1. Solution: Classical physics with Lagrangian

Transcription

1 1. Classical physics [5 pts.] A particle of mass m is subjected to the potential V = F x, where F is a constant. The particle travels from x = 0 to x = L in a time interval T. Find the motion of the particle x(t) in terms of L, T, F, and m, such that the action is a minimum. 1

2 1. Solution: Classical physics with Lagrangian The minimum of the action is obtained when the particle follows the classical trajectory. In this case, the particle is subjected to a constant force F = ( V/ x) along x, and its classical motion is a motion with constant acceleration a = F/m. From elementary kinematics, x(t) = A + Bt + Ct, with C = a/ = F/(m). Since the particle is at x = 0 when t = 0, one immediately finds A = 0. The value of B is obtained by imposing that the particle is at x = L when t = T, namely L = BT +CT, from which B = (L CT )/T = L/T F T/(m). To summarize, ( L x(t) = T F T ) t + F m m t or x(t) = Lt T + F t(t T ) m.

3 Problem : Electricity and Magnetism. An infinite wire carrying current I lies in the x-axis. Space is field with two different magnetic dielectrics, divided by xy-plane. The magnetic permeability is µ 1 at z > 0 and µ at z < 0. Use Maxwell s equations and Ampere s law to find vectors B and H in all space, following steps outlined below 1.[3 points] Write down two Maxwell s differential equation for H and B in a static case and integral Ampere s law..[5 points] Find fields H 0 and B 0, created by the given current in a vacuum (µ = µ 0 ) everywhere). 3.[5 points] Formulate boundary conditions for tangential and normal components of vectors B and H at the plane z = 0 using Maxwell s equations for H and B in the case of given magnetic medium 4.[1 points] Express field H and B in the medium with given permeability through H 0. Hints: 1. Differential equation for H does not contain permeability and it is homogenous outside the current.. Pay attention to the boundary conditions for normal components of B and H at z = 0. 1

4 Solution to Problem 1. Maxwell s equation for vectors H and B have a form curlh = J, (1) where J is a current density. Using the Stokes theorem we get an integral equation H l dl = I, () that is called Ampere s law. Here I is the total current through the wire. The integration is taken over any close loop with the current inside. H l is the projection of a field on a loop. Field B obeys equation divb = 0 (3) The two fields are connected by the relation B = µh (4). Because of the symmetry it is reasonable to choose a loop as a circle with radius r with the center in the wire. The loop is in any plane x = const. Eq. (3) means the absence of a magnetic charge (monopole). Then both fields has only angular components in the polar coordinates in any plane x = const. Thus H 0 = H l, which is independent of the angle. Then Eq. () gives H 0 πr = I, H 0 = I πr. (5) and B 0 = µ 0 H 0.

5 3. From the second Maxwell s equation divb = 0 it follows that the normal component of B should be continuous at z = 0. From the symmetry of the problem B has only normal component at z = 0 in the loop we have chosen. Note that B = µh. It follows from Eq.(1) that tangential component of H is continuous at z = 0, but it is not important for this problem. Important is that normal component of H might be discontinuous. 4. Eq. (1) does not contain permeability. So H 0 obeys this equation both at z > 0 and z < 0, but because permeability is different in these regions, this solution does not obey the boundary condition for normal component of B at z = 0. However in both these regions equation for H is homogeneous (I=0). Thus solution can be multiplied by a constant factor. Assume that H = a 1 H 0 at z > 0 and H = a H 0 at z < 0. From Ampere s law and continuity of B we get two equations for coefficients a 1,a : The solution is Finally Field B is axillary symmetric a 1 +a =, (6) µ 1 a 1 = µ a. (7) a 1 = µ µ 1 +µ, (8) a = µ 1 µ 1 +µ. (9) H 1 = H = µ H 0, µ 1 +µ (10) µ 1 H 0. µ 1 +µ (11) B = µ 1µ µ 1 +µ H 0. (13) At µ 1 = µ = µ we get H 1 = H = H 0, B = µh 0. 3 (1)

