Hover over the menu bar to pick a physics animation.

This will show how a gas's energy and pressure are partitioned regardless of particle mass.
The second law of thermodynamics states, in a closed system, that the motional (kinetic) energy cannot spread from a less energetic region to a more energetic region.
This is just another way of saying that diffusion causes paticle energies to spread away from high energy regions. For a gas,
the particles in high energy regions are moving faster so the spreading occurs faster.
The second law of thermodynamics is a result of diffusion (spreading) and all diffusion is due to particle interaction. Most people are familiar
with diffusion of particle **position** but, for the present animation, the diffusion is that of **particle speeds and energies**.
In this case, the particle interaction is by way of hard sphere collisions but in reality there is no such thing as a hard sphere collision:
All collisions are mediated by potentials between the particles. Potential-mediated collisions are handled under other menu items.
The equiparition concept just states that the average energy of smaller mass particles will be the same as that of a
higher mass particles. The smaller mass particles' average speeds are just enough higher to compensate for their smaller mass and
result in the same average energy. The same principle is true for the pressure exerted per particle. The pressure, `P_s`, due to the smaller mass
particles is just proportional to their number i.e. `P_s = P n_s/n_t` where `n_s` is the number density of smaller mass particles, P is the total pressure,
and `n_t` is the number density of all particles.
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 `kT` where `k` is Boltzmann's constant and `T` is temperature.
This page includes an active plot of the kinetic energy distribution versus energy. For this automation I have
chosen the number density of small and large mass particles to be the same. **For this animation the particles all start out with equal speeds, `v_0`,
and when energy, `mv^2/2`, bins, are plotted, they merge toward their final equilibrium energy distribution which**,
for a simple monoatomic gas like this, is the Maxwell-Boltzmann distribution:

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.#### Deriving the Equations for Final Velocities After Collision

In the following equations please be aware that letters with arrows over them 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:

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:

Canvas with Moving Gas Spheres

Single Step

Red Sphere Mass 1.5

Blue Sphere Mass 0.5

Plot `mv^{2}/2` bins
Plot `v^{2}/2` bins

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.

`p_{3D}(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.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`

When the `v^2/2` bins are plotted, the distributions stay separate on the x axis and have quite different widths that are proportional to the inverse of their masses. One might ask why the probabilty of zero energy is maximal in 2 dimensions and is 0 in 3 and more dimensions . The reason is that, in 3 dimensions, the result of a collision can almost never have all three velocity components of one of the particles be zero. To see this we need to get into the dynamics of 2 and 3 dimensional collisions.`m_1 vec bb"v"_{1f}+m_2 vec bb"v"_{2f}=m_1 vec bb"v"_{1i}+m_2 vec 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 vec bb"v"_{1f} cdot vec bb"v"_{1f} +m_2 vec bb"v"_{2f} cdot vec bb"v"_{2f}=m_1 vec bb"v"_{1i} cdot vec bb"v"_{1i} +m_2 vec bb"v"_{2i} cdot vec 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`vec 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 vec bb"v"_1=M delta v hat bb"u"=-m_2 delta vec 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 vec bb"v"_1+M delta v hat bb"u") cdot (m_1 vec bb"v"_1+M delta v hat bb"u")+ (m_2 vec bb"v"_1-M delta vhat bb"u") cdot (m_2 vec bb"v"_2-M delta vhat bb"u") =m_1 vec bb"v"_{1} cdot vec bb"v"_{1} +m_2 vec bb"v"_{2} cdot vec 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)(vec bb"v"_1-vec bb"v"_2) cdot hat bb"u"`

We can now make the identifications:`M=(2m_1m_2)/(m_1+m_2)`

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

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