6 and thus the negative derivative of the free energy (Eq. 3.18) is portion of the free energy per unit area is Here Z$_{S}$ = tanh( βJ$_{S}$ ) with J$_{S}$ the in-layer coupling on the surface and τ is the reduced temperature. In a change of notation, we now denote the thickness of the film by N . For simplicity, we assume that bulk couplings in both x and y directions are equal. Note that the re- duced temperature in the above equation is equal to 1 /N of the reduced temperature defined in Li et. al. [19]. To find the Casimir force, we take the negative of the deriva- tive of the free energy above with respect to N . Since N appears only in the exponent, we need to consider the following derivative Now we need to subtract from the above expression the ”background” casimir forces which is obtained when the thickness of the film is infinite. As N → ∞ , the free energy given in (3.18) becomes dominated by the term going as (1+2 τ/ √ τ 2 + ω $^{2}$) $^{N}$, and thus the corresponding Casimir force per unit area is − ∂F$_{singular}$ ∂N = 2 πN 2 ∫ 0 d Ω $^{√}$x 2 + Ω $^{2}$× $^{(}$1 + x √ x $^{2}$+Ω2 $^{)}$e 2 √ x $^{2}$+Ω2 − $^{(}$1 − x √ x $^{2}$+Ω2 $^{)}$e − 2 √ x $^{2}$+Ω2 $^{(}$1 + x √ x $^{2}$+Ω2 $^{)}$e 2 √ x $^{2}$+Ω2 + $^{(}$1 − x √ x $^{2}$+Ω2 $^{)}$e − 2 √ x $^{2}$+Ω2 (3.21) F$_{singular}$ = − 1 π ∫ 1 0 dω √ 1 − ω 2 { 2 ln(1 − Z$_{S}$ ) 2 + ln$^{[}$ 1 2 ( 1 + τ √ τ 2 + ω 2 ) $^{(}$1 + 2$^{√}$τ 2 + ω 2 ) N + 1 2 ( 1 − τ √ τ 2 + ω 2 ) $^{(}$1 − 2 $^{√}$τ 2 + ω 2 ) N ]} (3.18) ∂ ∂N $^{(}$1 + 2$^{√}$τ 2 + ω 2 ) N = $^{[}$ln $^{(}$1 + 2$^{√}$τ 2 + ω 2 $^{)] (}$1 + 2$^{√}$τ 2 + ω 2 ) N (3.19) 1 π ∫ 1 dω √ 1 − ω 2 ln $^{(}$1 + 2$^{√}$τ 2 + ω 2 ) (3.22) 0 Inserting the scaling variables x/N = τ and ω = Ω /N as before, we obtain 2 πN 2 ∫ 0 d Ω $^{√}$x 2 + Ω 2 (3.23) Subtracting this from (3.21), we are now allowed to ex- tend the upper limit of integration to ∞ , and thus we end up with the final expression for the Casmir force per unit area which helps us introduce the relevant scaling variables. If we replace the reduced temperature, τ , by x/N and the integration variable, ω , by Ω /N , the right hand side of (3.19) in the limit of large N becomes 2 N $^{√}$x 2 + Ω $^{2}$e 2 √ x $^{2}$+Ω2 (3.20) F = 1 πN 2 ∫ ∞ −∞ d Ω $^{√}$x 2 + Ω 2   $^{(}$1 + x √ x $^{2}$+Ω2 $^{)}$e 2 √ x $^{2}$+Ω2 − $^{(}$1 − x √ x $^{2}$+Ω2 $^{)}$e − 2 √ x $^{2}$+Ω2 $^{(}$1 + x √ x $^{2}$+Ω2 $^{)}$e 2 √ x $^{2}$+Ω2 + $^{(}$1 − x √ x $^{2}$+Ω2 $^{)}$e − 2 √ x $^{2}$+Ω2 − 1$^{}$  (3.24) IV. ALTERNATIVE APPROACH TO THE TWO DIMENSIONAL ISING MODEL It is also possible to obtain all the expressions above by a set of direct calculations of the partition function of the two dimensional Ising model in zero magnetic field. We make use of the method introduced by Schultz, Mattis and Lieb [15], which will hereinafter be referred to as the LSM method. The version of this model we will address is wrapped on a cylinder with free boundary conditions on the ends. Circumferentially, we will allow for either pe- which is plotted in Fig. 2 (long dashed line) and as ex- pected is negative giving rise to an attractive Casimir force. It is interesting to note that the strength of the Casimir force in the case of free boundary condition is zero in one dimension but in two dimensions is compa- rable in amplitude to the Casimir force with periodic boundary conditions. We also anticipate the results of the next section by plotting the Casimir force for fixed spin boundary conditions, previously obtained by Evans and Stecki [20], which is related to the Casimir force for free boundary conditions by duality.