TITLE: What is the intensity distribution behind the first beam splitter on an observation screen at the same distance like behind the second beam splitter?
QUESTION [0 upvotes]: Mach-Zehnder interferometer has been used, among other things, to measure phase shifts between the two beams caused by a sample or a change in length of one of the paths. My simple question is: What is the intensity distribution behind the first beam splitter on an observation screen at the same distance like behind the second beam splitter?

REPLY [1 votes]: If the beam splitters are all 50% reflectors and perfectly flat, etc.  then whatever spatial energy profile the incoming beam has will be seen in both paths behind the first splitter.  A typical interferometer is set up so that the paths are short enough that self-interference/diffraction does not happen.
Assuming your "second splitter" is the recombiner, then the intensity pattern there is the combination of the interference effects with the input spatial intensity pattern.  That is, if you had a "dim" region in the input beam,  you'd see the same fringe pattern as elsewehre in the beam but with lower peak intensity.


TITLE: Why is kinetic energy defined as the measure of work an object can do due the virtue of its motion?
QUESTION [0 upvotes]: Saying that KE=1/2mv^2 holds even when magnitude or direction of force changes, so the expression is valid irrespective of how the body acquires the velocity shouldn't be correct as work done depends on the direction of force relative to the displacement. If displacement and force are perpendicular work down by the object is 0. So how can the expression of kinetic energy hold under any direction of change in force?

REPLY [0 votes]: Kinetic energy is a scalar quantity. Velocity is a vector quantity, i.e. it has a direction. Likewise momentum is a vector quantity. Vector quantities can be resolved into components along coordinate lines. Scalar quantities like kinetic energy and potential are just single numbers at any spacetime location. You essentially loose the vector information when you square the velocity vector.
You should be cautioned that the definition you are using in the title is only valid if you measure the work in a "conservative interaction", i.e. one where the energy of the system is not dissipated in an unmeasured manner. You would need to account for any heat or sound when trying to sum up the "work" done in the process of decelerating a moving body.


TITLE: Physical meaning of Legendre transformation
QUESTION [87 upvotes]: I would like to know the physical meaning of the Legendre transformation, if there is any? I've used it in thermodynamics and classical mechanics and it seemed only a change of coordinates?

