8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
8 жыл бұрын
Пікірлер
@ritishasingh6867 22 күн бұрын
Crisp short and wonderful! Thank you 👌🏻
@abhijeetkale-qj9ww Ай бұрын
Great teaching! What does symbol 3+ or 3- or 3bar+ or 3+_1(different from rotational axis 3) and tA or tC symbol indicates? Reference:- img.chem.ucl.ac.uk/sgp/large/196az2.htm and img.chem.ucl.ac.uk/sgp/large/142az2.htm
@TIWERTER Ай бұрын
my apprentice said that the most common screw axis is 6,3 and I can hardly distinguish it from I6 symmetry axis 😢
@FrankHoffmann1000 28 күн бұрын
Hmmm, personally I think that 2(1) are far more common than 6(3); note that P2(1)/c is the most frequently space group.
@chemist4108 Ай бұрын
You save my life , love from Austria .can I get your email address please
@hikguru Ай бұрын
Hi Frank. At 3:44 you show how 2 shapes can form from the same cubic unit cell. The first is just a magnified version of the unit cell but the second shape (octahedron) looks different from it. Is it correct to say that both these shapes do not have the same symmetry elements? If so then it would mean that based just on the symmetry of the unit cell you cannot predict what the symmetry of the final crystal formed from it would look like because that depends on the other parameters such as the conditions under which the crystal is growing? Thanks.
@FrankHoffmann1000 Ай бұрын
Hi Rahul, interestingly, a cube and an octahedron have both the _same_ symmetry. But that also means that you cannot predict in an ad hoc manner, which of the many possible shapes of the same symmetry a certain crystalline compound wil develop. Of course, if you do have information about the chemical composition and the conditions under which the crystal growth then you can calculate the surface energy of the facets and the crystal will form a shape with the lowest total surface energy.
@hikguru Ай бұрын
@@FrankHoffmann1000 Understood. So does that mean that if a crystal exhibits a certain symmetry then we really cannot say for sure that the unit cell for this crystal will also have that symmetry? I am imagining an extreme case with a cubic unit cell and if the crystal growth proceeds only in 1 direction (say X) then you would end up with a rather long and skinny tetrahedron. This has 4 fold symmetry along axis parallel to X but only 2 fold symmetry along axis parallel to Y and Z. But the cubic unit cell has 4 fold symmetry in X, Y and Z, in other words the crystal and unit cell symmetries are different? Thanks.
@FrankHoffmann1000 Ай бұрын
@@hikguru I think, you mean rather a tetragonal prism instead of a tetrahedron, but that is not important for your question: Yes, in principle, this is possible, in particular, if certain compounds are present that can act as so-called capping agents; they can prevent that certain faces/facets of a crystal will develop leading to an overall shape that doesn't fit the symmetry of the unit cell.
@Beero.diaries 2 ай бұрын
Thankyou very much for sharing in so much detail and simpler way. It helps a lot in understanding the nets and topology.
@hikguru 2 ай бұрын
Hi Frank. In the first example of the car wheel with the 5 fold axis of symmetry is it true in general that if there is a 5 fold symmetry then there has to be 1 mirror plane as well? I am guessing no, because crystallographers don't like redundancies and they would then just say 5 and not 5m!
@FrankHoffmann1000 2 ай бұрын
Hi Rahul, you are guessing completely correct! 🙂
@abdulsami6779 2 ай бұрын
Thanks a lot Sir, It really so informative video. Stay blessed!
@johnnyq4260 3 ай бұрын
Apparently these definitions vary a bit. For example, on page 59 of Structure of Materials, 2ed., by Graef, trigonals (aka rhombohedrals) are such that a = b = c.
@FrankHoffmann1000 3 ай бұрын
The problem is that there is no rhombohedral crystal system. A lattice can be rhombohedral and a unit cell, too, but all crystals with a (primitive) rhombohedral unit cell belong to the _trigonal_ crystal system. And because there is a second kind of trigonal crystals for which the unit cell is _not_ rhombohedral one shouldn't use the definition by De Graef. For more details I would like to refer to unit 2.9: kzbin.info/www/bejne/gaSUhYGhfJunjrs
@abs5894 3 ай бұрын
1) Related to 0:36 . How to distinguish objects from crystal? Doesn't the crystal undergoes translation symmetry operation to create its object? Then, why there are no (limited) space groups or point groups for objects too? 2) Not related to this video: Does two-fold axis of rotation perpendicular to mirror plane generate glide plane?
@FrankHoffmann1000 3 ай бұрын
First of all, it should be clear that crystals are also objects, however, they are characterized by their strictly periodic arrangement of their constituents. This leads, for instance, to a characteristic _diffraction_ pattern when they are exposed to X-rays.
