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        <h2>Calculation</h2>

<h3>Destroy the liposome</h3>

Following mathematical formula reveals proves that smaller liposomes are more stable. From this calculation result, we think it is possible to destroy liposomes if we give them energy by trigger. </br> Calculation </br></br>

・Thought</br> 　Assume that a vesicle of large radius existing underwater changes to the plural vesicles of small radius (fig 1). The free energy of each sates F1: large radius, and F2: small radius, probably have following relationship (fig 2).  In this case, the vesicle is basically small states because being a small radius is more stable, but there is the energy gap “δ” that must exceed to change the size of vesicle. In this time we do not calculate δ, but calculate the free energy in each radius. From these energy and radius, we demand the relationship of them.</br>

<table> <tr> <td> <img src="http://openwetware.org/images/7/73/Fig.1.jpg" width="400"> </td> <td> <img src="http://openwetware.org/images/2/2e/Fig.2.jpg" width="400"> </td> </tr> <tr> <td> Fig.1 The assumed state of vesicle in the system </td> <td> Fig.2 The assumed relation of each free energy </td> </tr> </table>

・Setting</br> 　Surfactant molecules of the N0 units exist underwater, and N units participate in the formation of the vesicle. Assume that surface area of one molecule A=0.6 [nm^2] and the number of molecules to be included in one vesicle with n units. The volume of the water V=10^(-4) [m^3]. This time, we use DOPC (C44H84NO8P) so the molecular weight m=785 [g/mol]. The temperature of the system T=300[K], and Boltzmann’s constant and Planck's constant use the follows.</br>

</br>

・Calculation result</br> 　Free energy “F” of the vesicle is given as follows.</br>

In this time,</br>

(pf)bulk is a partition function of the single discrete molecule underwater, and (pf)vesicle is a partition function of the vesicle.</br> Therefore, free energy “F”,</br>

In this calculation, we assume that N, T, and V are constant,</br>

α= 0 at the moment of N0 – N = 0, so simplify,</br>

We do not know the from of (pf)vesicle, so this time we assume,</br>

N/n=X means the number of the vesicle of the whole system. So we calculate free energy by using this,</br>

Here, a vesicle which radius is 100μm changes to each 10 and 100 vesicles, calculate each free energy,</br>

From these results, we get the follow (fig 3).</br>

<img src="http://openwetware.org/images/c/c1/Fig.3.jpg" width="400"></br> Fig.3 The relation of each free energy by calculation<br>

From fig 3, free energy is smaller when the radius of vesicle is small.</br></br>

・Discussion</br> When all molecules participate in the formation of the vesicle, the free energy becomes a one-tenth time when the number of vesicle increase 10 times.  In addition, we think that the vesicle of the small radius is more stable.<br><br>

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