Thank you very much sir for making this video. While studying crystallography, this particular part was driving me crazy . I just could not get it. But you have made it super easy by using the animation and stuff. Thank you so much.
Great video! It's exactly what I needed - the relation between rhombohedral, trigonal and hexagonal lattices. The animations and to-the-point explanations reinforce each other perfectly.
Your series of lectures are not only very clear but also very fascinating to an undergrad. Hats off for the effort you put into each of these videos. It is helping me a lot through a course on crystal symmetry.
I was working with ZIF-7 and different places mentioned hexagonal, rhombohedral, and trigonal. I was confused and this explained the concept 100%. Thank you very much
Thank you so much Frank. I almost went through all your videos and they help me a lot. I was always looking for some systematic beginners' guide to the space group notations and their visual illustration. But I found all textbook I encountered before overlooked such beginners' guide. In my opinion, your videos are definitely the most helpful materials for researchers setting foot in crystallography. Of course, your book Introduction To Crystallography is also very helpful.
So this how a R unit cell is made of😂 Thanks for the explanation. I can't find anything like this on the internet for a week. You rock man ! I'll be damm sure checking out your new book.
Cool vid bro. There's a lot of people in this field who think that as soon as they've learnt something they can write a book about it ;-0 ;-0 However, I can tell that you really do know what you are talking about as no one has explained the rhombohedral/hexagonal conversion to me properly in an embarassingly long time (almost 40 years).
A perfect lecture as always. I also would like to ask you if you are planning to record new lectures about unit-cell transformations, like from Rhombohedral to pseudo-cubic unit cells. It is extremely hard to find a comprehensive explanation for this or a software for this purpose. Thank you very much for share this kind of content on your YT channel.
Dear Leonardo, thank you very much for your kind message! Sorry, no, I have no plans to record new lectures in the near future, but if you have a specific question you are invited to write me an email. From my understanding a pseudo-cubic unit cell is just a primitive rhombohedral one with alpha = beta = gamma = 90°. best wishes! Frank
Hi Frank. May I ask a question. When we try to determine the space group whose crystal system is trigonal, how can we determine whether it is a primitive trigonal or Rhombohedral Hexagonal? For example, space group R-3m and P3m1 both are trigonal crystal system. Why do we sometimes use R... and sometimes use P...?
Hi! P is used for those crystals of the trigonal crystal system that have a _hexagonal lattice system_, although its maximum rotational symmetry is 3 (not 6) This means that these crystals can be described with a primtive unit cell that looks loke a primtive hexagonal one. R is used for those crystals of the trigonal crystal system that have an underlying rhombohedral lattice, meaning that we have to use a rhombohedron, if we were forced to describe the crystal with a primitive unit cell. Or to express this the other way round: we have to apply an R-centering, if we want to derive at a unit cell with hexagonal metric, in which the 3-fold axis of rotation runs along the c-axis. Best! Frank
Hi Frank. Would it be correct to say that if the rule that the axis which gives the largest rotation order must be parallel to the c direction did not exist, then the trigonal crystal system would not have the same metric as the hexagonal system (in other words there would be 7 crystal families instead of 6)? Stated in another way the trigonal metric would now simply be a=b=c, and alpha=beta=gamma not equal to 90o? Thanks.
Hi! I think that I know what you mean, but well, no, I don't think that this would be justified, because this rule follows from the fact that it is simply possible to redefine the lattice in a way that the metric itself is identical to the one of the hexagonal crystal system, and this is independent of the nomenclature of axes. Imagine, that we call the former c axis the b axis, does it really change something? There are two axis that have the same length and another one which can be different and the angles are two times 90° and one time 120°, regardless of how you name the axes. Different opinion? Curious to hear about it! best Frank
Would you please tell me if what I understood is correct? The three-fold axis of rotation of trigonal crystal system or rhombohedral crystal lattice doesn't pass through (or parallel to) the c-axis. When we assign the 3-fold axis of rotation as the c-axis in trigonal crystal system or rhombohedral crystal lattice, it transform to R-hexagonal lattice (satisfying the condition that the rotation axis with highest order should pass through the c-axis). Thanks a lot for these wonderful insights.
You are welcome! This issue exists only for the rhombohedral lattice type; for this case your thoughts are correct. For the trigonal system, in which the primtive unit cell already looks like the hexagonal primtive one, there is no need for the change of the axes.
Hello Frank, I found it very difficult to understand from 2:57 to 5:04 in the video. I am really willing to have a detailed explanation of that bit in the video. Will you please explain it to me?
Hello Manusha, yes, but I think you have to be a little more specific. Can you describe, which aspect you do not understand? I think the difficulty is not so much in the content itself, but the geometrically relation of the original rhombohedral cell to the new (larger) cell with hexagonal metric - this is more a matter of the capability to 'think' in 3D. The first step should be clear, or not? Take a rhombohedron and orient it in such a way that it stands on one of the corners. If we look along the body diagonal (marked as c-prime) we see a threefold axis of rotation. In the next steps we draw a 2D projection of it, the three green and blue atoms are 'connected' by lines to build a triangle to emphasize the three-fold rotational symmetry, and then several such rhombohedrons are assembled together, because to identify the larger cell we need more of one rhombohedron. Look also at the relation between the rhombohedron and the new cell at 6:01! best Frank
@@FrankHoffmann1000 Hi Frank, So at 2:57 the rhombohedrans are being attached face to face and after that you have said that there is a problem with such arrangement because the crystallographers want the axis of rotation to be parallel to C- axis. So you have said this phrase specifying that ," the axis of rotation with the highest order should run parallel to c-axis." I am wondering what is meant by axis if highest order of rotation? And I can see how c' is drawn. But I can't picture how a' and b' are drawn and at 4:40 we get another figure describing how it looks when it is translated. I am wonndering whether it is still touching face to face as you have shown at 2:57. If possible may I please have the 3-D diagram of the figure which appeared at 4:40? Thanks a lot for answering my previous question. I really enjoyed your videos so far. Thanks a lot for these videos.
