Elastic Response of Wire Frame Glasses. II. Three Dimensional Systems
EElastic Response of Wire Frame Glasses. II. 3D Systems
Elastic Response of Wire Frame Glasses. II. Three Dimensional Systems
David A. King, a) Masao Doi, and Erika Eiser Cavendish Laboratory, University of Cambridge, J. J. Thomson Ave., Cambridge CB3 0HE,UK Centre of Soft Matter and its Applications, Beihang University, Beijing 100191, China (Dated: 5 February 2021)
We study the elastic response of rigid, wire frame particles in concentrated, glassy suspensions to a step strain byapplying the simple, geometric methods developed in part I. The wire frame particles are comprised of thin, rigid rodsof length L and their number density, ρ , is such that ρ L (cid:29)
1. We specifically compare rigid rods to L-shapes made oftwo equal length rods joined at right angles. The behaviour of wire frames is found to be strikingly different from thatof rods. The linear elasticity scales like ρ L for L-shaped particles, whereas it scales proportional to ρ for rods and thenon-linear response shows a transition from shear hardening to shear softening at a critical density ρ c ∼ (cid:112) K / k B T L ,where K is the bending modulus of the particles. For realistic particles made of double stranded DNA, this transitionoccurs at densities of about ρ L ∼
10. The reason for these differences is that wire frames can be forced to bend by theentanglements with their surroundings, whereas rods always remain straight. This is found to be very important evenfor small strains, with most particles being bent above a critical strain γ c ∼ ( ρ L ) − . I. INTRODUCTION
In the previous paper (part I) , we introduced a simple ge-ometric method for determining the initial stress response toa step strain of a dense, glassy suspension of wire frame par-ticles. In part I, the method was applied to a two dimensionalmodel system which could be treated easily. That modelshowed that there was a significant difference between straightrod-like particles and bent or branched wire frames in both themagnitude of the linear elastic response and character of thenon-linear behaviour. Here, we extend our discussion to threedimensional systems.The wire frame particles are comprised of connected, thin,rigid rods of length L . The joints between the rods are taken tobe effectively rigid. We are specifically interested in compar-ing suspensions of straight rods, as in Fig.(1a), to suspensionsof bent and branched wire frames, such as L-shapes and 3DCrosses shown in Figs.(1b&c) respectively. For simplicity,in this paper we will often only consider simple wire frameshapes, such as L-shapes, but the general conclusions will ap-ply to other, more complicated, bent and branched shapes.We consider number densities, ρ , in the range 1 / L (cid:28) ρ (cid:28) / V exc , where V exc is the extremely small excluded volumeof the particles. This means the most important interactionsbetween the particles are their topological entanglements, andthere is no long range orientational order. In this concentrationrange, the entanglements lead to glassy dynamical behaviourfor the wire frames , where the particles are effectively immo-bile, constrained to only a small set of positions and orienta-tions at any given moment. This is not the case for rods whichcan still diffuse via the reptation mechanism . This crucialdifference is expected to lead to markedly different flow be-haviour, which has been observed in simulations . The strik-ing differences in the 2D model of part I are also expected tocarry over to 3D. Understanding and predicting the differencesin behaviour of suspensions of these particles has relevance a) Electronic mail: [email protected] for the design of DNA nanostars as functional materials . Itis also an interesting theoretical problem since the well knownmethods for finding the stress in concentrated suspensions ofrods based on the tube model may not be used, as they relyon the rods being able to diffuse via reptation. As such, a newmethod is required which determines the stress directly fromthe the kinetic constraints placed on the wire frame particles.We briefly review the general methods introduced in part I be-fore applying them in three dimensions.Our goal is to calculate the initial elastic response of a densesuspension of wire frame particles in a glassy state to a step,shear strain. We do this by considering the change in free en-ergy, F , under shear deformation. This is related to the shearelastic stress, σ , by the virtual work principle, σ = ∂ F ∂ γ , (1)where γ is the shear strain.In the glassy state, the translational diffusion coefficient forthe wire frames goes to zero. Therefore we treat their centresof mass as fixed. We denote the orientation of a test particleby Γ , which in three dimensions may be represented by threeEuler angles. The entanglements between the particles meanthat each one may only access a small region of the config-uration space. The volume of this region is for a given testparticle in orientation Γ is Ω ( Γ , C ) , where C specifies the con-figuration of the constraints. The entropy associated with thetest particle follows from the