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STONY BROOK UNIVERSITY DEPARTMENT OF PHYSICS AND ASTRONOMY Comprehensive Examination, August 31, 2011 General Instructions: Twelve problems are given; you should solve four problems. If you do more than four problems, you must choose which four should be graded, and only submit those four. You may do two problems from the same ๏ฌeld only once, except for Astronomy, for which you can do up to four problems. Each problem counts 20 points, and the solution should typically take less than 45 minutes. Some of the problems cover multiple pages. Make sure you do all the parts of each problem you choose. Use one exam book for each problem, and label it carefully with the problem topic and number and your name. You may use a one page help sheet, a calculator, and with the proctorโs approval, a foreign language dictionary. No other materials may be used. Some potentially useful information: The atomic mass of hydrogen is 1.00794 amu. The atomic mass of helium is 4.002602 amu. 1 amu is 1.66×10โ27 kg. ๐ = 2.998 × 108 msโ1 . ๐บ = 6.673 × 10โ11 m3 kgโ1 sโ2 . The solar luminosity is 3.85×1026 W. The mass of the Sun is 1.989×1030 kg The radius of the Sun is 7.00 × 108 meters โ = 1.055 × 10โ34 J ๐ = 1.602 × 10โ19 C ๐๐ต = 1.38 × 10โ23 J/K ๐๐2 โ 0.5 MeV โ๐ โ 197 MeV-fm. 1 Astronomy 1 Many astrophysical objects may be approximated as blackbodies, ideal emitters that absorb all energy incident on them and re-radiate it with a characteristic spectrum. At the turn of the 20th century, Max Planck synthesized earlier results with the new quantum theory and found the Planck function, ๐ต๐ (๐ ) = 2โ๐2 /๐5 , ๐โ๐/๐๐๐ โ 1 for power per unit area per unit solid angle per wavelength that gave good agreement with experimental results. In the expression, ๐ is the temperature, ๐ is the frequency, โ is Planckโs constant, and ๐ is Boltzmannโs constant. There are two important aspects of this expression, one concerning the peak emission and one concerning the total emission. a. (3 pts) Derive Weinโs law for the peak emission as a function of wavelength. Hint: you will have to ๏ฌnd a solution to a transcendental equation. b. (3 pts) By integrating the Planck function, derive an approximate expression for the total emission of a blackbody and how it depends on the temperature. c. (5 pts) With the aid of a sketch, combine these to describe the behavior of blackbody radiation as a function of wavelength and temperature. d. (3pts) Show that the Rayleigh-Jeans law, ๐ต๐ (๐ ) โ 2๐๐๐ , ๐4 is an approximation to Planck function at long wavelengths. e. (3 pts) RXJ 185635-3754 is a bright x-ray source observed by the ROSAT X-ray observatory that has been identi๏ฌed as an isolated neutron star. The distribution in x-rays is well-๏ฌtted by a blackbody with a peak at a wavelength of 4.39 × 10โ9 m. Determine the temperature of the corresponding blackbody. f. (3 pts) Subsequent observations found that the optical ๏ฌux is about seven times larger than the Rayleigh-Jeans estimate. How can this be reconciled? SOLUTION: (a.) To ๏ฌnd the maximum wavelength, maximize the function. ๐ ๐ 2โ๐2 /๐5 ๐ต๐ (๐ ) = =0 ๐๐ ๐๐ ๐โ๐/๐๐๐ โ 1 2 This gives 0 = 2โ๐ 2 [ โ5๐ โ6 ( โ๐/๐๐๐ ๐ โ1 )โ1 +๐ โ7 ( โ๐/๐๐๐ (โ1) ๐ โ1 )โ2 ( ( )โ1 )โ2 โ๐/๐๐๐ โ๐ + ๐โ7 ๐โ๐/๐๐๐ โ 1 ๐ = โ5๐โ6 ๐โ๐/๐๐๐ โ 1 ๐๐ )โ1 โ๐/๐๐๐ โ๐ ( โ๐/๐๐๐ โ1 ๐ = 5โ๐ ๐ ๐๐ ) ( โ๐/๐๐๐ โ๐ โ 1 โ ๐โ๐/๐๐๐ = 5 ๐ ๐๐๐ Let ๐ฅ = โ๐ ๐๐๐ โ๐/๐๐๐ ๐ ( โโ๐ ๐2 ๐๐ )] and with some algebra we may write ๐โ๐ฅ = 1 โ ๐ฅ/5 A little trial and error gives ๐ฅ โ 4.956 in addition to the trivial solution (๐ฅ = 0). Now solve for ๐๐ to have the form of Weinโs law: ๐๐ = โ๐ ๐ (4.956) Apply SI units โ = 6.626 × 10โ31 Js ๐ = 1.381 × 10โ23 J/K ๐ = 2.998 × 108 m/s we have ๐๐ โ 2.9 × 10โ3 mK , which is Weinโs law. (b.) To derive an expression for the total power emitted, integrate the Planck function, here in terms of frequency. โซ โ โซ โ 2โ๐ 3 /๐2 ๐๐ ๐ต๐ (๐ )๐๐ = ๐โ๐/๐๐ โ 1 0 0 โซ โ 2 (โ๐ 3 ) ๐๐ = 2 2 โ ๐ 0 ๐โ๐/๐๐ โ 1 Substitute ๐ฅ = โ๐ ๐๐ and ๐๐ฅ = โ ๐๐ ๐๐ and โซ 2 (โ๐ )4 โ ๐ฅ3 = ๐๐ฅ โ3 ๐2 ๐๐ฅ โ 1 0 โซ โ ๐ฅ3 2 (โ๐ )4 ๐ 4 ๐4 = where ๐๐ฅ = โ3 ๐2 15 ๐๐ฅ โ 1 15 0 5 4 ๐ 4 2๐ ๐ = ๐ , where ๐ = ๐ 15โ3 ๐2 3 Figure 1: Plot of energy density vs. wavelength that illustrates the behavior of the Plank function. Two things to note- as the temperature increases, the emission increases at all wavelengths (the lines do not cross) and the maximum emission occurs at shorter wavelengths. Image from Wikipedia. Note that the key is that power โ ๐ 4 . The integral is a constant. (c.)Sketch and description should resemble this: (d.) Start with the Planck function ๐ต๐ (๐ ) = With ๐๐ฅ โ 1 + ๐ฅ for ๐ฅ โช 1, we have 2โ๐2 /๐5 . ๐โ๐/๐๐๐ โ 1 2โ๐2 /๐5 1 + โ๐/๐๐๐ โ 1 2โ๐2 /๐5 = โ๐/๐๐๐ 2๐๐๐ = ๐4 ๐ต๐ (๐ ) โ Note that the expression is independent of โ and thus classical. (e.) First, how I came to this: The observation was a โ57 eV blackbody.โ Accordingly, 57 eV โ 6.6 × 105 K. 8.6173 × 10โ5 eV/K Applying Weinโs law (๐max ๐ โ 0.0029 mK), ๐max โ 0.0029 mK = 4.39 × 10โ9 m 5 6.6 × 10 K 4 ๐ โ 0.0029 mK 0.0029 mK = 6.6 × 105 K = ๐max 4.39 × 10โ9 m (f.) The Rayleigh-Jeans law applies to the long wavelength tail in which the emitted power is proportional to temperature. One obvious solution is that the temperature is not uniform across the neutron star. Another is an atmosphere on the NS that will redistribute energy of the radiation, so the assumption of a blackbody is not valid over the range from X-ray to optical. 5 Astronomy 2 a. (16 points) The Sun shines by nuclear fusion. Estimate the main sequence lifetime of the Sun. Show all steps and clearly state all assumptions. Use only the following information: โ The solar constant, the radiant ๏ฌux from the Sun measured at the top of Earthโs atmosphere, is 1.38 ×106 erg/cm2 /s. This is a direct observation. โ The orbital velocity of the Earth is 30 km/s. This can be determined from the aberration of starlight. โ The length of the year is 365.2422 days. โ m1 ๐ป = 1.0078 amu โ m4 ๐ป๐ = 4.0026 amu โ 1 amu = 1.66 ×10โ24 gm โ ๐ผ = 1/137 โ c = 2.99792×105 km/s โ G = 6.67×10โ8 cm3 secโ2 gmโ1 โ h = 6.62×10โ27 erg sec b. (4 pts) Write down the nuclear reaction(s) involved in the PP I chain. SOLUTION: a. Use the Earthโs orbital velocity to determine the distance from the Earth to the Sun. 2๐D = vt, where t is the length of the year b. Use the inverse-square law to determine the solar luminosity c. Use Newtonian gravity to determine the mass of the Sun; m๐ธ v2 /r = Gmโ m๐ธ /r2 d. The nuclear reaction is 41 H โ 4 He. Use the mass de๏ฌcit to determine the energy per reaction, and the number of reactions/sec needed to drive the solar luminosity. e. The lifetime of the sun depends on the number of H atoms. f. For full credit, realize that only about 10% of the solar H is fused into He For the PP I chain: 6 โ 1 H + 1 H โ 2 D + e+ + ๐๐ โ 2 D + 1 H โ 3 He + ๐พ โ 3 He + 3 He โ 4 He + 1 H + 1 H 7 Astronomy 3 Consider a neutron star which is undergoing a burst. The energy source for the burst is thermonuclear energy release underneath the surface layer of the star. Assume the con๏ฌguration during the burst remains spherically symmetric. The energy released by the burst is in the form of photons. What critical luminosity of photons could just begin to raise an optically thick but physically thin surface layer? What is the name associated with this luminosity? Assume the opacity of the material is due to electron scattering, so that it is ๐ ๐ = 0.2(1 + ๐) cm2 gโ1 , where ๐ is the hydrogen mass fraction. Observed from Earth, the maximum ๏ฌux observed from a bursting neutron star is often taken to be associated with this critical luminosity. During the later stages of the burst, the observed ๏ฌux and temperature both steadily decrease. Assume that the radiation during the later stages can be well-modeled by a blackbody. Show that with these assumptions, measurements of the maximum ๏ฌux and the ๏ฌux and temperature late in the burst can be used to determine the mass and radius of the neutron star if the distance to the star is known. SOLUTION: The critical luminosity is called the Eddington