Unit 4.5 - Space Groups and Space Group Symbols

[@KSubruu]

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]

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

[@jeoh93]

This video helped me pass my qualifying exams for graduate school. Thank you, Dr. Hoffmann!

[@FrankHoffmann1000]

Welcome! Congratulations for passing your exams!

[@joshuajoshua6814]

very good explanation. you voice makes it really nice to listen and follow what you are saying. i wish you would read something to me when i go to sleep

[@mafeflorez28]

I don't think I can thank you enough, you're great! keep the good work

[@FrankHoffmann1000]

Thank you very much for your kind comment. Glad that you like the videos.

[@shaneyaw4542]

This was a very clear explanation of a complex topic. Thanks

[@BenjaminPullicino-qx8fh]

Heartfelt thanks from a desperate PhD student of Chemistry

[@KPLIUC]

Thank you so much, you just saved my final exam!!!

[@DeshmukhDevki]

Thanks a lot for the videos. You made crystallography comprehensible.

[@doorash]

You are awesome. I’m watching it in Korea.

[@워루-h8x]

저 몇개만 물어봐도되나요 ㅠㅠ

[@pravinkadhane8851]

Very helpful information about Space group and specially for tetragonal perovskite structure.

[@FrankHoffmann1000]

Thanks! We are glad that you find this unit helpful!

[@olliebudd8098]

Better than my lecturer

[@senemdemirci6143]

Thank you!!! You are the reason I passed my exam for my Masters degree!!!

[@emteemte3985]

i will learn crystalography in this winter semester 2021/22 for my masters in chemistry and i must pass the exam but this topic is too coplex for me could you tell me please how can i master this crystalography ?

[@C4pungMaster]

Thank You for the explanation about the viewing axes for different crystal systems :^)

[@FrankHoffmann1000]

Very welcome!

[@v.vigneshvenkatasubramania3456]

Thanks for the video. I have a question. To obtain space groups, we add the following elements: 1) Bravais lattice - Translation symmetry only 2) Point groups - which are applicable to macroscopic shapes of crystals 3) Glide plane and screw axis Why don't we talk about the symmetries of a unit cell in particular? It seems point groups must be applicable for unit cells as well. If yes, then how do you combine glide plane and screw axis with the unit cell? Also, regarding the viewing directions mentioned in this video: It seems to me to be very difficult to obtain the planes' miller indices of a macroscopic crystal shape. For example, let's say a macroscopic crystal has a hexagonal crystal system, how do you determine that it has a hexagonal shape when you cannot identify its [210] viewing direction.

[@FrankHoffmann1000]

Welcome! 1) We do speak about the symmetry of unit cells! The unit cell contains all symmetry elements and the diagrams of the symmetry elements depict all these for each of the space groups in the International Tables. But perhaps there is a misunderstanding concerning the term combine? Glide planes and screw axes are symmetry elements with translation components that are smaller then a whole unit cell. And because they act on the motif _inside_ the cell they always have to _compatible_ with the translation lattice! And I think, it is then clear that there can't be an infinite numbers of combinations (Bravais + glide/screw). If unit cells contain screw axes or glide planes - how should point groups be applicable to unit cells? The term point group implies that there are no translation components, no symmetry elements with a translation component. And in this sense point groups are not applicable to unit cells. Of course, you can disregard from all what is inside a unit cell and can consider only the outer shape of one unit cell. Then there are some relationships between point groups and the symmetry of the _outer- shape of _one_ unit cell, but not always. Take the primitive hexagonal unit cell. If you consider only one, you will never find a 6-fold axis of rotation. 2) I am not sure, if I understand the second part of your second question. Concerning the first part: That is true! It can be difficult. Here I would refer to this excellent web site: www.tulane.edu/~sanelson/eens211/axial_ratios_paramaters_miller_indices.htm To the second part: The hexagonal shape (or 6-fold rotational symmetry) can be immediately recognized, if you look at the outer crystal shape, completely independent of any viewing directions. regards Frank

[@aryan5015]

@5:00 sir, you said 2 mirrors automatically generate 2 fold axis. and there are much more like this. can you please tell what are others. what are suh possible cases. please list all. thank you

[@annavladislava4291]

This helped me a lot, thank you for this amazing work!