Hi Manusha, to the first question: There is no problem with the _arrangement of the rhombohedra_ - this is a misunderstanding! The crystal itself doesn't know anything about unit cells! The "problem" with the choice of a rhombohedral unit cell is that it doesn't follow the crystallographic rule you mentioned. To explain the axes and the crystallographic rule a bit more: Take a cube with its three orthogonal axes a, b, and c as usually defined. What kind of rotational axes are there? Four-fold, three-fold and two-fold axes. The four-fold is the axis of rotation of highest order (four is the largest number) - ok? And in a cube this c-axis of the coordinate system is indeed running parallel to c-direction. Now, for the rhombohedron the highest rotational order is three (there is no four- or six-fold axis of rotation). And this axis is running along the body-diagonal (from one tip to the other, shown at 3:07 and as you said as c-prime from 3:39 on) - but this is not the original c-axis of a rhombohedron! A rhombohedron is - as shown in the beginning - derived from a cube, when it is stretched or squished along the body-diagonal but has a coordinate system analogously to a cube (shown at 1:48, as three green lines). So, and this is the crystallographic problem. Now, the purpose is to find another unit cell, in which this three-fold axis, this c-prime axis is the new c-axis of another unit cell and is the question is how the new unit cell looks like. Okey, second issue, the a' and b' directions: Here, the three blue and three green atoms are connected to build triangles (three-fold rotational symmetry!) The a-prime direction is defined in such a way that the bottom-left blue atom is the 'origin' and the atom at the tip of the blue triangle is the end of this a-prime vector; the b-prime direction: again from the bottom-left atom now to the bottom-right blue atom of this triangle, ok? I hope it is now a bit clearer how the drawing on the right at 4:40 is constructed? The projection of one rhombohedron is multiplied and translated in the a' and b' direction - but remember that the rhombohedron stands on one of its corner atoms! You said: "I am wondering whether it is still touching face to face as you have shown at 2:57?" Yes, if you refer to the 'old' rhombohedra, but no if you refer to the new a' and b' prime axes, they are not defining axes of faces of the rhombohedra and the new a and b directions are only parallel to these directions (i.e. the new orange lines define the new faces!)! Unfortunately, I do not have a 3D model of this 2D projection! I can only refer once again to the perspective drawing at 6:09 - look carefully in which way the atoms of the original rhombohedral and the new cell with hexagonal metric are related to each other. best regards Frank
Hi Frank. Great lecture! Just a couple of questions. At 1:10 you show 2 possible primitive unit cells for the trigonal system (rhombohedral and hexagonal). Is it correct to say that if the motif is just a single point for both of these cells (so lattice and crystal is the same), then the rhombohedral cell will have a 3 fold symmetry, but the hexagonal cell (also with the point motif) will have a 6 fold symmetry? Of course in such a case the hexagonal unit cell cannot represent the trigonal crystal system because it violates the 3 fold symmetry requirement. In other words the thing that makes the hexagonal UC have the 3 fold symmetry is the motif? But in case of rhombohedral unit cell even if we use a single point as a motif it will still have a 3 fold symmetry? Also, if I understand correctly the defining symmetry of a cubic crystal system is a 3 fold symmetry along the body diagonal. However the body diagonal is not oriented along the a, b or c axes, so why does that not cause the same type of problems as the rhombohedral lattice? Thanks.
Hi Rahul! You answered all your question correctly! Concerning the cubic crystal system: Well, the rule is that the axis of rotation of highest order should run parallel to one of the axes - this is the case for the _four-fold_ rotational axis in the cubic system! Best! Frank
Hello sir, I'm from korea. so my English is not good, but please be generous I'am always grateful for your great lectures. Can you confirm that what I understand is correct? First, trigonal system is divided into 1. primitive rhombohedral lattice and 2. primitive hexagonal lattice, and this rhombohedral cell can be converted to R-centered hexagonal lattice rather than primitive hexagonal lattice. And the primitive hexagonal lattice has a 3-fold rotational symmtry because of its motif. Second, Is there only hexagonal lattice in hexagonal crystal system? Third, according to the 14 Bravais lattice classifications, R-centered hexagonal lattice belong to I- trigonal rather than the hexagonal?
Your English is completely sufficient!! To your questions: First - correct! Second - yes, there is only the hexagonal lattice! Third - exactly! Best! Frank
@@FrankHoffmann1000 Hi Frank. Thank you very much for your excellent videos. Following up on this question, we say if we have to represent all the crystals of the trigonal crystal system with primitive unit cells, there are two kinds: one is the primitive rhombohedral unit cell and the other one is primitive hexagonal unit cell. Then, we say the primitive rhombohedral unit cell can be converted to an R-centered hexagonal unit cell. Now, my question is about that primitive hexagonal unit cell with trigonal system symmetry. Could you please provide an example with a primitive hexagonal unit cell that has 3-fold rotational symmetry ( trigonal system). Where is the rotational axis located in such a unit cell? I suppose it should not be at the lattice point since in that case, it would be a 6-fold rotational axis, not a 3-fold rotational axis? Am I right?
@@mahdibeedel8390 Hi Mahdi, no, you are wrong. The three-fold axis of rotation is again at the lattice point. Please note that the lattice can have a higher symmetry than the crystal structure, because the lattice point has always the highest symmetry possible. But the question is always: what dows the lattice point represent? If it represents a trigonal motif, then you end up only with an overall 3-fold rotational symmetry. You might want to check out unit 4.4: th-cam.com/video/WuZVV2jtkKo/w-d-xo.html&ab_channel=FrankHoffmann At 4:47 the element Tellurium is presented; it has a 3(1) screw axis...
Hello! The ilmenite-type structure does not belong to the hexagonal crystal system, but to the trigonal. This is reflected in the space group symbol with a highest order of the rotational axis of 3, rather than 6.
Thank you very much for this helpful video. They are so clear and much more easier to understand than most of the books. But I have a question on this video at about 4:21, you said that if we look from the c-prime axis, there will be new axes a- prime and b-prime, as I can see from the picture, do you mean the axes along the two blue points? Or do I misunderstand your explenation? According to the next slide you have shown, the planes a-prime and b-prime should be the plane where the blue points and green points located. So I guess if the b-prime should represent the green points? This just makes me confused. I really appreciate if you could help to explain that.
Dear Lu, thank you very much for your kind comment. Concerning you question: Well, this was indeed perhaps a little confusing. The new axes a-prime and b-prime in the slides indicate only their relative orientation (i.e. internal angle of 60 degrees) and nothing is said a) about their origin (which is not necessarily one of the blue nor necesserily one of the green points) and b) about the height (along c-prime) of the plane that is defined tby these two axes. In the new orange centered cell the axes a and b are running _parallel_ to the formerly a-prinma and b-prime axes but the origin is shifted. Furthermore, the (a,b) _plane_ of the new centered cell is neither at the height of c' = 1/3 nor of 2/3, but at zero (and one). I have drawn another slide to try to make this clearer: 1drv.ms/b/s!ArTbwWHXPrwehNM72G8fLYIoxgugvg Hope, this helps! best! Frank
Thanks for the great video!! I got a basic question. Can we say that both rhombodedral and hexagonal unit cells are describing the same structure? Is there any exceptive cases that a structure can only be described with rhombodedral or hexagonal unit cell while another is not applicable? Thank you very much.
Well...after checking the other comments I may have understood above question. Only "R-centerd" hexagonal unit cell equals to rhombodedral one, while there is also a normal hexagonal unit cell that has no lattice points occupying the body diagonal and thus has a six-fold rotational symmtry. So, "R-centerd" hexagonal unit cell belongs to Trigonal crystal system and "normal" hexagonal unit cell belongs to hexagonal crystal system. Is it correct? Thank you~
@@xugao9119 Almost - but please note that those trigonal crystals that can be decribed with a normal, i.e., primitive hexagonal unit cell do not have six-fold rotation symmetry! The _lattice_ itself has six-fold rotational symmetry, that's true, but not the trigonal crystals, they have only 3-fold rotational symmetry - this means that the lattice (sometimes) has a higher symmetry than the actual _structure_ .