6 3 Thermodynamics Heat capacities are normally positive, but there is an important class of exceptions: systems of particles held together by gravity, such as stars and star clusters. a) [6 pts.] For any system of particles held together by mutual gravitational attraction, U potential = U kinetic, where each U refers to the total energy (of that type) for the entire system, averaged over some sufficiently long time period. This result is known as the virial theorem. Consider a system of two identical particles on circular orbits about their center of mass. Show that the gravitational potential energy of this system is - times the total kinetic energy. Now suppose that you add some energy to such a system and wait for the system to equilibrate. Does the average total kinetic energy increase or decrease? b) [4 pts.] A star can be modeled as a gas of particles that interact with each other only gravitationally. Assuming these particles are monatomic, and that the average temperature over the star is T, what is the average kinetic energy of a particle in the star in terms of T? c) [10 pts.] Express the total energy of the star in terms of the average temperature, T, and calculate the heat capacity. Note the sign. d) [5 pts.] The gravitational potential energy of a star of uniform density, total mass, M, and radius, R, is 3GM 5R. Using this simple model, estimate the average temperature of the sun, whose mass is kg, and whose radius is m. You may assume that the sun is made entirely of protons and electrons.

7 3 Thermodynamics Heat capacities are normally positive, but there is an important class of exceptions: systems of particles held together by gravity, such as stars and star clusters. a) [6 pts.] For any system of particles held together by mutual gravitational attraction, U potential = U kinetic, where each U refers to the total energy (of that type) for the entire system, averaged over some sufficiently long time period. This result is known as the virial theorem. Consider a system of two identical particles on circular orbits about their center of mass. Show that the gravitational potential energy of this system is - times the total kinetic energy. Now suppose that you add some energy to such a system and wait for the system to equilibrate. Does the average total kinetic energy increase or decrease? b) [4 pts.] A star can be modeled as a gas of particles that interact with each other only gravitationally. Assuming these particles are monatomic, and that the average temperature over the star is T, what is the average kinetic energy of a particle in the star in terms of T? c) [10 pts.] Express the total energy of the star in terms of the average temperature, T, and calculate the heat capacity. Note the sign. d) [5 pts.] The gravitational potential energy of a star of uniform density, total mass, M, and radius, R, is 3GM 5R. Using this simple model, estimate the average temperature of the sun, whose mass is kg, and whose radius is m. You may assume that the sun is made entirely of protons and electrons. Solution a) Each particle has a kinetic energy of 1 mv, so the total kinetic energy of the system is U kinetic = mv. If the distance of each particle from the center of mass is r, then the total potential energy of the two-particle system is U potential = Gm r. The gravitational force on each particle is Gm = ma, 4r where a is just the centripetal acceleration, a = v /r. So Then so mv r = Gm 4r. U kinetic = mv = Gm 4r U potential = U kinetic. = 1 U potential, The total energy is U total = U kinetic + U potential = U kinetic U kinetic U total = U kinetic, so increasing the total energy decreases the kinetic energy. b) According to the equipartition theorem, the average kinetic energy of each particle is 3 kt.

8 c) A star of N particles has a total kinetic energy of U kinetic = 3 NkT. The total energy is therefore U total = 3 NkT. The heat capacity is C = du total dt, so C = 3 Nk. d) Using the fact that U kinetic = 1 U potential, 3 NkT = 1 ( ) 3GM 5R T = 1 GM 5 NkR = 1 GM 5 kr ( ) M N M/N is the mass per particle in the star. Since the star should be made of equal numbers of protons and electrons, and m p m e, the mass of each particle is approximately 1 m p. T = 1 5 ( ) ( 10 kg 30 kg ) ( kg ) ( J/K) ( m) 11 N m T 10 6 K

9 Problem 4: Quantum Mechanics A semi-infinite square quantum well, shown in the figure is given by the following potential: x 0 V(x) = V 0 0 < x < a 0 x a A particle of mass m is in the ground (bound) state in this well. It has energy E, where E < 0. (a) [8 points] Write down the time-independent Schrödinger Equations for region 1 (0 < x < a ) and region (x a) shown in the figure, separately. Then write down their general solutions φ 1 (x) and φ (x). Remember to include separate normalization constants for the two regions. You don t need to normalize the functions. (b) [5 points] State the boundary conditions at (i) x = 0 for φ 1 (x), and at (ii) x for φ (x), respectively, and simplify the two functions from their most general form accordingly (c) [6 points] State the two boundary conditions for φ 1 (x) and φ (x) at x = a, and write down the corresponding relationships between the simplified functions from part (b). (d) [6 points] You are given that ma V 0 /ħ = 4. Using this fact, and the relationships between φ 1 (x) and φ (x) from part (c), find the ratio ε = E /V 0 for the ground state energy E to ONE significant digit. This is a numerical solution that requires you to use your calculator (hint: Make a table of ε and the relevant quantity that is supposed to be zero, and start with ε = 0.5).

