Lecture 8

Bacteria Forward Motion

How do bacteria achieve forward motion?

source

Bacteria have flagella that drive their propulsion. Bacteria flagella move in a corkscrew pattern. Flagella do not flap back and forth like a paddle. Why do flagella corkscrew instead of flapping in order to achieve forward motion? The last two minutes of this video illustrates why a bacteria would want to have its flagella move in a corkscrew instead of in a rowing motion.

This has to do with the Reynolds Number associated with the bacteria’s environment. Reynolds number may be expressed as
$$R = \frac{\rho v l}{\eta} .$$
$\rho$ is the density of the fluid,$\eta$ is the viscosity of the fluid, $l$ is the length scale of the object in the fluid, and $v$ is its velocity.

Reynolds Number Discussion

We can gain some insight into the dynamics of a bacteria’s motion by calculating its Reynolds number and comparing it to other cases.

In Water

$\rho = 10^3 \text{ kg}/\text{m}^3 \ \ $ and $\ \ \eta = 10^{-3} \text{ kg}/(\text{m s})$

• Fish

  • $l \sim 1 \text{cm}$
  • $v \approx 10^{-1} \text{m/s}$
    $\Rightarrow R \approx 10^3 >> 1$

• Bacteria

  • $l \sim 1 \mu\text{m}$
  • $v \approx 30\mu\text{m/s}$
    $\Rightarrow R \approx 3 \cdot 10^{-5} << 1$

In Molasses

$\eta \approx 10\text{ kg}/(\text{m s})$

• Fish
  $\Rightarrow R \approx 10^{-1} << 1$

From the bacteria’s perspective moving through water is analogous to a macroscopic object moving through syrup. Hence, observing macroscopic objects in very viscous fluids (like molasses) can give us insight into bacteria motion since they also experience an environment with very low Reynolds number.

Reynolds Number Experiment

Rods in Viscous Fluids

Observing how rods fall in a viscous fluid can give us insight into how bacteria move. The video above also gives examples of rods falling through viscous fluids (skip the first minute to get to the relevant parts). The video shows examples of rods falling horizontally, vertically and diagonally.

The drag force on an object is $F_\text{drag} = bv$; $\ b$ is the drag coefficient and depends on the length scale of the object and the fluid that it is in.

A rod falling vertically feels half of the drag froce that an identical rod falling horizontally experiences. This is because of the difference in the drag coefficients between the two scenarios.
$$b_\text{horz} = 4\pi\eta l \\ b_\text{vert} = \frac{1}{2}b_\text{horz} = 2\pi\eta l \\ $$

A rod falling diagonally will feel a drag force which has a component due to horizontal drag as well as vertical drag. The drag force in the perpendicular direction will be greater in magnitude than that in the parallel direction, $F_\perp > F_\parallel$. This will result in a net motion of the rod moving both down and to the side because of the inequality in the two drag components.

Comparison to Bacteria

We want to examine how the flagellum of a bacteria compares to the falling rod.

For our purposes lets only focus on the flagellum, not on the accompanying cilia diagram.

Assume that the flagellum is rotated a a frequency $f$. Lets consider a small section of the flagellum with length $l$. The whole flagellum moves in a corkscrew with angular velocity $\omega$ the motion traces out a circle of diameter $D$ and the flagellum makes an angle $\theta$ with the passive part. The angle $\theta$ is then the angle the small section makes normal to the basal body.

The forces acting on the small section is analogous to the forces acting on the diagonally falling rod except that gravity is negligible in this case. The velocity $v$ of the small section is radially outwards. It also experiences two drag forces, one point parallel to the section $F_\parallel$ and one point perpendicular $F_\perp$.
$$F_\perp = b_\perp v_\perp \\ F_\parallel = b_\parallel v_\parallel$$
$v_\perp$ and $v_\parallel$ are the components of the velocity vector pointing in the respective directions.
$$v_\perp = v\cos\theta \\ v_\parallel = v\sin\theta$$
Likewise, the drag coefficients are the same as those of the rod falling vertically or horizontally.
$$b_\perp = 4\pi\eta l \\ b_\parallel = 2\pi\eta l$$
The propulsive force $F_\text{p}^l$ that this small section exerts is equal to the components of the two drag forces that point in the direction of forward motion (normal to the basal body).
$$F_\text{p}^l = F_\perp \sin\theta - F_\parallel \cos\theta \\ F_\text{p}^l = 2\pi \ \eta \ l \ v\cos\theta\sin\theta$$