@abs5894 3 ай бұрын
Can the conventional unit cell possess higher symmetry than primitive cell of the same crystal?
@FrankHoffmann1000 3 ай бұрын
Well, it depends. It is possible to choose a primtive unit cell that is very unappropriate in terms of the underlying symmetry of the lattice. However, if you choose the most appropriate primitive unit cell, then it has the same symmetry as the conventional (centered) cell. But the symmetry will be more obvious, more apparent in the centered cell, because symmetry axes run parallel to the crystallographic axes and symmetry planes are oriented perpendicular to the crystallographic axes. See also Unit 2.4.
@AbyPaulBenny-fr1xv 3 ай бұрын
I can't believe how you made all basics of crystallography in a single channel. Nicely explained by images. What an effort
@UnrealogyTutorials 4 ай бұрын
Hello sir, I'm in high school currently so my understanding of solid state physics is quite rudimentary. I stumbled upon this because I was confused about HCP, and how it seemed to fit in with none of the bravais lattices. After watching your video I realised the lattice points could have a motif of more than one particle, so am I right in thinking that HCP is actually the hexagonal primitive lattice with a motif of 2 atoms - One being directly placed on the lattice point and 1/8th of 8 particles which are awkwardly placed inside the unit cell?
@FrankHoffmann1000 3 ай бұрын
Hi, almost: the lattice of the hexagonal-closest packing is indeed primitive and their are two atoms in the unit cell, however, both of them do _not_ coincide with one of the lattice points. Further information can be found here: crystalsymmetry.wordpress.com/2018/12/02/wurtzite-in-solid-state-chemistry-textbooks/ Best Frank
@BenjaminPullicino-qx8fh 4 ай бұрын
Heartfelt thanks from a desperate PhD student of Chemistry
@abhijeetkale-qj9ww 4 ай бұрын
Why is the name (glide) so?
@FrankHoffmann1000 4 ай бұрын
I don't who invented this name, but I personally fin dthis name not too bad, because we can imagine that the objects glides along the plane when we apply the respective symmetry operation.
@captainmichu 4 ай бұрын
Thank you, you are saving my life, at least if I will be able to see all your films by tomorrow
@user-sh8su1by5l 4 ай бұрын
Wonderful lecture Dr. Hoffmann
@GAUTAMSHARMA-xu7bl 4 ай бұрын
Beautiful explanation...........................I loved it....................................................
@user-mk5jg8kb1l 4 ай бұрын
Dear Prof. Hoffmann! Could you please give the example of real structure triclinic system with a = b = c α = β =γ= 90°?
@FrankHoffmann1000 4 ай бұрын
I don't know an example for the triclinic case. As I mentioned in the lecture, it is indeed not very likely that a crystal belonging to the triclinic crystal system shows that metric, but the point here is that it is mathematically not forbidden, it cannot be excluded. If you like, there is an interesting example for the monoclinic crystal system with an angle beta of 90°, see here: crystalsymmetry.wordpress.com/2015/03/02/the-monoclinic-crystal-system-and-the-skew-angle-beta/
@Ideal-physics 4 ай бұрын
Perfect explanation. Weldone
@nidanadia7267 4 ай бұрын
hi frank, I have one question. i am confused you told in this videos for atom 2, you said it might be 0. how can i know about this. i mean how can i find absolute coordinates. because its easy method to give half value from a,b,c but i am confused when you told about the atom 2
@FrankHoffmann1000 4 ай бұрын
Hi Nida, absolute coordinates (and therefore fractional coordinates) can only be specified if the crystal structure, the lengths and angles of the unit cell and all positions of the atoms are known - usually the result of a single-crystal structure analysis inferred from a X-ray diffraction experiment.
@nidanadia7267 4 ай бұрын
@@FrankHoffmann1000 thank you frank
@nidanadia7267 4 ай бұрын
@@FrankHoffmann1000 frank please tell me how to find atoms positions. i am stuck from a month in this mystery...my structure is cubic.a,b,c is 10.2 and alpha, beta, gamma is 90.space group is 225. i have 5 elements with CN bond. mean total 7 atoms. please make a video and tell the easiest way to find atoms positions.i will eagerly wait for your further response and video. try to find but failed again and again and no any related information in COD. so need to create CIF file but first need atoms position so need your precious guidance
@FrankHoffmann1000 4 ай бұрын
@@nidanadia7267 Sorry, but I think, I do not understand your question. How should I know, where your atoms (of what kind) are? By the way: 7 Atoms in the space group 225 is not possible, the Wyckoff positions a and b have multiplicities of 4, Wyckoff position c has already a multiplicity of 8.