Boltzmann definition, S ( Γ , C ) = k B T log Ω ( Γ , C ) . (2)When the step strain is applied, all of the surrounding particlesmove, and the volume of configuration space accessible to thetest particle changes to ˜ Ω ( γ ) . This will change the entropy,and therefore the free energy, which leads to the elastic stress.A similar method was introduced by Edwards to study flexiblepolymers with topological constraints and crosslinks . Asthe surrounding particles relax to equilibrium after the appli-cation of the strain, the test particle can explore more config-urations and the stress in turn relaxes. We are interested in the a r X i v : . [ c ond - m a t . s o f t ] F e b lastic Response of Wire Frame Glasses. II. 3D Systems 2 FIG. 1. Sketches of the systems considered. (a) A dense suspension of rod like particles. A test rod is shown in cyan. The motion of the testrod is restricted by the surrounding red rods, but it can still diffuse along its length, so the suspension is still in a fluid state. Panels (b) and (c)show examples of dense suspensions of wire frame particles. (b) shows a test L-shaped particle in cyan and a 3D cross particle is shown in(c). These particles are completely trapped by their surroundings, shown in red. If they moves along the length of one of their legs, the otherbecomes entangled. The system is frozen in a glassy state. magnitude of the elastic stress immediately after the deforma-tion. Therefore we only need to consider the instantaneouschange in the free energy.If there are ρ particles per unit volume, the contribution tothe free energy per unit volume from the instantaneous changein entropy is, F S ( γ ) = − k B T ρ (cid:90) d Γ ψ ( Γ ) (cid:68) log ˜ Ω ( Γ , C ; γ ) (cid:69) , (3)where ψ ( Γ ) is the orientational distribution function, whichmay be taken to be uniform since the system in a disorderedstate, and (cid:104)· · · (cid:105) denotes averaging over all realisations of thesurrounding constraints, C .Equation (3) is only the entropic contribution to the freeenergy and does not include any change to the internal energywhen the shear is applied. As was shown in part I, this is onlyvalid for the linear elasticity and the change in internal energymust be considered for the non-linear response. We considerthe system in the absence of an external potential, thereforethe internal energy only changes if the particles themselves arebent. In part I, we showed that this is only possible for wireframes, not rods. This bending mechanism must be includedto understand the non-linear elasticity of wire frame glasses.To account for the possibility of bending, we introduce thefunction, P ( γ ; Γ ) . This is the probability that a particle in con-figuration Γ has not bent at strain γ . In part I we argued fromsimple principles that this function should be of the form P ( γ ; Γ ) = − γ γ c ( Γ ) , (4)for small γ . The quantity γ c ( Γ ) should be interpreted as thecritical strain above which all particles in configuration Γ havebent. The bending contribution to the free energy can be de-termined using (4).At an applied strain of γ , a particle in orientation Γ con-tributes bending energy if it first bent at a strain γ (cid:48) < γ . The probability of this happening is, − ∂ P / ∂ γ evaluated at γ (cid:48) . In this case, the particle will have bent through an angle ϑ = ( γ − γ (cid:48) ) ∆ ( Γ ) , where ∆ depends on the particle geome-try. The elastic energy associated with this bending is K ϑ / K is the bending modulus of the particle. The bendingcontribution to the free energy can then be determined by av-eraging over the orientations of the particle and the strains atwhich it first bends, F B ( γ ) = − K ρ (cid:90) d Γ ψ ( Γ ) ∆ ( Γ ) (cid:90) γ d γ (cid:48) ( γ − γ (cid:48) ) ∂ P ( γ ; Γ ) ∂ γ (cid:12)(cid:12)(cid:12)(cid:12) γ (cid:48) . (5)Using the general from (4), the γ (cid:48) integral can be taken to give, F B ( γ ) = K ργ (cid:90) d Γ ψ ( Γ ) ∆ ( Γ ) γ c ( Γ ) . (6)As this is proportional to γ , this contribution is only rele-vant for the non-linear elasticity. The total free energy includ-ing both the entropic and the bending contributions, valid to O ( γ ) then follows, F ( γ ) = − k B T ρ (cid:90) d Γ ψ ( Γ ) P ( γ ; Γ ) (cid:68) log ˜ Ω ( Γ ; γ ) (cid:69) + K ργ (cid:90) d Γ ψ ( Γ ) ∆ ( Γ ) γ c ( Γ ) . (7)It is important to note that the first term in (7) now must con-tain a factor of P compared with (3). This is because when theparticle bends, its orientation is completely determined by theconstraints and therefore cannot contribute entropically.With equation (7) for the free energy we may determinethe elastic stress for the wire frame glass from purely geomet-ric considerations. In three dimensions it is difficult to makeprogress exactly without resorting to heavy algebra. In thispaper we aim to explain how the key features of the 2D modelfound in part I carry over to 3D, using simple arguments basedlastic Response of Wire Frame Glasses. II. 3D Systems 3on the understanding gained from part I. We construct a sim-ple model of the 3D systems in the next section. This is used tounderstand the bending mechanism in section III, which is thecrucial difference between wire frame particles and straightrods. This understanding is used to compute the linear elasticresponse for rods and wire frames in section IV. The consis-tency of the model is tested by comparing the result for rigidrods to the well known Doi & Edwards theory and we findexact agreement to leading order in density. The non-linearelastic response is considered in section V before our resultsare discussed in reference to real DNA nanostar systems insection VI. II. 3D MODEL
