# Calculating the average kinetic energy (expectation value) of gas particles from the Maxwell Boltzmann distribution

From what I already know, to calculate the expectation value/average from a probability distribution, you use the formula:

$$\langle x \rangle \ = \int_{-\infty}^{\infty} x f(x) \,\mathrm{d}x \tag{1}$$

where $$f(x)$$ is the probability distribution of some variable $$x$$.

However, in my lecture notes, the average kinetic energy $$\langle E \rangle$$ of gas particles in thermal equilibrium using the Maxwell Boltzmann distribution $$f(u)$$ has been given as:

$$\langle E \rangle \ = \frac{\int_{-\infty}^{\infty} \frac{1}{2}mu^2 f(u) \,\mathrm{d}u}{\int_{-\infty}^{\infty} f(u) \,\mathrm{d}u} \tag{2}$$

where $$u$$ is the velocity.

I don't understand why there is a division by $$\int_{-\infty}^{\infty} f(u) \,\mathrm{d}u$$ in Equation (2), unlike what I've previously learned as in Equation (1), and I can't find the same form elsewhere.

If anyone has seen this and would be able to explain it to me, I would really appreciate it!

• The eq. (1) implies that $f(x)$ is true probability measure, i.e. the integral over all possible values of random variable is unity. The eq. (2) contains division by normalization constant and it is two-step procedure: 1) we compute normalization constant, 2) we compute average with true (normalized) probability measure Commented Dec 31, 2023 at 13:34
• @ArtemAlexandrov Ah okay that makes sense, thanks so much! And thank you for answering so quickly as well! Commented Dec 31, 2023 at 14:07

The formula given for $$\langle E\rangle$$is useful for probability distribution that is not normalized. The formulas can be made to look like each other by defining the normalization constant $$N$$ as
$$\int_{-\infty}^\infty f(u) du =N$$ Which we can since this integral is just a number and then we can write the normalized $$f(u)/N = g(u)$$ so then the formula provided in (1) applies for $$g(u)$$.
$$\langle E\rangle = \int \frac{1}{2} mu^2 g(u) du$$