Hi Sir Hoffman, thank you smooch for this excellent course, which solved so much of my puzzles! I have a question, a primitive hexagonal unit cell has a six-fold rotation axis, and we can use it to describe a trigonal crystal lattice, which just has a 3-fold rotation axis. But if we use primitive hexagonal unit cell, it means this trigonal lattice has 6-fold axis and it then belongs to hexagonal crystal system, right?
It’s a little tricky: the primitive hexagonal lattice has always a six-fold axis of rotation, no matter if the crystal at hand belongs to the hexagonal or trigonal crystal system. If the crystal belongs to the trigonal crystal system then the motif reduces the maximum order of rotation to three, because the motif is incompatible with a six-fold axis of rotation, but compatible with a three-fold axis of rotation. This means that in this case the symmetry of the pure lattice is actually higher than that of the crystal.
wonderful video. If rhombohedral is written down in a hexagonal structure, it has 2 extra lattice points in the unit cell. Does that mean it has 6 extra points in the full hexagonal excluding the basal planes? That means it is tighter packed than a HCP which is impossible? HCP has 3 atoms between the basal planes in the full structure What am I missing? thank you
Hi Bob, well, I think it begins with speaking about "extra lattice points". This is a common misconception: that in centered cell there are lattice points that are addionally present. This is not the case. All the lattice points are exactly on the same place as before in the non-centered cell. The only thing that changed is a certain border around some lattice points. It is correct that the rhombohedral-centered cell hat 3 lattice points, but the volume is also three times the volume of the formerly primtive rhombohedral one! (You have to compare the new centered cell with it's primtive one, not with a different (primtive) hexagonal unit cell.) Therefore, the density of lattice points doesn't change, regardless, what kind of centering you choose, which means that the rhombohedral-centrered hexagonal cell has exactly the same density as the primitive rhombohedral cell. best wishes Frank
@@FrankHoffmann1000 thank you, i understand what you mean. very well explained. My mistake was comparing the HCP unit cell (2 atoms/unitcell) with the rhombohedral-centered unit cell (3atoms/cell), but taking the same volume! HCP is the tightest possible packing for hexagonal because the volume is much smaller.
@@FrankHoffmann1000 Sir I have question: If you know the miller index of a plane of the rhombohedral unit cell, how do you calculate to get the miller index of that plane for the R-centered hexagonal unit cell? I cant find sources for this. It makes sense visually that the 111 plane in the rhombo unit cell is equal to the 0001 plane in the hexagonal unit cell, but what for example about the 001 plane? Do you know the formulas for converting? thank you.
@@bobslee3089 No, I don't know such a formula. The (001) face in the rhombohedral setting should be the (0 -1 1) face in the hexagonal setting. I would recommend using VESTA for the conversion: Choose first the rhombohedral setting, place a lattice plane of your choice into the cell and then convert it to the hexagonal setting.
Hello, as usual, It's such a great lesson, my question is: Is it for this reason that the Quartz has a hexagonal lattice but belongs to the trigonal system? Or is it simply because the trigonal and the hexagonal crystal systems have the same metrics? Thanks for your answer
Sorry, but I think you have forgotten something in your question: "Is it for this reason that the Quartz has a hexagonal lattice but belongs to the trigonal system?" - what do you mean by "this reason"?
@@FrankHoffmann1000 yes, sorry for being unclear, I will ask my question in another way, but I am confused, so I will try to sum up what I understood and please correct me, 1) In the trigonal crystal system, we can Have a primitive rhombohidral unit cell, with such a cell we can have rhombohidral lattice. 2) as shown at the first of the video, we have a primitive hexagonal unit cell, with this cell we can form a hexagonal lattice, but in this case it doesn't mean that we are in the hexagonal crystal system? I know the crystal system never changes, so I wish if you could please make this point clearer. 3) and we also have hexagonal-R centered unit cell which is just a larger unit of the rhombohidral unit cell (can we call it Rhombohidral R centered too?) with the metric of the hexagonal but with a 3-fold symmetry axis. And talking about quartz, does he made by this hexagonal R centered unit cell? I know I've made a lot of mistakes in this comment, I am a bit lost and your help will be appreciated. Thanks again.
@@waliddjani5978 1) - correct 2) - the decisive point here is that a hexagonal lattice doesn't always mean that we have a six-fold rotational symmetry. The symmetry of the lattice can be higher than the symmetry of the crystal because of the motif. If the motif has only three-fold rotational symmetry, you can place the motif on a hexagonal lattice, but the maximum rotational symmetry of the crystal is only three-fold. 3) The R-centered unit cell is only an alternative representation of (some) of the trigonal crystals that build a rhombohedral lattice, when you choose a primtive unit cell. Quartz is not rhombohedral. It is of the "other" sort of triginal crystals, i.e. it has a primtive hexagonal lattice, again only with three-fold rotational symmetry.
@@FrankHoffmann1000 thank you so much, now it's much clearer. Please tell me if I am right: 1) Both Rhombohidral and R centered Unit cells form a Rhombohidral lattice, the difference between them, is just we choose the centered for symmetry reasons. 2) the primitive hexagonal unit cell form a hexagonal lattice, but depending on the symmetry of the motif, the global symmetry of the crystal will be either Hexagonal or Trigonal. I have 2 simple questions: For the R centered unit cell, do we call it hexagonal R centered, Rhombohidral R centered, or both? With the naked eye, can we differ crystals with the R centered unit cell, or can we do it the X ray crystallography? Or it's just a representation? Thank you
@@waliddjani5978 1) - No! Rhombohedral unit cells form a rhombohedral lattice, while R-centered hexagonal cells form a (inner-centered) hexagonal lattice with two additional lattice points inside the cell. 2) - correct Question 1: hexagonal R-centred Question 2: you can never see a centering with the naked eye, because a centering is a _mental_ concept describing a symmetry-adapted version of the repeat unit in crystals.
Hello sir, I would like to know if Wikipedia explains correctly the distinction between Crystal Families, Crystal Systems and Lattice Systems? From what i can understand in your videos you are explaining the 7 crystal systems, which include the trigonal (subdivided to the rhombohedral and hexagonal), but Wikipedia labels Rhombohedral as a lattice system instead of a crystal system, which makes it more confusing to me whether it is important to distinguish these 3 terms.
