Finding the Incompressibility Coefficient of Nuclear Matter

Jennifer KachelMarietta College, Marietta OH

Texas A&M University, Cyclotron REU Program

Advisors: Dr. Shalom Shlomo, and Mason Anders

MotivationThe incompressibility coefficient of nuclear matter is largely important to the area of physics because it allows scientists to have a better understanding of properties of neutron stars, supernova explosions, and heavy ion collisions. By measuring the incompressibility coefficients of nuclei, nuclear matter’s incompressibility coefficient can be extracted using an A-⅓ expansion analogous to that of the semi-empirical mass formula.

Classical idea of Compressibility

In introductory physics, when studying springs: F = kx and U = ½kx2.

If you exert the same force on two different springs what happens to the potential energy of the spring with the smaller spring constant?

Take:F1 F2 k1x1 k2x2

k1 2k2 x1 12

x2

Classical Idea of Compressibility

Substituting in these values:

Conclusion: Springs with smaller spring constants can store more potential energy

Supernova explosion

U2 12

k2x22 1

212

k1

(2x1)

2 k1x121 2U1

Nuclear Matter’s Equation of State

Equation of State defined is as the binding energy per nucleon(E/A) as a function of the matter density(ρ) [1].

Saturation point (E/ρ0 , ρ0)

ρ0= 0.16 fm-3 is the density at saturation point

E/A(ρ0)= -16MeV, extrapolated from the mass formula

E/A

[M

eV]

ρ[fm-3]

Equation of State

Definition: The compressibility coefficient K is directly related to the curvature of the EOS

Using the Taylor series expansion, the first derivative of the EOS is zero and we get:

EA

[] EA

[0 ] 118

K 0

0

2

...

K k f2

d 2 EA

dkf2

k f

92d 2 E

A

d 2

0

The Giant Monopole Resonance The density oscillates as a

function of time

Isoscalar monopole: both protons and neutrons oscillate in phase

Classical picture of the breathing mode.

(r, t) 0 (r)cos(t)

EGMR

Measuring KA

Before the compressibility coefficient of nuclear matter(Knm) can be found, the compressibility coefficients of a set of nuclei(KA) must be measured

The approximation of KA is given by the expression:

Where m is the mass of the nucleon, EGMRis the Energy of the monopole and rm is the mass radius

222 mGMRA rEmK

Determining the Mass Radius The proton radius

is accurately determined by electron scattering[2].

We approximate the difference between the proton(rp) and the neutron(rn) radii by

Solving for the neutron radius:

The mass radius(rm) is then obtained from:

rp rp2

rn rp r

r rn rp 0.01Z 1.2 N ZA

A rm2 Z rp

2 N rn2

The Mass RadiusNucleus rp rp error Δr Δr error rn rn error rm rm error40Ca 3.371 0.001 ‐0.045 0.010 3.326 0.011 3.349 0.00648Ca 3.368 0.001 0.155 0.043 3.523 0.044 3.460 0.02648Ti 3.489 0.002 0.053 0.027 3.542 0.028 3.518 0.01656Fe 3.639 0.001 0.035 0.024 3.674 0.026 3.658 0.01458Ni 3.678 0.001 ‐0.012 0.017 3.666 0.018 3.672 0.01060Ni 3.716 0.001 0.027 0.023 3.743 0.025 3.730 0.01490Zr 4.184 0.001 0.070 0.032 4.254 0.033 4.223 0.019110Cd 4.495 0.002 0.083 0.035 4.578 0.038 4.542 0.022116Cd 4.550 0.002 0.138 0.044 4.687 0.047 4.631 0.028112Sn 4.515 0.002 0.058 0.031 4.573 0.033 4.547 0.019116Sn 4.548 0.001 0.095 0.038 4.643 0.039 4.602 0.023124Sn 4.598 0.001 0.162 0.049 4.760 0.050 4.695 0.030144Sm 4.871 0.006 0.088 0.038 4.959 0.044 4.921 0.027208Pb 5.435 0.001 0.163 0.052 5.598 0.053 5.534 0.032

The Mass Radius

Calculating KA

A search of the literature was required to find the energies and the error bars of the giant monopole resonance for each nucleus (data from Professor Youngblood group at the Cyclotron Institute)[4-10].

