Ambiguity of the Pole Figure to ODF Reconstruction Problem

A pole figure is a two-dimensional projection of an orientation distribution function (ODF). Reconstructing the three-dimensional ODF from those projections is therefore an inverse problem without a unique answer.

This page assumes the definitions of an ODF and mrd from ODF Theory, and the projection developed in Pole Figures of an ODF. The practical reconstruction workflow is introduced in ODF Reconstruction.

Three sources of ambiguity must be kept apart because only the first can be reduced by measuring more:

Friedel's law underlies the last two losses, but they appear differently. The first is a symmetry ambiguity, while the second is the null space of the pole-figure transform. The latter produces the ghost effect and motivates ghost correction.

Too few pole figures

Experiments measure only a handful of lattice planes. The number needed generally grows as the texture becomes weaker and the crystal symmetry becomes lower. A classical sampling argument puts it on the order of the square root of the number of measured directions in each pole figure, which is far beyond routine measurements.

What that costs is best seen on two ODFs built to be different. The first has three components, rotations by 90 degrees about the three coordinate axes:

plottingConvention.default('y↑→x');
setMTEXpref('defaultColorMap','LaboTeX')
cs = crystalSymmetry('mmm');

orix = orientation.byAxisAngle(xvector,90*degree,cs);
oriy = orientation.byAxisAngle(yvector,90*degree,cs);
oriz = orientation.byAxisAngle(zvector,90*degree,cs);

odf1 = unimodalODF([orix,oriy,oriz]);

The second has three as well, rotations about (1,1,1) by 0, 120 and 240 degrees:

ori = orientation.byAxisAngle(vector3d(1,1,1),[0,120,240]*degree,cs);
odf2 = unimodalODF(ori);

The two share no component. Their sigma sections make the different peak positions visible.

plot(odf1,'sigma')

plot(odf2,'sigma')

Both textures put their bright peaks in the sigma = 0 and sigma = 90 degree sections, but at different positions inside them: where one has a single peak at the centre, the other has four at the rim. This confirms that the two model textures have no common component.

Yet seven of their pole figures are identical - (100), (010), (001), (110), (101), (011) and (111). The eighth drawn here, (120), is not, which is how one can tell them apart at all.

figure(1)
h = Miller({1,0,0},{0,1,0},{0,0,1},{1,1,0},{1,0,1},{0,1,1},{1,1,1},{1,2,0},cs);
plotPDF(odf1,h,'contourf')


figure(2)
plotPDF(odf2,h,'contourf')

The first seven panels have the same maxima and contours in both figures. Only the (120) panel changes, at the lower right.

So if only those seven were measured, no algorithm could decide which ODF produced them. The question worth asking is what MTEX returns in that situation.

% 1. step: simulate pole figure data
pf = calcPoleFigure(odf1,h(1:7),'upper');

plot(pf)

These seven simulated pole figures contain no panel that distinguishes the two model ODFs.

% reconstruct an ODF
odf = calcODF(pf,'silent');

plot(odf,'sigma')

% compare the mean density at the two sets of component orientations
densityAtModes7 = [mean(odf.eval([orix,oriy,oriz])), ...
  mean(odf.eval(ori))]

densityAtModes7 =
   66.1665   66.1665

The two values printed above are the mean density at the components of odf1 and odf2. Both are 66.1665 mrd, although only odf1 generated the data. Any mixture would fit the seven pole figures equally well. MTEX returns an admissible ODF close to uniform, which is a defensible choice but still not the true answer.

Adding five more lattice planes to the measurement changes the picture.

% 1. step: simulate pole figure data for all crystal directions
h = [h,Miller({0,1,2},{2,0,1},{2,1,0},{0,2,1},{1,0,2},cs)];
pf = calcPoleFigure(odf1,h,'upper');

% reconstruct an ODF
odf = calcODF(pf,'silent');

plot(odf,'sigma')

% compare reconstructed and true densities at both sets of modes
densityAtModes12 = [mean(odf.eval([orix,oriy,oriz])), ...
  mean(odf.eval(ori)),mean(odf1.eval([orix,oriy,oriz])), ...
  mean(odf1.eval(ori))]

densityAtModes12 =
  128.4721    5.3985  129.6016         0

The four values are the reconstructed densities at the odf1 and odf2 modes, followed by the two true densities. The unwanted modes fall from 66.1665 to 5.3985 mrd, while the wanted modes rise from 66.1665 to 128.4721 mrd. The true values are 129.6016 and 0 mrd. Five more lattice planes have nearly resolved this ambiguity. This is the one ambiguity that more measurement can cure.

Friedel's law

Under Friedel's law, a diffraction peak is the same for a lattice plane and its opposite. An ordinary diffraction pole figure is therefore antipodally symmetric whether the crystal point group is or not.