The total propulsive force of the whole flagellum is the sum of all the contributions from each small section.
$$F_\text{p} = N F_\text{p}^l = \frac{L}{l} F_\text{p}^l \\ F_\text{p} = 2\pi \ \eta \ L \ v\cos\theta\sin\theta \\ v = \pi D f \\ \Rightarrow F_\text{p} = 2\pi^2 \eta L D\cos\theta\sin\theta$$

Lets estimate the terminal velocity of a bacteria.

Approximate the bacteria as a rod of length $L$ moving at speed $u$. At terminal velocity, $F_\text{p} = F_\text{drag}$.
$$F_\text{drag} = 2\pi\eta Lu \\ \Rightarrow u = \pi Df\cos\theta\sin\theta$$
For a typical bacteria this equates to a terminal velocity $u_t \approx 50 \mu\text{m/s}$ which is very close to observed velocities $\sim 30 \mu\text{m/s}$.

Flagellar Efficiency

Purcell wrote a paper deriving the efficiency of flagellar propulsion.

source

In this case we assume that cell body is spherical and rotates with angular velocity $\Omega$ and the flagellum rotates at angular velocity $\omega$. The relative anglular velocity between the body and flagellum is $\omega_0 = \omega + \Omega$.

Next we want to calculate to forces and torques acting on both the body and the flagellum.

Forces and Torques

Cell Body
$$F=-Bv \\ \tau = D\Omega$$
$B$ and $D$ are drag coefficients. $B=6\pi\eta R$ and $D=8\pi\eta R^3$.

Flagellum
$$F = -bv + a\omega \\ \tau = cv-d\omega$$
Rotation of the flagellum causes a drag force and forward motion also causes a drag torque. In Purcell’s paper it was shown that $a=c$. So our system of equations simplifies to the following:
$$\begin{bmatrix} F \\ \tau \end{bmatrix} = \begin{bmatrix} -b & c \\ c& -d \end{bmatrix} \begin{bmatrix} v \\ \omega \end{bmatrix}$$

The constants scale as… $b\sim l$, $c\sim l^2$, $d\sim l^3$.

Balance Forces and Torques

If the bacteria’s angular and linear velocities constant then the net force and torque on the whole bacteria must be zero. This leads to the following two equations:
$$a\omega = bv+Bv \\ d\omega = cv + D\Omega$$
We can equate $a\omega$ with the propulsive $F_\text{p}$ that we calculated earlier and likewise we can think of $cv$ as the analogous propulsive torque.

In summary we have three equations

1. $\omega_0 = \omega + \Omega$
2. $a\omega = bv+Bv$
3. $d\omega = cv + D\Omega$

$$\Rightarrow v = \frac{cD}{(B+b)(D+d)-c^2} \ \omega_0$$

If we assume that $c^2 << bd$ and $D>>d$ then the above equation simplifies to
$$v \approx \frac{c}{B+b} \ \omega_0 .$$

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Lecture 8

Bacteria Forward Motion

How do bacteria achieve forward motion?

source

Bacteria have flagella that drive their propulsion. Bacteria flagella move in a corkscrew pattern. Flagella do not flap back and forth like a paddle. Why do flagella corkscrew instead of flapping in order to achieve forward motion? The last two minutes of this video illustrates why a bacteria would want to have its flagella move in a corkscrew instead of in a rowing motion.

This has to do with the Reynolds Number associated with the bacteria’s environment. Reynolds number may be expressed as
$$R = \frac{\rho v l}{\eta} .$$
$\rho$ is the density of the fluid,$\eta$ is the viscosity of the fluid, $l$ is the length scale of the object in the fluid, and $v$ is its velocity.

Reynolds Number Discussion

We can gain some insight into the dynamics of a bacteria’s motion by calculating its Reynolds number and comparing it to other cases.