10 Problem 4 Solution: (a) The time-independent Schrödinger Equation is: ħ d φ(x) + V(x)φ(x) = Eφ(x) m dx (i) In region 1, V(x) = V 0, and we have (noting that V 0 < E < 0): ħ d m dx φ 1(x) V 0 φ 1 (x) = E φ 1 (x) d dx φ 1(x) = m(v 0 E ) ħ φ 1 (x) This gives us a sinusoidal solution of the form φ 1 (x) = A cos kx + B sin kx = A e ikx + B e ikx, k = m(v 0 E )/ħ (ii) In region, V(x) = 0, and we have (noting that E < 0): ħ d m dx φ (x) = E φ (x) d dx φ (x) = m E ħ φ (x) And we have an exponential solution of the form φ (x) = Ce qx + De qx, q = m E /ħ (b) Boundary conditions at x = 0 and x : (i) Because the potential is infinite at x = 0, we have φ 0 (x) = A = 0, and so φ 1 (x) = B sin kx (ii) Because the particle is bound to the well near x = 0, we must have φ (x) 0 as x. Note that Ce qx, and De qx 0 as x. So we must have C = 0 and φ (x) = De qx

11 (c) The wave function and its derivative must both be continuous at x = a. (i) Continuity of the wave function at x = a: φ 1 (a) = φ (a) B sin ka = De qa [1] (ii) Continuity of dφ/dx at x = a gives us : dφ 1 dx = dφ x=a dx x=a kb cos ka = qde qa [] (d) Dividing [] by [1] yields k cot ka = q Multiplying both sides by the well width a gives us a transcendental equation that is dimensionless: x cot x + y = 0 where x = ka = ma (V 0 E )/ħ and y = qa = ma E /ħ We are given that ma V 0 /ħ = 4 ma V 0 /ħ = so that x = (1 E /V 0 ) = (1 ε) and y = E /V 0 = ε Trying different values of e starting at 0.5: e x y xcot(x)+y The answer to one significant digit is 0.4 (0.38 to significant digits).

12 5. General/Modern Physics The Sun is powered by nuclear fusion in its core. The reaction is called proton-proton chain (PP chain), which converts hydrogen nuclei (protons p) into helium nuclei ( 4 He). The overall reaction can be thought as that every four protons are converted into one 4 He nucleus, with an energy release Q. The masses of proton and 4 He nucleus are m p = m u and m4 He = 4.006m u, respectively, where m u = kg is the atomic mass unit. The mass and radius of the Sun are M = kg and R = m, respectively. The Sun-Earth distance (called astronomical unit) is d = m. [N.B. Use the constants given in this problem.] (a) [5 pts.] What is the energy release Q in units of MeV for each 4p-to- 4 He reaction? (b) [6 pts.] For the reaction to occur, one would naively argue that two protons should reach a separation of the order of R = m for the nuclear force to overcome the Coulomb repelling force. The core temperature of the Sun is about T = 10 7 K. For two protons both moving at the most probable thermal speed, what is the closest separation they can reach? Your answer should be much larger than R. What effect can help to overcome the Coulomb barrier, which then can make the nuclear fusion slowly happen in the core of the Sun? (c) [7 pts.] The majority of the released energy Q is in the form of photons and a small fraction is in the form of electron neutrinos and kinetic energy of particles. The photon-electron interaction cross-section is σ T = m. The neutrino-electron interaction crosssection is on the order of σ eν = m. By making the simplification that the Sun is composed of fully ionized hydrogen gas with uniform density, estimate the mean free paths of photons and neutrinos in the Sun, in units of the radius of the Sun R. Based on the results, which can be used to better probe the core of the Sun, photons or neutrinos? (d) [7 pts.] In the Sun, about 600 million tons of hydrogen are converted into helium per second. Each 4p-to- 4 He reaction produces two electron neutrinos. What is the expected solar neutrino flux (in units of m s 1 ) at the Earth distance? Neutrino experiments only detect about one third of the electron neutrino flux predicted by the solar model. To the best of your knowledge, what is the solution to such a discrepancy? 1

