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Next: 3. Reciprocal Space and Dual Space Up: 2. Crystallographic Definition Previous: 2.1 Definition

2.2 Fundamental law of the reciprocal lattice

(a) with each node of the reciprocal lattice whose numerical coordinates have no common divider can be associated a set of direct lattice planes

Let M be a reciprocal lattice point whose coordinates h, k, l have no common divider (M is the first node on the reciprocal lattice row OM), and P a point in direct space. We may write:
\begin{displaymath}
\textbf{OM} = h\textbf{a*} + k\textbf{b*} + l\textbf{c*} \quad
\textbf{OP} = x\textbf{a} + y\textbf{b} + z\textbf{c} \end{displaymath} (2.5)

Let us look for the locus of all points P of direct space such that the scalar product $\textbf{OP} \cdot \textbf{OM}$ should be constant. It is a plane normal to O and passes through the projection H of P on OM (Fig. 2). Using 2.4, we find easily that the equation of this plane in direct space is given by
\begin{displaymath}
\textbf{OP} \cdot \textbf{OM} = \textbf{OH} \cdot \textbf{OM} = hx + ky + lz = C\end{displaymath} (2.6)

Let us now assume that P is a node of the direct lattice:

\begin{displaymath}
\textbf{OP} = u\textbf{a} + v\textbf{b} + w\textbf{c}\quad (u, v, w \hbox{ integers})\end{displaymath}

The locus of P is a lattice plane of the direct lattice. Its equation is:

\begin{displaymath}
\textbf{OH} \cdot \textbf{OM} = hu + kv + lw = C\end{displaymath} (2.7)


 
Figure 1:
\begin{figure}
\includegraphics {fig2.ps}
\end{figure}

Since all numbers in the left hand side are integers, we find that C is also an integer. With each value of C we may associate a lattice plane and thus generate a set of direct lattice planes which are all normal to the reciprocal vector OM (Fig. 3). The distance of one of these planes to the origin is given by:

\begin{displaymath}
OH = C/OM \quad (C = -2, -1, 0, 1, 2, 3, \dots)\end{displaymath}

If OH1 is the distance of the first plane to the origin, we may write:

\begin{displaymath}
OH = C \times OH_1\end{displaymath}

The lattice planes have, as expected, an equal spacing:
\begin{displaymath}
d_{hkl} = OH_1 = 1/OM = 1/{\cal{N}}_{hkl}\end{displaymath} (2.8)
where $\cal{N}_{hkl}$ is the parameter along the reciprocal lattice row OM. Equation 2.8 may be rewritten:

\begin{displaymath}
d_{hkl} \cdot {\cal{N}}_{hkl} = 1\end{displaymath} (2.9)

This is the fundamental relation of the reciprocal lattice which shows that with any node M of the reciprocal lattice whose numerical coordinates have no common divider we may associate a set of direct lattice planes normal to OM. Their spacing is inversely proportional to the parameter along the reciprocal row OM.

In order that the correspondence between direct and reciprocal lattice should be fully established, the converse of the preceding theorem should also be demonstrated. This will be done in paragraph 2.2(c).


 
Figure 2:
\begin{figure}
\includegraphics {fig3.ps}
\end{figure}

It is interesting at this point to give an interpretation to the reciprocal lattice points whose numerical coordinates have a common divider. Let us consider such a point for which:

\begin{displaymath}
\textbf{OM} = h\textbf{a*} + k\textbf{b*} + l\textbf{c**}\end{displaymath}

where

\begin{displaymath}
h = nh_1; \quad k = nk_1: \quad l = nl_1\end{displaymath}

h1, k1, l1, have no common divider. We may write:

\begin{displaymath}
\textbf{OM} = n\textbf{OM}_1\end{displaymath}

where M1 is the first node on the reciprocal lattice row OM.

Let dh1k1l1 be the spacing of the direct lattice planes associated with M. The fundamental law of the reciprocal lattice may be written:

\begin{displaymath}
d_{h_1k_1l_1} \cdot {\cal{N}}_{h_1k_1l_1} = 1\end{displaymath}

We may also write:

\begin{displaymath}
\frac{1}{n}d_{h_1k_1l_1} \cdot n{\cal{N}}_{h_1k_1l_1} = 1\end{displaymath}

(2.10)

\begin{displaymath}
\frac{1}{n}d_{h_1k_1l_1} \cdot OM = 1\end{displaymath}

In other words, with the reciprocal lattice node M may be associated a set of fictitious planes in direct space whose spacing is n times smaller than the real lattice spacing. We shall see that in diffraction by crystal lattices a reciprocal lattice point may be associated with each Bragg diffraction: if the coordinates of this point have no common divider, Bragg`s law is satisfied to the first order (2d sin $\theta$ = $\lambda$); if they have a common divider, n, Bragg`s law is satisfied to the nth order (2d sin $\theta$ = n$\lambda$), one may also say it is satisfied to the first order for the fictitious lattice planes of spacing d/n (2d/n sin $\theta$ = $\lambda$) and this is what is actually always done in practice.


(b) Miller indices

Let us consider one particular lattice plane of equation

hx + ky + lz = C

and let Q, R and S be its intersections with the three axes, respectively (Fig. 4); we have:

\begin{displaymath}
x = C/h; \quad OQ = a \cdot C/h \qquad
y = C/k; \quad OR = b \cdot C/k \qquad
z = C/l; \quad OS = c \cdot C/l\end{displaymath}

We conclude that the lattice plane intercepts, along the three axes, lengths which are inversely proportional to three integers which have no common divider. This is the so-called Law of Rational Indices or Hauy Law. The three indices are called the Miller indices.

The planes which are crystallographically the most important ones are the densest ones, that is those with the largest spacing. Equation (2.9) tells us that they are associated with the shortest vectors in reciprocal lattice and that their Miller indices are therefore small. This is the reason why Hauy's law was also called the law of simple rational indices.


(c) The reciprocal law: to each set of direct lattice planes corresponds a reciprocal lattice vector

Let us consider a set of direct lattice planes of equation:

hx + ky + lz = C

Since x, y, z may be integers, h, k, and l are also integers. If C = 1, corresponding to the first plane in the family, h, k and l have no common divider. Let us now consider the reciprocal lattice vector

\begin{displaymath}
\textbf{ON}_{hkl} = h\textbf{a*} + k\textbf{b*} + l\textbf{c*}\end{displaymath}

Its scalar products with the vectors QR and RS (Fig. 4) are respectively equal to:

\begin{displaymath}
\textbf{ON}_{hkl}\cdot \textbf{QR} = (h\textbf{a*} + k\textb...
 ...\textbf{C}}{k}\textbf{b}-
\frac{\textbf{C}}{h}\textbf{a}\right)\end{displaymath}

\begin{displaymath}
\textbf{ON}_{hkl}\cdot \textbf{RS} = (h\textbf{a*} + k\textb...
 ...\textbf{C}}{l}\textbf{c}-
\frac{\textbf{C}}{k}\textbf{b}\right)\end{displaymath}

They are both equal to zero, which shows that the reciprocal lattice vector is normal to the set of direct lattice planes; the scalar product of ONhkl by OP where P is any direct lattice node in a plane of the set can be written in the form of equation (2.6). The reciprocal theorem is thus demonstrated.


 
Figure 3:
\begin{figure}
\includegraphics {fig4.ps}
\end{figure}


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Next: 3. Reciprocal Space and Dual Space Up: 2. Crystallographic Definition Previous: 2.1 Definition

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