Physics 161 Homework 2 - Solutions Wednesday August 31, 2011

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1 Physics 161 Homework 2 - s Wednesday August 31, 2011 Make sure your name is on every page, and please box your final answer. Because we will be giving partial credit, be sure to attempt all the problems, even if you don t finish them. The homework is due at the beginning of class on Wednesday, September 7th. Because the solutions will be posted immediately after class, no late homeworks can be accepted! You are welcome to ask questions during the discussion session or during office hours. 1. Draw the Feynman diagram representing electron-electron scattering to first order in the coupling constant note: there are two such diagrams). Draw three distinct second order-diagrams. A first-order Feynman diagram in electrodynamics) involves the exchange of a single photon. There are two diagrams involving the exchange of the photon between two electrons. The first is seen in the first diagram below, where the two electrons move towards each other, exchange a photon, then move apart. However, since the electrons are identical particles, and indistinguishable, we can t tell which electron ends up where. So, the second diagram differs by the exchange of the final electrons, as seen in the diagram to the right in the figure. e A second-order diagram involves the exchange of two photons. There are many such diagrams, four of which are seen below. The first diagram involves the emission of a photon by an electron, before it exchanges one with the other electron. The first electron then interacts with the second electron, and only then reabsorbs the original photon. This diagram has the net effect of renormalizing the electron s magnetic moment. The second diagram has the first electron

2 e emitting and reabsorbing a photon, before it interacts with the other electron. This diagram renormalizes the mass of the electron. In both diagrams, the electron interacts with its own field. e The third diagram above simply shows the interaction between two electrons via the exchange of two photons. The final diagram shows the interaction between the electrons via the exchange of one photon, which then forms a virtual electron-positron pair. The pair recombines back into a photon, only then being absorbed by the second electron. This final diagram leads to a renormalization of the electron s charge. Note, of course, that there are plenty of variations on these diagrams differing by the exchange of the final electrons, or by putting the photon in the second diagram on a different electron, etc.

3 2. a) How many different meson combinations can you make with 1, 2, 3, 4, 5, or 6 different quark flavors recalling that there are three different colors for each quark and that any observable particle must be colorless)? What s the general formula for n flavors? b) How many different baryons combinations can you make with 1, 2, 3, 4, 5, or 6 different quark flavors? What s the general formula for n flavors? a) Suppose that we start with one photon, say u which, of course, includes its antiparticle, ū). There is only one meson that can be made from this combination, uū. With two quarks u and d), we can make four mesons, uū, u d, ūd, and d d. With three mesons u, d, s), then the possibilities are uū, u d, u s, dū, d d, d s, sū, s d, and s s, for a total of nine. It s clear that the general case for n different quarks is n 2, and so four quarks gives 16 possibilities, five gives 25, and six gives 36 mesons. b) With one quark, we can only have one baryon, uuu, where each u is a different color. With two quarks, u and d, then we can form uuu, uud, udd, and ddd, or four baryons. With three quarks u, d, and s), then we can form uuu, uud, uus, udd, uds, dus, ddd, dds, dss, and sss. This gives a total of ten baryons. This pattern is harder to simply see, so we ll figure it out. For n quarks, for all three different, there are n different ways that we can make a baryon uuu, ddd, etc.). For two the same and one different there are nn 1) different ways, uud, uus, dds, etc. n ways of picking the first one, and n 1 ways of picking the second, since the first quark is used up). Finally, for all three different, there are n ways of picking the first, n 1 of picking the second, and n 2 of picking the third. However, because dsb is the same as bsd, for example, we need to divide by all six different permutations, for a grand total of nn 1)n 2)/6 possibilities. So, the total number is the sum n + nn 1) + nn 1)n 2) 6 = 1 6 6n + 6nn 1) + nn 1)n 2)) = 1 6n + 6 6n2 6n + n 3 3n 2 + 3) = 1 6 n3 + 3n 2 + 3n) = n n + 1) n + 2). 6 As a check, for n = 1, then the total is 1 2)3) = 1, while for n = 2, then we 6 have 23)4) = 4, and for three we get, 3 4)5) = 10, which all check out. So, for 6 6 four, five, and six quarks we have 45)6) = 20, 56)7) = 35, and 6 7)8) = possibilities, respectively.

