PHYS 6610: Graduate Nuclear and Particle Physics I H. W. Grießhammer INS Institute for Nuclear Studies The George Washington University Institute for Nuclear Studies Spring 2018 II. Phenomena 1. Shapes and Masses of Nuclei Or: Nuclear Phenomeology References: [PRSZR 5.4, 2.3, 3.1/3; HG 6.3/4, (14.5), 16.1; cursorily PRSZR 18, 19] PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.0
(a) Getting Experimental Information ⇒ M ≫ E ′ ≈ E ≫ m e → 0 heavy, spinless, composite target = � � 2 � � � � E ′ d σ Z α cos 2 θ 2 � � � q 2 ) = � F ( � � (I.7.3) � lab 2 E sin 2 θ d Ω 2 E 2 � �� � Mott when nuclear recoil negligible: M ≫ E ≈ E ′ ⇒ q 2 = − q 2 = − 2 E 2 ( 1 − cos θ ) = � translate q 2 ⇐ ⇒ θ ⇐ ⇒ E max. mom. transfer: θ → 180 ◦ min. mom. transfer: θ → 0 ◦ For E = 800 MeV (MAMI-B), 12 C ( A = 12 ): � ∆ x = 1 − q 2 θ | � q | 180 ◦ 0 . 15 fm 1500 MeV 90 ◦ 0 . 2 fm 1000 MeV 30 ◦ 400 MeV 0 . 5 fm 10 ◦ 1 . 5 fm 130 MeV [PRSZR] PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.1
“Typical” Example: 40 , 48 Ca measured over 12 orders 12 orders of magnitude. . . θ -dependence, multiplied by 10 , 0 . 1 !! [PRSZR] q | -dependence fm − 1 | � [HG 6.3] 40 Ca has less slope = ⇒ smaller size PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.2
Characterising Charge Densities ∞ � q 2 ) : = 4 π d r r q 2 = 0 ) = 1 F ( � q sin ( qr ) ρ ( r ) , normalisation F ( � Ze 0 q 2 → ∞ ) : impossible In principle: Fourier transformation = ⇒ need F ( � = ⇒ Ways out: (i) Characterise just object size: sin qr � �� � � � ∞ � qr − ( qr ) 3 q 2 → 0 ) = 4 π d r r + O (( qr ) 5 ) F ( � ρ ( r ) Ze q 3! 0 ∞ ∞ � � − q 2 q 2 → 0 ) = 4 π 4 π d r r 2 ρ ( r ) d r r 2 × r 2 ρ ( r ) + O (( qr ) 5 ) F ( � Ze 3! Ze 0 0 � �� � � �� � = 1 mean of r 2 operator � � q 2 ) ⇒ � r 2 � = − 3! d F ( � � = (square of) root-mean-square (rms) radius � � q 2 d � q 2 = 0 PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.3
Ways Out: (ii) Compare to FF of Assumed Charge Density Better parameterise as sum of a few Gauß’ians: b i exp − ( r − R i ) 2 ρ charge ( r ) = ∑ δ 2 i = ⇒ “line thickness” = parametrisation uncertainty [Mar] [PRSZR] ρ charge ( r = 0 ) decreases with A , but ρ N = A Z ρ charge ( r = 0 ) constant for large A : ρ N ≈ 0 . 17 nucleons 1 = 3 × 10 11 kg litre = 300 tons mm 3 = 3 space shuttles ≈ ( 1 . 4 × ( 2 × 0 . 7fm )) 3 � fm 3 mm 3 ρ N plateaus in heavy nuclei = ⇒ Saturation of Interaction : attraction long-range, repulsion up-close!?! Nucleons “separate but close”: Distance between nucleons in nucleus ≈ 1 . 4 × nucleon rms diameter. PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.4
Matter Densities in Nuclei (ii) Assume a certain ρ N ( r ) , calculate its FF, fit. 1st min at qR ≈ 4 . 5 [Tho 7.5] Very simple form for heavy nuclei: Fermi Distribution ρ 0 = A ρ N ( r ) = Z ρ charge 1 + exp r − c a experiments: radius at half-density c ≈ A 1 / 3 × 1 . 07 fm : � 5 3 � r 2 � ≈ A 1 / 3 × 1 . 21 fm translate hard-sphere: R = experiments: a ≈ 0 . 5 fm : independent of A !; [PRSZR p. 68] related to surface/skin thickness t = a × 4ln3 ≈ 2 . 40 fm PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.5
[HG 14.5; nucl-th/0608036; (b) Spin & Deformation: Deuteron nucl-th/0102049] q 2 ) angle-independent is the exception, not the rule. So far: no spin, no deformation = ⇒ F ( � Example Deuteron d ( np ) with J PC = 1 + − : � L = � J − � S , S = 1 L = 1 J ⊗ 1 S = 0 ⊕ 1 ⊕ 2 = ⇒ L = 0 (s-wave) or L = 2 d-wave – Parity forbids L = 1 . angular momentum = ⇒ mag. moment = ⇒ spin-spin interaction = ⇒ helicity transfer 1 G C ( q 2 ) = e � m J | J 0 | m J � ∑ avg. of hadron density, G C ( 0 ) = 1 Charge FF : 3 m J = − 1 G M ( 0 ) = : M d G M ( q 2 ) κ d , mag. moment κ d = 0 . 857 Magnetic FF : M N G Q ( q 2 ) Quadrupole FF : � d 3 r ( 3 z 2 − r 2 ) ρ charge ( r ) = 0 . 286 fm 2 quadrupole moment Q d : = G Q ( 0 ) = Ze � � � 2 � � � E ′ � M + 8 τ 2 M 4 d σ Z α cos 2 θ C + 2 τ 4 τ ( 1 + τ ) M tan 2 θ � G 2 3 G 2 G 2 G 2 d = × + � lab Q 2 E sin 2 θ d Ω 2 E 9 3 2 � �� � 2 � �� � spin = ⇒ helicity transfer Mott with τ = − q 2 as before 4 M 2 d = ⇒ Dis-entangle by θ & τ dependence. PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.6
