What Is the Shortest Reciprocal Vector for a BCC Lattice?

  • Thread starter Thread starter malawi_glenn
  • Start date Start date
  • Tags Tags
    Reciprocal Vector
AI Thread Summary
The discussion focuses on finding the shortest reciprocal vector G for a body-centered cubic (BCC) lattice, using a specific formula involving integer variables v_1, v_2, and v_3. The user attempts to minimize the length of the vector l(v_1, v_2, v_3) by taking its derivative and setting it to zero, leading to a linear equation system. However, the only solution found is the trivial case where v_1, v_2, and v_3 are all zero, which is incorrect. The user notes that valid solutions should yield reciprocal vectors resembling ±(2π/a)(x ± y). The goal is to correctly construct the first Brillouin zone for the BCC lattice.
malawi_glenn
Science Advisor
Messages
6,735
Reaction score
2,431

Homework Statement



Find the shortest reciprocal vector G, given below, v_1,...,v_3 are integers.

\vec{G} = \frac{2 \pi}{a}\left( (v_2 + v_3 )\vec{x} + (v_1 + v_3 )\vec{y} + (v_1 + v_2 )\vec{z} \right)

Homework Equations



x,y,z are ortonogal, length 1

l = l(v_1, v_2, v_3) = \vert \vec{G} \vert = \sqrt{\vec{G}\cdot \vec{G}}

l = \sqrt{ (v_2 + v_3 )^{2} + (v_1 + v_3 )^2 +(v_1 + v_2 )^2 }

The Attempt at a Solution



I want to minimize l(v_1, v_2, v_3)

\dfrac{\partial l}{\partial v_1} = \dfrac{2 \pi \left( (v_1 + v_3 ) + (v_1 + v_2 ) \right) }{\sqrt{ (v_2 + v_3 )^{2} + (v_1 + v_3 )^2 +(v_1 + v_2 )^2 }} = 0

etc. Gives me following linear equation system, it has only trivial solutions

<br /> \left( \begin{array}{ccc|c} 2 &amp; 1 &amp; 1 &amp; 0 \\ 1 &amp; 2 &amp; 1 &amp; 0 \\ 1 &amp; 1 &amp;2 &amp; 0 \end{array}\right)

v_1 = v_2 = v_3 = 0

And that is not true, they should be something like

\frac{2 \pi}{a} \left( \pm \vec{x} \pm \vec{y} \right)

etc.


Now what have I do wrong

by the way, this is the general reciprocal lattice vetctor for bcc lattice. I want to construct the first Brillouion zone.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top