[@Herngg]

Hi Frank, just to clarify, shouldn't the blue mirror plane be perpendicular to the a-axis since we are viewing it in the a direction? 10:33

[@FrankHoffmann1000]

Hi T JH, yes, that would be absolutely correct and the video is wrong at 10:33 - as I noticed this at some point I interchanges the a and b axis at 10:37 (pssst, it's a secret and a test for the viewers of the video :-))

[@Herngg]

Frank Hoffmann thanks for the clarification Frank! I realised it after posting my questions too haha

[@creatorlesedi]

Well explained! Thank you so much

[@jackcollings7257]

Dear Frank, Thank you so much for these videos - they are amazing! I have a question: Previously in the video on glide planes, glide plane "a" was described as a reflection in the "ab" plane, and translation in the a-direction by x+1/2. However in the example at 4:14, it says glide plane "a" is perpendicular to "b". However the "ab" plane is not perpendicular to "b". So is glide plane "a" free to reflect in any plane, as long as that plane is the one perpendicular to the axis given by the index of the HM notation? Can glide plane "a" be specified in the first index, such that there is a reflection in "bc", then translation by x+1/2?

[@FrankHoffmann1000]

Dear Jack, thanks for your comment, I'm glad that you like the videos. To your questions: (1) So is glide plane "a" free to reflect in any plane, as long as that plane is the one perpendicular to the axis given by the index of the HM notation? Yes, with the small constraint that it can never be perpendicular to the a-direction, which leads to the answer of your second question: (2) Can glide plane "a" be specified in the first index, such that there is a reflection in "bc", then translation by x+1/2? No, see above. The direction or orientation of the (glide) plane and the translation direction of the glide part cannot coincide. This is not possible, because it would mean that the movement caused by the reflection part is identical to the direction of the translation part; this leads to incompatibilities.

[@jackcollings7257]

@@FrankHoffmann1000 That makes perfect sense! Thanks you!

[@johndeng403]

I am confused about the last example, why don't you combined the two diffrent motif as a new motif. Then the crystal system will be P4/m m m.

[@FrankHoffmann1000]

Hi John, first of all, with motif we usually refer to a chemical unit, for instance, a molecule. Assume we have no square pyramids but pyridine molecules (a benzene derivative in which one carbon atom is substituted by a nitrogen atom), and assume further they would assemble in an analogue manner, 8 pyridine molecules with the N atom pointing upwards at the corners and one pyridine molecule in the center with the nitrogen pointing downwards - what do you mean then by combining the motif? Second, even if we virtually combine the motif (but the places at which each constituent of the motif is located should be the same) - is there then an additional mirror plane perpendicular to the 4-fold axis of rotation?

[@arijitsinha2263]

Sir i have a question im a little confused does symmetry of space group changes if i change my description from unit cell to primitive cell for a specific material? Suppose i have a material which has Cmcm symmetry in its unit cell does this means its primitive cell will also have same space group ??? Btw thanks a lot for this set of lectures they r really helpful for a person like me

[@FrankHoffmann1000]

No, the space group will be not the same. Why? Remember, the first letter of the space group specifies the Bravais lattice type, C stands for a C face centering, and if you decide to describe the symmetry of your material according to a primitive lattice then at least the Bravais type changes from C to P. Note that usually also the other generating symmetry elements are affected if you change the Bravais type.

[@arijitsinha2263]

@@FrankHoffmann1000 first of all thank you very much for your reply i was waiting for your answer. 1. In one of your previous lecture unit 2.4 you said that symmetry doesn't depend on unit cell choice as it is not unique, there could be some advantages for some specific choice of cell but symmetry doesn't change with different choice of cells. Why sir that thing is not applicable to space groups? They also tell us the symmetries right? Or sir at that video you mean symmetry by only translational symmetry?? 2. In some paper i saw people referring a phase of a material (viz Ta2NiS5) as orthorhombic Cmcm, and they are not explicitly mentioning the type of unit cell (primitive or conventional unit) they are referring for that space group so i am confused.I thought symmetries are absolute.is it wrong sir? Can you please help me sir from where does that misunderstanding is coming from i have thoroughly gone through your lectures till space group.

[@FrankHoffmann1000]

1. That's completely correct: you can't change the underlying symmetry of the crystal structure regardless of the choice of a more or less appropriate unit cell. As you mentioned unit 2.4: This is exactly the case for the example in which all primitive unit cells are oblique. There is one primitive unit cell that also shows the two mirror planes, but it is favourable to choose a centered one, because then mirror planes are perpendicular to cell axes; but, of course, you can also choose a unit cell which is even less favourable (the upper left one), in which no mirror plane is apparent. That doesn't mean that you change the symmetry of the lattice itself. It is there and has intrinsic symmetry properties. However, it is right that the symmetry of the unit cell can vary depending on which unit cell (small(est) part of the lattice) of a (infinite) lattice you choose. The goal is always, so to say, to choose a unit cell that has the full symmetry of the lattice. 2. The given space group Cmce is a centered one; how could it be then a primitive one? For what reason should it be important to state explicitly whether the cell is primitive or conventional? For te relation between (absolute) symmetry of the lattice and the apparent symmetry of the unit cell, see point 1.