@nidanadia7267 4 ай бұрын
@@FrankHoffmann1000 ok frank I have last question. can you tell me please what test/characterization/technique should i do for determination of atoms positions for crystal structure?
@dhoonygo 4 ай бұрын
In 2:44, why does the glide plane a lie in the ab plane and not in the ac plane?
@FrankHoffmann1000 4 ай бұрын
For no special reason.
@dhoonygo 4 ай бұрын
Thank you for your explanation! I have a couple of questions in 1:26. 1. Say we know a unit cell of a crystal. If we don't know the external shape of the crystal, we cannot determine the point group of the crystal? 2. Experimentally, how is the external shape of a crystal observed? Should we grow the crystal big enough to observe by a microscope anyway?
@FrankHoffmann1000 4 ай бұрын
1. Correct, a unit cell is not sufficient in order to determine a crystal class. 2. Yes. But in practice, many crystals build shapes that cannot be determined unambigeously, some also develop shapes that do not perfectly reflect their underlying symmetry (defects, twins etc.). But once the structure has been solved (regardless of the external shape of the crystal) we can derive the crystal class from its space group.
@dhoonygo 4 ай бұрын
Great video! In 6:30, why don't you consider 2 fold rotation in a and [210] direction? Is it not unique?
@FrankHoffmann1000 4 ай бұрын
Yes, because these 2-fold axes of rotation are generated by the mirror planes (each two of them which are oriented perpendicular generate one of these).
@dhoonygo 4 ай бұрын
@@FrankHoffmann1000 Let me check whether I understood. I thought there was a 2-fold axis of rotation perpendicular in a and [210] direction, respectively. So, I thought the point group was 6/m 2/m 2/m. But you didn't consider these perpendicular 2-fold axes of rotation. I can't get why 2-fold axes of rotation are generated by the mirror planes.
@FrankHoffmann1000 4 ай бұрын
@@dhoonygo You are completely correct that the _full_ name or symbol of the crystal class is 6/m 2/m 2/m, meaning that there are additional 2-fold axes of rotation _parallel_ (not perpendicular) to the a- and [210] direction; so the short symbol is simply 6/mmm. And to be precise, there is, of course, a 2-fold axis of rotation along the c-direction, because a 6-fold axis of rotation automatically contains a 2-fold axis of rotation. Concerning the 'uniqueness': There are many symmetry laws that refer to the generation of further symmetry elments on the condition of symmetry elements that are already present. The most famous one is: Two mirror planes perpendicular to each other automatically generate a 2-fold axis of rotation, namely one that runs along the intersection of the two mirror planes.
@dhoonygo 4 ай бұрын
@@FrankHoffmann1000 Thank you, I now understand more clearly. But why did you refer to these additional 2-fold axes of rotation to be _parallel_ to the a- and [210] direction? I thought a plane surface perpendicular to a direction vector was considered *_perpendicular_* , although the normal vector of the plane must be *_parallel_* to the direction vector.
@FrankHoffmann1000 4 ай бұрын
@@dhoonygo That is exactly the point: The orientation of a plane is given by the direction of the normal vector, which is geometrically a line! An axis is also a line! This is the reason why crystallographic axes are always specified as being parallel to a certain dirrection (also a line!) and why crystallographic planes are always given as being perpendicular to a certain direction, because then the normal vector is again parallel to that direction. Look at the first mirror plane in this example: We look along the crystallographic c-direction and the plane of the mirror plane is of course perpendicular to that direction (this is why it is written as '6 _over_ m') as the normal vector of this plane is _parallel_ to the c-direction.
@cheekocho 5 ай бұрын
Your video saved me a great amount of time, was really having trouble understanding this part. Thanks from South Korea!
@spin3488 5 ай бұрын
Hi Frank, Sorry to bother you again. I have a question regarding the multiplicity on special positions (for instance at 4:16 min). The multiplicity decreases when an atom falls precisely on a symmetry element, but only if the symmetry element does not have translation component (inversion center, mirror plane and fold axis). Is this correct? Or it applies to any symmetry element (including screw axis and glide planes). Thanks again and thank you again for all the videos.
@FrankHoffmann1000 5 ай бұрын
Always welcome! You are completely correct (only if the symmetry element does not have translation component)!
@Axel77430 5 ай бұрын
As a Mexican, it feels really nice to know that Naica is famous. Greetings and thanks for your videos, Professor(s).
