17
$\begingroup$

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, $K_u = {1 \over 2} m \overline {(v+u)^{2}}$).

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Your question seems related to some of the ideas behind the Unruh effect. $\endgroup$ – Brandon Enright Dec 16 '13 at 5:39
  • 1
    $\begingroup$ 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. $\endgroup$ – dmckee♦ Dec 16 '13 at 5:41
  • 1
    $\begingroup$ @dmckee this one? physics.stackexchange.com/q/83488 $\endgroup$ – user10851 Dec 16 '13 at 5:58
  • 1
    $\begingroup$ @ChrisWhite : That one asks about relativity. I'm just asking about plain pre-relativistic classical physics. $\endgroup$ – user114806 Dec 16 '13 at 6:35
  • 1
    $\begingroup$ 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. $\endgroup$ – jim Sep 24 '16 at 10:01
11
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.