In Water

$\rho = 10^3 \text{ kg}/\text{m}^3 \ \ $ and $\ \ \eta = 10^{-3} \text{ kg}/(\text{m s})$

• Fish

  • $l \sim 1 \text{cm}$
  • $v \approx 10^{-1} \text{m/s}$
    $\Rightarrow R \approx 10^3 >> 1$

• Bacteria

  • $l \sim 1 \mu\text{m}$
  • $v \approx 30\mu\text{m/s}$
    $\Rightarrow R \approx 3 \cdot 10^{-5} << 1$

In Molasses

$\eta \approx 10\text{ kg}/(\text{m s})$

• Fish
  $\Rightarrow R \approx 10^{-1} << 1$

From the bacteria’s perspective moving through water is analogous to a macroscopic object moving through syrup. Hence, observing macroscopic objects in very viscous fluids (like molasses) can give us insight into bacteria motion since they also experience an environment with very low Reynolds number.

Reynolds Number Experiment

Rods in Viscous Fluids

Observing how rods fall in a viscous fluid can give us insight into how bacteria move. The video above also gives examples of rods falling through viscous fluids (skip the first minute to get to the relevant parts). The video shows examples of rods falling horizontally, vertically and diagonally.

The drag force on an object is $F_\text{drag} = bv$; $\ b$ is the drag coefficient and depends on the length scale of the object and the fluid that it is in.

A rod falling vertically feels half of the drag froce that an identical rod falling horizontally experiences. This is because of the difference in the drag coefficients between the two scenarios.
$$b_\text{horz} = 4\pi\eta l \\ b_\text{vert} = \frac{1}{2}b_\text{horz} = 2\pi\eta l \\ $$

A rod falling diagonally will feel a drag force which has a component due to horizontal drag as well as vertical drag. The drag force in the perpendicular direction will be greater in magnitude than that in the parallel direction, $F_\perp > F_\parallel$. This will result in a net motion of the rod moving both down and to the side because of the inequality in the two drag components.

Comparison to Bacteria

We want to examine how the flagellum of a bacteria compares to the falling rod.

For our purposes lets only focus on the flagellum, not on the accompanying cilia diagram.

Assume that the flagellum is rotated a a frequency $f$. Lets consider a small section of the flagellum with length $l$. The whole flagellum moves in a corkscrew with angular velocity $\omega$ the motion traces out a circle of diameter $D$ and the flagellum makes an angle $\theta$ with the passive part. The angle $\theta$ is then the angle the small section makes normal to the basal body.

The forces acting on the small section is analogous to the forces acting on the diagonally falling rod except that gravity is negligible in this case. The velocity $v$ of the small section is radially outwards. It also experiences two drag forces, one point parallel to the section $F_\parallel$ and one point perpendicular $F_\perp$.
$$F_\perp = b_\perp v_\perp \\ F_\parallel = b_\parallel v_\parallel$$
$v_\perp$ and $v_\parallel$ are the components of the velocity vector pointing in the respective directions.
$$v_\perp = v\cos\theta \\ v_\parallel = v\sin\theta$$
Likewise, the drag coefficients are the same as those of the rod falling vertically or horizontally.
$$b_\perp = 4\pi\eta l \\ b_\parallel = 2\pi\eta l$$
The propulsive force $F_\text{p}^l$ that this small section exerts is equal to the components of the two drag forces that point in the direction of forward motion (normal to the basal body).
$$F_\text{p}^l = F_\perp \sin\theta - F_\parallel \cos\theta \\ F_\text{p}^l = 2\pi \ \eta \ l \ v\cos\theta\sin\theta$$

The total propulsive force of the whole flagellum is the sum of all the contributions from each small section.
$$F_\text{p} = N F_\text{p}^l = \frac{L}{l} F_\text{p}^l \\ F_\text{p} = 2\pi \ \eta \ L \ v\cos\theta\sin\theta \\ v = \pi D f \\ \Rightarrow F_\text{p} = 2\pi^2 \eta L D\cos\theta\sin\theta$$

Lets estimate the terminal velocity of a bacteria.

Approximate the bacteria as a rod of length $L$ moving at speed $u$. At terminal velocity, $F_\text{p} = F_\text{drag}$.
$$F_\text{drag} = 2\pi\eta Lu \\ \Rightarrow u = \pi Df\cos\theta\sin\theta$$
For a typical bacteria this equates to a terminal velocity $u_t \approx 50 \mu\text{m/s}$ which is very close to observed velocities $\sim 30 \mu\text{m/s}$.

