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REVIEW 3 major objections 5 minor 106 references

Three-nucleon force effects in polarization transfers from the doubly spin-polarized initial neutron-deuteron state to the outgoing neutron in neutron-deuteron scattering

T0 review · 3 major / 5 minor · reviewed 2026-08-12 · deepseek-v4-flash

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Pith's one-line read The paper claims that a double polarization transfer coefficient in neutron-deuteron scattering shows about 40% three-nucleon-force effects at 135 MeV, roughly four times the effects on its single-transfer constituents.

desk verdict First predictions for double spin-transfer observables in nd scattering, with a plausible and experimentally motivated claim that K_{y,y}^{y'} is a sensitive 3NF probe, though the headline 40% rests on a single (cD,cE) point. read the letter →

arxiv 2411.15834 v2 pith:KKHYJPTV submitted 2024-11-24 nucl-th

H. Wita{l}a , J. Golak , R. Skibiński , H. Sakai , K. Sekiguchi This is my paper · ORCID
classification nucl-th
keywords three-nucleonforcepolarizationtransferneutron-deuteronscatteringspinobservablesFaddeevequationschiraleffectivefieldtheorydeuteronbreakupelastic
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper proposes that a newly accessible class of spin observables in nucleon-deuteron scattering, the polarization transfer from a doubly spin-polarized initial state to the outgoing neutron, is far more sensitive to three-nucleon forces than the single-polarization transfers studied before. In elastic neutron-deuteron scattering at 135 MeV, the coefficient $K_{y,y}^{y'}$ changes by about 40% when a three-nucleon force is added to a chiral two-nucleon potential, while the constituent single-transfer coefficients change by only about 10%. The paper argues that the effect is concentrated at center-of-mass angles near 110 degrees, where the observable is large enough to measure with currently achievable beam and target polarizations. This makes the observable a sharper experimental test of three-nucleon dynamics than the single-transfer coefficients.

What carries the argument

The machinery is the Faddeev equation for three-nucleon scattering, solved in momentum-space partial waves, with the transition operator $T$ built from a two-nucleon $t$-matrix and a three-nucleon force $V_{123}$. From the density-matrix formalism, the double polarization transfer tensors are defined by traces of $T \rho_{\mathrm{in}} T^\dagger$ with two initial-state polarization tensors and one final-state tensor; the Cartesian coefficient $K_{y,y}^{y'}$ is a specific linear combination of these spherical tensors listed in Appendix A. The calculation uses the chiral N$^4$LO+ nucleon-nucleon potential with regulator $\Lambda = 450$ MeV combined with the chiral N$^2$LO three-nucleon force with strength parameters $c_D = 2.0$ and $c_E = 0.2866$ fixed from the triton binding energy and the 65 MeV proton-deuteron cross-section minimum. Parity conservation selects which of the 81 Cartesian coefficients can be nonzero.

What would settle it

A direct measurement of $K_{y,y}^{y'}$ in elastic proton-deuteron or neutron-deuteron scattering at $E = 135$ MeV at center-of-mass angles 90 to 130 degrees would settle the claim: finding a 3NF-induced deviation close to 10 percent rather than about 40 percent, or finding the effect at different angles, would contradict the prediction. A much cheaper check is to recompute the same observable with a different three-nucleon force regulator or with the N$^2$LO three-nucleon force fitted differently and compare the size and location of the effect.

Watch

Extended reading notes

Core claim

The central claim is that the double spin-polarization transfer coefficient $K_{y,y}^{y'}$ in elastic neutron-deuteron scattering is a sensitive probe of three-nucleon forces. At incoming neutron laboratory energy $E = 135$ MeV, adding a chiral N$^2$LO three-nucleon force to the N$^4$LO+ chiral nucleon-nucleon potential changes $K_{y,y}^{y'}$ by about 40% near $\Theta_{\mathrm{c.m.}} \approx 110^\circ$, roughly four times the about 10% effect seen in the constituent single polarization transfers $K_{0,y}^{y'}$ and $K_{y,0}^{y'}$. Similar enhancements appear for $K_{x,y}^{x'}$ and $K_{z,y}^{x'}$ at 135 MeV, and for the special coefficient $K_{z,x}^{y'}$, whose two constituent single transfers both vanish, with three-nucleon-force effects reaching about 50%. In deuteron breakup, the same double transfers show sizable three-nucleon-force effects in final-state-interaction kinematics at 135 MeV but are nearly insensitive in quasi-free-scattering kinematics. With neutron polarization 0.2 and deuteron polarization 0.6, the outgoing neutron polarization is predicted to be large enough for a feasible measurement of $K_{y,y}^{y'}$.

