Internal ("thermal") energy must be some combination of kinetic energy and potential energy, although most discussions of internal energy mention only the kinetic energy. However you also have potential energy -- as particles collide, they reach a minimum separation when they momentarily stop and all the energy is potential -- like when a bouncing ball hits the floor and, for an instant, all the energy is elastic potential energy in the squeezed ball. Then the particles get pushed apart, as that potential energy converts to kinetic. So, is the internal energy evenly split between kinetic and potential?

Or, should we be considering the virial theorem here and say that internal energy consists of twice as much potential energy as kinetic energy?

  • $\begingroup$ The combination of kinetic energy and potential energy at the molecular level is properly called "internal energy". The term "thermal energy" is ambiguous. It is often confused with heat and temperature. Recommend you read following dokumen.tips/documents/… $\endgroup$
    – Bob D
    Dec 3, 2022 at 17:19
  • $\begingroup$ Thanks, Bob. I have edited the question. $\endgroup$
    – Matt
    Dec 3, 2022 at 17:25
  • $\begingroup$ Your statement about the virial theorem does not add up. Something is badly off. I partially addressed that in my answer. Two sources for virial theorem discussion: Kevin Brown virial theorem, (English translation of) original article by R. Clausius Among the resources I looked at I found the one by Kevin brown to be the most informative. $\endgroup$
    – Cleonis
    Dec 4, 2022 at 9:16
  • $\begingroup$ This is a good question. A great answer would not just say "it depends" (which it does), but would also explain why we don't need to know the answer in order to reason about the thermal properties of a material. It's kind of amazing that we don't. $\endgroup$
    – N. Virgo
    Dec 5, 2022 at 10:12

6 Answers 6


In thermal equilibrium, the energy is equally distributed among different types. More precisely, any quadratic term in the energy per particle has average energy of $kT/2$ where $T$ is the temperature of the system and $k$ is the Boltzmann constant. This is the so-called equipartition theorem.

A diatomic molecule that translates, rotates and vibrates has average energy of $7kT/2$, being $3/7$ for translations, $2/7$ for rotations, $1/7$ for vibration and $1/7$ for elastic potential.

The only caveat about this is that depending on the energy scale, some degrees of freedom are "frozen". For example, at sufficiently low temperature a diatomic molecule may not vibrate which means there is no energy in this mode. Any "active" mode has the same energy per molecule.

  • $\begingroup$ If my crappy thermodynamics is correct, "In thermal equilibrium" is an entropic claim; if the energy wasn't evenly distributed, you could get useful work out of making it evenly distributed! And "thermal" description of energy is saying "no useful work unless you find a heat sink". By partitioning the energy up this way, you have no "internal gradients" to move energy around and get work out. $\endgroup$
    – Yakk
    Dec 5, 2022 at 15:12

This is not a case for the virial theorem.

The virial theorem

John Baez discusses the virial theorem on a page on his website, the virial theorem made easy

In general the force between particle A and particle B tends to be in proportion to some power n of the distance between A and B.

In the case of celestial mechanics n is -2, the inverse square law of gravitational attraction.

John Baez shows how the inverse square law of gravity gives rise to a particular conversion rate of kinetic energy and potential energy in the course of eccentric orbital motion.

That conversion rate generalizes from the case of two celestial bodies case to entire galaxies.

(Historically: the virial theorem was first stated in the context of statistical mechanics, but the validity of the conversion rate is independent of the number of particles involved.)

John Baez then generalizes to arbitrary power of n, with an expression for how the conversion rate comes out for each value of n. (Again independent of the number of particles involved.)

Other than that:
In the case of very low density gas the probability of a collision is low. I assume that for a given average velocity of the gas molecules the duration of a collisional interaction is a given. I would expect: the larger the density the larger the amount of collisions per unit of time, which would have the molecules spending a larger proportion of the time engaged in collisional interaction.

So if you would want to model how much proportion of time (on average) the internal energy is in the form of some potential energy, then you would need take the amount of collisional interactions per unit of time into account. So I expect it won't be a fixed ratio; I expect the density of the gas to be a factor.


It depends on the particular system. An ideal gas, for example, has no potential energy contributions to its internal energy.