13 5. General/Modern Physics Solution (a) [5 pts.] What is the energy release Q in units of MeV for each 4p-to- 4 He reaction? Q = (4m p m4 He)c = 0.078m u c = 5.9MeV. (1) (b) [6 pts.] For the reaction to occur, one would naively argue that two protons should reach a separation of the order of R = m for the nuclear force to overcome the Coulomb repelling force. The core temperature of the Sun is about T = 10 7 K. For two protons both moving at the most probable thermal speed, what is the closest separation they can reach? Your answer should be much larger than R. What effect can help to overcome the Coulomb barrier, which then can make the nuclear fusion slowly happen in the core of the Sun? From energy conservation, the closest separation r of the two protons satisfies where v p = kt/m p. We then have r = 1 m pv p = 1 4πɛ 0 e r, () e 8πɛ 0 kt = m, (3) much larger than R. Quantum tunneling is the main effect to overcome the Coulomb barrier for the fusion to happen. (c) [7 pts.] The majority of the released energy Q is in the form of photons and a small fraction is in the form of electron neutrinos and kinetic energy of particles. The photon-electron interaction cross-section is σ T = m. The neutrino-electron interaction crosssection is on the order of σ eν = m. By making the simplification that the Sun is composed of fully ionized hydrogen gas with uniform density, estimate the mean free paths of photons and neutrinos in the Sun, in units of the radius of the Sun R. Based on the results, which can be used to better probe the core of the Sun, photons or neutrinos? The number density of electrons in the Sun is n e = The mean free path of photons is The mean free path of neutrinos is M m p 4πR 3 /3 = m 3. (4) l γ = 1 n e σ T = m = R. (5) l ν = 1 n e σ eν = m = R. (6) While photons come out of the Sun after many interactions, neutrinos can escape almost directly from the Sun. Therefore, neutrinos are a better probe of the core of the Sun.

14 (d) [7 pts.] In the Sun, about 600 million tons of hydrogen are converted into helium per second. Each 4p-to- 4 He reaction produces two electron neutrinos. What is the expected solar neutrino flux (in units of m s 1 ) at the Earth distance? Neutrino experiments only detect about one third of the electron neutrino flux predicted by the solar model. To the best of your knowledge, what is the solution to such a discrepancy? Let ṀH = 600 million tons per second = kg s 1. The number of 4p-to- 4 He reactions per second is ṀH/(4m p ) and the number of electron neutrinos produced per second is ṀH/(m p ). The expected electron neutrino flux at the Earth distance is F = ṀH 1 m p 4πd = m s 1. (7) The discrepancy is explained by a quantum mechanical phenomenon called neutrino oscillation electron neutrinos, after produced in the core of the Sun, have a probability of transforming into other types of neutrinos (i.e., µ and τ neutrinos) during their propagation to the Earth. 3

15 Problem 6: Electricity and Magnetism: Pressure of Electromagnetic Radiation: This effect was originally predicted by J. C. Maxwell (1871) and demonstrated experimentally by P. Lebedev (1900) in Russia. A monochromatic, linearly polarized (in the x-direction) electromagnetic wave is incident normally on (i.e. traveling in a direction perpendicular to) a surface at the xy plane. (a) [5 points] Write down both the electric and magnetic fields as functions of coordinate z and time t, in terms of the electric field amplitude E 0, angular frequency ω, and the speed of light c. (b) [5 points] Write down both S, the instantaneous, and S, the time-averaged Poynting vectors, in terms of the electric field amplitude E 0, and c, μ 0 and/or ε The vector g, representing the time-averaged flow (per unit area per unit time) of momentum, is given by g = S (1) c (Poynting vector represents flow of energy, and for photons, momentum is given by p = E/c). If the wave is traveling originally in vacuum and the medium on the other has refractive index n, then the coefficient of reflection is given by (n 1) R = (n + 1) () With this assumption: (c) [5 points] Further assume that the refractive medium does not absorb radiation, find an expression for the momentum transferred to the medium per unit area over time interval t. Now express the radiation pressure P on the medium in terms of the radiation intensity I (the intensity is the power carried by the wave per unit area). (d) [5 points] Calculate the radiation pressure P (in appropriate SI units) for an incident wave with intensity of I = 10.0 W/m, and a receiving medium with refractive index n = (e) [5 points] Alternately, for a non-refractive medium, there is neither reflection nor transmission. All energy is absorbed. For such a medium, find the radiation pressure P in terms of intensity I.

