Professor, you said crystal systems are classified based on point groups (rotation and reflection symmetry) but you categorized different crystal systems only based on their rotation axes and didn't use reflection symmetry, so why did you say crystal systems are categorized based on both rotation and reflection symmetry?
Now I can say that if crystal systems, not carefully and fully explained as you have done, might lead to so many misconceptions about the science involved. I know this because I have been in a misconception all these days about what a crystal really is..!! Thanks prof.
sir, please explain about body centered tetragonal. why it is present in the table, from symmetry point of view. You explained about face centered cubic, from symmetry point of view, but not body centered tetragonal.
1. Cubic F is there in the list because it has 4 3 fold symmetries. 2. Tetragonal I is also there in the list because it has 1 4 fold symmetry. 3. A Cubic F can be seen as Tetragonal I and vice versa. 4. What decides that a lattice corresponding to a particular solid compound is to be considered as Cubic F or Tetragonal I or is it that we can take it either way. Please reply @Rajesh Prasad Sir.
Cubic F has 4 three-fold axes. This is characteristic of cubic crystal. So we will calli it cubic: Cubic F. Although tetragonal I unit cell is also possible, calling it tetragonal will imply it does not have 4 three-fold axes. This will be a misrepresentation.
@@rajeshprasad101 Thank You so much for your reply sir🙏 But again sir, in that case why don't we remove the Tetragonal one from the list if cubic has got the priority??
@@deepak.9994 I think I know the answer to your question. Because there ARE other Tetragonal I crystal that cannot be interpreted as a Cubic F. In fact, Cubic F is a special case of Tetragonal I where c=a*sqrt2, and for any other c values, the crystal only satisfies the conditions of a tetragonal structure.
@@westories3461 Thanks dear for your response, but what I think is the reason for discarding Cubic F is this that, we are supposed to consider a unit cell as small as possible and with its repeated placement (translational), it must create the whole lattice and since tetragonal I would be a smaller geometry than the cubic F, hence we consider the tetragonal I and discard cubic F.
@@deepak.9994 The idea of shrinking the unit cell to the smallest volume possible is based on the premise of identical symmetry. You can even say choice of unit cell is not quite important provided that symmetry is the same since it's just the way we view the crystal. But sending a wrong message about the symmetry of the structure really does matter. Let's say we're considering what planes belong to the {110} family of planes, then the answer is different for cubic and tetragonal structures. The point is no matter what unit cell we choose to represent the crystal, the symmetry of it is invariant, so when a crystal has a higher level of symmetry, in this case cubic, we should always point it out or else it could be confusing.
Loss of 3 fold axis----> not cubic ----> tetragonal loss of single 4 fold axis ----> Not tetragonal ----> orthorhombic All this by using only simple symmetry approach. Nice.
we can save CUBIC-F from Tetragonal-I because CUBIC-F have cubic symmetry (considering the 3-fold rotation), so CUBIC-F indeed belong to the cubic family. We can't save Tetragonal-C from Tetragonal-P because Tetragonal-C just have Tetragonal symmetry (with the 4-fold rotation). We get to Tetragonal-P just to shrink unit cell size. I'm not totally sure though.
The vertical faces of a tetragonal unit cell are rectangles. Thus the symmetry perpendicular to them are 2-fold axes. My drawing is not very good, the rectangle appears as sqaure, which is what, I suppose, is giving you the impression of a 4-fold axis.
By a change of unit cell it can be represented as an end-centred lattice which is already accounted for in the 14 Bravais lattices. In other words, a body-centred monoclinic is always equivalent to an end-centred monoclic. The choice between the two is conventional. By convention it has has been decided to represent it as an end-centred lattice.
In the last example of tetragonal A, two fold axis are also possible through the edge centers. Therefore it has two fold axis more than 3. So, now will it be called orthorhombic A??
@Rajesh Prasad Sir, In your next lecture "ssymetry base approach to bravis lattice" At 7.32 min. analysis of 2 fold assymetry axis passing through centre of faces not done for Cubic C?
Cubic C does have 2-fold axes perpendicular to the four-fold axis. However, this fact is not important in deciding the crystal system. Absence of 3-fold axes indicates that it is not cubic. Presence of a single 4-fold axis indicates that it is tetragonal.
Sir while checking cubic C structure we got 1 4 fold and 1 2 fold,now why did u select 1 4 fold and took tetragonal why not take 1 2 fold and select monoclinic cystal system?
