Chapter 5. Liquid crystal cell alignment The static LC cell alignment is determined by the boundary conditions (on the glass surfaces) and the elastic deformation energy of the LC molecules. 5.1 LC director LC optics is determined by it director. The LC director is the direction of the rod shaped LC molecule. z Director φ θ y x The director angles are θ and φ. They are usually expressed as functions of z, the distance inside the LC cell. Azimuthal angle: Polar angle: θ θ(z) φ φ(z) (Note: 90 o - θ is called the tilt angle α.) 1
This is called the one-dimensional case (1D). The variation of θ and φ is only along the cell direction z. There is no x,y dependence. This is good enough for all TN, STN modeling. For active matrix displays, and for small pixel sizes, it is necessary to consider θ(x, y, z) and φ(x, y, z). They are called 3D cases, and are obviously much more complicated. Commercial softwares are available for both 1D and 3D simulations of LCD. 1D software: DIMOS from Autronics. LCDSoft from HKUST. Shintec 3D software: LIQUID from TMS. 3DIMOS from Autronics. Shintec Once θ(z) and φ(z) is known, the optical properties such as transmission etc of the LCD, can be calculated. So it is necessary to understand how θ(z) and φ(z) change as a function of applied voltage. We shall try to understand the director distribution, i.e. θ(z) and φ(z), under situations: 1. No voltage. The director distribution is determined by boundary conditions.. Applied voltage. The director distribution will deform, and the optical properties will change. With no voltage, the alignment of the LC cell is determined by the boundary conditions on the cell surfaces.
5. LC director alignment on the boundary surface: The alignment or director distribution inside the LC cell is determined by the boundary conditions of θ(z) and φ(z) at z0 and zd. Liquid crystals near the surface of the alignment layer are fixed in direction by the alignment layer. Different kinds of boundary conditions: (1) Homogeneous alignment. It is obtained by rubbing the alignment layer. It can also be obtained by UV photoalignment. The director is nearly flat with a small pretilt angle. The alignment layer is usually a layer of polyimide. The rubbing direction determines the alignment direction and the direction of the pretilt angle. Rubbing direction Pretilt angle Director The type of PI determines the pretilt angle. It is usually in the range of 1 8 o. Low pretilt are used in TN and high pretilt are used in STN. 3
() Homeotropic alignment: The director is nearly perpendicular to the cell surface. It is obtained by applying a surfactant to the alignment surface. This type of alignment condition is not used often. (3) It is also possible to obtain any pretilt angle, especially near 45 0 by evaporation of SiO at oblique angles. Obviously it is not amenable to mass production and is used mostly in research. Different methods of alignment: 1. Rubbing of PI. Linear polarized light photopolymerization (LPP) 3. SiO evaporation 4. Langmuir-Blodgett film: This is a layer of surfactant. Can be used to control the pretilt angle. Not amenable to manufacturing. Given the boundary conditions of θ(z) and φ(z), it is straightforward to obtain the static alignment condition of the LC cell. For example: To obtain left-handed 90 o twist TN: 4
90 o Rubbing direction of top plate Rubbing direction of bottom plate The director will have an anticlockwise twist as you look down at the cell. As will be explained later, this particular alignment will give the best 6 o clock viewing. Most alignments are obtained by rubbing the alignment layer. Alignment can also be achieved by other means, such as SiO evaporation, and more recently UV light irradiation of the surfaces. However, the most reliable method in manufacturing is still rubbing of cured PI. 5
5.3 Elastic deformation of the LC director: The director of a LC cell is like a spring. It has a preferred alignment which will minimize its elastic energy. There are 3 kinds of elastic deformations: bend, splay and twist. They affect important LC properties such as threshold voltage and response time. In the continuum theory, the Frank s free energy density for a deformed LC is given by F 1 K 11 ( n) + 1 K (n xn) + 1 K 33 nx xn n director K 11 splay distortion elastic constant K twist distortion elastic constant K 33 bend distortion elastic constant Note that this elastic energy is similar to a stretched spring: F 1 kx Usually, K 33 > K 11 > K and they are all ~ 10-1 N. For MBBA, K 33 /K 11 1.3 and K 33 /K.9. 6
