Ideal Gas Simulation and the Maxwell–Boltzmann Distribution

Kinetic Energy (Joules)

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Temperature (Kelvin)

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Maxwell-Boltzmann Distribution for a microcanonical ensemble with no potential energy

Equation for 3d elastic collissions of spheres from here.

From the Boltzmann Distribution, when entropy is maximized, the probability of a particle being in any microstate is

P (\vec{x}, \vec{p}) = \frac{e^{- \frac{H (\vec{x}, \vec{p})}{k_{B} T}}}{\int_{- \infty}^{\infty} \int_{V} e^{- \frac{H ({\vec{x}}^{'}, {\vec{p}}^{'})}{k_{B} T}} d {\vec{x}}^{'} d {\vec{p}}^{'}} = \frac{e^{- \frac{H (\vec{x}, \vec{p})}{k_{B} T}}}{Z}

If the Hamiltonian has no potential energy, then for a non-relativistic particle it would just be

H (\vec{x}, \vec{p}) = \frac{{\vec{p}}^{2}}{2 m} = \frac{\vec{p} \cdot \vec{p}}{2 m} = \frac{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}{2 m}

So since $\vec{x}$ doesn't matter

P (\vec{p}) = \frac{e^{- \frac{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}{2 m k_{B} T}}}{Z}

And to calculate the partition function

Z = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- \frac{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}{2 m k_{B} T}} d p_{x} d p_{y} d p_{z}

And this is just a 3D gaussan so

= (τ m k_{B} T)^{3 / 2}

then substituting gives

P (\vec{p}) = (τ m k_{B} T)^{- 3 / 2} e^{- \frac{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}{2 m k_{B} T}} = (τ m k_{B} T)^{- 3 / 2} e^{- \frac{{\vec{p}}^{2}}{2 m k_{B} T}}

and switching to $\vec{p} = m \vec{v}$ while preserving the probability density

P (\vec{v}) = P (\vec{p} (\vec{v})) (\frac{d {\vec{p}}_{x}}{d {\vec{v}}_{x}} \frac{d {\vec{p}}_{y}}{d {\vec{v}}_{y}} \frac{d {\vec{p}}_{z}}{d {\vec{v}}_{z}}) = m^{3} P (\vec{p} (\vec{v}))

P (\vec{v}) = (\frac{m}{τ k_{B} T})^{3 / 2} e^{- \frac{m {\vec{v}}^{2}}{2 k_{B} T}}

Now if we want to find the speed we can create a bijection into polar coordinates

s = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}, θ, ψ

we get a Jacobian of

s^{2} \sin (θ)

So we can write

P (s, θ, ψ) = P (⟨ r \sin (θ) \cos (ψ), r \sin (θ) \sin (ψ), r \cos (θ) ⟩) s^{2} \sin (θ)

then to find the probability of a speed

P (s) = \int_{0}^{τ} \int_{0}^{τ / 2} P (⟨ r \sin (θ) \cos (ψ), r \sin (θ) \sin (ψ), r \cos (θ) ⟩) s^{2} \sin (θ) d θ d ψ

since each velocity coordinate is independent

\int_{0}^{τ} \int_{0}^{τ / 2} P (v_{x} = r \sin (θ) \cos (ψ)) P (v_{y} = r \sin (θ) \sin (ψ)) P (v_{z} = r \cos (θ)) s^{2} \sin (θ) d θ d ψ

And since it doesn't depend on angle

\int_{0}^{τ} \int_{0}^{τ / 2} P ({\vec{v}}^{2} = s^{2}) s^{2} \sin (θ) d θ d ψ

And thats just the surface area of a sphere

P (s) = 2 τ s^{2} P ({\vec{v}}^{2} = s^{2})

= 2 s^{2} \sqrt{\frac{1}{τ}} (\frac{m}{k_{B} T})^{3 / 2} e^{- \frac{m {\vec{v}}^{2}}{2 k_{B} T}}

Finding temperature

To find the temperature of a microcanonical system (NVE ensemble) with no potential energy we start with the Hamiltonian equal to a constant $E$ . Since there is only kinetic energy the position $\vec{x}$ doesn't matter.

H (\vec{x}, \vec{p}) = \sum_{i} \frac{p_{i}^{2}}{2 m} = E

What we want to find the temperature defined by

\frac{1}{T} = \frac{\partial S (N, V, E)}{E}

where the entropy $S = - k_{B} \sum_{i} P (x_{i}, p_{i}) l n (P (x_{i}, p_{i}) = - k_{B} l n (\frac{1}{W (E)}) = k_{B} l n (W (E))$ when entropy is maximized (so all states are the same probability) and $W (E)$ is the number of accessable states between $E$ and $E + Δ E$ .

