Problem 1:
A slowly moving antiproton is captured by a deuteron at rest producing a neutron and a neutral pion.
The rest masses of the particles involved are mpc2 » mnc2 » mDc2/2 » 939 MeV and mp0c2 = 135 MeV. Find the total energy of the emitted p0.
Solution:
 | Concepts, principles, relations
that apply to the problem: Relativistic collisions, energy and momentum conservation |
| Why do they apply? In relativistic collisions energy and momentum are always conserved. |
| How do they apply? We assume that the initial particles have no kinetic energy. Energy conservation: E = 3mpc2 = En + Ep0. (1) Momentum conservation: pn = pp0. (2) En2 = pn2c2 + mp2c4, Ep02 = pn2c2 + mp02c4, En2 - Ep02 = (En + Ep0)(En - Ep0) = mp2c4 - mp02c4. (3) (En - Ep0) = (mp2c4 - mp02c4)/(3mpc2). (4) = (3)/(1) 2Ep0 = 3mpc2 - (mp2c4 - mp02c4)/(3mpc2). (1) - (4) Ep0 = (8mp2c4 - mp02c4)/(3mpc2) = 1255 MeV. |
| Details of the calculation: None |
Problem 2:
A pion (mass Mp) decays into a muon (mass Mm)
and a neutrino (massless).
(a) Find the kinetic energy of the muon in the rest frame of the pion.
(b) The muon is unstable, with a lifetime t1
in the pion's rest frame. Find the lifetime t0
of the muon in its own rest frame.
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic "collisions", energy and momentum conservation |
| Why do they apply? The decay of a particle is a relativistic problem. In relativistic "collisions" energy and momentum are always conserved. |
| How do they apply? (a) Energy conservation: Mpc2 = (Mm2c4 + p2c2)1/2 + En. Here p is the momentum of the muon. Momentum conservation: pc = En. Combine: (Mpc2 - En)2 = Mp2c4 + En2 - 2Mpc2En = Mm2c4 + En2 En = (1/2)Mpc2 - (1/2)Mm2c2/Mp. Em = Mpc2 - En = (1/2)Mpc2 + (1/2) Mm2c2/ Mp. T = Em - Mmc2 = (Mp - Mm)2c2/2Mp. |
| Details of the calculation: (b) t0 = t1/g. We need to find g. T = (Mp - Mm)2c2/2Mp = (g-1)Mmc2 (g-1) = (Mp - Mm)2/2MmMp, g = ( Mp2 + Mm2)/2MmMp, t0 = t12MmMp/( Mp2 + Mm2). |
Problem 3:
The PEP-II collider at Stanford Linear Accelerator Center creates
electron-positron head-on collisions. In the laboratory frame the combination
of electron energy E- = 9 GeV and positron energy E+ = 3.1
GeV is resonant for the production of a single
¡ particle.
(a) What is the rest mass of the produced particle
¡?
(b) What is the speed of
¡ in units of c?
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic collisions, energy and momentum conservation |
| Why do they apply? In relativistic collisions between free particles energy and momentum are always conserved. |
| How do they apply? (a) E = E+ + E-, p = p+ + p-. p+2c2 = E+2 - me2c4 = (3.1 GeV)2 - (0.511 MeV)2 = 9.61 GeV2, p+c = -3.1 GeV. p-2c2 = E-2 - me2c4 = (9 GeV)2 - (0.511 MeV)2 = 81 GeV2, p-c = 9 GeV. E = 12.1 GeV, pc = 5.9 GeV m2c4 = E2 - p2c2, mc2 = 10.56 GeV. |
| Details of the calculation: (b) E = gmc2. g = 12.1/10.56 = 1.15. v2/c2 = 1 - 1/g2 = 0.24, v = 0.49c. |
Problem 4:
Let reference frame K' move with velocity v with respect to reference frame K. In K a sinusoidal electromagnetic plane wave has an angular frequency w and a wave vector k. Find w' and k' in reference frame K'.
Solution:
| Concepts, principles, relations that apply to the
problem: The Doppler shift |
| Why do they apply? We are asked to derive the relativistically correct expression for the Doppler shift. |
| How do they apply? If we picture the wave as a series of crests and troughs moving through space, then the phase of any labeled point is a relativistically invariant quantity. Observers in K and K' will agree on the phase of the a labeled point, i.e. they will agree if the point is on a crest, in a trough, etc. We label the phase by f = k×r-wt = k'×r'-w't'. Let w = k0c, then k0ct - k×r = k0'ct' - k'×r', (k0,k)×(ct,r) = ( k0',k')×(ct',r') for any 4-vector (ct,r). The dot product of (k0,k) with any 4-vector yields a Lorentz invariant quantity. Therefore (k0,k) is itself a 4-vector. |
| Details of the calculation: Since (k0,k) is a 4-vector,we know how it transforms under a Lorentz transformation. With b = v/c, |k| = k0 = w/c, and b×k = (vw/c2)cosq, where q is the angle between the directions of v and k, we have k0' = g(k0 - b×k) = g(k0 - (vw/c2)cosq) , k'|| = (k|| - bk0) = (k|| - vw/c2), k'^ = k^. Since w' = k0'c we have w' = g(w - v×k) = gw(1 - vcosq/c). w' = w[(1 - v/c)/(1 + v/c)]1/2 if k and v are parallel to each other. w' = w[(1 + v/c)/(1 - v/c)]1/2 if k and v are anti-parallel to each other. w' = gw if k and v are perpendicular to each other. There is a transverse Doppler shift, even if q = p/2. |