#### [SOLVED] 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, $K_u = {1 \over 2} m \overline {(v+u)^{2}}$). #### @Klodd 2013-12-16 09:33:25

The definition of temperature in the kinetic theory of gases emerges from the notion of pressure. Fundamentally, the temperature of a gas comes from the amount, and the strength of the collisions between molecules or atoms of a gas.

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.

This calculation is independent of frame translation, as it will add the same velocity component to both velocities, and the previous equation relies only on the difference in velocities.

The second step uses the ideal gas law to get to $T \propto \frac{1}{2}mv^2$.
For more detail you can check this Wikipedia article.

So the invariance with frame translation of the temperature is due to the invariance of pressure, which only considers relative velocities.

### [SOLVED] Why does cold air feel even colder when you are moving through it quickly?

• 2016-03-13 19:06:51
• Bruno KM
• 2194 View
• 1 Score
• Tags:   thermodynamics