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Calculating particle velocity along crystal direction

  1. Oct 29, 2015 #1
    I am currently looking at setting up a particle, with energy 5keV, travelling along the [11bar2 0] direction in Zr but I am struggling to get my head around how to resolve the velocity into the 3 directions to make sure it travels along this specific crystal direction.

    Could anyone help point me in the direction to calculate this?

    Thanks in advance.
     
  2. jcsd
  3. Oct 29, 2015 #2

    Daz

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    You can simply drop the redundant index (that's why it's redundant) and what's left are the three conventional Miller indices that you're familiar with. I'm a bit hazy on the notation but I think the third index is normaly the redundant one. Try googling "redundant miller index" or "Miller-Bravais index".
     
  4. Oct 29, 2015 #3
    Ahh okay.
    I looked it up and the equivalent of [11bar20] is [110] but I'm not sure how to convert this into equivalent Vx, Vy and Vz values if my V is 1107ang/ps ? i.e. what kind of angles or conversions do I need to do so that vx^2+vy^2+vz^2 = V ?

    Thanks in advance
     
  5. Oct 29, 2015 #4

    Daz

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    Are Vx,Vy and Vz orthogonal (cartesian) co-ordinates? Wouldn't it be easier to work with basis vectos along (100), (010) and (001)? The first two lie in the basal plane and the third is perpendicular to it. By the way, do you mean [1 , 1bar , 2 , 0] or [1 , 1 , 2bar , 0] I guess it's the second as the first wouldn't make much sense?

    EDIT: Found this on the web: http://www.labosoft.com.pl/hex.pdf - that should help?
     
  6. Oct 29, 2015 #5
    It's the [1,1,2bar,0]. I wasn't sure if you could resolve the velocity using the basis vectors? If so, how do you resolve the original V into Vx etc? Is it V * cos (pi/2) = Vx?
     
  7. Oct 29, 2015 #6

    Daz

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    I edited my reply to include an illuminating link after you replied. Try this: construct the basis vectors of the Bravais lattice in cartesian co-ordinates. (Link above gives you the required info') Next take the vector sum of those basis vectors to get your required direction. Scale the length to the required value and you're there.

    EDIT: By the way, this ought to be in the solid state forum - maybe a moderator would oblige?
     
    Last edited: Oct 29, 2015
  8. Oct 29, 2015 #7
    Okay, I'll give that a go. What do you mean by scaling the length to the required value?
     
  9. Oct 29, 2015 #8

    Daz

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    Starting with a speed, V (5keV) and a direction (110) you want to find the components of V along 3 Cartesian axes, right? Combining the basis vectors gives you a vector pointing in the right direction, then you need to multiply by a scalar to make it’s length equal to V (scale it) which gives you the components.

    By the way, what is the structure of Zr? If it’s hexagonal, you might do better to sketch the unit cell on a scrap of paper and mark the principal axes. Draw a vector of length V along (110) and you will see that it’s decomposition into Vx and Vy is trivial. (If it’s rhombbohederal on the other hand, then stick to the method above.)
     
  10. Oct 29, 2015 #9
    Yes, Zr is hexagonal. So I can use the (110) as my direction instead of converting it into x,y and z.
     
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