[@arijitsinha2263]

@@FrankHoffmann1000 Thank you very very much professor for attending my question i was getting confused between the "(absolute) symmetry of the lattice and symmetry of the unit cell".and some online crystal databases like OQMD,aflow.org shows the conventional cell space group and primitive cell space group same! Which is insane .... Anyway sir thanks again. now things are clear to me.

[@fabeehalaiq]

Dr. Hoffman , Hi sir, i really like your explanation of crystallography its help me a lot in understanding my research work in materials. i have a bit of confusion. In case of primitive tetragonal Bravice lattice we have its space group P4mmm. But when we put a different motif at center of Bravice lattice space group changed but you didnt changed P to I, i.e. in PbTiO3, Pb are all at corner and Ti is there at Center of tetragonal unit cell but the space group is P4mm not I4mm what specifically define theire Bravice lattice Pb atoms or Ti atoms or O atoms. mean how to define lattice motif for this system. thanks

[@FrankHoffmann1000]

Thanks, Laiq! Well, the perovskite structure is a good example to illustrate the difference between lattice points/motifs and atoms: the lattice points are representing (usually) one molecule or a formula unit, here all(!) atoms of the formula ABX3 (sometimes, particular in structures with low symmetry, also multiple formula units). In a body centered cell the lattice points are at the corners and the centre of the cell. This is obviously not the case for the ABX3 structure. The A atoms are at the corners, but the B atom is at the center of the cell (and the X atoms are at the face centers). best! Frank

[@fabeehalaiq]

@@FrankHoffmann1000 Sir, I understand your point your are saying that in ABO3 pervoskite complete formula unit can be considered as lattice for P4mm. So PbTiO3 as a whole can be regarded as a motif for tetragonl base center symmetry.

[@renatosilva9795]

Thank you again for the videos. I have one question. i did not understand very much the viewing directions for the triclinic system.

[@FrankHoffmann1000]

In the triclinic system we do not have viewing directions, because the maximum symmetry for that system is a center of inversions - and a point does not have an extension, so there is nothing like a direction.

[@renatosilva9795]

@@FrankHoffmann1000 Thank you.

[@lorenzoscarpelli7807]

Hi Frank, thank you for this wonderful playlist. can you explain the axes label change from slide twelve to thirteen please; in addition i have a doubt about mirror plane along -a axes, following the axes label reported into slide twelve .

[@FrankHoffmann1000]

Hi Lorenzo, sorry for the confusion, the coordinate system is wrong in slide 10 to 12, but I forget to correct this mistake. From slide 13 onward it is correct. So the change was unintentionally. best Frank

[@spin3488]

Hello Frank I have a question that is not directly related to the video, although this series spurred my curiosity, to be fair. I tried to do some research but I was not able to conclude anything, so maybe it is not obvious at all. Suppose we have a racemic mixture that separates into a conglomerate (i.e. xtals with only enantiomer R and xstals with only enantiomer S) These xstals should crystalize in in one of the 65 Sohncke groups. I think that if they xtalise in, let’s say P212121, you would have R xtals and S xtals both of them in this space group. However, if one enantiomer crystalizes in one of the 11 enantiomorphic pairs, for instance P41, would the other enantiomer also xtalize in P41, or it should do it in the enantiomorphic pair p43. Sorry for the question, but I would appreciate if you would have any clue. Cheers

[@FrankHoffmann1000]

Hello spin34, good question. I have no definite clue, but I would also guess that the other enantiomer should crystallize in the space group P4(3). Maybe you can find some examples of enantiomeric pairs and their space groups? I already tried to look for some amino acids and sugars, but found only P212121 (for both enantiomers)... best Frank

[@spin3488]

@@FrankHoffmann1000 Thanks a lot, Frank. Yes, I also tried to do the search but I got the same results. I will keep trying and see if I find one involving enantiomorphic pairs. Thanks again for your time and help!

[@shivamkansara1968]

Dear Prof. Please help me to understand in Pm-3m space group for CsPbI3. Here, P stands for Primitive so why associated crystal structures (CsPbI3) has an atom in center ? and same as for Pnma space group.

[@FrankHoffmann1000]

Well, this is a question that is often asked. The background is that the difference between a lattice point and an atom is often unclear. Only in very few cases are they identical (this is the case with many metals). A lattice point _represents_ an entire chemical motif, e.g. a formula unit. In the case of centring, we now look at whether the motif is centred, not whether any of the atoms of the overall motif happens to be in the centre of the cell. The entire formula unit would have to be located in the centre as well as at the corners of the unit cell for it to be a centered structure. In other words, the atoms at the corners and in the centre of the unit cell in CsPbI3 are different! They would have to be the same if there were centered. The most simple example in that regard is the structure of CsCl. This is also not a body-centered cubic structure, because the atom at the center is different from the atoms at the corners of the unit cell.