16 Problem 6 solution: (a) Electric field of wave traveling in the ±z direction (normally incident on the xy plane): and the magnetic field: E (z, t) = +x E 0 cos ω z ωt (3) c B (z, t) = ±y E 0 c cos ω z ωt (4) c The cosine function here can be replaced by sine and an arbitrary, but COMMON phase factor can be added to the two fields. For the remainder of this problem we will assume the +z direction for propagation. (b) The Poynting Vector, which gives both a direction and magnitude of the flow of energy (power per unit area) is defined by S = E H. In vacuum (which is the incident medium for this problem) it becomes S = E 1 μ 0 B (5) Substituting (3) and (4) into (5), assuming +z direction of propagation, we get S = 1 (x y ) E 0 μ 0 c cos ω c z ωt = z ε 0E 0 cos ω z ωt c and taking the time average at any location, we have cos ω z ωt = 1, and thus the time averaged c Poynting vector is S = 1 z ε 0E 0 The magnitude of this vector is known as the intensity: I = 1 ε 0E 0 (c) By conservation of energy, and assuming no absorption, Figure 6-1 an incident wave of intensity I would give a reflected wave of intensity RI and transmitted (will eventually pass out of the medium) wave of intensity (1 R)I, as illustrated by figure 6-1. Thus only fraction R of the energy flow is reflected. The magnitude of the momentum flow, by the definition g = S /c is given by g = I/c. The reflected part of the wave then undergoes a change of momentum of p = RgA t over area A and time interval t. The momentum transferred per unit area over time t is therefore p A = Rg t = R S (n 1) 1 t = c (n + 1) c ε 0E (n 1) ε 0 0 t = (n + 1) c E 0 t In other words the radiation pressure on the surface is P = 1 A p (n 1) ε 0 = t (n + 1) c E 0 (n 1) = (n + 1) I c = R I c

17 (d) n =.00 R = (n 1) /(n + 1) = (1.00) /(3.00) =1/9. We are given also I = 10.0 W/m. P = I 9 c = kg m s 1 s 1 m kg m m 9 = s 1 m = 7.41 N 10 9 m = Pa s where the symbol Pa stands for pascals. (e) With complete absorption, all the energy flow and momentum flow of the wave stops at the surface. The medium absorbs all the intensity, and all of the momentum flow is transferred to the surface. The momentum transferred per unit area over time t is therefore p A = g t = S c t = 1 c ε 0E 0 t In other words the radiation pressure on the surface is now P = 1 p A This is the same pressure as in the case where R = 1. t = ε 0 c E 0 = I c

18 Problem 7: Statistical Mechanics An ensemble of N weakly interacting atoms on a fixed lattice is placed in a uniform external magnetic field B = 15 T (tesla), directed along the z direction. Each atom (they are distinguishable by their lattice location) can be either in a S z = +1/ (spin-up) or S z = 1/ (spin-down) state, and the magnetic potential energy of atom i is given by E i = µb(s z ) i, where µ = J/T (joule/tesla). The ensemble is immersed in a thermal bath at temperature T. You might find it useful (making the math less cumbersome) to use the abbreviation x = μb/kt, where k is the Boltzmann Constant. (a) [8 points] Find general expressions for the fractions f + and f - of atoms in the spin-up and spin-down states in terms of x. (b) [7 points] In one mole (N = N A = ) of atoms, we find a total of N = atoms in the spindown state. Find the numerical value of the temperature T in kelvins. Note k = J/K. (c) [10 points] Find an expression for the entropy of the ensemble, in terms of x, N, and k. There are a number of different approaches to part (c). In one of the several cases, you might find the Stirling Approximation useful: ln(n!) n ln n n for very large n.