Diagram to illustrate three kinds of director distortions: ( B. Bahadur Vol. 1 pg 166) If there is a natural twist (chirality) for the LC, then the twist term is biased by a permanent twist q o, and Frank s free energy becomes F 1 K 11 ( n) + 1 K (n xn - q o ) + 1 K 33 nx xn 1 K11 ( n) + 1 K (n xn) + 1 K 33 nx xn - q o K n xn + 1 Κ q o Noting that n (sin θ cos φ, sin θ sin φ, cos θ),it is easy to show that F can be written as 1 F k1 (θ) 1 θ + k (θ) φ + 1 k3 (θ) φ + 1 Κ q o 7
where k 1 (θ) K 11 sin θ + K 33 cos θ k (θ) (K sin θ + K 33 cos θ) sin θ k 3 (θ) q o K sin θ F depends on θ, φ which are functions of z. Such F is called a functional. The director distribution is determined by a minimization of F by varying the functions θ(z) and φ(z). This is the calculus of variations. Theorem: For any functional F which can be written as F F( θ, θ, φ, φ) d Fdz 0 can minimized by solving the Euler-Lagrange equations F θ d dz F θ F φ d dz F φ which are equations for θ(z) and φ(z). This is a general result and appears in many other branches of physics and mechanics. 8
d In the present case, Fdz is the total elastic energy stored in 0 the LC cell. Minimizing the integral means finding a stable equilibrium configuration. The Euler-Lagrange equations in general have to be solved numerically. They can be solved for some special cases. (1) One constant approximation. K 11 K K 33 K Here k 1 (θ) K k (θ) K sin θ k 3 (θ) q o K sin θ () No twist. Only splay-bend deformation. φ 0. (3) Twist only. No splay-bend. θ 0. 5.4 Some special examples: (1) Homogenous cell (H-cell) Also called the N n alignment. It is obtained by anti-parallel rubbing of alignment layers. Here θ(0) θ(d) 90 o α 9
Assume that there is no doping (q o 0). From the BC, it is obvious that the solution will have no twist. Therefore above case () applies. 1 F k1 (θ) θ Assume further that K 11 K 33 Then there is only one Euler- Lagrange equation for θ d dz ( K11 θ ) 0 The solution is θ(z) Az + B where A, B are constants to be determined by the BC. Using the BC given, we find that θ(z) 90 o - α for all z. This solution is obvious actually. However, if we do not make the one constant approximation, or if there is doping, even this simple case can become very complicated and needs numerical solution. 10
z α (). Homeotropic cell: It is also called the N p alignment. It is obtained by a homeotropic treatment of the alignment layers. The biundary condition is like a homogeneous cell with θ 90 ο, φ 0 for all z. 11
(3) Splay cell (S-cell) Obtained by parallel rubbing of the alignment layers. Here θ(0) 90 o α; θ(d) 90 o + α; z α The Euler-Lagrange equation is the same as before and the solution the same θ(z) Az + B for some A.B. Using the BC here, the solution is θ(z) 90 o + α( z d) d This is a splay cell as shown. There is only splay deformation. 1
4) Bend cell (B-cell) Also called the π-cell. The boundary conditions are the same as the S-cell. The difference is that the tilt angles are larger, so that a bend deformation will be preferred over the splay deformation. The bend deformation can also be obtained if we apply a constant voltage to a S-cell. The general solution is the same as the S-cell. θ(z) Az + B But now the BC can be written as Here θ(0) 90 o α; θ(d) 90 o + α 180 ο ; 13
The solution is therefore θ ( z) o (90 α)(1 z ) d Difference between the B-cell and the S-cell: θ 90 o -α 90 o +α z -90 o +α 14
(5) Twist cell (T-cell) The boundary conditions are φ(0) 0, φ(d) φ ο θ(0) θ(d) 90 o α For simplicity, we can assume that θ constant throughout the cell. (This is actually not strictly correct.) Then above case (3) is valid. Also, we shall use the one constant approximation. Therefore F 1 K (θ) φ + 1 K3 (θ) φ + 1 Κ q o The Euler-Lagrange equation reduces to d dz ( φ) 0 The solution is given by a linear variation φ(z) Az + B where A,B are to be determined by the BC. For a 90 o twist TN LCD, the director distribution is given by 15
θ θ ο φ(z) πz d This is the most common LCD. φ is twisted by 90 o. The tilt angle (90 o -θ) is constant. (6). Hybrid aligned cell (HAN cell): The boundary conditions are θ(0) 0; θ(d) 90 o φ(z) 0. 16
