Unit 2.5 - Bravais Lattices (II)

[@cheekocho]

Your video saved me a great amount of time, was really having trouble understanding this part. Thanks from South Korea!

[@3339LuXz]

This is so a detailed course ! I LOVE IT! Greets from Austria :)

[@safwanshahriar9579]

In an earlier response to a question, you said non bravais lattices don't exist. But the honeycomb lattice structure of graphene is said to be non bravais. Could you explain the conundrum? Thank you.

[@FrankHoffmann1000]

There is always the confusion between a _structure_ and the underlying _lattice_ It is simply wrong to said that the honeycomb _structure_ has a non-Bravais lattice because - again - there are no non-Bravais lattices. The underlying lattice of the honeycomb structure (or you can say point pattern) is the well known hexagonal lattice. The _motif_ of this lattice consists of _two_ points/atoms as correctly described here: en.wikipedia.org/wiki/Hexagonal_lattice

[@loopgarcia3862]

Hi Frank, Great video! After watching it I have a question. Primitive lattice is always the smallest choice but sometimes is better to use a larger cell (with more lattice points) to get a better representation of the symmetry of the lattice. I also know that the “reduced cell” term exists. Is it just a primitive cell? Or it is a completely different concept. Cheers!

[@FrankHoffmann1000]

Hi Albert, these terms are related to each other. The reduced cell is always a primitive cell, yes, but additionally the three latice vectors are chosen in a way that the relations a < b < c (lengths are in ascending order) and alpha > beta > gamma (angles are in descending order) hold. In this way the unit cell is 'unique'. This makes it easy to compare unit cells with each other, in particular for those cases in which several equivalent representations exist, say P2(1)/a versus P2(1)/c.

[@jimh1996]

Sorry, I know you've addressed this already but I have to ask as it is still not clear: in the cubic F, the face centered points are distance (sqrt(2)/2)*a from the corner points yet these same face centered points are also distance "a" from a point on the opposite face...an apparent violation equivalent to your example for cubic C! Using simular logic: why is orthorhombic C not a symmetry violation, also equivalent to your cubic C example?

[@FrankHoffmann1000]

Hi! That's no problem :-) The shortest possible answer: because in the orthorhombic case, the three directions are completely unrelated to each other, they are independent! This is different to the cubic case, where - for symmetry reasons - the a, b, and c directions are identical; in terms of lattice constants and in terms of symmetry elements! Okey, to add some more words: In the forgoing units we saw that we choose centerings, if we can achieve a higher number of orthogonal (or hexagonal) angles of the unit cell (but leaving the underlying symmetry of the cell, of course, unchanged!). So, in the orthorhombic C case, probably all primitive unit cells are oblique, although the symmetry is orthorhombic! I made an additional sketch, see the following URL: crystalsymmetry.files.wordpress.com/2020/04/orthorhombic_c.pdf But as I mentioned, the key consideration is that the crystallographic directions are independent of each other in the orthorhombic case (and monoclinic and triclinic as well). Does this help? best wishes Frank

[@jimh1996]

@@FrankHoffmann1000 Perfect, thank you! And thank you for the quick reply!

[@bohanxu6125]

From my understanding, there is another approach purely from symmetry. I find that approach more systematic. Cubic symmetry include a 3-fold rotation around the (111) direction. (You can google cubic symmetry, and you would know which 3-fold I'm talking about) Cubic-F preserve the 3-fold rotation, so it still belong to the cubic family. Cubic-C doesn't have the 3-fold, so it doesn't belong to the cubic family. Another example of the same situation is tetragonal-C where the "C" is in the "rectangular-face". The extra "C" breaks the 4-fold symmetry of tetragonal, so {tetragonal-C where C is in the rectangular-face} doesn't belong to the tetragonal family (in fact it has all the symmetries of orthorhombic family so it belongs in there). Now consider tetragonal-C where the "C" is in the "square-face". This extra "C" preserve the 4-fold symmetry of tetragonal, so {tetragonal-C where C belong to the square-face} still belong to the tetragonal family. Actually, we can reduce it to tetragonal-P...(but it still belongs to the tetragonal family nevertheless)

[@mariamhasany5250]

thank you for the helpful content Prof. Frank

[@FrankHoffmann1000]

You are welcome!

