# Indices of Miller

The indices of Miller are a manner of indicating the plans in a Cristal.

A crystal is an ordered stacking of Atome S, Ion S or Molécule S, called hereafter “reasons”. Because of this order, one can define directions and plane private individuals, for example of the directions or plans having a great density of reasons.

The reasons being located on nodes of the network, the plans and directions are qualified the “nodal ones” (plane nodal, nodal direction). In Metallurgy, one frequently works with monoatomic crystals; one thus speaks about “atomic plan”, “atomic direction” or “line of atoms”, but they are only particular cases.

One can define a elementary Maille, and starting from this mesh three axes. The indices of Miller are coordinates of vectors in this bases.

## Importance of the plans and directions dense

The crystal is not isotropic, it does not have there reason which its properties it are. The lines and plans of large Densité will present particular properties:

• Optical S: the propagation of a light wave in the crystal (Refraction) is done by Diffusion Rayleigh gradually, between the atoms; the propagation velocity can thus differ according to the density of the direction, causing the phenomenon of Biréfringence;
• related to the surface Tension: if the material condenses in the shape of a crystal, it is that a reason is more stable when it is surrounded by other reasons;
• propagation of a Crack and plan of Cleavage: the reasons for a free face are exposed to the air; the free face is more stable if it corresponds to a plan of great density, because then each reason is surrounded by a maximum of reasons;
• form of a pore, for the same reason as above;
• Adsorption and reactivity: the number of sites of adsorption, and thus the chemical reactivity, depend on the density of atoms;
• Dislocation S
• the heart of a Dislocation more will extend in a dense plan, that reduces the Frottement during the displacement of dislocation (Force of Peierls-Nabarro during the plastic Déformation); the slips are thus done preferentially according to dense plans;
• the disturbance which a dislocation (Vecteur of Burgers) represents is a dense direction: indeed, a shift of a reason in a dense direction represents a weak ditorsion (reasons being brought closer);
• the line of a dislocation also will tend to being a dense direction, in order to decrease the Tension of line (a loop of dislocation will thus tend to be a Polygone).

## Location of a direction

A direction can be represented by a Vecteur. One chooses for that a vector having whole coordinates: a direction contains at least two reasons so that one can define his density, and the reasons are shifted of a whole linear combination of the vectors of the base. This direction is noted where, and are the whole coordinates. The negative numbers are noted with a feature above ( is read “less”):

• (or the opposite vector) indicates the large diagonal of the mesh;
• (or opposite) and (or opposite) indicates the two other diagonals;
• , and indicates the vectors of the base.
If one notes $\ vec \left\{has\right\}$, $\ vec \left\{B\right\}$ and $\ vec \left\{C\right\}$ the vectors of the base, then the indices of direction correspond to the vector
$u \ vec \left\{has\right\} + v \ vec \left\{B\right\} + W \ vec \left\{C\right\}$
, and are of the coordinates of the type contravariantes. ; Note
In the general case, the base $\left(\ vec \left\{has\right\}, \ vec \left\{B\right\}, \ vec \left\{C\right\}\right)$ is unspecified. It is a base Orthogonal E in the case of a network with orthorhombic symmetry or tétragonale, and orthonormal in the case of a network with cubic symmetry (see the article Réseau of Faced ).

## Location of a plan

Let us consider the plan nearest to the origin but which does not pass by the origin. If one takes the intersection of this plan with the three axes, one obtains the three coordinates of three points:

• the intersection of the plan with the axis of the X ;
• the intersection of the plan with the axis of the there ;
• the intersection of the plan with the axis of Z ;
then the reverse of the coordinates of the intersections gives the indices of Miller , with convention 1/∞ = 0 (the index is 0 if the axis is parallel with the plan). These indices are noted between bracket:
• ;
• ;
• .

If neither, neither, nor are null, then the plan thus passes by the points, therefore the following vectors are in the plan:

• $\ overrightarrow \left\{A_1 A_2\right\} \ left \left(- \ frac \left\{1\right\} \left\{H\right\}, \ frac \left\{1\right\} \left\{K\right\}, 0 \ right\right)$;
• $\ overrightarrow \left\{A_1 A_3\right\} \ left \left(- \ frac \left\{1\right\} \left\{H\right\}, 0, \ frac \left\{1\right\} \left\{L\right\} \ right\right)$;
• $\ overrightarrow \left\{A_2 A_3\right\} \ left \left(0, - \ frac \left\{1\right\} \left\{K\right\}, \ frac \left\{1\right\} \left\{L\right\} \ right\right)$.
These vectors not being colinéaires, two of these vectors form a base of the plan.

If one of the indices is null, then one of the vectors of the base of the mesh is also a vector of the plan, that whose nonnull component is the null index of the plan:

• if, the vector 0 0 (vector of coordinates (1,0,0)) is in the plan;
• if, the vector 1 0 is in the plan;
• if, the vector 0 1 is in the plan.

