Modern Physics Solutions to Homework 4 Handout

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1 Modern Physics Solutions to Homework 4 Handout Alvaro Núñez nunez@cooper.edu Cooper Union, School of Engineering, 51 Astor Pl., New York, NY 13. Three problems and Taylor, Zafiratos, Dubson Chapter,.4. 1 A particle of mass M = 5GeV/c decays into two particles, m 1 = 1.GeV/c and m = 1.5GeV/c. a) What are the speeds of the decay products? b) At what speed of the beam of decaying particles no particles move backward after the decay? (We can safely assume that when the decaying particle rests all directions are equally probable.) Solution A decaying particle in its CM reference frame is tacitly assumed for part a. That is, I imagine they want the speeds relative to the CM system. a.) Instead of writing the energies and momenta through the velocities, and having to solve two equations, notice how easy is to write the equation M = E P (1) for the first particle and use conservation of momentum: P =, which also implies p 1 = p for the magnitude of the products momenta. Lets call the magnitudes of these momenta just p, since they are equal. Then we rewrite the above equation ( ) M = m 1 + p + m + p () that is, as an equation for p. Knowing p will instantly give us the velocities. Below is the algebra: ( ) M = m 1 + p + m + p. (3) 1

2 Open the square: M = m 1 + p + m 1 + p m + p + m + p. (4) Leave the roots on one side: M m 1 m p = m 1 + p m + p. (5) Now, square again (M m 1 m p ) = 4(m 1 + p )(m + p ) (6) and don t expand too much, just the necessary to isolate powers of p, (M m 1 m ) 4(M m 1 m )p +4p 4 = 4m 1 m +4(m 1 +m )p +4p 4. (7) The quartic term cancels, (M m 1 m ) 4(M m 1 m )p = 4m 1m + 4(m 1 + m )p, (8) and the equation is linear and elementary in p : p = (M m 1 m ) 4m 1m 4M, (9) giving, after applying a b = (a + b)(a b) a couple of times, (M (m 1 + m ))(M p = (m 1 m ) ). (1) M It is good to write the answer for momentum this way as you get to appreciate that M has to be greater than the sum and the difference of m 1 and m. Plug the numbers here. Then, to find the velocities, invert p = m iv i 1 v i (11) for both i = 1 and i = ; remember, p is the same for both product particles; you should get v i = p/m i. (1) 1 + p /m i

3 b.) Now the equation M = E P (13) has to be written in the frame where the decaying particle has a momentum P such that p 1 =. Well, this means P = p, that is, the momentum of the decaying particle is taken completely by only one of the products. We have to exploit this fact. M = ( ) m 1 + m + P P. (14) We again find P and then find the velocity of particle M. M = m 1 + m 1 m + P + m + P P, (15) P = m 1 m + P = M m 1 m, (16) 4m 1 (m + P ) = (M m 1 m ), (17) P = (M m 1 m ) m (18) 4m 1 (M (m 1 + m ))(M (m 1 m ) ) m 1. (19) Plug the numbers. The velocity of particle M is now V = P/M 1 + P /M () As we have seen, a free electron cannot absorb a photon. Can two resting electrons absorb one photon? (Hint: Consider a situation when electrons are scattered at an angle to each other.) Solution Two resting electrons can t absorb a photon. After the reaction the total mass is m, while the total energy before the reaction is m+m+e, and the total momentum E. So we would have (m) = (m + m + E) E (1) 3

4 or which implies 4m = 4m + 4mE + E E () 4mE = (3) Since the mass of the electron is not zero, this means the energy of the photon is zero. But a photon without energy is no photon. 3 Consider a vehicle of mass m moving with constant speed v along a straight line inside a spaceship of mass M. a) Find the classical expression for the speeds of the spaceship relative to the center of mass. b) Find the relativistic expression for the speeds of the spaceship. Is the relativistic correction positive or negative? Solution a.) Suppose the spaceship has velocity V relative to the center of mass system. Then the vehicle has velocity V + V relative to the center of mass system. The definition of the c.m. system is that in it the total momentum is zero, e.g.: Solving for V, we find This is the classical answer. 1 MV + m(v + V ) = (4) V = 1 V 1 + M (5) m 1 It is easy to fall in the mistake of thinking that the correct equation is MV + mv = (6) and therefore V = m M V (7) Think about the result. For example, if the spaceship and the vehicle have the same mass, M = m, which is clearly possible, this would imply that the spaceship has to move with a speed opposite in direction and equal in magnitude to that of the vehicle relative to the spaceship. That would put the vehicle at rest relative to the c.m., while the spaceship is moving relative to the c.m., which is non-sense. Instead, for this case the right answer says that the velocity of the spaceship is half by magnitude that of the vehicle relative to the spaceship. This way they move opposite to each other at equal speeds, as expected for the c.m. system. 4

5 b.) The relativistic case is considered in complete analogy, but the vehicle s velocity relative to the center of mass is now so we end up with the equation Rewriting it as V + V 1 + V V (8) MV + m V+V 1+V V ( ) =. (9) V+V 1 1+V V MV = m V+V 1+V V ( ), (3) V+V 1 1+V V squaring and reducing the fractions brings this to a quadratic equation in V with solutions V = 1 ± M m ( 1 + M m ) (1 M m ). (31) We have to choose the minus sign in the numerator, as this is the one which gives the correct classical limit. The existence of two solutions is an artifact of having squared one of the equations during the solution. Be careful here. So 1 V =. (3) 1 + M m.4 Solution Again, m π = Eπ Pπ = ( m µ + p + p), (33) since the muon and neutrino momenta are the same, and the neutrino s momentum is its energy, like for any massless particle. Therefore m π = m µ + p + p (34) 5

6 and Plug this in p = m π m µ m µ. (35) v = p p + m µ (36) to get v = m π m µ. (37) m π + m µ If you divide the numerator and the denominator by m µ, you get the answer in the book. 6