[@ArifulIslam-qw6lf]

incredibly helpful, thank you very much

[@hikguru]

Hi Frank, you explained point groups in a future section, but would it be correct to say that the same set of symmetry operations that are possible for the non centered unit cell (say primitive simple cubic) are also possible for the centered version (eg. FCC)? I believe that FCC, BCC and SC all have 4 fold rotation symmetries viewed perpendicular to a face, plus 4 3 fold symmetries along the body diagonal i.e. no symmetry operation is excluded by going from the non centered to centered but I don't know if it is true as a general rule? It certainly seems that way. Thanks.

[@FrankHoffmann1000]

Hi! You are almost correct. Why only almost? If you speak about the maximum symmetry that is achievable for a given crystal system (i.e. the lattice only, without a motif or when the lattice point only represent a single atom) you are completely correct, but if we consider all space groups of a given crystal system then not all symmetry elements that you mentioned are present. Take for instance the space group P23, belonging to the cubic crystal system. Here, we have - surprisingly and somehow counterintuitive - no 4-fold axis of rotation perpendicular to the faces. But indeed: the four 3-fold axes of rotation along the body diagonals are _the_ characteristic_ feature of the cubic lattice, independent of the concrete space group. I think in one of the other units I talk a little bit about the characteristic symmetry elements of the other crystal systems.

[@alijutt3675]

i could not understand the last example. the length you measured as sqrt2/2 is also exist in cubic F. that would also be sqrt2/2 then why you just consider it on cubic C?

[@FrankHoffmann1000]

That is not the point if a certain length exist anywhere else. The point is that the length between _any_ two neighboring lattice points in the cubic unit cell have to be identical. And this is clearly not the case for this scenario shown in the video.

[@alijutt3675]

@@FrankHoffmann1000 thanks a alot :)

[@bohanxu6125]

From my understanding, there is another approach purely from symmetry. I find that approach more systematic. Cubic symmetry include a 3-fold rotation around the (111) direction. (You can google cubic symmetry, and you would know which 3-fold I'm talking about) Cubic-F preserve the 3-fold rotation, so it still belong to the cubic family. Cubic-C doesn't have the 3-fold, so it doesn't belong to the cubic family. Another example of the same situation is tetragonal-C where the "C" is in the "rectangular-face". The extra "C" breaks the 4-fold symmetry of tetragonal, so {tetragonal-C where C is in the rectangular-face} doesn't belong to the tetragonal family (in fact it has all the symmetries of orthorhombic family so it belongs in there). Now consider tetragonal-C where the "C" is in the "square-face". This extra "C" preserve the 4-fold symmetry of tetragonal, so {tetragonal-C where C belong to the square-face} still belong to the tetragonal family. Actually, we can reduce it to tetragonal-P...(but it still belongs to the tetragonal family nevertheless)

[@dmwaters]

Question: Wurtzite is described as having a primitive hexagonal unit cell, but its unit cell is made up of four atoms not all located at the corners of the unit cell. Why is it that it is defined as primitive then?

[@FrankHoffmann1000]

Hi Dale, sorry for answering quite lately, but I wasn't notified about your comment and saw it only just now. Well, the first thing is that you have to differentiate between lattice points and atoms. Primitive means not that there are only atoms at the corners, but it means that only lattice points are at the corners. And a lattice point _represents_ a certain motif. A motif can consist of only one atom but also of several atoms. And none of these atoms themselves have to be located at the corners, but only the mass center of that motif. It is sometimes not easy to see which atoms build the motif or of which atoms the motif is composed of. For rock salt (NaCl) the motif consists of one Na cation and one chlorine anion. For wurtzite the motif is already more complicated, it consists of one Zn, one S and 1/8 of a further ZnS pair (the one inside the cell, which has an inverse orientation). We can put it also the other way round: To have a centered cell, all of the motifs of the cell must have pure translational symmetry, meaning that by translational operations all motifs can be congruently mapped onto each other. This is not the case for the ZnS pair inside the cell of wurtzite, so ist has to be a primtive cell. I would also recommend to watch unit 4.5: kzbin.info/www/bejne/hJK1m3qcppZsfac best Frank