REPLY [24 votes]: There are already some nice answers regarding intuitive interpretations of the Legendre transform. What I want to contribute here, is a more physical reason/motivation for why they appear. I.e., instead of focusing on their physical interpretation, I will focus on the physical requirements that uniquely define Legendre transforms. This explains why you would inevitably end up inventing them, even if you had never heard of them before.
Variational principles are fundamental to physics. Legendre transforms naturally arise when requiring that variational principles carry over when changing variables; in fact, they turn out to be the unique solution to this physical requirement. I will illustrate this in the thermodynamic context. (A main conceptual difference from the other explanations on this page, is that according to this view, it is important that our function also depends on other variables in order to naturally arive at Legendre transforms!)
One important instance of a variational principle is the principle of minimum energy: a closed system with a fixed entropy will minimize its energy (note that 'closed' means that there is no mass transfer, but energy transfer is allowed!). This can be derived from the second law of thermodynamics; for a derivation, see here. To state this more precisely: if we denote our internal energy $U(S,X)$ where $S$ is the entropy and $X$ is some macrosopic variable that can thermalize (i.e., it is not fixed; e.g., a density profile, or the position of a sliding partition), then in equilibrium, we have
$$ \boxed{ \left. \frac{\mathrm dU(S,X)}{\mathrm dX} \right|_{S} = 0 } \quad \textrm{(for the equilibrium value $X= X_\textrm{eq}$).} $$
(I will be suppressing other thermodynamic variables that might be held fixed, such as volume.) Here I am following the notational convention that the bar indicates what one is holding fixed when taking the derivative.
To each extensive thermodynamic variable, such as entropy, we can associate an intensive variable, in this case the temperature:
$$ T(S,X) \equiv \left. \frac{\mathrm dU(S,X)}{\mathrm dS} \right|_{X}.  $$
Such associated extensive and intensive variables are said to be conjugate.
Depending on the physical context, it can be more natural to keep the intensive parameter $T$ fixed, rather than the entropy $S$ (e.g., for a system connected to a heat bath). In principle (if $U(S,X)$ is convex as a function of $S$), we can invert the above relationship to obtain $S(T,X)$. We can thus express the internal energy as a function of $T$ and $X$, i.e., $U(T,X) \equiv U(S(T,X),X)$. However, we have lost our variational principle: equilibrium no longer corresponds to a zero derivative! Indeed, by the chain rule,
$$ \left. \frac{\mathrm dU(T,X)}{\mathrm dX} \right|_{T} = \underbrace{ \left. \frac{\mathrm dU(S,X)}{\mathrm dS} \right|_{X}}_{= \; T} \left. \frac{\mathrm dS(T,X)}{\mathrm dX} \right|_{T} +  \boxed{ \left. \frac{\mathrm dU(S,X)}{\mathrm dX} \right|_{S}  }.  $$
Note that the boxed term is the one that is zero at equilibrium. Hence, when expressing the internal energy as a function of $T$ instead of $S$, our equilibrium condition has become
$$ \boxed{ \left. \frac{\mathrm dU(T,X)}{\mathrm dX} \right|_{T} = T \left. \frac{\mathrm dS(T,X)}{\mathrm dX} \right|_{T} } \quad \textrm{(for the equilibrium value $X= X_\textrm{eq}$).} $$
This is clearly annoying: it would be much nicer to have a single function which is extremized at equilibrium, rather than having to match the gradients of different functions! However, since the derivatives are with respect to fixed $T$, we can move it inside the derivative. Hence, moving everything to one side, we have
$$ \boxed{ \left. \frac{\mathrm d \left( U(T,X) - T \;S(T,X) \right)}{\mathrm dX} \right|_{T} = 0 } \quad \textrm{(for the equilibrium value $X= X_\textrm{eq}$).} $$
Defining $F(T,X) \equiv U(T,X) - T \; S(T,X)$ (which you might recognize as the (negative) Legendre transform!), we thus see that this is the appropriate quantity to variationally extremize when keeping temperature fixed! More generally, the above actually shows that
$$ \boxed{ \left. \frac{\mathrm dF(T,X)}{\mathrm dX} \right|_{T} = \left. \frac{\mathrm dU(S,X)}{\mathrm dX} \right|_{S} }, $$
for any value of $X$. If we enforce this as our physical requirement, then the above can be read as a proof that this uniquely gives the above $F(T,X)$ (called the Helmholtz free energy) up to an arbitrary function $f(T)$ (which is independent of $X$). To kill this remaining arbitrariness, we can in addition note that if we go for the aforementioned minimal choice of $F(T,X)$, then we have the pleasant property that (exercise)
$$ S(T,X) \equiv -\left. \frac{\mathrm dF(T,X)}{\mathrm dT} \right|_{X}.  $$
In other words: just as the derivative of the internal energy w.r.t. entropy gives us temperature, the derivative of the free energy w.r.t. temperature gives us the entropy (note that requiring this property forces our above unknown function to be constant: $f(T) = \textrm{cst}$). The significance of this is that it makes the whole process an involution! I.e., if we use $F(T,X)$ as our original starting point and feed it in at the top, we will end up with $U(S,X)$ as the result (possibly up to an irrelevant constant). In particular, this means that $F(T,X)$ and $U(S,X)$ carry the same information!
In summary, when the derivative w.r.t. the original variable is taking as the new variable, the Legendre transform naturally arises as the (virtually) unique function satisfying the constraints that (1) the derivatives w.r.t. other variables agree and that (2) the derivative of the new function w.r.t. the new variable gives back the original variable!


TITLE: Convexity of the ratio of the standard normal PDF by its CDF
QUESTION [1 upvotes]: Is there some way to show that the following function $\psi$ is concave or convex? Here, $\phi$ and $\mathbf{\Phi}$ are the standard normal PDF and CDF, respectively.
$$\psi\left(u\right)=u+\frac{\phi\left(u\right)}{\Phi\left(u\right)}$$
There is of course the calculus way, but the resulting second derivative is fairly complex. Hence, please advice if there is any alternate way or some simplification using the derivative approach itself that can establish this.

REPLY [0 votes]: This graph of the various functions should help demonstrate the convexity visually. Hope this helps. Thanks for all the pointers and useful suggestions.


TITLE: What obstacles prevent three-valued logic from being used as a modal logic?
QUESTION [6 upvotes]: I am familiar with many of the surveys of many valued logic referenced in the SEP article on many valued logic, such as Ackermann, Rescher, Rosser and Turquette, Bolc and Borowic, and Malinowski. It is asserted in the article that "Many-valued logic as a separate subject was created by the Polish logician and philosopher Łukasiewicz (1920), and developed first in Poland. His first intention was to use a third, additional truth value for “possible”, and to model in this way the modalities “it is necessary that” and “it is possible that”. This intended application to modal logic did not materialize. " My question is, why didn't it?
Edited: to move summarized known objections to an answer.

