A Rigorous Proof on the Crystallographic Restriction Theorem to Establish Human Being Zhang Yue Department of Physics.
Hunan Normal University.
Changsha.
PeopleAos Republic of China, 410081 phys_zhangyue@126.
Abstract: It is significant to find a more rigorous and satisfactory proof of the crystallographic restriction theorem.
The inexistence of C5 axis of symmetry is equivalent of that pentagons are impossible to fill all the space with a connected array of pentagons, on the basis of this viewpoint, using a purely mathematical approach the paper rigorously proves that C5 and Cn .
Ou7 ) axes of symmetry cannot exist , and one Ae, two Ae, three Ae, four Ae and six Ae fold axes of rotational symmetry are allowable.
Therefore, the axes of symmetry of the crystal can merely exist C1.
C2.
C3.
C4 and C6 .
Keywords: the crystallographic restriction theorem, pentagons, n-sided.
Ou.
polygons, proper rotation.
PACS : 61.
Ah .
IaoeOoaIoEAuAoa sOUEOaIUuioU410.
oA] IaoeOoaIUuOIAieUaaNuOueUUaoEAoaACsou C5 O asEiueUioEiOAuOoEUuoACCAU unNOioEiniuuAoaIsouoeoE C5 eU Cn .
Ou7 )UAUIAo u 1U2U3U4 uaO 6 NsOUaAUAoaIoeoEOaEou C1U C2UC3UC4 eU C6 AC Iiso oeOoaIUiUnOnOu7OouOO .
Introduction It is well - known that the axes of symmetry of the crystal can merely exist C1.
C2.
C3.
C4 and C6, this is the so - called crystallographic restriction theorem [ 1- 4 ].
Among various proofs of this theorem, there is a famous proof which is generally concurred by those people who are familiar with the solid state physics.
Although the well - known proof of the theorem has been applied in many famous books [ 2- .
, it is not satisfactory at least due to the following two reasons.
First, considering an n-fold .
is an intege.
rotation of the crystal in the two dimensional space, as shown as Fig .
Fig.
The point BAo is one of the points generated by an n-fold rotation axis through point A Budapest International Research and Critics Institute-Journal (BIRCI-Journa.
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operating on lattice point B with angle , and the point AAo by a similar axis through point B inversely operating on lattice point A with angle .
The value of angle is equal to u 1 BAoCoC B iAoNC A EoEA n - NsOO eioEACUAAoC oC A iAoNC B EoEAoEacuasoEnacO eioEACAC e in terms of the proof , because of the periodicity of lattice structure , the length of BAoAAo must be equal to the integral multiples of that of AB, namely.
BAoAAo= m t .
However, in the viewpoint of mathematics, eq.
is not clear as an argument for the 14 different Bravais lattice structures of real crystals , but not the supposing Bravais lattice, all of the 14 lattices should be respectively demonstrated in order to support eq.
Second, the calculation from eq.
in accordance with Fig .
1 demonstrates that the possible values of m are s1, 0, 1, 2, and 3, nevertheless, if m takes the value of s1, neither of the lengths of BAoAAo and t in eq.
can be significant to be negative .
Therefore, it is necessary to find a rigorous proof of the crystallographic restriction II.
Review of Literature The inexistence of C5 axis is equivalent of that pentagons are impossible to fill all the space with a connected array of pentagons [ .
, and this can be easily generalized to all the cases of Cn .
Ou7 ) axis .
At first , let us consider two congruent regular pentagons such as A5 and B5 in the two - dimensional space, as shown as Fig.
Fig.
Two congruent regular pentagons fit together .
u 2 IOoEiUauuaC In Fig.
2, 5 and 5 respectively note the interior angle and the exterior angle of the pentagon, and 5 is the clipped angle between the side of A5 and the side of B5.
Because the sum of all the exterior angles of a polygon is always equal to 360 A , thus, it can be written
= 72 A ,
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5 = 180 A s5 =108A .
It is clear from Fig.