Flagellar Efficiency

Purcell wrote a paper deriving the efficiency of flagellar propulsion.

source

In this case we assume that cell body is spherical and rotates with angular velocity $\Omega$ and the flagellum rotates at angular velocity $\omega$. The relative anglular velocity between the body and flagellum is $\omega_0 = \omega + \Omega$.

Next we want to calculate to forces and torques acting on both the body and the flagellum.

Forces and Torques

Cell Body
$$F=-Bv \\ \tau = D\Omega$$
$B$ and $D$ are drag coefficients. $B=6\pi\eta R$ and $D=8\pi\eta R^3$.

Flagellum
$$F = -bv + a\omega \\ \tau = cv-d\omega$$
Rotation of the flagellum causes a drag force and forward motion also causes a drag torque. In Purcell’s paper it was shown that $a=c$. So our system of equations simplifies to the following:
$$\begin{bmatrix} F \\ \tau \end{bmatrix} = \begin{bmatrix} -b & c \\ c& -d \end{bmatrix} \begin{bmatrix} v \\ \omega \end{bmatrix}$$

The constants scale as… $b\sim l$, $c\sim l^2$, $d\sim l^3$.

Balance Forces and Torques

If the bacteria’s angular and linear velocities constant then the net force and torque on the whole bacteria must be zero. This leads to the following two equations:
$$a\omega = bv+Bv \\ d\omega = cv + D\Omega$$
We can equate $a\omega$ with the propulsive $F_\text{p}$ that we calculated earlier and likewise we can think of $cv$ as the analogous propulsive torque.

In summary we have three equations

1. $\omega_0 = \omega + \Omega$
2. $a\omega = bv+Bv$
3. $d\omega = cv + D\Omega$

$$\Rightarrow v = \frac{cD}{(B+b)(D+d)-c^2} \ \omega_0$$

If we assume that $c^2 << bd$ and $D>>d$ then the above equation simplifies to
$$v \approx \frac{c}{B+b} \ \omega_0 .$$

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Lecture 7

We will start be continuing in our answer to the question:

What shapes emerge from beinding energy considerations alone?

Pancake

A pancake is a cylinder with rounded sides.

Let the rounded sides have height $2r$ and the radius of the enclosed cylinder be $R$.

For $R>>r$:
$$\Delta E \approx \pi^2 B \frac{R}{r} \\ V \approx \pi R^2 (2r) \\ A \approx 2 \pi R^2$$
Knowing the volume and surface area, we can calculate the reduced volume of the pancake.
$$\bar{V} = \frac{6r}{\sqrt{2} R} \\ \text{let } \rho \equiv \frac{R}{r} \ \Rightarrow \bar{V} = \frac{6}{\sqrt{2}}\frac{1}{\rho}\\ \Delta \bar{E} = \pi^2\rho \\ \bar{V} \approx \frac{6\pi^2}{\sqrt{2}\Delta \bar{E}}$$

Doublet

This is a sphere with a smaller sphere growing out of it. It can be thought of as simple model of cell division where a circular cell is being pinched out of another. Let the radius of the large sphere be $R$ and that or the smaller, pinched sphere be $r$.
$$V = \frac{4}{3}\pi (R^3+r^3) \\ A = 4\pi (R^2+r^2)$$
We can use this to calculate the reduced volume.
$$\Rightarrow \bar{V} = \frac{\rho^3 + 1}{(\rho^2+1)^{3/2}}$$
Again, $\rho \equiv R/r$. Lets examine some limiting cases for different values of $\rho$.

$\rho$		$\bar{V}$
$\rightarrow \infty$		$=1$
$=1$		$=\frac{1}{\sqrt{2}}$
$\rightarrow 0$		$=1$

Also note that since we are dealing with two spheres $\Delta E$ is simply the bending energy of a sphere which is a constant, independent of radius.
$$\Rightarrow \Delta \bar{E} = 4\pi(2B+G)$$

Stomatocyte

This shape is the opposite of a doublet. Instead of a sphere growing out from the edge of another sphere, this is a sphere with the volume of a smaller sphere being removed from it. The larger sphere has radius $R$, and the smaller sphere that is removing volume has radius $r$.
$$V = \frac{4}{3}\pi (R^3-r^3) \\ A = 4\pi (R^2+r^2)$$
Notice that the total volume is the difference rather than the sum of the two spheres whereas the surface area remains the same as that of a doublet. This means that $\Delta \bar{E}$ is the same as the doublet case and the reduced volume is very similar.
$$\bar{V} = \frac{\rho^3 - 1}{(\rho^2+1)^{3/2}}$$
As in the doublet case, lets examine some limiting cases.

