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In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules. If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to distinguish the random vibration KE and KE in one direction?

Furthermore, if we accelerate a block of metal with ultrasonic vibrator so that the metal is vibrating in very high speed with cyclic motion, can we say the metal is hot when it is moving but suddenly become much cooler when the vibration stop?

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  • $\begingroup$ What do you mean with "average" in formulas? Are you using the equipartition theorem? $\endgroup$ – falematte Apr 19 '15 at 13:17
  • $\begingroup$ physics.stackexchange.com/q/96327 and a couple more in the "Linked" sidebar therein. $\endgroup$ – dmckee Aug 23 '15 at 3:18
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In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules.

In the kinetic theory of gases random motion is assumed before deriving anything.

If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to distinguish the random vibration KE and KE in one direction?

The temperature is still defined by the random motion, subtracting the extra energy imposed . This is answered simply by the first part of @LDC3 's answer. Does your hot coffee boil in the cup in an airplane?

Furthermore, if we accelerate a block of metal with ultrasonic vibrator so that the metal is vibrating in very high speed with cyclic motion, can we say the metal is hot when it is moving but suddenly become much cooler when the vibration stop?

This is more complicated, because vibrations may excite internal degrees of freedom and raise the average kinetic energy for that degree of freedom. It would then take time to reach a thermal equilibrium with the surroundings after the vibrations stop. If one supposes that this does not happen, then the answer is the same as for the first part, it is the random motions of the degrees of freedom that define the kinetic energy which is connected to the definitions of temperature. So no heat will be induced by the vibrations.

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  • $\begingroup$ thanks for your answer. I have no problem in understanding cases like why hot coffee doesn't boil in an airplane. But for periodic motions like vibrations with high frequency and small amplitude, how does the specimen knows which part of its motion is random and which part is not? The motion of atoms in solid is also some sort of vibration. How to estimate temperature of a solid in such kind of motion? $\endgroup$ – Kelvin S Feb 2 '16 at 12:53
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    $\begingroup$ As I have stated in my answer, vibrations may change the temperature of the solid if they excite vibrational degrees of freedom in the lattice. This has to be studied : what frequency, what amplitude, frictional forces etc. If the frequency is such that no levels are excited, the temperature will not change, because the solid moves as a whole at each instant. Randomness will be introduced by quantum mechanical probabilities of interaction, if the frequencies etc are such that interactions are important. $\endgroup$ – anna v Feb 2 '16 at 13:02
  • $\begingroup$ Very good. One last question: Instead of uniform, regular periodic motion, if we impose irregular, random vibration to the object, would it be more likely to excite vibrational degrees of freedom in the lattice? $\endgroup$ – Kelvin S Feb 2 '16 at 13:09
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    $\begingroup$ If the randomness is also in the frequency spectrum, most probably yes, because of the probability of exciting internal degrees of freedom . $\endgroup$ – anna v Feb 2 '16 at 15:43
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There is a simple way of looking at this. Would a container of gas have a change in temperature if the container was given a different velocity?

For your second question, the vibrating membrane acts like a spring pendulum which transfers energy into the surroundings. The membrane does not have a change in temperature until it absorbs the energy back from the surrounds.

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In first place, the temperature is a quantity that measures thermal equilibrium by the zeroth law of thermodynamics. We have the contact with this quantity with a thermal equilibrium can do. For example, the Celsius units is constructed by define $0°~\rm C$ as the volume of mercury in contact with freezing water and $100 °~\rm C$ as the volume of mercury in contact with boiling water.

With more refinement, we may found a better scale for temperature, the Kelvin scale. In this scale the temperature is always positive and the energy in heat channel is expressed by:

$$ T\cdot \mathrm{d}S $$ where $S$ is the entropy (some mysterious function of state).

Now, with statistical mechanics, the entropy is identified by a measure of information ignored in your description of the system in units of a tiny constant value (in front with macroscopic units) $k_b$, the Boltzmann's constant, in a Napierian basis.

$$ S=k_bI_e \\ I_e=-\sum_{i=1}^{N}p_i \ln(p_i) $$ where $I_b$ is a Shannon entropy with $b=e\;.$

If we change again the unit of temperature in units of energy per $k_b$ (you can do this by send $k_b=1$), the temperature is now the energy per unit of information ignored. This means that when we ignore information, the mean energy increase by the ratio of temperature. $$d\langle E \rangle=T\cdot \mathrm{d}I_e$$ where $\langle E \rangle$ is the mean energy.

Note that now we can define a lot of units for temperature in terms of $\mathrm{\frac{Energy}{constant}}\,,$ when this constant is defined by the connection of $I_b$ and $S\,,$ for different basis. For canonical ensemble, the best basis is in fact is the Napierian. For microcanonical ensemble, the better basis is the basis that respect the decomposition of the system in subsystems.

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  • $\begingroup$ Does that mean temperature only relates to KE of random motion? $\endgroup$ – Kelvin S Apr 19 '15 at 15:01
  • $\begingroup$ Is simply! Divide your system by parts,by degrees of freedon. And apply the canonical ensemble to find the equipartition theorem. $\endgroup$ – Nogueira Apr 20 '15 at 19:19
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    $\begingroup$ @KelvinS Yes. is related to random motion. $\endgroup$ – Nogueira Feb 2 '16 at 19:10

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