8

I know that macroscopic temperature is a measure of kinetic energy of particles at very low scales (let's call it microscopic kinetic energy).

But how can we derive which part of this microscopic kinetic energy gives rise to temperature, and which part instead gives rise to macroscopic kinetic energy?

David Z
  • 76,371
Federico Toso
  • 711
  • No, temperature is not proportional to kinetic energy, not when quantum effects play a role. See for example https://physics.stackexchange.com/questions/413376/is-zero-point-energy-real/433824#433824 –  Aug 25 '19 at 19:40
  • Somewhat related: https://physics.stackexchange.com/q/218418/59023 – honeste_vivere Sep 11 '19 at 19:43

3 Answers3

10

The macroscopic kinetic energy of a system of particles is the kinetic energy due to the velocity of the center of mass of the collection of particles with respect to an external frame of reference.

For example suppose you have a container filled with an ideal gas. The temperature of the gas is a measure of the average kinetic energy of the randomly moving gas particles. That is its internal kinetic energy.

Let the container be moving at constant velocity with respect to an external frame of reference (e.g., the room where the container is located). The external (macroscopic) kinetic energy of the gas in the container is $\frac{mv^2}{2}$ where $m$ is the mass of the gas and $v$ is the velocity of the container with respect to the room. This kinetic energy is independent of the internal kinetic energy, not a part of it.

The total kinetic energy of the gas is the sum of its internal and external kinetic energies.

Hope this helps.

Bob D
  • 71,527
  • 4
    @Aaron Stevens: It's the defintion of the center of mass. If ${\bf v}_i$ is velocity in center of mass frame, then $\sum m_i {\bf v}_i=0$. – mike stone Aug 25 '19 at 13:07
  • @mikestone Ah yes, of course :) – BioPhysicist Aug 25 '19 at 13:25
  • Rather than "external" and "internal" would it be better to say "center of mass" and "thermodynamic"? It sounds especially weird to talk about the "external" KE of something. – abalter Aug 25 '19 at 20:53
  • @abalter I believe it is customary in thermodynamics to talk about the total energy of a system as the sum of its internal and external kinetic and potential energy. Taking kinetic energy, the internal kinetic energy of the gas (system) is based on the velocities of the particles with respect to an internal frame of reference (in this example, the container). The kinetic energy of the center of mass of the gas is based on the velocity of the COM with respect to an external frame of reference (the room where the gas is located). The term "external" helps to differentiate the two KE components – Bob D Aug 25 '19 at 21:09
  • @abalter The same goes for potential energy. The internal potential energy is based on the intermolecular forces between the molecules, usually ignored for an ideal gas. The external potential energy is based on the position of the COM in the gravitational with respect to an external frame of reference (e.g., the floor of the room). – Bob D Aug 25 '19 at 21:11
  • @abalter If your more comfortable with terms such as center of mass KE and PE, that's OK. These are merely labels. As long as we are consistent in using them its ok. – Bob D Aug 25 '19 at 21:13
4

I know that macroscopic temperature is a measure of kinetic energy of particales at very low scales (let's call it microscopic kinetic energy).

This is not generally true. The only case where this is true is for an ideal monoatomic gas. For all other materials there are more internal degrees of freedom than merely the kinetic energy.

For the remainder of your question, to determine which portion of the total energy is due to which parts, you have to distinguish between internal and external degrees of freedom. Then the thermal energy is the portion of the total energy contained in all internal degrees of freedom and the kinetic energy is the portion contained in the external rotation and translation degrees of freedom.

Dale
  • 99,825
  • 1
    Equipartition is valid for any classical system. –  Aug 25 '19 at 19:38
  • 1
    Wouldn't internal degrees of freedom (e.g. rotation, vibration) still count as part of kinetic energy? Or is kinetic energy only translation? – gardenhead Aug 26 '19 at 14:09
  • @gardenhead Vibration moves between moving and not moving? At the "end" of the vibration the vibrating thing has stopped; it is the "elasticity" (whatever force causes it to vibrate and not just move forever or stop) that holds the energy. – Yakk Aug 26 '19 at 14:14
  • 1
    @gardenhead as mentioned by Yakk much of the energy in the internal degrees of freedom is in potential energy, not internal kinetic energy. For example, if there is energy in a vibrational mode then at any moment of time that energy may be found in kinetic or potential energy or a combination of the two. – Dale Aug 26 '19 at 21:23
0

À heat engine is a device for converting heat energy into mechanical energy. Conceptually, the situation can be seen as interacting perfectly elastic spheres, like billiard balls. An accurate description of the system would involve Van der Waals forces, internal degrees of freedom for poly atomic gases and quantum effects but all these are secondary to the answer to the question.

Consider the cylinder of a heat engine. One wall can move, a piston. The pressure the piston feels is due to the momentum transfer of the molecules within a mean free path and whose momenta have components normal to the piston. It is the concerted effort of these molecules that causes the total force on the connecting rod behind the piston. When the piston moves, energy (F.ds) is extracted from the gas and the gas cools. In effect the heat engine sorts out molecules that have a common direction and converts some of their heat energy to work.

Charles Crummer
  • 59