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MD simulation of crystal oriented along 110, 111 planes

  1. Apr 9, 2010 #1
    Dear friends,
    I am doing a Molecular dynamics simulations of a crystal. For this the initial configuration is FCC (face centered cubic) lattice. My problem is that I have to study the system for different orientations of the crystal i.e. (100) plane, (110) plane and (111) planes. For (100), the initial configurations is simple. However, I would like to know, how I can create an initial configuration where the crystal is oriented along (110) plane or (111) plane?
     
  2. jcsd
  3. Apr 9, 2010 #2
    Are you creating a surface or interface or something? If it's just a bulk crystal it shouldn't matter what the orientation is.

    For FCC, your vectors are (0,a/2,a/2) (a/2,0,a/2) (a/2,a/2,0). You just need to rotate them so that you have the orientation you want. For instance if you want the (110) plane to be perpendicular to your x direction, then you need to rotate the vectors by 45 degrees. Construct a rotation matrix:
    [tex]
    \left[ \begin{array}{ccc}
    \cos \theta & -\sin \theta & 0 \\
    \sin \theta & \cos \theta & 0 \\
    0 & 0 & 1 \end{array} \right]
    [/tex]
    and multiply it by each of your vectors. Then if you want to construct a surface at (R,0,0) simply cut the generation of atoms where x > R.
     
  4. Apr 9, 2010 #3
    Hi Kanato,
    thanks for your reply. It has been very helpful.
    I actually want to create an interface and the z-axis should be perpendicular to the (110) plane. In another simulation, I want the z-axis to be perpendicular to the (111) plane.

    I still have three more questions.

    1.) So to generate the crystal with the desired orientation along z-axis, should I just rotate the unit cell vectors (0,0,0), (a/2,a/2,0),(0,a/2,a/2),(a/2,0,a/2) by the required angle and repeat them in x,y,z direction by the same lattice constant a or should the lattice constant be different? What exactly should be the basis vectors for (i) z-axis perpendicular to (110) plane and (ii) z-axis perpendicular to (110) plane

    2.)Secondly, for the (111) plane is the rotation matrix the same as what you have written or is it something else?

    3.)Thirdly, can you suggest some reference book or article which gives detailed information about this topic.

    Thanks in advance.

    Ronald
     
  5. Apr 9, 2010 #4
    The procedure is to take the vector (plane normal) you start with, and imagine a series of rotations that will bring it to the z direction. The rotation matrix I gave you [tex]R_z(\theta)[/tex] rotates about the z axis to bring the vector (1,1,0) to point in the (1,0,0) direction. Next you need a rotation matrix [tex]R_y(\theta)[/tex] about the y axis of 90 degrees to rotate the (1,0,0) vector to be (0,0,1), so the combination of these matrices would be [tex]R_y(\theta_2) R_z(\theta_1)[/tex] if you are using column vectors. These matrices don't commute, so the order they are applied is important. I suggest getting out Matlab or Octave and playing around with different rotation matrices to get a feel for it and to get the results you want.

    1) The lattice constant won't be any different. All we're talking about is orienting a coordinate system so you can create the cleavage you want.

    2) It will be a different matrix, and probably a little bit harder to find. Probably it will take two rotations, one to get it into the y-z plane (so its x-component is zero) and one to rotate it up to the z axis. For the first step you can take the product
    [tex]R_z(\theta)\left[ \begin{array}{c}1\\1\\1 \end{array} \right][/tex]
    and look at the resulting equation for the x component, solve for theta.

    3) I don't really know of one. The Arfken and Weber mathematical methods book has some review of it. Also some of the basics are here:
    http://en.wikipedia.org/wiki/Rotation_matrix#Dimension_three
     
  6. Apr 10, 2010 #5
    Hi Kanato,
    I guess I need to find the basis vectors and also the number of basic vector for the interface to be along (110) or (111) plane and the unit cell which I again guess will be cubic. Then I need to repeat these basis vectors along all three axis to generate the crystal with the required orientation.

    Thanks a lot for your help and detailed explanation. Hopefully, I will be able to build on your explanation and be able to solve the problem.
     
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