Quantum Physics III (8.06) Spring 2008 Solution Set 1

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1 Quantum Physics III (8.06) Spring 2008 Solution Set 1 February 12, Natural Units (a) (4 points) In cgs units we have the numbers a, b, c giving the dimensions of Q as [Q] = [gm] a [cm] b [s] c. (1) Using equation (2) from the Problem Set we can directly write [s] = [ ][ev ] 1, [cm] = [c][ ][ev ] 1, [gm] = [ev ][c] 2. (2) Inserting these into (1) we obtain α = b 2a β = b + c γ = a b c (3) One can also invert the above expressions to get a = β + γ, b = α + 2β + 2γ, c = α β 2γ (4) (b) (10 points) Below, we specify the values of a, b, c according to (1) and then calculate α, β, γ according to (3) [force]: The dimensions of force can be read from Newton s second law as [force] =[mass][acceleration] =[mass][length][time] 2. This gives a = 1, b = 1, c = 2 = α = 1, β = 1 γ = 2 = [force] = [c] 1 [ ] 1 [ev ] 2 [pressure]: The dimensions of pressure can be read from its definition [pressure] =[force]/[area] =[gm][cm] 1 [sec] 2. This gives a = 1, b = 1, c = 2 = α = 3, β = 3 γ = 4 = [pressure] = [c] 3 [ ] 3 [ev ] 4 1

2 [conductivity]: Conductivity for free electrons is given by ne 2 τ/m, where n is the number of electrons per unit volume, e is the electron charge, τ is the relaxation time and m is the electron mass. In order to find the units of conductivity we first need to find the units of charge in cgs. Using the Coulomb force law in cgs (F = q 1 q 2 /r 2, q 1,2 =charge, r=distance between charges), we find [charge] = [gm] 1/2 [cm] 3/2 [sec] 1. (5) Using the expression for the conductivity of free electrons given above we find [conductivity]=[sec] 1 and hence [conductivity] = [ ] 1 [ev ]. [magnetic moment]: The magnetic moment has dimension of [Energy]/[magnetic field]. Using the expression for the Lorentz force, F = q(v/c)b, we find [B]=[gm] 1/2 [cm] 1/2 [sec] 1 and hence [magnetic moment] = [c] 3/2 [ ] 3/2 [ev ] 1. [Note: If one uses the definition µ = IA where I is current and A is area, one would get an extra factor c for the unit for magnetic moment compared with the above answer. The difference is due to the choice of unit for magnetic field B. Full credit for this answer.] [viscosity]: Using the definition given in the Problem Set for viscosity we find [viscosity] =[gm][cm] 1 [sec] 1 = [viscosity] = [c] 3 [ ] 2 [ev ] 3 2. Planck scales (a) (3 points) In cgs and natural units, [G N ] = [gm] 1 [cm] 3 [s] 1 = [c] 5 [ ][ev ] 2. This gives [ev ] = [c] 5/2 [ ] 1/2 [G N ] 1/2. Hence using the results from problem 1 we can construct Notice that this allows us to express M P l =, (6) G N GN l P l = c 3, (7) GN t P l = c 5, (8) G N = M 2 P l (9) (b) (5 points) In natural units G N = GeV 2 5. Using (6) M P l = GeV c 2. In cgs units, M P l = gm. 2

3 The compton wavelength of a Planck mass particle is M P l c = GN c 3 = l P l. (10) t P l = l P l /c that is light travels a distance l P l in time t P l. In cgs units l P l = cm and t P l = s. (c) (4 points) m e = 0.511MeV/c 2 and the compton wavelength for the electron is λ e = /m e c. Then (taking positive values) E grav m e c 2 = G N m 2 e/λ e m e c 2 = G Nm 2 e = ( me M P l Thus for electrons, the gravitational effect is very small. (d) (3 points) E grav M P l c 2 = G NMP 2 l /l P l M P l c 2 = G NM 2 P l ) 2 = (11) = 1. (12) The above equation implies that for a Plank mass particle gravity becomes strong at a distance scale of its own Compton wavelength. Thus in understanding the quantum mechanical behavior of such a particle, one can no longer ignore the gravitational effects. In other words, one needs quantum gravity. [Note: full credit if h is used instead of in various expressions above.] 3. The Accelerating Universe (a) (3 points) The cosmological constant has units of energy density. The natural scale for the cosmological constant Λ is thus M Plank /lplank 3. M Plank and l Plank are the natural scales of mass and length in the unit system of, c and G N, which are M P l = G N GN l P l = We can now find the natural value of Λ: c 3 M P l /lp ev l = cm 3 (b) (2 point) In Planck units, the observed value is Λ = , which is an extremely tiny number. 4. Fermi energy, velocity and temperature of copper (a) (2 points) The Fermy energy of copper is E F = 2 2m (3ρπ2 ) 2/3 3

