Method

Our calculation method generalizes that of Ref.[7], which we follow closely. We first find the depletion force, per unit area between two infinite parallel plates, at separation h; the force between two large spheres of radius then follows by the Derjaguin approximation [14]:

 

from which the depletion potential can be obtained:

 

To find we invoke the ``pressure sum rule" [15], which expresses p, the (osmotic) pressure exerted on a hard wall by a solution of thin rods, as where , calculated below, is the density of rod ends in contact with the wall. For hard rod particles, this relation is exact to all orders in the virial expansion; it is the analogue of Henderson's formula for the pressure in terms of the contact density for the hard sphere case [16]. It may be proved rigorously (for rods in vacuo) by elementary kinetic theory (see Appendix).

For two hard parallel plates immersed at separation h in a solution of depletant, the force per unit area on one of the plates is simply the differential pressure on its two sides:

 

Here , the density of rod ends in contact with a plate, separated by h from another, is given by:

 

where denotes the number per unit volume of midpoints of rods with orientation , evaluated on the plane of contact at distance from the boundary.

Now we present an argument that will enable us to obtain : we use essentially the same idea as lies behind the potential distribution theorem due to Widom [17]. Suppose we put our system of rod particles and parallel plates in contact with a hypothetical reservoir, in which rods are exempt from mutual excluded-volume interactions. We first calculate the relation between the particle density in the reservoir and the bulk density . Imagine an infinitesimal volume centered at position r in the bulk solution, in which we consider placing (the midpoint of) a rod of orientation . This would exclude from a volume the midpoint of another rod with relative orientation defined with respect to coordinate axes in the first rod. We define our angular coordinates so that (hence ); the number density of rods in the reservoir with orientation in the range ( ) is then .

The presence of our proposed rod at r is permitted only if the volume is empty for all orientations ; this probability is simply the product of the probabilities of all sub-volumes being empty:

 

which can be multiplied out to give terms of different orders in concentration, cf. Ref.[5]:

 

here denotes the volume excluded by rod 1 that is otherwise available to rod 3, given the presence of (hence the exclusion due to) the rod 2. Now the picture of rod 3 touching and revolving around rod 2, which touchs rod 1, allows us to deduce that the chance of rod 1 and 2 excluding rod 3 simultaneously is smaller than that of rod 1 alone excluding 3 by roughly a factor of D/L, which is very small for thin rods. In other words, very thin rods are unlikely to form any loops; or mathematically, m mutually touching thin rods are likely to have just m-1 touching points. So we make the mean field approximation of , which, in the bulk, is essentially the Onsager approximation, valid for . (Notice that although, as we confirm below, the Onsager theory leads to truncation of the virial expansion for the pressure at second order [18], the same does not apply to the depletion force.)

In the limit of large aspect ratio, eq.(6) then becomes:

 

(This is, of course, independent of the orientation of the rod we are considering at r.) To find the density of rod centres at r (which must equal , since r is an arbitrary point in the bulk) we argue that, if a rod is allowed there (probability ) the density is ; otherwise it is zero. Hence where is the reduced density. Inverting this relation gives:

 

This completes our derivation of the relation between and . Of course, equating the chemical potential in the bulk (with mutual excluded volume interaction between two rods) for , with that in the reservoir (without such interaction), gives the same result. However, the argument given above is directly generalizable to the depletion problem as we now show.

For use in Eq.(3), we need to calculate , the end density contacting a confining plate separated by h from another. Consider a rod touching the plate with orientation and a second rod touching this rod, with relative orientation . (See Fig.1.) The height from the wall of the two ends of the first rod are respectively and ; those of the second rod, and ; and locate the position of the contact point of the rods as shown. The variables can be expressed in terms of the others (using elementary geometry) as , and . The infinitesimal volume excluded by the first rod to the (midpoint of the) second rod, both at fixed orientations and with their contact point specified in the range ( to , to ), is [18]:

 

We can now write , the density of rods with orientation at distance away from the wall, as:

 

with where H's, the Heaviside unit step functions, are needed to eliminate all configurations forbidden by the hard interaction with the confining walls. is the prior probability that the volume excluded by such a rod, at a distance away from the wall, contains no other rods. So Eq.(11) states that all positions and orientations that are allowed to a rod are equally populated with an occupation of , which is the same rod density per unit solid angle as in the reservoir; the probability of a chosen position and orientation being allowed is simply .

As before (cf.Eqs.(5-7)), the probability can be expressed as the product of the probabilities of all sub-volumes being empty:

 

The orientation denotes that of the second rod in a coordinate system relative to the plates rather than that of the first rod (with respect to which is defined). Making the same approximation of ignoring loop configurations as earlier, is then given by:

 

This, together with Eqs.(9,11), leads to the following self-consistent integral equation for the density distribution function :

 

with defined by Eq. (13) and (10) as:

 

Notice the density is exactly the same if we reverse the direction of rod 1 (this swaps the value of and and change to ). We can therefore choose in the calculation of the function without loss of generality. Setting , we have, by definition, the contact density , which will lead to the depletion force via Eq.(4). Note that, if we write down the grand thermodynamic potential of the system as a functional of the density distribution , then the minimization of this potential with respect to the function gives rise to exactly the same self-consistent integral equation (14). This was pointed out (for a single plate) in Ref.[19], though the resulting equation was not solved there. An approximate solution for a single plate is provided in Ref.[20]. In this paper, we numerically solve the equation with high precision for the geometry of two parallel plates.

The generalization of Eq.(14) to include an arbitrary enthalpic interaction between rods is rather straightforward: we simply need to insert a term with an appropriate Boltzmann factor in Eq.(11). In this way we could, in principle, treat cases with any pairwise interaction potential between rods; however we do not pursue this here.