The model we will use for these 3D systems is constructedby analogy to that used in part I for 2D systems. We considera single test particle whose centre is fixed at the origin. Therotations of the particle about its centre are restricted by legsof surrounding particles which pierce a sphere of radius L cen-tred at the origin, surrounding the test particle. We representthese intersections by points on the surface of the sphere, theposition vectors of which are given by L v i , where v i is a unitvector. The configuration of the constraints is determined bythe set of these unit vectors, C = C ( v , v , · · · ) . The intersec-tion points may be connected to each other to form triangu-lar “cells” tessellating the surface of the sphere. Each leg ofthe test particle is then constrained to lie within one of thesecells. This is sketched in Fig.(2), with an L-shaped, test parti-cle shown in blue, the intersection points shown in red joinedby red lines to form the cells. The cell occupied by one legof the test particle is shown shaded in red. The unit vectors v i determine the vertices of the cells and can be used to find theaccessible volume of configuration space, which is written, Ω ( Γ ; C ( { v i } )) . The orientations of each leg of the particle areidentified as the centres of the cells which they respectivelyoccupy, which determines Γ .When the shear transformation is applied, we assume thateach intersection point is moved affinely so that the unit vec-tors v i transform according to, v i → ˜ v i = ( I + κ ) · v i | ( I + κ ) · v i | , (8)where κ is the strain tensor. Each cell then changes shapeaccordingly and the accessible volume of configuration spacetransforms; Ω (cid:0) Γ ; C ( { v i } ) (cid:1) → ˜ Ω (cid:0) Γ ; ˜ C ( { ˜ v i } ) (cid:1) (9)This formulation of the problem can be used for detailed cal-culations but it also allows for the important physical featuresto be determined from simple considerations. III. BENDING PROBABILITY
In part I it was shown that the most important difference be-tween straight, rod-like particles and general wire frames was
FIG. 2. A sketch of the 3D model we consider. The intersectionsof legs of surrounding particles with the sphere encircling the blue,L-shaped, test particle are shown as red points on that sphere. Theposition vectors of these points are, L v i . These points are connectedtogether by red lines to make triangular cells which tessellate thesurface of the sphere. The legs of the test particle are constrainedto lie in one of these cells. The cell occupied by one leg of the testparticle in this case is shown shaded in red. The identification ofthe cells occupied by the particle totally determines its orientationas well as Ω , and all properties of the cells are determined by thevectors of its vertices, v i . that the wire frames can be forced to bend by the surroundingswhen the shear stress is applied, whereas rods always remainstraight. This is quantified by the function P ( γ ; Γ ) , as intro-duced in (4). This is always one for rigid rods, because thetransformation (8) is one to one. This means all the vertices ofeach cell tessellating the surface of the sphere are transformedto distinct points, and no cell vanishes or overlaps with an-other. The value of Ω for a rod is always given by the surfacearea of one of these cells, and is therefore never zero or neg-ative; bending is impossible. We wish to estimate the criticalstrain γ c above which a general wire frame particle will beginto bend.To make this estimation we may consider a simple particleshape. A particularly illustrative choice is a ‘kinked’ particle,a generalisation of an L-shape where the two equal length legsare joined at an angle χ . This kind of particle was studied indetail in part I. Before the strain is applied, each leg lies ina separate cell. Let us define the two vectors pointing to thecentre of each cell as u and u . Initially these make an angleof χ to each other, cos χ = u · u . (10)This situation before the shear is sketched as a cross sectionin the plane of the test particle in Fig.