Deuteron Form Factors [Garçon/van Orden: nucl-th/0102049] does not include most recent data (JLab), but pedagogical plot high-accuracy data up to Q 2 � ( 1 . 4 GeV ) 2 Theory quite well understood: embed nucleon FFs into deuteron, add: deuteron bound by short-distance + tensor force: one-pion exchange N † � σ · � q π ( q ) N = ⇒ photon couples to charged meson-exchanges and many π ± π ± π ± more!! open issues: zero of G Q , form for Q � 3 fm − 1 ,. . . PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.7
(c) Nuclear Binding Energies (per Nucleon) 100 Mean Field Models Densit y F un tional Shell Model(s) Effective Microscopic 10 Interactions Ab Initio Proton Number χ EFT, EFT( / π ) 3 He 4 QCD He 1 3 p d H QCD Vacuum Quark-Gluon QCD Interaction Vacuum n 50 1 5 10 100 Neutron Number B / A rapidly increases from deuteron ( A = 2 ): 1 . 1 MeV / A to about 12 C: 7 . 5 MeV / A for A � 16 (oxygen) remains around 7 . 5 ... 8 . 5 MeV / A maximal for 56 Fe– 60 Co– 62 Ni: 8 . 5 MeV / A small decrease to A ≈ 250 (U): 7 . 5 MeV / A = ⇒ “typically”, fusion gains (much) energy up to Fe; fission gains (some) energy after Fe. = ⇒ Fe has relatively large abundance: fusion and fission product. PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.8
Bethe-Weizsäcker Mass Formula & Interpretation: Liquid Drop Model Know < 3000 nuclei = ⇒ roughly parametrise ground-state binding energies, not only for stable nuclei Z 2 ( N − Z ) 2 δ Total binding energy Semi-Empirical Mass Formula B = a V A − a s A 2 / 3 − a C A 1 / 3 − a a − A 1 / 2 (1935/36) 4 A cf. ρ N ≈ 0 . 17 fm − 3 = a V = 15 . 67 MeV ⇒ saturation volume term = ⇒ Well-separated, quasi-free nucleons, next-neighbour interactions like in liquid. a s = 17 . 23 MeV less neighbours on surface = ⇒ less B surface tension a C = 0 . 714 MeV of protons = ⇒ tilt to N > Z Coulomb repulsion a a = 93 . 15 MeV (a)symmetry/Pauli term Pauli principle = ⇒ tilt to N ∼ Z − 11 . 2 MeV Z & N even opposite spins have net attraction δ = wf overlap decreases with A 0 Z or N odd pairing term + 11 . 2 MeV = ⇒ A -dependence Z & N odd PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.9
Valley of Stability around N � Z Z 2 ( A − 2 Z ) 2 δ B = a V A − a s A 2 / 3 − a C A 1 / 3 − a a − A 1 / 2 4 A 2 ( a a + a C A 2 / 3 ) � A a a A = ⇒ Parabola in Z at fixed A with Z min = 2 [HG 16.2/3] PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.10
Valley of Stability Probe nuclear interactions A = 56 by pushing to ”drip-lines” : Facility for Rare Isotope Beams FRIB at MSU A = 150 PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.11
(d) Application: Nuclear Fission [PRSZR 3.3] For A > 60 , fission can release energy, but must overcome fission barrier = ⇒ assume fission into 2 equal fragments Estimate: infinitesimal deformation into ellipsoid (egg) with excentricity ε at constant volume = ⇒ surface tension ր , Coulomb ց : E ( sphere ) − E ( ellipsoid ) � 2 a s A 2 / 3 − a C Z 2 A − 1 / 3 � ∼ ε 2 5 Fission barrier classically overcome when ≤ 0 : Z 2 A ∼ 2 a s ≈ 48 e.g. Z > 114 , A > 270 � a C � below: QM tunnel prob. ∝ exp − 2 2 M ( E − V ) between points with r ( E = V ) Induced Fission: importance of pairing energy n + 238 92 U → 239 92 U : even-even → even-odd ⇒ invest pairing E δ = δ = 11 . 2MeV = √ = 0 . 7 MeV 239 n + 235 92 U → 236 92 U : even-odd → even-even δ = ⇒ gain pairing energy √ = − 0 . 7 MeV = ⇒ can use thermal neutrons (higher σ !) 236 PHYS 6610: Graduate Nuclear and Particle Physics I, Spring 2018 H. W. Grießhammer, INS, George Washington University II.1.12
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