I'm in undergraduate stat mech/thermo. In the context of the Maxwell-Boltzmann distribution, the mean kinetic energy of a gas particle is $\langle KE \rangle = \frac{1}{2}m \langle v^2 \rangle$.

I do not see why we use $\langle v^2 \rangle$, and not $\langle v \rangle ^2$. I understand that they are different terms mathematically, just can't figure out what necessitates the use of one over the other. Thank you.

  • 8
    $\begingroup$ By definition, kinetic energy is $K = \frac{mv^2}{2}$, so the average, $\langle K \rangle$ is $\langle mv^{2}/2\rangle = \frac{m}{2}\langle v^2\rangle$, because the ideal gas is thought to be of one component so the mass is just a constant. Just that. $\endgroup$
    – Verktaj
    Commented Feb 21 at 5:38
  • 2
    $\begingroup$ Assuming we are in the rest frame of the system $\langle v \rangle$ is zero so it wouldn't be very useful for calculating the internal energy. $\endgroup$ Commented Feb 21 at 7:38
  • $\begingroup$ @JohnRennie These comments together are the answer. $\endgroup$
    – my2cts
    Commented Feb 22 at 11:35

1 Answer 1


In the kinetic theory of gases, the average velocity of a particle is derived by considering the molecules of gas to be point particles with velocities $v_{i}$, where $i\, =\, 1,\,2, \, \ldots,\,N$ is the particle index and $N$ is the total number of particles. In this context, $ \langle v \rangle$ is the average velocity of the collection:

$$ \langle v \rangle^2 = \left( \frac{1}{N} \sum_{i=1}^{N} v_i \right)^2 $$


$ \langle v^{2} \rangle$ is:

$$ \langle v^2 \rangle = \frac{1}{N} \sum_{i=1}^{N} v_i^2. $$

Expanding a few terms reveals the problem:

$$ \langle v \rangle^2 = \frac{1}{N^2} (v_1 + v_2 + v_3 + \ldots + v_N)^2, $$


$$ \langle v^2 \rangle = \frac{1}{N} \sum_{i=1}^{N} v_i^2 = \frac{1}{N} (v_1^2 + v_2^2 + v_3^2 + \ldots + v_N^2). $$

I hope this helps!


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