Modern Physics for Engineers

Problem Set 9

Problem Set - Due in class, Wednesday, Nov. 24, 1999.

Problems:

  1. SM&M 15-13

    2. The effective mass of a system, E_cm^2 = E_tot^2-p_tot^2 c^2 does not change from reference from to reference frame. This mass is, obviously, the center of mass energy, or the energy available to make new particles. For part A of this problem E_cm = 4 m_p c^2 = 3.7532 GeV. For part B) E_cm = (2 m_p + 2 m_n) c^2 = 3.7557 GeV. In either case 

    	E_tot-lab = E_beam + m_p c^2 = KE_beam + 2 m_p c^2
	P_tot-lab = Sqrt[E_beam^2 - m_p^2 c^4]/c
	          = Sqrt[(KE + m_p c^2)^2 - m_p^2 c^4]/c
	          = Sqrt[KE^2 + 2 KE m_p c^2]/c
	E_cm^2 = E_lab-tot^2 - p_tot-lab^2 c^2
	       = KE^2 + 4 KE m_p c^2 + 4 m_p^2 c^4 - KE^2 - 2 KE m_p c^2
	       = 2 KE m_p c^2 + 4 m_p^2 c^4
	KE = (E_cm^2 - 4 m_p^2 c^4)/(2 m_p c^2)

Part A) KE = 5.63 GeV ---> p_beam = 6.50 GeV/c
Part B) KE = 5.64 GeV ---> p_beam = 6.51 GeV/c
  2. SM&M 15-16

    3. Part A) In the laboratory, E_cm is given above for one particle hitting a stationary target. Thus, 

    	E_cm = Sqrt[2 KE m_p c^2 + 4 m_p^2 c^4] = 43.36 GeV
	Mass_X = (E_cm - 2 m_p c^2)/c^2 = 41.48 GeV/c^2

    Part B) This is the center of mass (total momentum is zero). Thus: 

    	E_cm = 2 KE + 2 m_p c_2 = 1001.9 GeV/c
	Mass_X = E_cm - 2 m_p c^2)/c^2 = 1000 GeV/c^2
  3. SM&M 15-20

  4. SM&M 15-25
    Plus, what is the momentum of the pion in the lab if the initial momentum of the Lambda^0 in the lab is 15 GeV/c and the pion and proton appear in the lab on trajectories colinear with the parent particle?

    5. Solution to book problem:
    Conservations of Energy: 

    	M_L c^2 = E_p + E_pi

    Conservation of Momentum: 

    	0 = p_p + p_pi ---> p_p = - p_pi = p

    Constituent equations: 

    	E_p  = Sqrt[p^2 c^2 + m_p^2 c^4]
	E_pi = Sqrt[p^2 c^2 + m_pi^2 c^4]

    First solve for the energy of the proton: 

    	E_p = (m_L^2 + m_p^2 - m_pi^2)c^2/(2 m_L)
	    = (1115.6^2 + 938.3^2 - 139.6^2)/(2 1115.6)MeV
	    = 943.65 MeV 
	KE_p = E_p - m_p c^2 = 5.35 MeV
	E_pi = (m_L^2 + m_pi^2 - m_p^2)c^2/(2 m_L)
	     = 171.9 MeV
	KE_pi = E_pi - m_pi c^2 = 32.3 MeV

    Solution to added part:
    Equations are the same except that the momentum conservation equation equals the momentum of the lambda instead of zero and you add a constituent equation for the Lambda. The algebra gets a little messy. The solution is in a Mathematica Notebook.

  5. SM&M 15-27
    Plus how far will the kaon travel if it has a momentum of 20 GeV/c? (Remember time dihilation for both of these parts!) 
Book:
	L = v t_lab = c beta t_lab = c beta gamma t_rest  
	Beta = .96 --> gamma = 3.57
	t_rest = .9e-10 sec
	L = 3e8 * .96 * 3.57 * .9e-10 = .0925 m = 9.25 cm
Added:
	Gamma = E/m = Sqrt[20^2 + .498^2]/.498= 40.2
	Beta = Sqrt[1 - 1/gamma^2] = .9997
	L = 3e8 * .9997 * 40.2 * .9E-10 = 1.085 m