Talk:Diamond cubic

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[edit] Greater clarity

The article describes the positions of atoms in a unit cell as follows:

Atomic placement in unit cell of side length a is given by the following placement vectors.

\mathbf{r}_0 = \vec{0}

\mathbf{r}_1 = (a/4)(\hat{x} + \hat{y} + \hat{z})

\mathbf{r}_2 = (a/4)(2\hat{x} + 2\hat{y})

\mathbf{r}_3 = (a/4)(3\hat{x} + 3\hat{y} + \hat{z})

\mathbf{r}_4 = (a/4)(2\hat{x} + 2\hat{z})

\mathbf{r}_5 = (a/4)(2\hat{y} + 2\hat{z})

\mathbf{r}_6 = (a/4)(3\hat{x} + \hat{y} + 3\hat{z})

\mathbf{r}_7 = (a/4)(\hat{x} + 3\hat{y} + 3\hat{z})

No explanation is given for what the \hat{x}, \hat{y}, or \hat{z} mean.

Just in case they represent standard unit basis vectors, wouldn't it be a lot clearer if the article presented the coordinates of the atoms in a unit cell of side length = 1, without inventing mysterious symbols for the unit basis vectors? Also, they could be ordered in an organized manner. The list of points would then be as follows:

( 0, 0, 0),

(1/4, 1/4, 1/4),

(1/2, 1/2, 0),

(1/2, 0, 1/2),

( 0, 1/2, 1/2),

(3/4, 3/4, 1/4),

(3/4, 1/4, 3/4),

(1/4, 3/4, 3/4).

Daqu (talk) 06:23, 2 April 2008 (UTC)