@ron_nist 5 ай бұрын
Thank you so much. I finally understand it
@user-ph3ih6fe2r 6 ай бұрын
Grüezi Herr Prof. Hoffmann Ich und meine Freunde besuchen momentan eine Vorlesung in diesem Bereich und haben uns folgende Frage gestellt: Die unit cell muss definitionsgemäss die Symmetrieelemente des Kristalls besitzen, doch im Falle von einem Hexagon ist die Unit cell trotzdem ein Rhomboid, der ja offensichtlich die C6 Achse nicht in sich trägt. Wie ist dies mit der Definition vereinbar?
@user-ph3ih6fe2r 5 ай бұрын
Unit 2.8 erklärt dieses Problem, danke!
@loopgarcia3862 6 ай бұрын
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 6 ай бұрын
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.
@bigproton 6 ай бұрын
Dear Frank, thanks a lot for the videos, they are great! I was wondering if the distances between planes 100 correspond to a, and similarly d010 to b and d001 to c (where the lengths of a, b, c are the unit cell dimensions)
@FrankHoffmann1000 6 ай бұрын
That's correct!
@nano605 6 ай бұрын
Thank you for your wonderful lecture. I've really learned a lot of fundamental concepts from your video. I have tiny questions that are unclear to me. When we mention a unit cell, do we use the name of its crystal system? For example, as in your example, even if the unit cell parameters look like tetragonal (a = b = 12 Å, c= 12.1Å, α = β = γ = 90°), if its crystal system is triclinic, should we call it triclinic unit cell"? Or the unit cell is "tetragonal", but the crystal system is "triclinic"? / Another thing in my mind is that I can imagine a unit cell having a = b = c, α = β = 90°, and γ ≠ 90°. But I cannot find this metric, so I don't know how to call this unit cell. But now I understand that its crystal system is based on restrictions that you showed in the video. I would appreciate it if you would explain these points.
@FrankHoffmann1000 6 ай бұрын
Hi! To your first question/example: The classification according to crystal systems is based on the symmetry of the crystal. This means that also the smallest repeat unit of the crystal, the unit cell, has that symmetry. So a triclinic unit cell means it has triclinic symmetry. The lattice parameter are only the metric of the unit cell. To your second question: Yes, it is possible that a unic cell has the metric you mentioned, a = b = c, α = β = 90°, and γ ≠ 90°. You don't find this restriction, because none of the possible 7 kind of the symmetries leads to such a restriction, but there is at least one restriction that covers your metric (triclinic). If you would redefine you axes in such a way that instead α = β = 90°, and γ ≠ 90° --> α = γ = 90°, and β ≠ 90° then also the monoclinic system is possible. Best Frank
@howardbai9918 6 ай бұрын
It's the end of the lecture!! Thank you so much Frank, this is the most helpful course I found in youtube.
@FrankHoffmann1000 6 ай бұрын
Thank you very much for your kind comment!
@spin3488 6 ай бұрын
Hi Frank Thanks a lot for your videos! They are great and well explained. I have two questions regarding the image in the video at 3:28. 1) I guess that unit cell has 4-fold rotational symmetry and could be considered to be each of the rectangles shown on the video (I.e the image shows 4 unit cells). Is that correct 2) Where are the 4 fold axes located? Are they in the center of each rectangle and also on each of the 4 vertices? Thank you very much for all your time and work, Frank!!
@FrankHoffmann1000 6 ай бұрын
Hi Spin3488! Correct, there are 4 rectangular unit cells shown. But interestingly, the unit cell has no rotational symmetry at all (apart from the 1-fold axis of rotation). The only symmetry element present is a (vertical) reflection line that runs through the middle of the a-axis. The plane group is pm. Best! Frank
@spin3488 6 ай бұрын
Thank you very much, Frank! You have to take into account the symmetry of the pentagon as well to check the symmetry of the unit cell, that is why there is no rotational symmetry in the unit cell. Is this correct?@@FrankHoffmann1000
@FrankHoffmann1000 6 ай бұрын
@@spin3488 correct!
@user-lp2xc8dj3i 6 ай бұрын
8.05 Are the planes really triangular?
@FrankHoffmann1000 6 ай бұрын
This is only for illustration purpose, to make clear, where the points of interception are - the planes are of course infinitely extended.
@MatterMadeMoot 7 ай бұрын
I think you're the only person on the internet who's explained these concepts at a graduate level so cleanly and concisely. Can't thank you enough.
@FrankHoffmann1000 7 ай бұрын
Thank you very much for your kind comment! I'm glad that you like the videos.
@alanthayer8797 7 ай бұрын
Africa Channel with a NON KINKY HAIR reporter! She Isa Foreigner with NO Melanin + Silky hair = mutated genetics !