$\rho$		$\bar{V}$
$\rightarrow \infty$		$=1$
$\rightarrow 0$		$=0$

Shape phase space

If we were to plot $\Delta\bar{E}$ as a function of $\bar{V}$ we would find a phase space of where different shaped membranes exist. In certain regimes of reduced volume varying shapes are energetically favorable.

However, our arguments breakdown if we examine bacteria instead of ordinary cells. Bacteria are single cell organisms which have a hard cell wall. The phase space does not apply to them since their shape is defined by a much more rigid cell wall.

Bacteria Cell Walls

Bacteria keep thier internal pressure much higher than the external pressure in the environment. This is why they need something much harder than a cell membrane in order to keep their shape intact.

Lets examine the forces a cell wall must sustain for walls of different shapes. In the following examples let the outward pressure the cell wall feels be $P$ and the thickness of the wall be $d$.

It is also useful to define the quantities stress and tension.
$$\text{Stress } = \sigma \equiv \frac{\text{Force}}{\text{Area}} \\ \text{Tension } = \tau \equiv \frac{\text{Force}}{\text{Length}}$$

Spherical Cell Wall

In the spherical case we can imagine cutting the sphere in half and consider the force $F$ needed to keep these two halves together.
$$F = \int da \ P \cos\theta \\ da = R^2 \sin\theta \ d\theta \ d\phi \\ \Rightarrow F = \int_0^{2\pi} d\phi \int_0^{\pi/2} d\theta \ R^2\sin\theta \ P \cos\theta \\ F = \pi R^2 P$$
And so the stress and tension the spherical cell wall feels are the following:
$$\sigma = \frac{RP}{2d} \\ \tau = \frac{RP}{2}$$

Spherocylinder

Spherocylinder is a very common shape of cell walls. There are two ways of pulling apart a spherocylinder.

Again, in our notation $h \rightarrow L$.

1. Pulling from either spherical end. This equates to forces pulling in the z-direction. This force will be denoted as $F_z$.
2. Pulling it apart through the z-axis. This equates to equal, opposite forces pushing in the x-y plane. this force will be denoted as $F_\phi$ since it is azimuthaly symmetric.

$F_z$ will be identical to that of spherical wall, and so will the stress and tension.
$$\sigma_z^\text{cyl} = \sigma^\text{sph}$$
We can calculate $F_\phi$ by imaging we cut the spherocylinder in half down the z-axis and consider the amount of force it would take to keep the two halves together.
$$F_\phi = \int da \ P\sin\phi \\ F_\phi = \int_0^L dz \int_0^{2\pi} d\phi \ R^2 P \sin\phi \\ F_\phi = 2RLP$$
And so the stress in the $\phi$ direction on the cell wall is the following:
$$\sigma_\phi = \frac{PR}{d} \\ \rightarrow \sigma_z = \frac{1}{2} \sigma_\phi$$
The stress on the spherical ends of the cell wall are half that of the cylindrical sides.

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1v5cells_comparison

Comparison of 1 and 5 cell Simulations

Simulation Set-Up

• Reflective boundary on the left.
• Absorbing boundary on the right.
• For the 5 cell simulation:
  
  • Cells are not connected.
  • Cells may not occupy the same lattice point.
  • Cell 1 is initialized on the reflective boundary
  • Cell 5 is initialized 5 lattice points to the right of cell 1.

Results (Updated 2/9/15)

The results can be found online here.

Passage Time

Update: Plots for the time-averages passage times have been updated. Added a plot of the average first passage time for each cell.

Lets compare the passage times of the 5 cells to that of the single cell diffusion simulation.

The following plots are created by making histograms of the simulation output data. In order to keep the data normalized throughout, a factor dependent on the bin size is added to the probability density. The probability density $P(t_i)$ at certain time is expressed as
$$P(t_i) = \frac{1}{\Delta t} \frac{c(t_i)}{\sum c(t_i)} \\ \Delta t \equiv \text{bin size }$$
The following plots are the histograms of each cell overlayed. One cell refers to the data from the single cell simulation.

