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Description
To diagnose CPO texture we have 1) angles between any kind of mean orientation in SO(3) and a reference direction, 2) texture strength and 3) texture symmetry.
Getting the mean angle (without crystalline symmetry considerations) is fairly simple using Bingham averaging. Getting the basic symmetry diagnostics (Vollmer 1990) uses the same inertia matrix, also pretty simple. Texture strength not so much.
There are a few options in the literature (checkboxes for when they are implemented with tests):
- J-index (texture index), requires calculating the continuous ODF first, see Add orientation distribution function estimation #113.
- M-index, in [0, 1], but requires calculating n_grains choose 2 combinations (multiplied by the square of the number of symmetry operations for the lattice system! fix: Complete elastic decomposition and M-index and simplify API #139 ) of misorientation angles (499500 for our "minimum" default of 1000 grains)
- Chi-squared misorientation test, Schaeben 2007 (Tectonophysics 441), same problem as M-index and also requires plotting on a log scale because values seem to blow up quickly at large strain
- Jm-index (texture index calculated from an misorientation distribution function), requires calculating an MDF, as difficult as J-index (spherical harmonics) and as expensive as M-index
- S-index (entropy of an ODF), Mainprice 2014 for a formula, from Schaeben 1988
- "Pole density", Jpf, see Mainprice 2014 (equations are not numbered...), not clear how this differs from the J-index
- Reduced M-index, i.e. M-index calculated after throwing out grains at random (but with the chance modulated by the grain size, smaller grains get thrown out easier). Don't reuse the GBS threshold or anything for this, it's not based on any particular physics, but the normal M-index procedure is simply too expensive when we get to 5000+ grains per aggregate.
C = ln(λ₁/λ₃) using eigenvalues of the inertia (scatter) matrix as used for the Bingham avg., can't add symmetry considerations so can yield "unexpected results" (Mainprice 2014)Let's not do this, we have enough Vollmer-based diagnosics with the PGR symmetry numbersI = (15n/2) (λ₁² + λ₂² + λ₃² - 1/3) for n = n_grains, equation from Mainprice 2014, apparently from Mardia 1972 and/or Mardia & Jupp 2000 (book) but I can't find it, no idea where the 15 comes from, can have negative values which is a bit weird
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