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Bravais lattices in 2 dimensions (and 3 dimensions)

  1. Aug 21, 2015 #1
    I'm reading M. Omar Ali's Elementary Solid State Physics and in it, in Subsection 1.4 The Fourteen Bravais Lattices and the Seven Crystal Systems he says that "..., but one cannot place many such pentagons side by side so that they fit tightly and cover the whole area. In fact, it can be demonstrated that the requirement of translation symmetry in two dimensions restricts the number of possible lattices to only five (see the problem section at the end of this chapter)."

    However, the problem section does not explain it either.

    So, in simple terms, a Bravais lattice is just a mathematical way to describe all solid single-crystal structures. It is an idealization that depends on being able to describe every point in terms of appropriately scaled (by integers!) basis vectors. In 2D, there should be just two vectors because two noncolinear basis vectors are all that is needed to span two dimensions. The fact that there must be translational symmetry given by integer multiples of the basis vectors greatly reduces the total number of possible 2D Bravais lattices.

    But, how do you get 5? And in 3D, how do you get 14? I think starting with 2D for now makes more sense as I can't quickly sketch a rough proof in my head.

    Thanks, as always, for the help.
     
  2. jcsd
  3. Aug 26, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Sep 3, 2015 #3
    You start with the most general case, for 2D the angle is something odd (neither 90 nor 120 deg) and the two lattice parameters are different.

    Such a lattice has no special symmetry.

    Then you check what happens when you introduce rotational and mirror symmetries. What constraints do you impose on the lattice parameters?
     
  5. Sep 8, 2015 #4
    I think I don't know how to impose those constraints in a way.

    I'll try my best to interpret it.

    So, you have two lattice basis vectors ##a_1## and ##a_2## with an angle ##\theta## between the two.

    Constraint 1: Rotational symmetry. Meaning if the entire crystal is rotated by an angle ##\theta = 360^{\circ}/n## around an axis perpendicular to a lattice point in 2D, the same translational symmetry that existed before to get to any other point in the crystal remains i.e. given a
    transformation for each lattice point (need help with this part, sadly), the following is maintained.

    \begin{align*}\vec{r} &= c_1\vec{a_1} + c_2\vec{a_2}\end{align*}

    Constraint 2: Mirror symmetry. Meaning if the entire crystal is inverted through a line of mirror symmetry, the same translational symmetry that existed before to any other point in the crystal remains i.e. given a transformation for each lattice point (need help with this part, sadly), the following is maintained.

    \begin{align*}\vec{r} &= c_1\vec{a_1} + c_2\vec{a_2}\end{align*}

    But, I'm not sure firstly how to translate the constraints into transformations acting on some basis vectors. Second, while I can see how that would cull the set of valid lattices, I'm not sure how exactly that process happens mathematically.

    Thanks for giving me something to work with though. It is valued and appreciated.
     
  6. Sep 8, 2015 #5
    In general, the condition is that if you have a lattice vector G and a transformation C, then C(G) must be again a lattice vector (the same or a different one).

    For the 2D case you can write G in cartesian coordinates, and C as a 2x2 matrix. In the 3D case C can be written as 3x3 matrix.

    Since G = c1 a1 + c2 a2, it is sufficient to show that C(a1) and C(a2) are lattice vectors, i.e.

    C . a1 = c1' a1 + c2' a2
    C . a2 = c1'' a1 + c2'' a2
     
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