Past exam of the nuclear and particle physics course of the University of Oxford / 2021 / 2

Solution

The magnetic rigidity gives the momentum, the bending direction gives the sign of electric charge, and the calorimetric energy then gives the mass. For planar tracks perpendicular to a uniform magnetic field, the Lorentz force and circular motion give

p = ∣ q ∣ B r, p [GeV / c] = 0.2998 ∣ q / e ∣ B [T] r [m] .

Trace each track away from its production or decay vertex. With the magnetic field into the page, a positive track bends counterclockwise and a negative track bends clockwise. For example, a positive particle initially travelling left accelerates downwards. The known incoming negative pion fixes the sign convention experimentally. A radius alone does not distinguish electric charge signs.

For a calibrated total energy

E

, the relativistic energy-momentum relation gives

m^{2} c^{4} = E^{2} - p^{2} c^{2} .

If the calorimeter instead supplies only kinetic energy

T

, solve

(T + m c^{2})^{2} = p^{2} c^{2} + m^{2} c^{4}

to obtain

m c^{2} = \frac{p ^{2} c ^{2} - T ^{2}}{2 T} .

(3)

Compare with candidate pion, kaon, and proton masses, together with the electric charge sign and shower behavior. An electron makes an electromagnetic shower; a muon is penetrating; a hadron generally requires a hadronic calorimeter and its own calibration. If either energy escapes or the response is uncalibrated, simply substituting a deposited signal for

E

need not identify the particle correctly. The absent figure prevents attaching identities to individual drawn tracks, but the requested identification procedure does not require its geometry.

c

Solution

The labels X and Y cannot be uniquely assigned from the surviving text: their decay vertices and daughter tracks were contained in the missing figure. The available conservation laws nevertheless specify how the assignment would be obtained.

The initial

π^{-} p

state has

(Q, B, S) = (0, 1, 0)

, with

S

denoting strangeness. At a primary strong interaction vertex all three numbers must be reproduced. A final Lambda baryon or Sigma baryon has

B = 1

S = - 1

and therefore needs accompanying

S = + 1

, for example a neutral kaon

K^{0} = d \overset{s}{ˉ}

or a positive kaon

K^{+} = u \overset{s}{ˉ}

, with the remaining electric charge balanced. At later weak interaction decay vertices, strangeness need not be conserved; it would be wrong to enforce it there.

A displaced

p π^{-}

pair with the appropriate invariant mass identifies

Λ^{0} \to p π^{-}

. A displaced

π^{+} π^{-}

pair can identify

K_{S}^{0} \to π^{+} π^{-}

after an invariant mass and kinematic check. For a candidate radiative Sigma-to-Lambda transition, the neutral parent satisfies

p_{Σ} = p_{Λ} + p_{γ}, p_{Λ} = p_{p} + p_{π^{-}}, p_{Σ}^{2} = m_{Σ}^{2}, p_{γ}^{2} = 0

in natural units. Its photon energy in the parent rest frame is

(m_{Σ}^{2} - m_{Λ}^{2}) / (2 m_{Σ})

. With known charged four-momenta, each unobserved neutral momentum is the difference between the incoming and visible outgoing momenta at its vertex; its inferred direction must agree with the dashed segment.

This information cannot establish which dashed segment was called X. For a concrete ambiguity, both

π^{-} p \to K^{0} Λ^{0}, π^{-} p \to K^{0} Σ^{0} \to K^{0} Λ^{0} γ

obey the primary electric charge, baryon number, and strangeness constraints and can yield a neutral kaon decay plus a displaced Lambda baryon decay. Their distinct neutral segments and the extra photon would be resolved by the missing topology and kinematics. The supplied possible Sigma baryon decay alone is insufficient to label X and Y. A labelled answer is therefore a source gap, rather than a completed identification.

d

Solution

The conversion from a measured beam peak to a particle resonance mass is fixed, but the missing graph supplies neither the peak position nor its width. Let

T_{R}

be the incoming pion kinetic energy at the peak and

Δ T

the full width at half maximum after allowing for instrumental resolution. With a stationary proton and natural units,

s = m_{p}^{2} + m_{π}^{2} + 2 m_{p} (m_{π} + T) = (m_{p} + m_{π})^{2} + 2 m_{p} T .

Thus the peak invariant mass is

M = (m_{p} + m_{π})^{2} + 2 m_{p} T_{R} .

The width must also be transformed to the centre-of-mass frame. To first order about the peak,

\frac{d s}{d T} = \frac{m _{p}}{s}, Γ ≃ \frac{m _{p}}{M} Δ T, τ = \frac{ℏ}{Γ} .

(3)

For a broad feature, transform the two half-height positions separately: if they are

T_{-}

and

T_{+}

, then the corresponding width in invariant mass is

(m_{p} + m_{π})^{2} + 2 m_{p} T_{+} - (m_{p} + m_{π})^{2} + 2 m_{p} T_{-}

. An energy-dependent numerator or width can also displace the observed maximum from the resonance pole parameter.

The low-energy

π^{+} p

channel is normally dominated by

Δ^{++} (1232)

, a Delta baryon with

Q = 2

B = 1

and

I = 3/2

. As an external reference example, not a reading of the absent plot,

M = 1232 MeV

and

Γ = 117 MeV

give

T_{R} ≃ 190 MeV

Δ T ≃ 154 MeV

and

τ ≃ 5.63 \times 1 0^{- 24} s

.^[ref] The actual graphical estimates remain unavailable. In particular, using the laboratory width directly in

τ = ℏ/Δ T

misses the factor

m_{p} / M

e

Solution

A constant-width Breit-Wigner resonance cross section is the simplest nonrelativistic curve; a P-wave scattering width and relativistic kinematics explain important expected distortions. A comparison to specific observed deviations cannot be made without the missing plot.

Write

E

for the centre-of-mass frame kinetic energy,

μ = m_{p} m_{π} / (m_{p} + m_{π})

for the reduced mass, and

k

for the relative wave number. The nonrelativistic relations are

E ≃ \frac{m _{p}}{m _{p} + m _{π}} T, k^{2} = \frac{2 μ E}{ℏ ^{2}} .

For an isolated particle resonance of spin

J

, averaging over the initial proton spin and using the zero pion spin gives

σ_{el} (E) = \frac{π}{k ^{2}} \frac{2 J + 1}{2} \frac{Γ _{π p}^{2}}{( E - E _{R} ) ^{2} + Γ ^{2} /4} .