12 Forces and Interactions in Soft Matter

12.1 Interactions involving polar molecules

Many molecules exhibit a dipole or even higher moments due to the fact that the charges are not evenly distributed over the molecular structure. Some of the atoms exhibit a stronger tendency to accept charges than others. This is typically measured by electronegativity and provides an idea of whether atoms rather donate or accept a charge when binding to other atoms. While homo-atomic bonds therefore do not have dipole moments, hetero-atomic bonds do (see table).

Bond	Dipole moment [D]
C-C	0
C-N	0.22
O-H	1.51
F-H	1.94
N=O	2.0

Molecule	Dipole moment [D]
hexane	0
water	1.85
ethanol	1.7
acetone	2.9

The dipole moment of a molecule is measured by the displacement of two charges $±q$ from each other:

\[\begin{equation} \vec{u}=q\vec{l}. \end{equation}\]

Its direction is from the negative to the positive side. It creates an electric field that is given by

\[\begin{equation} \vec{E}=\frac{3(\vec{u}\cdot \hat{r})\hat{r}-\vec{u}}{4\pi \epsilon_{0}\epsilon r^{3}}, \end{equation}\]

where $\hat{r}=\vec{r}/|r|$. The dipole self-energy, i.e., the energy to create the dipole in a solvent is given by

\[\begin{equation} \mu^{0}=\frac{1}{4\pi \epsilon_{0}\epsilon}\left [ \frac{q^2}{2a}+\frac{q^2}{2a}-\frac{q^2}{l}\right ]. \end{equation}\]

For $l=2a$ this results in $\mu^0=q^2/(8\pi\epsilon_0 \epsilon a)=u^2/(4\pi\epsilon_0 \epsilon l^3 )$ and thus yields a similar dependence of the chemical potential on the dielectric function $\epsilon$ as in the case of a single charge. The result is a similar dependence of the solubility on the dielectric function.

The plot below shows the electric field of a dipole:

12.2 Ion–Dipole interaction

The interaction energy of a dipole with a charge can be calculated by

\[\begin{equation} w(r)=-\frac{qQ}{4\pi\epsilon_{0}\epsilon}\left [ \frac{1}{r-\frac{1}{2}l \cos(\theta)}-\frac{1}{r+\frac{1}{2}l \cos(\theta)}\right ]=-\frac{qQ}{4\pi\epsilon_{0}\epsilon r^2}\cos(\theta)=-uE\cos(\theta) \end{equation}\]

where the last two equations are assuming that the distance between both objects $r$ is much larger than the extent of the dipole $l$ itself. From the last equation, we see that the interaction can be either attractive or repulsive. An angle $θ=0^\circ$ results in an attractive interaction, while $θ=180^\circ$ yields repulsive interaction. Using a single charge (e.g., an $\mathrm{Na^+}$ ion) and a dipole of $u=1.85\, \mathrm{D}$ (water molecule) results in an interaction energy of about $39\, k_\mathrm{B}T$. Ions align and bind polar molecules like water, for example. The alignment is caused by the torque $\tau=\vec{u}\times \vec{E}$. For arbitrary polar molecules this is called solvation, while for water the term hydration is used. The strength of the hydration can effect the mobility of ions in solution as it makes them effective charges and is of interest, for example, in the study of ion transport through ion channels, as this requires the stripping of the hydration shell.

Example: Sodium Ion Hydration

The plot below shows the ion dipole interaction for a sodium ion and a water molecule as a function of distance.

The table below shows some selected hydration properties of ions. The hydrated radius determines the diffusion of the ion in water. The hydration number is the number of orientationally bound water molecules. Typically, smaller ions have a larger hydration number.

Ion	Bare Ion radius (nm)	Hydrated radius (nm)	Hydration number
Na+	0.095	0.36	4
Mg2+	0.065	0.43	6
Cl-	0.181	0.33	1
OH-	0.176	0.3	3

12.3 Dipole-Dipole Interactions

The interaction of two dipole will depend on three angles, the two angles of the dipoles with the connecting axis $\theta_1, \theta_2$ and the angle $\phi$ between the two planes which contain the individual dipoles and the axis. The figure below shows the corresponding geometry.

The energy of the two dipoles at a distance $r$ follows then without a detailed calculation with

\[ w(r,\theta_2,\theta_2,\phi)=-\frac{u_1 u_2}{4\pi \epsilon_0 \epsilon r^3}\left [ 2\cos(\theta_1) \cos(\theta_2)-\sin(\theta_1)\sin(\theta_2)\cos(\phi)\right ] \]

The image below (taken from the book by Israelachvili) shows the dipole dipole interaction energy for the parallel and the in-line configuration as indicated. As shown, the in-line configutation with facing opposite charges is more favorable.

