# Does the same temperature imply the same translational kinetic energy?

So I basically have three inter related questions.

Is temperature a measure of average kinetic energy (both components) or specifically the translational kinetic energy component?

Is it correct (for the most part) to say that the atoms of a solid generally have less kinetic energy than that of a liquid than that of a gas?

And finally if the answer to the two previous questions are correct (for classic cases of course), considering a a liquid solid and gas at the same temperature, do they all have the same average translational kintetic energy?

• What do you mean by "both components"? which components are you referring to? and what do you mean exactly by translational kinetic energy? Jan 4, 2020 at 22:29
• from my understanding internal kinetic energy can either be (i) translational or (ii) vibrational and rotational...... translational kinetic energy refers to (at least from what i understand) is the kinetic energy associated with the atoms actual moving a "considerable" distance away from a "rest/initial" position Jan 4, 2020 at 22:33

As you say, this is only true in the classical approximation, when quantum effects are negligible. It is not true for water and ice at $$0 ^\circ$$C, where quantum effects are significant for the proton.

But in melting lead, the classic approximation should be good. Then the atoms in the solid have the same kinetic energy as atoms in the liquid. Similar for boiling lead.

Here data for solid and liquid argon from inelastic neutron scattering by Fradkin et al: In a classical system, temperature, an intensive quantity, does not measure directly the average kinetic energy (translational or not), which is an extensive quantity. The correct statement is that each kinetic energy term of the Hamiltonian contributes by $$\frac12 k_B T$$ to the average energy of the system (equipartition theorem). Therefore, the contribution to internal energy per molecule of all the translational degrees of freedom is always $$\frac{3}{2} k_B T$$. This result is independent of the kind of molecules (mono-, di-, tri-,... atomic) and on the specific thermodynamic phase (of course, provided temperature and density are such that quantum effects on translational degrees of freedom can be ignored.

Things are usually more complicated for the rotational and vibrational energy of the molecules. Even at room temperature, the equipartition theorem cannot be used for some of the corresponding degrees of freedom, and for those degrees of freedom proportionality between contribution to the internal energy per molecule and temperature may not be valid.

In conclusion, in the classical regime, solids, liquids, and gases at the same temperature will always have the same average kinetic energy per molecule. The result is even stronger: not only the average kinetic energy per molecule but also the velocity distribution function (the Maxwellian distribution) of the molecules is precisely the same at the same temperature.

• from what i gather, please correct me if i am wrong, you are saying that it is technically incorrect to say that "the average kinetic energy per molecule of a solid is generally less than that of a liquid (or gas)" because internal kinetic energy is largely dependent on temperature Jan 6, 2020 at 1:03
• @user3602727 In general the amount of kinetic energy depends on the thermodynamic state. The phase (solid, liquid, gas) depends , of course, on the thermodynamic state as well, but the phase alone does not tell unambiguously anything about the ordering of temperatures. The simplest case is that of phase coexistence. For instance, at the triple point gas, liquid and solid phases coexist at the same temperature. The energy per molecule is exactly the same in the three phases (always in a classical treatment). Jan 6, 2020 at 8:48

Is temperature a measure of average kinetic energy (both components) or specifically the translational kinetic energy component?

Vibration of atoms in a solid, vibration of atoms in a molecule of a liquid or a gas, and translational movement of molecules in liquid and gases are all translational kinetic energy in my opinion. What change is the mean free path.

Rotating molecules seems a really different mode of kinetic energy. But in the cases where it is relevant it also contributes to temperature.

Is it correct (for the most part) to say that the atoms of a solid generally have less kinetic energy than that of a liquid than that of a gas?

And finally if the answer to the two previous questions are correct (for classic cases of course), considering a a liquid solid and gas at the same temperature, do they all have the same average translational kintetic energy?

If a solid is in thermal equilibrium with a liquid, (or a gas) the kinetic energy of its atoms are being transferred all the time, so they must have the same average energy.

In the case of a solid, the atoms oscillate around an equilibrium position. In a sense, it is similar to internal vibration of molecules of gases or liquids, only that here it is the sole mode of kinetic energy.

• The vibrational modes all involve a potential energy as well as a kinetic energy.
– Dale
Jan 5, 2020 at 0:03
• @Dale you are right. In one extreme of the model, the molecules have a constant velocity (and constant kinetic energy) between collisions. In the other extreme (vibrations), the velocity is not constant, and the kinetic energy should be taken from the maximum velocity of the path. Jan 5, 2020 at 1:08