  • $\begingroup$ How can that be, considering that, when two particles collide head-on, at closest approach they are (instantaneously) stopped and have no kinetic energy? $\endgroup$
    – Matt
    Dec 3, 2022 at 17:56
  • 1
    $\begingroup$ Do we not assume that the time spent in collisions by a molecules an ideal gas is negligible, so the time-average of the molecules' potential energy due to repulsive collision forces is zero? $\endgroup$ Dec 3, 2022 at 19:30
  • 4
    $\begingroup$ @Matt An ideal gas is composed of non-interacting particles, with the exception of elastic collisions of a timescale we can ignore. $\endgroup$
    – agaminon
    Dec 4, 2022 at 1:19
  • $\begingroup$ @agaminon However, the ideal gas is an excellent model even for real gases at normal conditions. And not just the simple kinetic theory but also Chapman-Enskog with more detailed predictions for various coefficients. It is not just a toy model. So the question would be why is like that. Phillip's comment seems to be more relevant. $\endgroup$ Dec 4, 2022 at 17:46
  • $\begingroup$ It's certainly much more than a toy model, but the theory of real gases is also very much needed in ton of cases. For example, Joule-Thompson heating is not an effect that can happen for an ideal gas. $\endgroup$
    – agaminon
    Dec 4, 2022 at 20:40

There is kinetic energy (KE) and potential energy (PE) at the macroscopic level, which is the "mechanical energy" of an object and its KE and PE at the microscopic, or molecular level, which is the "internal energy" of the object. In general one should avoid the term "thermal energy" because it is ambiguous, being confused with heat, temperature, and internal energy.

That said, in your example of the bouncing ball, I think it's best to think in terms of its mechanical KE and PE rather than its internal (molecular) KE and PE. The reason is the increase in elastic PE during compression, like the compression of a spring, is mechanical PE stored in the configuration of a material due to elastic deformation. It is not due to a decrease in internal (molecular) KE, but due to a decrease in mechanical KE of the center of mass (COM) of the ball when it comes to a stop. Then during decompression, the loss of elastic PE is not due to an increase in internal KE but due to an increase in mechanical KE of the COM on rebound.

Keep in mind that the internal molecular kinetic energy is primarily composed of the rapid random motions of the molecules of the object. The motion of the molecules due strictly to expansion is relatively organized and the increase in speeds involved during the expansion are insignificant compared to the speed of the random motions. Any increase in the internal KE during compression and decompression will be due primarily to inelastic collision behavior (internal friction) during compression and decompression rather than due to the molecules "moving apart" during decompression. The greater the inelasticity, the greater the increase in internal KE.

Bottom line: In answer to your question "So, is the internal energy evenly split between kinetic and potential?" the answer is, for a perfectly elastic collision, the decrease in elastic PE during decompression equals the increase in mechanical KE of its rebound velocity, not an increase in internal KE.

Hope this helps.


agaminon answered your question. I will add some detail.

First, in the thermodynamics we need to think averages. We may think of it as a time average over a very long time, or more appropriately as a an ensemble average over all possible microscopic states (positions and velocities) could have.

In this average sense internal energy is the total energy of the system, and it is the sum of kinetic and potential: $$ E = E_K + E_P $$ In the simple case of point molecules with three transnational degrees of freedom the kinetic energy at equilibrium is always $$E_K = \tfrac{3}{2} N k T,$$ and the heat capacity is $C_V = \tfrac{3}{2}k$.

How much potential energy is in the system depends on interactions. Think of interactions as springs between particles that store this energy. Its amount depends on the stiffness of the springs (magnitude of interactions) and the distance between particles (pressure).

In this ideal-gas particles are on average at such large distance from each other that the potential energy is negligible. In terms of time averages, the fraction of time that two molecules come close enough to interact is infinitesimal when pressure is sufficiently small and this case $E_P=0$.

At higher pressure, as well as in liquids and solids, interactions are very strong and the potential energy can be significant. An easy way to estimate its amount via the heat capacity, $C_V$. If we add $Q=C_V N \Delta T$ amount of heat under constant volume, kinetic energy increases by $\tfrac32k\Delta T$ and the rest goes to potential energy. Therefore, the fraction of energy that goes to kinetic and potential is \begin{align} & \text{fraction to Kinetic} & \frac{C_V^\text{ig}}{C_V} \\ & \text{fraction to Potential} & 1-\frac{C_V^\text{ig}}{C_V} \end{align} This is true not just for point molecules but for molecules of any structure. The ideal-gas heat capacity $C_V^\text{ig}$ is a tabulated property and can be found for most pure components.


The part that boils is kinetic energy, the parts that are left, you know (they've got potentials)

Once it becomes free, its potential shows. The more freedom something has, the more it's brought out, in infinite detail.

However, if there're other entropy introduced either through directly injecting it or imagining a part of it is separate now it's got potentials again.

The becoming of what it is, is the process of unfolding its potential, before it happens completely it remains more or less uncertain. That is because it's to also serve another purpose that's conductivity.

Generally you can tell how the mechanical part and the potential parts is being split, by excessively picturing which part of it is conductive / obedient which part of it is not conductive, some extreme examples are super conducting conditions and ideal gas (an example in the middle is water boiling on the stove, the part that's boiled away is the mechanical part, the part that's left is the part that's got the potential)

It might not be what your teachers are teaching, in that case, ask your teachers before down voting. And if you do, I would love to learn.


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