Consider point group -43m, which has no fourfold axis, and two orientations that differ by 90 degrees about the third Euler axis.

cs = crystalSymmetry('-43m');

ori1 = orientation.byEuler(30*degree,60*degree,10*degree,cs);

ori2 = orientation.byEuler(30*degree,60*degree,100*degree,cs);

h = Miller({1,0,0},{1,1,0},{1,1,1},{1,2,3},cs);
plotPDF(ori1,h,'MarkerSize',12)
hold on
plotPDF(ori2,h,'MarkerSize',8)
hold off

The large and small markers do not coincide in the (111) and (123) panels. The two orientations are therefore distinct under -43m.

Now impose antipodal symmetry, as ordinary diffraction does.

plotPDF(ori1,h,'MarkerSize',12,'antipodal')
hold on
plotPDF(ori2,h,'MarkerSize',8,'antipodal')
hold off

Every small marker now lies inside a large marker. The measurement cannot distinguish these orientations.

Imposing antipodal symmetry on all pole figures is the same as adding the inversion to the point group, that is, replacing it by its Laue group. Doing that explicitly gives the same picture:

ori1.CS = ori1.CS.Laue;
ori2.CS = ori2.CS.Laue;
h.CS = h.CS.Laue;

plotPDF(ori1,h,'MarkerSize',12)
hold on
plotPDF(ori2,h,'MarkerSize',8)
hold off

The marker pairs still coincide, confirming the equivalence between antipodal pole figures and Laue symmetry.

So an ODF reconstructed from diffraction pole figures is always centrosymmetric: its point group is a Laue group, and the noncentrosymmetric part of the texture is not lost by the algorithm but was never measured. No amount of additional ordinary pole figures recovers it. If the crystal point group is already a Laue group, as it is for most materials measured this way, this symmetry step costs nothing.

This statement assumes ordinary kinematic diffraction. Anomalous or resonant scattering can distinguish opposite directions and has been used to determine the odd part of a texture.

The odd order harmonics

The third ambiguity survives even a complete set of pole figures and a centrosymmetric crystal. Consider triclinic symmetry and a weak unimodal ODF at the identity:

cs = crystalSymmetry('-1');

odf1 = 2/3 * uniformODF(cs) + 1/3 * ...
  unimodalODF(orientation.id(cs),'halfwidth',30*degree);

plotPDF(odf1,Miller({1,0,0},{0,1,0},{0,0,1},cs),'antipodal')

The three pole figures contain broad, weak maxima. They are the complete diffraction data used for the harmonic comparison below.

Written as a harmonic series it is the same function in another representation.

odf1 = FourierODF(odf1,10);

plotPDF(odf1,Miller({1,0,0},{0,1,0},{0,0,1},cs))

The pole figures are unchanged by converting the representation.

Its coefficients decay quickly, which is why cutting the series at degree 10 loses nothing here.

close all
plotSpektra(odf1,'linewidth',2,'figSize','small')

The spectrum falls rapidly towards zero by degree 10.

Now build a second ODF that differs only in the odd order coefficients - all of them set to zero.

A = mod(1:11,2)';
odf2 = conv(odf1,A);

hold on
plotSpektra(odf2,'linewidth',2,'figSize','small')

hold off
legend('odf1','odf2')

The second spectrum agrees at every even degree and is zero at every odd degree. Nevertheless, all pole figures of odf2 are identical to those of odf1:

plotPDF(odf2,Miller({1,0,0},{0,1,0},{0,0,1},cs),'antipodal')

The odd order coefficients simply do not appear in a pole figure. Nothing in the data distinguishes the two functions, and they are not the same function - along the alpha fibre:

alphaFibre = orientation.byAxisAngle(zvector,(-180:180)*degree,cs);

close all
plot(-180:180,odf1.eval(alphaFibre),'linewidth',2)
hold on
plot(-180:180,odf2.eval(alphaFibre),'linewidth',2)
hold off
legend('odf1','odf2')
xlim([-180,180])

The two curves differ along the fibre even though their pole figures do not. Setting the odd degrees to zero has changed the ODF, not merely its representation.

Flipping the sign of the odd coefficients instead of zeroing them makes the point sharper.

odf1 = 4/5 * uniformODF(cs) + 1/5 * unimodalODF(orientation.id(cs),'halfwidth',30*degree);

A = (-1).^(0:10)';
odf2 = conv(odf1,A);

close all
plot(-180:180,odf1.eval(alphaFibre),'linewidth',2)
hold on
plot(-180:180,odf2.eval(alphaFibre),'linewidth',2)
hold off
legend('odf1','odf2')
xlim([-180,180])

One ODF has a single preferred orientation at the identity; the other has preferred orientations at every 180 degree rotation. They have the same pole figures. No reconstruction method can prefer one over the other on the evidence.

