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Consider a volume $V$ with $5$ particles each of mass $m$ at positions $\mathbf{q}_i=(x_i,y_i,z_i) \in V$ and with velocities $\mathbf{v}_i=(u_i,v_i,w_i)$. The speeds of the particles are between $0$ and $v_{max}$.

If we assume that all the microstates are equally likely can we calculate the average energy of the system.

I believe that the number of microstates is $V^{15}\left(\frac{4\pi}{3}v_{max}^3 \right)^5$. Apparently the average energy is $$\langle E \rangle = \frac{4\pi \displaystyle{\int_0^{v_{max}}}\left( 5\frac{mv^2}{2} \right)v^2 dv}{\frac{4\pi}{3}v_{max}^3}$$ but I cannot see how this quantity has been calculated. More generally I am having problems understanding how to calculate average quantities of gas with $N$ particles.

I believe that for average energy we want $$\langle E \rangle = \frac{\displaystyle{{\int \text{function for particle energy}}}{\text{?}}$$ but I cannot see past this.

From a guess I believe that the correct answer might actually be $$\langle E \rangle = \frac{4\pi \displaystyle{\int_0^{v_{max}}}\left( 5\frac{mv^2}{2} \right)v^2 dv}{(\frac{4\pi}{3}v_{max}^3)^5}$$

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When calculating expectation values, you need to know a few things:

  • What is my random variable?
  • What is my distribution function?
  • What is my desired quantity in terms of the random variable?

The general form in one dimension would look like this

$$ \langle G(x) \rangle = \frac{\displaystyle{\int_{x_{min}}^{x_{max}}} f(x) G(x) dx}{\displaystyle{\int_{x_{min}}^{x_{max}}} f(x) dx} \, ,$$

where $x$ is the random variable, which can take values between $x_{min}$ and $x_{max}$, $f(x)$ is the distribution function that assigns weights to different values of of $x$, and $G(x)$ is the quantity whose expectation value we seek.

In your case,

$$ x = \vec{v} \\ f(\vec{v}) = 1 \\ G(\vec{v}) = \frac{1}{2} m v^2 \times 5 \, .$$

This means that the expectation values of the total kinetic energy of the five particles is given by

$$ \langle E \rangle = \langle \frac{5}{2} m v^2 \rangle = \frac{\displaystyle{\int_{0}^{2\pi}} d\phi \int_{0}^{\pi}d\theta \sin\theta \int_{0}^{v_{max}} dv \, v^2 \, 1 \cdot \frac{5}{2} m v^2}{\displaystyle{\int_{0}^{2\pi} d\phi \int_{0}^{\pi}d\theta \sin\theta \int_{0}^{v_{max}}} dv \, v^2 \, 1} \, , $$

where I've chosen to use spherical coordinates to express the integral. Evaluating the angular integrals and also the $v$ integral in the denominator should get you the first expression you wrote.

You may wonder what about the position integrals? Since your expectation value of interest does not depend on the positions, the integrals over the position domain in both the numerator and denominator will cancel each other.

Edit: Rereading your question, the source of confusion seems to be the denominator. It is often referred to as the "normalization" of $f$. In probability theory, integrating a distribution function over all possible values of the random variable should result in a value of $1$. If its integral is not 1, then you need to make it so by dividing by whatever $\int f$ actually is.

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