REPLY [0 votes]: I have identified a few obstacles.

The prefix notation developed and used by Łukasiewicz is far less popular than the infix type of notation used in conventional and boolean algebra and by most logicians, where the "box" and "diamond" notation used by C.I. Lewis for modal logic is typically not used in investigation of  Łukasiewicz logic.

This objection is one of familiarity.

The 3-valued Łukasiewicz logic can be used to distinguish between necessary and impossible propositions on the one hand, and contingent, doubtful, or equivocal propositions on the other, but does not distinguish between contingently true and contingently false.

It would need require at least a four valued logic to make this
distinction, but it is not so clear why this is essential to a first
step. There is some reason to suspect that it would be useful to
clarify the principles of the the three valued case first.

The definitions of "possibility" and "necessity" in Łukasiewicz  differ from those conventionally used in modal logic.
In the usual semantics of modal logic, the formula:

A: L(p v q) & M~p & M~q
is desirable. This would assert that the disjunction of two statments may be necessarily true, while the individual statements themselves are not necessarily true.  As an example, this might be interpreted as a claim that it is necessarily the case that Wanda is a fish or Wanda is a bicycle, but it is possible that Wanda is not a fish and Wanda is not a Bicycle. But in the Łukasiewicz semantics,  no assignment gives this formula the value t.
Similarly, the formula
B: (Mp & Mq) & L~(p&q)
is also thought sometimes desirable. This would claim that "It is possible that Wanda is a fish and Wanda is a bicycle, but it is necessary not the case that Wanda is both fish and bicycle".

This objection may be partly addressed by noting that if p and q are
mutually exclusive so that q = ~p, this amounts to  L(p v ~p) on the
left side hand of expression A, and M~p & Mp on the right.  Likewise,
we have Mp & M~p on the left of expression B and L~(p&~p) on the other
right.  The Lukasewicz-consistent semantics would have it that that
this is equivalent to asserting the law of the excluded middle on the
one hand and denying it on the other, and furthermore, that this is a
contradiction.
This objection has to do with a conflict between the dictates of the
truth functional de Morgan algebra of the logic, and the requirements
of conventionalized intution. It also has to do with a strong tendency
to adhere to the law of the excluded middle, in spite of the numerous
and venerable challenges to it. Although the conventional
interpretations of modal logicians are long-standing and even well
formalized, it would not be the first time that the algebra has been
more consistent than  intuition. It might be useful to compare and
contrast intuitive objections in more detail.

Another objection, raised by C.I. Lewis, is that in  Łukasiewicz logic, the conditional Cpq does not support the most common and valuable rules of inference such as modus ponens and transitivity of the conditional.

This objection can be addressed by noting that this is because Cpq
allows doubtful conditionals (those that have the value of u), and
that these should should not be allowed in a system of valid
deduction.  Łukasiewicz himself overlooked this, and so did his
critics. Adopting a definite or strict conditional LCpq is a simple
definition that more than repairs the deficiency. It is apparently
much too simple to be believed or investigated.

It has been noted that that there is a proof that none of the Lewis systems of Modal logic can be reduced to a system of three valued logic: Dugundji, James. Note on a Property of Matrices for Lewis and Langford's Calculi of Propositions. Journal of Symbolic Logic 5 (1940), no. 4, 150--151. jstor, doi: 10.2307/2268175

From "many valued logic" by Reshner (1969). (Gregg Revivals page 192)  "There exist no finitely many valued logic that is characteristic of any of the Lewis systems S1 to S5 , because any finitly many-valued logic will contain tautologies that are not theorems of S5 (and fortiory not of S1 to S4 either)

This objection is specious. The system of modal logic based on
Łukasiewicz logic is not one of the Lewis systems. It is truth
functional, which the Lewis systems are not, it rejects the law of the
excluded middle as a universally truth, while the Lewis systems accept
it, and it uses a different conditional than the Lewis systems.   None
of the Lewis systems reduce to it, it is not "characteristic of any of
the lewis systems S1-S5", and it does contain tautologies that are not
theorems of S5.


