TITLE: Nonlinear optics as gauge theory
QUESTION [15 upvotes]: the widely used approach to nonlinear optics is a Taylor expansion of the dielectric displacement field $\mathbf{D} = \epsilon_0\cdot\mathbf{E} + \mathbf{P}$ in a Fourier representation of the polarization $\mathbf{P}$ in terms of the dielectric susceptibility $\mathcal{X}$:
$\mathbf{P} = \epsilon_0\cdot(\mathcal{X}^{(1)}(\mathbf{E}) + \mathcal{X}^{(2)}(\mathbf{E},\mathbf{E}) + \dots)$ .
This expansion does not work anymore if the excitation field has components close to the resonance of the medium. Then, one has to take the whole quantum mechanical situation into account by e.g. describing light/matter interaction by a two-level Hamiltonian.
But this approach is certainly not the most general one.
Intrinsically nonlinear formulations of electrodynamics
So, what kind of nonlinear formulations of electrodynamics given in a Lagrangian formulation are there?
One known ansatz is the Born-Infeld model as pointed out by Raskolnikov. There, the Lagrangian density is given by
$\mathcal{L} = b^2\cdot \left[ \sqrt{-\det (g_{\mu \nu})} - \sqrt{-\det(g_{\mu \nu} + F_{\mu \nu}/b)} \right]$
and the theory has some nice features as for example a maximum energy density and its relation to gauge fields in string theory. But as I see it, this model is an intrinsically nonlinear model for the free-space field itself and not usefull for describing nonlinear matter interaction.
The same holds for an ansatz of the form
$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \lambda\cdot\left( F^{\mu\nu}F_{\mu\nu} \right)^2$
proposed by Mahzoon and Riazi. Of course, describing the system in Quantum Electrodynamics is intrinsically nonlinear and ... to my mind way to complicated for a macroscopical description for nonlinear optics. The question is: Can we still get a nice formulation of the theory, say, as a mean field theory via an effective Lagrangian?
I think a suitable ansatz could be
$\mathcal{L} = -\frac{1}{4}M^{\mu\nu}F_{\mu\nu}$
where $M$ now accounts for the matter reaction and depends in a nonlinear way on $\mathbf{E}$ and $\mathbf{B}$, say
$M^{\mu\nu} = T^{\mu\nu\alpha\beta}F_{\alpha\beta}$
where now $T$ is a nonlinear function of the field strength and might obey certain symmetries. The equation $T = T\left( F \right)$ remains unknown and depends on the material.
Metric vs. $T$ approach
As pointed out by space_cadet, one might ask the question why the nonlinearity is not better suited in the metric itself. I think this is a matter of taste. My point is that explicitly changing the metric might imply a non-stationary spacetime in which a Fourier transformation might not be well defined. It might be totally sufficient to treat spacetime as Lorentzian manifold.
Also, we might need a simple spacetime structure later on to explain the material interaction since the polarization $\mathbf{P}$ depends on the matter response generally in terms of an integration over the past, say
$\mathbf{P}(t) = \int_{-\infty}^{t}R\left[\mathbf{E}\right](\tau )d\tau$
with $R$ beeing some nonlinear response function(al) related to $T^{\mu\nu\alpha\beta}$.
Examples for $T$
To illustrate the idea of $T$, here are some examples.
For free space, $T$ it is given by $T^{\mu\nu\alpha\beta} = g^{\mu\alpha}g^{\nu\beta}$ resulting in the free-space Lagrangian $\mathcal{L} = -\frac{1}{4}T^{\mu\nu\alpha\beta}F_{\alpha\beta}F_{\mu\nu} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$
The Lagrangian of Mahzoon and Riazi can be reconstructed by
$T^{\mu\nu\alpha\beta} = \left( 1 + \lambda F^{\gamma\delta}F_{\gamma\delta} \right)\cdot g^{\mu\alpha}g^{\nu\beta}$.
One might be able to derive a Kerr nonlinearity using this Lagrangian.
So, is anyone familiar in a description of nonlinear optics/electrodynamics in terms of a gauge field theory or something similar to the thoughts outlined here?
Thank you in advance.
Sincerely,
Robert
Comments on the first Bounty
I want to thank everyone actively participating in the discussion, especially Greg Graviton, Marek, Raskolnikov, space_cadet and Willie Wong. I am enjoying the discussion relating to this question and thankfull for all the nice leads you gave. I decided to give the bounty to Willie since he gave the thread a new direction introducing the material manifold to us.
For now, I have to reconsider all the ideas and I hope I can come up with a new revision of the question that should be formulated in a clearer way as it is at the moment.
So, thank you again for your contributions and feel welcome to share new insights.