[@shivamkansara1968]

@@FrankHoffmann1000 Ohh now I understood the difference. Thank you so much for quick reply. I was stuck here. After drawing, I am able to understand the 3 fold rotation. So Pm-3m is Primitive Bravais lattice, mirror, three fold rotation followed by inversion, mirror. Is this right description ? If no, Could you please help me to make a sentence for this?

[@FrankHoffmann1000]

@@shivamkansara1968 This is correct :-)

[@shivamkansara1968]

@@FrankHoffmann1000 Thank you🙂

[@MyKrevet]

Hi! Thanks for the videos, they are very helpful! And one question, is there a center of inversion when viewed from the c direction?

[@FrankHoffmann1000]

Hi! Thanks for your comment! Concerning your question: No. First of all, a center of inversion (i.e. a point) has no direction; but independent from that: A center of inversion would lead to pyramids with their tips pointing downwards. best Frank

[@MyKrevet]

@@FrankHoffmann1000 That's a fast reply! Thank you very much :)

[@mvilatusell]

Hi Frank! Awesome video and series of videos! I'm extremely grateful. I have a couple of questions: 1) When you reduce from space group to crystal classes, is the viewing direction maintained? 2) In the example at 10:25, why is not just P4mm with a hat above the 4 to indicate inversion? In other words, why do we need to specify 4-fold rotation + perpendicular mirror insteada of 4-fold rotation + inversion? Thanks!!!

[@FrankHoffmann1000]

Hi mvilatusell, first of all, the given space group symbol is not a complete list of all symmetry elements that are present, but only the so-called generating symmetry elments. This means that all other symmetry elements that might be present as well can be derived from this generating set. Indeed, the space group P4/mmm conatins also a 4-bar (4-fold rotoinversion axis), however it is not constitutional, so to speak. Secondly, yes, the viewing directions are maintained when specifying the crystal class. Third, I am not sure, if I understand your question correctly, why the space group P4/mmm is not specified as P-4mm. The short answer would be two-fold: 1) There is simply no such space group, 2) these are not equivalent specifications: (i) a -4 is not identical with 4/m, (ii) where is the last mirror plane (perpendicular to [110]?

[@mvilatusell]

@@FrankHoffmann1000 Hi! Sorry if that was not clear. Let me rephrase in two parts. We have to specify the symmetry operations along the three axis: c, a, [110] and the Bravais lattice. So to me, the logical notation would be P4mm, one symmetry element per axis. But instead, the correct space group is P4/mmm. Why are there 2 symmetry elements for the c axis? I guess it's because these ones are the minimum to generate all other symmetries of the group. But, how do I know which ones are the generators? For example, this system also has inversion symmetry, so I could write P\overline{4}mm. Does mirrors take priority over inversion? I hope now is clearer. Thank you a lot!!

[@FrankHoffmann1000]

@@mvilatusell P4mm (No. 99) is a completely different space group compared to P4/mmm (No. 123). For instance P4mm is a so-called polar space group, meaning that there is no mirror plane perpendicular to the 4-fold axis of rotation. P4mm is therefore a non-centrosymmetric space group. It doesn't have a center of inversion, while P4/mmm has such center of inversion. So, this is one reason, why in this case two symmetry elements are specified for one viewing direction. Or to put in another way: A -4 doesn't necessarily imply that there is a mirror plane perpendicular to that 4-fold rotoinversion axis, see for instance the space groups P-42m or P-42c. And in this respect, yes, we can say that "mirrors take priority over inversion". You always have to ask: Which symmetry elements imply the presence of (all) others. So, if you ask "But, how do I know which ones are the generators?", the answer is: You have tro trust the International Tables for Crystallography. Even for professionals it is far from being simple to derive all interdependencies of the symmetry elements.

[@mvilatusell]

@@FrankHoffmann1000 Thank you very much. Now everything is clear. I was just trying to find a logic to write down the space group of any given system that is presented to you. I mean, there is a logic on all the concepts you have explained in the video, but as you just said, you also need to know which symmetry elements imply the presence of all others, and I guess that is not trivial. I guess with some experience you get used to the particular examples and then it becomes more straightforward. A final question, thought. I want to add magnetism and spins on top of all this, i.e. understanding the magnetic space groups. Could you suggest me a reference for that? Thank you!

[@FrankHoffmann1000]

@@mvilatusell No, sorry, nothing beyond the references in the Wikipedia article on magnetic space groups. (I am really bad at magnetism!).

[@PanchoKnivesForever]

Question: I must have completely missed this in the previous videos, but is there any chance you could possibly define for me or refer me to the video with the definitions of face-diagonal and space-diagonal, please? They were said to be perpendicular to the indices [110] and [111], respectively.