Hello! I do not see a contradiction. What I stated (from 1:12 on) is that the term 'trigonal' refers indeed to a _crystal_ system (not to a lattice system) and that these trigonal crystals can be subdivided to crystals with a rhombohedral or hexagonal _lattice_. This is in complete accordance to what is stated at wikipedia: en.wikipedia.org/wiki/Hexagonal_crystal_family or also in the online dictionary of the International Union of crystallography: reference.iucr.org/dictionary/Lattice_system ok? best wishes Frank
@@FrankHoffmann1000 Thank you, I rewatched that session and understood it better now. So basically, trigonal itself is a crystal system because it defines the morphology type of a crystal, but rhombohedral is only a lattice system because it only covers the unique symmetrical property of a trigonal subgroup right? Therefore, while both the rhombohedral and hexagonal share the same essential 3-fold symmetry, they are differentiated by how this symmetry is distributed (as in, the lattice or repeating pattern and its' subsequent parameters, e.g. hexagonal gamma must be 120 degrees) It's difficult because sometimes I hear these terms used together and I didn't learn that they were not interchangeable (lattice system vs crystal system) and the reason for it, so it's easy to mix them up. Also, lectures I went to did not really make it clear the relationship between trigonal, hexagonal and rhombohedral, so I used to assume they were somewhat interchangeable as well.
The volume of _every_ parallelepiped can be calculated by the following formula: V = abc * sqrt[1 + 2 cos (alpha) * cos (beta) * cos (gamma) - cos^2 (alpha) - cos^2 (beta) - cos^2 (gamma) ]
Well, this is not dependent on the crystal system and the value is, so to say, "unique" for every concrete crystalline compound. Usually, you carry out a single-crystal X-ray diffraction measurement, determine the space group and the reciprocal lattice parameters, calculate the real lattice parameters and put them into the given formula.
Hello Frank, Thanks a lot for your videos and explaining the concepts so fluidly. I could understand in this particular video that the trigonal cell characterized by the 3-fold symmetry can be represented as two different primitive unit cells, namely, rhombohedral and hexagonal. You showed the rhombohedral cell to clearly posses 3-fold symmetry and therefore it belongs to the trigonal crystal system. However I am not able to understand the hexagonal primitive unit cell of the trigonal crystal system. How does this cell posses only 3-fold symmetry and not 6 fold? As you explained in the previous video (2.8) that the hexagonal crystal system has 6 fold symmetry and the rotational axis lies along the "c" direction. In my opinion if we translate the hexagonal primitive unit cell of the trigonal crystal system in 3 directions we will end up getting a 6 fold symmetry. Also the metrics of the hexagonal crystal system and hexagonal unit cell of the trigonal system are the same.
Dear bonny, thanks for your question! This is only understandable if you take into account the _motif_! All the crystals belonging to the triginal crystal system which can be represented by this lattice which looks like (and indeed have the same metric!) the primitive hexagonal lattice have a motif which is incompatible with a six-fold rotational symmetry (along the c axis), but are only compatible with a three-fold rotational symmetry. The impression that a six-fold rotational axis is present is generated by the circumstance that the lattice points are usually points or circles. But these points or circles _represent_ the motif - substitute these lattice points, for instance, with triangles! Now, it should be clear that only a three-fold axis of rotation is present (i.e. rotation by 120° leads to an indistinguishable configuration). And now it should also be clear why a trigonal and hexagonal _lattice_ can look identical - this is because a two-fold rotation by 60 degrees is compatible with a three-fold rotational symmetry. See for illustration purposes also this PDF: crystalsymmetry.files.wordpress.com/2018/01/trigonal_hexagonal_primtive_lattice.pdf Hope this helps! best! Frank
Dear Frank, Thanks a lot for the explanation. I could understand this now. If I get the whole concept clearly, then is it correct to say that, given a type of atomic arrangement, we first determine the symmetry of the system, which decides the crystal system that atomic arrangement should belong to. Then we have the freedom to chose the lattice/unit cell. So incase our atomic arrangement has the -3m symmtery, trigonal crystal system is ascribed to it and then we can chose to represent it in either rhombohedral lattice or hexagonal lattice. Is my inference correct?
Dear bonny, sorry for the late reply! Well, I think your statement is almost correct. But note that we do not have the freedom to choose between different lattices for all cases. For the trigonal crystal system: for all the space groups which contain a "R" in the space group symbol you can choose between a rhombohedral (= primtive cell) or hexagonal (= R-centered) representation; these are space groups no. 146, 148, 155, 160, 161, 166, and 167. For all other space groups of the trigonal crystem you do not have this choice. What I don't understand is your expression "So in case our atomic arrangement has the -3m symmetry". The atomic arrangement is not detached from the translational symmetry component. -3m symmetry is not specific for the one or the other space group. You have these symmetry elements for instance in the trigonal space group P-31m (no. 162) and also in R-3m (no. 155). best! Frank
Dear Frank, Thanks a lot for your reply. Now I am able to get more clarity as earlier I was not considering different spacegroups for the same crystal system. Kind Regards, Bonny
Hey J B, first of all, a rhombohedral cell is characterized by a different symmetry than a cubic cell/lattice. Secondly, a parallelepiped has 8 corners, not 4. But you are right that there is no centered variant of a rhombohedral lattice, only primitive ones exist.
@@FrankHoffmann1000 Mr Hofmann Thank you for sharing your knowledge. In the case of simple cubic unit cell, slip is not possible because there is no close packed plane. Is it true that for the same reason sllip is not possible for rhombohedral unit cell? This means that materials comprising of rhombohedral will fracture in the elastic regime?
@@user-kl4oh2co2y Hey J B, never thought about this question. But I think, this could be true: Most of the metals which adopt a rhombohedral structure are brittle. best wishes Frank
Thank you very much sir for making this video. While studying crystallography, this particular part was driving me crazy . I just could not get it. But you have made it super easy by using the animation and stuff. Thank you so much.
Very welcome!
Great video! It's exactly what I needed - the relation between rhombohedral, trigonal and hexagonal lattices. The animations and to-the-point explanations reinforce each other perfectly.
Your series of lectures are not only very clear but also very fascinating to an undergrad. Hats off for the effort you put into each of these videos. It is helping me a lot through a course on crystal symmetry.
Thank you very much for your kind words!
SO HELPFUL! Often instructors and videos just flash the photos at you and say "what don't you get about this?".
I was working with ZIF-7 and different places mentioned hexagonal, rhombohedral, and trigonal. I was confused and this explained the concept 100%. Thank you very much
Thank you so much Frank. I almost went through all your videos and they help me a lot. I was always looking for some systematic beginners' guide to the space group notations and their visual illustration. But I found all textbook I encountered before overlooked such beginners' guide. In my opinion, your videos are definitely the most helpful materials for researchers setting foot in crystallography. Of course, your book Introduction To Crystallography is also very helpful.
Thank you very much for your kind comment. I'm glad that you find these videos so helpful.
So this how a R unit cell is made of😂 Thanks for the explanation. I can't find anything like this on the internet for a week. You rock man ! I'll be damm sure checking out your new book.