REPLY [9 votes]: Just a few random thoughts.
There is something important in your observation that the Born-Infeld model is essentially a free-space model. It is known to Boillat and Plebanski (separately in 1970) that the Born-Infeld model is the only model of electromagnetism (as a connection on a $U(1)$ vector bundle) that satisfies the following conditions

Covariance under Lorentz transformations
Reduces to Maxwell's equation in the small-field strength limit
$U(1)$ gauge symmetry
Integrable energy density for a point-charge
No birefringence (speed of light independent of polarization).

(the linear Maxwell system fails condition 4.) (See Michael Kiessling, "Electromagnetic field theory without divergence problems", J. Stat. Phys. (2004) doi:10.1023/B:JOSS.0000037250.72634.2a for an exposition on this and related issues.)
Now, since you are interested in nonlinear optics inside a material, instead of in vacuum, I think conditions 1 and 5 can safely be dropped. (Though you may want to keep 5 as a matter of course.) Condition 4 is intuitively pleasing, but maybe not too important, at least not until you have some candidate theories in mind that you want to distinguish. Condition 3 you must keep. Condition 2, on the other hand, really depends on what kind of material you have in mind.
In any case, a small suggestion: personally I think it is better to, from the get-go, write your proposed Lagrangian as
$$ L = T^{abcd} F_{ab}F_{cd} $$
instead of $M^{ab}F_{cd}$. I think it is generally preferable to consider Lagrangian field theories of at least quadratic dependence on the field variables. A pure linear term suggests to me an external potential which I don't think should be built into the theory.
If you want something like condition 2, but with a dielectric constant or such, then you must have that $T^{abcd}$ admit a Taylor expansion looking something like
$$ T^{abcd} = \tilde{g}^{ac}\tilde{g}^{bd} + O(|F|) $$
where $\tilde{g}$ is some effective metric for the material. Birefringence, however, you don't have to insert in explicitly: most likely a generic (linear or nonlinear) $T^{abcd}$ you write down will have birefringence; it is only when you try to rule it out that you will bring in some constraints.
An interesting thing is to consider what it means to have an analogous notion to condition 1. In the free-space case, condition 1 implies that the Lagrangian should only be a function of the Lorentz invariant $B^2 - E^2$ (in natural units) and of the pseudo-scalar invariant $B\cdot E$. In terms of the Faraday tensor these two invariants are $F^{ab}F_{ab}$ and $F^{ab}{}^*F_{ab}$ respectively, where ${}^*$ denote the Hodge dual. The determination of the linear part of your theory (of electromagnetic waves in a material) is essentially by what you will use to replace condition 1. If you assume your material is isotropic and homogeneous, then some similar sort of scalar + pseudo-scalar invariants is probably a good bet.


TITLE: Prove condition number of a invertible matrix is atleast one
QUESTION [3 upvotes]: Show that the condition number of an invertible matrix must be at least 1. What matrices have
condition number equal to 1.
If someone could help me with this and give an explanation that would be very helpful. I do not know where to start

REPLY [3 votes]: Hint: note that $\|AB\| \leq \|A\|\cdot \|B\|$ (that is, $\|\cdot\|$ is "sub-multiplicative").
So, $\|A\|\cdot \|A^{-1}\| \geq \cdots ?$


TITLE: Find the derivative of the following function w.r.t $x:$ $y=(\sin x)^{(\cos x)^{(\cos x)^{(\cos x)^{\cdots\infty}}}}$
QUESTION [7 upvotes]: Find $\frac{dy}{dx},$ if
$y=(\sin x)^{(\cos x)^{(\cos x)^{(\cos x)^{\cdots\infty}}}}$
My attempt:
If the base and power were the same function,then we could find it's derivative with respect to $x$ as $y=x^y.$ But,here in the question,$\sin x$ and $\cos x$ are two different functions.So,I have no idea how to differentiate it with respect to $x.$
Also,I don't know if this function exist or not.If exist,then find its derivative.
Any help would be appreciated.Thank you!

