2010/02/15 P2.1 [4 + 4 = 8 Points] (A) Show That the Combination of Two Su

NIU PHYS 684, Spring 2010 Introduction to High Energy Physics

Solution to HW2 [25 points]

Assigned: 2010/02/08 Due: 2010/02/15

P2.1 [4 + 4 = 8 points]

(a) Show that the combination of two successive Lorentz transformations with parallel velocities β1 and β2 is equivalent to a single Lorentz β1 + β2 transformation with velocity β = . 1 + β1β2 (b) Show that the set of all Lorentz transformations with this composition rule forms a group.

S2.1 (a) For parallel Lorentz boosts, the corresponding rapidities add like angles in successive 3-d Euclidean rotations about the same space axis (except, of course, rapidities are unbounded hyperbolic “angle”s):

η = η1 + η2. (1)

So,

tanh η1 + tanh η2 β1 + β2 β = tanh η = tanh (η1 + η2) = = . (2) 1 + tanh η1 tanh η2 1 + β1β2

(b) Since {η} clearly form the group of real numbers under addition, so must {β} with the above composition rule.

P2.2 [4 + 3 = 7 points]

(a) A particle of mass m is produced with energy E and decays after traveling a distance ℓ. How long did the particle live in its own rest frame? (b) Use the result of part (a) to determine the separation between the production and decay vertices of a B0 meson, of mass 5.3 GeV, car- rying an energy of 65 GeV. The ability to tell the two vertices apart in such decays is of tremendous value to the studies of the top quark and to the searches for Higgs boson(s) at the Tevatron, the LHC, and the ILC.

S2.2 (a)

E 1 ℓ = γβτ; γ = m ; β = q1 − γ2 ; (3) ⇒ τ = ℓ = ℓ . γβ E 2− q( m ) 1

1 NIU PHYS 684, Spring 2010 Introduction to High Energy Physics

(b) The mean lifetime of a B0 meson is 1.54 ps. Therefore, the mean decay length of a B0 meson with E = 65 GeV is

E 2 ℓ 0 = τ 0 − 1 B B q m

2 = (1.54 × 10−12 s) × (3 × 108 ms−1) × 65 − 1 q 5.3

≈ 5.65 mm.

P2.3 [4 points] Determine the maximum energy which can be carried oﬀ by any one of the decay particles, when a particle of mass m0 at rest decays into three particles with masses m1, m2, and m3.

S2.3 The decay particle in question has its maximum energy when the system of the other two has the least possible mass, which happens when the latter are moving with the same velocity. Say, particle 1 moves at a speed β1max when particles 2 and 3 move together with speeds β′. The kinematics of particle 1 is the same as it would be if it were recoiling against a single sister particle of mass m2 +m3. Since there is no motion in 2 of the 3 space dimensions, we can ignore those. Conservation of momentum dictates µ µ pi = pf , (4) where the subscripts i and f denote the initial and the ﬁnal states, re- spectively. For µ = 0 we have ′ m0 = γ1maxm1 + γ (m2 + m3), (5) and, for µ = 1, ′ ′ 0 = γ1maxβ1maxm1 − γ β (m2 + m3). (6) Now, 1 γ = ⇒ γ2β2 = γ2 − 1. (7) 1 − β2 p Thus Eq. 6 yields ′ ′ γ β (m2 + m3) = γ1maxβ1maxm1 ′2 2 2 2 or, (γ − 1)(m2 + m3) = (γ1max − 1)m1 (8) ′2 2 2 2 2 or, γ (m2 + m3) = (γ1max − 1)m1 + (m2 + m3) . We rearrange the terms in Eq. 5, square both sides, and use Eq. 8 to get the maximum energy of particle 1: ′ m0 − γ1maxm1 = γ (m2 + m3) 2 2 2 2 2 2 2 ⇒ m − 2γ1maxm1m0 + γ m = (m2 + m3) + γ m − m 0 1max 1 1max 1 1 (9) 2 2 2 m0 + m1 − (m2 + m3) ⇒ E1max = γ1maxm1 = . 2m0

2 NIU PHYS 684, Spring 2010 Introduction to High Energy Physics

P2.4 [3 points] How many times per second does a proton, fully accelerated to 980 GeV, go around the Tevatron? The radius of the Tevatron is 1 km and the mass of the proton 931 MeV.

S2.4 E 980 γ = = ≈ 1052.6 ⇒ β = 1 − γ−2 ≈ 0.99999955 m 0.931 p i.e. the proton is traveling essentially at the speed of light (the fractional diﬀerence is small compared to uncertainties in the path length). So, the rate of revolutions is (3 × 108 ms−1)/(2π × 1000 m) = 47746.5 s−1.

P2.5 [3 points] Explain why the processes γ → e+e− and e− → γe− cannot occur in free space. The ﬁrst process represents the decay of a photon into an electron and a positron, while the latter represents the emission of a photon by an electron.

S2.5 The Feynman diagram representing the process of a photon decaying into a electron=positron pair γ → e+e− in free space is shown below.

e+ γ

e−

Conservation of momentum requires

pµ(k) = pµ(l) (10) X i X f k l where k and l are particle indices. The above condition requires all 3 particles to lie on a plane, say (x,y). Since the electron and the positron have the same mass me, their directions of ﬂight must make the equal but opposite angles (say, θ) with the photon direction (say, x). Putting µ = 2 (i.e., the y component) in Eq. 10 gives

+ − py(γ) = 0 = py(e ) + py(e ). (11)

Hence, + − E(e ) = E(e ) ≡ Ee (12) and 1 − + 2 2 2 |p(e )| = (Ee − me) = |p(e )| ≡ pe. (13) Putting µ = 1 (i.e., the x component) in Eq. 10 gives

+ − px(γ) = E(γ) = px(e ) + px(e ) = 2pe cos θ, (14)

3 NIU PHYS 684, Spring 2010 Introduction to High Energy Physics

while µ = 0 (i.e., the t component) gives

E(γ) = 2Ee. (15)

Combining Eqs. 14 and 15 leads to the condition

Ee = pe cos θ ≤ pe (16)

which is impossible to reconcile with the requirement

1 2 2 2 Ee = (pe + me) >pe. (17)

Hence, a photon cannot decay into an electron-positron pair in free space. The process e− → γe− is merely a rotation of the above process in the tx plane and, therefore, likewise forbidden. That any process involving the decay of a single massless particle into any number of massive particles must be forbidden is easy to see even without elaborate algebra. If all particles in the final state are massive, then one can always transform to the center-of-mass frame of the final state particles where, by definition, the total 3-momentum of the final state is zero. However, no such “rest frame” exists for the massless particle in the initial state.