Intuition behind classical virial theorem 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?
 A: 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.
A: 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.