19 Solution to Problem 7: (a) A spin-down atom has energy E = + 1 μb, which is at a higher energy than a spin-up atom at E + = 1 μb. The Boltzman distribution gives us that N = e (E E +) kt = e x N + where x = μb/kt The fraction of spin-down atoms is then given by Dividing top and bottom by N + then gives Similarly: f = f + = N + N + + N = N f = N + + N N /N + N + /N + + N /N + = N /N N /N + = e x 1 + e x N + /N + N + /N + + N /N + = N /N + = e x μb exp f = kt exp μb, f + = kt 1 + exp μb kt Alternatively we can express the answers in terms of hyperbolic functions: f = e x 1 + e x = e x/ cosh(x ) sinh(x e +x/ = ) + e x/ cosh(x ) 1 f + = 1 + e x = e +x/ cosh(x ) + sinh(x e +x/ = ) + e x/ cosh(x ) Yet a different Approach is to use the Partition function: Z = exp E i kt = exp E + i kt + exp E μb μb = exp + + exp kt kt kt = e+x/ + e x/ The fraction of occupation in each state is equal to the probability of finding an atom in that state: We have f i = p i = 1 Z exp E i kt exp E kt = exp +1 μb kt = e x/ e x/ f = p = e +x/ + e x/ = e x 1 + e x and exp E + kt = exp 1 μb kt = e +x/ e+x/ f + = p + = e +x/ + e x/ = e x

20 (b) N = N A = , N = N + = N N = N / N + =1/3 But and so N N + = f f + = e x e x = 1 3 ex = 3 x = ln 3 T = μb k ln 3 = J/T 15T J/K ln 3 = 9.9 K μb kt = ln 3 (c) In many thermal physics textbooks, the most generalized expression/definition for entropy is (for quantized energy states of energy E i and N total atoms): S = Nk p i ln p i where p i is ther probability of occupation of state i with energy E i. Applied to our case, the we have e x e x S = Nk (p ln p + p + ln p + ) = Nk ln 1 + e x 1 + e x e x ln e x Expanding, we get: S Nk = e x 1 + e x ln(e x ) + e x 1 + e x ln(1 + 1 e x ) 1 + e x ln(1) e x ln(1 + e x ) = xe x e x e x 1 + e x e x ln(1 + e x ) i And so we have xe x S = NK ln(1 + e x ) e x Alternatively: We really should be using the formalism for Canonical Ensemble (system in thermal bath), which is formulated based on the Partition Function (again for N atoms): S = F T = + T (NkT ln Z) = Nk ln Z + T Z Z dx x dt ln Z = ln e +x + e x = ln e +x (1 + e x ) = x + ln(1 + e x ) Z x = x x e+ + e x = 1 e+x e x dx, dt = d dt μb μb = kt kt = x T so 1 Z = 1 e +x + e x T Z Z dx x dt = T e +x e x x e +x + e x T = x 1 e x 1 + e x S = Nk x + ln(1 + e x ) x 1 e x 1 + e x = Nk x 1 + e x 1 + e x x 1 e x 1 + e x + ln(1 + e x ) = Nk x 1 + e x 1 + e x 1 + e x + ln(1 + e x ) = Nk xe x 1 + e x + ln(1 + e x ) which is the same answer as before.

21 Yet another (perhaps more familiar) approach: For a micro-canonical ensemble, entropy is given by Boltzmann s relation: S = k ln Ω, where Ω is the number of micro-states that corresponds to a given macro-state specified by energy E instead of temperature T. In our case, we have a canonical ensemble the macro-state is completely specified by the temperature T (taking µ and B to be constants), which in turns specifies the fraction of atoms in the spin-down (or spin-up) state. However, the calculation of average entropy S yields the same results for either approach. But as Prof. Gondolo pointed out, the fluctuations would be completely wrong. Counting the number of ways the N spin-down atoms can be arranged out of N total atoms (lattice sites) Ω = N (N 1) (N ) (N N + 1) N (N 1) (N ) 1 Here the numerator denotes the choice of N possible sites for the first spin-down atom, times (N 1) for the second.etc. and (N N + 1) sites for the last. The denominator is the number of different equivalent re-arrangements (permutations) of the N spin-down atoms. The numerator can be rewritten as N! N (N 1) (N ) (N N + 1) = (N N )! = N! N +! and the denominator is obviously N (N 1) (N ) 1 = N! and so we have Ω = N!/N +! N! = N! N +! N! Alternatively, we can distribute spin-up atoms and get the same expression Ω = N (N 1) (N ) (N + + 1) = N!/N! N! = N + (N + 1) (N + ) 1 N +! N +! N! Either way the entropy is then given by: N! S = k ln Ω = k ln N +! N! = k(ln N! ln N +! ln N!) Applying Stirling s approximation (and noting that N = N + + N ): S k = (N ln N N) (N + ln N + N + ) (N ln N N ) = N ln N N + ln N + N ln N N = N ln N N + ln 1 + e x N ln Ne x 1 + e x = N ln N N + ln N + N + ln(1 + e x ) N ln N N ln(e x ) + N ln(1 + e x ) Again the same answer as before. = (N N + N ) ln N + (N + + N ) ln(1 + e x ) + Nxe x 1 + e x = N ln(1 + e x ) + Nxe x 1 + e x xe x S = Nk ln(1 + e x ) e x