[@FrankHoffmann1000]

Dear Francisco, both the face and space diagonal are explained for instance here: en.wikipedia.org/wiki/Space_diagonal I have to admit that 'indices' is not the best term I could used in this context. These numbers in brackets are not to be mixed up with Miller indices, but are specifying lattice directions, see for instance the last paragraph of the following web page: www.doitpoms.ac.uk/tlplib/crystallography3/parameters.php best! Frank

[@ManjuVV]

Dear sir, i have a question. I have recorded XRD for cotton fibers, from the literature i had taken lattice parameters and the space group was P21/n, the expert raised a question, why there is non standard monoclinic is chosen? Could you kindly help me out sir..

[@FrankHoffmann1000]

There are two main reasons: 1. Sometimes a non-standard setting is chosen, if the skew angle beta is then more closely to 90° 2. If you deal with several space groups that occur, for instrance, at phase transitions and you do not want to change the axes (system of coordinates). I do not know the reason for the specific case you asked for; for this you have to ask the author of the study.

[@bohanxu6125]

About the P4mm at 12:14 It seems like there is another interesting symmetry, which is {reflection c -> -c} followed by a transtionation {0.5, 0.5, 0.5}. is this the diagonal glide plane "/n" ? should it be P4/nmm? I understand /n can only do translation along the a-b-plane. But we can always arbitrarily chose the position (the c component..or the "altitude" of) of the reflection plane. Let's chose the plane to be at z=c*1/4. Then there should be a "/n"... no?

[@FrankHoffmann1000]

That's a very interesting and sharp observation! In fact, it is only almost true: We can not choose an arbitrary value for the 'altitude' for a glide plane, they have fixed positions, depending on the concrete space group, however, for P4/nmm the height is indeed at c = 1/4. A glide plane n perpendicular to the 4-fold axis of rotation has the translation components (1/2a, 1/2b). Together with the mirroring along c at height c = 1/4 the pyramids at the corners are converted indeed to the ones in the center and vice versa! So, why it is only almost true: this would only be the case if the pyramid in the center were inverted at the exact center of the unit cell (0.5, 0.5, 0.5), but it is inverted at a point a little below that point. But, once again, congratulation for your perspicacity!

[@bohanxu6125]

​@@FrankHoffmann1000 haha "but it is inverted at a point a little below that point. " I feel like I just got trolled lol~ "We can not choose an arbitrary value for the 'altitude' for a glide plane, they have fixed positions, depending on the concrete space group". Ah, I didn't know that. Although this raises another practical question. suppose we have a structure where the inverted pyramid is not in the middle, but at the bottom (0c, 1/2 a, 1/2 b). Not knowing what you told me, I would just think it's P4/nmm. Knowing what you told me, I can always google P4/nmm and double check and see that my guess is wrong, which is good. But... Q1 how do I know what the space group it should be called? (i never see space group names explicitly specifying "center" of reflection, rotation, and such) Q2 I googled "P4/nmm". It's not what I'm expecting. It doesn't seem to have a simple 4-fold rotation...ie, I don't see {y, -x ,z}...although there is a {y, -x ,-z} which is a 4-fold rotoreflection? I'm confused. Thanks for the previous answer by the way :]

[@FrankHoffmann1000]

@@bohanxu6125 "I feel like I just got trolled lol~" Why that? There is no reason for this. Q1 --> A1: That's the difficulty with the International Tables for Crystallography; they give only the location of symmetry elements and the coordination transformation. You have to figure out what symmetry operation leads to which coordination transformation by yourself (as I and every crystallographer have to do!). Q2 --> A2: P4/nmm has a simple 4-fold axis of rotation, but the location is dependent on the choice of origin; check this page out... img.chem.ucl.ac.uk/sgp/large/129a.htm ...and choose option (d) if you want to see (y, -x, z). There are indeed also 4-fold rotoinversions present.

[@bohanxu6125]

@@FrankHoffmann1000 wow this is harder than I expected~ but I got it now (more or less). Thank you so much for you patient explanations!

[@nezizquierdo9968]

Awesome videos!

[@FrankHoffmann1000]

Thanks very much!

[@rameezmalik890]

Greetings Professor, You have explained P21/n symmetry from space group to 2/m point group. I am finding it hard to practically imagine to identify in a unit cell. For example, in the last example, where does it show 4/m screw axis in p4/mmm? Thanks

[@FrankHoffmann1000]

Greetings! First of all, the identification of symmetry elements needs, of course, some practice. But regarding the mentioned example, the 4 fold axis of rotation should be recognizable (marked with the red line in the c-direction with a rhombus at the top) as well as the mirror plane perpendicular to it (watch from 9:56 again). Don't be confused by the slightly perspective drawing, the base should be a square (a = b!). Note, that the point group for this example is not 4/m but 4/mmm. And note, too, that 4/m is not a screw axis! regards Frank

[@segovemoc4776]

In slide 6 you are saying that: "2 perpendicular mirror planes automatically generate a two fold axis of rotation and there are many many more of such relations ". Could you please specify some source where all of these relations are given? E.g. I know that two intersecting rotational axis generate third one, but I guess there must me many more rules as well.