In the one constant approximation, the solution is the same as the B-cell or S-cell: θ(z) Az + B Substituting in the BC, we obtain: θ(z) πz d In general, the solution can be quite complicated. It is shown below: 90 o θ K 11 >K 33 K 11 <K 33 d z In general, K 11 <K 33. It is better to use numerical solutions: 17
General classification: Twist cell (T-cell): Picture of T-cell has been shown before. It is the most common kind. It should be noted that the T-cell can be made without any internal stress by adding chiral dopants, so the LC has a natural twist already. This is not the case for other kinds of distortions (S-cell and B-cell), where the internal elastic energy cannot be compensated. ECB cell All non-twist cells operate as a birefringent plate. Therefore they are all electrically-controlled birefringent (ECB) LCDs. A twist cell operates by polarization rotation. 18
The total birefringence of a non-twist cell (S-cell, B-cell, HAN cell) is given by: δ d π n( θ ( z)) dz λ 0 where we have emphasized the fact that the birefringence depends on the tilt angle. 5.5 Details on T-cells: 1. Pitch: Pitch is the distance covered when the director twists by 360 o (one complete cycle). If thickness d, twist angle φ, then pitch p is given by p πd φ e.g. a 5µm thick TN cell has a pitch of 0µm. If the twist is induced by the boundary conditions, then this is a forced pitch. The LC can be doped with a chiral dopant to have a natural pitch of 0 µm as well. In that case, there will be no elastic deformation energy. 19
. Right twist and left twist amibiguity: A RH 90 o twist cell can also have a 70 o LH twist under the same rubbing conditions. Similarly, a LH 180 o twist cell can also be a 180 o RH cell. This leads to anti-twist domains and grain boundaries defects. These defects can be eliminated by (1) doping and/or () having high pretilt angles. 3. Doping: For a twist cell, it is possible to add a chiral dopant to give the director a natural twist. There are dopants that give RH twist and there are some for LH twist. Pitch 1/(dopant concentration x HCP) HCP helical twist power e.g. S811 from Merck. HCP -10.1 to -11.6 /µm depending on the LC a 0.4% doping concentration > p 90 µm, LH twist In order to prevent reverse twist in TN displays, the induced helical pitch by the dopant should be 10-30 times the cell thickness d. e.g. d 6 µm > p 60 180 µm. Therefore 0.4% is OK. 0
Commercial dopants: Merck: S811 (LH): HTP -10.9 /µm CB15 (RH): HCP +6.5 /µm Roche (Merck): LH: CM5715A, CM9111S, CM909F, CN RH: CM5815C, CM907E For sophisticated use, have to mix + and dopants to compensate for temperature dependence. 4. d/p ratio: Here p is the pitch induced by the dopant, not the pitch induced by the cell boundary conditions. For the TN cell, p is 10-30 times d. Therefore d/p 0.03 0.1. For STN, it is better for the pitch to match the twist angle. For example, if the twist angle is 180 o, then the pitch should be 360 p d d 180 Therefore d/p 0.5 In actual practice, a d/p value of 0.5 0.5 is used for STN. This d/p affects the steepness of the electro-optic curve for the STN. The actual d/p value used also depends on K 11, K, and K 33, and other parameters. It affects not only reverse twist dislocations, it also affects the hysteresis and slope of the voltage response curve. Actually, it is possible to play with the d/p value to obtain bistable phenomenon the bistable twisted nematic (BTN) display. 1
Reference: H S Kwok, Z L Xie, T Z Qian and P Sheng, Study of Switching Behavior in Bistable Nematic Liquid Crystal Displays, Liquid Crystals, Proceedings of SPIE, pp -30, 1997. 5. 6 o clock and 1 o clock viewing: 90 o TN cells are made mostly for 6 o clock viewing. There are 4 possible ways to rub the alignment layers to give a LC director distribution that is symmetrical about the 6-1 o clock axis: Bottom glass Top glass Left handed twist Right handed twist The top configurations will give better 6 o clock viewing while the bottom will give good 1 o clock viewing. There are 4 other possible rubbing combinations, which will give director distributions that are symmetrical about the 3-9 o clock axis. Check these as an exercise! These configurations are used generally.
