# Coarse Grained

Mathematics and physics through TCM, in a non-representative stream

## Higher derivative theories

We tend to not use higher derivative theories. It turns out that there is a very good reason for this, but that reason is rarely discussed in textbooks. We will take, for concreteness, $L\left(q,\dot q, \ddot q\right)$, a Lagrangian which depends on the 2nd derivative in an essential manner. Inessential dependences are terms such as $q\ddot q$ which may be partially integrated to give ${\dot q}^2$. Mathematically, this is expressed through the necessity of being able to invert the expression $$P_2 = \frac{\partial L\left(q,\dot q, \ddot q\right)}{\partial \ddot q},$$ and get a closed form for $\ddot q \left(q, \dot q, P_2 \right)$. Note that usually we also require a similar statement for $\dot q \left(q, p\right)$, and failure in this respect is a sign of having a constrained system, possibly with gauge degrees of freedom.

In any case, the non-degeneracy leads to the Euler-Lagrange equations in the usual manner: $$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q} + \frac{d^2}{dt^2}\frac{\partial L}{\partial \ddot q} = 0.$$ This is then fourth order in $t$, and so require four initial conditions, such as $q$, $\dot q$, $\ddot q$, $q^{(3)}$. This is twice as many as usual, and so we can get a new pair of conjugate variables when we move into a Hamiltonian formalism. We follow the steps of Ostrogradski, and choose our canonical variables as $Q_1 = q$, $Q_2 = \dot q$, which leads to \begin{align} P_1 &= \frac{\partial L}{\partial \dot q} - \frac{d}{dt}\frac{\partial L}{\partial \ddot q}, \\ P_2 &= \frac{\partial L}{\partial \ddot q}. \end{align} Note that the non-degeneracy allows $\ddot q$ to be expressed in terms of $Q_1$, $Q_2$ and $P_2$ through the second equation, and the first one is only necessary to define $q^{(3)}$.

We can then proceed in the usual fashion, and find the Hamiltonian through a Legendre transform: \begin{align} H &= \sum_i P_i \dot{Q}_i - L \\ &= P_1 Q_2 + P_2 \ddot{q}\left(Q_1, Q_2, P_2\right) - L\left(Q_1, Q_2,\ddot{q}\right). \end{align} Again, as usual, we can take time derivative of the Hamiltonian to find that it is time independent if the Lagrangian does not depend on time explicitly, and thus can be identified as the energy of the system.

However, we now have a problem: $H$ has only a linear dependence on $P_1$, and so can be arbitrarily negative. In an interacting system this means that we can excite positive energy modes by transferring energy from the negative energy modes, and in doing so we would increase the entropy — there would simply be more particles, and so a need to put them somewhere. Thus such a system could never reach equilibrium, exploding instantly in an orgy of particle creation. This problem is in fact completely general, and applies to even higher derivatives in a similar fashion.

## Central Limit Theorem

There's a proof of the Central Limit Theorem which I am very fond of, which is not often seen in textbooks. It is a sort of renormalisation argument. In a way, it's not even a rigorous proof — often the way with renormalisation — but in conjunction with the more classical proofs it lends a powerful insight.

Without loss of generality, let's consider a whole bunch of identical, univariate variables $X_n$, each with zero mean. Thus we have quite trivially that a sum of $N$ of them will still have zero expectation, and a variance of $N$.

Now, rather than summing all of them at the same time, we do it in steps, and renormalise along the way to see what happens. So concretely, let $X$ have a distribution given by $f$, which is assumed to be sufficiently well-behaved for whatever we need to do. Then $X+X \sim f \star f$, where the multiplication is a convolution. This is our coarse-graining step, so we still need to re-scale, so that we get back a univariate distribution: $$f^\prime(x) = \sqrt 2 (f \star f)(\sqrt 2 x).$$ Convolutions are easiest to take in Fourier space: $$\widetilde{f^\prime}(k) = \left[ \tilde{f}(k/\sqrt{2}) \right]^2.$$ It is then fairly trivial to check that the univariate Gaussian $\widetilde{f^*}(k) = e^{-k^2/\sqrt{2}}$ is a fixed point.

