, there should be an electric field that points inward at any point of the surface surrounding the charge. Such an electric field has a nonzero flux through the chosen surface: d S E < 0, d S pointing along the outer normal. By Gauss s law, this would require a negative charge inside the surface, which contradicts the assumption that there are no other charges besides the test one inside the surface. Hence the described situation is impossible, and an isolated charge cannot be in equilibrium under the action of only electrostatic fields. Problem 8 a) The stationary Schrödinger equation can be written simply as ˆp 2 2m ψ = Eψ, where ψ is the particle s wave function, and ˆp = i r is the momentum operator. Since motion of the particle is restricted to the azimuthal direction, the only component of ˆp that enters the Schrödinger equation is the one that FIG. 1: Force diagram

22 Problem 8 Quantum Mechanics A particle is constrained to move along a circular ring of radius R. The width of the ring is very small, so that there is no motion in the radial direction. (a) [4 pts.] Write down the Schrödinger equation for the particle in terms of the relevant coordinate. (b) [5 pts.] Determine the energy levels of the particle, and their degeneracies. (c) [6 pts.] Write down the expression for the angular momentum operator L z along the axis perpendicular to the plane of the circle (call it z). Find its commutator L z, H with the Hamiltonian (does L z commute with the Hamiltonian?). What are the eigenvalues of the angular momentum? (d) [10 pts.] The ring is placed in a constant, uniform magnetic field of magnitude B, perpendicular to its plane. The particle has charge e (elementary charge). Write down the new Hamiltonian, and find the new energy levels of the modified system. Also find the values of the magnetic flux through the ring, for which the spectrum of energies is unchanged. Hint: When a magnetic field is included in the system, one needs to replace the momentum operator p by p qa in the Hamiltonian. Here q is the charge of the particle and A is the magnetic vector potential.

23 Problem 9 Modern Physics / Special Relativity A particle of rest mass m and charge q > 0 is initially traveling in free space along the positive y direction at speed of 0.60c, where c is the speed of light in vacuum. (a) [6 pts.] Find the initial total energy, and the relativistic momentum of the particle in terms of m, c, along with a numeric factor. (b) [4 pts.] If the particle were to decay after 2.0 microseconds in its rest frame, how far would it travel in the laboratory frame (in which it is traveling at 0.60c) before it decays? Use the value for the speed of light of c = m/s, and express your answer in meters. The particle does NOT decay in the rest of this problem. At time zero, the particle is at the origin (and moving at 0.60c in the +y direction), and a constant, and uniform electric field is turned on with magnitude E 0 and pointing in the +x direction. Even though the particle is relativistic, Newton s 2 nd Law still holds in its original form dp dd = F (c) [5 pts.] Find the x- and y-components of the particle s momentum, p x (t) and p y (t), as functions of time. Express your answers in terms of m, c, q, E 0, t, along with numeric factor(s). You can write the initial momentum as just p 0 if you did not get an answer for it in part (a). (d) [4 pts.] Find the total energy of the particle as a function of time for t > 0. Express your answer in terms of m, c, q, E 0, t, along with numeric factor(s). (e) [6 pts.] Find the x- and y-components of the particle s velocity, v x (t) and v y (t), as function of time for t > 0. Express your answers in terms of m, c, q, E 0, t, along with numeric factor(s). Find also the limiting values of v x and v y as t.

24 Solution Problem 9 Modern Physics / Special Relativity A particle of rest mass m and charge q > 0 is initially traveling in free space along the positive y direction at speed of 0.60c, where c is the speed of light in vacuum. (a) [6 pts.] Find the initial total energy, and the relativistic momentum of the particle in terms of m, c, along with a numeric factor. Initial (as indicated by the subscript i) total energy is given by E i = γ 0 mc 2, γ i = β i, β i = v i c In the given case, we have v i = 0.60c, so β i = 0.60, and 1 γ i = 1 (0.60) = = = = 1.25 E i = 1.25mc 2 The initial momentum is given by p i = γ i mv i = γ i m(β i c) = γ i β i mm For this case, γ 0 β 0 = = 0.75 And so p i = 0.75mm (b) [4 pts.] If the particle were to decay after 2.0 microseconds in its rest frame, how far would it travel in the laboratory frame (in which it is traveling at 0.60c) before it decays? Use the value for the speed of light of c = m/s, and express your answer in meters. Decay time interval in rest frame: τ = s Time dilation in the lab frame means the decay time interval seen in the lab frame is t = γ τ In this case then in the lab frame: t = (γ i τ), s = v i t = (β i c) (γ i τ) = γ i β i c τ Substituting in actual values: s = (0.75)( m s)( s) = 450m The particle does NOT decay in the rest of this problem. At time zero, the particle is at the origin (and moving at 0.3c in the +y direction), and a constant, and uniform electric field is turned on with magnitude E 0 and pointing in the +x direction. Even though the particle is relativistic, Newton s 2 nd Law still holds in its original form dp dd = F (c) [5 pts.] Find the x- and y-components of the particle s momentum, p x (t) and p y (t), as functions of time. Express your answers in terms of m, c, q, E 0, t, along with numeric factor(s). You can write the initial momentum as just p 0 if you did not get an answer for it in part (a). dp dd = F, F = qe 0 x + 0y