Notice that the slope of the one cell and that of the 1st cell in the 5 cell simulation are very similar.

The average first passage times for each respective cell are plotted below.

An interesting ting to note, for the one cell case the first passage time is $\sim 10^4$ and the cell had to diffuse $10^2$ lattice points to get there.

Now lets plot the data while scaling the probability densities and the number of time steps by the mean passage time for each respective cell.
$$\text{let } \tilde{t}_i \equiv \frac{t_i}{\left<t_i\right>} \\ \rightarrow P(\tilde{t}) = \left<t_i\right>P(t)$$
The following plots are with respect to these new quantities, $P(\tilde{t})$ and $\tilde{t}_i$.

Looking more closely at the relationship between the first, last and individual cell:

Position

The following plots compare the probability densities of the position of each cell. The probability of the 5 cells and the one, individual cell are plotted together at different time steps. The probability densities are also plotted with respect to the averaged position.

We would expect that the first cell of the multiple cell simulation to have a normal distribution but it does not appear that way. Boundary conditions tell us that the slope of the first cell’s probability density should be zero at the reflecting boundary. Zooming in on the first cell for small positions shows that the slope approaches zero but the probability density is clearly not flat.

I believe this plot is at the 2500th time step. Looking at different time steps shows very much the same pattern. At small time steps the curve is slightly steeper and as time increases the slope very gradually decreases.

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Lecture 6

What determines the shape of grouped lipids?

Lipid Geometry

Micelle

A good guess is that conical shaped lipids are more likely to form micelles. Let a cone-shaped lipid have a length $l$, circular surface area $a$ and total volume $v$. Using these assumptions we can calculate the number of lipids $N$ that fit in a micelle of radius $R$.
$$N = \frac{4\pi R^2}{a} = \frac{\frac{4}{3}\pi R^3}{v} \\ \Rightarrow R = \frac{3v}{a}$$
Lets restrict $R<l$ since we want the lipids packed tightly together such that no water is in the center of the micelle.
$$\rightarrow \frac{3v}{a} < l \\ \text{Sphere Regime:} \\ \frac{v}{al} < \frac{1}{3}$$
This dimensionless quantity $v/al$ is referred to as the shape factor of the lipid. Different shape factors results in different shaped lipids which in turn tells us what kind of form a group of those lipids will take.

Cylinder

Imagine grouping lipids together such that they form a cylinder. We can calculate the number of lipids $N$ in that cylinder of length $L$ and radius $R$.
$$N = \frac{2\pi RL}{a} = \frac{\pi R^2 L}{v} \\ \Rightarrow R = \frac{2v}{a} < l \\ \text{Cylinder Regime:} \\ \frac{1}{3} \lesssim \frac{v}{al} \lesssim \frac{1}{2}$$

Bilayer

For a bilayer with surface area $L^2$ and depth $2R$:
$$N = \frac{2 L^2}{a} = \frac{L^2 2R}{v} \\ \Rightarrow \frac{v}{al} = 1 \\ \text{Bilayer Regime:} \\ \frac{1}{2} \lesssim \frac{v}{al} \lesssim 1$$

Inverted Micelle

An inverted micelle occurs in the regime with the largest shape factor.
$$\text{Inverted Micelle Regime:} \\ 1 \lesssim \frac{v}{al}$$

Typical Lipid

$$a \approx 0.5 \text{nm}^2 \\ l \approx 1.3 \text{nm} \\ v \approx 0.6 \text{nm}^3 \\ \Rightarrow \frac{v}{al} \approx 0.8$$

Therefore the typical grouped lipid shape is a bilayer.

Problems with Bilayers

• In the form of a single sheet, the edges are exposed which is energitically unfavorable.
  
  • This is quantified with an Exposure Energy.
• In the form of a spherical surface, there are no edges so this penalty is avoided. However, now there is an energy penalty due to the bending of the lipids.
  
  • This is quantified with a Binding Energy.

Exposure Energy

Define $\lambda$ as the line tension of the exposed side of a bilayer. $\lambda$ has units of energy per length. So we can set the exposure energy as the following:
$$\rightarrow E=\lambda L$$

Bending Energy

Define $C$ as the curvature of the bilayer. We can relate the radius of the curved bilayer to the curvature through the following expression:
$$C = \frac{1}{R}$$
The curvature can lie within the range $0 \leq C \leq \infty$, in which $0$ corresponds to a flat bilayer and $\infty$ corresponds to a sharp kink in the bilayer.