This makes so much more sense of the two types of unit cells of the trigonal crystal system. thanks!
Thankyou sir for the explanation. I was not getting this particular thing anywhere else. Thankyou so much sir!!
Cool vid bro. There's a lot of people in this field who think that as soon as they've learnt something they can write a book about it ;-0 ;-0
However, I can tell that you really do know what you are talking about as no one has explained the rhombohedral/hexagonal conversion to me properly in an embarassingly long time (almost 40 years).
Thanks Jon, much appreciate your comment!
Awesome explanation.....
Thank you so much...
Love from India
Thank You very much! I am going to watch all your videos!
very welcome!
A perfect lecture as always.
I also would like to ask you if you are planning to record new lectures about unit-cell transformations, like from Rhombohedral to pseudo-cubic unit cells.
It is extremely hard to find a comprehensive explanation for this or a software for this purpose.
Thank you very much for share this kind of content on your YT channel.
Dear Leonardo,
thank you very much for your kind message!
Sorry, no, I have no plans to record new lectures in the near future, but if you have a specific question you are invited to write me an email.
From my understanding a pseudo-cubic unit cell is just a primitive rhombohedral one with alpha = beta = gamma = 90°.
best wishes!
Frank
Hi Frank. May I ask a question. When we try to determine the space group whose crystal system is trigonal, how can we determine whether it is a primitive trigonal or Rhombohedral Hexagonal? For example, space group R-3m and P3m1 both are trigonal crystal system. Why do we sometimes use R... and sometimes use P...?
Hi!
P is used for those crystals of the trigonal crystal system that have a _hexagonal lattice system_, although its maximum rotational symmetry is 3 (not 6) This means that these crystals can be described with a primtive unit cell that looks loke a primtive hexagonal one.
R is used for those crystals of the trigonal crystal system that have an underlying rhombohedral lattice, meaning that we have to use a rhombohedron, if we were forced to describe the crystal with a primitive unit cell. Or to express this the other way round: we have to apply an R-centering, if we want to derive at a unit cell with hexagonal metric, in which the 3-fold axis of rotation runs along the c-axis.
Best! Frank
@@FrankHoffmann1000 I understood, Sir. Thank you very much, Frank. Great explanation!
Hi Frank. Would it be correct to say that if the rule that the axis which gives the largest rotation order must be parallel to the c direction did not exist, then the trigonal crystal system would not have the same metric as the hexagonal system (in other words there would be 7 crystal families instead of 6)? Stated in another way the trigonal metric would now simply be a=b=c, and alpha=beta=gamma not equal to 90o? Thanks.
Hi!
I think that I know what you mean, but well, no, I don't think that this would be justified, because this rule follows from the fact that it is simply possible to redefine the lattice in a way that the metric itself is identical to the one of the hexagonal crystal system, and this is independent of the nomenclature of axes. Imagine, that we call the former c axis the b axis, does it really change something? There are two axis that have the same length and another one which can be different and the angles are two times 90° and one time 120°, regardless of how you name the axes.
Different opinion? Curious to hear about it!
best
Frank
This was AMAZING
Yo big fan...😁
Would you please tell me if what I understood is correct? The three-fold axis of rotation of trigonal crystal system or rhombohedral crystal lattice doesn't pass through (or parallel to) the c-axis. When we assign the 3-fold axis of rotation as the c-axis in trigonal crystal system or rhombohedral crystal lattice, it transform to R-hexagonal lattice (satisfying the condition that the rotation axis with highest order should pass through the c-axis). Thanks a lot for these wonderful insights.
You are welcome!
This issue exists only for the rhombohedral lattice type; for this case your thoughts are correct. For the trigonal system, in which the primtive unit cell already looks like the hexagonal primtive one, there is no need for the change of the axes.
Hello Frank,
I found it very difficult to understand from 2:57 to 5:04 in the video. I am really willing to have a detailed explanation of that bit in the video. Will you please explain it to me?
Hello Manusha,
yes, but I think you have to be a little more specific. Can you describe, which aspect you do not understand? I think the difficulty is not so much in the content itself, but the geometrically relation of the original rhombohedral cell to the new (larger) cell with hexagonal metric - this is more a matter of the capability to 'think' in 3D.
The first step should be clear, or not? Take a rhombohedron and orient it in such a way that it stands on one of the corners. If we look along the body diagonal (marked as c-prime) we see a threefold axis of rotation. In the next steps we draw a 2D projection of it, the three green and blue atoms are 'connected' by lines to build a triangle to emphasize the three-fold rotational symmetry, and then several such rhombohedrons are assembled together, because to identify the larger cell we need more of one rhombohedron.
Look also at the relation between the rhombohedron and the new cell at 6:01!
best
Frank
@@FrankHoffmann1000
Hi Frank,
So at 2:57 the rhombohedrans are being attached face to face and after that you have said that there is a problem with such arrangement because the crystallographers want the axis of rotation to be parallel to C- axis. So you have said this phrase specifying that ," the axis of rotation with the highest order should run parallel to c-axis."
I am wondering what is meant by axis if highest order of rotation?
And I can see how c' is drawn. But I can't picture how a' and b' are drawn
and at 4:40 we get another figure describing how it looks when it is translated. I am wonndering whether it is still touching face to face as you have shown at 2:57. If possible may I please have the 3-D diagram of the figure which appeared at 4:40?
Thanks a lot for answering my previous question.
I really enjoyed your videos so far. Thanks a lot for these videos.
Hi Manusha,
to the first question:
There is no problem with the _arrangement of the rhombohedra_ - this is a misunderstanding! The crystal itself doesn't know anything about unit cells! The "problem" with the choice of a rhombohedral unit cell is that it doesn't follow the crystallographic rule you mentioned.
To explain the axes and the crystallographic rule a bit more: Take a cube with its three orthogonal axes a, b, and c as usually defined. What kind of rotational axes are there? Four-fold, three-fold and two-fold axes. The four-fold is the axis of rotation of highest order (four is the largest number) - ok? And in a cube this c-axis of the coordinate system is indeed running parallel to c-direction.
Now, for the rhombohedron the highest rotational order is three (there is no four- or six-fold axis of rotation). And this axis is running along the body-diagonal (from one tip to the other, shown at 3:07 and as you said as c-prime from 3:39 on) - but this is not the original c-axis of a rhombohedron! A rhombohedron is - as shown in the beginning - derived from a cube, when it is stretched or squished along the body-diagonal but has a coordinate system analogously to a cube (shown at 1:48, as three green lines). So, and this is the crystallographic problem.
Now, the purpose is to find another unit cell, in which this three-fold axis, this c-prime axis is the new c-axis of another unit cell and is the question is how the new unit cell looks like.