The energies that are typically found are also larger than the thermal energy at room temperature and thefore lead to ordering effects.

12.4 Rotating Dipoles, angle averaged Potential

The equations we derived so far deliver the energy for a specific fixed orientation of dipoles for example. However, molecules with dipoles may rotate and undergo rotational Brownian motion driven by thermal energy. This rotational diffusion can be very fast and the rotational sampling creates an average interaction that may scale differently with the distance. To get an effective distance dependence in the interaction, we have to integrate the Boltzmann factor over the orientational degrees of freedoms, i.e.

\[ \exp\left ( -\frac{w(r)}{k_B T} \right)=\frac{\int \exp\left (-\frac{w(r,\Omega)}{k_B T} d\Omega\right )}{\int d\Omega}=\left \langle \exp\left (-\frac{w(r,\Omega)}{k_B T}\right ) \right \rangle_{\Omega} \]

with

\[d\Omega =\sin(\theta)d\theta d\phi\]

and

\[\int d\Omega=4\pi\]

\[ \exp\left ( -\frac{w(r)}{k_B T} \right) =\frac{1}{4\pi}\int_0^{2\pi}d\phi\int_0^{\pi}\exp\left ( -\frac{w(r,\phi\theta)}{k_B T} \right)\sin(\theta)d\theta \]

When $w(r,\theta)<< k_B T$ then we can expand the Boltzmann factor

\[ \exp\left ( -\frac{w(r)}{k_B T} \right) \approx 1-\frac{w(r)}{k_B T}+\frac{1}{2}\left (\frac{w(r)}{k_B T} \right )^{2}+\ldots \]

and therefore obtain

\[ w(r)=\langle w(r,\Omega)- \frac{w(r,\Omega)^2}{2k_B T}+\ldots \rangle_{\Omega} \]

which we can readily use to calculate the interaction energy of a charge and a dipole

\[ \begin{aligned} w(r)=&\langle -\frac{Q u \cos(\theta)}{4\pi\epsilon_0\epsilon r^2} - \left ( \frac{Q u }{4\pi\epsilon_0\epsilon r^2}\right)^2 \frac{\cos(\theta)^2}{2k_B T}+\ldots \rangle\\ =& \frac{Q^2 u^2}{6(4\pi\epsilon_0\epsilon)^2k_B T r^4} \end{aligned} \]

when $k_B T>\frac{Q u }{4\pi\epsilon_0\epsilon r^2}$. Interestingly, this interaction energy for the freely rotating dipole in the electric field of an ion decays as $r^{-4}$, while the static interation decayed as $r^{-2}$. Also, the interaction of the freely rotating dipole is always attrative as compared to the static dipole charge interaction.

The previous result can also be obtained when using $G=-k_B T \ln(Z)$, where

\[ Z=\int \exp(-\frac{w(r,\Omega)}{k_B T}) d\Omega \]

is the partition function.

12.5 Dipole-Dipole interaction with rotating dipoles

We can extend the above calculation now also to two dipoles interacting when freely rotating. The interaction energy was given by

\[ w(r,\theta_2,\theta_2,\phi)=-\frac{u_1 u_2}{4\pi \epsilon_0 \epsilon r^3}\left [ 2\cos(\theta_1) \cos(\theta_2)-\sin(\theta_1)\sin(\theta_2)\cos(\phi)\right ] \]

and therefore the partition function can be calculated as

\[ Z=\int_0^{\pi}\int_{0}^{\pi}\int_{0}^{2\pi}e\exp(-\beta w(r,\theta_1,\theta_2,\phi))\sin(\theta_1)\sin(\theta_2)d\phi d\theta_1 d\theta_2 \]

The result of this calculation is the Keesom interaction energy

\[ w(r)=-\frac{u_1^{2}u_{2}^2}{3(4\pi\epsilon\epsilon_0)^2 k_B T r^6} \]

when

\[ k_BT >\frac{u_1 u_2}{4\pi\epsilon_0 \epsilon r^3} \]

The Keesom interaction tells on one side that permanent dipoles align themselves to be always attractive. It further belongs to a set of 3 different electrostatic interactions, which scale as $r^{-6}$ and which are termed van der Waals interactions.

12.6 Interactions involving polarizability

All atomic and molecular systems carry the potential to get polarized in an external electric field. This polarization gives rise to a dipole moment $u_{ind}$, that is in linear response proportional to the external electric field. The dipole itself also creates an electric field. There are different types of polarizability. We will address two of them, i) the electronic polarizability, i.e. shifting the center of mass of positive and negative charges of a molecule or atom, ii) the reorientation of a permanent electric dipole in and external electric field which is an orientational polarization.