(3a).When the shear transformation is applied, the two unit vec-tors change according to the rule (8). This changes the anglebetween them to χ + ∆ χ , defined by,cos ( χ + ∆ χ ) = ˜ u · ˜ u . (11)The change in the angle will be a function of the applied strain, ∆ χ ( γ ) , as well as the orientation of the particle which is de-lastic Response of Wire Frame Glasses. II. 3D Systems 4 FIG. 3. (a) A sketch of configuration of a kinked, test particle withopening angle χ . Shown before the step shear is applied and as across section in the plane of the particle. Both legs of the particleare constrained to lie in the shaded red cells which each have angularsize ∆ θ . The central vectors of each cell are u and u which makean angle of χ to each other. (b) After the strain is applied, the centralvectors are transformed to ˜ u and ˜ u and the angle between them isnow χ + ∆ χ . The average angular size of a cell after the shear is thesame as before, (cid:102) ∆ θ ≈ ∆ θ . If the change ∆ χ is bigger than ∆ θ onaverage, then the test particle cannot maintain its original shape andit must bend. scribed by the two unit vectors u and u . If, on average, thesize of this change in angle is larger than the average angularsize of a cell, ∆ θ , then the particle cannot maintain its origi-nal shape. This is shown in Fig.(3b). Let us define the averagesize of ∆ χ as its root mean square value, ∆ χ ( γ ) = (cid:113) (cid:104) ∆ χ ( γ ) (cid:105) , (12)the average here is taken over an isotropic distribution for thevectors u and u . The particle will be forced to bend if ∆ χ > ∆ θ , hence the critical strain for bending is estimated from, ∆ χ ( γ c ) ∼ ∆ θ . (13)The average squared change in angle is found by assuming ∆ χ ∼ γ , expanding both sides of (11) and matching them tofirst order in the strain so that, ∆ χ sin χ = κ αβ ( u α u β + u α u β ) cos χ − ( κ αβ + κ βα ) u α u β , (14)here and henceforth summation is implied over repeated in-dices. From which it may be shown that (see the appendix fordetails), (cid:104) ∆ χ ( γ ) (cid:105) = γ sin χ . (15)The average angular size of a cell will be approximately thesame both before and after the shear transformation. This canbe estimated straightforwardly by using an argument similarto that used to find the tube radius for the reptation of rods .We define ∆ θ as the average angle through which the particlemust be rotated to first come into contact with a surroundingparticle. When the particle is rotated about any axis each of its legs sweep out a plane. The constraints on the test parti-cle are imposed by particles intersecting this plane. The area, a , of the region swept out by rotating the particle by ∆ θ isapproximately, a ∼ L ∆ θ . (16)Therefore, the average number of particles intersecting thisregion can be approximated as, N ∼ ρ La ∼ ρ L ∆ θ . (17)Choosing ∆ θ so that N ∼
1, determines the average angularsize of each cell, ∆ θ ∼ ρ L . (18)Then using (12), (15) and the condition (13) for the criticalstrain, we have γ c ∼ ρ L | sin χ | . (19)This is a very illuminating result, showing explicitly the dif-ference between straight rods and bent wire frame particles.When the rod is straight, χ → , π , this critical strain diverges.This shows that rods never bend, no matter the applied strain.On the other hand, if the particle is even slightly kinked,then the critical strain decreases to an extremely small value ∼ ( ρ L ) − . We can estimate for which particle shapes thebending mechanism becomes important by finding the open-ing angle χ c where γ c ∼
1. In the concentrated limit, this trans-lates to, χ c ∼ ρ L . (20)So for particles with opening angle χ (cid:38) χ c , L-shapes or 3Dcrosses for example, the suspensions behaviour will be dom-inated by the bending mechanism. In the next sections it willbe shown that this is responsible for markedly different be-haviour of suspensions of wire frame particles compared tosuspensions of rods. IV. LINEAR RESPONSE
Initially we focus on the linear response of these systems.This requires computing the entropic free energy as in (3).The additional factor of P ( γ ; Γ ) included in equation (7) is notneeded here as it only effects the non-linear stress. To checkthe accuracy of the model we have constructed, we compareits results for rods to the well known Doi & Edwards theory ,which is based on the tube model. We then discuss the linearelasticity for wire frame particles, by using the simple exam-ple of an L-shaped particle.lastic Response of Wire Frame Glasses. II. 3D Systems 5 FIG. 4. A sketch of a test rod, shown in blue, with orientationparallel to u which is constrained by the surroundings to lie in the redcell whose area is Ω ( u ; C ) . The vertices of the cell are at positionsgiven by v , v and v . The area Ω can be determined by first findingthe volume, V of the green polyhedron whose vertices are the originand those of the red cell. A. Rods
For rod-like particles, the model can be used to calculatethe linear elastic stress exactly, to leading order in the den-sity. The orientation of the rod is specified by the unit vector u running parallel to its length. The constraints on the parti-cle are defined by the cell, C , which it occupies. This cell isdefined by the three vectors of its vertices, v , v and v . Theaccessible volume of configuration space is written, Ω ( u ; C ) .When the shear transformation is applied, the vertices of C move according to the transformation rule (8) and the acces-sible volume changes to ˜ Ω ( u ; ˜ C ) .To determine the entropy, we must compute the ratio ˜ Ω / Ω .This can be calculated by considering the change in volumeof the polyhedron whose vertices are at the origin, v , v and v . The polyhedron is sketched in Fig.(4). Using simple ge-ometry, the volume of this polyhedron before the shear is, V = (cid:90) d Ω (cid:90) | u | dr r ≡ | u | Ω ( u ; C ) . (21)After the shear is applied the volume changes by definition to,˜ V = det ( I + κ ) V . (22)This can also be written as,˜ V = | ( I + κ ) · u | ˜ Ω ( u ; ˜ C ) . (23)Dividing (23) by (21) and using (22) yields the ratio,˜ Ω ( u ; ˜ C ) Ω ( u ; C ) = | u | | ( I + κ ) · u | det ( I + κ ) . (24)To calculate the entropy, we need to take the logarithm of thisratio,log ˜ Ω ( u ; ˜ C ) Ω ( u ; C ) = | u | − | ( I + κ ) · u | + log det ( I + κ ) . (25) The final term of this can be treated using the identity,log det A = Tr log A , which holds for any square, non-singularmatrix A . This yieldsTr log ( I + κ ) = −