REPLY [4 votes]: Here's how I would do it:
As @Blitzer said, define:
$$f(x) = (\cos x)^{(\cos x)^{(\cos x)^{\dots}}} = (\cos x)^{f(x)}$$
such that $y=(\sin x)^{f(x)}$
Now differentiate this like so:
$$\ln (f(x)) = f(x) \ln (\cos x) \implies \frac{f'(x)}{f(x)} = -f(x) \tan x + f'(x) \ln (\cos x)$$
$$\implies f'(x) = -(f(x))^2 \tan x + f(x) f'(x) \ln (\cos x)$$
So upon solving for $f'(x)$ we have:
$$f'(x) = \frac{(f(x))^2 \tan x}{f(x) \ln (\cos x)-1}$$
Using the same trick:
$$\ln (y) = f(x) \ln (\sin x) \implies \frac{y'}{y} = f(x) \cot (x) + f'(x) \ln (\sin x)$$
$$\implies \frac{dy}{dx} = f(x) (\sin x)^{f(x)} \cot(x)+f'(x) (\sin x)^{f(x)} \ln (\sin x)$$
Upon plugging in our expression of $f'(x)$, we have:
$$\frac{dy}{dx} =  f(x) (\sin x)^{f(x)} \cot(x)+\left(\frac{(f(x))^2 \tan x}{f(x) \ln (\cos x)-1}\right) (\sin x)^{f(x)} \ln (\sin x)$$
Thus finally, by plugging in $f(x)$, we obtain the rather horrible expression
$$\frac{dy}{dx} =  \left((\cos x)^{(\cos x)^{(\cos x)^{\dots}}}\right) (\sin x)^{\left((\cos x)^{(\cos x)^{(\cos x)^{\dots}}}\right)} \cot(x)+\left(\frac{\left((\cos x)^{(\cos x)^{(\cos x)^{\dots}}}\right)^2 \tan x}{\left((\cos x)^{(\cos x)^{(\cos x)^{\dots}}}\right) \ln (\cos x)-1}\right) (\sin x)^{\left((\cos x)^{(\cos x)^{(\cos x)^{\dots}}}\right)} \ln (\sin x)$$


TITLE: Different ways of having infinite global dimension
QUESTION [4 upvotes]: Is there any ring $R$ of infinite global dimension such that any $R$-module is a retract (i.e. direct summand) of some $\oplus_{i\in I}M_i$ where each $M_i$ has finite projective dimension?
I ask this because in the easy examples of rings of infinite global dimension I have in mind, there is always a simple $R$-module with infinite projective dimension. I wanted to know if this happens in general, or if pathological examples exist. In the situation I ask for above, the global dimension would be infinite, morally for asymptotic reasons.

REPLY [4 votes]: This doesn't quite answer the question but shows that the real question is for an example where all cyclic modules have finite but unbounded projective dimension.  So I will leave this here.
In https://projecteuclid.org/download/pdf_1/euclid.nmj/1118799684 Auslander shows that the global dimension of $R$ is the supremum of the projective dimensions of cyclic modules.
If a cyclic module is a retract of a direct sum of modules of finite projective dimension, then it would be a retract of a finite direct sum of such modules and hence have finite projective dimension.  So what you want can't happen: there is no such $R$.
Added for clarity:  the projective dimension of a module $M$ is the largest $n$ such that $Ext^n(M,-)$ is non-zero. Since Ext commutes with direct sums a direct summand of a finite direct sum of modules of finite projective dimension has finite projective dimension.