We consider a very simple classical model for the electronic polarizability. An electron is orbiting a positive nucleus and is subjected to en electric field according to the above image. In the stationary state, the forces on the charges exerted by the external electric field and the internal forces balance and give rise to the polarized state with a dipole moment.

The external force on each charge is given by

\[ F_{ext}=-eE \]

while the internal force between the two charges is

\[ F_{int}=-\frac{e^{2}}{4\pi \epsilon_0 R^{2}}\sin(\theta)\approx \frac{e^{2}l}{4\pi \epsilon_0 R^{3}}=-\frac{e u_{ind}}{4\pi \epsilon_0 R^3}=F_{ext} \]

We can transform the latter equation to yield the induced dipole moment

\[ u_{ind}=4\pi\epsilon_0 R^3 E=\alpha_0 E \]

Accordingly, the electronic polarizability of the atom is

\[ \alpha_0=4\pi\epsilon_0 R^3 \]

which just says that the polarizability scales with the volume of the object, which is found in many occations, including the scattering and absorption cross sections of nanoparticles.

12.7 Ion-Induced Dipole Interaction

Ion-induced dipole interactions occur when an ion distorts the electron cloud of a nearby neutral molecule, creating an induced dipole. This interaction can be described through the following key points:

12.7.1 Field of the Induced Dipole

The electric field $E$ created by an ion at a distance $r$ induces a dipole moment $\mu_{\text{ind}}$ in a neutral molecule:

\[ \mu_{\text{ind}} = \alpha E \]

where $\alpha$ is the polarizability of the molecule, and $E$ is given by:

\[ E = \frac{q}{4 \pi \epsilon_0 r^2} \]

The resulting induced dipole field $E_d$ is:

\[ E_d = -\frac{2 \mu_{\text{ind}}}{4 \pi \epsilon_0 r^3} = -\frac{2 \alpha E}{4 \pi \epsilon_0 r^3} = -\frac{2 \alpha q}{(4 \pi \epsilon_0)^2 r^5} \]

12.7.2 Force and Potential Energy

The force $F$ between the ion and the induced dipole is given by:

\[ F(r) = -\frac{\partial U}{\partial r} = -2 \alpha \left( \frac{q}{4 \pi \epsilon_0 r^2} \right)^2 = -\frac{2 \alpha q^2}{(4 \pi \epsilon_0)^2 r^5} \]

Integrating this force gives the potential energy $U$:

\[ U(r) = \int F \, dr = -\frac{\alpha q^2}{(4 \pi \epsilon_0)^2 r^4} \]

This shows that the interaction energy decreases rapidly with the fourth power of the distance.

12.7.3 Polarizability by Polar Molecules

For polar molecules, an electric field $E$ induces a dipole moment that can be averaged over all orientations, leading to an additional orientational polarization. The total induced dipole moment $\mu_{\text{ind}}$ for a polar molecule with permanent dipole moment $\mu_0$ is:

\[ \mu_{\text{ind}} = \mu_0 \cos(\theta) + \alpha E \cos^2(\theta) \]

Averaging over all orientations gives the mean induced dipole moment:

\[ \langle \mu_{\text{ind}} \rangle = \frac{\mu_0^2 E}{3 k_{\text{B}} T} \]

Therefore, the overall polarizability $\alpha_{\text{total}}$ is a combination of the intrinsic polarizability $\alpha_0$ and the orientational contribution:

\[ \alpha_{\text{total}} = \alpha_0 + \frac{\mu_0^2}{3 k_{\text{B}} T} \]

For example, for an ion-induced dipole interaction at 300 K with $\mu_0 = 4 \times 10^{-30} \, \text{Cm}$ and $\alpha_0$:

\[ \alpha_{\text{total}} = 4 \pi \epsilon_0 \times 8 \times 10^{-30} \, \text{m}^3 + \frac{\mu_0^2}{3 k_{\text{B}} T} \]

These principles explain how ion-induced dipole interactions influence the behavior of molecules in various environments, including solutions and biological systems.

