Why are there only seven crystal systems and not 11

  • Thread starter Thread starter foges
  • Start date Start date
  • Tags Tags
    Crystal Systems
foges
Messages
52
Reaction score
0
I mean, the system is clearly related to seeing if the sides are of equal length and if the angles equal 90 degrees. If we first exclude the hexagonal, we have three sides and three angles each of which could be either equal/90degree. That gives us 3^2 = 9 possibilities.

Then looking at the hexagonal and with the same logic, the last length can either equal the two other or it can not, which gives another two possibilities.

The total then being 11 possibilities.. why is there for example no structure with a = b \neq c, \;\; \alpha = \beta = 90 \neq \gamma?
 
Physics news on Phys.org
The crystal systems are more precisely defined by their symmetry elements than by the sides and angles. You probably know that the monoclinic system has angles a=b=90, c=/=90 and sides A,B,C not equal. In fact I think it's better to say A,B,C not necessarily equal. The defining characteristic of the monoclinic system is actually that it has one diad, ie if you consider an axis normal to the plane of the angle c, you can rotate the crystal by 180 around that axis with no change. Now imagine making the lengths A and B equal (or even A, B and C). With a little thought you can convince yourself that the crystal does not gain any additional symmetry elements due to this change, since the angle c prevents any diads along the other axes, or otherwise. So it's still monoclinic. Note, however, that if you make the angle c equal to 90, you will be in the orthorhombic system. So that one is a strict criteria.
 
Ok, thanks makes sense now.
 
From the BCS theory of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly increase the temperature, we must observe a decrease in the actual supercurrent, because the density of electron pairs and total supercurrent momentum decrease. However, this supercurrent...
Hi. I have got question as in title. How can idea of instantaneous dipole moment for atoms like, for example hydrogen be consistent with idea of orbitals? At my level of knowledge London dispersion forces are derived taking into account Bohr model of atom. But we know today that this model is not correct. If it would be correct I understand that at each time electron is at some point at radius at some angle and there is dipole moment at this time from nucleus to electron at orbit. But how...
Back
Top