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This will be analogous to a conductor-resistor-conductor-battery circuit. The battery of voltage `V` gives rise to an electric field `vec E=V/L bb hat x` in the x direction in the container (resistor) of length `L`. The resistor's host particles gain energy from ion collisions. Since the ions lose energy, the sum of the host energy plus the ion energy slowly grows. We may choose to re-inject the ions that leave the output end with the defined energy into the input end. That avoids the need for the complicated conductor and battery path. The green host particles will provide an impedance for the red particles. We will discuss the flow of ion current in a conductor due to an electric field, `vec E`. The electric field provides an acceleration to the ions

`(d vecv)/dt=e/m_i vec E`

where `e` is the ion charge and `m_i` is its mass. The rms average thermal speed of the ions and host particles is zero. The acceleration continues until the ion reaches its mean free path `lambda_i` which, on average, occurs in time`tau=lambda_i/v_(rms)`

where `v_(rms)` is the root-mean-square thermal velocity of the ions:`v_(rms)=sqrt(1/Nsum_1^N(v_i^2))`

where `N` is the total number of ions.We expect the hard sphere collision cross section to be `sigma=2pi r^2`

where `r` is the particle radius which results in a simple value of `lambda_i`:`lambda_i=1/(n sigma)\ \ \ \ (1)`

where n is the particle density (particle/Volume). When the ion has accelerated a time `lambda_i/v_(rms)` it has gained an average velocity in the `bb hat x` direction`delta vec v=e vec E/m_i lambda_i/v_(rms)bb hat x \ \ \ \ (2)`

A simple treatment would conclude that `delta vec v` is a good approximation of the average drift velocity of the ions. The reasoning is that the ion's memory of its velocity gain is lost due to its collisions with a much more massive particle. However the numerical scattering results for average drift speed are larger than equation 1 would suggest. So a more functional value for the mean free path would be one where the collision energy change is sigificant. The energy change can be written in terms the initial velocity difference `bb "v"_1-bb "v" _2` and the unit vector `bb hat "u"` betweem the sphere centers:`delta E=E-E_0=(bb "p"'cdot bb "p"')/(2m)-(bb "p"cdot bb "p")/(2m)` ``

From Appendix 1 we obtain:`(deltaE_1)/E_1=(-2m_1Mbb hat"u"cdot(bb"v"_1-bb"v"_2)bb"v"_1cdotbbhat"u"+M^2(bb hat "u"cdot("v"_1-bb"v"_2))^2)/(m_1^2v_1^2)` ``

If we have a significant energy change upon scatter, then it is reasonable to use use the numerical result for `lambda_i`. Otherwise we should use equation 1 for `lambda_i`. So I have decided that if, after scatter `(deltaE_1)/E_1lt1/4`, then the mean free path will be that given by equation 1. Using these criteria, the drift speed given by equation 2 agrees quite well with the numerical drift speed. So, if the density of ions is `n_e` we can conclude that the ion current density, `vec J`, is the following`vec J_i=n_e e delta vec v=n_e e^2 vec E/m_i lambda_i/v_(rms)\ \ \ \ (amps)/("unit-area")`

or the number of ions per unit area and unit time flowing across a plane is:`vec (dn_i)/dt=n_e delta vec v=n_e e vec E/m_i lambda_i/v_(rms)\ \ \ \ ("ions")/("unit-area-time")`

The diffusing particles (ion spheres) are initially located in a very narrow band (patch) at the center of the container. Both host and ion spheres start with three dimensional (3D) random velocities and all have the same initial energy.

The red ion spheres are accelerated by the electric field due to the voltage, `V`, across the length, `L`, of the container. The electric field is expressed algebraically as `vec E=V/Lbbhat "x"` They also collide with the inner walls of the container and with other spheres of their own family as well as a much larger family of green host spheres. They eventually reach an average of the drift velocity given by equation 2. And this gives rise to an ion current. This is similar to the way a resistor works. Canvas 1 shows the particle x and y positions. The particles also have free movement (except for collision) in the z dimension (into the screen) but I have chosen not to show them in 3D. Plots of the ion density and ion x speed, `v_x` are shown in Canvas 2. Printed results for both physics equation mean free path and mean free path from numerical 3D scatter are shown on Canvas 2. Printed results for both physics equation drift speed and drift speed from numerical 3D scatter are also shown on Canvas 2. The animation runs until a small number of ions reach the right hand container wall where they would bounce backward and reduce the calculated drift speed which would invalidate the numerical drift speed result.

