Crystallography, point groups, Lecture 2 of 9

[@hikguru]

Hello Dr. Bhadeshia. I had a basic question, probably reflects my lack of understanding but would really appreciate your help. You mentioned that the crystal classes have certain sets of point groups/ symmetry elements associated with each of them. So a cubic class will always have 4 3 fold symmetry axes along its body diagonal. It will have other elements depending on the motif used but it must have 4 triads irrespective of the motif. However what if the motif was some irregular object, say for example a potato? In other words what if I put a potato at the corners of the cube, of course making sure that they are oriented identically at all corners. In this case the environment at each corner is the same. However this does not have any type of rotation symmetry (or does it?) despite the fact that it's lattice representation is still a cube. Practically you will not have these type of motifs but in theory could it not be possible to choose a motif that eliminates all rotational symmetries which would otherwise have been possible with a more symmetrical motif? Thanks.

[@bhadeshia123]

The operation of the symmetry elements, including the triads, would be preserved irrespective of the symmetry of the motif. In other words, the potato would locate at a different lattice point by a rotation of 120° about , in precisely the same orientation as the potato located there.

[@Passingthroughbushes]

Thank You So Much Sir!!

[@bhadeshia123]

Glad it is useful. You can download (free) a book on the subject, www.phase-trans.msm.cam.ac.uk/2020/Crystallography_book.pdf

[@SFYN..]

Much appreciated ❣️

[@bhadeshia123]

You are welcome.

[@alirezafarhadizadeh5674]

superb

[@bhadeshia123]

Thank you

[@MangoResearcher]

Sir, If the triangle is equilateral or isosceles triangle, would the fold of symmetry change?

[@bhadeshia123]

Yes, because an equilateral triangle has a three-fold axis normal to its plane, which an isosceles does not.

[@AnupKumar-wm1qp]

What determine the point group of the particular crystal? Do we determine it after looking the external symmetry of crystal i.e. its external morphology? For ex- We see that the crystal (ulmannite) found in different forms (cubic, octahedral, or pyritohedral) in nature, all belong to single point group ‘23’. So, how can we say that ulmannite belongs to point group '23' ? If we determine using external morphology then (if it is of cubic form) it should belong to the point group that has highest symmetry of cube. Please clarify this, or do I missing some basic concepts?

[@bhadeshia123]

Have a look at this freely available book: www.phase-trans.msm.cam.ac.uk/2020/Crystallography_book.pdf

[@ancyjohn1121]

Thank you Sir.

[@bhadeshia123]

You are most welcome

[@jennawardini2557]

At 29:50, isn't there another mirror plane perpendicular to the 4 fold rotation axis? If so how do you include that in the 4mm symbol?

[@bhadeshia123]

You are quite right, it should be 4/mmm. Well done!

[@danieldossantosavila7431]

@@bhadeshia123 at 28:40, isn't there also a mirror plane parallel to (110)?

[@bhadeshia123]

@@danieldossantosavila7431 4/mmm covers that

[@danieldossantosavila7431]

@@bhadeshia123 sorry, I meant at 26:40 (epsomite)

[@bhadeshia123]

@@danieldossantosavila7431 No, if you look at page 24 of the accompanying book (free download) www.phase-trans.msm.cam.ac.uk/2020/Crystallography_book.pdf the angles are not 90° to permit that.

[@sanjitmetallurgy1205]

very nice video sir

[@bhadeshia123]

Thank you.

[@akeedasharma4249]

Sir,At 15:47, for primitive cubic ,you said that each lattice point has 1 Al and 3 Ni atoms. How did you get this? Also, does that mean each lattice point has 2 motifs ( Al and Ni) ?

[@bhadeshia123]

A primitive cube has only one lattice point (each point at the corner is shared by eight cells, so each corner only contributes one eighth of a point to the cell). Therefore, all the atoms in the cell are in the motif. The motif consists of Al at 0,0,0, and each of the three nickel atoms at the face centres.

[@Dibyendu.M]

thank you sir

[@bhadeshia123]

Most welcome.