[@FrankHoffmann1000]

Dear Sego, yes, there are indeed many more such rules, however, unfortunately, I am not aware of such reference. The most important symmetry relationship apart from the already mentioned rule is: A mirror perpendicular to a two-fold axis of rotation generates a center of inversion (at their point of intersection). Note, that this rule implies also the following relationships: (a) An inversion center at a mirror generates a two-fold axis of rotation (being perpendicular to the mirror) and (b) A cneter of inversion at a two-fold axis of rotation generates a mirror plane (perpendicular to the axis of rotation). best Frank

[@jackcollings7257]

Dear Frank, I have one more question which relates to the last example. Would P 4_{2}/m m m not work? If a 4_{2} screw is undertaken, and then a reflection is undertaken in [110] plane halfway up the unit-cell, does that combination of operations not leave the unit-cell unchanged? Or does the notation restrict these operations from being combined?

[@jackcollings7257]

I mean the plane defined by the "a" and "'[110"] axes in this question, not just "[110]"*

[@FrankHoffmann1000]

Or does the notation restrict these operations from being combined? Yes, so to say. Additionally: Each single of these symmetry elements should leave the unit cell and orientation and positions of the motif unchanged.

[@jackcollings7257]

@@FrankHoffmann1000 Would this be a benefit of classifying by crystal class? The crystal class indicates a point group, and I imagine that the motif must obey that point group (where the point group operations occur w.r.t. the lattice point with which the motif is associated)? Thanks again by the way :)

[@FrankHoffmann1000]

@@jackcollings7257 I am not sure, if I completely understand your question. In which regard should the classification of (or by) the point group be helpful?

[@jackcollings7257]

@FrankHoffmann1000 Ah I don't think what I said is quite correct, as I think things become difficult if atoms in the motif lie on symmetry elements (excepting the lattice point). Nevermind!

[@smashgambits]

Mirror planes take precedence over rotational symmetry, right? So your "simple tetragonal" crystal is P4/mmm because the mirror does not guarantee a 4-way rotational symmetry. But if you had "simple orthorhombic" it would have space group Pmmm? Even though there are also three sets of 2-way rotational symmetries, since a mirror guarantees 2-way symmetry, it takes precedence? So Pmmm would be simple orthorhombic, and P222 would need something else which breaks mirror symmetry?

[@FrankHoffmann1000]

>Mirror planes take precedence over rotational symmetry, right? You cannot say that in this form. Each kind of symmetry is present or not. You are, of course, right, that the presence of certain symmetry elements automatically generate other symmetry elements that are present, too. And it also correct that certain elements are not present in the space group symbol, at least in their short notation form. But it is not true that a 2-fold rotational symmetry generates a mirror plane. One of the most important symmetry rules is: _Two_ _perpendicular_ mirrors generate a 2-fold axis of rotation. Therefore, Pmmm is actually P 2/m 2/m 2/m, because you have three pairs of mutually orthogonal mirrors. If you have, for instance, only two perpendicular mirror planes you will end up with... ...Pmm2 (and in fact, the 2-fold axis of rotation is generated by the two mirrors). In the end you are right again, stating that in P222 the mirror symmetry is broken along all three direction.

[@smashgambits]

@@FrankHoffmann1000 Okay, thanks :) It's been a while since I took a class in structures, haha. So an atom at each corner of an orthorhombic cell would be Pmmm, but if you changed those atoms to something like water molecules (which all have the same orientation and don't realign) then you would get P222?

[@FrankHoffmann1000]

@@smashgambits If the mirror planes of the water molecule _itself_ don't align with the crystallographic axes - yes! Otherwise... it depends :-)

[@smashgambits]

@@FrankHoffmann1000 Thanks so much! Also, I visited your website and I have definitely seen your book cover before (don't think I read it though). Pretty cool that a big-shot in the crystallography world is responding to my comments on KZbin :)

[@FrankHoffmann1000]

@@smashgambits big-shot --> blush :-)

[@ahmadmajed7374]

Excuse me, in the 2nd example, what would make the difference in space group, when the same exact structure doesn't have the motif in the middle? because I believe it would be the same space group of P4mm

[@FrankHoffmann1000]

You are right, the overall motif would be different but the space group would be indeed P4mm, too.