[@chrisharrison3180]

Hi again Frank - I have another question, this time on Bravais Lattices. With real world structures can a structure have multiple Bravais lattices? For example, if you take the perovskite structure (ABX3) as an example, could you say that this is cubic lattice (with interstitial atoms at the B and X site?) or a face centred cubic (with an interstitial atom at the B site?) or even a body centre cubic (with interstitial atoms at the X sites?). Or do we pick one in particular? And how do we decide which one to choose? I know that the lattice points don't necessarily need to be atoms but, from what I can see in examples, it tends to be atoms which sit at the lattice points - which made me wonder about how you determine the system of the more complex structures.Thanks, Chris

[@FrankHoffmann1000]

Hi Chris! Yes, a structure can have multiple Bravais lattices, because for centered lattices you can always choose the primitive lattice, too. For the triclinic crystal system, usually all structures are represented by a primtive lattice, but it is also possible to choose a centered cell; C1 is a possible space group in this crystal system - although it doesn't make really sense to do so, because the reason for choosing a larger, centered cell should be: the resulting cell should have angles of 90° or 120°, and this can't be achieved by centerings in the triclinic case. The cubic perovskite structure has a cubic lattice, yes - but a primitive lattice, neither a face nor a body centered cell - the space group is Pm-3m! Your question is rather related to the choice of the origin, but not to different Bravais lattices. The cubic perovskite structure is a good example at which the difference between lattice points and atoms become evident: the lattice points are representing one molecule or a formula unit, here all(!) atoms of the formula ABX3. In a face centered cell the lattice points are at the corners and the middle of all faces. This is obviously not the case for the ABX3 structure. Only the A atoms are at the corners, but the B atom is in the center and the X atoms are at the face centers. You can also take CsCl as an illustrative example: it is a cubic structure, but it is not body-centered. Why? Because the Cs atoms are at the corners and the Cl atom at the center (or the other way round, if you choose a different origin.), so it is a primtive lattice. best Frank

[@chrisharrison3180]

@@FrankHoffmann1000 Thanks very much for your response, Frank. I apologize as this has sparked some more questions which are perhaps quite basic but I wanted to clarify!: 1. In the perovskite case then, if each lattice point is an ABX3 molecule....should we not expect to see more atoms in the perovskite unit cell (we actually have 4 A atoms, 1 B atom, 6 X atoms which does not offer enough atoms for 4 ABX3 molecules - you'd expect 4 A atoms, 4 B atoms and 12 X atoms)? I know that we say that each A atom is actually only 1/4 of an A atom because of neighboring cells so perhaps this is why? In reality, are we not showing ALL of the B and X atoms which are associated with each A-atom at the corner of the unit cell? I would assume that these B and X atoms would extend out beyond the boundaries of a single unit cell. I know that the unit cell is simply the smallest repeating unit so I assume this is why we follow this convention (i.e, for simplicity)? 2. Related to this (and an answer to 1 may make this question redundant), I have until now been assuming that A molecules are actually bonded to MORE than 1 B molecule in (i.e, they are shared out such that 1 A atom is associated with 4 B atoms and 6 X atoms) - is this incorrect? Should it instead be viewed that there are individual molecules of ABX3 which are bonded by inter-molecular forces? Many thanks, Chris

[@FrankHoffmann1000]

No worries, Chris. Sparking new questions while answering one, is very common :-) (probably you have expect some delay regarding my answers as I am currently on holidays) 1a. We have a primtive cell (note the space group is Pm-3m). This means we have exactly one lattice point per unit cell (not four!). Now, let's count the atoms: we have 8 A atoms at the corners, each atom at a corner of a unit cell belongs simultaneously to eight unit cells, so we have 8 times 1/8 = 1 A atom. The B atom is inside the cell and belongs only to this specific cell, so we have 1 B atom, and, finally, we have 6 X atoms at the center of the faces; atoms at the faces belong to two cells simultanesoulsy... So we have ABX3, exactly one lattice point per unit cell, exactly one 'formula unit' per unit cell. 1b. Not quite sure, what the meaning of "inreality, are we not showing ALL..." - but, yes, if we want to _show_ the coordination environment of all atoms correctly we would have to display more than one unit cell (or sometimes we can choose another origin). Yes, not only for simplicity, but also because of this defintion of a unit cell... 2. Well, the perovskite structure is best described as composed of ions, not molecules, although there are some minor covalent parts of the interaction forces. So the answer is 'No', there are not really ABX3 molecules. Does this help? Frank