5.6 Bistability of LC alignment 1. BTN: bistable Twisted Nematic cells: For a twist cell with boundary conditions that favor a twist of φ, twist angles of φ±nπ should also be allowed, for any N. This ambiguity is usually eliminated in practical cells by doping the LC so that the natural pitch and the LC cell pitch are agreeable. For a cell with thickness d, the d/p ratio for the of φ±nπ twist state is d/p φ/π ± N Consider the states φ and φ+π, the elastic energy as a function of the natural twist (helicity) q o is given by F 1 k (θ) 1 K φ + 1 k3 (θ) φ + 1 Κ q o φ + q o K φ + 1 Κ q o 1 K (q o - d Φ ) where we have used the result for a T-cell z φ d Φ 3
Also d/p q o d/π F φ φ+π φ+π d/p Hence if the d/p ratio of the cell is between the values, bistability will occur, with both twist states equally possible. Known BTN with (-90 o, 70 o ), (0 o, 360 o ), (90 o, 450 o ) have been demonstrated by us.. Pi-cells: Bistable Bend/splay cells (BBS): A bend cell can also be bistable with a splay cell. For a large pretilt angle, the splay deformation may have the same elastic energy as a bend deformation. Hence bistability will occur. The total elastic energy for the S-cell or B-cell is 4
F d 1 dθ ( K11 sin θ ( z) + K33 cos θ ( z)) dz 0 dz If we take the one constant approximation, then the B-cell and the S-cell have the same elastic energy if α 45 o. This makes sense. In general, the integral has to be evaluated numerically to find the tilt angle for which the bend and splay energies are equal. In both the BTN and the BBS cells, the addition of a bias voltage will alter the energy balance and give interesting effects. 5
Chapter 6. Electromechanics of LC cells - Director deformation under an electric field: 6.1 Introduction When a voltage is applied to a LCD, effects can happen current effect and field effect. In early days with LCD operating in the dynamic scattering (DS) mode, there is current that generates turbulence to scatter light. Nowadays all LCD operates with field effect. Resistivity of LC material: ρ ~ 10 10 10 13 Ω cm with the low end used in TN displays and high end used in AMLCD. For example, a 6 µm TN segment display with a segment area of 3 mm will have a resistance of 00 MΩ. It is practically an insulator. The RC decay time constant is independent of the size of the segment or pixel. τ ρε r For ε r 10, and ρ 10 10, the RC time constant is 8.85 ms. Any voltage applied will decay quite rapidly, and constant refresh is needed. For AMLCD, this RC decay is too fast. 6. LC dielectric anisotropy In a LC, the dielectric constant is anisotropic, just like the refractive index: 6
ε ε x 0 0 ε 0 y 0 ε 0 0 z For LC, ε x ε y. It is customary to write ε ε x ε y, and ε // ε z. Then ε ε 0 0 0 ε 0 ε 0 0 // We can define the dielectric anisotropy as ε ε // ε. For most LC, -6 < ε < 50 Positive anisotropy: ε // > ε In this case, the LC would like to align parallel to the electric field at high voltages. (Homeotropic alignment). Most LCD are of this type. This can be seen in the above example for TN cell. At high fields, the midplane tilt is all 90 o. At the boundaries, the LC cannot move much due to strong anchoring effects. Negative anisotropy: ε // < ε 7
In this case, the LC would like to align perpendicularly to the electric field at high voltages. (Homogeneous alignment). The data sheet from Merck details the dielectric constants of various fluids. The dielectric anisotropy depends on T, λ, and ω. Same as birefringence n. But note that their dependence can be quite different as they are at very different frequencies. Below is an example of the T dependence of ε. 8
Note that it is possible to have a negative ε and a positive n. There is no conflict since they are defined for difference frequencies. Usually, large ε means large n as well. 6.3 Dielectric energy The dielectric energy stored in the LC cell is dependent on ε, which is dependent on the director distribution due to dielectric anisotropy. Hence the alignment of the LC cell depends on the applied field. The optical properties of the LC cell is determined by its director. Thus the optical properties of an LC cell can be controlled by the application of an external electric field. This is the electro-mechanical-optic effect. The effect of the electric field on the LC alignment is through the dielectric term. Recall from elementary electromagnetics that the electrostatic energy in an isotropic medium is given by U 1 ε E where ε is the dielectric constant. ε of a material is highly dependent on the polarizability of the molecules. Polar molecules are more polarizable. For example, ε of H O is 80. glass (silica) has a dielectric constant of 3.8. For an anisotropic medium, the electrostatic energy is better written as u 1 D E 9