We have to look at all symmetry elements of lattice/crystal. So cubic C has one 4-fold and two 2-fold axes. This will qualify as tetragonal due to presence of one 4-fold axis. Presence or absence of additional 2-fold axes does not matter. But the absence of other 4-fold axes matter. As there can be cubic lattices having three 4-fold axes. When we say requirement of one 2-fold axis for monoclinic we mean ONLY one two-fold axis. If it has more than one two-fold axes it is NOT monoclinic. It can be tetragonal, as in the example above, or orthorhombic (three 2-fold axes), trigonal (three 2-fold axes with a 3-fold axis), hexagonal (six 2-fold axes with a 6-fold axis), or cubic (six 2-fold axes with four 3-fold axes)
Bu Sir, when cubic C was rotated it lost its cubic nature. So it was transformed to tetragonal P. But tetragonal C when rotated preserved its tetragonal nature. Then why do we have to transform it into tetragonal P ?
This is a good question. It is clear that Cubic C is not possible as it lacks cubic symmetry (Four 3-fold axes). It has required symmetry (one 4-fold axis) of the tetragonal system. Within the tetragonal system tetragonal P and tetragonal C are equivalent. So we could have chosen tetragonal C instead of tetragonal P. However, in such cases, we conventionally choose the smaller unit cell. So we refer to it as tetragonal P.
But sir, In Cubic F we have tetragonal I which is smaller when compared to Cubic F but we have chosen both Cubic F and Tetragonal I. Why can't we eliminate Cubic F?
@@harikrishnan3112 thats because, as mentioned by sir in last example, we should prioritize symmetry classification above unit cell classification. in case of cubic c, it already met the symmetry condition of cubic crystal. so we dont need to further investigate for other crystal types. however, in last example, fact that it was a tetragonal crystal type was already established. We only needed to find the type of tetragonal bravias lattice. so we went an extra mile for constructing another adjacent unit cell and later finding out a smaller tertagonal p lattice as a replacement for tetragonal c.
@@AKASHYADAV-fb7po Thanks bro. But if we look at Tetragonal-C it has also meet the crystal symmetry requirement of one four fold axis. But we have eliminated Tetragonal-C stating that we have a smaller unit cell than that which is Tetragonal-P. I'm completely ok with that. But the same applies for cubic-f right as you quoted cubic-f has meet the 4 three fold axis requirement but it has smaller Tetragonal-I inside that, if we are eliminating Tetragonal-C due to availability of smaller volume Tetragonal-p in it. why can't we do the same by eliminating cubic-f since we have smaller Tetragonal-I inside that or why can't we have Tetragonal-C also in the lattice
@@harikrishnan3112 If you think about it, every Tetragonal-P is equivalent to a Tetragonal-C with twice its unit cell volume. We then of course choose to represent it with one single and simpler way. However, same does NOT apply to Tetragonal-I and Cubic-F. The condition for a crystal to be "cubic" requires higher level of symmetry; namely, Cubic F is merely a special case of Tetragonal I where c=a*sqrt2. So these two are not equivalent, and it is essential to differentiate them.
C-centring means placing a lattice point at the centre of the a-b faces of the unit cell, but not at the centres of b-c or a-c faces. In orthorhombic, this results in a possible Bravais lattice. In cubic, it does not result in a Bravais lattice. I have considered Cubic C to discuss this point, i.e., why it is not there as a Bravais lattice.
Love your lectures a lot.
You offer Crystal clear concepts.
Thank you very much.
I have studied that a thousand times but first now understood that. Thank you, sir. You are the best. Now I am happy as well :)
I don't know about others but that smilie was hilarious!
😹😹😹and scary
Absolutely :)
Horondous
Thank you sir, it just solved a small and fascinating problem which grew a lot difficult in my mind.
Professor, you said crystal systems are classified based on point groups (rotation and reflection symmetry) but you categorized different crystal systems only based on their rotation axes and didn't use reflection symmetry, so why did you say crystal systems are categorized based on both rotation and reflection symmetry?
Now I can say that if crystal systems, not carefully and fully explained as you have done, might lead to so many misconceptions about the science involved. I know this because I have been in a misconception all these days about what a crystal really is..!! Thanks prof.
Just amazing.... The way you teach is superb.. I am your fan now
sir, please explain about body centered tetragonal. why it is present in the table, from symmetry point of view. You explained about face centered cubic, from symmetry point of view, but not body centered tetragonal.
1. Cubic F is there in the list because it has 4 3 fold symmetries.
2. Tetragonal I is also there in the list because it has 1 4 fold symmetry.
3. A Cubic F can be seen as Tetragonal I and vice versa.
4. What decides that a lattice corresponding to a particular solid compound is to be considered as Cubic F or Tetragonal I or is it that we can take it either way.
Please reply @Rajesh Prasad Sir.