If the base is orthonormal , the scalar Produit vector with these vectors is null:

$\ left \left(- \ frac \left\{1\right\} \left\{H\right\}, \ frac \left\{1\right\} \left\{K\right\}, 0 \ right\right) \ cdot \left(H, K, L\right) = -1 + 1 + 0 = 0$
$\ left \left(- \ frac \left\{1\right\} \left\{H\right\}, 0, \ frac \left\{1\right\} \left\{L\right\} \ right\right) \ cdot \left(H, K, L\right) = -1 + 0 + 1 = 0$
$\ left \left(0, - \ frac \left\{1\right\} \left\{K\right\}, \ frac \left\{1\right\} \left\{L\right\} \ right\right) \cdot \left(H, K, L\right) = 0 -1 + 1 = 0$
Thus in the case of a cubic lattice, the vector is Perpendiculaire to surface, it in is a normal vector. In the general case, it is necessary to change basic so that the vector of coordinates is perpendicular to the plan (cf will infra ).

## Crystalline symmetries and permutation of the indices

Some crystalline structures have symmetry particular allowing the permutation of the indices.

### Crystal with cubic symmetry

For a crystal according to a Réseau of Faced cubic, the three diagonals are equivalent, the three faces of the cube are equivalent,…; one can thus permute the indices at will and to take the opposites of one or plusiers indices, that will représentra a direction immutably having the same properties.

The whole of the directions obtained by permutation is called “family of direction”, and is noted between hook:

indicates at the same time, as well as the combinations obtained by changing one or two signs.
For example
• <100> indicates the directions, and;
• <110> indicates the directions, and (like their opposite).

As for the directions, in the case of the crystals of cubic symmetry, one can permute the indices. A family of plans is then noted between accodances:

indicates the plans, as well as the combinations obtained by changing one or two signs.

### Crystal with hexagonal symmetry

In the case of the structures with symmetry hexagonal, or trigonal, one defines sometimes a fourth index to indicate the plans; it is the notation of Face-Miller. The index, placed in third position, is redundant (the three indices, and are alone enough to define a plan); it is defined by

This notation makes it possible to apply circular shifts of index to define families of plans.

In fact, if one considers the basic plan (001), this plan has a symmetry of order 3, i.e. it is invariant by a rotation of 1/3 of turn (2π/3 rad, 120 °). It thus contains three identical directions, and. If one takes the intersection of the plan with these three axes, the reverse of the X-coordinates of the intersections give the indices, and.

## Indexing of the peaks of diffraction

---- ; Note

By “peak”, we in the case of indicate not only the peaks of the diffractograms the numerical recordings, but also the tasks of diffraction in the case of diffraction on a monocrystal (Cliché of Laue, electronic Microscopie in transmission), as well as the rings of diffraction in the case of diffraction on a powder (Chambre of Debye-Scherrer).
See the article Theory of diffraction on a crystal.
----

In the experiments of diffraction with a Wavelength about the cell parameters (Diffraction of x-rays, Neutron diffraction, Electron diffraction in electronic Microscopy in transmission), the position of the peaks of diffraction can be calculated according to the distances interréticulaires, by the Loi of Bragg.

One can thus connect each peak to a plan; the indices of miller of the plan are the indices of the peak.

## Reciprocal space and diffraction

Let us consider the reciprocal Espace, i.e. the vector space formed by the vectors of wave; the use of this space makes it possible to determine the conditions of diffraction easily (see also the article Théorie of diffraction on a crystal ).

One defines the reciprocal base in it $\left(\ vec \left\{E\right\} ^*_1, \ vec \left\{E\right\} ^*_2, \ vec \left\{E\right\} ^*_3\right)$ by:

$\ vec \left\{e^*_1\right\} = \ frac \left\{1\right\} \left\{V\right\} \ cdot \ vec \left\{E\right\} _2 \ wedge \ vec \left\{E\right\} _3$
$\ vec \left\{e^*_2\right\} = \ frac \left\{1\right\} \left\{V\right\} \ cdot \ vec \left\{E\right\} _3 \ wedge \ vec \left\{E\right\} _1$
$\ vec \left\{e^*_3\right\} = \ frac \left\{1\right\} \left\{V\right\} \ cdot \ vec \left\{E\right\} _1 \ wedge \ vec \left\{E\right\} _2$
where V is the volume of the mesh $\left(\ vec \left\{e_1\right\}, \ vec \left\{e_2\right\}, \ vec \left\{e_3\right\}\right)$ which can be written:
$V = \ vec \left\{e_1\right\} \ cdot \left(\ vec \left\{e_2\right\} \ wedge \ vec \left\{e_3\right\}\right) = \ vec \left\{e_3\right\} \ cdot \left(\ vec \left\{e_1\right\} \ wedge \ vec \left\{e_2\right\}\right) = \ vec \left\{e_2\right\} \ cdot \left(\ vec \left\{e_3\right\} \ wedge \ vec \left\{e_1\right\}\right)$