--- execute: echo: false title: Forces and Interactions in Soft Matter jupyter: python3 --- ## Interactions involving polar molecules ```{python} #| include: false #| nbsphinx: hidden #| tags: [parameters, hide-input] import plotly import plotly.graph_objs as go from IPython.display import display, HTML import plotly.io as pio from ipywidgets import interact plotly.offline.init_notebook_mode() display(HTML( '<script type="text/javascript" async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.1/MathJax.js?config=TeX-MML-AM_SVG"></script>' )) import numpy as np import matplotlib.pyplot as plt from numpy.linalg import norm from scipy.constants import c,epsilon_0,e,physical_constants %config InlineBackend.figure_format = 'retina' # the lines below set a number of parameters for plotting, such as label font size, # title font size, which you may find useful plt.rcParams.update({'font.size': 14, 'font.family':'sans-serif', 'axes.titlesize': 16, 'axes.labelsize': 18, 'axes.labelpad': 14, 'lines.linewidth': 1, 'lines.markersize': 10, 'xtick.labelsize' : 18, 'ytick.labelsize' : 18, 'xtick.top' : True, 'xtick.direction' : 'in', 'ytick.right' : True, 'ytick.direction' : 'in',}) ``` Many molecules exhibit a dipole or even higher moments due to the fact that the charges are not evenly distributed over the molecular structure. Some of the atoms exhibit a stronger tendency to accept charges than others. This is typically measured by electronegativity and provides an idea of whether atoms rather donate or accept a charge when binding to other atoms. While homo-atomic bonds therefore do not have dipole moments, hetero-atomic bonds do (see table). | Bond | Dipole moment [D] | :------ | :----------: | | C-C | 0 | | C-N | 0.22 | | O-H | 1.51 | | F-H | 1.94 | | N=O | 2.0 | | Molecule | Dipole moment [D] | :------ | :----------: | | hexane | 0 | | water | 1.85 | | ethanol | 1.7 | | acetone | 2.9 | The dipole moment of a molecule is measured by the displacement of two charges $±q$ from each other: \begin{equation} \vec{u}=q\vec{l}. \end{equation} Its direction is from the negative to the positive side. It creates an electric field that is given by \begin{equation} \vec{E}=\frac{3(\vec{u}\cdot \hat{r})\hat{r}-\vec{u}}{4\pi \epsilon_{0}\epsilon r^{3}}, \end{equation} where $\hat{r}=\vec{r}/|r|$. The dipole self-energy, i.e., the energy to create the dipole in a solvent is given by \begin{equation} \mu^{0}=\frac{1}{4\pi \epsilon_{0}\epsilon}\left [ \frac{q^2}{2a}+\frac{q^2}{2a}-\frac{q^2}{l}\right ]. \end{equation} For $l=2a$ this results in $\mu^0=q^2/(8\pi\epsilon_0 \epsilon a)=u^2/(4\pi\epsilon_0 \epsilon l^3 )$ and thus yields a similar dependence of the chemical potential on the dielectric function $\epsilon$ as in the case of a single charge. The result is a similar dependence of the solubility on the dielectric function. The plot below shows the electric field of a dipole: ::: {.content-visible when-format="html"} ```{python} #| nbsphinx: hidden #| tags: [] plt.figure(figsize=(5, 5)) # generate grid x=np.linspace(-2, 2, 32) y=np.linspace(-2, 2, 32) x, y=np.meshgrid(x, y) def E(q, a, x, y): return q*(x-a[0])/((x-a[0])**2+(y-a[1])**2)**(1.5), \ q*(y-a[1])/((x-a[0])**2+(y-a[1])**2)**(1.5) # calculate vector field Ex1, Ey1=E(-1, [-0.1, 0], x, y) Ex2, Ey2=E(1, [0.1, 0], x, y) Ex=Ex1+Ex2 Ey=Ey1+Ey2 E_max=5 E=np.sqrt(Ex**2+Ey**2) Ex.flat[E.flat[:]>E_max]=np.nan Ey.flat[E.flat[:]>E_max]=np.nan # plot vecor field plt.quiver(x, y, Ex/E, Ey/E, pivot='middle', headwidth=3, headlength=5,scale=40) plt.xlabel('$x$') plt.ylabel('$y$') plt.title('dipole field') plt.tight_layout() plt.savefig("img/dipole_field.png") plt.show() ``` ::: ::: {.content-visible unless-format="html"} ![](img/dipole_field.png) ::: ## Ion–Dipole interaction ![Charge Dipole](img/dipole_q.png) The interaction energy of a dipole with a charge can be calculated by \begin{equation} w(r)=-\frac{qQ}{4\pi\epsilon_{0}\epsilon}\left [ \frac{1}{r-\frac{1}{2}l \cos(\theta)}-\frac{1}{r+\frac{1}{2}l \cos(\theta)}\right ]=-\frac{qQ}{4\pi\epsilon_{0}\epsilon r^2}\cos(\theta)=-uE\cos(\theta) \end{equation} where the last two equations are assuming that the distance between both objects $r$ is much larger than the extent of the dipole $l$ itself. From the last equation, we see that the interaction can be either attractive or repulsive. An angle $θ=0^\circ$ results in an attractive interaction, while $θ=180^\circ$ yields repulsive interaction. Using a single charge (e.g., an $\mathrm{Na^+}$ ion) and a dipole of $u=1.85\, \mathrm{D}$ (water molecule) results in an interaction energy of about $39\, k_\mathrm{B}T$. Ions align and bind polar molecules like water, for example. The alignment is caused by the torque $\tau=\vec{u}\times \vec{E}$. For arbitrary polar molecules this is called **solvation**, while for water the term **hydration** is used. The strength of the hydration can effect the mobility of ions in solution as it makes them effective charges and is of interest, for example, in the study of ion transport through ion channels, as this requires the stripping of the hydration shell. <div class="alert alert-info"> **Example: Sodium Ion Hydration** </div> ![](img/NA_hydration.png) The plot below shows the ion dipole interaction for a sodium ion and a water molecule as a function of distance. ```{python} #| nbsphinx: hidden #| tags: [] def ion_dipole(u,r,l,theta,epsilon): q=u/l f=e*q/(4*np.pi*epsilon*epsilon_0) return(-f*(1/(r-0.5*l*np.cos(theta))-1/(r+0.5*l*np.cos(theta)))) ``` ```{python} #| nbsphinx: hidden #| tags: [] D=physical_constants["atomic unit of electric dipole mom."][0] J2eV=physical_constants["electron volt-joule relationship"][0] ``` ```{python} #| nbsphinx: hidden #| tags: [] r=np.linspace(0.1e-9,0.5e-9,200) l=0.02e-9 ``` ::: {.content-visible when-format="html"} ```{python} #| nbsphinx: hidden #| tags: [] #plt.figure(figsize=(5,5)) #plt.plot(r*1e9,ion_dipole(1.8*D,r,l,0,1)/J2eV,'k--') #plt.plot(r*1e9,ion_dipole(1.8*D,r,l,np.pi,1)/J2eV,'k--') #plt.plot(r*1e9,ion_dipole(1.8*D,r,l*5,0,1)/J2eV,'k') #plt.plot(r*1e9,ion_dipole(1.8*D,r,l*5,np.pi,1)/J2eV,'k') #plt.ylim(-9.2,9.2) #plt.xlim(0,0.5) #plt.xlabel("distance [nm]") #plt.ylabel("energy [eV]") #plt.title("dipole charge interaction") #plt.savefig("img/dipole_charge_interaction.png") #plt.show() ``` ```{python} #| tags: [] # Create the figure fig = go.Figure() # Plot the curves (corresponding to the lines in the Matplotlib version) fig.add_trace(go.Scatter( x=r * 1e9, # Convert meters to nanometers y=ion_dipole(1.8 * D, r, l, 0, 1) / J2eV, # Plot 1 mode='lines', line=dict(color='black', dash='dash'), name="phi=0, l=l" )) fig.add_trace(go.Scatter( x=r * 1e9, y=ion_dipole(1.8 * D, r, l, np.pi, 1) / J2eV, # Plot 2 mode='lines', line=dict(color='black', dash='dash'), name="phi=pi, l=l" )) fig.add_trace(go.Scatter( x=r * 1e9, y=ion_dipole(1.8 * D, r, l * 5, 0, 1) / J2eV, # Plot 3 mode='lines', line=dict(color='black', dash='solid'), name="phi=0, l=5l" )) fig.add_trace(go.Scatter( x=r * 1e9, y=ion_dipole(1.8 * D, r, l * 5, np.pi, 1) / J2eV, # Plot 4 mode='lines', line=dict(color='black', dash='solid'), name="phi=pi, l=5l" )) # Set layout details fig.update_layout( xaxis_title="Distance [nm]", yaxis_title="Energy [eV]", title="Dipole Charge Interaction", xaxis_range=[0, 0.5], # Limit x-axis from 0 to 0.5 nm yaxis_range=[-9.2, 9.2], # Limit y-axis from -9.2 to 9.2 eV width=700, height=500 ) pio.write_image(fig, "img/dipole_charge_interaction.png", scale=2) # Show the figure fig.show() ``` ::: ::: {.content-visible unless-format="html"} ![](img/dipole_charge_interaction.png) ::: The table below shows some selected hydration properties of ions. The hydrated radius determines the diffusion of the ion in water. The hydration number is the number of orientationally bound water molecules. Typically, smaller ions have a larger hydration number. | Ion | Bare Ion radius (nm) | Hydrated radius (nm) | Hydration number | | :------ | :----------: | :----------: | :----------: | | Na+ | 0.095 | 0.36 | 4 | | Mg2+ | 0.065 | 0.43 | 6 | | Cl-| 