# Calculating average quantities in kinetic theory

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 $$\langle E \rangle = \frac{4\pi \displaystyle{\int_0^{v_{max}}}\left( 5\frac{mv^2}{2} \right)v^2 dv}{(\frac{4\pi}{3}v_{max}^3)^5}$$

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.

$$x = \vec{v} \\ f(\vec{v}) = 1 \\ G(\vec{v}) = \frac{1}{2} m v^2 \times 5 \, .$$
$$\langle E \rangle = \langle \frac{5}{2} m v^2 \rangle = \frac{\displaystyle{\int_{0}^{2\pi}} d\phi \int_{0}^{\pi}d\theta \sin\theta \int_{0}^{v_{max}} dv \, v^2 \, 1 \cdot \frac{5}{2} m v^2}{\displaystyle{\int_{0}^{2\pi} d\phi \int_{0}^{\pi}d\theta \sin\theta \int_{0}^{v_{max}}} dv \, v^2 \, 1} \, ,$$
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.
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.