Okey, second issue, the a' and b' directions: Here, the three blue and three green atoms are connected to build triangles (three-fold rotational symmetry!) The a-prime direction is defined in such a way that the bottom-left blue atom is the 'origin' and the atom at the tip of the blue triangle is the end of this a-prime vector; the b-prime direction: again from the bottom-left atom now to the bottom-right blue atom of this triangle, ok?
I hope it is now a bit clearer how the drawing on the right at 4:40 is constructed? The projection of one rhombohedron is multiplied and translated in the a' and b' direction - but remember that the rhombohedron stands on one of its corner atoms!
You said: "I am wondering whether it is still touching face to face as you have shown at 2:57?" Yes, if you refer to the 'old' rhombohedra, but no if you refer to the new a' and b' prime axes, they are not defining axes of faces of the rhombohedra and the new a and b directions are only parallel to these directions (i.e. the new orange lines define the new faces!)!
Unfortunately, I do not have a 3D model of this 2D projection! I can only refer once again to the perspective drawing at 6:09 - look carefully in which way the atoms of the original rhombohedral and the new cell with hexagonal metric are related to each other.
best regards
Frank
Thank you very much! This has been a hugely helpful video.
Thanks for nice words - we are happy to hear this!
Hi Frank. Great lecture! Just a couple of questions. At 1:10 you show 2 possible primitive unit cells for the trigonal system (rhombohedral and hexagonal). Is it correct to say that if the motif is just a single point for both of these cells (so lattice and crystal is the same), then the rhombohedral cell will have a 3 fold symmetry, but the hexagonal cell (also with the point motif) will have a 6 fold symmetry? Of course in such a case the hexagonal unit cell cannot represent the trigonal crystal system because it violates the 3 fold symmetry requirement. In other words the thing that makes the hexagonal UC have the 3 fold symmetry is the motif? But in case of rhombohedral unit cell even if we use a single point as a motif it will still have a 3 fold symmetry?
Also, if I understand correctly the defining symmetry of a cubic crystal system is a 3 fold symmetry along the body diagonal. However the body diagonal is not oriented along the a, b or c axes, so why does that not cause the same type of problems as the rhombohedral lattice? Thanks.
Hi Rahul!
You answered all your question correctly!
Concerning the cubic crystal system: Well, the rule is that the axis of rotation of highest order should run parallel to one of the axes - this is the case for the _four-fold_ rotational axis in the cubic system!
Best!
Frank
@@FrankHoffmann1000 Thanks! This is a deceptively confusing topic which you think you have understood, until you find out you haven't :)
Hello sir, I'm from korea. so my English is not good, but please be generous
I'am always grateful for your great lectures.
Can you confirm that what I understand is correct?
First, trigonal system is divided into 1. primitive rhombohedral lattice and 2. primitive hexagonal lattice, and this rhombohedral cell can be converted to R-centered hexagonal lattice rather than primitive hexagonal lattice. And the primitive hexagonal lattice has a 3-fold rotational symmtry because of its motif.
Second, Is there only hexagonal lattice in hexagonal crystal system?
Third, according to the 14 Bravais lattice classifications, R-centered hexagonal lattice belong to I- trigonal rather than the hexagonal?
Your English is completely sufficient!!
To your questions:
First - correct!
Second - yes, there is only the hexagonal lattice!
Third - exactly!
Best!
Frank
@@FrankHoffmann1000
Thank you very much!
It’s helping a lot
Thank you for always.
@@FrankHoffmann1000 Hi Frank. Thank you very much for your excellent videos. Following up on this question, we say if we have to represent all the crystals of the trigonal crystal system with primitive unit cells, there are two kinds: one is the primitive rhombohedral unit cell and the other one is primitive hexagonal unit cell. Then, we say the primitive rhombohedral unit cell can be converted to an R-centered hexagonal unit cell. Now, my question is about that primitive hexagonal unit cell with trigonal system symmetry. Could you please provide an example with a primitive hexagonal unit cell that has 3-fold rotational symmetry ( trigonal system). Where is the rotational axis located in such a unit cell? I suppose it should not be at the lattice point since in that case, it would be a 6-fold rotational axis, not a 3-fold rotational axis? Am I right?
@@mahdibeedel8390 Hi Mahdi, no, you are wrong. The three-fold axis of rotation is again at the lattice point. Please note that the lattice can have a higher symmetry than the crystal structure, because the lattice point has always the highest symmetry possible. But the question is always: what dows the lattice point represent? If it represents a trigonal motif, then you end up only with an overall 3-fold rotational symmetry.
You might want to check out unit 4.4: th-cam.com/video/WuZVV2jtkKo/w-d-xo.html&ab_channel=FrankHoffmann
At 4:47 the element Tellurium is presented; it has a 3(1) screw axis...
@@FrankHoffmann1000 Thank you for your clarification.
Hello sir,thanks for the amazing lesson.my question is
Why MnTiO3 ,a ilemnite structure,which stabilise in hexagonal structure has R3 space group?
Hello!
The ilmenite-type structure does not belong to the hexagonal crystal system, but to the trigonal. This is reflected in the space group symbol with a highest order of the rotational axis of 3, rather than 6.
Thank you very much for this helpful video. They are so clear and much more easier to understand than most of the books. But I have a question on this video at about 4:21, you said that if we look from the c-prime axis, there will be new axes a- prime and b-prime, as I can see from the picture, do you mean the axes along the two blue points? Or do I misunderstand your explenation? According to the next slide you have shown, the planes a-prime and b-prime should be the plane where the blue points and green points located. So I guess if the b-prime should represent the green points? This just makes me confused. I really appreciate if you could help to explain that.
Dear Lu,
thank you very much for your kind comment.
Concerning you question: Well, this was indeed perhaps a little confusing. The new axes a-prime and b-prime in the slides indicate only their relative orientation (i.e. internal angle of 60 degrees) and nothing is said a) about their origin (which is not necessarily one of the blue nor necesserily one of the green points) and b) about the height (along c-prime) of the plane that is defined tby these two axes. In the new orange centered cell the axes a and b are running _parallel_ to the formerly a-prinma and b-prime axes but the origin is shifted. Furthermore, the (a,b) _plane_ of the new centered cell is neither at the height of c' = 1/3 nor of 2/3, but at zero (and one).
I have drawn another slide to try to make this clearer:
1drv.ms/b/s!ArTbwWHXPrwehNM72G8fLYIoxgugvg
Hope, this helps!
best!
Frank
Dear Frank,
Thank you very much for your explanation. That helps a lot. I have understood this point.
Best,
Lu
Thank you so much for the helpful explanation!
Thanks for the great video!! I got a basic question. Can we say that both rhombodedral and hexagonal unit cells are describing the same structure? Is there any exceptive cases that a structure can only be described with rhombodedral or hexagonal unit cell while another is not applicable? Thank you very much.
