Unit 3.3 - Point Symmetry and Rotoinversions
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- Опубликовано: 26 окт 2024
- Unit 3.3 of our course The Fascination of Crystals and Symmetry
Additonal resources at: crystalsymmetr...
In this unit the center of inversion and the so-called rotoinversion axis will be explained. A rotoinversion is a combined symmetry operation, where two transformations have to be carried out: first a rotation by 360 divided by n degrees, where n is the order of this rotoinversion, followed by an inversion at the center of this object.
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I can't believe how you made all basics of crystallography in a single channel. Nicely explained by images. What an effort
These videos are brilliant, im going through your entire playlist. Im a biologist who ended up doing biochemistry - and I am really lacking a lot of this crystallography theory for doing protein crystallisation. These videos are perfect as they really have given me the intuitive understanding of how it works, and thus has made it much easier to dive into the math and statistics that I have to have some understand of in data collection and data processing.
Thank you very much, Heidi, for your kind words!
Great work, I appreciate this!
Great work! Thank you
I can imaging the mirror planes and 3-fold axes of rotation for tetrahedron. Can you tell me where is the diad?
The 2-fold axes are running along the midpoints of two opposite edges, see for instance here: de.m.wikipedia.org/wiki/Datei:Tetrahedron_with_2-fold_rotational_axes_RK01.png
Amazing graph. Perfectly showing the diad 🎉🎉🎉 thank you so much
I am puzzled by bar-6 --- since it is equivalent to 3-fold rotation plus a perpendicular mirror plane, why is it classified preferentially as a hexagonal symmetry element, but not a trigonal symmetry element? After all, intuitively it looks more like trigonal than hexagonal...
In former times, the classification of crystal classes or their belonging to a crystal system was based exclusively on the morphology of crystals. All crystals with a three-fold axis of rotation belonged to the trigonal system, and all crystals with a six-fold axis of rotation belonged to the hexagonal crystal system.
And you are indeed right that 6-bar has no 'genuine' 6-fold axis of rotation, but belongs to the hexagonal crystal system. However, the modern assignment of crystal classes to crystal systems is based on the criterion of the order of (simple) axes of rotation or rotoinversion axes: all crystal classes that have a six-fold axis of rotation or a six-fold rotoinversion axis belong to the hexagonal crystal system.
The advantage of this consideration may not be immediately apparent, but it arises from at least two points of view:
1. The first aspect is that the systematization should be as consistent as possible. Look at the classes of the tetragonal crystal system: here too, not all crystal classes have a four-fold axis of rotation. Two of the classes have only one two-fold axis of rotation. If only the criterion of the order of the rotation axes is to be used for classification, these classes should belong to the orthorhombic system. However, the presence of a two-fold axis of rotation is not a qualifying characteristic; keep in mind that the symmetry results in a restriction of the metric of the axis system: both a four-fold axis of rotation and a four-fold rotoinversion axis require a square base face of the unit cell (a = b) as well as an orthogonality of the three lattice vectors (α = β = γ = 90°). In the orthorhombic system, the first condition is not necessarily met.
2. Classification according to the order of the axes, regardless of whether it specifies a pure rotational symmetry or a coupled rotoinversion, has a further advantage, which you can infer by inspecting the classes of the trigonal crystal system: the characteristic feature of the trigonal crystal system is that symmetry only appears in two directions. If you add the classes with a six-fold rotoinversion axis to the trigonal crystal system, you would need a third viewing direction.
@@FrankHoffmann1000
Hi Frank, thank you very much for your time and effort in giving your answers! I enjoyed reading the arguments and will (probably) come back with additional questions in a few days or weeks (kind of busy right now...)😁
Dear Frank, I noticed that at 4.34 to 4.38 the mirroring at a point is not clear to me, as I've noticed that the length used are not proportional or as far from the other side as the balls are the at 4.35 was longer and at 4.38.
Hi David,
mirroring at a point, here in fact the origin of this object (i.e. the larger orange sphere with coordinates 0,0,0), means nothing more than moving an object (the smaller blue spheres) from coordinates x, y, z to -x, -y, -z. The absolute values of x and -x, y and -y, and z and -z are the same. If you see something different on the screen it is probably due to the perspective drawing; it's always difficult to illustrate 3D operation in 2D.
best
Frank
@@FrankHoffmann1000 ok! Thanks for making the video, I will like to add that it is very helpful.
You are so good, I want to study with you
Dear Frank, I'm starting my journey with crystallography. 3.50 I can't see why after rotation of the starting molecule around 360/4 and inversion at a center of symmetry give the same molecule.... I've got homework to find symmetry elements of CF4, PF5 and various conformers of cyclohexane and I'm lost :(
Well, spatial imagination is differently developed. Perhaps physical models that you can actually rotate in space would help? Note that SF4 or CF4 don't have a center of inversion - the operation inversion is carried out at the center of the molecule, that's correct, but only together with the rotation operation together it is a symmetry operation/element, here a -4.
@@FrankHoffmann1000 OK NOW I CAN SEE :D Thank you!!!