Cubic F has 4 three-fold axes. This is characteristic of cubic crystal. So we will calli it cubic: Cubic F. Although tetragonal I unit cell is also possible, calling it tetragonal will imply it does not have 4 three-fold axes. This will be a misrepresentation.
@@rajeshprasad101 Thank You so much for your reply sir🙏
But again sir, in that case why don't we remove the Tetragonal one from the list if cubic has got the priority??
@@deepak.9994 I think I know the answer to your question.
Because there ARE other Tetragonal I crystal that cannot be interpreted as a Cubic F. In fact, Cubic F is a special case of Tetragonal I where c=a*sqrt2, and for any other c values, the crystal only satisfies the conditions of a tetragonal structure.
@@westories3461 Thanks dear for your response, but what I think is the reason for discarding Cubic F is this that, we are supposed to consider a unit cell as small as possible and with its repeated placement (translational), it must create the whole lattice and since tetragonal I would be a smaller geometry than the cubic F, hence we consider the tetragonal I and discard cubic F.
@@deepak.9994 The idea of shrinking the unit cell to the smallest volume possible is based on the premise of identical symmetry. You can even say choice of unit cell is not quite important provided that symmetry is the same since it's just the way we view the crystal. But sending a wrong message about the symmetry of the structure really does matter. Let's say we're considering what planes belong to the {110} family of planes, then the answer is different for cubic and tetragonal structures. The point is no matter what unit cell we choose to represent the crystal, the symmetry of it is invariant, so when a crystal has a higher level of symmetry, in this case cubic, we should always point it out or else it could be confusing.
Loss of 3 fold axis----> not cubic ----> tetragonal
loss of single 4 fold axis ----> Not tetragonal ----> orthorhombic
All this by using only simple symmetry approach.
Nice.
thank u for the lecture sir, I just loved it
Sir,
At 7.32 min. analysis of 2 fold assymetry axis passing through centre of faces not done for Cubic C?
Very clumsy concept thankyou sir i get now
if we can save CUBIC-F from Tetragonal-I by the property of symmetry.... why can't we save Tetragonal-C from Tetragonal-P by the same concept...
we can save CUBIC-F from Tetragonal-I because CUBIC-F have cubic symmetry (considering the 3-fold rotation), so CUBIC-F indeed belong to the cubic family.
We can't save Tetragonal-C from Tetragonal-P because Tetragonal-C just have Tetragonal symmetry (with the 4-fold rotation). We get to Tetragonal-P just to shrink unit cell size.
I'm not totally sure though.
Exactly my question. But no explanation.
i am not able to understand the rotation part. can anyone tell me from where can i read that
Thank u prof
In the video at time 20:48, isn't that a 1 four fold symmetry?
The vertical faces of a tetragonal unit cell are rectangles. Thus the symmetry perpendicular to them are 2-fold axes. My drawing is not very good, the rectangle appears as sqaure, which is what, I suppose, is giving you the impression of a 4-fold axis.
@@introductiontomaterialsscience amazing sir..
Sir why is body centered monoclinic lattice not possible ?
By a change of unit cell it can be represented as an end-centred lattice which is already accounted for in the 14 Bravais lattices. In other words, a body-centred monoclinic is always equivalent to an end-centred monoclic. The choice between the two is conventional. By convention it has has been decided to represent it as an end-centred lattice.
@@introductiontomaterialsscience
Thank you Sir.
I am a big fan of you !!
Hi prof, U are aweome.
In the last example of tetragonal A, two fold axis are also possible through the edge centers. Therefore it has two fold axis more than 3. So, now will it be called orthorhombic A??
There are no two-fold axes through the edge centres.
Sir will it be correct to say that the tetragonal A has Single four fold character in the horizontal axis of rotation ?
Regards
In tetragonal A the centred faces are rectangular and not square. So you will have twofold and not fourfold axis.
@Rajesh Prasad
Sir,
In your next lecture "ssymetry base approach to bravis lattice"
At 7.32 min. analysis of 2 fold assymetry axis passing through centre of faces not done for Cubic C?
Cubic C does have 2-fold axes perpendicular to the four-fold axis. However, this fact is not important in deciding the crystal system. Absence of 3-fold axes indicates that it is not cubic. Presence of a single 4-fold axis indicates that it is tetragonal.
@@introductiontomaterialsscience
Sir,
Thanks
brillant!
Sir while checking cubic C structure we got 1 4 fold and 1 2 fold,now why did u select 1 4 fold and took tetragonal why not take 1 2 fold and select monoclinic cystal system?
