Spontaneous strain in BaTiO3

Glazer, A. M.; Cox, K. G.

doi:10.1107/97809553602060000633

International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

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International Tables for Crystallography (2006). Vol. D. ch. 1.6, pp. 174-175

Section 1.6.7.2. Spontaneous strain in BaTiO₃

A. M. Glazer^a ^* and K. G. Cox^b ^‡

^a Department of Physics, University of Oxford, Parks Roads, Oxford OX1 3PU, England, and ^bDepartment of Earth Sciences, University of Oxford, Parks Roads, Oxford OX1 3PR, England
Correspondence e-mail: glazer@physics.ox.ac.uk

1.6.7.2. Spontaneous strain in BaTiO₃

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As an example of the calculation of the relationship between spontaneous strain and linear birefringence , consider the high-temperature phase transition of the well known perovskite BaTiO₃. This substance undergoes a transition at around 403 K on cooling from its high-temperature $[Pm{\bar 3}m]$ phase to the room-temperature [P4mm] phase. The [P4mm] phase is both ferroelectric and ferroelastic. In this tetragonal phase, there is a small distortion of the unit cell along [001] and a contraction along $[\langle 100\rangle]$ compared with the unit cell of the high-temperature cubic phase, and so the room-temperature phase can be expected to have a uniaxial optical indicatrix .

The elasto-optic tensor for the $[m{\bar 3}m]$ phase is (Table 1.6.7.1) $[\pmatrix{ p_{11}&p_{12}&p_{12}&0&0&0\cr p_{12}&p_{11}&p_{12}&0&0&0\cr p_{12}&p_{12}&p_{11}&0&0&0\cr 0&0&0&p_{44}&0&0\cr 0&0&0&0&p_{44}&0\cr 0&0&0&0&0&p_{44} }. \eqno (1.6.7.5)]$ Consider the low-temperature tetragonal phase to arise as a small distortion of this cubic phase, with a spontaneous strain [S_3^s] given by the lattice parameters of the tetragonal phase: $[S_3^s = [(c - a)/a]. \eqno(1.6.7.6)]$ Therefore, the equations (1.6.7.4) for the dielectric impermeability in terms of the spontaneous strain component are given in matrix form as $[\eqalignno{\pmatrix{ \Delta\eta_1\cr \Delta\eta_2\cr \Delta\eta_3\cr \Delta\eta_4\cr \Delta\eta_5\cr \Delta\eta_6\cr} &= \pmatrix{ p_{11}&p_{12}&p_{12}&0&0&0\cr p_{12}&p_{11}&p_{12}&0&0&0\cr p_{12}&p_{12}&p_{11}&0&0&0\cr 0&0&0&p_{44}&0&0\cr 0&0&0&0&p_{44}&0\cr 0&0&0&0&0&p_{44}} \pmatrix{0\cr 0\cr S_3^s\cr 0\cr 0\cr 0} &\cr &= \pmatrix{ p_{12}S_3^s\cr p_{12}S_3^s\cr p_{11}S_3^s\cr 0\cr 0\cr 0 } &(1.6.7.7)}]$ so that $[\eqalignno{\Delta\eta_1 &= \Delta\eta_2 = p_{12}S_3^s &\cr \Delta\eta_3 &= p_{11}S_3^s &\cr \Delta\eta_4 & = \Delta\eta_5 = \Delta\eta_6 = 0. & (1.6.7.8)}]$ By analogy with equations (1.6.6.5) and (1.6.6.6), the induced changes in refractive index are then $[\eqalignno{\Delta n_1 &= \Delta n_2 = - {n_{\rm cub}^3\over 2} p_{12}S_3^s &\cr \Delta n_3 &= - {n_{\rm cub}^3\over 2} p_{11}S_3^s, & (1.6.7.9)}]$ where $[n_{\rm cub}]$ is the refractive index of the cubic phase. Thus the birefringence in the tetragonal phase as seen by light travelling along [x_1] is given by $[\Delta n_3 - \Delta n_2 = - {n_{\rm cub}^3\over 2} (p_{11} -p_{12})S_3^s .\eqno(1.6.7.10)]$ Thus a direct connection is made between the birefringence of the tetragonal phase of BaTiO₃ and its lattice parameters via the spontaneous strain. As in the case of the linear electro-optic effect , the calculation can be repeated using equation (1.6.3.14) with the susceptibilities $[\chi_{11}]$ and $[\chi_{12}]$ to yield the relationship $[p_{11} = {c_{1111}\varepsilon_o\chi_{11} \over n_o^4};\quad p_{12} = {c_{1122}\varepsilon_o\chi_{12} \over n_o^4 }. \eqno(1.6.7.11)]$