[@chrisharrison3180]

Hi @@FrankHoffmann1000 - firstly, thanks for the above (sorry it has taken me a while to respond - I have just been re-capping my questions and your answers!). I wonder if you could help me... I am studying a material with a layered structure (a Ruddlesden-Popper phase with perovskite layers separated by NaCl-type LaO layers). The literature suggests it (La2NiO4) has numerous possible space groups dependent on preparation conditions. In at least one case, it has a face centered unit cell. I cannot visualize how this can be the case given the complexity of the structure. I know the lattice points do not represent the atoms themselves but I am having a difficult time understanding what 'pattern'/molecule could be used to result in a face-centred unit cell. I can only visualise a cubic cell with the La2NiO4 molecule stretching into the perovskite layer and into the NaCl-type layer - can you help me understand how a face centred space group may occur? Is there a good way that this can be visualised?

[@FrankHoffmann1000]

@@chrisharrison3180 I think it is not that difficult. But I also think that at this point we should switch to another mode of exchange instead of using the comment section of KZbin. Would you mind to give me your eMail address?

[@VivekSingh-rj2ck]

Thanks for the video, I have a doubt, though. Why is C-centered unit cell preffered over the I-centered variant for a monoclinic system? Thanks.

[@FrankHoffmann1000]

Dear Vivek, two aspects have to be considered here: (a) one aspect is the number of principally different lattice types. Because every I-centered monoclinic cell can be transformed into a A-, B, or C-centered lattice, the I-centered lattice do not constitute an own lattice type. (b) The C-centered lattice is only preferred over the I-centered lattice in the sense that this is the standard lattice type. However, this rule or convention is not very strict. You are free to choose also the I-centered lattice. Usually the choice is guided by the skew angle beta: Commonly, the lattice in which beta is more closely to 90 degrees is chosen. best! Frank

[@VivekSingh-rj2ck]

Thanks.. Got it.

[@zrazra7856]

Sorry, I couldn't understand the example explanation of symmetry incompatibility. Side face centered cubic is not the primitive one. So what is the problem with unequally(what there is in all-side face-centered cubic too) of vector lengths in one side face-centered cubic? at the end of explanation was shown the same as previous example(redundancy) that the primitive cubic unit cell can be found! I think I didn't get the concept.

[@FrankHoffmann1000]

Dear Muhammad, the main point here is that for _both_ cells, i.e. the primitive _and_ the centered variant one, the restrictions of the cubic crystal system have to be applied: the cell lengths have to be equal, a = b = c! So, we can reverse this task: First, take a cubic primitive unit cell with a = b = c, assemble some of these together and now try to construct a _one_-side centered cell, in which also the same restriction is valid! I think it can be realized that this is not possible - do you agree? best! Frank

[@waliddjani5978]

Hello, there is something I don't get regarding the calculations from 2:30 I understood all the process and all the calculations but not the results, what do you mean by "lattice points/unit cell" for instance In the primitive unit cell, we have 8 lattice points, so why do we say 1 lattice point by unit cell?

[@FrankHoffmann1000]

Come on, this can't be too difficult: we do not have 8 _whole_ lattice points per unit cell. Place a lattice point at a corner and answer the question: which fraction is _inside_ the unit cell and which fraction is outside? Or, to express it in another way: to how many unit cells does one lattice point belong?...

[@waliddjani5978]

@@FrankHoffmann1000 oh so obvious, how can miss this! It's clearer now, thanks

[@srividyas6579]

I have a doubt @ 6.40, if a crystal system can be defined as I(monoclinic) as well as C(monoclinic), what is the criteria for choosing the bravais lattice as C(monoclinic) instead of I(monoclinic)?

[@FrankHoffmann1000]

In principle, you are free to choose the I or C centering, however, there is one guideline (but not a strict rule): the skew angle beta should be as close to 90 degrees as possible. best! Frank

[@moglibora]

Dear Frank, Is the existence of 14 bravais lattices an empirical result? Can we predict their existence by doing the mathematics only?