2 that if the other one or more pentagons just can fill the space within the scope of 5 with no Augaps Ay between pentagons, it requires that one or more interior angles can just fill 5 angle with no AugapsAy between them, or the size of 5 must be just equal to the integral multiples ( positive ) of the size of an interior angle, namely, 5= m5 .
= 1, 2, 3AA).
Nevertheless, using eq.
, the size of 5 is given by 5= 25=2y 72 A = 144 A , .
from eq.
and eq.
, it can be found 5= 5 , .
in terms of eq.
C5 axis can not exist .
Fig .
3 depicts the AugapsAy between pentagons in the scope of 5 in a close packing of pentagons in the two-dimensional space [ 3 ].
Fiq .
C5 axis of symmetry does not exist .
u 3 C5 souAC i.
Discussion Assuming to substitute two congruent regular n Aesided .
Ou.
polygons such as A n and B n respectively for A5 and B5 in Fig .
2, accordingly , n ( or n ) for 5 .
r 5 ), and n for 5.
Thus, an exterior and an interior angles of the n - sided ( nOu7 ) polygon are respectively written
360 A 2
n = = y 180 A , .
360 A ns2 n = 180 sn=180 s y 180 A , .
in accord with Fig.
360 A 4
n =2n =2y = y 180 A .
Comparing eq.
with eq.
, obviously, if nOu7, the inequality nun holds true, thus nOmn .
= 1, 2 , 3AA) , .
Cn .
Ou.
axis can not exist .
Budapest International Research and Critics Institute-Journal (BIRCI-Journa.
Volume I.
No 4.
December 2018.
Page: 25-28 e-ISSN: 2615-3076(Onlin.
, p-ISSN: 2615-1715(Prin.
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In the viewpoint of mathematics.
C1.
C2.
C3.
C4 and C6 axes must also be discussed .
first, it is clear that C1 axis represents an one-fold rotation with the rotation angle 0 A or 360 A , and will certainly remain the crystal invariant .
We separately consider an oblique Bravias lattice in the two-dimensional space, if a two-fold rotation with the rotation angle of 180 A through any lattice point in a primitive cell, the primitive cell will remain invariant, this is also true for equivalent points in other primitive cells [ 3 ].
Therefore.
C2 axis for a crystal based on such a primitive cell can exist .
Differing from the case of pentagons , in Fig.
If assuming respectively to substitute two congruent regular triangles such as A3 and B3 , tetragons such as A4 and B4 and hexagons such as A6 and B6 for A5 and B5, accordingly , k .
r k ) ( k=3,4, 6 ) for 5 .
r 5 ) , and k ( k=3, 4, 6 ) for 5, it is easy to calculate out :
3= 43 , 4 =24 , 6 =6 ,
In accordance with eq.
, it can be recognized that C3.
C4, and C6 axes are allowable for the crystal rotation .
IV.
Conclusion With respect to the crystallographic restriction theorem, the paper proposed different opinions on a proof applied in many famous books.
Due to the periodicity of the lattice structure, the inexistence of C5 axis is equivalent of that pentagons are impossible to fill all the space with a connected array of pentagons, for example, in Fig.
2, if one or more other congruent regular pentagons can fill all the space within the scope of 5 with no AugapsAy between them, the value of 5 must be the integral multiples of the value of an interior angle of the pentagon.
Nevertheless, from the calculation it can be found that the value of 5 is not an integral multiple of that of 5, therefore.
C5 axis do not exist.
Similarly, if assuming to substitute two congruent regular n-sided .
Ou7 ) polygons for the two pentagons in Fig.
2, the present calculation demonstrates that the value of n .
Ou7 ) is smaller than that of the interior angle of the n-sided .
Ou7 ) polygon, no possible to be its integral But differing from these cases, if assuming to substitute the two pentagons in Fig .
with two congruent regular triangles, or tetragons, or hexagons, it is easy to calculate out that3= 43, 4 =24, 6 =6 , they are consistent with eq.
Moreover, it is clear that C1 and C2 are compatible with translational symmetry .
Therefore, the possible axes of rotation of the crystal are merely C1.
C2.
C3.
C4 and C6 .
References