We have to look at all symmetry elements of lattice/crystal. So cubic C has one 4-fold and two 2-fold axes. This will qualify as tetragonal due to presence of one 4-fold axis. Presence or absence of additional 2-fold axes does not matter. But the absence of other 4-fold axes matter. As there can be cubic lattices having three 4-fold axes.
When we say requirement of one 2-fold axis for monoclinic we mean ONLY one two-fold axis. If it has more than one two-fold axes it is NOT monoclinic. It can be tetragonal, as in the example above, or orthorhombic (three 2-fold axes), trigonal (three 2-fold axes with a 3-fold axis), hexagonal (six 2-fold axes with a 6-fold axis), or cubic (six 2-fold axes with four 3-fold axes)
@@introductiontomaterialsscience thank you sir🙏i feel blessed and lucky to recive the knowledge u have provided here sir,god bless u sir
What is A in the tetragonal A
It is end-centred tetragonal with A faces centred. Faces containing b and c axes are called A faces.
Then shouldn't it (the picture you drew) be called tetragonal B ? Because you have centred the faces containing ( a and c ).
@@thewhiteshirtmanoksh2163 Yes actually prof was confused a little bit but concept holds same, if we change the side name convention.
Beautiful
Bu Sir, when cubic C was rotated it lost its cubic nature. So it was transformed to tetragonal P. But tetragonal C when rotated preserved its tetragonal nature. Then why do we have to transform it into tetragonal P ?
This is a good question. It is clear that Cubic C is not possible as it lacks cubic symmetry (Four 3-fold axes). It has required symmetry (one 4-fold axis) of the tetragonal system. Within the tetragonal system tetragonal P and tetragonal C are equivalent. So we could have chosen tetragonal C instead of tetragonal P. However, in such cases, we conventionally choose the smaller unit cell. So we refer to it as tetragonal P.
A stonishing lecture
Why can't we say that cubic FCC is body centered tetragonal ?
Becuase it has cubic symmetry, i.e. 4 three-fold axes.
@@introductiontomaterialsscience thank you,is it that we choose the unit cell to be most symmetric
Only youtuber who replied to everyone.
He is not a youtuber
Sir then why don't you choose tetragonal C as a bravais lattice instead of tetragonal P?
Because tetragonal P is smaller unit cell and has the same symmetry as tetragonal C.
But sir, In Cubic F we have tetragonal I which is smaller when compared to Cubic F but we have chosen both Cubic F and Tetragonal I. Why can't we eliminate Cubic F?
@@harikrishnan3112 thats because, as mentioned by sir in last example, we should prioritize symmetry classification above unit cell classification.
in case of cubic c, it already met the symmetry condition of cubic crystal. so we dont need to further investigate for other crystal types. however, in last example, fact that it was a tetragonal crystal type was already established. We only needed to find the type of tetragonal bravias lattice. so we went an extra mile for constructing another adjacent unit cell and later finding out a smaller tertagonal p lattice as a replacement for tetragonal c.
@@AKASHYADAV-fb7po Thanks bro. But if we look at Tetragonal-C it has also meet the crystal symmetry requirement of one four fold axis. But we have eliminated Tetragonal-C stating that we have a smaller unit cell than that which is Tetragonal-P. I'm completely ok with that. But the same applies for cubic-f right as you quoted cubic-f has meet the 4 three fold axis requirement but it has smaller Tetragonal-I inside that, if we are eliminating Tetragonal-C due to availability of smaller volume Tetragonal-p in it. why can't we do the same by eliminating cubic-f since we have smaller Tetragonal-I inside that or why can't we have Tetragonal-C also in the lattice
@@harikrishnan3112 If you think about it, every Tetragonal-P is equivalent to a Tetragonal-C with twice its unit cell volume. We then of course choose to represent it with one single and simpler way.
However, same does NOT apply to Tetragonal-I and Cubic-F. The condition for a crystal to be "cubic" requires higher level of symmetry; namely, Cubic F is merely a special case of Tetragonal I where c=a*sqrt2. So these two are not equivalent, and it is essential to differentiate them.
Sir pls explain hexagonal monoclinic symmetry also
Please explain your doubt in more detail.
What is cubic c?
C-centring means placing a lattice point at the centre of the a-b faces of the unit cell, but not at the centres of b-c or a-c faces. In orthorhombic, this results in a possible Bravais lattice. In cubic, it does not result in a Bravais lattice. I have considered Cubic C to discuss this point, i.e., why it is not there as a Bravais lattice.
16:50
sir isn't Tetragonal-a has 1 four fold symmetry about horizontal axis
Please see the reply to the same question raised by Prasson Singh.
No. tetragonal A will have a two-fold symmetry along a-axis because the faces being centred are rectangles.
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D