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Energy Distribution of a Three Dimensional Gas with Hard Sphere Collisions

The energy distribution of a gas leads to the evolution of the second law of thermodynamics. The second law states that the motional (kinetic) energy cannot go from a less energetic body to a more energetic body. Note that I have avoided the use of the word "temperature" because temperature is a very poorly defined concept. In this page we will see that the average kinetic energy takes the place of k*T where k is Boltzmann's constant and T is temperature. This page includes an active plot of the kinetic energy distribution versus energy. The particles (spheres since they are 3 dimensional) all start out with equal kinetic energies and tend toward their final equilibrium energy distribution which, for a simple monoatomic gas like this, for 2D the following expression is the Maxwell-Boltzmann distribution:
Canvas with Moving Gas Spheres
Single Step
Sphere Radius 5
Initial Energy 0.2

`p(E)=b sqrt(E/bar E)exp(-{3E}/{2bar E})`

where `E` is kinetic energy and `bar E` is the average kinetic energy of all particles.


Plot of Energy Distribution Histogram as Well as Exponential of Maxwell-Boltzmann Theory. Note that, after a few minutes, the red histogram matches the green theory curve quite accurately.
where `E` is kinetic energy and `bar E` is the average kinetic energy of all particles.
In this expression `p(E)` is the probability of a given energy and `b` is a normalization constant such that the integral of `p(E)` over all energies becomes 1.

`b int_{0}^{oo}p(E)dE=1`



Deriving the Equations for Final Velocities After Collision

In the following equations please be aware that bold letters denote vectors which have either (x,y) or (x,y,z) components depending on whether we are considering 2D or 3D calculations. For hard sphere collisions one must consider both conservation of momentum and energy. If the collision was mediated by a potential between the particles, one would only have to consider conservation of kinetic and potential energy. Conservation of momentum requires that the final momentum be the same as the initial momentum. For a collision between particles 1 and 2:

`m_1 bb"v"_{1f}+m_2 bb"v"_{2f}=m_1 bb"v"_{1i}+m_2 bb"v"_{2i} (1)`

where the subscript f stands for final value and the subscript i stands for initial value and the arrowed bold letter `vec bb"v"` stands for the velocity vector. The conservation of kinetic energy equation is as follows:

`m_1 bb"v"_{1f} cdot bb"v"_{1f} +m_2 bb"v"_{2f} cdot bb"v"_{2f}=m_1 bb"v"_{1i} cdot bb"v"_{1i} +m_2 bb"v"_{2i} cdot bb"v"_{2i} (2)`

where the `cdot` between the velocity vectors stands for the inner or dot product of the two vectors. When we combine equations 1 and 2 we get equations that relate the components of the velocity vectors. The most important concept in these calculations is that of the unit vector ` hat bb"u"` between the centers of the two particles that are colliding

`hat bb "u"=[(x_2-x_1)hat bb"x"+(y_2-y_1)hat bb"y"+(z_2-z_1)hat bb"z"]/r_{12}`

where the x, y, and z that have hats are unit vectors along their respective axes and

`r_{12}=sqrt[(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2]`

Note that the unit vector points from sphere 1 to sphere 2 so, if `x_2` is greater than `x_1`, the unit vector's x component is positive.
Applying equation 1, we know that the changes of momentum of the spheres are equal and opposite and are along the unit vector `hat bb"u"` so we can write:

`m_1 delta bb"v"_1=M delta v hat bb"u"=-m_2 delta bb"v"_2 (3)`

where `M` has units of mass and `M` and the scalar speed increment `delta v` are still to be determined. Now we may re-write equation 2 using equation 3 and obtain:

`((m_1 bb"v"_1+M delta v hat bb"u") cdot (m_1 bb"v"_1+M delta v hat bb"u"))/(m_1)+ ((m_2 bb"v"_1-M delta vhat bb"u") cdot (m_2 bb"v"_2-M delta vhat bb"u"))/(m_2) =m_1 bb"v"_{1} cdot bb"v"_{1} +m_2 bb"v"_{2} cdot bb"v"_{2}(3)`

After cancellations of the terms on its right side, equation 4 simplifies to:

`M delta v=(2m_1m_2)/(m_1+m_2)( bb"v"_1- bb"v"_2) cdot hat bb"u"`

We can now make the identifications:

`M=(2m_1m_2)/(m_1+m_2)`

and

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

where M is known as the "reduced mass" Finally using equation 3 again we can write the expressions for the final velocity vectors:

` bb"v"_{1f}= bb"v"_{1i}+M/m_1 [( bb"v"_1- bb"v"_2) cdot hat bb"u"] hat bb"u"`
` bb"v"_{2f}= bb"v"_{2i}-M/m_2 [( bb"v"_1- bb"v"_2) cdot hat bb"u"] hat bb"u"`

As always, the learner is encouraged to press F12 (Windows) and select "Sources" to examine the HTML and javascript code that was used to create this page.