For a two dimensional surface - like a bilayer - there are two ways it can curve and so the bending energy has to take that into account. For bilayers this is usually expressed in terms of energy per unit area in the following way:
$$\frac{E}{A} = \frac{B}{2} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^2 + G \frac{1}{R_1 R_2}$$
$B$ and $G$ are parameters with units of energy which depend on the type of lipid bilayer. We can simplify this expression for the case of a spherical and cylindrical bilayer.
$$\text{Spherical Surface: } R_1 = R_2 \\ \frac{E}{A} = 4\pi (2B+G) \\ \text{Cylindrical Surface: } R_1=R, \ R_2 = \infty \\ \frac{E}{A} = \frac{\pi BL}{R}$$

Disk vs Sphere Comparison

Let the surface area of a flat disk shaped bilayer be equal to the surface area of a spherical bilayer (vesicle).
$$A = \pi (2R)^2 = 4\pi R^2$$
The two bilayers will have the following binding energies:
$$E_{\text{disk}} = 4\pi\lambda R \\ E_{\text{sphere}} = 4\pi (2B+G)$$
Since $E_{\text{sphere}} < E_{\text{disk}}$ a vesicle will form because it is energetically favorable.

$$E_{\text{sphere}} < E_{\text{disk}} \\ \Rightarrow 4\pi (2B+G) < 4\pi\lambda R \\ \lambda > \lambda_\text{min} \equiv \frac{2B+G}{R}$$
$\lambda_\text{min}$ is the minimum line tension for which the above inequality holds. For typical bilayers at room temperature $\lambda_\text{min} = 10^{-13} \text{J}/\text{m}$ and line tension has been observed to be on the order of $10^{-11}\text{J}/\text{m}$.

However, in this discussion we have not the entropy cost of such configurations. The bilayer sheet has higher entropy than the sphere since the positions the lipids can take in a vesicle are much more constrained. In order to account for this lets assume that the line tension is actually a function of lipid length and temperature.
$$\rightarrow \lambda_\text{min} \approx f(b,T) \\ \lambda_\text{min} = \frac{kT}{b}$$
For typical lipid chain lengths of 4nm this leads to a $\lambda_\text{min} \approx 10^{-12}\text{J}/\text{m}$.

Closed Bilayer Shape

What shape does a closed bilayer (vesicle) take from binding energy considerations alone?

Consider a vesicle with volume $V$ which is less than the volume of a sphere $V_\text{sph}$ but that has surface area equal to a sphere, $A = A_\text{sph}$.

Reduced Volume

We create a unit less measure called the reduced volume. The goal is to have a unit less measure that compares an objects volume and surface area. Hence, the reduced volume should have the following form in order to be dimensionless.
$$\Rightarrow \frac{V}{A^{3/2}} \\ \text{For a sphere: } \frac{V}{A^{3/2}} = \frac{1}{6\sqrt{\pi}}$$
Lets normalize the measure by the resulting ratio for a sphere. So the reduced volume $\bar{V}$ is defined as
$$\bar{V} \equiv \frac{6\sqrt{\pi}V}{A^{3/2}}$$
Note that the reduced volume ranges $0 \leq \bar{V} \leq 1$. Now lets calculate the reduced volume $\bar{V}$ for different shapes.

Spherocylinder (Rod)

In our case let $h \rightarrow L$ and define $l \equiv L/R$.
$$\bar{V} = \frac{4+3l}{\sqrt{2}(2+l)^{3/2}} \\ \text{If } l=0 \Rightarrow \bar{V} = 1 \\ \text{If } l>>1 \ \ \ \Rightarrow \bar{V} \approx \frac{3}{\sqrt{2l}}$$
Lets define $\Delta\bar{E} \equiv \Delta E / B$. Now let us calculate the bending energy and compare it to that of a sphere.
$$\Delta E = E - E_\text{sph} = \frac{\pi BL}{R} \\ \Delta \bar{E} = \pi l \\ \Rightarrow \bar{V} \sim \frac{1}{\Delta \bar{E}^{1/2}}$$

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