[@FrankHoffmann1000]

Dear Physics Dept., interesting question! Well I would not say that this an empirical result. In 1848, Bravais systematically investigated all possible point ensembles for all different kind of symmetries that might be present in such lattices (symmetry axes, symmetry planes). He constructed such lattices as drawings on paper, recognized that only rotational symmetry of the order 2, 3, 4, and 6 is compatible with lattices and so on, recognized that certain primitive lattices at which the symmetry is not apparent can be turned into larger, cnetered one which reflect the full symmetry of the lattice and so on. So, he derived his results long before it was confirmed experientally by X-ray diffraction. Concerning mathematics, I am not very gifted, but yes, I think that there are at least some proofs for this hypothesis. But I do not know if there is a simple algorithm that would be able to predict these numbers for 3D but also higher dimensions. But you are of course right in the sense that this is basically a mathematical and not a chemical question! best Frank

[@moglibora]

@@FrankHoffmann1000Thank you very much professor. I was actually surfing the internet to find any mathematical proof for this. Also, I was looking for the original paper of Auguste Bravais from 1848. I could not find any of them. Again, Thanks a lot for your reply and these excellent presentations.

[@FrankHoffmann1000]

Very welcome! The original work of Bravais concerning lattices in its German translation is freely available here at this URL archive.org/details/abhandlungberdi01bravgoog however, I do not know if an English translation exists..

[@moglibora]

​@@FrankHoffmann1000I also found the book "Études cristallographiques" (1866) written by A. Bravais. But It's in French and has no English translation. I can only understand English :')

[@scummybabyelephant6094]

then shouldnt there b symmetry incompatibilities regarding face centered cubic unit cell too?? the lattice points on the faces have distance( root two* a) /2 between them too. also i dont understand why there arent any symmetry incompatibilites for the body centered cubic unit cell??? i hope you will explain this to me.

[@FrankHoffmann1000]

Hi Erza, I don't understand your first question, because for the cubic system you see a circle or empty entry for C(AB)-centered cells. C(AB) means C AND B AND A (but if you have only a single-face centered cell, by convention you name it C-centered). So, there is no contradiction between you and what is said in the video. Concerning the second question: The characteristic feature of the cubic crystal system is that there are four rotational axes of order three, precisely along the body diagonals. And because the additional lattice point of the cubic I-lattice is exactly in the center of the cell this point remains at its position when you carry out the respective symmetry operations (rotation along the body diagonals by 120°). best Frank

[@bohanxu6125]

From my understanding, there is another approach purely from symmetry. I find that approach more systematic. Cubic symmetry include a 3-fold rotation around the (111) direction. (You can google cubic symmetry, and you would know which 3-fold I'm talking about) Cubic-F preserve the 3-fold rotation, so it still belong to the cubic family. Cubic-C doesn't have the 3-fold, so it doesn't belong to the cubic family. Another example of the same situation is tetragonal-C where the "C" is in the "rectangular-face". The extra "C" breaks the 4-fold symmetry of tetragonal, so {tetragonal-C where C is in the rectangular-face} doesn't belong to the tetragonal family (in fact it has all the symmetries of orthorhombic family so it belongs in there). Now consider tetragonal-C where the "C" is in the "square-face". This extra "C" preserve the 4-fold symmetry of tetragonal, so {tetragonal-C where C belong to the square-face} still belong to the tetragonal family. Actually, we can reduce it to tetragonal-P...(but it still belongs to the tetragonal family nevertheless)

[@sayanjitb]

In C (AB) type unit cell it was called single side face centered unit cell. But rather it contains lattice points in two face centered sides. Why is it so?

[@FrankHoffmann1000]

These are not two faces, the "second" one is identical to the "first" one, it is only shifted by one whole unit cell length along one axis, so that they are actually the same.

