Slip systems in HCP metals

Chemist20

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!

Astronuc

{001}<100> would apply to cubic systems. One can slip on the cube faces, the diagonal in 2-dimensions (0,0,0) to (0,1,1), or the superdiagonal from (0,0,0) to (1,1,1).

Normally for hcp, there are 4 indices.

For hcp planes, there are the basal planes, then prismatic planes and pyramidal planes.
http://books.google.com/books?id=TX7...page&q&f=false

Chemist20

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.

M Quack

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).

Chemist20

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!!!!

Astronuc

[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}

M Quack

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.

Chemist20

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].....

M Quack

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.

Chemist20

I'm lost.. can you explain please?????

M Quack

(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.

Chemist20

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...

M Quack

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?

Chemist20

( 0 0 1) ?????????

M Quack

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?!?

Chemist20

woops okay. so im guessing the answer is 110 but i don't know why.!!!!!!

M Quack

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)