luminosity or the Eddington limit, and is given by 4๐๐บ๐ ๐ฟ๐ธ๐๐ = . ๐ ๐ This can be derived assuming a balance between the outward acceleration due to radiation pressure and gravitation. The energy ๏ฌux at a distance ๐ท from the starโs center is ๐ฟ ๐๐ธ = . ๐๐ด๐๐ก 4๐๐ท 2 The momentum of a photon is related to the energy by ๐ = ๐ธ/๐. Then the momentum transfer rate (the radiation force) to an electron is ๐๐ ๐๐ ๐๐ ๐ฟ๐๐ , = = ๐๐ก ๐๐ด ๐๐ก 4๐๐ท 2 where the Thomson cross section is ๐๐ . The gravitational force felt by a proton at the same distance is ๐บ๐๐๐ /๐ท 2 . Equating these, ๐ฟ ๐บ๐๐๐ = , 4๐๐ท 2 ๐ท2 ๐ฟ= 4๐๐บ๐๐๐ ๐ . ๐๐ Since the opacity is related to the cross section by ๐ ๐ = ๐๐ /๐๐ we have the desired result. 8 Observed from the Earth, the radiation is gravitationally redshifted, and the Eddington ๏ฌux is โ โ ๐ฟ๐ธ๐๐ 2๐บ๐ ๐๐บ๐ 2๐บ๐ 1 โ = 1 โ , ๐น๐ธ๐๐ = 4๐๐ท 2 ๐ ๐2 ๐ ๐ ๐ท 2 ๐ ๐2 where ๐ท is the distance and ๐ and ๐ are the neutron starโs mass and radius, respectively. Credit will be given if the Newtonian expression, without the redshift factor, is used. Late in the burst, if the radiation is approximately blackbody, the observed ๏ฌux ๐น๐๐๐ and temperature ๐๐๐๐ are related by ๐น๐๐๐ = 4๐๐ 2 2๐บ๐ โ1 4 ๐๐ต ๐๐๐๐ (1 โ ) . 2 4๐๐ท ๐ ๐2 Again, credit is given if the redshift factors are neglected. Forming the combination โ 4 ๐น๐ธ๐๐ ๐ ๐ท 2 ๐๐ต ๐๐๐๐ ๐บ๐ 2๐บ๐ = ๐ผ = (1 โ ), ๐บ๐3 ๐น๐๐๐ ๐ท 2 ๐ ๐2 ๐ ๐2 one can solve the quadratic for ๐บ๐/๐ ๐2 : ๐บ๐ 1 1โ ± = 1 โ 8๐ผ. ๐ ๐2 4 4 Credit is given for the Newtonian expression ๐บ๐/๐ ๐2 = ๐ผ. Then ๐ can be found from the formula for the Eddington ๏ฌux: ๐= 2๐บ๐ โ1/2 ๐ ๐ท 2 ๐น๐ธ๐๐ (1 โ ) . ๐๐บ ๐ ๐2 ๐ can be found from ๐ = ๐บ๐ ๐บ๐ โ1 ( ) . ๐2 ๐ ๐2 9 Astronomy 4 This problem derives the local instability condition of a rotating galactic disk, so called the Toomre instability. The Toomre instability is often characterized by the parameter ๐, which is โ ๐๐ /๐บฮฃ, where ๐, ๐ , ๐บ, and ฮฃ are the velocity dispersion, epicyclic frequency, gravitational constant, and surface density, respectively. We consider a small square region of in๏ฌnitesimal thickness in the galactic plane with the size ๐ฟ at a galactic distance of ๐ . The mass within the region is ๐ = ฮฃ๐ฟ2 . The region is on a circular orbit around the galaxy. The angular rotation speed and gravitational potential of the galaxy are ฮฉ(๐ ) and ฮฆ(๐ ), respectively. Four types of forces act on the boundaries of the region: the galactic tidal force, centrifugal force, self-gravity within the region, and internal pressure. Consider only forces parallel to the galactic radius vector, and neglect any internal pressure. For example, the galactic tidal force stretches the region; comparing the gravitational forces of the galaxy at radii ๐ + ๐ฟ/2 and ๐ โ ๐ฟ/2 the galactic tidal force is ( ) ( ) ๐ฮฆ ๐ฮฆ ๐2 ฮฆ ๐G = โ = ๐ฟ (1) ๐๐ ๐ +๐ฟ/2 ๐๐ ๐ โ๐ฟ/2 ๐๐ 2 Here we consider the changes of these forces when the region is perturbed in the direction of galactic radius by ฮ๐ฟ, where ฮ๐ฟ << ๐ฟ. A positive ฮ๐ฟ corresponds to a compression of the region. a. Derive the change of the galactic tidal force (๐G ). Does it increase or decrease? b. Derive the change of the centrifugal force (๐c = ฮฉ2 ๐ ). Does it increase or decrease? For this question, use the conservation of angular momentum (๐ฝ = ฮฉ๐ 2 ). c. Derive the change of the self-gravity of the region (๐g ). State whether the self-gravity increases or decreases. d. Compare the above three forces, ฮ๐G , ฮ๐c , and ฮ๐g , and derive the threshold size ๐ฟcrit for the region to become unstable. Use the epicyclic frequency ๐ 2 = ๐2 ฮฆ/๐๐ 2 + 3ฮฉ2 . e. In addition to this condition, the Jeans criterion provides another โ condition for instability. It is derived by comparing the free-fall time ๐ก๏ฌ = ๐ฟ/๐บฮฃ and ๐กcross = ๐ฟ/๐. The region is unstable when ๐ก๏ฌ < ๐กcross , which gives the Jeans length ๐J : ๐2 ๐๐ฝ = <๐ฟ (2) ๐บฮฃ From this condition and the condition from (d), derive the Toomre Q parameter and Toomreโs instability condition. 