12 Tr κ + O ( κ ) . (26)The first order term in κ in this expansion vanishes because ofthe incompressiblilty condition, Tr κ = κ ,log ˜ Ω / Ω = − u · κ · u − (cid:2) | κ · u | − ( u · κ · u ) (cid:3) −
12 Tr κ . (27)The entropic free energy can now be obtained by averagingover u , which is written in components as, F s ( κ ) k B T = ρκ αβ (cid:104) u α u β (cid:105) + ρκ µα κ µβ (cid:104) u α u β (cid:105)− ρκ αβ κ µν (cid:104) u α u β u µ u ν (cid:105) − ρκ αβ κ βα . (28)To lowest order in density, the distribution of u can be takento be isotropic. In which case the averages are, (cid:104) u α u β (cid:105) = δ αβ , (29a)and (cid:104) u α u β u µ u ν (cid:105) = ( δ αβ δ µν + δ αµ δ βν + δ αν δ β µ ) . (29b)So that the free energy becomes, F ( κ ) k B T = ρκ αβ κ αβ − ρκ αβ κ βα . (30)Where once again we have used the fact that κ is traceless. Wenow restrict our attention to the case of simple shear, where κ has only one non-vanishing component κ xy = γ . In this case,it is straightforward to show that, F ( κ ) = k B T ργ , (31)and the linear elastic stress follows immediately, σ = k B T ργ . (32)This result is in precise agreement with the well known Doi& Edwards result. This demonstrates the consistency of oursimple 3D model. B. Wire Frames
An explicit calculation of the stress for general wire frameparticles in 3D is very complicated. However, the understand-ing developed in part I can be used to determine the scalingof the stress from a simple argument. As discussed in sectionlastic Response of Wire Frame Glasses. II. 3D Systems 6III, the crucial difference between general wire frame parti-cles and rods is the possibility of bending. While rods cankeep their shape for arbitrary γ , wire frame particles cannotand begin to bend above a certain strain. We will exploit thisfact to estimate Ω for general wire frames.Let us imagine a perfectly flexible L-shaped wire frame par-ticle. The legs of such a particle are freely jointed so thatthe angle between the two legs can take any value withoutincurring any energetic cost. No matter the size of the ap-plied strain, this particle can always access a range of con-figurations. The accessible volume of configuration space forthis flexible particle after the strain is applied is defined to be,˜ Ω flex ( Γ ; ˜ C ) . Up until the strain where the rigid particle beginsto bend, it may access the same volume of configuration spaceas the imagined flexible particle. However, above γ c , the rigidwire frame particle bends and ˜ Ω rigid =
0. Hence we may write,˜ Ω rigid ( Γ ; ˜ C ) = P ( γ ; Γ ) ˜ Ω flex ( Γ ; ˜ C ) . (33)Substituting this into the expression for the free energy (3)gives, F s ( γ ) k B T = − ρ (cid:10) log P ( γ ; Γ ) (cid:11) − ρ (cid:10) log ˜ Ω flex ( Γ ; ˜ C ) (cid:11) , (34)where the angle brackets denote averaging over orientations Γ and constraint configurations C . The second term may be cal-culated in a similar manner as for rods in the previous section.Each leg of the L-shape is independent of the other, so the freeenergy associated with each is essentially the same as that ofa singe rod. Therefore, this term must be of the form, ρ (cid:10) log ˜ Ω flex ( Γ ; ˜ C ) (cid:11) = − c ργ , (35)where c is a positive constant of order unity. The first term in(34), however, is very different. Using (4) we have, ρ (cid:10) log P ( γ ; Γ ) (cid:11) = ρ (cid:10) log (cid:0) − γ / γ c ( Γ ) (cid:1)(cid:11) . (36)This may be expanded to O ( γ ) to obtain the free energy rel-evant for the linear elasticity of wire frame systems, F ( γ ) k B T = ρ (cid:28) γ γ c ( Γ ) (cid:29) + c ργ . (37)Replacing γ c ( Γ ) by its average, and using its scaling found in(19), this becomes, F ( γ ) k B T = a ρ ( ρ L ) γ + c ργ , (38)where a is another constant. For the dense suspensions weconsider, where ρ L (cid:29)
1, the first term dominates the secondand the free energy scales like, F ( γ ) ∼ k B T ρ ( ρ L ) γ . (39)It follows then that the linear elastic stress for concentratedsuspensions of L-shapes scales as, σ ∼ k B T ρ L γ . (40) This is far larger than that found for rods in (32), and muchmore sensitive to the density and size of the particle. This isqualitatively consistent with the simulation results for densesuspensions of 3D crosses, sketched in Fig.(1c). Those sim-ulations showed that the zero shear rate viscosity of a densesuspension of these particles increased by approximately threeorders of magnitude when the volume fraction was increasedfrom 0 to 0 .