We can view this step as a transform on the space of distributions, and so it makes sense to linearise about this fixed point and look at what happens to small perturbations: \begin{align*} \widetilde{f^*}(k) + \widetilde{g_n}(k) &\rightarrow \widetilde{f^*}(k) + 2 \widetilde{g_n}(k/\sqrt{2}) \widetilde{f^*}(k/\sqrt{2}) + \mathrm{h.o.t.} \\ &= \widetilde{f^*}(k) + \lambda_n \widetilde{g_n}(k) \end{align*} Which has solutions as $\widetilde{g_n}(k) = (ik)^n \widetilde{f^*}(k)$ with eigenvalues $\left|\lambda_n\right| = 2^{1-n/2}$; these correspond to mathematically meaningful perturbations if and only $n$ is an integer greater than 0, for reasons of convergence and normalisation. That still leaves $n=1$ or $n=2$ as being relevant and marginal; the former correspond to shifting the mean, the latter to changing the variance — and since those are not allowed by assumption, we find that the Gaussian is a stable fixed point.

Notice that this does not say anything about the size of basin of attraction, so if another fixed point existed it could cause finite perturbations to actually flow away. Indeed, this makes it not quite a proper proof. On the other hand, this procedure gives the actual rate of convergence to a Gaussian, something that the classical proofs do not give.

## Product representation

In dealing with complex analytic functions, it is often quite handy to represent them as their Taylor expansions, i.e. a summation representation. We can generalise a little bit and expand even around non-essential, isolated singularities with a Laurent series. However, quite often we really care about the zeros of a function (e.g. Yang-Lee circle theorem on zeros of the grand partition function for Ising-esque models), and extracting those out of summations is unwieldy. Therefore, it would be much nicer to have a product representation instead.

For polynomials, such a representation is obvious, and unique — the fundamental theorem of algebra guarantees the factorisation into linear factors: $$p(z) = p(0) \prod_{\mathrm{finite~}n} \left(1 - \frac{z}{z_n}\right).$$ The various $z_n$ are then the location of the zeros. We would like to extend this to more general functions.

However, in general this is difficult, and non-unique (see Weierstrass factorization theorem for an existence statement). For one particular subset however, we can create an effective procedure for manufacturing these: functions with only simple isolated zeros and no poles.

So let $f(z)$ be such a function. If $f$ has a zero of order $m$ at $z=0$, then we can divide out by $z^m$ and get something without a zero at the origin, and so without loss of generality that's what we'll assume. Let $z_n$ be the location of remaining (infinite number of) zeros. Then $g(z) = [\ln f(z)]^\prime = f^\prime(z)/f(z)$ has only simple isolated poles with unit residues at $z_n$; thus if we find a summation representation of $g$ we could integrate and exponentiate to find a product representation for $f$.

Now consider: $$\frac{1}{2\pi i} \oint_{\Gamma_n} \frac{g(z^\prime)}{z^\prime-z} dz^\prime = g(z) + \sum^n_j \frac{1}{z_j - z}$$ where $\Gamma_n$ is a contour which encloses the closest $n$ poles to the origin. Then $$g(z) - g(0) = \frac{z}{2\pi i} \oint_{\Gamma_n} \frac{g(z^\prime)}{z^\prime(z-z^\prime)} dz^\prime + \sum_j^n \left(\frac{1}{z-z_j} + \frac{1}{z_j}\right).$$ Thus if we can find a sequence of contours $\Gamma_n$ such that $g$ remains bounded on them, the integral will converge to zero as $n \rightarrow 0$. In that case, we find $$g(z) = g(0) + \sum_n \left(\frac{1}{z-z_n} + \frac{1}{z_n}\right).$$

So now we can return to factorising $f$. Integrating $g$ gives $$\ln f(z) - \ln f(0) = cz + \sum_n \left[ \ln\left(1-\frac{z}{z_n}\right) + \frac{z}{z_n} \right]$$ where $c = g(0)$; re-exponentiating gives $$f(z) = f(0) e^{cz} \prod_n \left(1 - \frac{z}{z_n}\right) e^{z/z_n}.$$

As an example, consider $f(z) = \sin(z)/z$. Then $g(z) = -1/z + \cot(z)$; we can pick contours $\Gamma_n$ which sit between the poles in $\cot z$, and our procedure will converge. A bit of limit work shows that $g(0) = 0$ and $f(0) = 1$. The zeros sit at $z_n = n \pi$, with $n$ being any non-zero integer. Thus we find \begin{align*} \frac{\sin(z)}{z} &= \prod_{n \neq 0} \left(1 - \frac{z}{\pi n}\right) e^{z/n\pi} \\ &= \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{\pi^2 n^2}\right). \end{align*}