[@akshitasharma5340]

I have doubt . Why monoclinic C was chosen over monoclinic I? one of the criteria to chose unit cell is to maximize the symmetry. How this condition is getting fulfilled ? Also, i am getting confused between the lattice points and atoms position in the unit cell? Thanks

[@FrankHoffmann1000]

The preference of monoclinic C over monoclinic I is just a convention. Please note that the symmetry is, of course, _identical_ for both variants! Please also note that it is not forbidden to choose the body-centered I variant. Usually, the choice is based on the skew angle beta: This angle should be as close to 90° as possible. Your second question is unclear to me. What do you mean? Here, only lattice points are shown. These points _represent_ motifs, but we don't know the motif, so we cannot say something about the atomic positions.

[@akshitasharma5340]

@@FrankHoffmann1000My first doubt is now cleared. regarding my second question. It is general question not related to the first one. My doubt is when we choose unit cell/ lattice - for example cubic face centered. is it necessary that on each lattice point of unit cell an atom has to be present ? Thanks and regards.

[@FrankHoffmann1000]

@@akshitasharma5340 Of course, not. For molecular crystals it is relatively unlikely that even one atom is located at a lattice point. But there are, of course, some structures where the lattice points coincide with atoms, for instance - as you mentioned face-centered cubic - all the metallic elements with fcc packing of their atoms, i.e. copper, silver, gold, nickel etc.

[@akshitasharma5340]

@@FrankHoffmann1000is it not odd that we considered lattice to be repeating unit of the crystal but atoms are not coinciding with lattice points ?? Then, what is the need to lattice and why are measuring lattice parameters. It seems that lattice parameters are just imaginary numbers as it is not the distance between two real entities(atoms) I know my questions are sounding stupid but am trying to understand this subject .

[@FrankHoffmann1000]

It seems that there are still some deficiencies concerning the concept of the term lattice and lattice point. Firstly, probably it is useful to watch unit 1.9 again (kzbin.info/www/bejne/kGecfZ-arbF_lbc). Secondly, maybe speaking about "does not coincide with lattice points" was a source of misunderstanding (but indeed lattice points are a construct, they are not physical entities). I would recommend to look at a fairly simple crystal structure, say benzene. You can find the CIF at my website: crystalsymmetry.wordpress.com/textbook/ Benzene crystallizes according to a primitive lattice, which usually means, lattice points are only at the corners of the unit cell. Now look at the CIF - do you see atoms at the corner? Third, lattice parameters are of course not imaginary numbers. Try to understand that all atoms of a crystal are subject of the same(!) translational principle/lattice even if the atoms are not placed (it is a question of defining the _origin_ of the cell!) _at_ a lattice point. Once again: A lattice point _represents_ a (whole) motif, but is rarely the motif itself. So, this means that the distances from lattice point to lattice point are of course identical to distances between some atoms (not necessarily the next-neighboring atom). Maybe this graphic is helpful: crystalsymmetry.files.wordpress.com/2020/03/lattice_motif.pdf The distance between atom a1 and a1' is of course identical with the distance from lattice point l1 to lattice point l1'.

[@Upgradezz]

This doubt is killing me can anyone help: Fcc cubic unit cell is equal to body centered tetragonal, then why do we keep both of them as Bravais lattices?

[@FrankHoffmann1000]

Hi Dawn, can you say a little bit more about the source of your doubt? Why should the face centered cubic lattice be identical with the body centered tetragonal? I can anticipate that they are not. best Frank

[@dr.jayantthakare8707]

Dear Dawn, for a crystal system to be called a bravais lattice, it should satisfy the classification based on symmetry. The cubic crystal should have four 3-fold symmetry axes, which is there in FCC Cubic and hence we keep it in bravis lattice. The same is not true with end centric cubic crystal system and it satisfies the symmetry criterion of BODY CENTERED TETRAGONAL .

[@kathleenhayes2171]

Super helpful! Thank you

[@rajatkamboj230]

very nicely explained

[@rajatkamboj230]

Please also make a video to show why honeycomb is not a Bravais lattice. I know it is possible to convert honeycomb with a 2 atom basis to hexagonal Bravais lattice. But I still lack the visualizaton.

[@a.m8659]

Thank you frank 💛

[@FrankHoffmann1000]

Very welcome :-)

[@BeeNotDismayed]

4:55 Bookmark

[@faultlady]

Thank you!

[@richardtsui4310]

Thanks a lot!