1. It is difficult to quantitatively compare theseresults to ours because the density dependence of the zeroshear rate viscosity is also determined that of the relaxationtimescale of the stress, which is not addressed in this series ofpapers. We hope that the methods we have introduced can pro-vide a useful framework for studying this problem and moreaccurate comparisons can be made in the future.The argument presented in this section shows that the bend-ing mechanism is responsible for the very large magnitudeof the linear elasticity in these systems. In section III itwas shown that γ c is very small and the bending mechanismbecomes important for kinked particles with opening angle χ (cid:38) ( ρ L ) − . Therefore, the very strong density dependenceof the stress found in (40) is expected for particles which areonly slightly bent away from a straight rod, through an angleof about χ ≈ ◦ say. It is clear also that branched particles,such as Y-shapes or 3D crosses (shown in Fig.(1c)), should ex-hibit the same scaling. This discussion shows that the elasticstress in concentrated suspensions of wire frame particles isextremely sensitive to the particle shape, displaying radicallydifferent behaviour compared to rods for even modestly bentparticles. The example of 3D crosses is especially noteworthy,since a dilute suspension of these particles has no elastic re-sponse for fundamental symmetry reasons . This means thatthe behaviour of suspensions of these particles will be mostsensitive to concentration, exhibiting a sharp transition from aNewtonian fluid to an elastic gel. V. NON-LINEAR RESPONSEA. Rods
For rod like particles, the extension non-linear elasticity isstraightforward because there is no probability of bending, P ( Γ ; γ ) =
1. Therefore all that needs to be done is expand theentropy to O ( γ ) . The exact calculation is long winded butwe can appeal to the result in part I, as well as results basedon the tube model for the general form. These show that thesuspension shear thins, with the stress to O ( γ ) being givenby, σ = k B T ρ ( a γ − b γ ) , (41)where a and b are positive, order unity and approximately in-dependent of density for ρ L (cid:29) B. Wire Frames
For wire frame particles, both the entropic and bending con-tribution to the free energy are required. In particular, forlastic Response of Wire Frame Glasses. II. 3D Systems 7L-shaped particles, the entropic free energy can be estimatedfrom the expression for the accessible configuration space vol-ume for a rigid wire frame (33) introduced in section IV B.From the first term in (7) we have, F s ( γ ) k B T = − ρ (cid:10) P ( γ ; Γ ) log P ( γ ; Γ ) (cid:11) − ρ (cid:10) P ( γ ; Γ ) log ˜ Ω flex ( Γ ; ˜ C ) (cid:11) . (42)For simplicity, we introduce the following pre-averaging ap-proximations, (cid:10) P ( γ ; Γ ) log P ( γ ; Γ ) (cid:11) ≈ (cid:10) P ( γ ; Γ ) (cid:11) log (cid:10) P ( γ ; Γ ) (cid:11) , (43a)and (cid:10) P ( γ ; Γ ) log ˜ Ω flex ( Γ ; ˜ C ) (cid:11) ≈ (cid:10) P ( γ ; Γ ) (cid:11)(cid:10) log ˜ Ω flex ( Γ ; ˜ C ) (cid:11) . (43b)These approximations make our analysis simpler, but do notchange the nature of our results.Due to the fact that legs of the flexible L-shape behave asindependent rods, we can write (cid:10) log ˜ Ω flex ( Γ ; ˜ C ) (cid:11) ∼ − a γ + b γ , (44)where a and b are positive and approximately independent of ρ , so that the stress arising from this term is the same form as(41) for rods. Along with the definition (4), this allows us towrite the entropic free energy as, F s ( γ ) k B T = − ρ (cid:18) − γ γ c (cid:19)(cid:20) log (cid:18) − γ γ c (cid:19) − a γ + b γ (cid:35) . (45)Which, when expanded to fourth order in γ gives F s ( γ ) k B T = ρ (cid:18) γ c + a (cid:19) γ − ρ (cid:18) γ c + a γ c + b (cid:19) γ . (46)Using the scaling of γ c with density, and dropping the sub-dominant terms we obtain the general scaling form of the en-tropic free energy for wire frame particles, F s ( γ ) k B T ∼ g ρ ( ρ L ) γ − h ρ ( ρ L ) γ , (47)where g and h are positive constants which depend on the par-ticle geometry. Note that in deriving this form, we have as-sumed that the function P ( γ ; Γ ) is exactly as given in (4) withno corrections O ( γ ) . This clearly need not be the case, how-ever including these terms does not alter the general from of(47). The term proportional to γ in the entropic free energy(47) must appear with a negative sign since at a strain ∝ γ c , theentropic free energy is zero. This strain also approximatelyrepresents the limit of the description presented here, becauseabove this strain other effects, such as the non-linear elasticityof the particles themselves, will begin to play a role.The bending free energy is calculated directly from (6), F B ( γ ) ∼ K ρ ( ρ L ) C B γ , (48) where we have defined the positive constant, C B = ( ρ L ) (cid:28) ∆ ( Γ ) γ c ( Γ ) (cid:29) , (49)which is order unity and depends on the particle geometry.The total free energy is therefore the sum of (48) and (47), F ( γ ) k B T ∼ g ρ ( ρ L ) γ + ρ ( ρ L ) (cid:18) Kk B T C B − h ( ρ L ) (cid:19) γ , (50)This can lead to two different behaviours for the stress, de-pending on the value of K relative to the density. This is dueto the competition between the positive and negative contribu-tions to the coefficient of the γ term. If the coefficient of γ ispositive, the suspension shear hardens. This occurs when thebending modulus satisfies, Kk B T (cid:38) ( ρ L ) . (51)This condition is non-trivial since, for our results to be validwe require both ρ L (cid:29)
1, and due to the rigidity of the parti-cles, K (cid:29) k B T . We can take, K ∼ k B T ( ρ L ) p , for any p ≥ p > ρ c ∼ (cid:114) Kk B T L , (52)the suspension will shear thin. The exact values of K and ρ where the cross over between these two behaviours occursdepends on the particle geometry through the ratio C B / h .This situation is qualitatively exactly the same as was foundin part I. This behaviour has an explanation at the level of themodel presented here. When a particle starts to bend, its ori-entation is completely determined by the surroundings, and assuch cannot contribute to the entropic free energy. This effectis captured by the P ( γ ; Γ ) factor in the first term of equation(7). As the applied strain is increased, more and more parti-cles begin to bend, so fewer and fewer contribute entropically.This deficit leads to the strong shear softening behaviour of F s ( γ ) . If, at a given strain, the bending contribution is notsufficient to make up this deficit, the total stress will be shearsoftening. Therefore, there is some critical value of the bend-ing modulus which must be exceeded to see a shear hardeningresponse. VI. DISCUSSION
We have discussed the elastic response of a dense suspen-sion of rigid rod like and L-shaped particles in three dimen-sions. A simple geometric method is used to calculate theentropy of the system by determining the volume of configu-ration space accessible to a particular particle, given the con-straints placed on it by its surroundings. The change in acces-sible volume under the transformation associated with an ap-plied shear leads to a change in the free energy of the system.lastic Response of Wire Frame Glasses. II. 3D Systems 8For the L-shaped particles, it is possible that they need to bendwhen the system is sheared in order for them to respect theconstraints placed on them by their surroundings. This bend-ing mechanism contributes to the free energy of the system.This is taken into account by introducing the function, P ( γ ) ,interpreted as the proportion of particles which have not bentat a strain γ . We find that, to lowest order in γ , this is givenby P = − ( γ / γ c ) , where γ c is the critical strain above whichmost particles have bent. We determined that γ c ∼ ( ρ L ) − ,and found the elastic stress up to O ( γ ) , valid for strains lessthan this critical value.As for the 2D model presented in part I, we find two veryinteresting results. First, the elastic stress in a system of L-shaped particles is significantly higher than that for a systemof rods, scaling proportional to ρ L as opposed to ρ . Sincewe focus on the concentrated regime with ρ L (cid:29)
1, this repre-sents a drastic change in the suspensions behaviour. We alsofind that this behaviour is very sensitive to the shape of thesuspended particles, with the new scaling present for particlesbent through any angle ∼ O ( ) . This conclusion also holds forany branched particle shape, e.g. 3D Crosses (see Fig.(1c)).Second, we find that there is a critical density above whichthe solution is shear softening. This density depends on thebending modulus of the particle and is approximately, ρ c ∼ ( K / k B T ) / L − . Conversely, if the particles are rigid enough,then the suspension shear hardens. This is in contrast to thebehaviour of a rigid rod system, which always shear thins.A potential realisation of the kind of system discussed hereare DNA nano-stars, where double stranded DNA legs arejoined together at prescribed angles. Taking the length of thelegs to be on the order of ten base pairs ( ∼ K ∼ k B T , which meansthat at concentrations ρ L (cid:38)
10, we expect shear softening.The analysis presented here is valid in the concentration range,1 (cid:28) ρ L (cid:28)
50, where the upper limit is set by the concentra-tion at which excluded volume effects become important. Wetherefore hope that the transition between the two behaviourscan be verified experimentally.