GREAT JOB!!!
it's an easy video to understand . Thx
Dear Frank,
When you are talking about even rotoinversions contain automatically an axis of rotation of the half order, is it about 2S4 = C2 ?
Kind regards
Dear 天奔 陈,
almost - even or uneven is related to the order, i.e. to the index of the rotoinversion, but not to the number of rotoinversions that might be present in a molecule. Please note that S4 is not a rotoinversion axis of order 4, but a rotary reflection plane of order 4.
However, you are right in the case of the order 4, because the order of S4 is identical to the order of -4. And this means exactly that
-4 ("4-bar") or S4 implies the presence of C2, and
-6 ("6-bar") or S3 implies the resence of C3.
And because S6 is equivalent to 3-bar, a 3-fold axis of rotation is also present here, as 3-bar = "3 + 1-bar".
best!
Frank
Dear Frank,
Thanks for your reply. I re-watch the structure of tetrahedron, I see you mean. if the structure undergoes S4 and remain unchanged, this structure must have C2 axis.
Best wishes
the 3bar example at 6:10 is strange to me
3 and 1bar are already symmetry elements...so it's not the best example.
am i missing something? it looks more like a 6fold rotation followed by a mirror that inverts vertical direction
Sometimes, symmetry is tricky :-)
Two aspects:
First, the interesting thing is that I cannot give you a different example as this is a law: a macroscopic body or a finite configuration of points in space that has a 3-bar has _automatically_ a 3-fold axis of rotation and a 1-bar. But please note that, of course, neither a body with 3-fold axis of rotation must contain a 1-bar, nor a body with 1-bar must contain a 3-fold axis of rotation.
Second, as you mentioned the 6-fold rotary reflection: It is up to you, if you want to express the symmetry (of the same(!) body) either with a 3-fold rotoinversion axis or a 6-fold rotary reflection, these two are completely equivalent, see also unit 3.4:
ruclips.net/video/3i7UvVd7bjs/видео.html
best wishes
Frank
@@FrankHoffmann1000 Got it. Thank you very much!
Thank you! the best..!
No problem about 3bar = 3 + 1bar. However, I think that 4 bar is not equivalent to 4 +1bar. I could not find some combination which can produce 4 bar symmetry.
Thank you....
Dear Sopaes,
the "+"-sign does indeed _not_ mean that both symmetry elements are present at the same time! Instead, it should rather read as a operation sign. First, apply a 4-fold axis of rotation, then carry out an inversion. It is explicitely stated in the example of the tetrahedron (starting from 2:56) that it shows neither a 4-fold axis of rotation nor a center of inversion. You are completely right that odd and even rotoinversions behave differently: odd rotoinversions indeed possess automatically a center of inversion, but this does not hold for even rotoinversions.
best wishes!
Frank
Hi Sir, how about the Rotoinversion exes of order 4?
Can I write as 4 ̅=4+m? Thanks
Hi,
well, if you take this 4+m as an combined operation, then the answer is yes, because a rotoinversion of order 4 is identical with a rotoreflection of order 4 (S4). But note that it does not mean that this implies the independent existence of neither a four-fold axis of rotation nor a mirror plane.
best
Frank
@@FrankHoffmann1000 Thank you.
Sir what about allenes ?do they have inversion?
Dear Sana, no they haven't, even if the substituents on the terminal carbon atoms are identical, because they are oriented perpendicularly to each other.
@@FrankHoffmann1000 thank u
Thank you a lot .
3 bar rotoinversion is I guess wrongly depicted. It should be 6 bar I think. Please clarify frank.
3-bar means: rotation by 120 degress followed by an inversion - I think this is exactly what the picture shows. Could you explain, why do you think that this is wrongly depicted?
best!
Frank
It is because the symmetry operation applied dosen't yield the same configuration as the original. Whereas a rotation by 60 degrees followed by inversion would give original one back. 5.34 of video,3rd example
Sorry, but I think your are wrong. Please see the follwoing PDF:
crystalsymmetry.files.wordpress.com/2018/04/crystal_mooc_yt_en_03_03_6_vs_3_bar.pdf
Here, I graphically depicted the operation you suggested with the point No. 6 - as you will see rotation by 60° (leading to 6') and inversion (leading to 6'') will not generate an indistinguishable configuration.
I wonder why 6 bar = 3 + m? Shouldn't it be 6 bar = 3 bar + m?
Dear Xinyu,
first of all, perhaps, have a look again at 3-bar (from 5:28 on): If you add a mirror perpendicular to the 3-bar axis the result would be a simple 6-fold axis of rotation, not a 6-fold rotoinversion. Secondly: Try to carry out the symmetry operations you suggested as being equivalent on the picture, which shows the sphere pattern of a 6-bar: 3-bar means rotation by 120° and then inversion - a new position is realized - if you now additionally reflect this sphere perpendicular to the axis, there is also not a sphere - so this is not a symmetry operation of this pattern.
best!
Frank
You could explain this in more better way
The first question is, why I haven't done so, although I could...
The second question is, why you don't ask a question, if you haven't understand a certain aspect...
The third question is, why you don't show me how to explain this in a better way...
Anyway, no offense.