[@ahmadmajed7374]

@@FrankHoffmann1000 Thank you so much, your videos were very informative and systematic .. I'm going to use this knowledge in my qualifying exam =))

[@amishamishra7641]

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]

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.

[@ivanbombana7282]

Excuse me, I have a question.... In 10:43.... the green plane is not perpendicolar to rotation axis. So why 4/mmm and not 4m/mm in tetragonal system ?

[@FrankHoffmann1000]

The symmetry elements are specified according to the three viewing directions (which are dependent on the crystal system). Here they are c, a, [110]. For the first direction we find a 4-fold axis of rotation _and_ a mirror plane perpendicular to it. This results in 4/m (4-over-m). Then you specify the symmetry elements in the second direction: simply one m (mirror planes are always specified perpendicular to a viewing direction). And you proceed with the third direction: again one m. Altogether: 4/m m m. In your notification we would have a 4-fold axis of rotation in the first direction, then a mirror plane in and a second mirror plane perpendicular to the second direction (m/m) - but this is already not possible. Please note also with respect to the green mirror plane that this is the mirror plane perpendicular to the _third_ viewing direction (i.e. [110]).

[@twrikbling7894]

very easy course for me because i am living in 2 dimension world

[@mahdibeedel8390]

When you are talking about a mirror plane along a certain direction in space group notation you consider it like the normal vector of that mirror plane is perpendicular to that certain direction (10:29), but when you were talking about a mirror plane in point group notation (unit 3.6 minute 1:30) you consider it like the normal vector is along that certain direction and the plane intersects that direction. I am a little confused; Could you please let me know if we have different convention for mirror planes in space group and point group or it was just a mistake?

[@FrankHoffmann1000]

Dear Mahdi, no, there are no different conventions between point and space groups concerning the orientation of planes. The normal vector of the planes are always parallel to the given viewing direction but the plane _itself_ is of course perpendicular to that direction (as the normal vector is perpendicular to the plane itself). best! Frank

[@mahdibeedel8390]

Thank you so much Frank.

[@pluralichroma9685]

I am sorry, but then I don't understand the mirror plane at 10:33, because the plane's normal vector is not parallel but perpendicular to the viewing direction a, is it? Oh I see. A few seconds ahead you changed the labeling of the axes and everything is correct again.

[@FrankHoffmann1000]

Dear Michael, oh yes, you are right - very attentive! I have just recognized that I forgot to change the coordinate system in the lower left for all the slides of the screencast- look at 10:37 and onwards.... Sorry for the confusion! best! Frank

[@karelinaleksandr4585]

Could you please explain why there are only simple monoclinic and base-centered monoclinic in the Bravais lattice?

[@FrankHoffmann1000]

Please first refer to units 2.4 and 2.5, Bravais lattices (I) and (II): kzbin.info/www/bejne/qWXVg4CjidR6eLM kzbin.info/www/bejne/pWfUfKWVhL9krNk If you have any remaining questions, please feel free to ask.

[@origamigek]

Awesome video, nice voice. One question: why did the space group go from I4mm to P4mm in the last example?

[@FrankHoffmann1000]

Thanks for your kind words! Because the pyramid in the center of the cell has a different orientation compared to those at the corners. Centering is a _pure_ translational symmetry operation. This means that it is not allowed to change the orientation, when we go from one motif (say at the corner) to another (say to the one in the center). The question which might arise: what is the motif/base in the given case? Well, it is a one pyramid pointing upwards plus 1/8 of the pyramid pointing downwards in the center. best! Frank

[@origamigek]

Thanks for the quick reply. So because the upside down pyramid violates the translational symmetry it is omitted? I.e. it is not taken into account for determining the Bravais lattice.

[@FrankHoffmann1000]

Yes, so to say - it is taken into account, but only in the sense that _no_ I centering is present

[@joelaw999]

@@FrankHoffmann1000 Hi Frank, I am confused with the same issue here. I know why the last example is different from I-centered bravias lattice, but primitive lattice shall be a quite simple one, isn't it? I mean, I can understand why it is P-lattice when there are ONLY 8 motifs at each corner of the unit cell, but with a another motif at the center( whatever its orientation), why isn't it different from the primitive one? THX a lot!