22 8 Special Relativity Two powerless rockets are heading towards each other on a collision course. As measured by Sarah, a stationary Earth observer, Rocket A has speed 0.800c, Rocket B has speed 0.600c, both rockets are 50.0 m in length, and they are initially.5 Tm = m apart. Rocket A is piloted by Ann, a stationary observer in Rocket A s frame, and Rocket B is piloted by Bob. It will take Ann and Bob 50 minutes (in their respective frames) to evacuate their rockets. [h] a) [5 pts.] What is the proper length of Rocket A? b) [5 pts.] What is the velocity of Rocket A according to Bob? c) [5 pts.] What is the length of Rocket A according to Bob? d) [5 pts.] From Sarah s perspective, how much time will pass before the rockets collide? e) [5 pts.] According to Ann, how much time will pass before the collision? Will she escape?

23 Solution a) The proper length of Rocket A is L 0 = γl, where L is the length perceived by an observer moving with speed v with respect to Rocket A, and γ = 1/ 1 (v/c). γ = = 5 3 L A 0 = 5 (50 m) = 83.3 m 3 b) In Sarah s frame, if we take Rocket A to be moving in the positive x direction, its velocity is v A x = 0.800c. Then Rocket B is moving in the negative x direction with velocity v B x = 0.600c. The velocity of Rocket A as observed by Bob is given by v A x = va x v 1 v A xv/c, where vx A is Rocket A s velocity in Sarah s frame, and v is the velocity of Bob s frame with respect to Sarah s, i.e. v = vx B. vx A 0.800c ( 0.600c) = 1 (0.800c)( 0.600c)/c v A x = 0.946c. Note that an observer on Rocket A would measure Rocket B to be approaching at this speed, as well. c) Just as in part 1, the length of Rocket A in Bob s frame is related to the proper length of Rocket A by L A 0 = γla, where now 1 γ = = So the length of Rocket A as observed by Bob is L A = 1 (83.3 m) = 7.0 m d) Sarah observes that, in the time before the rockets collide, Rocket A travels a distance of 0.800c t, while Rocket B travels 0.600c t. The total distance traveled by the two rockets is D = (0.800c+0.600c) t =.5 Tm. t = m 1.4 ( m/s) t = s = 100 mins. e) The proper time interval, t 0, is related to another observer s coordinate time interval by t = γ t 0. We already calculated the Lorentz factor in part 1: γ = 5 3, so the time before the collision is t 0 = mins. = 60 mins. 5 She and her crew will survive.

24 9. Quantum mechanics Consider a non-relativistic spinless particle in a potential V (r). (a) [8 pts.] Using the Schrödinger equation, prove Ehrenfest s theorem for the time derivative of the expectation value O = ψ O ψ of an operator O(r, p, t) in a state ψ(t) : d dt O = i O [H, O] +. t (b) [10 pts.] Prove that the rate of change of the expectation value of the orbital angular momentum L is equal to the expectation value of the torque: where d L = N, dt N = r ( V ). (c) [7 pts.] Show that d L /dt = 0 for any spherically symmetric potential, i.e. for such a potential angular momentum is conserved. 1

