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Slip systems in HCP metals

  1. Mar 27, 2012 #1
    Hello,

    I'm having looots of trouble trying to find out the operative slip systems of hcp metals. the slip system is {001}<100> and in my notes it says that there are 3 slip systems and I don't see why!! I'm guessing there's only one slip plane the (001) or (0001), and three directions. But the three directions I'm calculating are [100]. [010] and [001], which in the 4-indices system are: [2 -1 -1 0], [-1 2 -1 0] and [0 0 0 1]. Is this correct????

    If it is correct, then only the first two are contained in the (001) plane, so there should be only 2 SLIP SYSTEMS!!!

    I'm lost!
     
  2. jcsd
  3. Mar 28, 2012 #2

    Astronuc

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    Staff: Mentor

  4. Mar 29, 2012 #3
    Yes, i know that for hco you usually use 4 indices, but you can also talk about 3, since these can converted into the 4-inices system! Eg: {001} is the same as {0001}. what i dont see is how do I know which directions are equivalente in hcp and therefore belong to the same family?

    Regards.
     
  5. Mar 29, 2012 #4
    The fourth index is only added (between the 2nd and 3rd) to make like planes look like.

    (H K L) contains all the necessary information, but often (H K (-H-K) L) is written.

    (1 0 0) would then become (1 0 -1 0).

    (1 0 -1 0), (0 -1 1 0), (-1 1 0 0) are all in the same star, which is easy to see in the 4-index notation.

    (1 1 -2 0) is in a different star with (-2 1 1 0), (1 -2 1 0).
     
  6. Mar 29, 2012 #5
    right, so basically if i change the order or sign of the indices, they still belong to the same family???

    then why do [100], [010] and [110] (in the 3 indices system) belong to the same family in hcp?!?!?!?!?!?!?!??!?! don't seee why [110] should belong to the same one as the other two!!!!
     
  7. Mar 29, 2012 #6

    Astronuc

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    Staff: Mentor

    [110], [101], [011] is a different system than [100], [010], [001]. Look at the number of zero indices, and nonzero indices.

    In hcp, basal planes are {0001}, primary prism planes {10-10}, and pyramidal planes {10-11}. Another pyramidal plane is {11-21}
     
  8. Mar 29, 2012 #7
    Yes, just remember that the c-axis (00L) or (000L) has a completely different symmetry.
    For HCP, you can change the sign, but this is not true for all hexagonal crystal classes.

    It is counter-intuitive because the two in-plane vectors a* and b* are not at right angles.

    Just take a piece of paper, draw a hexagon and sketch the vectors. You will see.
     
  9. Mar 29, 2012 #8
    Yeap I see. But i still dont know why in my book it says that the three directions in the <100> family in hcp metals are [100] [010] and [110].....
     
  10. Mar 29, 2012 #9
    You actually get 6 directions corresponding to the 6 corners of the hexagon. But HCP systems have inversion symmetry, so (110) is the same as (-1 -1 0) and this reduces to just 3 directions.
     
  11. Mar 31, 2012 #10
    I'm lost.. can you explain please?????
     
  12. Mar 31, 2012 #11
    (1 1 0) you are looking at the top of the plane, (-1 -1 0) you are looking at the bottom of the plane.

    If the plane is a symmetry plane, then that makes no difference. This is the case in HCP metals.

    In lower symmetry systems that might make a difference - say above the plane there are OH groups attached, and below F ions.
     
  13. Mar 31, 2012 #12
    Yes, I do understand that. But why is the -1 -1 0 direction in the same family as 1 0 0 for hcp????? That's what I don't get...
     
  14. Apr 1, 2012 #13
    The defining feature of all hexagonal crystal structures is a 3-fold rotation axis along c, i.e. if you rotate the whole crystal by 120 deg about the c-axis you end up with the same crystal structure.

    If you rotate (1 0 0) by 120 deg, you get (0 1 0).
    If you rotate (0 1 0) by 120 deg, what do you get?
     
  15. Apr 1, 2012 #14
    ( 0 0 1) ?????????
     
  16. Apr 1, 2012 #15
    Nope. Try again :-)

    (0 0 1) is the c-axis. You are rotating about the c-axis. How can any vector perpendicular to the axis or rotation become parallel to it?!?
     
  17. Apr 2, 2012 #16
    woops okay. so im guessing the answer is 110 but i don't know why.!!!!!!
     
  18. Apr 2, 2012 #17
    Oh well.. :-)

    Let [itex]\vec{a}=\left(\begin{array}{c}0\\ 0\\1 \end{array} \right)[/itex]

    Let [itex]C_3 = \left( \begin{array}{ccc} c & -s & 0 \\ s & c & 0\\ 0 & 0 & 1 \end{array} \right) [/itex]

    With s=sin(120 deg) and c= cos(120 deg).

    Calculate [itex] \vec{b} = C_3 \vec{a} [/itex] and [itex] \vec{d} = C_3 \vec{b} [/itex]

    Write d as linear combination of a and b.

    Show C_3^3 =1 and C_3 c = c with c=(0 0 1)
     
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