text:
\section{Conclusions}
\label{sect:conclusion}

We presented a novel approach to extract a  probabilistic dynamic description of hazardous thunderstorm regions from state-of-the-art nowcast data. We integrated the stochastic thunderstorm model in a stochastic, optimal trajectory planning tool, which maximizes the probability of reaching a specified goal set while avoiding the hazardous thunderstorm regions. Simulations of different scenarios using historic nowcast files have proven the obstacle avoidance quality of the generated trajectories, as well as the accuracy of the computed safety levels. This information  could be presented to both pilots and controllers to aid them in decision making processes.
The computational performance is nevertheless not satisfactory for online implementation. The main reason is the curse of dimensionality inevitably resulting from the use of the dynamic programming approach in a 4D spatiotemporal space. Hence, the developed algorithm  could be used in pre-tactical scenarios and as a benchmark tool to compare trajectories generated by other optimization approaches with the safety-optimal trajectories. To improve runtime, future work should explore alternative, more efficient numerical optimization methods such as nonlinear programming or chance-constrained optimization, and parallelized GPU computing. More efficient computational performance would also enable to explore control frameworks such as receding horizon control. This would enable replanning of the trajectory as new forecasts come in. Another potential field of future research is the incorporation of other adverse meteorological phenomena, e.g., wind-shear, hail, or clean air turbulence, into the presented framework, further maximizing safety.

config: arxiv


TITLE: Elementary Number Theory: If $p$ and $a$ are natural numbers with prime $p$, and $p^4|a^3$, then $p^2|a$.
QUESTION [1 upvotes]: I'm reviewing the basic number theory from my undergrad program (so I can jump back into abstract algebra well enough), and I've come across this problem in the section regarding the Fundamental Theorem of Arithmetic. I'm sure the answer is sitting right in front of my face, but I can't figure out if I'm having an issue of not seeing the last step or going down the wrong path in the first place. Any help is appreciated; thanks!
Prove: Let $a,p\in\mathbb{N}$ with $p$ prime. If $p^4|a^3$, then $p^2|a$.
Proof (so far): Let $a,p\in\mathbb{N}$ be given, with $p$ prime, such that $p^4|a^3$. Then $p|a^3$, so by Euclid's Lemma, $p|a$, and thus $a=pn$ for some $n\in\mathbb{Z}$. Now, for some $k\in\mathbb{Z}$, $p^4k=a^3=ap^2n^2$, so $p^2\frac{k}{n^2}=a$. (Here's where I'm stuck; it would be great if I knew $n^2|k$)

REPLY [1 votes]: Another way
Let the highest exponent of prime $p$ that divides $a$ is $n$
So, the highest exponent of $p$ that divides $a^3$ will be $3n$
We need $3n\ge4\implies n\ge \dfrac43$
As $n$ is an integer, $n\ge2$