30 Since there is cylindrical symmetry, we can assume that the LC lies on the x-z place as shown: θ x z x z In the principal axes (x,z ) of the LC molecule ' ' 0 0 ' ' // z x z x E E D D ε ε In the laboratory frame (x, z), this relationship is transformed to z x zz zx xz xx z x E E D D ) ( ) ( ) ( ) ( θ ε θ ε θ ε θ ε The laboratory frame and the molecular frame are related by a coordinate rotation by θ. It is easy to show that
ε xx (θ) ε zz (θ) ε sin θ + ε // cos θ ε xz (θ) ε sin θ cos θ ε zx (θ) -ε xz (θ) It is also common to write ε zz (θ) as ε zz ( θ ) ε // ε sin θ ( z) E Notice that because 0 Ey E Therefore x 0 z z Applying the boundary conditions yields E x E y 0. Therefore, only the z-component of E is nonzero. E E z (z) z However, D x does not have to be zero because D x ε xz (θ) E z (z) and D z ε zz (θ) E z (z) Additionally, D 0 31
D implies that z 0 z or D z (z) D constant Therefore from above E z ( z) D ε zz ( θ ) and u 1 D ε zz ( θ ) Therefore the total dielectric energy stored in the LC due to the applied voltage is given by F E udv d AD dz ε z 0 // ε sin θ ( ) The physics is that if an external voltage is applied, θ(z) should be such that F E is minimized. Let us consider the simple case that there is no elastic energy. The effect of minimizing F E is easy to see. 3
(1) If ε > 0, then F E is minimum if θ(z) 0 o for all z. This corresponds to the homeotropic alignment. Therefore the external electric field tends to move the LC molecules parallel to the field for positive anisotropic LC. For such cases, it is better to have the LC homogeneously aligned in the V0 condition. () If ε < 0, then F E is minimum if θ(z) 90 o for all z. This corresponds to the homogeneous alignment. Therefore the external electric field tends to move the LC molecules perpendicular to the field for positive anisotropic LC. For such cases, it is better to have the LC homeotropically aligned in the V0 condition Therefore the dielectric force tends to rotate the LC molecules depending on the sign of ε. This force interacts with the elastic mechanical force of LC alignment to produce the electro-mechanico-optic effect. Notice that the dielectric force is proportional tot he square of D, and does not depend on the sign of V. 6.4 Euler-Lagrange equations The alignment of the LC cell in the presence of an electric field is obtained by minimizing the total F. F 1 K 11 ( n) + 1 K (n xn) + 1 K 33 nx xn + 1 D ε zz ( θ ) The calculus of variation will still lead to the Euler-Lagrange equations: 33
F θ d dz F θ F φ d dz F φ The E-L equations are:.. k dk 1θ + θ dθ dk 3 d 1 φ D ( ) dθ dθ ε zz θ 1 dk φ dθ 0 k φ + k3 cons tant Commercial softwares are available to solve the Euler- Lagrange equations numerically to give θ(z) and φ(z) at any voltage. We shall look at some simple cases here and derive some general features of the E-L equation. The applied voltage is related to D and E z (z) by V d E( z) dz 0 d dz D ε zz( θ ( z)) 0 For homogeneous LC cell, Dd V ε Ed, since D ε Ε 34
For homeotropic LC cell, Dd V ε // Ed, since D ε // Ε There are a few constants of integration which are useful in solving the E-L equations: F θ F constant θ There are powerful softwares to solve the E-L equations. The 1D versions give θ θ(z,v) and φ φ(z,v) while the 3D versions give θ θ(x,y,z,v) and φ φ(x,y,z,v) There are several ways to visualize the dependence of the director orientation on V: (1) Complete data plot, () molecular plot, and (3) a plot of just the midplane tilt angle. They are shown below: Example: LC alignment of a 90 o TN cell at various voltages. (Both θ(z), φ(z) plots) Alpha tilt angle beta twist angle 35
Molecular plots: 36
37