Well...after checking the other comments I may have understood above question. Only "R-centerd" hexagonal unit cell equals to rhombodedral one, while there is also a normal hexagonal unit cell that has no lattice points occupying the body diagonal and thus has a six-fold rotational symmtry. So, "R-centerd" hexagonal unit cell belongs to Trigonal crystal system and "normal" hexagonal unit cell belongs to hexagonal crystal system. Is it correct? Thank you~
@@xugao9119 Almost - but please note that those trigonal crystals that can be decribed with a normal, i.e., primitive hexagonal unit cell do not have six-fold rotation symmetry! The _lattice_ itself has six-fold rotational symmetry, that's true, but not the trigonal crystals, they have only 3-fold rotational symmetry - this means that the lattice (sometimes) has a higher symmetry than the actual _structure_ .
@@FrankHoffmann1000 I understand! Sir, thank you so much for your kind response and clarification!
@@xugao9119 Welcome!
Hi Sir Hoffman, thank you smooch for this excellent course, which solved so much of my puzzles!
I have a question, a primitive hexagonal unit cell has a six-fold rotation axis, and we can use it to describe a trigonal crystal lattice, which just has a 3-fold rotation axis. But if we use primitive hexagonal unit cell, it means this trigonal lattice has 6-fold axis and it then belongs to hexagonal crystal system, right?
It’s a little tricky: the primitive hexagonal lattice has always a six-fold axis of rotation, no matter if the crystal at hand belongs to the hexagonal or trigonal crystal system. If the crystal belongs to the trigonal crystal system then the motif reduces the maximum order of rotation to three, because the motif is incompatible with a six-fold axis of rotation, but compatible with a three-fold axis of rotation. This means that in this case the symmetry of the pure lattice is actually higher than that of the crystal.
@@FrankHoffmann1000 OK I see, thank you so much for your explanation and also thanks so much for this lecture, it's so good!
wonderful video. If rhombohedral is written down in a hexagonal structure, it has 2 extra lattice points in the unit cell. Does that mean it has 6 extra points in the full hexagonal excluding the basal planes? That means it is tighter packed than a HCP which is impossible? HCP has 3 atoms between the basal planes in the full structure What am I missing? thank you
Hi Bob,
well, I think it begins with speaking about "extra lattice points". This is a common misconception: that in centered cell there are lattice points that are addionally present. This is not the case. All the lattice points are exactly on the same place as before in the non-centered cell. The only thing that changed is a certain border around some lattice points.
It is correct that the rhombohedral-centered cell hat 3 lattice points, but the volume is also three times the volume of the formerly primtive rhombohedral one! (You have to compare the new centered cell with it's primtive one, not with a different (primtive) hexagonal unit cell.) Therefore, the density of lattice points doesn't change, regardless, what kind of centering you choose, which means that the rhombohedral-centrered hexagonal cell has exactly the same density as the primitive rhombohedral cell.
best wishes
Frank
@@FrankHoffmann1000 thank you, i understand what you mean. very well explained. My mistake was comparing the HCP unit cell (2 atoms/unitcell) with the rhombohedral-centered unit cell (3atoms/cell), but taking the same volume! HCP is the tightest possible packing for hexagonal because the volume is much smaller.
@@FrankHoffmann1000 Sir I have question: If you know the miller index of a plane of the rhombohedral unit cell, how do you calculate to get the miller index of that plane for the R-centered hexagonal unit cell? I cant find sources for this. It makes sense visually that the 111 plane in the rhombo unit cell is equal to the 0001 plane in the hexagonal unit cell, but what for example about the 001 plane? Do you know the formulas for converting? thank you.
@@bobslee3089 No, I don't know such a formula. The (001) face in the rhombohedral setting should be the (0 -1 1) face in the hexagonal setting. I would recommend using VESTA for the conversion: Choose first the rhombohedral setting, place a lattice plane of your choice into the cell and then convert it to the hexagonal setting.
@@FrankHoffmann1000 amazing. thank you for this tip
Hello, as usual, It's such a great lesson, my question is: Is it for this reason that the Quartz has a hexagonal lattice but belongs to the trigonal system? Or is it simply because the trigonal and the hexagonal crystal systems have the same metrics? Thanks for your answer
Sorry, but I think you have forgotten something in your question: "Is it for this reason that the Quartz has a hexagonal lattice but belongs to the trigonal system?" - what do you mean by "this reason"?
@@FrankHoffmann1000 yes, sorry for being unclear, I will ask my question in another way, but I am confused, so I will try to sum up what I understood and please correct me,
1) In the trigonal crystal system, we can Have a primitive rhombohidral unit cell, with such a cell we can have rhombohidral lattice.
2) as shown at the first of the video, we have a primitive hexagonal unit cell, with this cell we can form a hexagonal lattice, but in this case it doesn't mean that we are in the hexagonal crystal system? I know the crystal system never changes, so I wish if you could please make this point clearer.
3) and we also have hexagonal-R centered unit cell which is just a larger unit of the rhombohidral unit cell (can we call it Rhombohidral R centered too?) with the metric of the hexagonal but with a 3-fold symmetry axis.
And talking about quartz, does he made by this hexagonal R centered unit cell?
I know I've made a lot of mistakes in this comment, I am a bit lost and your help will be appreciated.
Thanks again.
@@waliddjani5978 1) - correct
2) - the decisive point here is that a hexagonal lattice doesn't always mean that we have a six-fold rotational symmetry. The symmetry of the lattice can be higher than the symmetry of the crystal because of the motif. If the motif has only three-fold rotational symmetry, you can place the motif on a hexagonal lattice, but the maximum rotational symmetry of the crystal is only three-fold.
3) The R-centered unit cell is only an alternative representation of (some) of the trigonal crystals that build a rhombohedral lattice, when you choose a primtive unit cell.
Quartz is not rhombohedral. It is of the "other" sort of triginal crystals, i.e. it has a primtive hexagonal lattice, again only with three-fold rotational symmetry.
@@FrankHoffmann1000 thank you so much, now it's much clearer. Please tell me if I am right:
1) Both Rhombohidral and R centered Unit cells form a Rhombohidral lattice, the difference between them, is just we choose the centered for symmetry reasons.
2) the primitive hexagonal unit cell form a hexagonal lattice, but depending on the symmetry of the motif, the global symmetry of the crystal will be either Hexagonal or Trigonal.
I have 2 simple questions:
For the R centered unit cell, do we call it hexagonal R centered, Rhombohidral R centered, or both?
With the naked eye, can we differ crystals with the R centered unit cell, or can we do it the X ray crystallography? Or it's just a representation?
Thank you
@@waliddjani5978 1) - No! Rhombohedral unit cells form a rhombohedral lattice, while R-centered hexagonal cells form a (inner-centered) hexagonal lattice with two additional lattice points inside the cell.
2) - correct
Question 1: hexagonal R-centred
Question 2: you can never see a centering with the naked eye, because a centering is a _mental_ concept describing a symmetry-adapted version of the repeat unit in crystals.