Load-bearing premise

One load-bearing premise is that the 40 percent effect is not an artifact of the single three-nucleon force model used: the strength parameters were fixed by matching the triton binding energy and the 65 MeV proton-deuteron cross-section minimum, and the paper does not test how the effect changes with a different regulator, chiral order, or fitting choice.

Editorial extensions

If this is right

  • At 135 MeV, $K_{y,y}^{y'}$ in elastic nd scattering becomes a far more sensitive three-nucleon-force probe than its single-transfer constituents, with a roughly 40% effect localized around $\Theta_{\mathrm{c.m.}} \approx 110^\circ$.
  • The coefficient $K_{z,x}^{y'}$, for which both constituent single transfers vanish, isolates the double-transfer contribution; its three-nucleon-force effects reach about 50% at 135 MeV.
  • In deuteron breakup, FSI(1-3) kinematics at 135 MeV show large three-nucleon-force effects in the double transfers, while QFS(1-2) kinematics are almost insensitive, so QFS configurations can serve as a three-nucleon-force-insensitive reference.
  • With neutron polarization 0.2 and deuteron polarization 0.6, the outgoing neutron polarization is predicted to be large enough (about 0.6) that the double-transfer contribution (about 0.15) is measurable with present-day polarized sources.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • Inference: if the 40% enhancement survives changes in the three-nucleon-force parametrization, $K_{y,y}^{y'}$ could discriminate among chiral potentials more sharply than the cross-section minimum or single-transfer observables.
  • Inference: the near-vanishing of three-nucleon-force effects in QFS kinematics suggests that comparing FSI and QFS double-transfer data in one experiment could separate three-nucleon-force contributions from final-state interactions.
  • Inference: the feasibility estimate assumes polarization products of about 0.12; experiments with lower beam polarizations would still be viable in the same angular region but would require more statistics.
  • Inference: extending the same double-transfer analysis to proton-deuteron scattering with the Coulomb interaction included at lower energies would test whether the large three-nucleon-force effect persists after Coulomb corrections.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

3 major / 5 minor

Summary. This manuscript studies polarization transfer coefficients from the doubly spin-polarized initial nucleon-deuteron state to the outgoing nucleon in elastic nd scattering and in selected nd breakup configurations. The authors define the new double spin-polarization transfer observables in terms of spherical tensor amplitudes, provide the full set of 81 Cartesian coefficients in Appendix A, and compute them by solving the three-nucleon Faddeev equations with the chiral SMS N4LO+ NN potential (Lambda=450 MeV) alone and combined with the N2LO 3NF at the parameter point (c_D=2.0, c_E=0.2866). They report that the 3NF sensitivity of K_{y,y}^{y'} in elastic nd scattering reaches about 40% at E=135 MeV, which they describe as the most promising observable for future measurement, and they estimate the outgoing neutron polarization to support experimental feasibility.

Significance. If the quantitative claims hold, the paper identifies a new class of spin observables in three-nucleon scattering and gives concrete motivation for their measurement at RIKEN. The formalism is a strength: the Faddeev equations, the spherical-tensor definitions of the transfer coefficients, the rotation to the outgoing-particle frame, and the complete Appendix A listing of Cartesian observables are all stated explicitly and would allow independent implementation. It is also a strength that the 3NF strengths are fitted to triton binding and the pd cross-section minimum, not to the polarization-transfer observable, so the reported sensitivity is not circular. The main weakness is that the headline 40% enhancement rests on a single Hamiltonian and a single parameter point, with no propagation of the quoted c_D uncertainty and no regulator or chiral-order variation; the paper also shifts between nd and pd framing despite neglecting the pp Coulomb force. These issues affect the central recommendation rather than the underlying formal derivation, so they are repairable in revision.