TITLE: Complicated Integral from an Integral Test
QUESTION [1 upvotes]: I have this infinite series
$$\sum_{n=1}^{\infty} \frac{n\sqrt{n}}{n^2 +2^{-n}}$$
I can only use Integral, Bounded-Sum, Comparison, and Divergence tests. Divergence Test is inconclusive on this. Integral Test seems promising since the series satisfies the conditions, but $$\int_{1}^{\infty} \frac{x\sqrt{x}}{x^2 +2^{-x}} $$
appears to be a difficult integral. Even if I resort to Comparison Test,
$$\frac{n\sqrt{n}-2^{-n}}{n^2}<\frac{n\sqrt{n}}{n^2 +2^{-n}}$$
It's easy to show $\displaystyle \sum_{n=1}^{\infty}\frac{n\sqrt{n}}{n^2}$ is divergent, but $\displaystyle \sum_{n=1}^{\infty} \frac{2^{-n}}{n^2}$ is (absolutely) convergent with ratio and root test, which I cannot use yet. Thus by Comparison Test, I know that $\displaystyle \sum_{n=1}^{\infty} \frac{n\sqrt{n}}{n^2 +2^{-n}}$ is divergent. The problem is I can only use the tests I mentioned above, or no tests beyond Comparison Test. So I'm thinking that I just need to evaluate the improper integral for the Integral Test to justify divergence. Any insights on how I should solve the integral or frame my answer in a different way?

REPLY [1 votes]: $\frac{n\sqrt{n}}{n^2 +2^{-n}}>\frac{n\sqrt{n}}{n^2 +1}\geq  \frac{n\sqrt{n}}{2n^2}$ and $\sum \frac 1 {2n^{1/2}}$ is divergent.

REPLY [0 votes]: You have$$\lim_{n\to\infty}\frac{\frac{n\sqrt n}{n^2+2^{-n}}}{\frac1{\sqrt n}}=\lim_{n\to\infty}\frac{n^2}{n^2+2^{-n}}=\lim_{n\to\infty}\frac1{1+n^{-2}2^{-n}}=1.$$Since the series $\sum_{n=1}^\infty\frac1{\sqrt n}$ diverges, so does your series.


TITLE: Level structure on relative Jacobian
QUESTION [1 upvotes]: Assume that $X/S$ is a family of smooth projective curves say in characteristic zero. Can the relative Jacobian of $X/S$ always be endowed with a level structure? In other words does a map $S\to A_{g,n}$ exists for $n>3$? I would appreciate any reference for this question.

REPLY [1 votes]: This is not always possible.
Consider an elliptic curve $E$ over $S= Spec \mathbb{Q}$ with trivial Mordell-Weil group, for instance. Can you take it from here?


TITLE: Is there a formula to solve this probability question?
QUESTION [0 upvotes]: Five students forget their backpacks in class. Their teacher returns the backpacks randomly. What is the probability that exactly two of them get their own backpacks?
I know that there are 5! ways in which the backpacks can be returned to the students and by a simple counting argument I found that there are 20 different ways in which two students get their own backpack returned. And so the answer is a probability of 1/6. My question is what is a formulaic way to come up with 20 different ways in which two students get their own backpack retuned instead of just literally counting the number of different ways?

REPLY [0 votes]: Count the number of ways a specific pair get their packs back first, say 1 and 2. Then you must identifycstudentc1 with pack 1, student 2 with pack 2, and thecrest cannot have their own packs. Thus student 3 gets pack 4 or pack 5. You should see that each of the last two cases leaves only one choice for students 4 and 5, so there are two permutations for 1 and 2 getting their own packs.
So then there would also be two permutations for 1 and 3, two for 1 and 4, and so on. Thus you multiply the two permutations for and specific pair by $\binom{5}{2}=10$ possible pairs getting 20 "correct" permutations in all and a probability of 1/6.


TITLE: Refrigerator Binomial Distribution problem
QUESTION [0 upvotes]: A company produces refrigerators and knows that the proportion of defective units is 0.05.
The refrigerators are sold in lots of 35 units. A large network of appliance stores accept the lot if it has a maximum of 2 refrigerators with defect in the batch.
Admitting that they've accepted the lot, what is the probability of having observed exactly one faulty refrigerator?
Answer: 0.4102
I've tried to do this using the binomial distribution "formula" using N (trials) as 35 and n (success) as 1, and 2, and probability of success as 0.05 and couldn't resolve this question.