25 So the equations for the components are dp x dd = qe 0, Using the initial conditions dp x dd = 0 p xx = 0 and p yy = p i = γ i β i mm And integrating with respect to time, then we have p x (t) = p xx + qe 0 t = qe 0 t p y (t) = p yy = p i = γ i β i mm ***If no answer for initial momentum was obtained in part a then p y (t) = p 0 (d) [4 pts.] Find the total energy of the particle as a function of time for t > 0. Express your answer in terms of m, c, q, E 0, t, along with numeric factor(s). Total Energy Can be related to Total momentum by E = p 2 c 2 + m 2 c 4 So for t > 0: E = p x 2 + p y 2 c 2 + m 2 c 4 p x 2 + p y 2 = γ i 2 β i 2 m 2 c 2 + q 2 E 0 2 t 2 So the total energy is therefore given by E = q 2 E 0 2 c 2 t 2 + γ i 2 β i 2 m 2 c 4 + m 2 c 4 = 1 + γ i 2 β i 2 m 2 c 4 + q 2 E 0 2 c 2 t 2 = mc γ i 2 β i 2 + q2 E 0 2 m 2 c 2 t2 Substituting in the value γ i β i = 0.75, we have 1 + γ i 2 β i 2 = E = mc q2 E 0 2 m 2 c 2 t2 (e) [6 pts.] Find the x- and y-components of the particle s velocity, v x (t) and v y (t), as function of time for t > 0. Express your answers in terms of m, c, q, E 0, t, along with numeric factor(s). Find also the limiting values of v x and v y as t. The momentum components are given by p x = γγv x, p y = γγv y So we have: γ = E mc = γ 2 i β 2 i + q2 2 E 0 m 2 c 2 t2 v x = p x γγ = qe 0 t m 1 + γ 2 i β 2 i + q2 2 E 0 m 2 c 2 t2 t qe 0 t m q2 E 0 2 m 2 c 2 t2 = c v y = p y γγ = γ i β i mm m 1 + γ 2 i β 2 i + q2 2 E 0 m 2 c 2 t2 0 t

26 Problem 10 Statistical Physics A system of N particles, which have spin ½ is placed in a uniform magnetic field B aligned in the +z direction. The energy of a spin parallel to magnetic field is u p = mb ; the energy of a spin anti-parallel to magnetic field is ua = mb, where m is a magnetic moment of the particle. The system is at temperature T. (a) [5 pts.] Find the total energy of the system in terms of N, m, B, T, and the Boltzmann Constant k. (b) [5 pts.] Find the entropy of the system in terms of N, m, B, T, and k. (c) [5 pts.] The magnetic susceptibility, χ m, is defined by M = χ m B, where M is the total magnetic moment of the system. Find χ m in the low magnetic field, high temperature limit (mb<<kt), and hence show that χ m is inversely proportional to the temperature in this limit (This is known as Curie s Law after Pierre Curie). (d) [10 pts.] Continuing to work in the mb<<kt limit, the system of spins described above is initially at temperature T0 and in a small magnetic field B 0 ( mb0 << kt0). Then, the system is thermally isolated and the external magnetic field is reduced from B0 to B 1 (the field direction does not change). We assume that although our system is thermally isolated, the spins come to thermal equilibrium with each other. Find the temperature T 1 of the isolated spin system after it reaches the new internal thermal equilibrium.

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