major comments (3)
  1. [Section II (formalism) vs Section IV (summary)] The headline 3NF enhancement of about 40% in K_{y,y}^{y'} at E=135 MeV is computed at the single parameter point (c_D=2.0, c_E=0.2866). The paper itself reports that the fit to the pd cross-section minimum gives c_D=2.10±0.24, so the adopted value is within the fit uncertainty, but the uncertainty is not propagated and no regulator variation (e.g., Lambda=500 MeV) or change in the chiral order of the 3NF is tested. Because the observable magnitude is about 0.3 near Theta_cm≈110°, a 40% effect is a shift of about 0.12, and a factor-of-two variation in this shift would change the stated enhancement to a range that no longer supports the recommendation of K_{y,y}^{y'} as the most promising observable. Please add a robustness study, or at least an explicit and quantified caveat that the enhancement is a single-model prediction.
  2. [Section II vs Section IV] The calculation omits the proton-proton Coulomb force, and Section II explicitly states that 'the obtained results and conclusions are restricted at these energies and angles to the nd system.' Section IV, however, presents the observables as 'presently accessible' in pd elastic scattering and cites the planned RIKEN pd experiment. This mismatch is load-bearing for the experimental recommendation. The authors should either estimate the Coulomb corrections to K_{y,y}^{y'} at E=135 MeV in the relevant angular region, or recast the recommendation as an nd prediction and note that a pd measurement requires a Coulomb-corrected calculation.
  3. [Section III vs Section IV] The quantitative claim about the relative size of the 3NF effect is stated inconsistently: Section III says the effect reaches about 10% for single transfer coefficients but about 40% for K_{y,y}^{y'}, which is a factor of about four, while Section IV says the effect is 'by a factor of about two larger' than for the constituent single transfers. The abstract follows Section III. Please correct the factor and ensure that the abstract, Section III, and Section IV quote the same quantitative comparison.
minor comments (5)
  1. [References] Reference [3] lists the journal as 'Phys. Phys. C'; this should be 'Phys. Rev. C'.
  2. [Author list and references] The text contains LaTeX artifacts such as 'Wita/suppress la' and 'Kuro´s-˙Zo/suppress lnierczuk'; these should be cleaned before publication.
  3. [Section IV] The phrase 'the magnitude of K_y,y^y′ is of the order of -0.3' appears to contain a notational slip; the symbol should match the defined observable K_{y,y}^{y'}.
  4. [Figure captions] Several figure captions use 'Theta_c,m.' instead of 'Theta_c.m.'; please correct the typo.
  5. [Equation (17)] Equation (17) would benefit from an explicit statement that the denominator and all numerator terms are taken from the NN-only calculation, since the feasibility estimate is then subject to the same single-model limitation as the rest of the study.

Circularity Check

0 steps flagged · score 0.0 of 10

No significant circularity: the 3NF sensitivity predictions are computed from an independent Faddeev solution with 3NF strengths fitted to unrelated data.

full rationale

The paper's central claim is that the double spin-polarization transfer coefficient K_{y,y}^{y'} in elastic nd scattering shows 3NF effects up to about 40% at E=135 MeV, larger than the effects in its constituent single transfer coefficients. The derivation chain is self-contained: the Faddeev equation (Eq. 1) is solved with an established NN potential (SMS N4LO+ with Lambda=450 MeV) and an N2LO 3NF, and the polarization transfer observables are computed from the resulting amplitudes via Eqs. (14)-(16). The 3NF strengths cD=2.0 and cE=0.2866 are determined from the triton binding energy and the pd elastic cross-section minimum at 65 MeV, which are independent of the polarization transfer observables being predicted. Thus the large 3NF effect in K_{y,y}^{y'} is not fitted to that observable and does not reduce to its input by construction. The paper's self-citations to the Faddeev formalism [6-9] and to earlier 3NF studies [19,20] are not load-bearing in a circular sense: these are independently established numerical methods and data-fitting procedures, not unverified premises invoked to force the conclusion. The paper's quoted uncertainty of cD=2.10 +/- 0.24 and its lack of propagation into the K_{y,y}^{y'} prediction is a robustness or uncertainty concern, not a circularity. No step in the derivation equates the prediction to a fitted parameter or to a self-citation chain.

Assumptions & free parameters 2 free parameters · 5 assumptions · 0 invented entities

The paper's predictions depend on the fitted 3NF strengths cD and cE from prior literature, but no new particles, forces, or conserved quantities are introduced. The theoretical inputs are the standard Faddeev formalism and the chiral Hamiltonian from the literature, so the free-parameter burden is limited to the two 3NF low-energy constants.

free parameters (2)
  • cD = 2.0
    Strength of the short-range N2LO three-nucleon force; determined from the correlation line using triton binding energy and the Nd cross-section minimum at about 70 MeV, with a quoted fit value of 2.10 ± 0.24 from pd data at 65 MeV. The central predictions depend on this value.
  • cE = 0.2866
    Strength of the short-range N2LO three-nucleon force; determined together with cD on the correlation line from the same fit. Combined with cD, it sets the 3NF used in all NN+3NF predictions.
assumptions (5)
  • standard math The Faddeev-type integral equation (Eq. 1) with the 2N t-matrix and 3NF provides an exact solution of the three-nucleon scattering problem.
    Invoked in Section II; this is the standard nonrelativistic formulation for three-nucleon scattering with pairwise and three-body forces.
  • domain assumption The chiral SMS N4LO+ NN potential with Lambda=450 MeV plus the N2LO 3NF adequately represents the nuclear interaction at the energies studied.
    Used throughout Section III; the validity of chiral effective field theory at these momenta and the representativeness of this single Hamiltonian are assumed.
  • domain assumption The partial-wave expansion is converged with jmax=5, Jmax=25/2 and 3NF acting up to J=7/2.
    Stated in Section II as the numerical truncation; convergence for the new observables is not demonstrated in the paper.
  • domain assumption The pp Coulomb force can be neglected for the energies and angles considered, allowing nd results to be applied to the pd system.
    Stated in Section II; the restriction is acknowledged, and the paper limits its conclusions at low energies and forward angles for the pd case.
  • standard math The spin-polarization transfer formalism of Ohlsen applies to the definition of double spin-transfer coefficients.
    Invoked in Section II via Refs. [1,2]; the Cartesian-to-spherical tensor relations and parity-selection rules follow from standard angular momentum algebra.