REPLY [1 votes]: The binomial distribution's pmf is: $p(x)=\binom{35}{x}(0.05)^x(0.95)^{35-x} \quad[0\leq x\leq 35, x\in\Bbb N]$
The conditional probability that there is exactly 1 faulty unit when given that there is a maximum of 2 is $$\dfrac{p(1)}{p(0)+p(1)+p(2)}$$


TITLE: Generalization of Lévy's continuity theorem for nuclear spaces
QUESTION [7 upvotes]: I am interested in a generalization of the following finite-dimensional results in infinite dimensional vector-space with nuclear structure, especially for the cases of the spaces of distributions $\mathcal{D}'(\mathrm{R}^N)$ and $\mathcal{S}'(\mathrm{R}^N)$.
Theorem 1:
Let $X_n$, $n\in\mathbb{N}$ and $X$ be random variables in $\mathbb{R}^N$ and $\Phi_n$ and $\Phi$ their characteristic functions. Then,
$$\left( X_n \overset{\mathcal{L}}{\rightarrow} X  \right) \Leftrightarrow \left( \forall \omega, \Phi_n(\omega) \rightarrow \Phi(\omega) \right).$$
Theorem 2: (Lévy's continuity theorem)
Again, the $X_n$'s are random variable and the $\Phi_n$'s are their characteristic functions. Assume that the limit $\lim \Phi_n(\omega)$ exists pointwise and is denoted by $\Phi(\omega)$. We then have the equivalence:

$(X_n)$ converges in law to some random variable $X$.
$\Phi$ is continuous at $0$.

Now I precise my question. Given a probability measure $\mu$ on $\mathcal{N}' = \mathcal{D}'(\mathrm{R}^N)$ or $\mathcal{S}'(\mathrm{R}^N)$, we can define its characteristic functional on $\mathcal{N}$ by
$$\hat{\mu}(\varphi) = \int_{\mathcal{N}'} \mathrm{e}^{\mathrm{j} \langle u , \varphi \rangle} \mathrm{d}\mu (u).$$
This generalizes the concept characteristic function of a random variable (Bochner's theorem being generalized by Minlos's theorem). Is the following result true?
Possible generalization of theorem 1:
Let $\mu_n$, $n\in\mathbb{N}$ and $\mu$ be probability measures on $\mathcal{N}'$ and $\hat{\mu}_n$ and $\mu$ their characteristic functionals. Then,
$$\left( \mu_n \overset{\mathrm{weakly}}{\rightarrow} \mu  \right) \Leftrightarrow \left( \forall \varphi, \hat{\mu}_n(\varphi) \rightarrow \mu(\varphi) \right).$$
Similarly, can we generalize the Lévy's continuity theorem?
Thanks for attention.

REPLY [0 votes]: A more abstract version of this theorem can be found in Le théorème de continuité de P. Lévy sur les espaces nucléaires where it is proven for Borel measures on strict inductive limits of nuclear Fréchet spaces.
The method of proof appears to be the one from Processus linéaires, processus généralisés as mentioned in another answer.


TITLE: Integrals and Area-element in Cylindrical coordinates
QUESTION [0 upvotes]: I was trying to solve for the moment of inertia of a solid and a hollow cylinder, and I faced a small problem. I looked through online resources and found many ways to approach the problem.
One of these methods was, integrating the following in cylindrical polar coordinates.
$I_{yy} = \int dm (x^2+z^2)$
In case of hollow cylinder, $x=Rcos\phi$ and $dm=\sigma Rd\phi dz$.Using this I was easily able to obtain the moment of Inertia.
Similarly, in case of solid cylinder, $x=rcos\phi$ and $dm=\rho rdrd\phi dz$. Again, I was able to obtain the correct expression using this.
My question is, what if I try to solve the above integrals forcibly in the cartesian coordinate system. How would the limits of integration, and the area and volume elements change?
I was able to do this for the solid cylinder, as follows :
$I_{yy} = \rho \int\limits_{-a}^{a} \int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \int\limits_{-l/2}^{l/2} (x^2+z^2)dxdydz$
However, I don't know how to write the integral for hollow cylinder for cartesian coordinates. What would be the surface area element of a cylinder, in cartesian coordinates? I'm inclined to say, $dm=\sigma dz\sqrt{(dx)^2+(dy)^2}$, however, I'm not sure if this is correct.
How would we write the integral then, or is it at all possible to write the integral for surface element purely in cartesian coordinates, or is it only possible in a cylindrical coordinate system?

