FCC has 3 fold symmetry but BCT doesn't. So why should we draw a BCT inside a FCC and name it as martensite. A reverse can also be done where we can draw BCT and they get a FCC out of it. If so in 14 bravais lattices there should be no BCT. How should we overcome this?
(1) This is the treatment of Bain correspondence. From `BCT` in FCC matrix, it can be transformed to BCC( sometimes BCT) martensite structure by pure strain (Bain strain or Bain correspondence), as shown in the attached figure. The draw is easy for us to understand how FCC structure transforms to BCC martensite.
(2) `T` in BCT stands for Tetragonal, it is one of the bravais lattices.
(3) For the `BCT` drawing from FCC, it is one of different way to draw the unit cell. The symmetry is unchanged and the same with FCC, and its symmetry is higher than the true BCT structure.
In the BCT drawn 3 fold symmetry is missing then how is the symmetry of FCC and this BCT is same?
You may be missing the point that there is an extra contraction on the vertical axes. Either you can make thta contraction until you get the same parameters for all sides of the cube (BCT) or you can get vertical cell parameter enough to accomodate tetragonality, either due to intrinsic characteristics of the material or to the presence of solutes, like Carbon.
I guess you say:
A reverse can also be done where we can draw BCT and they get a FCC out of it. If so in 14 bravais lattices there should be no BCT. How should we overcome this?
because you assumed it was just a mater of drawing lines between fixed points.
Yes Mr.Raúl Bolmaro a reverse can be done. When i connect the points in a BCT i can get FCC May i know the mistake in it?
Sorry but is not I who is in doubt about that, it is Mr. Xinfu Gu. I know a reverse process can be done.
Hi Bolmaro, I donot doubt about the statement (one have different way to plot different unit cell for the same structure), and just try to explain it. The drew *BCT* and FCC is the same lattice but different way of draw unit cell like I show in the first figure.
BTW, for general BCT structure, one can not draw FCC structure from it, because BCT has lower symmetry.
Oh, I see!. You mean the BCT you get just by drawing the cell different. But that is not what is known as the phase transformation on a steel for example. Any Bain transformation for a completely pure Fe material will contract along the vertical axes (vertical just to follow your drawing but it can be any of them) until it gets bcc. Going back will imply an elongation along the same axes. A high content of C will prevent that contarction to be enough to reach FCC symmetry and the cell will remain BCT but with cell parameters far away from the original ones (the ones you obtain just by drawing the cell different). Both BCT and FCC are undistinguishable only as far as you do not perform any change on that lengths. And that is true for many crystallographic Bravais lattices. If you do not contract that axes the real symmetry is the FCC no matter how you draw it. You are missing symmetries just because a wrong appreciation of the the real symmetry and wrong drawing. Imagine just 2 simpple cubic cells and draw a tetragonal cell including both of them, You might be confused and think that the cell is tetragonal with two atoms per cell. But it is not, you just missed the fact that one of the cell parameters is twice the other two and you have two SC cell.
Hi Mr.Raúl Bolmaro it's confusing. See initially you draw a BCT as shown by blue lines. Similarly draw many BCT in all directions. Now join the BCT cell to obtain FCC as shown by red lines. If my assumption to do like this correct. Then in bravais lattice there must not be BCT.
I am missing your poit Kedharnath Arumugakani. There are among the Bravais lattices one that is actually BCT. One of the conditions to remain as BCT is that the length of the longest axes should not be twice the length of the two shortest ones, otherwise you have a higher symmetry and you can index the structure as an FCC Bravais lattice.
You say that of when c>2a then a 3-fold symmetry is possible. No it is not. Only when c=a you can be able to draw an equilateral triangle inside a crystal structure or else you can draw only an isosceles triangle.
Sorry, I said:
One of the conditions to remain as BCT is that the length of the longest axes should not be twice the length of the two shortest ones, otherwise you have a higher symmetry and you can index the structure as an FCC Bravais lattice.
I missed a sqrt symbol.
If c remains equal to sqrt (2) * a´, where a´ is the new cell parameter of the new BCT, this new crystal is actually still an FCC crystal. The BCT is a missleading symmetry that will be better indexed as FCC because you are missing some symmetries, stemming from c=sqrt (a´). a´ meanwhile is equal to sqrt (2) * a/2, before you proceed to any contractio of the new c (vertical on the drawing) axes.
So, just make a drawing of two FCC side by side cells, draw an artificial cell as if it were BCC. If you do not contract the vertical axes, wich will become the new c axes, the diffraction will show the original FCC symmetry and not a BCT symmetry.
It is like taking a triclinic symmetry and making all angles = 90 deg, which is not forbiden a priori, in which case yo end up with a simple orthorrombic cell, but you will not see many of the expected peaks because you are imposing more symmetries by making the angles 90 deg.
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