0.181| 0.33 | 1 | |OH-|0.176 | 0.3 | 3 | ## Dipole-Dipole Interactions The interaction of two dipole will depend on three angles, the two angles of the dipoles with the connecting axis $\theta_1, \theta_2$ and the angle $\phi$ between the two planes which contain the individual dipoles and the axis. The figure below shows the corresponding geometry. ![Dipole](img/dipole_dipole.png) ::: {.content-visible when-format="html"} ```{python} #| tags: [] # Constants epsilon_0 = 8.854e-12 # Vacuum permittivity (F/m) epsilon = 1 # Relative permittivity (adjust as needed) u1 = 1e-30 # Dipole moment 1 (C·m) u2 = 1e-30 # Dipole moment 2 (C·m) r_dipoles = 1e-9 # Distance between the dipoles (1 nm) # Dipole setup dipole_length = 1 # Length of each dipole from midpoint to each charge (in arbitrary units) midpoint_distance = 2 # Distance between the two midpoints on the horizontal line (arbitrary units) # Define the dipole-dipole interaction energy function def dipole_energy(theta1, theta2, phi, u1=u1, u2=u2, r=r_dipoles, epsilon=epsilon): return - (u1 * u2) / (4 * np.pi * epsilon_0 * epsilon * r**3) * ( 2 * np.cos(theta1) * np.cos(theta2) - np.sin(theta1) * np.sin(theta2) * np.cos(phi) ) # Define the function to calculate the positions of dipoles based on theta1 and theta2 def dipole_positions(theta1, theta2): # Midpoints on the dotted line mid1_x, mid1_y = -midpoint_distance / 2, 0 # Midpoint for dipole 1 mid2_x, mid2_y = midpoint_distance / 2, 0 # Midpoint for dipole 2 # Calculate the positions of the positive and negative charges for dipole 1 x1_pos = mid1_x + dipole_length * np.cos(theta1) # Positive end of dipole 1 y1_pos = mid1_y + dipole_length * np.sin(theta1) x1_neg = mid1_x - dipole_length * np.cos(theta1) # Negative end of dipole 1 y1_neg = mid1_y - dipole_length * np.sin(theta1) # Calculate the positions of the positive and negative charges for dipole 2 x2_pos = mid2_x + dipole_length * np.cos(theta2) # Positive end of dipole 2 y2_pos = mid2_y + dipole_length * np.sin(theta2) x2_neg = mid2_x - dipole_length * np.cos(theta2) # Negative end of dipole 2 y2_neg = mid2_y - dipole_length * np.sin(theta2) return (x1_pos, y1_pos, x1_neg, y1_neg), (x2_pos, y2_pos, x2_neg, y2_neg) # Function to plot dipoles and display energy def plot_dipoles(theta1=np.pi/4, theta2=np.pi/4, phi=np.pi/4): # Calculate dipole positions (x1_pos, y1_pos, x1_neg, y1_neg), (x2_pos, y2_pos, x2_neg, y2_neg) = dipole_positions(theta1, theta2) # Calculate dipole-dipole interaction energy energy = dipole_energy(theta1, theta2, phi) # Create figure fig = go.Figure() # Add Dipole 1 fig.add_trace(go.Scatter(x=[x1_neg, x1_pos], y=[y1_neg, y1_pos], mode='lines+markers', name="Dipole 1", marker=dict(size=20, symbol="circle-open"))) fig.add_annotation(x=x1_pos, y=y1_pos, text="+", showarrow=False, font=dict(size=20)) fig.add_annotation(x=x1_neg, y=y1_neg, text="-", showarrow=False, font=dict(size=20)) # Add Dipole 2 fig.add_trace(go.Scatter(x=[x2_neg, x2_pos], y=[y2_neg, y2_pos], mode='lines+markers', name="Dipole 2", marker=dict(size=20, symbol="circle-open"))) fig.add_annotation(x=x2_pos, y=y2_pos, text="+", showarrow=False, font=dict(size=20)) fig.add_annotation(x=x2_neg, y=y2_neg, text="-", showarrow=False, font=dict(size=20)) # Add horizontal reference line (the dotted line in your image) fig.add_trace(go.Scatter(x=[-midpoint_distance, midpoint_distance], y=[0, 0], mode="lines", line=dict(dash="dot", color="black"), showlegend=False)) # Display the energy as text in the plot fig.add_annotation(x=0, y=1.5, text=f"Energy: {energy:.3e} J", showarrow=False, font=dict(size=18), xref="x", yref="y") # Set axis limits and layout fig.update_layout( xaxis=dict(range=[-midpoint_distance - 1, midpoint_distance + 1], zeroline=False), yaxis=dict(range=[-2, 2], zeroline=False), title=f"Dipoles with θ₁ = {np.degrees(theta1):.1f}°, θ₂ = {np.degrees(theta2):.1f}°, ϕ = {np.degrees(phi):.1f}°", width=700, height=500, showlegend=False, font=dict(size=16) ) # Show figure fig.show() # Create interactive sliders for theta1, theta2, and