@azice6034 7 ай бұрын
Great series. Wishing I knew about it this summer. It’s absolutely spectacular. I have a couple of questions, first of, how do you know that P2(1)/c is actually P12(1)/c1? That is, how do you know that there are 1’s in the a and c viewing directions and that the 2(1)/c is in the c direction specifically? Couldn’t you have something like P2(1)/c11 or P112(1)/c where it is in the a or c viewing directions?
@FrankHoffmann1000 7 ай бұрын
It is abolutely possible to have something like P2(1)/c11 or P112(1)/c. It depends only on your choice regarding the choice of the axis in which the symmetry appears. Note that in the monoclinic crystal system symmetry is present in only _one_ direction. Usually you choose the axes in such the way that this symmetry direction is the b-direction, menaing: it is purely a convention. In Eastern Europe is is not uncommon to choose the c-direction. The a direction is usually not chosen.
@3mro49 7 ай бұрын
Hi Frank. I was wondering if you have work sheets related to the topic to examine ourselves! Thanks for your explanation.
@FrankHoffmann1000 7 ай бұрын
No, sorry, I do not have work sheets. Best wishes! Frank
@shiwangi1222 7 ай бұрын
I'm Indian and I like your teaching skill with examples
@shiwangi1222 7 ай бұрын
Nice class sir
@bidyadharmahato4527 8 ай бұрын
Very nice, thank you
@sasukeuchiha998 9 ай бұрын
A lot easier than looking at the guide.
@amishamishra7641 9 ай бұрын
Sir can you please make a video on how the point group change with addition of a centre of symmetry and axis of symmetry and plane of symmetry.... please What will be the new point group when a centre of symmetry is added to m and 4mm?
@FrankHoffmann1000 9 ай бұрын
I think there is no need for another video: If you add a center of inversion to 4mm you will get 4/mmm, already included in this video (at around 11:00). And if you will watch Unit 3.5 (kzbin.info/www/bejne/eovbmpuwdrmrhpIsi=_UFC647lbxTKoXyV) you will see that i + m will lead to 2/m. Note that a center of inversion is only explicitely mentioned in the space group symbol, if it is the only symmetry element present.
@younesmecabih7959 9 ай бұрын
how do we know if the wyckoff site are occupied?
@FrankHoffmann1000 9 ай бұрын
Only by inspecting the concrete crystal structure of a given sample, usually solved by single-crystal X-ray diffraction experiments.
@KSubruu 9 ай бұрын
I came looking for a good video on space group symbols and stumbled upon this! I am glad that instead of just watching this and leaving, I went through all the videos from Chapter 1. You have done a truly amazing job making this lecture series with immaculate clarity, free materials and zero ads. It was remarkable to see you answering the questions in the comment section even after 7 years of posting the video. The doubts I had were posted in those comments and your responses helped me clear it. You are an amazing teacher and thank you for this series!
@FrankHoffmann1000 9 ай бұрын
Thank you very much for this big compliment and for taking the time to write this detailed, kind comment! I'm very pleased that you enjoyed the videos so much! I'm also glad that the comment section is helpful for some people! Best regards Frank
@mehmettasyagan2195 9 ай бұрын
I love these lectures, always recommend interesting resources, learned a lot from them. Also congrats on the new book, looks fantastic.
@FrankHoffmann1000 9 ай бұрын
Thank you very much for your kind comment! I'm glad that you find these videos helpful! Best wishes! Frank
@bladdnun3016 9 ай бұрын
Very interesting lecture, thank you very much! I'm currently working on software that is primarily interested in the local, geometrical side of things (distortions of polyhedra and molecules), but uses a topological approach to automatically identify those clusters. It's great to see how other people have tackled similar problems explained in such a clear way, without all the jargon that can make reading research papers so difficult sometimes if one is not trained in the right field.
@FrankHoffmann1000 9 ай бұрын
Thank you very much for your kind and very interesting comment! Glad, you like the video! Do you have a resource pointing to your approach/work? best wishes!
@bladdnun3016 9 ай бұрын
@@FrankHoffmann1000 I always appreciate feedback or advice :)
@shimjaejin810 10 ай бұрын
A really good voice and an even better lesson, thank you very much!
@davidli6931 10 ай бұрын
Can I say the e glide plane is basically a diagonal translation along a plane, and then followed by a 180 degree rotation axis parallel to the plane?
@FrankHoffmann1000 10 ай бұрын
No, these two operations you mentioned are not equivalent to the operations involving a glide plane e.
@davidli6931 10 ай бұрын
@@FrankHoffmann1000 Yeah rewatching the animation makes me understand why. Thanks for the clarification. 🙏🏻