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Cite this review

Pith. "Pith review of Three-nucleon force effects in polarization transfers from the doubly spin-polarized initial neutron-deuteron state to the outgoing neutron in neutron-deuteron scattering." pith.science (2026). https://pith.science/paper/KKHYJPTV

@misc{pith2026241115834,
  author       = {Pith},
  title        = {Pith review of: Three-nucleon force effects in polarization transfers from the doubly spin-polarized initial neutron-deuteron state to the outgoing neutron in neutron-deuteron scattering},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/KKHYJPTV}},
  note         = {Machine review of arXiv:2411.15834}
}
abstract

We discuss new spin observables presently accessible to measurement in the proton-deuteron (pd) system, namely polarization transfer coefficients from doubly spin-polarized initial state to the outgoing nucleon in the elastic nucleon-deuteron (Nd) scattering and in the nucleon-induced deuteron breakup reactions. The sensitivity of these observables to three-nucleon force (3NF) effects is investigated and compared to sensitivities of the constituent standard single polarization transfer coefficients in the neutron-deuteron (nd) system. $K_{y,y}^{y'}$ in elastic nd scattering, for which large 3NF effects, up to 40\%, have been found at higher energies, seems the most promising observable to measure.

Figures

Figures reproduced from arXiv: 2411.15834 by the authors.

Figure 1
Figure 1. FIG. 1. (color online) Polarization transfer coefficients fr [PITH_FULL_IMAGE:figures/full_fig_p021_1.png] view at source ↗
Figure 2
Figure 2. FIG. 2. (color online) The same as in Fig. 1 but for [PITH_FULL_IMAGE:figures/full_fig_p022_2.png] view at source ↗
Figure 3
Figure 3. FIG. 3. (color online) The same as in Fig. 1 but for the tensor p [PITH_FULL_IMAGE:figures/full_fig_p023_3.png] view at source ↗
Figures from the paper (4 more)
Figure 4
Figure 4. Figure 4: FIG. 4. (color online) Polarization transfer coefficients fr [PITH_FULL_IMAGE:figures/full_fig_p024_4.png]
Figure 5
Figure 5. Figure 5: FIG. 5. (color online) The same as in Fig. 4 but for [PITH_FULL_IMAGE:figures/full_fig_p025_5.png]
Figure 6
Figure 6. Figure 6: FIG. 6. (color online) Polarization transfer coefficients in [PITH_FULL_IMAGE:figures/full_fig_p026_6.png]
Figure 7
Figure 7. Figure 7: FIG. 7. (color online) The polarization of the outgoing neut [PITH_FULL_IMAGE:figures/full_fig_p027_7.png]

Discussion (0). Continue with ORCID to comment.

Reference graph

Works this paper leans on

106 extracted references · 75 canonical work pages

  1. [1]

    K z′ x,x = 2 3 × 1 2 √ 3 2(t10 1−1,1−1 − t10 1−1,1+1 − t10 1+1,1−1 + t10 1+1,1+1)

  2. [2]

    K z′ y,x = − 2 3 × i 2 √ 3 2(t10 1−1,1−1 − t10 1−1,1+1 + t10 1+1,1−1 − t10 1+1,1+1)

  3. [3]

    and QFS(1-2). While practically no 3NF effects are revealed in quas i-free-scattering configurations, significant 3NF effects are clearly seen for all four double spin-polarization 13 transfer coefficients in FSI(1-3) at E = 135 MeV. The most sizable effects are localized in FSI(1-3) configurations around their production angle θlab 1 ≈ 40o. We are convinced that...

    2024

  4. [4]

    K z′ z,x = 2 3 × √ 3 2 (t10 10,1−1 − t10 10,1+1)

  5. [5]

    K z′ x,y = − 2 3 × i 2 √ 3 2(t10 1−1,1−1 + t10 1−1,1+1 − t10 1+1,1−1 − t10 1+1,1+1)

  6. [6]

    K z′ y,y = − 2 3 × 1 2 √ 3 2(t10 1−1,1−1 + t10 1−1,1+1 + t10 1+1,1−1 + t10 1+1,1+1)

  7. [7]

    K z′ z,y = − 2 3 × i √ 3 2 (t10 10,1−1 + t10 10,1+1)

  8. [8]

    K z′ x,z = 2 3 × √ 3 2 (t10 1−1,10 − t10 1+1,10) 15

Show all 106 references
  1. [9]

