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Ian Bentley University of Notre Dame

Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

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Page 1: Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

Ian Bentley

University of Notre Dame

Page 2: Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

Wigner Energy

• E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where X=4, based on the supermultiplet formalism.

• More recently, work by Jänecke et al. has determined that X is typically 1 but in the region A>80 near N=Z, X is 4.

• We’ve done a different fit with a third order polynomial term, and found a few regions with different X values.

Page 3: Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

22/3 P

V S SYM C3/4 1/3

A-2Z Z Z-1aBE= a A -a A -a + -a

A A A

Semi-Empirical Mass Formula

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

340

360

380

400

420

440

EE OO

E(T

Z) [M

eV]

Isospin TZ

Subtracted Coulomb Fit from BE for A=44

22Ti

21Sc

20Ca

19K 18

Ar

17Cl

16S

15P

14Si

23V

24Cr

25Mn

Strong Energy Coulomb Energy

Z

N-Z A-2ZT =

2 2

Mirror Nuclei, have roughly the same Strong Components. Therefore, around N=Z one can fit the Coulomb Energy.

LBNL Isotopes Project Nuclear Data Dissemination Home Page. Retrieved March 11, 2002, from http://ie.lbl.gov/toi.html

Z Zs

s Z

Z

Exp avg Exp avgExp

Z

T( T +1)E =

θ∂ E T +1/2

ΔT≡ =∂T θ

E T +1 -E T -1∂EΔT≡ ≈

∂T 2

Page 4: Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

Experimental Analysis

40 50 60 70 80 90 100

-0.010

-0.005

0.000

0.005

0.010

0.015

0.020

0.025

EE OO EO OE

3rd Order Coefficient of Polynomial Fit Using 4 points

Curv

atu

re C

oeffi

cient

Atomic Number

40 50 60 70 80 90 100

-1.5

-1.0

-0.5

0.0

EE OO EO OE

X-Intercepts of Polynomial Fit Using 4 points

X-Inte

rcept o

f

Atomic Number

Page 5: Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

Reconciling the BCSA) Using the two BCS self consistency equations one can solve for

expectation value of the pure pairing Hamiltonian.

B) Compare the energy levels with those from diagonalizing an exact pairing Hamiltonian in matrix form.

+ + + +k k k k kk k k k

k>0 k>0

22 2 4

k k k kk>0

H= ε a a +a a -G a a a a

G ΔBCS H BCS =2 ε -λ-Gν ν + ν -

2 G

2 2k>0k

2pairs k

k>0

G ΔΔ=

2 ε -λ +Δ

N = ν

Example: 3protons 3 neutrons in 3 levels

nn

nn

nn

nn

nn

nn

nn

nn

nn

E -4G -G -G 0 -G 0 G 0 0 0 0 -G 0

-G E -3G 0 0 0 0 0 0 0 0 0 0 -G

-G 0 E -4G -G 0 0 G 0 -G -G 0 0 0

0 0 -G E -3G 0 0 0 0 0 0 -G 0 0

-G 0 0 0 E -4G -G G -G 0 0 -G 0 0

0 0 0 0 -G E -3G 0 0 -G 0 0 0 0

G 0 G 0 G 0 E -3G 0 G 0 G 0 G

0 0 0 0 -G 0 0 E -3G -G 0 0 0 0

0 0 -G 0 0 -G G -G E -4G 0 0 0 -G

0 0 -G 0

n

n

nn

nn

nn

0 0 0 0 0 E -3G -G 0 0

0 0 0 -G -G 0 G 0 0 -G E -4G 0 -G

-G 0 0 0 0 0 0 0 0 0 0 E -3G -G

0 -G 0 0 0 0 G 0 -G 0 -G -G E -4G

+ + +11 p v p v

v

11 p v p vv

+ + +1-1 n v n v

v

1-1 n v n vv

+ + + + +10 n v p v p v n v

v

10 p v n v n v p vv

P = c c

P = c c

P = c c

P = c c

1P = c c c c

21

P = c c c c2

Page 6: Ian Bentley University of Notre Dame. Wigner Energy E. Wigner’s work in the 1930’s indicated that the symmetry energy will proportional to T(T+X), where

Also (particle, photon) cross sections