ACKNOWLEDGMENTS
We are grateful to Prof. Daan Frenkel for a number of in-sightful, interesting and important discussions. D.A.K. ac-knowledges financial support from the UK Engineering andPhysical Sciences Research Council Ph.D. Studentship awardNo. 1948692.
Appendix: Calculation of (cid:104) ∆ χ (cid:105) In this appendix we give some details of the calculation ofthe mean squared value of ∆ χ . Our starting point is equation(14) of the main text, ∆ χ sin χ = κ αβ ( u α u β + u α u β ) cos χ − ( κ αβ + κ βα ) u α u β . (A.1) Taking the square of this and averaging over u and u gives, (cid:104) ∆ χ (cid:105) sin χ = cos χκ αβ κ γδ (cid:10) ( u α u β + u α u β )( u δ u δ + u γ u δ ) (cid:11) + ( κ αβ + κ βα )( κ γδ + κ δγ ) (cid:10) u α u β u γ u δ (cid:11) − χκ αβ ( κ γδ + κ δγ ) (cid:2)(cid:10) u γ u α u β u δ (cid:11) + (cid:10) u δ u α u β u γ (cid:11)(cid:3) . (A.2)Let us refer to the three terms on the right hand side of thisequation as I,II and III respectively so that,I = cos χκ αβ κ γδ (cid:2)(cid:10) u α u β u γ u δ (cid:11) + (cid:10) u α u β u γ u δ ) (cid:11) + (cid:10) u γ u δ u α u β (cid:11) + (cid:10) u α u β u γ u δ (cid:11)(cid:3) , (A.3a)II = ( κ αβ + κ βα )( κ γδ + κ δγ ) (cid:10) u α u γ u β u δ (cid:11) , (A.3b)III = − χκ αβ ( κ γδ + κ δγ ) (cid:2)(cid:10) u γ u α u β u δ (cid:11) + (cid:10) u δ u α u β u γ (cid:11)(cid:3) . (A.3c)The unit vectors are distributed isotropically but with a fixedangle χ between them. The averages of the fourth rank tensorsin the above equations can therefore be determined easily, (cid:10) u α u β u γ u δ (cid:11) = (cid:0) δ αβ δ γδ + δ αγ δ βδ + δ αδ δ βγ (cid:1) , (A.4a) (cid:10) u α u β u γ u δ (cid:11) = cos χ (cid:0) δ αβ δ γδ + δ αγ δ βδ + δ αδ δ βγ (cid:1) , (A.4b) (cid:10) u α u β u γ u δ (cid:11) = (cid:2) ( − cos χ ) δ αβ δ γδ + ( χ − ) (cid:0) δ αγ δ βδ + δ αδ δ βγ (cid:1)(cid:3) . (A.4c)Using these in (A.3) for the specific case of simple shear,where κ has only one non-zero component κ xy = γ , it isstraight forward to show that,I = γ
15 cos χ + γ
15 cos χ ( χ − ) , (A.5a)II = γ ( − cos χ ) + γ ( χ − ) , (A.5b)III = − γ
15 cos χ . (A.5c)Hence we have, (cid:104) ∆ χ ( γ ) (cid:105) sin χ = γ (cid:0) − χ + χ (cid:1) = γ sin χ . (A.6)From which, equation (15) of the main text is easily recovered.lastic Response of Wire Frame Glasses. II. 3D Systems 9 DATA AVAILABILITY
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