4 3. Sketch the lowest-order Feynman diagram representing Delbruck scattering, + +. This process, the scattering of light by light, has no analog in classical electrodynamics.) Photons do not interact with photons directly since they carry no electric charge), and so we can t simply draw a diagram with three photons coming together at a vertex, like we can with electrons and a photon. The only vertex that we really have in quantum electrodynamics is an incoming electron, emitting or absorbing) a photon, giving a final electron. The simplest diagram that can account for light scattered off of light is seen in the diagram below. In this diagram, two photons interact via the creation of virtual electron-positron pairs, which then reannihilate forming final photons. This is a second-order process, and so is suppressed at low energies.

5 4. Draw all the lowest-order diagrams contributing to the process e + + W + + W. One of them involves the direct coupling of Z 0 to W ± s, and the other the coupling of to W s, so if a positron-electron collider is ever built with sufficient energy to make two W s, these interactions will be directly observable.) There are three lowest-order diagrams for this process. In the first diagram, the electron and positron interact and directly produce the W ± pair. In the second case, the electron-positron pair mutually annihilate and form a virtual Z 0 particle, which then transforms into a W ± pair. W W W + e W + e Z 0 Finally, the last diagram has a similar process to the second diagram, only now the electron-positron pair annihilates to a virtual photon before decaying to a W ± pair. W W + e

6 5. Examine the following processes, and state for each one whether it is possible or impossible, according to the Standard Model of Particle Physics which does not include GUTs, with their potential violation of the conservation of lepton number and baryon number). If the reaction is possible, state which interaction is responsible - strong, electromagnetic, or weak; if it is impossible, cite a conservation law that prevents it from occurring. Look up any unfamiliar particles. a) p + p π + + π 0 b) η + c) + e + µ + + µ d) µ + ν e e) ν e + p n + e + f) p e + + g) p + p p + p + p + p h) π 0 + a) This reaction is impossible since it violates charge conservation. The reactants begin with zero charge, but the products carry a net positive charge. b) This reaction is possible via electromagnetic[ effects. The η] particle is a mixture of u, d, and s quarks and antiquarks, η = 1 6 uū + d d 2s s. Each of these pairs can interact electromagnetically. c) This reaction is possible via electromagnetic effects. The electron-positron pair annihilates, forming a virtual photon, which then decays into a µ ± pair. d) This reaction is impossible since it violates muon number conservation. The reactants have a net muon number of 1, but the products have a net muon number of zero, since none of the products are muons or muon-type neutrinos). e) This reaction is possible via weak effects. It is a variation of neutron decay, n p + + ν e, with the electron moved to the left-hand side of the reaction, changing it to a positron. f) This reaction is impossible since it violates baryon and lepton number conservation. The reactant starts with a baryon number of one, and lepton number of zero, while the reactants have baryon number of zero, and lepton number of -1. g) This reaction is possible via strong effects. The extra energy imparted by slamming the protons together can form quark-antiquark pairs, which can form the proton-antiproton pair. h) This reaction is possible via electromagnetic effects. The neutral pion is a combination of a uū and d d [ pair, π 0 = 1 2 uū d d], which can all interact electromagnetically, giving two photons.

