16.451 Lecture 14: Two Key Experiments News flash from Indiana University: 21/10/2003 1 Tour-de-force experiment: http://www.cerncourier.com/main/article/43/5/4

d



d

 4 He   o ???

2 • isospin-forbidden reaction since T = 0 for the d, 4 He, and T=1 for  °: “textbook case”

(technically speaking, this reaction breaks “charge symmetry” which is the symmetry under reversal of all up and down quarks in a wave function, or equivalently a quark “isospin flip”. The pion wave function is CS – odd; the others are CS even)

• Charge symmetry is broken by the electromagnetic interaction: up-down quark mass difference, and their electric charge differences • reaction could proceed with very low cross section compared to isospin-allowed cases, but there was never any convincing evidence published until a couple of weeks ago • compare similar cross-sections at reaction threshold: p + d  d + d  3 He +  ° 4 He +  °  = 13 

b

 = 13  2 pb (Isospin allowed) (forbidden, new result) • Rough estimate of cross section ratio :  ~   

f V



i d

3

r

 2   

allowed forbidden

~  

V s V em

  2    1 4 

o



c

  2  2  10 4 ?

Comparison of precise measurement and theory, accounting for all known CSB effects, tests our understanding of CS as a symmetry of the strong interaction

Cooler CSB

3

the search for d+d

a 0  0 slides courtesy of Dr. E. Stephenson, Indiana University

Ed Stephenson Physics Colloquium 9/24/03

full set: http://www.iucf.indiana.edu/Experiments/COOLCSB

CHARGE SYMMETRY BREAKING

4 atomic nucleus proton neutron Simple notion: charge symmetry The proton and neutron are the same except for electromagnetic properties.

Isospin: the quantum number for CS Proton and neutron have

T(I) = 1/2

quarks inside nucleons: CS says up and down are the same except for charge But they are different: m N – m P = 1.3 MeV (The neutron decays in 887 s: n  p + e – +

ν

e ) z = 2/3 u u d u d d z = –1/3 Nuclear charge symmetry breaking comes from: electromagnetic interactions among quarks m d > m u How much does each contribute?

5

Observation of the Isospin-forbidden d+d

 4

He+ π

0

Reaction near Threshold d + d

 4

He + π

0

isospin: 0 0 0 1

CHARGE SYMMETRY says that the physics is unchanged when protons and neutrons are swapped, or when up and down quarks are swapped.

The pion wavefunction   1 2 

uu



dd

 is not symmetric under up-down exchange.

Deuterons and helium reverse exactly. Thus, an observation of this process is also an observation of charge symmetry breaking.

Threshold Energy:

m 3 - m n m 1 m 2 For a reaction to occur in a fixed target experiment, m 1 has to hit m 2 with enough energy to make the particles in the final state. The minimum kinetic energy required is called the threshold energy: T thr = Q m 1 + m 2 2m 2 +

S

m f Q = m 1 + m 2 -

S

m f Examples: d + d p + d

a

+

 0

3 He +

 0

T thr = 225.4 MeV T thr = 198.7 MeV

6

Experimental approach:

Search just above threshold (225.5 MeV) (No other π channel open for d+d.) Capture forward-going 4 He.

Pb glass arrays for π 0  γγ.

Efficiency on two sides ~ 1/3.

Insensitive to other products ( γ beam = 0.51) Pb-glass measures photon energy via Cerenkov light from high energy e- produced in a ‘shower’ initiated by high energy photon collisions 7 Target density = 3.1 x 10 15 Stored current = 1.4 mA Luminosity = 2.7 x 10 31 /cm 2 /s Expected rate ~ 5 /day 6  bend in Cooler straight section Target upstream, surrounded by Pb-glass Magnetic channel to catch 4 He (~100 MeV) Reconstruct kinematics from channel time of flight and position.

d + d

a

+

 0

in the lab close to threshold:

8 g

d d .

 0 a g

For a fixed target experiment just above threshold, cone angle

a

particles emerge within a narrow cone about the 0-degree line.

(Spectrometer with small forward acceptance will catch every

a

.)

low-energy

 0

quickly decays into two photons which emerge nearly back to back in the lab.

Therefore, the apparatus must identify a forward

a

in coincidence with two photons that have a large opening angle between them.

