# Internal energy and thermal energy

I'm watching a lecture introduction internal energy and in it, the lecturer states the following for some system:

$$E_{internal} = E_{translational} + E_{vibrational} + E_{rotational} + E_{other}$$

where ## E_{other} ## could be chemical energy, magnetic energy, electrostatic energy etc.

Then he circles ## E_{translational} + E_{vibrational} + E_{rotational} ## and states that the RANDOM portions of these energy contributions represents the thermal energy. He says there is an ordered contribution from these terms and an unordered contribution and that this unordered (or random) contribution is the thermal energy.

What does the ordered vs. random contributions mean? I have no idea what that means.

## Answers and Replies

DrClaude
Mentor
What does the ordered vs. random contributions mean? I have no idea what that means.
Me neither! Can you post a link to the lecture?

$$E_{internal} = E_{translational} + E_{vibrational} + E_{rotational} + E_{other}$$
This doesn't bode well. Translational energy is definitely not a form of internal energy.

mjc123
Science Advisor
Homework Helper
I think it's talking about the motion of a body as a whole, i.e. the coordinated motion of the atoms that make it up. For example, if you throw a ball, the translational motion of the ball is not internal energy - if you throw it faster it is not hotter - but the random motions of the atoms (superimposed on their ordered motion with the ball) do constitute internal energy - if the ball is hotter these random motions are more energetic.

sophiecentaur
Science Advisor
Gold Member
Translational energy is definitely not a form of internal energy.
Is he not referring to the translational Energy of the individual molecule within the gas? After all, in elementary kinetic theory of gases, that is the only form of internal energy that's considered. The Rotational Energy that is also in the expression would normally be assumed the molecular rotation and not the coordinated rotation of the whole.
We would need to see the lecture to get the exact context of the expression. @mjc123 and I have both managed to get totally opposite messages from this; that speaks volumes.

Chestermiller
Mentor
I support mjc123's description of this very standard development. An example is an ideal monatomic gas. If ##\mathbf{v_i}## is the velocity of the i'th atom, then the average velocity of the gas is $$\bar{\mathbf v}=\frac{\sum_1^N{\mathbf{v_i}}}{N}$$If ##\mathbf{v'_i}## represents the deviations from the average velocity, then $$\mathbf{v_i}=\bar{\mathbf{v}}+\mathbf{v'_i}$$The total energy of the gas is the sum of the kinetic energies its particles: $$E=\sum_1^N\frac{m}{2}(\bar{\mathbf{v}}+\mathbf{v'_i})^2=N\frac{m}{2}\bar{v}^2+\frac{m}{2}\sum_1^Nv_i^2$$where the square on the left hand side of this equation symbolically represents the dot product of the vector velocity with itself. The first term on the right hand side of this equation is the "coordinated" kinetic energy of the gas, and the second term is the "random" kinetic energy, also called the internal energy of the gas. Thus,
$$E=(KE)+U$$U is typically calculated from the Boltzmann velocity distribution.

Lord Jestocost
Hey all!

Here is the lecture:

I set it to start at 1:25 but it's over the first few minutes that he discusses this.

In terms of the translational component to the internal energy, I think he is referring to the individual motions of the particles, as seen in the center of mass and as noted by @sophiecentaur.

I'll take a look at what you wrote @Chestermiller and respond soon!

DrClaude
Mentor
Well, I don't know what to say. After all these years, including many teaching thermodynamics, I had never heard of this nomenclature.

Guess it's never too late to learn...

Chestermiller
Mentor
I think we should add to this the fact that internal energy also includes the potential energy associated with molecular interactions.