## Calculation

### Destroy the liposome

Following mathematical formula reveals proves that smaller liposomes are more stable. From this calculation result, we think it is possible to collapse liposomes if we provide with energy by trigger.

・Hypothesis

We assume that a vesicle of large radius existing underwater changes to the plural vesicles of small radius (fig 1). The free energy of each states F1: large radius, and F2: small radius, probably have the following relationship (fig 2). In our case, the vesicle is basically small state because a small radius is more stable.The decisive factor is, however, that there is the energy gap “δ” that must exceed to change the size of vesicle. In our calculation we have not had the possibility to calculate δ, but to calculate the free energy in each radius. With this values of energy and radius the energy gap "δ" can be deducted.

Fig.1 Assumed state of vesicle in the system | Fig.2 Assumed relation of each free energy |

・Assumed situation

N

_{0}surfactant molecule units exist in a solution, and N units of them participate in the formation of the vesicle. One vesicle consists of n surfactant molecule units.

Here we have been working with DOPC (C

_{44}H

_{84}NO

_{8}P).

A : Surface area of one molecule =0.6 [nm

^{2}].

V : Volume of the water =10

^{-4}[m

^{3}].

m : Molecular weight of DOPC =785 [g/mol].

T : Temperature of the system T=300[K].

Boltzmann’s constant and Planck's constant use the following values.

・Calculation result

Free energy “F” of the vesicle is given as follows*.(*Reference)

In this time,

(pf)

_{bulk}is a partition function of the single discrete molecule underwater, and (pf)

_{vesicle}is a partition function of the vesicle.

Therefore, free energy “F”is defined as,

In this calculation, we assume that N, T, and V are constant,

Because of α= 0 at the moment of N

_{0}– N = 0

This time we approximate,

N/n=X means the number of the vesicles in whole of this system. Therefore we calculate free energy by using this,

Here, a vesicle which radius is 100μm changes to each 10 and 100 vesicles, calculate each free energy from,

Based on these results, we get the following figure 3 (fig 3).

Fig.3 The relation of each free energy by calculation

From fig 3,we can conclude free energy is smaller when the radius of vesicle is small.

・Discussion

When all molecules participate in the formation of the vesicle, the free energy becomes a one-tenth time and the number of vesicle increase 10 times. Smaller radius vesicles are more stable.Therefore, it is possible to destroy vesicles using trigger.

・Reference

Yukio Suezaki "Physics of lipid film"