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Aug 3, 2022 at 10:21 comment added Bjaam Right, I should have said they do not dictate the crystal system.
Aug 3, 2022 at 9:11 comment added gryphys Then let me add a comment also on this "I don't think if lattice parameters change, it implies translational symmetry breaking". It is obvious that a change in lattice parameters does not imply a symmetry breaking, but rather a change in their relationship. As I wrote before, a cubic primitive lattice cannot be considered a tetragonal primitive lattice with a=c; a lattice exists in the geometrical space and such an assumption would imply that some symmetry operations are disregarded, even though they are actually present.
Aug 3, 2022 at 9:06 comment added gryphys Sorry but I cannot agree with your statement "Lattice parameters does not dictate the lattice type". Bravais lattice systems are distinguished on the basis of lattice parameter relationships. How can you distinguish a tetragonal primitive lattice with a=c from a cubic primitive lattice? Your statement is correct if you change it in "Lattice parameters does not dictate the crystal structure type. Symmetry does". And this is exactly what I meant in my rhetorical question.
Aug 3, 2022 at 6:17 comment added Bjaam Unless the Bravais lattice type changes.
Aug 3, 2022 at 6:04 comment added Bjaam Also let me remind you that when people talk about the breaking of translation symmetry, they mean a lot more important case: the one of the crystallization where the uniformity of space descends onto a discrete one. I don't think if lattice parameters change, it implies translational symmetry breaking.
Aug 3, 2022 at 6:01 comment added Bjaam Lattice parameters does not dictate the lattice type. Symmetry does. You could still have a "square"-like lattice (the length remains the same), but the point group associated at each lattice point be those of a "tetragonal" one.
Aug 2, 2022 at 15:39 comment added gryphys Your explanation adjust somehow the statement in the paper that should be corrected in this light as "...but leaves both translation and SOME reflection symmetries unbroken”. Anyway there is still the problem of the translation symmetry. Please, explain how is it possible to have a D4h to D2h lattice transition with unbroken translation symmetries of the lattice. I cannot see how this is possible. In the D4h to D2h transition a square lattice becomes a rectangular one, so the translation symmetry is broken with nematic transition.
Aug 2, 2022 at 10:14 comment added Bjaam That's correct. The lattice cannot be anymore a square (its crystal system is changed). This is why the diagonal mirrors should also vanish. In the particular transition stated in this paper, the symmetry group of the order parameter loses its rotational symmetry, hence obviously, as you also mentioned, the transition is not endowed with the reduction to a merohedry. I am aware of what confuses you here. $D_{4h}$ to $D_{2h}$ transition is a perfectly good example for a nematic transition. Although its reflection elements subset is smaller, this does not mean the reflection symmetry is broken.
Aug 2, 2022 at 8:54 comment added gryphys First of all, if you reduce the holohedry, the 2-dimensional lattice cannot be anymore a square (if you reduce to a merohedry obviously the 4-fold rotation is still present). Then, as I wrote above, in the paper it is stated “The nematic phase breaks the four-fold rotation symmetry of the lattice, but leaves both translation and reflection symmetries unbroken". These unbroken reflection symmetries contrasts with your reply when you write " Of course, diagonal mirror planes have to disappear as well". Then if you cut out diagonal mirror the lattice will be rectangular not square
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S Aug 1, 2022 at 7:16 review First answers
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S Aug 1, 2022 at 7:16 history answered Bjaam CC BY-SA 4.0