Matthies' way out is a physical prior, not additional information in the data. A real texture is usually a uniform background plus a few components. Among the ODFs that fit the data, ghost correction therefore prefers the one with the largest uniform portion. MTEX applies this correction by default.

Ghost correction at work

Simulate seven distinct pole figures from the peaked ODF above.

h = Miller({1,0,0},{0,1,0},{0,0,1},{1,1,0},{0,1,1},{1,0,1}, ...
  {1,1,1},cs);
pf = calcPoleFigure(odf1,h);

plot(pf)

These smooth pole figures constrain the even-degree coefficients but contain no direct evidence for the missing odd degrees.

Reconstruct with the default, which includes ghost correction.

odf_rec1 = calcODF(pf,'silent');

Reconstruct once more without ghost correction.

odf_rec2 = calcODF(pf,'noGhostCorrection','silent');

Along the alpha fibre the corrected reconstruction recovers odf1 closely, while the uncorrected one lands between odf1 and odf2 - a peak that is too low sitting on a background that is too high.

close all
plot(-180:180,odf_rec1.eval(alphaFibre),'linewidth',2)
hold on
plot(-180:180,odf_rec2.eval(alphaFibre),'linewidth',2)
hold off
legend('odf rec1','odf rec2')
xlim([-180,180])

The harmonic coefficients say the same thing in the natural language of the problem. Without correction, the recovered odd-degree coefficients are far too small.

close all
plotSpektra(odf1,'linewidth',2,'bandwidth',10,'figSize','small')
hold on
plotSpektra(odf2,'linewidth',2)
plotSpektra(odf_rec1,'linewidth',2)
plotSpektra(odf_rec2,'linewidth',2)
hold off
legend('odf1','odf2','odf rec1','odf rec2')

The Santa Fe example

Historically, the ghost effect is tied to the Santa Fe model ODF. This standard texture gives different inversion programs the same known truth to reconstruct.

odf = SantaFe;
plot(odf,'contourf')

The sections show several broad components on a nonzero background. That combination makes the model sensitive to an incorrect uniform portion.

Simulate diffraction pole figures from it:

% crystal directions
h = Miller({1,0,0},{1,1,0},{1,1,1},{2,1,1},odf.CS);

% simulate pole figures
pf = calcPoleFigure(SantaFe,h,'antipodal');

% plot them
plot(pf,'MarkerSize',5)

The four pole figures contain the projections that both reconstructions below must reproduce.

Reconstruct twice.

% one with ghost correction
rec = calcODF(pf,'silent');

% one without ghost correction
rec2 = calcODF(pf,'noGhostCorrection','silent');

Both reproduce the measured pole figures. This is the crux of the whole page: agreeing with the data is not evidence of being right.

figure(1)
plotPDF(rec,pf.h,'antipodal','complete','upper')

figure(2)
plotPDF(rec2,pf.h,'antipodal','complete','upper')

The same maxima and contour shapes appear in both sets of recalculated pole figures. Agreement with the measured projections does not select the correct ODF.

The ODFs, however, differ - and the extra components in the second one are the ghosts.

close all
figure(1)
plot(rec,'gray','contourf','figSize','small')

figure(2)
plot(rec2,'gray','contourf','figSize','small')

Extra maxima appear in the uncorrected sections even though they did not spoil the pole-figure fit. Once more, the harmonic coefficients show where they come from.

close all;
% the harmonic coefficients of the sample ODF
plotSpektra(SantaFe,'bandwidth',32,'linewidth',2,'MarkerSize',10,'figSize','small')

% keep plot for adding the next plots
hold on

% the harmonic coefficients of the reconstruction with ghost correction:
plotSpektra(rec,'bandwidth',32,'linewidth',2,'MarkerSize',10)

% the harmonic coefficients of the reconstruction without ghost correction:
plotSpektra(rec2,'bandwidth',32,'linewidth',2,'MarkerSize',10)

legend({'true ODF','with ghost correction','without ghost correction'})
% next plot command overwrites plot
hold off
setMTEXpref('defaultColorMap','WhiteJet')

Cubic crystal and orthorhombic sample symmetry leave degree 9 as the only odd degree with weight here, and that is where the three curves part: the true value 0.29 is met by the corrected reconstruction and missed by the uncorrected one, which reaches 0.10. Every even degree is reproduced either way. Ghost correction has selected a physically plausible member of the solution family; it has not created new measured information.

Further reading

Next

Ghost Effect Analysis quantifies what the correction buys on a deliberately weak texture. The Santa Fe Example then adds counting noise and compares both reconstructions with the known model.