Using this data and having calculated the mass radius, we can now find the experimental values of KA

Experimental KA

Nucleus EGMR +ev ‐ev rm rm error KA + error ‐error40Ca 20.42 0.89 0.36 3.349 0.006 112.67 10.22 4.3848Ca 22.64 0.27 0.33 3.460 0.026 147.83 5.72 6.5148Ti 20.25 0.99 0.28 3.518 0.016 122.29 13.07 4.4956Fe 19.57 0.73 0.16 3.658 0.014 123.47 10.18 2.9958Ni 20.81 0.9 0.28 3.672 0.010 140.69 12.94 4.5660Ni 19.54 0.78 0.23 3.730 0.014 128.03 11.17 3.9690Zr 18.69 0.65 0.3 4.223 0.019 150.13 11.76 6.14110Cd 15.58 0.4 0.09 4.542 0.022 120.66 7.38 2.58116Cd 15.02 0.37 0.12 4.631 0.028 116.58 7.14 3.26112Sn 16.05 0.26 0.14 4.547 0.019 128.34 5.23 3.32116Sn 16.13 0.2 0.2 4.602 0.023 132.78 4.60 4.60124Sn 14.96 0.1 0.11 4.695 0.030 118.88 3.10 3.26144Sm 15.12 0.3 0.3 4.921 0.027 133.41 6.78 6.78208Pb 14 0.2 0.2 5.534 0.032 144.66 5.82 5.82

Experimental KA

100

110

120

130

140

150

160

170

0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.31

KA

[MeV

]

A-1/3

KA vs. A-1/3

The Semi-Empirical Mass Formula Binding energy per nucleon as an expansion

with terms based on the factors due to attractive and repulsive forces in the nucleus:

Using a fitting technique and comparing to experimental data on binding energies, scientists have found the values of the constants above.

There is a large amount of precise data to fit so the values for each constant are accurate.

BACVol

CSurf

A13

CCoulZ(Z 1)

A43

CSym (A 2Z)2

A2 Cp

A74

...

KA Expansion

The expansion for the compressibility constant is analogous with the empirical mass formula:

Also has parameters based on their attractive and repulsive forces in the nucleus

A statistical fit can be used to find the parameters of the expansion just as they did with the mass formula

KA KVol KSurf

A13

KCurve

A23

KSym KSS

A13

N ZA

2

KcoulZ 2

A43

...

Finding the Parameters

Each of the parameters were found by a method of least squares fit: minimizing reduced chi squared values.

Utilizing Microsoft Excel’s solver tool, the experimental data was fitted.

Also using solver, the errors on each parameter were calculated.

2 1NData NParam

KAExp KA

Theory

2

Solver

Set objective to min by changing variable cells

Calculated KA Expansion Constants

σ of KA is measuredNumber of Parameters Kvol ± Ksurf ± Kcurv ± Ksym ± Kss ± Kcoul ± Χ2

4 ‐82 10 555 44 0 0 489 380 0 0 19 2 4.255 ‐269 8 950 37 0 0 ‐2191 320 15140 1600 42 2 3.346 ‐2036 7 11313 34 ‐17293 150 1062 290 11197 1440 114 2 3.16

σ of KA is 104 128 11 2 48 0 0 0.01 606 0 0 0.01 3 1.655 119 12 23 50 0 0 2 520 807 2551 0.5 2 1.946 ‐1542 9 8852 39 ‐13796 159 ‐0.0005 421 12455 2075 88 2 1.45

Fit to microscopic theory gives: Kvol= 240 ± 20 MeV

Conclusions

More data points and better accuracy are needed to produce a better fit

EGMR for smaller nuclei needs to be measured more accurately

Solver Theory is incomplete meaning something

needs to be changed before incompressibility coefficients can be extracted

References[1]D.C. Fuls. Microscopic Description of the Breathing Mode and Nuclear Incompressibility (2005) [2]I. Angeli. Atomic Data and Nuclear Data Tables 87 (2004) 185-206. [3] Y. Tokimoto, Y-W Lui, H. L. Clark, B. John, X Chen, and D. H. Youngblood. Phys. Rev. C 74, 044308 (2006) [4]Y-W Luiet al. Phys Rev. C 83, 044327 (2011)[5]Y-W Lui et al. Phys. Rev. C 69, 034611 (2004)[6] Y-W Lui et al. Phys Rev. C 70, 014307 (2004) [7]D. H. Youngblood et al. Phys. Rev. C 69, 054312 (2004)[8]D. H. Youngblood et al. Phys. Rev. C 63, 067301 (2001) [9] Y-W Lui et al. Phys. Rev. C 73, 014314 (2006)[10] D.H. Youngblood et al. Phys. Rev. C 69, 034315 (2004)

Acknowledgements

Thanks to Dr. Sherry Yennello, Larry May and Leslie Spiekes for organizing the REU program and keeping it running smoothly this summer. Special thanks to Dr. Shalom Shlomo and Mason Anders for all of their help with my research. This program was funded by the National Science Foundation and the Department of Energy.