phys 3446 – lecture #6yu/teaching/fall16-3446-001/lectures/phys3446-fa… · wednesday, sept. 21,...

TWednesday, Sept. 21, 2016 PHYS 3446, Fall 2016 1

PHYS 3446 – Lecture #6Wednesday, Sept 21, 2016

Dr. Jae Yu

1. Relativistic Treatment2. Feynman Diagram3. Nuclear Phenomenology4. Properties of Nuclei

• Labeling• Masses• Sizes• Nuclear Spin and Dipole Moment• Stability and Instability of Nuclei


Announcement• Faculty expo #3

– 4pm today in SH100• Reading assignment

– Read and follow through Appendix A, special relativity


Reminder: Homework Assignment #31. Derive Eq. 1.55 starting from 1.48 and 1.49 (5 points)2. Derive the formulae for the available CMS energy (

) for• Fixed target experiment with masses m1 and m2 with

incoming energy E1. (5points)• Collider experiment with masses m1 and m2 with incoming

energies E1 and E2. (5points)3. End of chapter problem 1.7 ( 5points)• These assignments are due next Monday, Sept. 26• Reading assignment: Section 1.7

s


Some Quantities in Special Relativity• Fractional velocity:

• Lorentz g factor

• Relative momentum and the total energy of the particle moving at a velocity is

• Square of the four momentum P=(E,pc), rest mass energy

E =

2

11

gb

=-

2 T +Re

2st

E = 2 2 2 4P c M c+ = 2M cg

2 2 2 2 2P Mc E p c= = -

!β =!vc

!v =!βc!

P = Mγ !v = Mγ!βc


Relativistic Variables• Velocity of CM in the scattering of two particles

with rest mass m1 and m2 is:

• If m1 is the mass of the projectile and m2 is that of the target, for a fixed target we obtain

!βCM =

!vCMc

=

!P1 +!P2( )c

E1 + E2( )

!βCM =

!P1c

E1 + E2( ) =!P1c

P12c2 +m1

2c4 +m2c2


Relativistic Variables• At very low energies where m1c2>>P1c, the velocity

reduces to:

• At very high energies where m1c2<<P1c and m2c2<<P1c , the velocity can be written as:

Expansion

!βCM = m1

!v1cm1c

2 +m2c2 =

m1!v1

m1 +m2( )c

βCM =!βCM = 1

1+ m1c2

P1c⎛⎝⎜

⎞⎠⎟

2

+ m2c2

P1c

≈2

2 1

1 1

112

m c m cP P

æ ö- - ç ÷

è ø


Relativistic Variables• For high energies, if m1~m2,

•

• gCM becomes:CMg = ( ) 1/ 221

CMb

-- »

( )( ) 1/21 1CM CMb b

-é ù- + »ë û

1 2

2

12 m cP

-é ùæ ö

=ê úç ÷ê úè øë û

1

22Pm c

βCM ≈ 1− m2cP1

⎛⎝⎜

⎞⎠⎟


Relativistic Variables• In general, we can rewrite

• Thus the generalized notation of gCM becomes

21 CMb- =

( )21/ 22 1 2

2 4 2 2 41 1 2 2

12CMCME m c

m c E m c m cg b

- += - =

+ +

Invariant Scalar: s

( )2 4 2 2 41 1 2 2

221 2

2m c E m c m c

E m c

+ +

+


Relativistic Variables• The invariant scalar, s, is defined as:

• So what is this the CMS frame?

• Thus, represents the total available energy in the CMS

( )21 2s P P= + =2 4 2 4 21 2 1 22m c m c E m c= + +

2 4 2 4 21 2 1 22s m c m c E m c= + +

( )21 2CM CME E= + ( )2CMToTE=

0

s

= E1CM + E2CM( )2 − !P1CM +!P2CM( )2 c2

E1 + E2( )2 − !P1 +!P2( )2 c2


Useful Invariant Scalar Variables• Another invariant scalar, t, the four momentum transfer

(differences in four momenta), is useful for scattering:

• Since momentum and total energy are conserved in all collisions, t can be expressed in terms of target variables

• In CMS frame for an elastic scattering, where PiCM=Pf

CM=PCM and Ei

CM=EfCM:

t = P1f − P1

i( ) = E1f − E1

i( )2 − !P1 f −!P1i( )2 c2

t = P2f − P2

i( ) = E2f − E2

i( )2 − !P2f −!P2i( )2 c2

t = − PCMf 2 + PCM

i2 − 2!PCM

f ⋅!PCMi( )c2 = −2PCM

2 c2 1− cosθCM( )


Feynman Diagram• The variable t is always negative for an elastic scattering• The variable t could be viewed as the square of the invariant

mass of a particle with and exchanged in the scattering

• While the virtual particle cannot be detected in the scattering, the consequence of its exchange can be calculated and observed!!!– A virtual particle is a particle whose mass is different than the rest

mass

Time

t-channel diagramMomentum of the carrier is the momentum difference between the two particles.

!P2

f −!P2i

2 2f iE E-


Useful Invariant Scalar Variables• For convenience we define a variable q2,• In the lab frame, , thus we obtain:

• In the non-relativistic limit:

• q2 represents “hardness of the collision”. Small qCMcorresponds to small q2.