10 SOLUTION: (a) ๐2 ฮฆ ๐2 ฮฆ ๐2 ฮฆ (๐ฟ โ ฮ๐ฟ) โ ๐ฟ = โ 2 ฮ๐ฟ ฮ๐G = ๐๐ 2 ๐๐ 2 ๐๐ The tidal force that acts on the region decreases. (b) ฮ๐ฝ = ฮ(ฮฉ๐ 2 ) = ฮฮฉ๐ 2 + 2ฮฉ๐ ฮ๐ = 0 ฮ๐c = ฮ(ฮฉ2 ๐ ) = 2ฮฉฮฮฉ๐ + ฮฉ2 ฮ๐ = โ3ฮฉ2 ฮ๐ = โ3ฮฉ2 ฮ๐ฟ The centrifugal force decreases. (c) ๐บ๐ ๐บ๐ ๐บ๐ ฮ๐ฟ ฮ๐ฟ + 2 โ โ2 2 ฮ๐g = โ = โ2๐บฮฃ 2 (๐ฟ โ ฮ๐ฟ) ๐ฟ ๐ฟ ๐ฟ ๐ฟ (3) (4) (5) (6) The self-gravity decreases. (d) โฃฮ๐G + ฮ๐c โฃ < โฃฮ๐g โฃ, and ๐ฟ < ๐ฟcrit = 2๐บฮฃ ๐ 2 (7) (8) (e) and ๐J < ๐ฟcrit (9) 1 ๐๐ <1 ๐= โ 2 ๐บฮฃ (10) 11 AMO 1 The interaction of atoms with a laser ๏ฌeld can be described in a simple classical model (Lorentz model). In each atom, the bound electron subjected to an oscillating electric ๏ฌeld โฐ = โฐ0 ๐๐๐๐ก is assumed to respond like a damped harmonic oscillator with an undamped resonance frequency ๐0 , ๐ (11) ๐ฅ¨(๐ก) + ๐ฝ ๐ฅ(๐ก) ห + ๐02 ๐ฅ(๐ก) = โฐ(๐ก) ๐ a. (8 pts) By considering the atomic dipole moment ๐๐ฅ(๐ก) โก ๐ผโฐ(๐ก), show that the real and imaginary parts of the polarizability ๐ผ are given by 1 ๐2 ๐ฝ 2 2 2๐๐๐ ๐ฟ + ๐ฝ 2 /4 ๐2 ๐ฟ ๐ ๐(๐ผ) = โ , 2 2๐๐๐ ๐ฟ + ๐ฝ 2 /4 ๐ผ๐(๐ผ) = โ (12) (13) near resonance with ๐ฟ โช ๐ + ๐0 where ๐ฟ โก ๐ โ ๐0 is the detuning. b. (5 pts) What physical process(es) give(s) can give rise to the damping coe๏ฌcient ๐ฝ in a gas of atoms? Write an expression for ๐ฝ (gas density ๐, atomic mass ๐, collisional cross-section ๐, temperature ๐ ), accounting for all of the terms. c. (3 pts) Now consider the interaction ฮ = 12 ๐ผโจ๐ ๐(โฐ 2)โฉ. What does the real part of ฮ represent? What does it mean that ฮ is complex? d. (4 pts) Derive an expression for the rate at which an atom scatters photons from the laser beam. SOLUTION: a. Using the ansatz ๐ฅ(๐ก) = ๐ฅ0 ๐๐๐๐ก yields ๐ผ = (๐2 /๐)/ [๐02 โ ๐ 2 + ๐๐ฝ๐]. Sorting into real and imaginary parts, and making the approximation ๐02 โ ๐ 2 โ 2๐0 ๐ฟ, near resonance for ๐ฟ โช ๐ + ๐0 , lead to the desired results. b. 1. Radiative damping (i.e. spontaneous emission). Note: The radiative damping rate may be โ changed by putting the atoms โin a cavity. 2. Collisions between atoms. ๐ฝ = ฮ2 + (๐๐๐ฃ๐กโ )2 , where ๐ฃ๐กโ = ๐๐ต ๐ /๐. โ Note: The damping coe๏ฌcient also depends on the e๏ฌects absorption and stimulated emission (saturation; power broadening), which is however not part of the Lorentz model. c. Real: AC Stark Shift (Optical potential), Imaginary: damping. d. Calculate scattering rate ฮ๐ ๐ using 2โฃ๐ผ๐(ฮ)โฃ = โฮ๐ ๐ 12 AMO 2 An optical mug for atoms (โgravito-optical surface trapโ; cf. Yu. B. Ovchinnikov, I. Manek, R. Grimm, PRL 79, 2225 (1997)) can be implemented above a glass surface as illustrated in ๏ฌg.1. The surface is covered with an evanescent light ๏ฌeld, combined with a hollow laser beam. The evanescent ๏ฌeld, which is formed by total internal re๏ฌection at the glass surface, gives rise to a repulsive optical potential ๐(๐ง) = ๐0 ๐โ2๐ง/๐ , where ๐ง is the distance above the surface and ๐ is the decay length of the evanescent ๏ฌeld. The repulsive optical potential of the hollow laser beam (radius ๐ ), which passes through the prism vertically, forms the walls of the mug. a. (5 pts) The combined potential in ๐ง direction resulting from the combination of gravity and the repulsive wall is approximately wedge shaped. Derive an expression for the density distribution ๐(๐ง) for an atomic gas of ๐ atoms (mass ๐) at temperature ๐ in this potential. What is the peak density ๐0 of the gas? b. (5 pts) Calculate the corresponding phase space density ๐ท, and give an estimate for the critical temperature ๐๐ at which quantum degeneracy sets in. c. (5 pts) At low temperature, the details of the shape of the potential near the surface become important, as most atoms reside near the potential minimum. Calculate the frequency ๐๐ง for harmonic oscillations around the potential minimum. For which range of temperatures is the gas 2D?. d. (5 pts) How could you conveniently implement evaporative cooling in the setup, and how would it work? 13 SOLUTION: โ๐ง/๐ง0 , with ๐ง0 = ๐๐ต ๐ /๐๐. From a. Barometric density 0๐ โซ 3 distribution ๐(๐ง) = ๐๐๐ normalization ๐ ๐ ๐ = ๐ obtain ๐0 = ๐๐ 2 ๐๐ต (๐/๐ ) โ b. Phase-space density ๐ท = ๐0 ฮ3 with thermal de-Broglie wavelength ฮ = โ/ 2๐๐๐๐ต ๐ . Obtain ๐ท โผ ๐๐ โ5/2 , and ๐๐ from ๐ท โผ 1. 