# Thermal energy defintion

1. Aug 23, 2010

### hasan_researc

I am a first year Physics undergraduate. My question is with regrads to thermal energy.

Many people define thermal energy as kT and equate that with the kinetic energy.
In other instances, they equate 0.5kT with the kineitc energy.

What is thermale energy? Why has it been defined as kT? And why is 0.5 kT sometimes used?

2. Aug 23, 2010

### r_tea

The short story: So, in thermodynamics there's this thing called the "equipartition theorem", which says every "degree of freedom" in your system contributes 0.5kT to the total average energy. For example, if you have a one dimensional, ideal (non-interacting) gas, then you would be correct in saying K.E.=0.5kT. For a three dimensional gas your atoms have three directions they can move in, so K.E.=3*0.5kT. There's a little more to it if you want to consider vibrational degrees of freedom...

I've skipped a lot of details here. What is thermal energy? It's a measure of the average energy of a system, given that it's at some temperature T (..at least when considering the canonical ensemble--systems that are in equilibrium with a reservoir at temperature T). For ideal gases this happens to be a measure of the average kinetic energy, because they have no interactions hence no other forms of potential energy which store energy. The way you derive the equipartition theorem is using Boltzmann statistics: the probability of being in a given state is P(s)=exp(-E(s)/kT), where E(s) is the energy of state s. Taking expectation values yields the equipartition theorem (http://farside.ph.utexas.edu/teaching/sm1/lectures/node67.html)

Summary:
0.5kT for every degree of freedom. Each translational degree of freedom adds 0.5kT, each rotational is another 0.5kT, but each vibrational is 2(0.5)kT.

Also note that a lot of profs are lazy and just drop the factor of 0.5 because they're only looking for an order of magnitude estimate for something.

EDIT: I realize I may have been a bit unclear about the vibrational degrees of freedom. Each vibrational degree of freedom technically only adds 0.5kT, but because for vibration to be possible you also need a translational degree of freedom, that's where the second 0.5kT comes from. I.e., if I said "lets put a mass on a spring, but not allow the mass to move (i.e., allow no translational degree of freedom)", then it wouldn't make much sense to talk about the vibrational degree of freedom since nothing would ever vibrate (because the mass can't move).

Last edited: Aug 23, 2010
3. Aug 23, 2010

### hasan_researc

Thank you so much! Your answer's greatly helped me. I have just two more questions.

"I've skipped a lot of details here. What is thermal energy? It's a measure of the average energy of a system, given that it's at some temperature T":
Do you mean the average of the total internal energy (I mean excluding the energy due to ordered motion of the particles?

"I realize I may have been a bit unclear about the vibrational degrees of freedom. Each vibrational degree of freedom technically only adds 0.5kT, but because for vibration to be possible you also need a translational degree of freedom, that's where the second 0.5kT comes from. I.e., if I said "lets put a mass on a spring, but not allow the mass to move (i.e., allow no translational degree of freedom)", then it wouldn't make much sense to talk about the vibrational degree of freedom since nothing would ever vibrate (because the mass can't move). ":
But then trnaslation and vibration are one and the only thing, isn't it?

4. Aug 23, 2010

### r_tea

Yeah so internal energy is the total energy that arises due to random, or statistical events. I think I used the world "average" sloppily: The internal energy, U, is the total energy of the system of N particles, and if you divide the internal energy by the number of particles, U/n, that's now the average energy per particle (though whether or not U is an exact or average value I think depends on the kind of system you consider: in the microcanonical ensemble U is exact; in the canonical ensemble U is probabilistic...in any case, it's not a big deal for now). Among the things we don't consider to be part of internal energy are: center of mass motion (e.g., a tank of water moving at speed v has the same internal energy as a stationary tank), relativistic energy (mc^2), gravitational potential energy, etc..

No, translation is different from vibration (..is that what you're asking?..I'm not sure). You can have translational degrees of freedom without any vibrations. E.g. a 3D free, non-interacting gas has 3 degrees of freedom, while a 3D crystal lattice (where you model the atom-atom bonds as springs) as 6 degrees of freedom.

The slightly longer story to the equipartition theorem is that if you write out the total energy for your particle(s) in question, then for each quadratic term you get 0.5kT of energy. For example, the total energy of a single moving particle is $$E=p^2/2m$$, so you see there's a quadratic term, and so that means $$p^2/2m=0.5kT$$. A simple harmonic oscillator has $$E=p^2/2m+0.5kx^2$$, so $$p^2/2m=0.5kT$$, $$0.5kx^2=0.5kT$$, and $$E=kT$$. That's why it's called the "equipartition theorem", because the thermal energy is equally partitioned to every quadratic degree of freedom.

Last edited: Aug 23, 2010
5. Aug 23, 2010

### hasan_researc

Does the $$E=p^2/2m+0.5kx^2$$ correspond to the centre of mass motion of the oscillator?

6. Aug 23, 2010

### r_tea

Okay, I see where you're going with this. Yes, in that case I was talking about CM motion--it's just a mass on a spring, hence the location of the mass is the center of mass. Since I just said "CM motion doesn't count in temperature", and here were are talking CM motion and temperature, you may get a little upset.

When I said "we don't count CM motion", I mean we don't count "bulk" CM motion--or motion which is not due to statistical effects. I used the single particle example for simplicity, but real systems where temperature is relevant are a collection of lots of particles doing lots of different things, and the center of mass has no net motion--and if it did, we'd say it's because the bulk material is moving, not because it's "hotter" when it's moving.

7. Aug 23, 2010

### hasan_researc

If $$E=p^2/2m+0.5kx^2$$ does correspond to the centre of mass motion of the oscillator, as you have just said it does, then if the centre of mass of the vibrating molecule is stationary, the thermal energy of the molecule becomes 0.5KT, does not it?

And if the centre of mass of the molecule is moving, then only and only then do we add an extra 0.5kT to the thermal energy. Am I right?

[I am assuming that in both cases, the molecule is vibrating, so the molecule has a minimum of 0.5kT of thermal energy.]

8. Aug 23, 2010

### hasan_researc

I am sorry I have made a mistake. In my last two posts, when I wrote $$E=p^2/2m+0.5kx^2$$, I actually meant just $$E=p^2/2m$$.

9. Aug 23, 2010

### r_tea

Sorry, I missed your point--sorry if this became confusing. In the case of a molecule, i.e. two masses connected by a spring, then there are 3 degrees of freedom. 1. The CM motion of the molecule. 2. The relative motion of the molecules. and 3. The vibrational potential energy.

The single spring-mass system doesn't illustrate this, because it only has a single mass and hence, technically the CM motion is the motion of the mass.

For a linear molecule, you could write:
$$E=p_1^2/2m+p_2^2/2m + 1/2k(x_1-x_2)^2$$
or, equivalently
$$E=p_{CM}^2/2(m+m)+p_{rel}^2/2m_{reduced}+1/2kx_{rel}^2$$

In any case, there's 3 quadratic terms thus U=1.5kT.

So no, your last example was incorrect. If you have a linear molecule which as no total momentum, there are still two degrees of freedom (U=kT). If you add CM motion, then U=1.5kT.

Last edited: Aug 23, 2010
10. Aug 24, 2010

### hasan_researc

Thank you very much!