Hello sir,
I would like to know if Wikipedia explains correctly the distinction between Crystal Families, Crystal Systems and Lattice Systems?
From what i can understand in your videos you are explaining the 7 crystal systems, which include the trigonal (subdivided to the rhombohedral and hexagonal), but Wikipedia labels Rhombohedral as a lattice system instead of a crystal system, which makes it more confusing to me whether it is important to distinguish these 3 terms.
Hello!
I do not see a contradiction. What I stated (from 1:12 on) is that the term 'trigonal' refers indeed to a _crystal_ system (not to a lattice system) and that these trigonal crystals can be subdivided to crystals with a rhombohedral or hexagonal _lattice_. This is in complete accordance to what is stated at wikipedia:
en.wikipedia.org/wiki/Hexagonal_crystal_family
or also in the online dictionary of the International Union of crystallography:
reference.iucr.org/dictionary/Lattice_system
ok?
best wishes
Frank
@@FrankHoffmann1000 Thank you, I rewatched that session and understood it better now. So basically, trigonal itself is a crystal system because it defines the morphology type of a crystal, but rhombohedral is only a lattice system because it only covers the unique symmetrical property of a trigonal subgroup right?
Therefore, while both the rhombohedral and hexagonal share the same essential 3-fold symmetry, they are differentiated by how this symmetry is distributed (as in, the lattice or repeating pattern and its' subsequent parameters, e.g. hexagonal gamma must be 120 degrees)
It's difficult because sometimes I hear these terms used together and I didn't learn that they were not interchangeable (lattice system vs crystal system) and the reason for it, so it's easy to mix them up. Also, lectures I went to did not really make it clear the relationship between trigonal, hexagonal and rhombohedral, so I used to assume they were somewhat interchangeable as well.
Thank you! This was very helpful!
How to find out volume of rhombohedral crystal system
The volume of _every_ parallelepiped can be calculated by the following formula:
V = abc * sqrt[1 + 2 cos (alpha) * cos (beta) * cos (gamma) - cos^2 (alpha) - cos^2 (beta) - cos^2 (gamma) ]
That is I know but how to find out it
Well, this is not dependent on the crystal system and the value is, so to say, "unique" for every concrete crystalline compound. Usually, you carry out a single-crystal X-ray diffraction measurement, determine the space group and the reciprocal lattice parameters, calculate the real lattice parameters and put them into the given formula.
@@FrankHoffmann1000 So we identify the shape of a unit cell of a crystal by X-ray diffraction measurement?
@@37prakhar yes! Its shape and its dimension.
Hello Frank, Thanks a lot for your videos and explaining the concepts so fluidly.
I could understand in this particular video that the trigonal cell characterized by the 3-fold symmetry can be represented as two different primitive unit cells, namely, rhombohedral and hexagonal. You showed the rhombohedral cell to clearly posses 3-fold symmetry and therefore it belongs to the trigonal crystal system.
However I am not able to understand the hexagonal primitive unit cell of the trigonal crystal system. How does this cell posses only 3-fold symmetry and not 6 fold? As you explained in the previous video (2.8) that the hexagonal crystal system has 6 fold symmetry and the rotational axis lies along the "c" direction. In my opinion if we translate the hexagonal primitive unit cell of the trigonal crystal system in 3 directions we will end up getting a 6 fold symmetry. Also the metrics of the hexagonal crystal system and hexagonal unit cell of the trigonal system are the same.
Dear bonny,
thanks for your question!
This is only understandable if you take into account the _motif_! All the crystals belonging to the triginal crystal system which can be represented by this lattice which looks like (and indeed have the same metric!) the primitive hexagonal lattice have a motif which is incompatible with a six-fold rotational symmetry (along the c axis), but are only compatible with a three-fold rotational symmetry. The impression that a six-fold rotational axis is present is generated by the circumstance that the lattice points are usually points or circles. But these points or circles _represent_ the motif - substitute these lattice points, for instance, with triangles! Now, it should be clear that only a three-fold axis of rotation is present (i.e. rotation by 120° leads to an indistinguishable configuration). And now it should also be clear why a trigonal and hexagonal _lattice_ can look identical - this is because a two-fold rotation by 60 degrees is compatible with a three-fold rotational symmetry.
See for illustration purposes also this PDF:
crystalsymmetry.files.wordpress.com/2018/01/trigonal_hexagonal_primtive_lattice.pdf
Hope this helps!
best!
Frank
Dear Frank,
Thanks a lot for the explanation. I could understand this now.
If I get the whole concept clearly, then is it correct to say that, given a type of atomic arrangement, we first determine the symmetry of the system, which decides the crystal system that atomic arrangement should belong to. Then we have the freedom to chose the lattice/unit cell.
So incase our atomic arrangement has the -3m symmtery, trigonal crystal system is ascribed to it and then we can chose to represent it in either rhombohedral lattice or hexagonal lattice.
Is my inference correct?
Dear bonny,
sorry for the late reply!
Well, I think your statement is almost correct. But note that we do not have the freedom to choose between different lattices for all cases. For the trigonal crystal system: for all the space groups which contain a "R" in the space group symbol you can choose between a rhombohedral (= primtive cell) or hexagonal (= R-centered) representation; these are space groups no. 146, 148, 155, 160, 161, 166, and 167. For all other space groups of the trigonal crystem you do not have this choice.
What I don't understand is your expression "So in case our atomic arrangement has the -3m symmetry". The atomic arrangement is not detached from the translational symmetry component. -3m symmetry is not specific for the one or the other space group. You have these symmetry elements for instance in the trigonal space group P-31m (no. 162) and also in R-3m (no. 155).
best!
Frank
Dear Frank,
Thanks a lot for your reply. Now I am able to get more clarity as earlier I was not considering different spacegroups for the same crystal system.
Kind Regards,
Bonny
Could you please tell me which course are you pursuing related to this subject?
Great video! Thanks!
is it true that a rhombohedral cell can only have simple cubic lattice? so only atoms at the four corners? thank you sir hoffman
Hey J B,
first of all, a rhombohedral cell is characterized by a different symmetry than a cubic cell/lattice. Secondly, a parallelepiped has 8 corners, not 4. But you are right that there is no centered variant of a rhombohedral lattice, only primitive ones exist.
@@FrankHoffmann1000 Mr Hofmann Thank you for sharing your knowledge. In the case of simple cubic unit cell, slip is not possible because there is no close packed plane. Is it true that for the same reason sllip is not possible for rhombohedral unit cell? This means that materials comprising of rhombohedral will fracture in the elastic regime?
@@user-kl4oh2co2y Hey J B,
never thought about this question. But I think, this could be true: Most of the metals which adopt a rhombohedral structure are brittle.
best wishes
Frank
@@FrankHoffmann1000 sir do you know where to find a databse with all possible slip systems for every lattice system? thank you
@@user-kl4oh2co2y Nope.
dank
Bitte schön, geschehen :-)