4 where ρ is the number of free electrons per unit volume. For copper there is one free electron per atom, therefore ρ = N Ad A 8.96 gm/cm atoms = = atoms/cm gm/mole mole where N A is the Avogadro s number, d is the density of copper and A is the atomic weight of copper. Substituting numbers we obtain (b) (2 points) E F 7.1eV erg. The corresponding Fermi velocity can be found from the relationship E F = 1 2 mv2 F. We have 2EF v F = m = c = c = cm/sec Since the Fermi velocity is much smaller than the speed of light we can safely assume that electrons in the copper crystal are nonrelativistic. Note that the obtained Fermi v F velocity is about c/137, i.e. is of order of e 2 /, the electron velocity in hydrogen atom. This agrees with what one expects on dimensional grounds. (c) (2 points) The Fermi temperature is given by T F = E F k B K 7.1eV. Therefore we can approximate the electron gas to be at zero temperature. (d) (2 points) The degeneracy pressure is P = (3π2 ) m ρ gm cm sec atm. 5. Free fermion gas in two dimensional well (6 points) Let the size of the square well be a, the number of electrons be N and the Fermi wavevector be denoted as k F. Following similar steps as the calculation done in class for a 3-dimensional fermion gas: (spin degeneracy) (area in k-space)/(area per point in k-space) = number of electrons = πkf 2 ( π a )2 = N, (13) where the factor of 2 comes from the spin degeneracy and the factor of 1/4 comes from requirement that we only count positive wavevectors (in 2 dimensions this becomes a restriction to the first quadrant in k x -k y space). Denoting the electron number density as σ = N/a 2, we find k F = (2πσ) 1/2. (14) 4

5 The Fermi energy is given by where m e is the electron mass. E F = 2 k 2 F 2m e = 2 πσ m e, (15) 6. White dwarfs, Neutron stars and Black holes (6 points) In this problem we treat the star in a constant density approximation ignoring selfconsistency which leads to a more complicated density profile. In parts a and b we assume the particles are nonrelativistic. (a) (3 points) Note: The derivation of the gravitational energy and the energy of a degenerate electron gas was done in class. The total energy was found to be: where E tot = k 2N 2 R + k 1N 5/3 R 2 k 1 = 3 2 f 5/3 10m e ( ) 2/3 9π, k 2 = 3G Nm 2 p. (16) 4 5 The white dwarf s radius is obtained by minimizing the total energy with respect to R, resulting to R white dwarf = 2k 1 k 2 N 1/3 = 7150 km. This is in fact very close to Earth s radius (a coincidence of course). The ratio of the mass densities of the white dwarf and the sun is ( ) 3 ρ white dwarf R Sun = = ρ Sun R white dwarf (b) In a neutron star the pressure is so great that the electrons have merged with protons to form neutrons, so we may assume that the star consists entirely from neutrons. The formula (16) holds if we replace m e with m p and use f = 1. The radius of the neutron star with the mass of the sun is R neutron star = 12.3 km. The neutron Fermi energy is E F = 2 k 2 F 2m = 2 2mR 2 ( 9πfM 4m p ) 2/3. To find if the neutrons in the neutron star should be treated as relativistic particles we calculate the ration of the Fermi energy to the rest energy of the neutron E F m p c 2 = 0.06, 5

6 thus the neutrons are nonrelativistic. Note: Consider a black hole of mass M. The quantity MG N c 2 has dimensions of length. Since it is the only quantity with the dimension of length which can be constructed out of G N,M and c it should therefore equal to the Schwarzschild radius, r s, up to a numerical constant (any object lyingwithin this radius from the center of gravity of the black hole cannotescape and will be devoured by the hole, this defines the surface of no return ). For a star with mass equal to the mass of the Sun, we can estimate r s MG N c km. 7. Free electron gas (a) (4 points) Here, we will treat the electrons as a gas inside a box of volume V and try to estimate when the Coulomb interactions between the electrons become important. The average volume per electron is V/N e. We can ignore the Coulomb interactions if per electron E degeneracy E Coulomb. (17) The energy of a degenerate electron gas was calculated in class and it was found to be An estimate of the Coulomb energy is given by E degeneracy = 3 2 kf 2. (18) 5 2m e E Coulomb e2 r e 2 e 2 ( ) 1/3 = V (3π 2 ) k 1/3 F, (19) N e where k F = ( ) 3π 2 1/3 Ne V and r is the average distance between electrons. Hence the approximation is valid if 3(3π 2 ) 1/3 10 k F e 2 1. (20) m e c (b) (2 points) As discussed in lecture, for mass equal to the Sun s mass we have k F /m e c 1. Since the numerical prefactor in (20), 3(3π 2 ) 1/3 /10, is also of order 1 and /e 2 137, then inequality (20) is satisfied and we can ignore the Coulomb interactions. Note: For electrons in copper, considered in Problem 4, the left hand side of inequality (20) gives us 0.67 and thus ignoring Coulomb interactions between electrons in copper may be suspect. 6

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MIDTERM #1 PHYS 33 (MODERN PHYSICS II) DATE/TIME: February 16, 17 (8:3 a.m. - 9:45 a.m.) PLACE: RB 11 Only non-programmable calculators are allowed. Name: ID: Please read the following instructions: This