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I am continuing to brush up my statistical physics. I just want to gain a better understanding. I have gone through the derivation of the classical virial theorem once more. I have thought about it, googled it and slept about it. The statement:

$\langle x_i \frac{\partial \cal H}{\partial x_j} \rangle= kT \delta_{ij}$

is still just counter-intuitive to me. So I am at a fixed position in phase space and I'm looking at my Hamiltonian. Then I step away from my current position and watch how the Hamiltonian changes and multiply that knowledge by how far away from my initial position I have moved. I do this a lot in a random way and then I take an average. Et voilá, I have arrived at the equilibrium temperature of a system.

Right now this is just some math to me (which I totally get) to calculate the temperature of a system of particles in thermal equilibrium. Is there more to it? Am I not getting it? What is the intuition behind this?

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up vote 4 down vote accepted

The conclusion – the claim of the virial theorem – is not "just some math" because all the objects in the claim have a physical interpretation. So it's physics and it has big implications in theoretical physics as well as applied physics.

The derivation is a mathematical derivation but it's not right to attach the disrespectful word "just" to a mathematical derivation. Mathematical derivations are the most solid and the only truly solid derivations one may have in science. On the contrary, it's derivations and intuitions that are not mathematical that should be accompanied by the word "just" because they are inferior. Instead, the right way is to adjust one's intuition so that it's compatible with the most solid results in physics – and they're the mathematically formulated results. Incidentally, there are various derivations – dealing with the microcanonical ensemble, canonical ensemble etc. The details of the proof differ in these variations but the overall physical conclusion is shared and important.

The exact proof of the theorem can't be simplified too much – otherwise people would do so – but one may offer heuristic, approximate proofs for approximate versions of the virial theorem and its special cases. For example, the quantity in the expectation value contains the derivative of $H$ with respect to a coordinate. The larger the derivative is, the more the Hamiltonian increases with the coordinate, and the more the Boltzmann factor $\exp(-H/kT)$ of the canonical distribution decreases with the coordinate which makes the expectation value of the coordinate smaller. So if we multiply the quantity by the coordinate again, we get something that behaves constantly, independently of the slope. And indeed, the expectation value of the product only depends on the temperature.

This theorem is important in statistical physics because statistical physics is all about the computation of statistical averages of various quantities, the theorem allows us to express some expectation values in a simpler way, and $x_i \cdot \partial H / \partial x_j$ are among the simplest and most important quantities whose statistical averages may be computed or interesting. So we should better know how they behave.

An important special case of the theorem you mentioned deals with the calculation of the expectation value of the kinetic energy and the potential energy. The former is $n/2$ times the latter for power-law potentials of the form $ar^n$, for example. So we know how big a percentage of the energy is stored in the kinetic one and how big portion is the potential energy. For example, both the kinetic and potential energy contribute 50% for harmonic-oscillator-like $r^2$ potentials. For the Keplerian or Coulomb $-C/r$ potential, i.e. $n=-1$, the potential energy is negative, $-|V|$, and the kinetic energy is $+|V|/2$, reducing the potential one by 50% while keeping the total energy negative. There are many other things we may learn from the theorem in various situations – and in classes of situations.

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Thanks for the answer. I am not a native speaker, so "just" doesn't have a negative co-notation to me. I could change it or include smth in parenthesis like (not a disrespectful just). If I remove the "just" now, your answer would not make so much sense anymore. suggestions? – seb Mar 27 '13 at 9:03
Well, if you remove the just, my answer may be meaningless, but so will be your question. It will just say that the mathematical proof of the theorem is a mathematical proof, a tautology, and it will ask what you're missing. One can't determine what you're missing if the only information we get is a vacuous tautology. You may be missing nothing, you may be missing a lot, and you may be missing everything. – Luboš Motl Mar 27 '13 at 9:05
ok, I won't remove it then. maybe I'm over-thinking this one. I could rephrase the question to: how in the world did anybody come up with the derivation. Each step in the derivation is totally valid. All steps taken together are just not an obvious way (at least to me) to connect both sides of the equal sign. – seb Mar 27 '13 at 9:08
People were calculating the average values of kinetic energy, potential energy, and other functions of $x,p$, because they were just interested in them. That's the point of the statistical approach to be interested in such things. They noticed some patterns, and they generalized these observations up to the form you wrote down - and the proof wasn't too hard to be found, either. It wasn't at all manifest that you were asking about the history - you could have mentioned the word "history" if it is the case. The answer would be found e.g. at – Luboš Motl Mar 27 '13 at 9:10
Otherwise this theme "it's bad that the steps aren't obvious" appears very often here. Maybe there should be in some FAQ here? There's nothing wrong about the steps' not being obvious a priori. If the validity of the claim were obvious, we wouldn't need a proof and we wouldn't call the claim a "theorem" because it would be just an obvious triviality. The theorem - and every theorem - is important exactly because its validity is not obvious, at least not from the beginning. But it's still true. – Luboš Motl Mar 27 '13 at 9:16

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