    K z′ y,z = − 2 3 × i √ 3 2 (t10 1−1,10 + t10 1+1,10)

  2. [10]

    K z′ z,z = 2 3 × √ 3 2t10 10,10

  3. [11]

    K z′ xx,x = 1 2[−(t10 20,1−1 − t10 20,1+1) + √ 3 2(t10 2−2,1−1 − t10 2−2,1+1 + t10 2+2,1−1 − t10 2+2,1+1)]

  4. [12]

    K z′ yy,x = 1 2[−(t10 20,1−1 − t10 20,1+1) − √ 3 2(t10 2−2,1−1 − t10 2−2,1+1 + t10 2+2,1−1 − t10 2+2,1+1)]

  5. [13]

    K z′ xy,x = − 3 2 × i√ 6(t10 2−2,1−1 − t10 2−2,1+1 − t10 2+2,1−1 + t10 2+2,1+1)

  6. [14]

    K z′ xz,x = 3 2 × 1√ 6(t10 2−1,1−1 − t10 2−1,1+1 − t10 2+1,1−1 + t10 2+1,1+1)

  7. [15]

    K z′ yz,x = − 3 2 × i√ 6(t10 2−1,1−1 − t10 2−1,1+1 + t10 2+1,1−1 − t10 2+1,1+1)

  8. [16]

    K z′ zz,x = (t10 20,1−1 − t10 20,1+1)

  9. [17]

    K z′ xx,y = 1 2[i(t10 20,1−1 + t10 20,1+1) − i √ 3 2(t10 2−2,1−1 + t10 2−2,1+1 + t10 2+2,1−1 + t10 2+2,1+1)]

  10. [18]

    K z′ yy,y = 1 2[i(t10 20,1−1 + t10 20,1+1) + i √ 3 2(t10 2−2,1−1 + t10 2−2,1+1 + t10 2+2,1−1 + t10 2+2,1+1)]

  11. [19]

    K z′ xy,y = 3 2 × 1√ 6(−t10 2−2,1−1 − t10 2−2,1+1 + t10 2+2,1−1 + t10 2+2,1+1)

  12. [20]

    K z′ xz,y = − 3 2 × i√ 6(t10 2−1,1−1 + t10 2−1,1+1 − t10 2+1,1−1 − t10 2+1,1+1)

  13. [21]

    K z′ yz,y = − 3 2 × 1√ 6(t10 2−1,1−1 + t10 2−1,1+1 + t10 2+1,1−1 + t10 2+1,1+1)

  14. [22]

    K z′ zz,y = −i(t10 20,1−1 + t10 20,1+1)

  15. [23]

    K z′ xx,z = 1 2[− √ 2t10 20,10 + √ 3(t10 2−2,10 + t10 2+2,10)]

  16. [24]

    K z′ yy,z = 1 2[− √ 2t10 20,10 − √ 3(t10 2−2,10 + t10 2+2,10)]

  17. [25]

    K z′ xy,z = − 3 2 × i√ 3(t10 2−2,10 − t10 2+2,10)

  18. [26]

    K z′ xz,z = 3 2 × 1√ 3(t10 2−1,10 − t10 2+1,10)

  19. [27]

    K z′ yz,z = − 3 2 × i√ 3(t10 2−1,10 + t10 2+1,10)

  20. [28]

    K z′ zz,z = √ 2t10 20,10

  21. [29]

    K x′ x,x = 2 3 × √ 3 4 (t1−1 1−1,1−1 − t1+1 1−1,1−1 − t1−1 1−1,1+1 + t1+1 1−1,1+1 − t1−1 1+1,1−1 + t1+1 1+1,1−1 + t1−1 1+1,1+1 − t1+1 1+1,1+1)

  22. [30]

    K x′ y,x = − 2 3 × i √ 3 4 (t1−1 1−1,1−1 − t1+1 1−1,1−1 − t1−1 1−1,1+1 + t1+1 1−1,1+1 + t1−1 1+1,1−1 − t1+1 1+1,1−1 − t1−1 1+1,1+1 + t1+1 1+1,1+1)

  23. [31]

    K x′ z,x = 2 3 × 1 2 √ 3 2(t1−1 10,1−1 − t1+1 10,1−1 − t1−1 10,1+1 + t1+1 10,1+1)

  24. [32]

    K x′ x,y = − 2 3 × i √ 3 4 (t1−1 1−1,1−1 − t1+1 1−1,1−1 + t1−1 1−1,1+1 − t1+1 1−1,1+1 − t1−1 1+1,1−1 + t1+1 1+1,1−1 − t1−1 1+1,1+1 + t1+1 1+1,1+1)

  25. [33]