COOLER-CSB MAGNETIC CHANNEL and Pb-GLASS ARRAYS

•separate all 4 He for total cross section measurement •determine 4 He 4-momentum (using TOF and position) •detect one or both decay g ’s from  0 in Pb-glass array Scintillators D E-2 E Veto-1 Veto-2 scintillators measure flight time (ToF)  velocity  momentum Pb-glass array 256 detectors from IUCF and ANL (Spinka) + scintillators for cosmic trigger Target D 2 jet MWPCs Scintillator D E-1 Focussing Quads MWPC 228.5 or 231.8 MeV deuteron beam 20  Septum Magnet Separation Magnet removes 4 He at 12.5

 from beam at 6  Note: MWPC = multiwire proportional chamber – gives tracking information for trajectory angles 9

 0  gg from p+d  3 He+  0 LEFT RIGHT Pb-glass Hit Patterns beam goes into X cosmic ray muon 10 color scale: red > pink > blue

`Missing Mass’ measured with proton beam:

p + d 3 He +

 o conservation of energy: W  E p + E d – E( 3 He) = m  • E p from beam energy • deuteron at rest in target • E( 3 He) from energy and momentum measured with the magnetic channel • calculate W from data, should find a peak at the pion mass for reaction at threshold.

• then check in Pb glass array to see if pion was observed

Mass of

 o

134.98 MeV/c Resolution is ~ 100 KeV.

: 2 Mass (Mev/c 2 * 100)

11

SEPARATION OF a 0 AND agg EVENTS 12 IDEA: Calculate missing mass from the four-momentum measured in the magnetic channel, using time-of-flight for z-axis momentum and MWPC X and Y for transverse momentum. Should see a peak for a ° reaction and a broad background from agg MWPC spacing = 2 mm Need very good resolution so that the peak is detectable!

[Monte Carlo simulation for illustration. Experimental errors included.] needed TOF resolution  GAUSS = 100 ps MWPC 1 X-position (cm) a 0  TOT peak = 10 pb agg background (16 pb) Time of Flight ( ΔE 1 ΔE 2 ) (ns) missing mass (MeV) Background: d + d  a .

+ 2 g

RESULTS

(measured at two different beam energies)

228.5 MeV 66 events σ TOT = 12.7 ± 2.2 pb 13 Events in these spectra must satisfy: correct pulse height in channel scintillators usable wire chamber signals good Pb-glass pulse height and timing 231.8 MeV 50 events σ TOT = 15.1 ± 3.1 pb First ever convincing observation of both the a ° and agg reactions!

Peaks give the correct π 0 mass with 60 keV error.  missing mass (MeV) Bottom line: time to revise all the textbooks!!!

Back to the neutron: isospin  14 Lifetime: 885.7 

n

 0.8 sec (world average)

p



e

  

e

Case study: “state of the art” neutron lifetime measurement 15

Outline of the method:

n



p



e

  

e

decay rate:

dN dt

 

N

 measure rate by counting decay protons in a given time interval (dN/dt) and normalizing to the neutron beam flux (N) decay proton, to detector transmitted beam incoming n beam 16 neutron detector decay volume, length L Ideally done with “cold neutrons”, e.g. from a reactor, moderated in liquid hydrogen...

Issues: 1. precise decay volume ? 2. proton detection ? 3. beam normalization ? ...

Neutron beam distribution – definitely not monoenergetic: 17 • ~ MeV neutrons from a reactor are “moderated” by scattering in a large tank of water (“thermal”) or liquid hydrogen (“cold”) • after many scatterings, they come to thermal equilibrium are extracted down a beamline to the experiment with the moderator and • velocity distribution is “Maxwellian”: energies in the meV range (kT = 26 meV @ 293K) Krane, Fig 12.4

Neutron detection at low energy:

(Krane, ch. 12)

• several light nuclei have enormous neutron capture cross sections at low energy: (recall, cross sectional area of a nucleus, e.g. 10 B is about 0.2 barns, lecture 4) • key feature: cross sections scale as 1/velocity at low energy 18 10 B + n  a + 7 Li + 2.79 MeV  =  o v o /v ;  o = 3838 b, v o = 2.2 km/s kinetic energy of ionized fragments can be converted into an electrical signal  detector

 Put this together for detector counting rates as shown: 

N dN

/

dt



N

beam

N

decay

N

decay 

dN

beam

dt

decay proton, to detector 19 transmitted beam 10 B neutron detector

N

 det 

N

det v ( 

o

v

o

) ~ neutrons must be captured to be detected!

N

beam

N

beam  (

const

) 

N

det

L T

incoming n beam

N

beam decay volume, length L

L

 v

T

(DC beam, thermal energy spread) Decays and transmitted detector counts accumulated for time T    (

const

) 

N

det

N

decay 

L

Experimental details (all in vacuum, at ILL reactor, France): 20 • use Penning trap to confine decay protons • let them out of the trap after accumulation interval T  ~

N N

det decay  • measure the ratio N det /N decay as a function of trap length L  slope gives 

L N

decay proton detector

N

det a particle detectors for capture products thin 10 B foil to capture beam n’s

L

variable length Penning trap (16 electrodes) protons spiral around field lines when let out of the trap

Amazing results: Rate versus trap length L 21 proton energy from timing: about 34 keV (after kick – raw energy is only 0.75 keV) Result: 893.6  5.3 seconds (1990)

PDG average: 885.7



0.8 (2003)