2๐0 . c. Total 1D potential ๐(๐ง) = ๐0 ๐โ2๐ง/๐ + ๐๐๐ง has minimum at ๐ง = ๐ง0 = 2๐ ln ๐๐๐ 1 1 โฒโฒ 2 2 2 Harmonic expansion around ๐ง0 , i.e. 2 ๐๐๐ป๐ (ฮ๐ง) = 2 ๐ (๐ง0 )(ฮ๐ง) , yields โ ๐๐ป๐ = 2๐/๐. Gas is 2D for ๐๐ต ๐ < โ๐๐ป๐ . a. Evaporative cooling relies on the escape of atoms with more than the average energy, and subsequent rethermalization of the remaining atoms in the gas. This can be implemented by reducing the laser power of the evanescent ๏ฌeld such that those atoms can escape toward the surface. It can also be done by reducing the power of the hollow beam, but the e๏ฌciency will be lower. 14 Condensed Matter 1 The lower left plot below is the phonon dispersion relation of neon, measured by Leake et al. (Phys. Rev. 1969) by inelastic neutron scattering. The crystal has ๐ ๐๐ structure. The lattice constant is ๐ = 4.43ห ๐ด, and the primitive vectors are a = ๐(0, 1/2, 1/2), b = ๐(1/2, 0, 1/2), c = ๐(1/2, 1/2, 0). a. What is the velocity of longitudinal sound waves in the (111) direction? b. Consider the LA mode at the k = (1/2, 1/2, 1/2) point of the Brillouin zone, designated โLโ in the picture and at the top of the phonon graph. Its energy is 6.7 meV. What is its group velocity? (Note that the notation is an abbreviation for k = (2/๐)(1/2, 1/2, 1/2).) c. Consider the same LA mode. What is the normal mode displacement pattern? (Explain carefully in words, and provide at least a simpli๏ฌed picture.) d. Explain why three โbranchesโ are shown to the left of k = (0, 0, 0) and only two to the right. e. What are the frequencies of the phonons at k = (1, 1, 1) and at k = (0, 0, 1) (abbreviated notation; neither is shown explicitly on the graph.) 15 SOLUTION: (a.) The slope in the limit ๐ โ 0 is 9.5๐๐๐ /โ ฮ๐ = 2๐ 1 1 1 = 1.17 × 103 ๐/๐ ฮ๐ ( , , ) ๐ 2 2 2 (b.) ๐๐/๐๐ = 0 (c.) The (111) planes displace perpendicular to the planes parallel to (111) direction by alternate amounts ±๐ข. (d.) 120 degree rotational symmetry around the (111) direction (to the โrightโ of (000) in the graph) requires that the two transverse branches are degenerate. In the (110) direction (to the left of (000)) there is only 180 degree rotational symmetry so transverse branches have di๏ฌerent velocities. (e.) Since (2๐/๐)(111) is a reciprocal lattice vector, it is โthe sameโ state as ๐ = 0 and the frequencies are 0. But ๐ = (2๐/๐)(001) is not a reciprocal lattice vector. Therefore we can subtract the reciprocal lattice vector (2๐/๐)(โ1, โ1, 1) to convert ๐ to (2๐/๐)(110), the โX pointโ, where the frequencies are 7 meV (LA branch) and 4.8 meV (doubly degenerate TA) 16 Condensed Matter 2 The free energy density of a superconductor is approximated in the Ginzburg-Landau theory by ,( ) ,2 2 , ๐โ 1 ,, โ 2 4 , +โ . โ โ ๐จ ฮจ ๐๐ (๐) โ ๐๐ = ๐ โฃฮจโฃ + ๐ โฃฮจโฃ + , 2๐โ , ๐ ๐ 8๐ Here ๐โ , ๐โ are phenomenological charge and mass of supercurrent carriers, ๐(๐ ) and ๐(๐ ) are temperature dependent constants and ฮจ(๐) is the โwave functionโ of the superconducting condensate and ๐๐ is a free energy of the normal state at the same temperatures. ๐จ(๐) is vector potential of an external magnetic ๏ฌeld โ. a. Assuming that close to some temperature ๐๐ the temperature dependence of the coe๏ฌcient ๐(๐ ) is not signi๏ฌcant, ๐(๐ ) โ ๐ > 0, and ๐(๐ ) โ ๐ผ(๐ โ ๐๐ ), show that ๐๐ is the critical temperature of the superconducting phase transition in the absence of external ๏ฌelds. Find the temperature dependence of the โsuperconducting order parameterโ, de๏ฌned as ฮ โก โฃฮจโฃ, close to the critical temperature ๐๐ . b. A strong external magnetic ๏ฌeld โ > ๐ป๐ (๐ ) destroys superconductivity. Under the assumption that the surface e๏ฌects can be neglected, and that the magnetic ๏ฌeld is uniform, ๏ฌnd the dependence ๐ป๐ (๐ ) close to ๐๐ by comparing the condensation energy (the minimum of the above free energy without โ) with the energy of magnetic ๏ฌeld ๐ป๐2 /8๐. ๐ป๐ (๐ ) is called the critical magnetic ๏ฌeld of a superconductor. c. Let us look for the solution ฮจ(๐) under the following conditions: (i) ฮจ(๐) = ๐ (๐ฅ), โ = 0, ฮจ(๐ฅ < 0) = 0, ฮจ(๐ฅ โ +โ) = ฮ with ฮ found in a. These conditions describe the interface between a superconductor and the vacuum (vacuum is at ๐ฅ < 0). Write down an equation for ๐ (๐ฅ). Extract from this equation the typical length ๐ (known as the coherence length) at which order parameter ฮจ can change signi๏ฌcantly. Solve the equation or at least sketch the solution qualitatively. What is the temperature dependence of ๐? d. Using the free energy density ๐๐ describe qualitatively the reason why weak magnetic ๏ฌeld does not penetrate the superconductor (โMeissner e๏ฌectโ). Apply this reasoning and the standard properties of the vector potential to derive โ๏ฌux quantizationโ condition for allowed values of magnetic ๏ฌux threading a superconducting loop. Take into account that the microscopic BCS theory shows that the supercurrent is carried by โCooper pairsโ, so that ๐โ = 2๐. 17 SOLUTION: (a.) Let us assume that an external ๏ฌeld is absent โ = 0 and ๐จ = 0. We look for a uniform solution ฮจ = ๐๐๐๐ ๐ก. Minimizing the free energy with respect to โฃฮจโฃ2 we have ( ๐ )1/2 ( ๐ผ )1/2 โ = 2๐ ๐๐ โ ๐ . ๐ + 2๐โฃฮจโฃ2 = 0 which gives ฮ = โ 2๐ 2 (b.) The minimum of the free energy at the optimal value of ฮ derived in a. is โ ๐4๐ . This is so called โcondensation energyโ. Together with the energy of diamagnetic 2 โ2 currents in the presence of an external ๏ฌeld we have ๐๐ โ ๐๐ โ โ ๐4๐ + 8๐ . This energy ( )1/2 โ 2 = 2๐๐ผ/๐ (๐๐ โ ๐ ). This means that for is positive for โ > ๐ป๐ (๐ ) = 2๐ ๐๐ โ > ๐ป๐ (๐ ) it is favorable to have a normal state and superconductivity is destroyed. 2 โ2 โฒโฒ (c.) โ 2๐ + ๐๐ + 2๐๐ 3 = 0 or โ๐ 2 ๐ โฒโฒ + ๐ 3 = ๐, where ๐ = ๐ /ฮ and ๐ 2 = 2๐โโ โฃ๐โฃ so โ๐ that ๐ โผ (๐๐ โ ๐ )โ1/2 . ( ๐ )1/2 . Let us (d.) As we saw the condensation energy argument favors โฃฮจโฃ = ฮ = โ 2๐ consider the free energy where the absolute value of the wave function is ๏ฌxed while the phase can change in space, i.e., ฮจ = ฮ๐๐๐ : ( )2 ๐โ โ2 ฮ2 โ๐ โ ๐จ . ๐ฟ๐ = 2๐โ โ๐ We can see that the presence of magnetic ๏ฌeld inside the superconductor the vector potential can not be compensated by the gradient of the phase (as โ × ๐จ โ= 0) and the weak magnetic ๏ฌeld is repelled from โฎ the superconductor. โ โฎ โ Considering the loop integral 2๐๐ = โ๐ โ ๐๐ = ๐โ๐ ๐จ๐ โ ๐๐ = ๐โ๐ ฮฆ we see that the total magnetic ๏ฌux ฮฆ through a loop is quantized in units of ฮฆ0 = โ๐ โ๐ . = ๐โ 2๐ 18 Nuclear 1 Consider the scattering of a proton with energy ๐ธ = 250 GeV o๏ฌ a Gold nucleus at rest. Estimate the total and elastic cross sections of proton-nucleus scattering, and sketch the di๏ฌerential cross section as a function of the scattering angle. Will the di๏ฌerential cross section have a minimum? SOLUTION: 1. The mean free path of a proton in the nucleus can be estimated as ๐ โ (๐๐)โ1 , where ๐ โ 0.18 fmโ3 is the nuclear density, and ๐ โ 3.4 fm2 is the protonnucleon inelastic cross section. Note that a rough estimate for this cross section can be obtained by simply writing ๐ โ ๐๐ ๐2 and using ๐ ๐ โ 1 fm as a proton radius. Numerically, we get for the mean free path ๐ โ 1.6 fm which is much smaller than the radius of the Gold nucleus, ๐ ๐ด โ 1.1 × (197)1/3 fm โ 6.4 fm. Because the mean free path is short, the nucleus acts as a black disk. The ๐ด 2 total, elastic and inelastic cross sections are thus ๐๐ก๐๐ก โ 2๐๐ ๐ด โ 2.6 barn, ๐ด ๐ด 2 โ13 ๐๐๐ โ ๐๐๐๐๐ โ ๐๐ ๐ด โ 1.3 barn. Units: 1 fm = 10 cm; 1 barn = 10โ28 cm2 = 2 100 fm . 