!P2Labi = 0

2 2q c =

( )2 22 2 22 f

Labm c E m c= - 22 22 f

Labm c T=

2fLabT »

( ) ( )2 222 2 2f fLab LabE m c P cé ù- - -ê úë û

22 2

12m v

2 2q c = t-


Relativistic Scattering Angles in Lab and CMS • For a relativistic scattering, the relationship between the

scattering angles in Lab and CMS is:

• For Rutherford scattering (m=m1<<m2, v~v0<<c):

• Divergence at q2~0, a characteristics of a Coulomb field

Resulting in a cross section

β sinθCMγ CM β sinθCM + βCM( )tanθLab =

2dq =

2ddqs=

( )22 cosP d q- =2P dpW

( )22

2 4

4 ' 1kZZ e

v q

p


Nuclear Phenomenology• Rutherford scattering experiment clearly demonstrated the

existence of a positively charged central core in an atom• The formula deviated for high energy a particles (E>25MeV),

especially for low Z nuclei.• 1920’s James Chadwick noticed serious discrepancies

between Coulomb scattering expectation and the observed elastic scattering of a particle on He.

• None of the known effects, including quantum effect, described the discrepancy.

• Clear indication of something more than Coulomb force involved in the interactions.

• Before Chadwick’s discovery of neutron in 1932, people thought nucleus contain protons and electrons. è We now know that there are protons and neutrons (nucleons) in nuclei.


• The nucleus of an atom X can be labeled uniquely by its:– Electrical Charge or atomic number Z (number of protons).– Total number of nucleons A (=Np+Nn)

• Isotopes: Nuclei with the same Z but different A– Same number of protons but different number of neutrons– Have similar chemical properties

• Isobars: Nuclei with same A but different Z– Same number of nucleons but different number of protons

• Isomers or resonances of the ground state: Excited nucleus to a higher energy level

• Mirror nuclei: Nuclei with the same A but with switched Np and Nn

Properties of Nuclei: Labeling

A ZX


• A nucleus of has Np=Z and Nn=A-Z• Naively one would expect

• Where mp~938.27MeV/c2 and mn=939.56MeV/c2

• However measured mass turns out to be

• This is one of the explanations for nucleus not falling apart into its nucleon constituents

Nuclear Properties: Masses of NucleiA ZX

( ),M A Z =

( ) ( ), p nM A Z Zm A Z m< + -

( )p nZm A Z m+ -


• The mass deficit

• Is always negative and is proportional to the nuclear binding energy

• How are the BE and mass deficit related?

• What is the physical meaning of BE?– A minimum energy required to release all nucleons from a

nucleus – So B= -BE is the energy required to keep a nucleus

Nuclear Properties: Binding Energy

( ),M A ZD =

( ) 2. ,B E M A Z c= D

( ),M A Z ( )p nZm A Z m- - -


• BE per nucleon is

Nuclear Properties: Binding Energy

B BEA A

-=

( ) 2,M A Z cA

-D=

( ) ( )( ) 2,p nZm A Z m M A Z c

A

+ - -=

• Rapidly increase with A till A~60 at which point B/A~9MeV.

• A>60, the B/A gradually decrease è For most the large A nucleus, B/A~8MeV.

9MeV


• de Broglie’s wavelength:– Where is the Planck’s constant– And is the reduced wave length

• Assuming 8MeV was given to a nucleon (m~940MeV), the wavelength is

• Makes sense for nucleons to be inside a nucleus since the size is small.

• If it were electron with 8MeV, the wavelength is ~10fm, a whole lot larger than a nucleus.

Nuclear Properties: Binding Energy! = "

p!!

! = "p= "

2mT= !c

2mc2T∼ 197MeV − fm

2 ⋅940 ⋅8∼1.6 fm


• At relativistic energies, the magnetic moment of electron also contributes to the scattering– Neville Mott formulated Rutherford scattering in QM and

included the spin effects– R. Hofstadter, et al., discovered the effect of spin, nature of

nuclear (& proton) form factor in late 1950s • Mott scattering x-sec (scattering of a point particle) is

related to Rutherford x-sec:

• Deviation from the distribution expected for point-scattering provides a measure of size (structure)

Nuclear Properties: Sizes

Mott

ddsæ ö

ç ÷Wè ø= 24cos

2 Rutherford

dd

q sæ öç ÷Wè ø


• Another way is to use the strong nuclear force of sufficiently energetic strongly interacting particles (pmesons, protons, etc)– What is the advantage of using these particles?

• If the energy is high, Coulomb interaction can be neglected• These particles readily interact with nuclei, getting “absorbed” into

the nucleus• Thus, probe strong interactions directly

– These interactions can be treated the same way as the light absorptions resulting in diffraction, similar to that of light passing through gratings or slits



• The size of a nucleus can be inferred from the diffraction pattern

• All these phenomenological investigation provided the simple formula for the radius of the nucleus to its number of nucleons or atomic number, A:


1 30R r A= »

How would you interpret this formula?

13 1 31.2 10 A cm-´ = 1 31.2 fmA


• Both protons and neutrons are fermions with spins• Nucleons inside a nucleus can have orbital angular

momentum• In Quantum Mechanics orbital angular momenta are integers• Thus the total angular momentum of a nucleus is

– Integers: if even number of nucleons in the nucleus– Half integers: if odd number of nucleons in the nucleus

• Interesting facts are– All nucleus with even number of p and n are spin 0.– Large nuclei have very small spins in their ground state

• Hypothesis: Nucleon spins in the nucleus are very strongly paired to minimize their overall effect

Nuclear Properties: Spins1 2

phys 3446 – lecture #6yu/teaching/fall16-3446-001/lectures/phys3446-fa… · wednesday, sept. 21,...

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