25 9. Quantum mechanics Consider a non-relativistic spinless particle in a potential V (r). (a) Using the Schrödinger equation, prove Ehrenfest s theorem for the time derivative of the expectation value O = ψ O ψ of an operator O(r, p, t) in a state ψ(t) : d dt O = i O [H, O] +. t Using Schrödinger s equation i ( Ψ/ t) = Hψ, d dt Ψ O Ψ = Ψ t OΨ O + Ψ t Ψ + Ψ O Ψ t = 1 HΨ i OΨ + 1 O Ψ i OHΨ + t = i HO OH + O t. (b) Prove that the rate of change of the expectation value of the orbital angular momentum L is equal to the expectation value of the torque: where Component by component, d L = N, dt N = r ( V ). d dt L x = i [H, L x]. [H, L x ] = 1 [ ] p, L x + [V, Lx ]. m The first term is zero, because p is invariant under rotations; the second term would be zero too if V were a function of r = r only [this is part (b)]. In general, using [V, p x ] = i V x, [V, p y] = i V y, [V, p z] = i V z,

26 we find So [H, L x ] = [V, yp z zp y ] = y[v, p z ] z[v, p y ] = yi V z zi V y = i [r ( V )] x. d L x dt = [r ( V )] x. The same goes for the other two components, so d L dt = r ( V ) = N. (b) Show that d L /dt = 0 for any spherically symmetric potential, i.e. for such a potential angular momentum is conserved. Either proceed as above and notice that [H, L] = 0, or reason as follows. If V (r) = V (r), then V = ( V/ r)ˆr, and r r = 0, so d L /dt = 0. 3

27 10. Classical Physics A thin, incompressible rod with length l and negligible mass has two point masses A and B (each with mass m) attached at the two ends. The system is initially put at rest on a smooth horizontal surface. Another point mass C (with mass m and moving on the surface with initial velocity v 0 ) has an elastic collision with mass A at an angle θ (see the figure). After the collision, mass C is observed to bounce back (i.e., it moves in the opposite direction to its initial motion), with a velocity v 0. The motion of the AB system can be decomposed into a center-of-mass velocity V cm and a rotation with angular velocity ω around the center of mass. [N.B. The expressions in (a), (b), and (c) should be in terms of m, l, θ, v 0, v 0, V cm, and ω.] (a) [6 pts.] Write down the expression of momentum conservation of the whole system(a+b+c). (b) [6 pts.] Write down the expression of energy conservation of the whole system. (c) [6 pts.] Write down the angular momentum conservation of the whole system about the position of mass A (its position right before the collision). (d) [7 pts.] Based on the above three conservation equations, solve V cm, ω, and v 0. That is, express them in terms of m, l, θ, and v 0. 1

28 10. Classical Physics Solution (a) [6 pts.] Write down the expression of momentum conservation of the whole system. Before the collision, the total momentum is mv 0, along the CA direction. After the collision, the momentum from C is mv 0 (in the CA direction) and that from the system AB is mv cm (which should be in the CA direction). We then have mv 0 = mv 0 +mv cm. (1) (b) [6 pts.] Write down the expression of energy conservation of the whole system. In this problem, the total kinetic energy of the whole system is conserved. Before the collision, the kinetic energy is mv0 /. After the collision, the kinetic energy from C is mv 0 /. The kinetic energy from the system AB is the sum of the center-of-mass kinetic energy (m)vcm/ and the kinetic energy in the center-of-mass reference frame, [m(ωl/) /]. We then have 1 mv 0 = 1 mv (m)v cm + [ 1 m(ωl/) ]. () (c) [6 pts.] Write down the angular momentum conservation of the whole system about the position of mass A (its position right before the collision). Before the collision, the angular momentum of the whole system about point A is zero. After the collision, the angular momentum from C is zero. For the system AB, the angular momentum is the sum of the center-of-mass angular momentum (m)v cm (l/)sinθ and that in the center-of-mass reference frame, m(ωl/)(l/). We have 0 = (m)v cm (l/)sinθ m(ωl/)(l/). (3) (d) [7 pts.] Based on the above three conservation equations, solve V cm, ω, and v 0. The three equations in (a), (b), and (c) reduce to The solutions for V cm, ω, and v 0 are v 0 = v 0 +V cm, (4) v0 = v 0 +V cm + 1 (ωl), (5) ωl = V cm sinθ. (6) V cm = ω = v 0 = 3+sin θ v 0, (7) 4sinθ v 0 3+sin θ l, (8) cos θ 3+sin θ v 0. (9)