    K x′ y,y = 2 3 × √ 3 4 (−t1−1 1−1,1−1 + t1+1 1−1,1−1 − t1−1 1−1,1+1 + t1+1 1−1,1+1 − t1−1 1+1,1−1 + t1+1 1+1,1−1 − t1−1 1+1,1+1 + t1+1 1+1,1+1)

  26. [34]

    K x′ z,y = − 2 3 × i 2 √ 3 2(t1−1 10,1−1 − t1+1 10,1−1 + t1−1 10,1+1 − t1+1 10,1+1)

  27. [35]

    K x′ x,z = 2 3 × 1 2 √ 3 2(t1−1 1−1,10 − t1+1 1−1,10 − t1−1 1+1,10 + t1+1 1+1,10) 16

  28. [36]

    K x′ y,z = − 2 3 × i 2 √ 3 2(t1−1 1−1,10 − t1+1 1−1,10 + t1−1 1+1,10 − t1+1 1+1,10)

  29. [37]

    K x′ z,z = 2 3 × √ 3 2 (t1−1 10,10 − t1+1 10,10)

  30. [38]

    K x′ xx,x = 1 2[− 1√ 2(t1−1 20,1−1 − t1+1 20,1−1 − t1−1 20,1+1 + t1+1 20,1+1) + √ 3 2 (t1−1 2−2,1−1 − t1+1 2−2,1−1 − t1−1 2−2,1+1 + t1+1 2−2,1+1 + t1−1 2+2,1−1 − t1+1 2+2,1−1 − t1−1 2+2,1+1 + t1+1 2+2,1+1)]

  31. [39]

    K x′ yy,x = 1 2[− 1√ 2(t1−1 20,1−1 − t1+1 20,1−1 − t1−1 20,1+1 + t1+1 20,1+1) − √ 3 2 (−t1−1 2−2,1−1 + t1+1 2−2,1−1 + t1−1 2−2,1+1 − t1+1 2−2,1+1 − t1−1 2+2,1−1 + t1+1 2+2,1−1 + t1−1 2+2,1+1 − t1+1 2+2,1+1)

  32. [40]

    K x′ xy,x = − 3 2 × i 2 √ 3(t1−1 2−2,1−1 −t1+1 2−2,1−1 −t1−1 2−2,1+1 +t1+1 2−2,1+1 −t1−1 2+2,1−1 +t1+1 2+2,1−1 +t1−1 2+2,1+1 − t1+1 2+2,1+1)

  33. [41]

    K x′ xz,x = 3 2 × 1 2 √ 3(t1−1 2−1,1−1 − t1+1 2−1,1−1 − t1−1 2−1,1+1 + t1+1 2−1,1+1 − t1−1 2+1,1−1 + t1+1 2+1,1−1 + t1−1 2+1,1+1 − t1+1 2+1,1+1)

  34. [42]

    K x′ yz,x = − 3 2 × i 2 √ 3(t1−1 2−1,1−1 −t1+1 2−1,1−1 −t1−1 2−1,1+1 +t1+1 2−1,1+1 +t1−1 2+1,1−1 −t1+1 2+1,1−1 −t1−1 2+1,1+1 + t1+1 2+1,1+1)

  35. [43]

    K x′ zz,x = 1 √ 2(t1−1 20,1−1 − t1+1 20,1−1 − t1−1 20,1+1 + t1+1 20,1+1)

  36. [44]

    K x′ xx,y = 1 2[ i√ 2(t1−1 20,1−1 − t1+1 20,1−1 + t1−1 20,1+1 − t1+1 20,1+1) − i √ 3 2 (t1−1 2−2,1−1 − t1+1 2−2,1−1 + t1−1 2−2,1+1 − t1+1 2−2,1+1 + t1−1 2+2,1−1 − t1+1 2+2,1−1 + t1−1 2+2,1+1 − t1+1 2+2,1+1)]

  37. [45]

    K x′ yy,y = 1 2[ i√ 2(t1−1 20,1−1 − t1+1 20,1−1 + t1−1 20,1+1 − t1+1 20,1+1) + i √ 3 2 (t1−1 2−2,1−1 − t1+1 2−2,1−1 + t1−1 2−2,1+1 − t1+1 2−2,1+1 + t1−1 2+2,1−1 − t1+1 2+2,1−1 + t1−1 2+2,1+1 − t1+1 2+2,1+1)]

  38. [46]

    K x′ xy,y = 3 2 × 1 2 √ 3(−t1−1 2−2,1−1 +t1+1 2−2,1−1 −t1−1 2−2,1+1 +t1+1 2−2,1+1 +t1−1 2+2,1−1 −t1+1 2+2,1−1 +t1−1 2+2,1+1 − t1+1 2+2,1+1)

  39. [47]