phi interact(plot_dipoles, theta1=(0, 2*np.pi, 0.1), theta2=(0, 2*np.pi, 0.1), phi=(0, 2*np.pi, 0.1)); ``` ::: The energy of the two dipoles at a distance $r$ follows then without a detailed calculation with $$ w(r,\theta_2,\theta_2,\phi)=-\frac{u_1 u_2}{4\pi \epsilon_0 \epsilon r^3}\left [ 2\cos(\theta_1) \cos(\theta_2)-\sin(\theta_1)\sin(\theta_2)\cos(\phi)\right ] $$ The image below (taken from the book by Israelachvili) shows the dipole dipole interaction energy for the **parallel** and the **in-line** configuration as indicated. As shown, the in-line configutation with facing opposite charges is more favorable. ![Dipole Energy](img/dipole_energy.png) The energies that are typically found are also larger than the thermal energy at room temperature and thefore lead to ordering effects. ## Rotating Dipoles, angle averaged Potential The equations we derived so far deliver the energy for a specific fixed orientation of dipoles for example. However, molecules with dipoles may rotate and undergo rotational Brownian motion driven by thermal energy. This rotational diffusion can be very fast and the rotational sampling creates an average interaction that may scale differently with the distance. To get an effective distance dependence in the interaction, we have to integrate the Boltzmann factor over the orientational degrees of freedoms, i.e. $$ \exp\left ( -\frac{w(r)}{k_B T} \right)=\frac{\int \exp\left (-\frac{w(r,\Omega)}{k_B T} d\Omega\right )}{\int d\Omega}=\left \langle \exp\left (-\frac{w(r,\Omega)}{k_B T}\right ) \right \rangle_{\Omega} $$ with $$d\Omega =\sin(\theta)d\theta d\phi$$ and $$\int d\Omega=4\pi$$ $$ \exp\left ( -\frac{w(r)}{k_B T} \right) =\frac{1}{4\pi}\int_0^{2\pi}d\phi\int_0^{\pi}\exp\left ( -\frac{w(r,\phi\theta)}{k_B T} \right)\sin(\theta)d\theta $$ When $w(r,\theta)<< k_B T$ then we can expand the Boltzmann factor $$ \exp\left ( -\frac{w(r)}{k_B T} \right) \approx 1-\frac{w(r)}{k_B T}+\frac{1}{2}\left (\frac{w(r)}{k_B T} \right )^{2}+\ldots $$ and therefore obtain $$ w(r)=\langle w(r,\Omega)- \frac{w(r,\Omega)^2}{2k_B T}+\ldots \rangle_{\Omega} $$ which we can readily use to calculate the interaction energy of a charge and a dipole $$ \begin{aligned} w(r)=&\langle -\frac{Q u \cos(\theta)}{4\pi\epsilon_0\epsilon r^2} - \left ( \frac{Q u }{4\pi\epsilon_0\epsilon r^2}\right)^2 \frac{\cos(\theta)^2}{2k_B T}+\ldots \rangle\\ =& \frac{Q^2 u^2}{6(4\pi\epsilon_0\epsilon)^2k_B T r^4} \end{aligned} $$ when $k_B T>\frac{Q u }{4\pi\epsilon_0\epsilon r^2}$. Interestingly, this interaction energy for the freely rotating dipole in the electric field of an ion decays as $r^{-4}$, while the static interation decayed as $r^{-2}$. Also, the interaction of the freely rotating dipole is always attrative as compared to the static dipole charge interaction. The previous result can also be obtained when using $G=-k_B T \ln(Z)$, where $$ Z=\int \exp(-\frac{w(r,\Omega)}{k_B T}) d\Omega $$ is the partition function. ## Dipole-Dipole interaction with rotating dipoles We can extend the above calculation now also to two dipoles interacting when freely rotating. The interaction energy was given by $$ w(r,\theta_2,\theta_2,\phi)=-\frac{u_1 u_2}{4\pi \epsilon_0 \epsilon r^3}\left [ 2\cos(\theta_1) \cos(\theta_2)-\sin(\theta_1)\sin(\theta_2)\cos(\phi)\right ] $$ and therefore the partition function can be calculated as $$ Z=\int_0^{\pi}\int_{0}^{\pi}\int_{0}^{2\pi}e\exp(-\beta w(r,\theta_1,\theta_2,\phi))\sin(\theta_1)\sin(\theta_2)d\phi d\theta_1 d\theta_2 $$ The result of this calculation is the **Keesom interaction energy** $$ w(r)=-\frac{u_1^{2}u_{2}^2}{3(4\pi\epsilon\epsilon_0)^2 k_B T r^6} $$ when $$ k_BT >\frac{u_1 u_2}{4\pi\epsilon_0 \epsilon r^3} $$ The Keesom interaction tells on one side that permanent dipoles align themselves to be always attractive. It further belongs to a set of 3 different electrostatic interactions, which scale as $r^{-6}$ and which are termed **van der Waals** interactions. ## Interactions involving polarizability All atomic and