Rearange to

\sum_{i} p_{i}^{2} = 2 m E

If we want to find $W (E)$ then assuming all particles are indistunguishable for the factorial and using Plank's constant as a measure of the phase space then

W (E) = \frac{1}{h^{3 N} N!} \int_{V} \int_{- \infty}^{\infty} 𝟙 (E \leq H (\vec{x}, \vec{p}) \leq E + Δ E) d^{3 N} \vec{p} d^{3 N} \vec{x}

With no potential energy the position doesn't matter and can be factored out

= \frac{V^{3 N}}{h^{3 N} N!} \int_{- \infty}^{\infty} 𝟙 (2 m E \leq \sum_{i} p_{i}^{2} \leq 2 m (E + Δ E)) d^{3 N} \vec{p}

= \frac{V^{3 N}}{h^{3 N} N!} \int_{- \infty}^{\infty} 𝟙 (\sum_{i} p_{i}^{2} \leq 2 m (E + Δ E)) - 𝟏 (\sum_{i} p_{i}^{2} \leq 2 m E) d^{3 N} \vec{p}

= \frac{V^{3 N}}{h^{3 N} N!} (\int_{- \sqrt{2 m (E + Δ E)}}^{\sqrt{2 m (E + Δ E)}} \int_{- \sqrt{2 m (E + Δ E) - p_{1}^{2}}}^{\sqrt{2 m (E + Δ E) - p_{1}^{2}}} \dots \int_{- \sqrt{2 m (E + Δ E) - p_{1}^{2} - \dots p_{n - 1}^{2}}}^{\sqrt{2 m (E + Δ E) - p_{1}^{2} - \dots p_{n - 1}^{2}}} d^{3 N} \vec{p}

- \int_{- \sqrt{2 m E}}^{\sqrt{2 m E}} \int_{- \sqrt{2 m E - p_{1}^{2}}}^{\sqrt{2 m E - p_{1}^{2}}} \dots \int_{- \sqrt{2 m E - p_{1}^{2} - \dots p_{n - 1}^{2}}}^{\sqrt{2 m E - p_{1}^{2} - \dots p_{n - 1}^{2}}} d^{3 N} \vec{p})

And recognize these are just the equation for a $3 N$ dimensional hypersphere of radius $\sqrt{2 m E}$ and $2 m (E + δ E)$ .

= \frac{V^{3 N}}{h^{3 N} N!} (\frac{(\frac{τ}{2})^{3 N / 2} (2 m (E + Δ E))^{3 N / 2}}{Γ (\frac{3 N}{2} + 1)} - \frac{(\frac{τ}{2})^{3 N / 2} (2 m E)^{3 N / 2}}{Γ (\frac{3 N}{2} + 1)})

= \frac{V^{3 N} (\frac{τ}{2})^{3 N / 2} 2 m^{3 N}}{h^{3 N} N! Γ (\frac{3 N}{2} + 1)} ((E + Δ E)^{3 N / 2} - E^{3 N / 2})

And now

\frac{1}{T} = k_{B} \frac{\partial l n (W (E))}{\partial E} = k_{B} \frac{1}{W (E)} \frac{\partial W (E)}{\partial E}

substitute

= k_{B} \frac{1}{\frac{V^{3 N} (\frac{τ}{2})^{3 N / 2} 2 m^{3 N}}{h^{3 N} N! Γ (\frac{3 N}{2} + 1)} ((E + Δ E)^{3 N / 2} - E^{3 N / 2})} \frac{V^{3 N} (\frac{τ}{2})^{3 N / 2} 2 m^{3 N}}{h^{3 N} N! Γ (\frac{3 N}{2} + 1)} \frac{3 N}{2} ((E + Δ E)^{(3 N / 2) - 1} - E^{(3 N / 2) - 1})

cancell

= k_{B} \frac{3 N}{2} \frac{(E + Δ E)^{(3 N / 2) - 1} - E^{(3 N / 2) - 1}}{(E + Δ E)^{3 N / 2} - E^{3 N / 2}}

If we take the limit then

\lim_{Δ E \to 0} k_{B} \frac{3 N}{2} \frac{(E + Δ E)^{(3 N / 2) - 1} - E^{(3 N / 2) - 1}}{(E + Δ E)^{3 N / 2} - E^{3 N / 2}}

= k_{B} \frac{3 N}{2} \frac{\frac{3 N}{2} - 1}{\frac{3 N}{2} E}

\frac{1}{T} = k_{B} \frac{\frac{3 N}{2} - 1}{E}

making

k_{B} T = \frac{E}{\frac{3 N}{2} - 1}

This equation is consistent with the equipartition theorem as each of the $3 N - 1$ degrees of freedom contribute $\frac{3}{2} k_{B} T$ to the total energy.