According to the properties of the vector Product, one a:

$\ vec \left\{e_1^*\right\} \ cdot \ vec \left\{e_2\right\} = \ vec \left\{e_1^*\right\} \ cdot \ vec \left\{e_3\right\} = 0$, either $\ vec \left\{e_1^*\right\} \ club-footed \ vec \left\{e_2\right\}$ and $\ vec \left\{e_1^*\right\} \ club-footed \ vec \left\{e_3\right\}$
$\ vec \left\{e_2^*\right\} \ cdot \ vec \left\{e_3\right\} = \ vec \left\{e_2^*\right\} \ cdot \ vec \left\{e_1\right\} = 0$, or $\ vec \left\{e_2^*\right\} \ club-footed \ vec \left\{e_3\right\}$ and $\vec \left\{e_2^*\right\} \ club-footed \ vec \left\{e_1\right\}$
$\ vec \left\{e_3^*\right\} \ cdot \ vec \left\{e_1\right\} = \ vec \left\{e_3^*\right\} \ cdot \ vec \left\{e_2\right\} = 0$, is $\ vec \left\{e_3^*\right\} \ club-footed \ vec \left\{e_1\right\}$ and $\ vec \left\{e_3^*\right\} \ club-footed \ vec \left\{e_2\right\}$

In addition, if ( m , N , p ) is a circular shift of (1, 2,3), one a:

$\ vec \left\{e_m\right\} \ cdot \ vec \left\{e_m^*\right\} = \ frac \left\{1\right\} \left\{V\right\} \ cdot \left(\ vec \left\{e_m\right\} \ cdot \ vec \left\{e_n\right\} \ wedge \ vec \left\{e_p\right\}\right) = 1$

Let us note $\ vec \left\{K\right\}$ the vector having the coordinates ( H , K , L ) in this reciprocal base:

$\ vec \left\{K\right\} = H \ cdot \ vec \left\{e^*_1\right\} + K \ cdot \ vec \left\{e^*_2\right\} + L \ cdot \ vec \left\{e^*_3\right\}$
then this vector is normal in the plan ( HKL ): if neither H , neither K , nor L is null, then
$\ vec \left\{K\right\} \ cdot \ overrightarrow \left\{A_1 A_2\right\} = \left(H \ cdot \ vec \left\{e^*_1\right\} + K \ cdot \ vec \left\{e^*_2\right\} + L \ cdot \ vec \left\{e^*_3\right\}\right) \ cdot \ left \left(- \ frac \left\{1\right\} \left\{H\right\} \ cdot \ vec \left\{e_1\right\} + \ frac \left\{1\right\} \left\{K\right\} \ cdot \ vec \left\{e_2\right\} \ right\right)$
that is to say
$\ vec \left\{K\right\} \ cdot \ overrightarrow \left\{A_1 A_2\right\} = - \ vec \left\{e^*_1\right\} \ cdot \ vec \left\{e_1\right\} + \ frac \left\{H\right\} \left\{K\right\} \ cdot \ vec \left\{e^*_1\right\} \ cdot \ vec \left\{e_2\right\} - \ frac \left\{K\right\} \left\{H\right\} \ cdot \ vec \left\{e^*_2\right\} \ cdot \ vec \left\{e_1\right\} + \ vec \left\{e^*_2\right\} \ cdot \ vec \left\{e_2\right\} - \ frac \left\{L\right\} \left\{H\right\} \ cdot \ vec \left\{e^*_3\right\} \ cdot \ vec \left\{e_1\right\} + \ frac \left\{L\right\} \left\{K\right\} \ cdot \ vec \left\{e^*_3\right\} \ cdot \ vec \left\{e_2\right\}$
thus
$\ vec \left\{K\right\} \ cdot \ overrightarrow \left\{A_1 A_2\right\} = -1 + 0 - 0 + 1 - 0 + 0 = 0$
thus $\ vec \left\{K\right\} \ club-footed \ overrightarrow \left\{A_1 A_2\right\}$. One can show just as $\ vec \left\{K\right\} \ club-footed \ overrightarrow \left\{A_2 A_3\right\}$ and than $\ vec \left\{K\right\} \ club-footed \ overrightarrow \left\{A_3 A_1\right\}$. The $\ vec \left\{K\right\}$ is perpendicular to two vectors not colinéaires plan, it is thus normal in the plan. If H , K or L is null, one shows orthogonality while basing oneself on the fact that one or two of the axes is parallel to the plan.

However, the vectors having whole coordinates in the reciprocal base correspond to the conditions of diffraction. As follows:

• in the case of diffraction on a monocrystal (Stereotyped of Laue, electronic Microscopy in transmission), one can associate a task of diffraction with a crystallographic plan;
• in the case of a powder (Room of Debye-Scherrer, Diffractometer Bragg-Brentano), one can associate a ring of Debye or a peak of diffraction with a crystallographic plan.
One speaks thus about task, ring or peak ( HKL ). This association is called “the indexing”.

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