REPLY [1 votes]: I don't exactly understand how you assigned your coordinates, however it seems Y-axis is the principal axis here about which you have been calculating the Moment of Inertia. So I'll assume the same.
We can indeed calculate MoI of a hollow cylinder using cartesian coordinates.
$$I=\int dm.r^2=\int dm.(x^2+z^2)$$
Since, $x^2+z^2=R^2$ for a hollow cylinder, where R is the radius of the cylinder; and $dm=\sigma dA=\sigma (dy)\sqrt{(dx)^2+(dz)^2}$
$$\Rightarrow I=\sigma R^2\int(dy)\sqrt{(dx)^2+(dz)^2}$$
Now, since we can't integrate such a function without removing the square root, we will use the relation between $x$ and $z$ i.e. $x^2+z^2=R^2$.
Differentiating the above relation w.r.t $x$, we get $2xdx+2zdz=0$
$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\Rightarrow dz=-\frac xz dx$
We can substitute the above relation in $I$, and using appropriate limits, we get:
$$\Rightarrow I=\sigma R^2\int_{-R}^R\int_{-\frac L2}^{\frac L2}\frac {\sqrt{x^2+z^2}}{z}dx\quad \quad ...(1)$$
$$\Rightarrow I=\sigma R^2\int_{-R}^R\int_{-\frac L2}^{\frac L2}(dy)\frac R{\sqrt{R^2-x^2}}dx \quad \quad (since,x^2+z^2=R^2, we \;substituted\; z.)$$
$$=\sigma R^3L \int_{-R}^R\frac {dx}{\sqrt{R^2-x^2}}= \sigma R^3L\biggl[sin^{-1}\Bigl(\frac xR\Bigr)\biggr]_{-R}^R=\sigma R^3L(\pi)$$
Now, we replace $\sigma$ with $\sigma=\frac M{2\pi RL}$, and we get:$$\Rightarrow I= \frac 12MR^2$$
However, in this process we did not cover the entire $Z$-axis, instead only half of it. This is because in $eq(1)$, $z=\sqrt{R^2-x^2}$ allows only positive $z$. Therefore, we carry out the same process for the other half of the cylinder i.e. $z=-\sqrt{R^2-x^2}$,and we get:
$$I_{hollow,yy}=\frac 12 MR^2+\frac 12 MR^2=MR^2$$


TITLE: $\sigma$-algebra generated by all countable and co-countable sets
QUESTION [2 upvotes]: Let $S=[0,1]$.
Let $\Sigma$ be a $\sigma$-algebra generated by all countable and co-countable subsets of $S$.
I know that a set of all countable and co-countable subsets itself is a $\sigma$-algebra.
But what happens if this set generates a new $\sigma$-algebra $\Sigma$?
Will $\Sigma$ be bigger, will it be $\mathcal{P}(S)$?

REPLY [0 votes]: If I understand correctly, first we take $\mathcal P(S)$, and then the set $\Sigma$ of all countable and cocountable sets.
Then you already know that $\Sigma$ is a $\sigma$-algebra, and ask what is $\sigma(\Sigma)$, the generated $\sigma$-algebra.
And the answer is that $\sigma(\Sigma) =\Sigma$, since $\Sigma$ is the smallest $\sigma$-algebra containing $\Sigma$.

REPLY [0 votes]: As it has already been answered that $\Sigma$ will exactly be the sigma algebra of countable and co-countable sets, I will give a brief explanation of why is that the case?
As you have already pointed out that the set of all countable and co-countable sets form a sigma algebra. Let's call this sigma algebra $\Sigma_0.$ And, let the sigma algebra generated by countable and co-countable sets be denoted by $\Sigma.$
It is obvious that $\Sigma_0\subseteq \Sigma.$ But, by definition of $\Sigma$ we have that if any sigma algebra $\mathcal{F}$ contains all countable and co-countable sets then $\Sigma\subseteq \mathcal{F}.$ In particular, for $\mathcal{F}=\Sigma_0,$ we get that $\Sigma\subseteq \Sigma_0.$ It thus follows that $\Sigma=\Sigma_0.$