2. The angle at which the di๏ฌerential cross section for the scattering o๏ฌ a black disk has a di๏ฌractive minimum is ๐๐ โ 1/(๐ ๐ ๐ด ). Using the protonโs momentum ๐ โ 250 GeV โ 1250 fmโ1 and the Gold radius ๐ ๐ด โ 6.4 fm we estimate ๐๐ โ 10โ4 rad. 19 Nuclear 2 Consider three amplitudes of pion-nucleon scattering: ๐ + โก ๐(๐ + ๐ โ ๐ + ๐), ๐ โ โก ๐(๐ โ ๐ โ ๐ โ ๐), and ๐ 0 โก ๐(๐ โ ๐ โ ๐ 0 ๐) โ Assuming isosping invariance, ๏ฌnd the relation between these scattering amplitudes โ When the center-of-mass pion-nucleon energy is close to the mass of the ฮ resonance (isospin ๐ผ = 3/2), it dominates the scattering process. Find the ratio of the cross sections of the three scattering processing above in this energy range. SOLUTION: a) The isospin of the pion is ๐ผ = 1, with ๐โ+, ๐ 0 and ๐ โ having the isospin projections ๐ผ๐ง = +1, 0, โ1 respectively. The isospin of the nucleon is ๐ผ = 1/2, with proton and neutronโs isospin projections ๐ผ๐ง = +1/2, โ1/2. Therefore the three pion-nucleon states that enter as the initial and/or ๏ฌnal states can be decomposed in the isospin space as โฃ๐ + ๐ > = โฃ3/2, 3/2 > โ โ 1/3โฃ3/2, โ1/2 > โ 2/3โฃ1/2, โ1/2 > โฃ๐ โ ๐ > = โ โ โฃ๐ 0 ๐ > = 2/3โฃ3/2, โ1/2 > + 1/3โฃ1/2, โ1/2 > Isospin invariance implies that the scattering amplitudes depend only on the total isospin, but not on the isospin projection. We can thus de๏ฌne two isospin amplitudes ๐3/2 and ๐1/2 , and express the scattering amplitudes of interest as ๐+ = ๐3/2 ๐โ = (1/3)๐3/2 + (2/3)๐1/2 โ โ ๐0 = ( (2)/3)๐3/2 โ ( 2/3)๐1/2 Now it is easy to ๏ฌnd the relation that we are looking for โ 2๐0 + ๐โ = ๐+ b) The dominance of the ฮ(๐ผ = 3/2) resonance allows us to neglect the ๐ผ = 1/2 scattering amplitude. In this case the scattering amplitudes are related as โ ๐+ : ๐โ : ๐0 = 1 : 1/3 : 2/3 20 Neglecting the small mass di๏ฌerences (in accord with the assumed isospin invariance) then yields the ratio of the cross sections: ๐+ : ๐โ : ๐0 = 9 : 1 : 2. This prediction agrees quite well with the experimental data. 21 High Energy 1 Consider Compton scattering, the process of electron-photon scattering, ๐โ (๐) + ๐พ(๐) โ ๐โ (๐โฒ ) + ๐พ(๐ โฒ ) a. (5 pts) Write down the Feynman diagrams which contribute at the tree graph level, and the corresponding amplitudes in terms of momenta, spinors and polarization vectors ๐๐ (๐) and ๐๐ (๐ โฒ ). b. (5 pts) Prove gauge invariance: show by explicit calculation that the sum of the amplitudes vanishes if one replaces one of the two polarization vectors of the photons by its 4-momentum. c. (5 pts) Now consider quark-gluon scattering. Let the incoming gluon have color index a, and the outgoing gluon color index b. There is a new Feynman diagram that contributes. What is it? How are the previous amplitudes modi๏ฌed to take color into account? d. (5 pts) Now prove again gauge invariance. First write the amplitude for the new diagram in terms of the 3-gluon vertex V. Then substitute the expression for V. Hint: the 3-gluon vertex is equal to: ๐๐๐ ๐๐๐๐ (๐, ๐, ๐) = ๐๐ ๐๐๐ [๐ ๐๐ (๐ โ ๐)๐ + ๐ ๐๐ (๐ โ ๐)๐ + ๐ ๐๐ (๐ โ ๐)๐ ] if all momenta are incoming. (a.) For diagram I we have ] โ๐(/๐ + ๐/ ) + ๐ ๐ข¯(๐ )/๐(๐ ) /๐(๐)๐ข(๐) (๐ + ๐)2 + ๐2 โฒ โฒ [ (14) and for diagram II we have ] โฒ / โ๐( ๐ โ ๐ ) + ๐ / ๐ข¯(๐โฒ )/๐(๐) /๐(๐ โฒ )๐ข(๐) โฒ 2 2 (๐ โ ๐ ) + ๐ [ (15) (See ๏ฌgures (2) and (3) for the associated Feynman diagrams.) (b.) Replacing /๐(๐) by ๐/, and substituting ๐/ = (๐/ + /๐ โ ๐๐) โ (/๐ โ ๐๐) in (14), and ๐/ = (๐/ โ /๐โฒ + ๐๐) + (/๐โฒ โ ๐๐) with ๐ โ ๐โฒ = โ๐ + ๐ โฒ in (15), and using (/๐ โ ๐๐)๐ข(๐) = 0 ๐ข¯(๐โฒ )(/๐โฒ โ ๐๐) = 0 ] [ ][ ] [ โ๐(/๐ + ๐/) + ๐ /๐ + ๐/ โ ๐๐ = โ๐ (๐ + ๐)2 + ๐2 22 (16) (17) one can see that the propagators cancel, and one ๏ฌnds ๐ข¯(๐โฒ )/๐โฒ (๐ โฒ )(โ๐)๐ข(๐) for (14) and ๐ข¯(๐โฒ )(๐)/๐โฒ (๐ โฒ )๐ข(๐) for (15). The sum clearly vanishes. (c.) In (14) and (15) one must insert color matrices ๐๐ ๐๐ and ๐๐ ๐๐ , respectively, giving an extra factor ๐๐๐ ๐ ๐๐ , so that the sum of diagrams (2) and (3) becomes equal to ๐ข¯(๐โฒ )(๐)/๐โฒ (๐ โฒ )๐ข(๐)๐๐๐ ๐ ๐๐ . (18) (d.) The new Feynman diagram is in ๏ฌgure (4). The new amplitude is ๐ฟ ๐๐ (โ๐) ๐๐๐ ๐๐๐๐ (๐, โ๐ โฒ , ๐ โฒ โ ๐)๐๐ (๐)๐๐ (๐ โฒ ) ๐ข¯(๐ )๐พ ๐๐ ๐ข(๐) โฒ 2 (๐ โ ๐ ) (19) ๐๐๐ (๐, โ๐ โฒ , ๐ โฒ โ ๐) = ๐๐ ๐๐๐ ๐๐ (๐ โฒ ) [๐๐ (โ๐ + ๐ โฒ )๐ โ 2๐ โ ๐ โฒ ๐ฟ๐ ๐ ] ๐ ๐ ๐๐ (๐ โฒ )๐๐๐๐ (20) โฒ ๐ One ๏ฌnds The term with (โ๐ +๐ โฒ )๐ = (๐โ๐โฒ )๐ gives no contribution due to current conservation at the quark-gluon vertex, while the term with ๐ โ ๐ โฒ cancels the contribution of the ๏ฌrst two graphs (again the propagator (๐ โ ๐ โฒ )โ2 = (โ2๐ โ ๐ โฒ )โ1 cancels). ๐โฒ ๐ ๐+๐ ๐ ๐โฒ Figure 2: Diagram I 23 ๐ ๐โฒ ๐ โ ๐โฒ ๐โฒ ๐ Figure 3: Diagram II ๐ ๐โฒ ๐ โ ๐ โฒ = ๐โฒ โ ๐ ๐ ๐โฒ Figure 4: Diagram III 24 High Energy 2 Jets are collimated sprays of hadrons emitted in high-energy collisions. They are the direct o๏ฌspring of high-energy quarks and gluons that are โkicked outโ in a highenergy collision. Such collisions occur, for instance, at the Large Hadron Collider (LHC), a proton synchrotron accelerator/collider of 27 km circumference near Geneva, Switzerland. At the LHC counter-rotating proton beams of 1014 protons /beam are accelerated to 3,500 GeV/proton. The beams circulate and are brought into collision at several interaction regions around the ring where sophisticated particle detectors are located. a. Describe how protons are accelerated to energies in this range in a proton synchrotron. b. A jet is created when a quark or gluon is ejected from a proton during a violent collision a. List all types of quarks that have been discovered so far and their charges and mass estimates b. Have free quarks or gluons been observed? What mechanism has been proposed to describe this? c. Describe the process of jet creation, i.e. the creation of a collimated spray of hadrons (pions, kaons, protons, neutrons and many other types of hadrons) c. The simplest Feynman diagrams that describe the interactions between the quarks and gluons in two colliding protons are the so-called โtwo-to-twoโ processes. a. Sketch two Feynman diagrams for processes where two gluons (one from either proton) produce a pair of gluon jets. b. Sketch two Feynman diagrams for processes where a quark and antiquark (one from either proton) produce two gluon jets. c. Sketch two Feynman diagrams for processes where a gluon and a quark (one from either proton) produce two jets. d. Assuming that a quark carries - on average - one-sixth of the protonโs energy, calculate the maximum transverse energy a quark jet from an average collision may have. e. We now discuss measuring the direction and energy of a jet in a collider detector. 25 a. The primary energy loss mechanism for particles in a jet which interact with a detector is the strong nuclear interaction. Given that a typical cross section for a hadronic interaction of a pion on a nucleon is around 40 mbarn (1 barn = 10โ28 m2 ), estimate the mean free path of a pion in iron (๐ด๐น ๐ = 55, ๐๐น ๐ = 7.87 g/cm3 , ๐๐ด = 6.022 × 1023 ) b. Describe and motivate the layout and structure of a typical collider detector. SOLUTION: (a.) The protons are accelerated by radio-frequency electric ๏ฌelds. The beams are kept at in a circle of constant radius by increasing the strength of magnetic ๏ฌelds in bending magnets at a rate proportional to the proton momentum. (b.) Quark Charge (units of e) Mass (GeV/c2 ) up down +2/3 -1/3 0.05 0.05 charm +2/3 1.7 strange top bottom -1/3 +2/3 -1/3 0.5 175 5 Free quarks have not been observed because of โcolor con๏ฌnementโ, the idea that the strength of the strong force increases with separation distance between quarks. The only free states are bound states of two or three quarks with no net color charge. Jet creation arises as quarks (or gluons) with signi๏ฌcant relative momentum move apart. The strong interaction force increases to the point it becomes enegetically favorable (๐ธ = ๐๐2 ) to create new quarks โin betweenโ the initial quarks. This process continues until the quarks have low enough relative momentum that they form colorless bound states. (c.) ๐๐ โ ๐๐ processes: 26 ๐๐ โ ๐๐ processes: ๐๐ โ ๐๐ processes: (d.) The largest transverse momentum occurs when the jet is at 90๐ to the beam, so โ that all of the momentum appears in the transverse plane. For this case, with ๐ = ๐ฅ1 ๐ฅ2 ๐1 ๐2 with ๐ฅ๐ the fraction of the proton momentum ๐1 = ๐2 = ๐ equal to ๐ฅ๐ = (1/6)๐, ๐๐ = ๐/6 (e.) The mean free path โ is given by โ = (๐ × ๐ ๐๐๐๐)โ1 with ๐ the nucleon density. The nucleon density is related to the given constants by ๐ = ๐๐๐ด /๐ with ๐ โ 55, the molar mass. The result is โ โ 2 mm. 27