# Why isn't temperature frame dependent?

In (non-relativistic) classical physics, if the temperature of an object is proportional to the average kinetic energy ${1 \over 2} m\overline {v^{2}}$of its particles (or molecules), then shouldn't that temperature depend on the frame of reference - since $\overline {v^{2}}$ will be different in different frames?

(I.e. In the lab frame $K_l = {1 \over 2} m\overline {v^{2}}$, but in a frame moving with velocity $u$ relative to the lab frame, $K_u = {1 \over 2} m \overline {(v+u)^{2}}$).

• Your question seems related to some of the ideas behind the Unruh effect. Dec 16, 2013 at 5:39
• There is an earlier instance of the question floating around. Or at least, I'd have sworn it was, but so far I can't run it down. Dec 16, 2013 at 5:41
• @dmckee this one? physics.stackexchange.com/q/83488
– user10851
Dec 16, 2013 at 5:58
• If not making use of relativity then $$K_u = {1 \over 2} m \overline {(v+u)^{2}} = {1 \over 2} m (\overline v^2 + 2 \overline {vu} + \overline{u^2} ).$$ Since $u$ is a constant speed have $\overline{vu} = {\overline v}u = 0$, since the gas a whole is stationary. This gives $$K_u = {1 \over 2} m \overline {(v+u)^{2}} = {1 \over 2} m (\overline v^2 + u^2)$$ showing that the energy is made up of a "random bit" and a "translation bit". The temperature is only due to the random bit. Energy is always undefined to within a constant.
– jim
Sep 24, 2016 at 10:01

The first step considers an (elastic) impact between two particles, and writes $\Delta p = p_{i,x} - p_{f,x} = p_{i,x} - ( - p_{i,x}) = 2\,m\,v_x$ where the direction $x$ denotes the direction of the collision. This, of course, is considering that the two particles have opposing velocities before impact, which is equivalent to viewing the impact in the simplest frame possible.
The second step uses the ideal gas law to get to $T \propto \frac{1}{2}mv^2$.