7 6. Muons can be created by cosmic rays in the upper atmosphere. A muon at rest has a mean lifetime of about 2.2 microseconds. Suppose that a muon is created 10 km in the air, and is produced with a velocity of 0.99c. a) According to classical physics, how far does the muon travel before decaying according to an observer on the ground)? b) According to Special Relativity, how far does the muon travel before decaying according to an observer on the ground)? c) In it s own rest frame, the muon still lasts only 2.2 microseconds. How can it make it to the point given in part b)? a) According to classical physics, the muon travels a distance d = vτ, where v is the velocity of the muon 0.99c), and τ is the lifetime of the muon. So, d classical = vτ = ) ) 654 meters. The muons only make it about 650 meters. b) According to Special Relativity the muons last longer, due to time dilation effects. Their lifetime, as measured by stationary observers on the ground, is τ = τ = 1 v2 /c = seconds. During this time, the muons can move a distance d SR = vτ, where τ is the new lifetime. Thus, d SR = ) ) = 4630 meters, which is more than seven times further than the classical result. c) From the muon s point of view, it is stationary, and the Earth is rushing up towards it. This means that the muon sees the distance between it and the Earth contracted by an amount d = 1 v 2 /c 2 d. This is contracted by exactly the same factor by which the lifetime was dilated from the stationary observer s point of view. Thus, both the observers on the Earth and the muon agree on the location that the muon reaches.

8 7. Show that the interval, ds 2, is invariant under Lorentz transformations. The interval reads Under a Lorentz transformation ds 2 = c 2 dt 2 + dx 2 + dy 2 + dz 2. t = t vx/c2 x = x vt y = y z = z, then the interval changes to ds 2 = c 2 dt 2 + dx 2 + dy 2 + dz 2. Now, taking the differentials of the Lorentz transformations gives and so Thus, dt = dt vdx/c2 dx = dx vdt dy = dy dz = dz, dt 2 = dt2 2vdxdt/c 2 +v 2 dx 2 /c 4 dx 2 = dx2 2vdxdt+v 2 dt 2 dy 2 = dy 2 dz 2 = dz 2. ds 2 = c 2 dt 2 2vdxdt/c 2 +v 2 dx 2 /c 4 + dx2 2vdxdt+v 2 dt 2 + dy 2 + dz 2 1 = c 2 dt 2 + 2vdxdt v 2 dx 2 /c 2 + dx 2 2vdxdt + v 2 dt 2 ) + dy 2 + dz 2 1 = c 2 1 v 2 /c 2 ) dt v 2 /c 2 ) dx 2 ) + dy 2 + dz 2 = c 2 dt 2 + dx 2 + dy 2 + dz 2 = ds 2, ) which shows the invariance of the interval under the Lorentz transformations.

9 8. A pion at rest decays into a muon and a neutrino π µ + ν µ ). On the average, how far will the muon travel in vacuum) before disintegrating? The muon is ejected with some velocity, v that we need to determine), with respect to the rest frame of the pion. By time dilation the muon lasts a time τ = τ 0, where = 1 v 2 /c 2 ) 1/2, and τ 0 = 2.2 µs is the average lifetime of the muon. So, in this time the muon can travel a distance d = vτ = vτ 0. Thus, we just have to determine v, which we can do using energy and momentum conservation. We could get our answer using four-momenta, but we ll simply use energy and momentum conservation and a bit of extra algebra. Energy conservation says that E i = E π = E µ + E µ = E f. Now, since the pion is initially at rest, it s initial energy is simply its rest-mass energy, E i = m π c 2. The energy of the muon is E µ = m µ c 2, while the neutrino has E ν = p ν c, since it is assumed massless. Now, by momentum conservation, since the pion is at rest we have 0 = p µ + p ν. Thus, p ν = p µ = m µ v, which gives for our energy conservation expression, m π c 2 = m µ c 2 m µ vc = m µ c 2 1 v ) = m µ c 2 1 v/c c 1 + v/c, or 1 v/c 1 + v/c = m π m µ. So, we just need to solve this expression for the velocity, v. Squaring both sides and rearranging gives 1 v ) 2 c = mπ 1 + v ) ) 2 ) ) 2 mπ mπ v 1 = 1 + m µ c m µ m µ c. Solving gives m 2 π m 2 ) µ v = c. m 2 π + m 2 µ From this we find that 1 = 1 v 2 /c 2 ) ) m 2 = 1 π m 2 2 1/2 µ m 2 π +m2 µ ) m 2 2 ) ) π = +m2 µ m 2 m π m 2 2 1/2 µ 2 π+m 2 µ m 2 π+m 2 µ m = 4 π+2m 2 πm 2 µ+m 4 µ m 4 µ+2m 2 µm 2 π m 4 µ m 2 π +m2 µ) 2 ) 1/2 = 4m 2 π m2 µ m 2 π+m 2 µ) 2 = m2 π +m2 µ 2m πm µ. ) 1/2

