Modern Physics I Solutions to Homework 4 Handout

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1 Moern Physis I Solutions to Homework 4 Hanout TA: Alvaro Núñez an33@sires.nyu.eu New York University, Department of Physis, 4 Washington Pl., New York, NY Bernstein, Fishbane, Gasiorowiz: Chapter III, problems The gravitational potential energy of the lifte mass M woul be where m = g. Hene we oul lift a mass M = m gh 0 3 kg (3 0 8 m/s) 0 m/s 0 3 m 9. By onservation of energy Therefore Mgh = m () 0 0 kg = = 0 million tons. () E p + E = E 3 He + E γ. (3) E 3 He = m p + m E γ. (4) On the other han, the helium nuleus taks a momentum E γ / its mass (m p + m E γ ) ( ) Eγ m 3 He = = kg. (5)

2 0. Pressure is fore per unit area. The magnitue of the fore exerte on the wall by one pion equals aoring to Newton s seon law the magnitue of the hange in its momentum P per unit time t, F = P t (6) The hange in the momentum is from some P, enoe in the energy E given, to 0: E P = 0 m 4 E = m 4. (7) where m is the pion s mass. The number of pions hitting an area A of the wall in a time t is N = IA t (8) where I is the intensity of the pion beam given. Overall we will have a pressure q E p = N F m 4 (E ) A = IA t t = I m A 4 = P a. (9) Moral: partile beam pressure=beam intensity times partile momentum. The photon eay γ e + + e (0) woul imply by onservation of energy E γ = m 4 + p + m 4 + p + () an by onservation of momentum p γ = p + p. () But p γ = E γ /. Diviing the onservation of energy equation by, squaring both equations an subtrating one from eah other will lea to m + m 4 + p m 4 + p + + p p + = 0 (3) but this is impossible for all quantities on the left are non-negative an m > 0.

3 . Energy onservation or expliitly m ψ = an momentum onservation E ψ = E J/ψ + E γ, (4) m J/ψ 4 + p J/ψ + E γ, (5) 0 = p J/ψ E γ / (6) allow to fin E γ = m ψ m J/ψ = 0.49 MeV. (7) m ψ 3. For the total mass, total energy an total momentum after the reation, we have (m π 0 + m p ) 4 = E p. (8) Lets write E an p in terms of the unknown E γ an the masses an then use this equation. By onservation of energy, the total final energy equals the total initial E = E γ + E p (9) or expliitly E = E γ + m p (0) sine the proton was at rest. The momentum is onserve as well, or expliitly p = p γ + p p () p = E γ () sine the proton was at rest. Plugging E an p in the first equation we solve for E γ to get (m π 0 + m p ) 4 = (E γ + m p ) E γ (3) E γ = (m π 0 + m p) m ( p = m π 0 + m ) π 0. (4) m p m p 3

4 4. Conservation of energy an momentum give Solving for E γ leas to M = E γ + (M + M) 4 + p, (5) 0 = E γ E γ = M ( + M M 5. This is one of the few instanes were p. (6) ) (7) E = m β (8) leas to the esire answer faster than the E p = m equation sine we know the ratio an the numerator an we are looking for a veloity. This allows to fin ( ) m β =. (9) E an then the time t = l = 4.05 yrs. (30) β 6. Aoring to the previous problem, a range of energies [E, E ] implies a range of spees, between ( ) m γ β, = (3) E, whih implies a range of arrival times t = β β = = ( ) mγ E ( ) mγ E (3) 4

5 The mass of the photon must be small ompare to its energy sine they are ultrarelativisti. It means we an expan in series to get t = m γ 3 E E (33) or m γ = t 4 4 E E. (34) Given the arrival time interval τ, this implies a onstraint on the photon mass of m γ < τ. (35) 4 4 We an go further an use E = ev an E = MeV, in the re-aste expression m γ < τ E 4 E E E E (36) the ratio E /E is small, giving after expansion ( ) τ E m γ < + E. (37) E In the smallest approximation m γ < τ E (38) or τ m γ < E ev = 0 5 ev. (39) 0 9 Compare this to the eletroni rest mass 0.5MeV. 7. In just any two-boy ollision mass is not neessarily onserve. Take for example, the eletron-positron annihilation to γ-photons. That woul be an inelasti ollision. The authors mean in this problem an elasti ollision, e.g. one in whih no new partiles are proue, or one in whih the total kineti energy is onserve. 5

6 8. In the same spirit as in problem 3, (m K 0 + m Λ 0) = (E π + m p ) ( E π m π ) (40) 9. p is at least zero, so E π = (m K 0 + m Λ 0) m π m p m p. (4) E γ = m e + p. (4) E γ = m e. (43) 30. In the enter of mass frame E e = m p. Compare this, by the way, with the lab frame alulation: same approah as always tells us to write 3. whih leas to (m p ) = (E m e ) E + m e (44) E = m p m e m e, (45) whih is muh greater than m p. an = βt = 3. The total energy an total momentum t = γτ (46) γ βγτ = βτ γ γ = 8.6 µm. (47) E + E = m + p (48) E E = p (49) m + p E = + p, (50) m + p E = p. (5) 6

7 33. Their momenta are equal by magnitue, lets all it p. On the other han the mass shell equation gives M = ( ) m 0 + p + m + + p (5) p = (M m 0 m +) 4m 0m + 4(M + (m 0 +. (53) m +)) 34. E = m + p (54) an we an use the result from problem 33 for p fin the energy in terms of the masses. 35. (a) The mass shell equation in the enter of mass system gives p = M = M ( m + p ) (55) 4 m = M m 4M. (56) (b) The mass shell equation in the lab system (another name for the fixe-target system) leas to M = (E + m) E + m (57) E = M m (58) m Comment: Consier the ultrarelativisti ase, m is small. Then (a) says that the energy in the enter of mass system must be of orer M, while (b) tells us that the energy in fixe-target frame must be of orer M /m. Its easier in general to aelerate two partiles to energies of orer M than one to an energy of orer M /m. 7