    K x′ xz,y = − 3 2 × i 2 √ 3(t1−1 2−1,1−1 −t1+1 2−1,1−1 +t1−1 2−1,1+1 −t1+1 2−1,1+1 −t1−1 2+1,1−1 +t1+1 2+1,1−1 −t1−1 2+1,1+1 + t1+1 2+1,1+1)

  40. [48]

    K x′ yz,y = 3 2 × 1 2 √ 3(−t1−1 2−1,1−1 + t1+1 2−1,1−1 −t1−1 2−1,1+1 + t1+1 2−1,1+1 −t1−1 2+1,1−1 + t1+1 2+1,1−1 −t1−1 2+1,1+1 + t1+1 2+1,1+1)

  41. [49]

    K x′ zz,y = − i√ 2(t1−1 20,1−1 − t1+1 20,1−1 + t1−1 20,1+1 − t1+1 20,1+1)

  42. [50]

    K x′ xx,z = 1 2[−(t1−1 20,10 − t1+1 20,10) + √ 3 2 (t1−1 2−2,10 − t1+1 2−2,10 + t1−1 2+2,10 − t1+1 2+2,10)]

  43. [51]

    K x′ yy,z = 1 2[−(t1−1 20,10 − t1+1 20,10) − √ 3 2 (t1−1 2−2,10 − t1+1 2−2,10 + t1−1 2+2,10 − t1+1 2+2,10)]

  44. [52]

    K x′ xy,z = − 3 2 × i√ 6(t1−1 2−2,10 − t1+1 2−2,10 − t1−1 2+2,10 + t1+1 2+2,10)

  45. [53]

    K x′ xz,z = 3 2 × 1√ 6(t1−1 2−1,10 − t1+1 2−1,10 − t1−1 2+1,10 + t1+1 2+1,10)

  46. [54]

    K x′ yz,z = − 3 2 × i√ 6(t1−1 2−1,10 − t1+1 2−1,10 + t1−1 2+1,10 − t1+1 2+1,10)

  47. [55]

    K x′ zz,z = (t1−1 20,10 − t1+1 20,10)

  48. [56]

    K y′ x,x = 2 3 × i √ 3 4 (t1−1 1−1,1−1 + t1+1 1−1,1−1 − t1−1 1−1,1+1 − t1+1 1−1,1+1 − t1−1 1+1,1−1 − t1+1 1+1,1−1 + t1−1 1+1,1+1 + 17 t1+1 1+1,1+1)

  49. [57]

    K y′ y,x = 2 3 × √ 3 4 (t1−1 1−1,1−1 + t1+1 1−1,1−1 − t1−1 1−1,1+1 − t1+1 1−1,1+1 + t1−1 1+1,1−1 + t1+1 1+1,1−1 − t1−1 1+1,1+1 − t1+1 1+1,1+1)

  50. [58]

    K y′ z,x = 2 3 × i 2 √ 3 2(t1−1 10,1−1 + t1+1 10,1−1 − t1−1 10,1+1 − t1+1 10,1+1)

  51. [59]

    K y′ x,y = 2 3 × √ 3 4 (t1−1 1−1,1−1 + t1+1 1−1,1−1 + t1−1 1−1,1+1 + t1+1 1−1,1+1 − t1−1 1+1,1−1 − t1+1 1+1,1−1 − t1−1 1+1,1+1 − t1+1 1+1,1+1)

  52. [60]

    K y′ y,y = − 2 3 × i √ 3 4 (t1−1 1−1,1−1 + t1+1 1−1,1−1 + t1−1 1−1,1+1 + t1+1 1−1,1+1 + t1−1 1+1,1−1 + t1+1 1+1,1−1 + t1−1 1+1,1+1 + t1+1 1+1,1+1)

  53. [61]

    K y′ z,y = 2 3 × 1 2 √ 3 2(t1−1 10,1−1 + t1+1 10,1−1 + t1−1 10,1+1 + t1+1 10,1+1)

  54. [62]

    K y′ x,z = 2 3 × i 2 √ 3 2(t1−1 1−1,10 + t1+1 1−1,10 − t1−1 1+1,10 − t1+1 1+1,10)

  55. [63]

    K y′ y,z = 2 3 × 1 2 √ 3 2(t1−1 1−1,10 + t1+1 1−1,10 + t1−1 1+1,10 + t1+1 1+1,10)

  56. [64]

    K y′ z,z = 2 3 × i √ 3 2 (t1−1 10,10 + t1+1 10,10)

  57. [65]

    K y′ xx,x = 1 2[− i√ 2(t1−1 20,1−1 + t1+1 20,1−1 − t1−1 20,1+1 − t1+1 20,1+1) + i √ 3 2 (t1−1 2−2,1−1 + t1+1 2−2,1−1 − t1−1 2−2,1+1 − t1+1 2−2,1+1 + t1−1 2+2,1−1 + t1+1 2+2,1−1 − t1−1 2+2,1+1 − t1+1 2+2,1+1)]