10 Thus, the total distance that the muon travels will be m 2 π + m 2 ) µ m 2 π m 2 ) ) µ m 2 π m 2 ) µ d = vτ = c τ 2m π m µ m 2 π + m 2 0 = cτ 0. µ 2m π m µ Plugging in the numbers gives m 2 π m 2 ) ) µ 139.6) ) 2 3 d = cτ 0 = ) ) = 186 meters. 2m π m µ ) 105.7)

11 9. Show that the Minkowski metric is invariant under Lorentz transformations. The Minkowski metric reads with our choice of metric signature) η µν = , while the Lorentz transformation matrix reads Λ µ ν = v c 0 0 v c Under a Lorentz transformation the Minkowski metric becomes η µν = Λ α µλ β νη αβ = Λ T ηλ, where the last expression is written in matrix notation and the superscript T denotes the transpose of the matrix). Thus we just have to multiply the three matrices together, v v 0 0 c Λ T ηλ = v c 0 0 c v 0 0 c = = = v 0 0 c v 0 0 c v 0 0 c v 0 0 c v 2 /c 2 ) v 2 /c 2 ) , where we have used the fact that = 1 v 2 /c 2 ) 1/2. Thus, we see that the Lorentz transformation has returned the same Minkowski metric; in other words, the Minkowski metric is invariant under Lorentz transformations.

12 10. A photon of wavelength λ collides elastically with a charged particle of mass m. If the photon scatters at an angle θ, find its outgoing wavelength, λ. We will solve this problem using fourvectors. The photon is incident along the x axis, and strikes the stationary electron. The photon scatters upwards at an angle θ, with respect to it s original direction. The electron is scattered downwards by the same angle, in order to conserve momentum, as seen in the figure to the right. The initial four-momentum of the photon is p µ, while that of the electron is p µ e. The final four-momenta we ll call p µ and p µ e. From the picture, and noting that E = p c for a photon, we can write p µ = E c 1, 1, 0, 0) p µ = E c 1, cos θ, sin θ, 0) p µ e = m e c, 0, 0, 0) p µ e = Ee c, p e cos θ, p e sin θ, 0 ). = p µ + p µ e, which, upon rear- Conservation of momentum and energy gives p µ + p µ e ranging gives p µ p µ = p µ e p µ e. Now, squaring both sides gives p 2 + p 2 2pµ p µ = p 2 e + p2 2p µ e p µe. Recalling that p 2 = m 2 c 2 for a particle of mass m, and that a photon is massless, then we have 2p µ p µ = 2m 2 ec 2 2p µ e p µe p µ p µ = m 2 ec 2 + p µ e p µe. Now, from our expressions for the four-momenta, then p µ p µ = EE c 2 1 cos θ) p µ e p µe = m e E e, giving E E c 2 1 cos θ) = m 2 ec 2 m e E e.

13 Now, by energy conservation, E e = E + E E = E E + m e c 2, and so E E c 2 1 cos θ) = m 2 ec 2 m e E E + m e c 2). Canceling off the m 2 ec 2 terms and dividing through by m e E E gives cos θ) = m e c2 E 1 E. Now, recalling that the energy of a photon is E = hν = hc λ, then λ hc λ hc = 1 1 cos θ). m e c2 Finally, solving for λ we get our final answer, λ = λ + h 1 cos θ). m e c

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