# Free Surface Energy and Hansen Solubility Parameter Vector Field. Interface Thickness
## 3 Data and Discussion
### The Hansen Solubility Parameters Vector Field  
#### 3.1.1 Concepts and Definitions  
Hansen solubility parameters correspond to a three-dimensional vector field in which the dimensions are London dispersion forces, \(D\), polarity forces due to permanent dipoles, \(P\), and forces due to hydrogen bonding, \(H\). The said vector field dictates the physical phenomena where intermolecular forces (like solubility, polymer swelling and cracking), or free surface energies (surface tension, interfacial tension, contact angle and others) come into play. Inside that vector field, any given substance is a vector, \(\overset{\rightarrow}{S}\), composed of the sum of three vectors:  
\[\overset{\rightarrow}{S}=\overset{\rightarrow}{S}_{D}+\overset{\rightarrow}{S} _{P}+\overset{\rightarrow}{S}_{H} \tag{7}\]  
where their components are the multiplication of the scalar solubility parameters by the direction vectors, \(\overset{\rightarrow}{i}\), \(\overset{\rightarrow}{j}\) and \(\overset{\rightarrow}{k}\), of the base of vector space. \[\overset{\rightarrow}{S}_{D}=\delta_{D}\overset{\rightarrow}{i}\quad \overset{\rightarrow}{S}_{P}=\delta_{P}\overset{\rightarrow}{\hat{S}}_{H}= \delta_{H}\overset{\rightarrow}{k} \tag{8}\]  
The particularity of the Hansen vector field is that the magnitude of the direction vectors of the field base, \(\overset{\rightarrow}{j}\) y \(\overset{\rightarrow}{k}\), is 1 (like in Euclidean fields); meanwhile, the magnitude of the vector of the base \(\overset{\rightarrow}{i}\) is two times bigger. In other words, their magnitudes are as follows:  
\[\left|\overset{\rightarrow}{i}\right|=2\quad\left|\overset{\rightarrow}{j} \right|=1\quad\left|\overset{\rightarrow}{k}\right|=1 \tag{9}\]  
The vector field base is orthogonal. The vector that represents a substance in the vector field is a position vector and it is expressed as the following:  
\[\overset{\rightarrow}{S}=\delta_{D}\overset{\rightarrow}{i}+\delta_{P} \overset{\rightarrow}{j}+\delta_{H}\overset{\rightarrow}{k} \tag{10}\]  
Likewise:  
\[\overset{\rightarrow}{S}=(\delta_{D},\delta_{P},\delta_{H}) \tag{11}\]  
This vector starts in the origin and has its destination at a certain point of the field \((\delta_{D},\delta_{P},\delta_{H})\). The magnitude of vector \(\overset{\rightarrow}{S}\) is as follows:  
\[\left|\overset{\rightarrow}{S}\right|=\sqrt{\left(\delta_{D}\overset{ \rightarrow}{i}\right)^{2}+\left(\delta_{P}\overset{\rightarrow}{j}\right)^{2 }+\left(\delta_{H}\overset{\rightarrow}{k}\right)^{2}} \tag{12}\]  
Because \(\left|\overset{\rightarrow}{i}\right|=2\), this leads to the following:  
\[\left|\overset{\rightarrow}{S}\right|=\sqrt{4\delta_{D}^{2}+\delta_{P}^{2}+ \delta_{H}^{2}} \tag{13}\]  
The coefficient 4 that goes with the dispersion component is obtained by squaring the direction vector \(\overset{\rightarrow}{i}\). Figure 1 shows three substances, namely 1, 2 and 3, with their vectors \(\overset{\rightarrow}{S}_{1}\), \(\overset{\rightarrow}{S}_{2}\) and \(\overset{\rightarrow}{S}_{3}\). It can be observed that, in the base of the space, the direction vector \(\overset{\rightarrow}{i}\) is twice as large as vectors \(\overset{\rightarrow}{j}\) and \(\overset{\rightarrow}{k}\) and, therefore, the scale of \(\delta_{D}\) is twice the scales of \(\delta_{P}\) and \(\delta_{H}\). The interaction between substances 1 and 2 is represented by a new vector \(\overset{\rightarrow}{S}_{12}\) called the interaction vector, which is defined by the difference between both said vectors:  
\[\overset{\rightarrow}{S}_{12}=\overset{\rightarrow}{S}_{2}-\overset{ \rightarrow}{S}_{1} \tag{14}\]  
\[\overset{\rightarrow}{S}_{12}=(\delta_{D2}-\delta_{D1})\overset{\rightarrow }{i}+(\delta_{P2}-\delta_{P1})\overset{\rightarrow}{j}+(\delta_{H2}-\delta_ {H1})\overset{\rightarrow}{k} \tag{15}\]Vector \(\overset{\rightarrow}{S}_{21}\) is directed from 2 to 1. Additionally, it is true that the magnitudes of the interaction vectors \(\overset{\rightarrow}{S}_{12}\) and \(\overset{\rightarrow}{S}_{21}\) are idempotent.