  58. [66]

    K y′ yy,x = 1 2[− i√ 2(t1−1 20,1−1 + t1+1 20,1−1 − t1−1 20,1+1 − t1+1 20,1+1) − i √ 3 2 (t1−1 2−2,1−1 + t1+1 2−2,1−1 − t1−1 2−2,1+1 − t1+1 2−2,1+1 + t1−1 2+2,1−1 + t1+1 2+2,1−1 − t1−1 2+2,1+1 − t1+1 2+2,1+1)]

  59. [67]

    K y′ xy,x = 3 2 × 1 2 √ 3(t1−1 2−2,1−1 + t1+1 2−2,1−1 − t1−1 2−2,1+1 − t1+1 2−2,1+1 − t1−1 2+2,1−1 − t1+1 2+2,1−1 + t1−1 2+2,1+1 + t1+1 2+2,1+1)

  60. [68]

    K y′ xz,x = 3 2 × i 2 √ 3(t1−1 2−1,1−1 + t1+1 2−1,1−1 − t1−1 2−1,1+1 − t1+1 2−1,1+1 − t1−1 2+1,1−1 − t1+1 2+1,1−1 + t1−1 2+1,1+1 + t1+1 2+1,1+1)

  61. [69]

    K y′ yz,x = 3 2 × 1 2 √ 3(t1−1 2−1,1−1 + t1+1 2−1,1−1 − t1−1 2−1,1+1 − t1+1 2−1,1+1 + t1−1 2+1,1−1 + t1+1 2+1,1−1 − t1−1 2+1,1+1 − t1+1 2+1,1+1)

  62. [70]

    K y′ zz,x = i√ 2(t1−1 20,1−1 + t1+1 20,1−1 − t1−1 20,1+1 − t1+1 20,1+1)

  63. [71]

    K y′ xx,y = 1 2[− 1√ 2(t1−1 20,1−1 + t1+1 20,1−1 + t1−1 20,1+1 + t1+1 20,1+1) + √ 3 2 (t1−1 2−2,1−1 + t1+1 2−2,1−1 + t1−1 2−2,1+1 + t1+1 2−2,1+1 + t1−1 2+2,1−1 + t1+1 2+2,1−1 + t1−1 2+2,1+1 + t1+1 2+2,1+1)]

  64. [72]

    K y′ yy,y = 1 2[− 1√ 2(t1−1 20,1−1 + t1+1 20,1−1 + t1−1 20,1+1 + t1+1 20,1+1) − √ 3 2 (t1−1 2−2,1−1 + t1+1 2−2,1−1 + t1−1 2−2,1+1 + t1+1 2−2,1+1 + t1−1 2+2,1−1 + t1+1 2+2,1−1 + t1−1 2+2,1+1 + t1+1 2+2,1+1)]

  65. [73]

    K y′ xy,y = − 3 2 × i 2 √ 3(t1−1 2−2,1−1 +t1+1 2−2,1−1 +t1−1 2−2,1+1 +t1+1 2−2,1+1 −t1−1 2+2,1−1 −t1+1 2+2,1−1 −t1−1 2+2,1+1 − t1+1 2+2,1+1)

  66. [74]

    K y′ xz,y = 3 2 × 1 2 √ 3(t1−1 2−1,1−1 + t1+1 2−1,1−1 + t1−1 2−1,1+1 + t1+1 2−1,1+1 − t1−1 2+1,1−1 − t1+1 2+1,1−1 − t1−1 2+1,1+1 − t1+1 2+1,1+1) 18

  67. [75]

    K y′ yz,y = − 3 2 × i 2 √ 3(t1−1 2−1,1−1 + t1+1 2−1,1−1 + t1−1 2−1,1+1 + t1+1 2−1,1+1 + t1−1 2+1,1−1 + t1+1 2+1,1−1 + t1−1 2+1,1+1 + t1+1 2+1,1+1)

  68. [76]

    K y′ zz,y = 1 √ 2(t1−1 20,1−1 + t1+1 20,1−1 + t1−1 20,1+1 + t1+1 20,1+1)

  69. [77]

    K y′ xx,z = 1 2[−i(t1−1 20,10 + t1+1 20,10) + i √ 3 2(t1−1 2−2,10 + t1+1 2−2,10 + t1−1 2+2,10 + t1+1 2+2,10)]

  70. [78]

    K y′ yy,z = 1 2[−i(t1−1 20,10 + t1+1 20,10) − i √ 3 2(t1−1 2−2,10 + t1+1 2−2,10 + t1−1 2+2,10 + t1+1 2+2,10)]

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  74. [82]

    K y′ zz,z = i(t1−1 20,10 + t1+1 20,10) 19

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Pith tools

Reviewed August 12, 2026 · model on record in the stance chip above.