6.5 Threshold voltage and Frederick transition: As the voltage is increased from 0 volts, the LC alignment does not always change immediately. There is a finite voltage when the alignment starts to change. This is called the threshold voltage. We can qualitiatively derive the threshold voltage for the case of H-cell with positive anisotropy. First, φ 0 for a ll z and V. Using the one constant approximation, the E-L equation for θ becomes K d dz θ D d dθ ε// 1 ε sin θ This can be simplified for small ε << ε // to K d θ D dz ε // ε sinθ cosθ Now d θ d θ θ dz d θ So we have a first order differential equation in θ with a solution given by 38
θ D ε Kε // cos θ m cos θ where we have applied the boundary condition that θ(d/) θ m. 90 o - θ m is called the midplane tilt angle. Hence, the solution for θ(z) is given by the integral: θ θm dθ D ε d z cos θ cos ε// K m θ In particular, at z d, we have the equation for θ m π / θm dθ D ε d cos θ cos ε// K m θ This equation can be used to solve for θ m for any applied voltage. If D 0, θ m 90 o which is always true. If D is large, then θ m should be near 0. As D decreases, θ m should approach 90 o. It can be shown that π / dθ lim θ 90 m θ m cos θ m cos θ π Hence the behavior of θ m is schematically like 39
θ m 90 o D Now for a H-cell, V ε Dd Hence the threshold voltage is given by ε π ε V th // K ε Above the threshold voltage, the LC director become distorted more and more and eventually becomes homeotropic. But there is always some residual anisotropy. This is the Frederick transition voltage. In general, for other deformations, ε π ε V // th Kii ε where K ii K 11 for B-cell K ii K for T-cell K ii K 33 for S-cell 40
or chiral nematic LCD, the threshold voltage is given by V th d K π p ε Interesting fact The HAN cell has no threshold. e.g. K 11 1.6x10-1 N, ε // 108x10-1 F/m, ε 4x10-1 F/m, V th 3.53 volts. Typical threshold for TN LCD is 1-1.5V. Usually it is listed on the catalogs, together with other LC parameters. Example of the mid-plane tilt angle dependence on voltage for a TN cell. Midplane tilt angle 90 60 30 0 0.00 1.00.00 3.00 4.00 5.00 Voltage 41
Notice that the midplane tilt has a lower threshold than the capacitance dependence on voltage. Also, the LC cell is just like a capacitor. It is possible to measure or calculate its capacitance, as a function of voltage. At low voltage, At high voltage, C ε A d ε A d C // Thus one can simply measure the capacitance as a function of V to get an idea of the tilt angle change, and also measure the threshold voltage. For example: Capacitance 3.00E-05.00E-05 1.00E-05 0.00E+00 0.00 1.00.00 3.00 4.00 5.00 Voltage 4
6.6 Response times: Another LC property that is closely related to the elastic constants is the viscosity ν, η. η ρ ν where η is the dynamic viscosity, ρ is the density and ν is the kinematic viscosity. LC density is typically 0.98 1.0 gm/cm 3, same as water. For high speed operation, need small viscosity. General rule More polar molecules > larger ν Longer Y, Z groups > larger ν More benzene rings > larger ν More side chains > larger ν Low temperature > large ν Also notice that large n requires more polar molecules. So large birefringence and higher viscosity goes hand in hand. Typical ν 5-100 mm /s at room temp. It can be larger than 5000 at low temp. The dynamics of LC molecule realignment due to external fields involves solving the Erickson-Leslie equations. The results for the H-cell and low twist TN cells are τ on ε ηd V Kii π and τ off ηd Kiiπ 43
The total response time is defined as τ τ on + τ off For high twist cells and cholesteric cells, the formulas are modified to: τ on η ε V K iiπ d p and τ off ηp Kiiπ where p is the pitch of the twist. Notice that the response time is proportional to d. Therefore very thin cells are needed for fast response time, e.g. for video applications. Also, strictly speaking, ν and η are also tensors dependent on the flow directions. Example: ν 100 mm /s, ρ 1 gm/cm 3, K 11 1.6x10-1 N, d 6 µm. Substitute into formula, get τ d 30 ms. This is a typical number for TN. Typical response times for LCD: STN 100-00 ms TN 10-50 ms H-cell -5 ms B-cell 0.5 ms Ferroelectric 0 µs The response time of ferroelectrics are very fast because it is based on a dipole force rather than a dielectric energy force, i.e. the force is first order in E rather than E. 44