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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:

$$\langle x^i \frac{\partial \cal H}{\partial x^j} \rangle= kT \delta^i_j$$

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?

Qmechanic
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seb
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2 Answers2

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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.

Luboš Motl
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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 09:03
  • 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 09:05
  • 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 09:08
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    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 http://en.wikipedia.org/wiki/Virial_theorem#History – Luboš Motl Mar 27 '13 at 09:10
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    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 09:16
  • ok, it is not about the history of the virial theorem. the derivation is completely clear to me. however, i do lack the intuition of what the theorem is actually trying to tell me. i have edited the question to "what is the intuition behind it". – seb Mar 27 '13 at 09:17
  • After this debate, thank you, it is even harder for me to understand what you may be asking. If you understand the concept of expectation value, the coordinates on the phase space, the notion of partial derivatives, and the temperature, then you must understand what the theorem is saying, right? – Luboš Motl Mar 27 '13 at 09:18
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The point of using expressions such as $$\langle x^i \frac{\partial H}{\partial x^j} \rangle $$ is not to necessarily obtain information about any general system, but to obtain a tool to study specific systems or at least classes of systems case by case.

For instance, in Newtonian/Galilean dynamics, most systems of interest will be possible to express in coordinates such that their Hamiltonian has the separable form $$H = T(p_i) + V(x^i) $$ Now let us consider the vicinity of a non-degenerate equilibrium, that is a point $x^i_0$ such that $\partial V/\partial x^j (x^i_0)$, but the all the eigenvalues of the matrix $V_{ij} \equiv \partial^2 V/ \partial x^j \partial x^k (x^i_0)$ are positive (i.e. the second-derivative matrix is non-degenerate and positive-definite). Then we can make a transform to $\delta x^i = x^i-x^i_0$ and the Hamiltonian can be reexpressed in the form $$H = T(p_i) + \frac{1}{2}V_{ij}\delta x^i \delta x^j + \mathcal{O}(\delta x^3)$$(Assuming Einstein summation convention.) In that case we can say that near equilibrium $x^i_0$ $$\delta x^k\frac{\partial H}{\partial x_l} = \delta x^k V_{lj}\delta x^j$$ (Automatically dropping $\mathcal{O}(\delta x^3)$ from now on.) If you put $k=l$ and sum over the indices you get $$\delta x^l\frac{\partial H}{\partial x_l}(x^i) = \delta x^l V_{lj}\delta x^j \approx 2 (V(x^i) - V(x^i_0))$$ In other words, $\delta x^l\frac{\partial H}{\partial x_l}$ has the meaning of roughly twice the difference of potential energy as compared to equilibrium.

In principle, if you know the matrix $V_{ij}$, knowing each $\delta x^k \partial H/\partial x^l$ also allows you to figure out the approximate distance $\delta x^i$ of the system away from the equilibrium of the potential. You can also use some dimensional analysis for a rule of thumb estimate of the complete release of the system from equilibrium. Assume that from the physics of the system you understand that the potential is associated with a binding energy scale $E_{\rm b}$ and that the second-derivative matrix goes as $V_{ij} \sim E_{\rm b}/L_{\rm V}^2$ where $L_{\rm V}^2$ is a variability length. You can then estimate that the system stays bound as long as $$\delta x^l\frac{\partial H}{\partial x_l}(x^i) \lesssim E_{\rm b}$$ We also see that and equivalent condition is $|\delta x| \lesssim L_{\rm V} $, which is also the condition for the small-$\delta x$ expansions above to be valid.


So far I have just discussed classical mechanics, no statistical physics involved. Let us now, for simplicity, shift our coordinate system so that $x^i_0 = 0$ and then $$\langle x^k \frac{\partial H}{\partial x^l}\rangle \approx 2 \langle V_{lj}x^j x^k \rangle$$ The statement that for $k\neq l$ this is zero means simply that fluctuations in "energy-orthogonal" directions are uncorrelated. This can be understood particularly well if you rotate into a basis where $x^i$ are eigenvectors of $V_{ij}$ (i.e. a basis where $V_{ij}$ is diagonal). The $k=l$ case (without Einstein summation!) gives you the correlation of "energy-related" fluctuations about the potential equilibrium.

For example, once using the virial theorem and the rule of thumb estimate for bound systems near equilibrium, we get a condition for the system staying bound as $$\langle x^l \frac{\partial H}{\partial x^l}\rangle \approx 2n\langle V - V_{\rm eq}\rangle = n k_{\rm B} T \lesssim E_{\rm B}$$ (here $n$ is the number of degrees of freedom.) I.e. $\langle x^l \frac{\partial H}{\partial x^l}\rangle$ allows you to analyse in detail whether the system stays in equilibrium, possibly reaches other local equilibria etc. etc.

Of course, this is just an example of a class of systems. There are systems with degenerate equilibria for which the discussion is changed in a few details but the general meaning of the term $\langle x \partial H/\partial x\rangle$ is similar. In quantum mechanics one actually has to use a similar analysis to answer whether the temperature is sufficient to excite a degree of freedom at least by a single quantum jump and whether it thus has to be included in the state sum. In astrophysics one also often discusses systems where the virial theorem is very important but the gravitational potential is $\sim 1/|x - x'|$ between every two particles. However, the meaning of the term $\langle x \partial H/\partial x\rangle$ is not quite universal and becomes particularly murky in relativistic physics. So it is as I said in the beginning, the virial theorem provides a useful tool for specific classes of systems but perhaps not all systems.

Void
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