In an actual resistor, the neutral particles are partially bound so their reaction to collision is less than we have here. Here the neutral particles are mobile. So this animation is closer to ions moving through an electrolyte due to an applied potential as would exist in electro-plating of a metal surface i.e. anodizing a metal.

For hard sphere collisions, we need to use both momentum and energy conservation to find the final velocities of particles. The change of momentum of one particle is the negative of the change of momentum of the other:

`m_1deltabb"v"_1=Mdeltavbbhat"u"=-m_2deltabb"v"_2\ \ \ \ \(A1.1)`

where bold characters indicate vectors and where `M` and `deltav` are presently unknown and `bbhat"u"` is a unit vector pointing between the centers of the particles. Now we'll insert equation A1.1 into an expression for the conservation of kinetic energy.`((m_1bb"v"_1+Mdeltavbbhat"u")cdot (m_1bb"v"_1+Mdeltavbbhat"u"))/m_1+ ((m_2bb"v"_2-Mdeltavbbhat"u")cdot (m_2bb"v"_2-Mdeltavbbhat"u"))/m_2= m_1bb"v"_1cdotbb"v"_1+m_2bb"v"_2cdotbb"v"_2`

Here the centered dot between vector quantities indicates the dot or inner product defined as follows:`bb"a"cdot bb"b"=a_xb_x+a_yb_y+a_zb_z`

where the subscripts `x y z` indicate the cartesian components of the `a` and `b` vectors. Cancelling the terms on the left with the terms on the right we obtain:`2bb"v"_1cdot(Mdeltav bbhat"u")-2bb"v"_2cdot(Mdeltav bbhat"u")+ (Mdeltav)^2/m_1+ (Mdeltav)^2/m_2=0`

`2(Mdeltav bbhat"u")cdot(bb"v"_1-bb"v"_2)+(Mdeltav)^2(m_2+m_1)/(m_1m_2)=0`

Solving for `Mdeltav` we obtain:`Mdeltav=-2(m_1m_2)/(m_1+m_2)bb hat "u"cdot(bb"v"_1-bb"v"_2)`

So the results for `M` and `deltav` are:`M=(2m_1m_2)/(m_1+m_2)`

and`deltav=-bb hat"u"cdot(bb"v"_1-bb"v"_2)`

Thus `bb"v'"_1` becomes:`m_1bb"v'"_1=m_1bb"v"_1+Mdeltavbbhat"u"=m_1bb"v"_1-Mbb hat"u"cdot(bb"v"_1-bb"v"_2)bbhat"u"\ \ \ \ \(A1.2)`

and `bb"v'"_2` becomes:`m_1bb"v'"_2=m_2bb"v"_2-Mdeltavbbhat"u"=m_2bb"v"_2+Mbb hat"u"cdot(bb"v"_1-bb"v"_2)bbhat"u"\ \ \ \ \(A1.3)`

where `bb"v"'` indicates the value of the velocity vector
`deltaE_1=(m_1bb"v"_1'cdotbb"v"_1')/2-(m_1bb"v"_1cdotbb"v"_1)/2`

`deltaE_1=((m_1bb"v"_1+Mdeltavbbhat"u")cdot(m_1bb"v"_1+Mdeltavbbhat"u"))/(2m_1)-(m_1bb"v"_1cdotbb"v"_1)/2`

`deltaE_1=((2m_1bb"v"_1cdotMdeltavbbhat"u")+(Mdeltav)^2(bbhat"u"cdotbbhat"u"))/(2m_1)`

`deltaE_1=((-2m_1Mbb hat"u"cdot(bb"v"_1-bb"v"_2)bb"v"_1cdotbbhat"u")+(Mdeltav)^2)/(2m_1)`

`deltaE_1=(-2m_1Mbb hat"u"cdot(bb"v"_1-bb"v"_2)bb"v"_1cdotbbhat"u"+M^2(bb hat "u"cdot("v"_1-bb"v"_2))^2)/(2m_1)`