[@FrankHoffmann1000]

Hi Joe, you shouldn't confuse lattice points with motifs. And, of course, the orientation is of utmost importance! I know that this centering thing is one of the hardest things to understand, if you go from a circle/sphere (i.e. _lattice point_) to real motifs, but perhaps it might be helpful to think from the reverse: The first important thing is to realize and to remind yourself that the translation principle is strongly valid. This means, if we go from one lattice point to another in a primitive cell, then we are only allowed to translate the motif, rotations are not allowed, ok? Now, the next important step: In a centered cell that or these additional lattice points are, of course, always the same lattice points! The lattice point in the center of a centered cell is not distinguishable from any other lattice point! Why? Because - in reverse - you can turn any centered call back to its primitive variant! But this would mean that you have one pyramid pointing upwards as one lattice point/motif and an inverted pyramid as another - this can't be the case! Therefore: all lattice points are the same! And this, in turn, means, that the inverse pyramid 'destroys' in a way the possibility to have a centered cell. Is this understandable? best Frank

[@MrCooldude4172]

In the "a" viewing direction for the tetragonal cell, isn't there also a 2-fold rotational symmetry?

[@FrankHoffmann1000]

For the tetragonal cell with the spheres as motifs, yes. But not for the body-centered tetragonal cell with the pyramids as motifs, because a rotation by 180° along the a direction would invert the pyramids.

[@MrCooldude4172]

Frank Hoffmann I see, so with the tetragonal cell with the spheres as motif shouldn't it be 2/m instead of just m for the "a" viewing direction?

[@FrankHoffmann1000]

Yes! Here, only the short notation of the space group was used, the full symbol is indeed P 4/m 2/m 2/m!

[@arijitsinha2263]

Wonderfully explanation. Sir I have a question : symmetry for a crystal structure is uniquely determined by the space group. Now for some structure let say NaCl structure is described by space group Fd-3m which is cubic and face centered, which is very clear from conventional cell of NaCl. Now sir if I take the primitive cell of NaCl it should also have same symmetries ( symmetries doesn't depend on choice of unit cell as you correctly explained) but the primitive cell neither look cubic nor centered in this case sir can we still assign Fd-3m to primitive cell of NaCl? Again for some crystals space group is given wrt to the primitive cell (e.g Pmnm and so on) and for some space group is given wrt conv. Cell (like NaCl) Why is that sir? Is there any rule on intuitive way to assign space group? How to know which cell to take? Please help me sir if i am wrong somewhere kindly let me know.

[@FrankHoffmann1000]

The answer to your first question is: No, because a primitive cell/lattice is a non-centered cell and is usually abbreviated with P, not with F. However, there is also the special case of rhombohedral cells abbreviated by R. By the way: The space group of the face-centered variant is Fm-3m, not Fd-3m. With regard to your second question I would recommend to watch unit 2.4: kzbin.info/www/bejne/qWXVg4CjidR6eLM To comment further on a potential misunderstanding: It is correct that the symmetry of the lattice/structure doesn't change, regardless of the unit cell you choose, but you can, of course, choose completely inappropriate unit cells, in which the cell doesn't reflect the symmetry of the lattice at all (see also unit 2.4). best Frank

[@arijitsinha2263]

@@FrankHoffmann1000 Thank you very much sir for your reply. Now it is clear to me space group of a crystal structure is not unique depends on the choice of unit cell. Thanks again sir and sorry for the NaCl incorrect space group.

[@savagegirl6318]

Sir, Only one four fold symmetry is there, how?? Is it not possible to have four fold symmetry in front and back planes as well as right and left planes?if it is not possible why?? Please reply

[@FrankHoffmann1000]

Look carefully again, think about it again - I believe, you can figure it out by yourself. Hint: Does a rectangle have four-fold symmetry?

[@savagegirl6318]

@@FrankHoffmann1000 thanks a lot for the thought provoking spark and prompt and timely response!got it !

[@alexdavidson7785]

Hope the books a big success Now explain quasicrystals...

[@bhavatharani7412]

2 screw axis is parallel to c...... I can't understand 👆... pls explain

[@FrankHoffmann1000]

Is there a screw axis in this unit? And please explain more specifically what aspect you don’t understand .

[@PrinceGupta-be1jv]

add more videos regarding miller indices mam......

[@FrankHoffmann1000]

yo man, you mean additionally to that one? kzbin.info/www/bejne/a5a5hJpqn8SDo5o

[@mennaelmanzalawy2275]

Thank you!

[@FrankHoffmann1000]

Very welcome

[@matslundemo803]

Hello! I've tried sending you an e-mail on: [e-mail] But I was unable to forward the e-mail, for some reason. I've also tried to send you a message on KZbin, with no luck so far. Is there any way for me to contact you? Would like to talk to you about using you as a scholar-reference for my bachelor-theory. Best regards, Mats L.

[@FrankHoffmann1000]

Dear Mats, I answered already your message on KZbin, however, obviously something went wrong :-) The second "n" in "hoffmann" was missing, therefore I haven't received your eMail. You can contact me either via my institutional eMail "[e-mail]" or via my private gmail account "[e-mail]" best! Frank

[@matslundemo803]

I've contacted you via [e-mail] :)

[@studychannel8345]

ich feiere die stimme