molecular systems carry the potential to get polarized in an external electric field. This polarization gives rise to a dipole moment $u_{ind}$, that is in linear response proportional to the external electric field. The dipole itself also creates an electric field. There are different types of polarizability. We will address two of them, i) the electronic polarizability, i.e. shifting the center of mass of positive and negative charges of a molecule or atom, ii) the reorientation of a permanent electric dipole in and external electric field which is an orientational polarization. ![](img/polarizability.png) We consider a very simple classical model for the electronic polarizability. An electron is orbiting a positive nucleus and is subjected to en electric field according to the above image. In the stationary state, the forces on the charges exerted by the external electric field and the internal forces balance and give rise to the polarized state with a dipole moment. The external force on each charge is given by $$ F_{ext}=-eE $$ while the internal force between the two charges is $$ F_{int}=-\frac{e^{2}}{4\pi \epsilon_0 R^{2}}\sin(\theta)\approx \frac{e^{2}l}{4\pi \epsilon_0 R^{3}}=-\frac{e u_{ind}}{4\pi \epsilon_0 R^3}=F_{ext} $$ We can transform the latter equation to yield the induced dipole moment $$ u_{ind}=4\pi\epsilon_0 R^3 E=\alpha_0 E $$ Accordingly, the electronic polarizability of the atom is $$ \alpha_0=4\pi\epsilon_0 R^3 $$ which just says that the polarizability scales with the volume of the object, which is found in many occations, including the scattering and absorption cross sections of nanoparticles. ## Ion-Induced Dipole Interaction Ion-induced dipole interactions occur when an ion distorts the electron cloud of a nearby neutral molecule, creating an induced dipole. This interaction can be described through the following key points: ### Field of the Induced Dipole The electric field $E$ created by an ion at a distance $r$ induces a dipole moment $\mu_{\text{ind}}$ in a neutral molecule: $$ \mu_{\text{ind}} = \alpha E $$ where $\alpha$ is the polarizability of the molecule, and $E$ is given by: $$ E = \frac{q}{4 \pi \epsilon_0 r^2} $$ The resulting induced dipole field $E_d$ is: $$ E_d = -\frac{2 \mu_{\text{ind}}}{4 \pi \epsilon_0 r^3} = -\frac{2 \alpha E}{4 \pi \epsilon_0 r^3} = -\frac{2 \alpha q}{(4 \pi \epsilon_0)^2 r^5} $$ ### Force and Potential Energy The force $F$ between the ion and the induced dipole is given by: $$ F(r) = -\frac{\partial U}{\partial r} = -2 \alpha \left( \frac{q}{4 \pi \epsilon_0 r^2} \right)^2 = -\frac{2 \alpha q^2}{(4 \pi \epsilon_0)^2 r^5} $$ Integrating this force gives the potential energy $U$: $$ U(r) = \int F \, dr = -\frac{\alpha q^2}{(4 \pi \epsilon_0)^2 r^4} $$ This shows that the interaction energy decreases rapidly with the fourth power of the distance. ### Polarizability by Polar Molecules For polar molecules, an electric field $E$ induces a dipole moment that can be averaged over all orientations, leading to an additional orientational polarization. The total induced dipole moment $\mu_{\text{ind}}$ for a polar molecule with permanent dipole moment $\mu_0$ is: $$ \mu_{\text{ind}} = \mu_0 \cos(\theta) + \alpha E \cos^2(\theta) $$ Averaging over all orientations gives the mean induced dipole moment: $$ \langle \mu_{\text{ind}} \rangle = \frac{\mu_0^2 E}{3 k_{\text{B}} T} $$ Therefore, the overall polarizability $\alpha_{\text{total}}$ is a combination of the intrinsic polarizability $\alpha_0$ and the orientational contribution: $$ \alpha_{\text{total}} = \alpha_0 + \frac{\mu_0^2}{3 k_{\text{B}} T} $$ For example, for an ion-induced dipole interaction at 300 K with $\mu_0 = 4 \times 10^{-30} \, \text{Cm}$ and $\alpha_0$: $$ \alpha_{\text{total}} = 4 \pi \epsilon_0 \times 8 \times 10^{-30} \, \text{m}^3 + \frac{\mu_0^2}{3 k_{\text{B}} T} $$ These principles explain how ion-induced dipole interactions influence the behavior of molecules in various environments, including solutions and biological systems.