TITLE: $\iiint \frac{1}{x^2+y^2+(z-2)^2}dA$ where $A=\{x^2+y^2+z^2 \leq 1\}$ check my answer!
QUESTION [2 upvotes]: I would like someone to review my solution please, the original question is to calculate
$\iiint \frac{1}{x^2+y^2+(z-2)^2}dA$ where $A=\{x^2+y^2+z^2 \leq 1\}$
What I did:
First I changed variables to polar coordinates in order to simplify the boundaries:
$x=r\sin\theta cos\phi$, $y=r\sin\theta \sin\phi$, $z=r\cos\theta$, $0\leq r \leq 1$, $0\leq \theta \leq \pi$, $0 \leq \phi \leq 2\pi$ and the jacobian is $r^2\sin\theta$.
to get the following result:
$$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta d\phi$$
We can see that $\phi$ doesn't appear in any integral, so we can just multiply by $2\pi-0=2\pi$ and forget about it:
$$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta d\phi =2\pi \int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta$$
now I used substitution, $\alpha=-cos\theta$, $d\alpha =sin\theta d\theta$, $-1 \leq \alpha \leq 1$
$$2\pi \int_{0}^{\pi}\int_{0}^{1} \frac{r^2sin\theta}{r^2-2rcos\theta+4}drd\theta=2\pi \int_{-1}^{1}\int_{0}^{1}\frac{r^2sin\theta}{r^2+2r\alpha+4}dr\frac{d\alpha}{sin\theta}$$
cancel the sine to get:
$$2\pi \int_{-1}^{1}\int_{0}^{1}\frac{r^2}{r^2+2r\alpha+4}drd\alpha$$
this integral is very difficult (at least wasn't apparent to me) if we integrate by $r$. luckily, Fubini's theorem states that we can integrate by $\alpha$ first and the result won't change:
$$2\pi \int_{-1}^{1}\int_{0}^{1}\frac{r^2}{r^2+2r\alpha+4}drd\alpha=2\pi\int_{0}^{1}\int_{-1}^{1}\frac{r^2}{r^2+2r\alpha+4}d\alpha dr $$
The antiderivative of $$\frac{1}{r^2+2r\alpha+4}$$ with respect to alpha is $$\frac{\ln(r^2+2r\alpha+4)}{2r}$$
since $\alpha$ goes from $-1$ to $1$: $$\frac{\ln(r^2+2r+4)}{2r}-\frac{\ln(r^2-2r+4)}{2r} = \frac{\ln(r+2)^2-\ln(r-2)^2}{2r}=\frac{\ln(\frac{r+2}{r-2})}{r}$$
so:
$$2\pi\int_{0}^{1}\int_{-1}^{1}\frac{r^2}{r^2+2r\alpha+4}d\alpha dr=2\pi \int_{0}^{1}r\ln(\frac{r+2}{r-2})dr$$
Now I integrated by parts $\int uv'=uv-\int u'v$ where $u=\ln(\frac{r+2}{r-2})$, $u'=\frac{-4}{r^2-4}$, $v'=r$, $v=\frac{r^2}{2}$:
$$\int r\ln(\frac{r+2}{r-2})dr=\frac{r^2ln(\frac{r+2}{r-2})}{2}-\int\frac{-2r^2}{r^2-4}dr =\frac{r^2ln(\frac{r+2}{r-2})}{2}+2(r+\ln(\frac{r-2}{r+2})+2)$$
when $r=1$:
$$\frac{1^2ln(-3)}{2}+2(1+\ln(\frac{-1}{3})+2)=\frac{1}{2}\ln(-3)-2\ln(-3)+6=-\frac{3}{2}\ln(3)-\frac{3\pi i}{2}+6$$
when $r=0$:
$$0+2(0+\ln(-1)+2)=2\ln(-1)+4=2\ln(1)+2\pi i+4=4+2\pi i$$
Subtract the 2 results to get: $$-\frac{3}{2}\ln(3)-\frac{3\pi i}{2}+6-4-2\pi i=2-\frac{3}{2}\ln(3)-\frac{7\pi i}{2}$$
Multiply this result by $2\pi$ to reach the final answer which is $$4\pi-3\pi \ln(3)-7\pi^2i$$
Is this indeed the correct answer? Is there a better way of solving this? this seems like a very difficult way to do the exercise.

REPLY [3 votes]: \begin{align*}
  \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{\pi} \frac{r^2\, \sin\theta}{r^2-4\, r\, \cos\theta+4}\, d\theta \, dr\, d\phi &= 2\, \pi\, \int_{0}^{1} \, \frac{r}{4}\, \log(r^2-4\, r\, \cos\theta +4) \Big|_0^\pi \, dr\\
&= 2\, \pi\, \int_{0}^{1} \, \frac{r}{4}\, \left(\log{(r+2)^2}-\log{(r-2)^2}\right)\, dr\\
&=  2\, \pi\, \Big(\frac{r^2}{8} \, \log(r+2)^2 - \frac{r^2}{8} \, \log(r-2)^2 + r - \log(r + 2) + \log(r - 2) \Big) \Big\vert_0^1\\
&= \pi\, \left(2-\frac{3}{2}\log{3}\right) \approx 1.1060968643
\end{align*}


TITLE: Galilean, SE(3), Poincare groups - Central Extension
QUESTION [11 upvotes]: After having learnt that the Galilean (with its central extension) with an unitary operator
$$ U = \sum_{i=1}^3\Big(\delta\theta_iL_i + \delta x_iP_i + \delta\lambda_iG_i +dtH\Big) +  \delta\phi\mathbb{\hat1} = \sum_{i=1}^{10} \delta s_iK_i + \delta\phi\mathbb{\hat1}  $$
This makes sure the commutation relations hold good in the group (especially for the boosts). However, in the case of Poincare group, the commutators still hold without a central extension. Similarly, is the case with SE(3) (there are no central extensions involved).
My question is why is there a necessity for central extensions in the first case, but not later ??
PS: This answer is somewhat related to the question, but am not able to get to the bottom of this thing.

REPLY [16 votes]: There are 3 actions of the Galilean group on the free particle: On the configuration space, on the phase space and on the quantum state space (wave functions). The Galilean Lie algebra is faithfully realized on the configuration space by means of vector fields, but its lifted action on Poisson algebra of functions on the phase apace and on the wave functions (by means of differential operators) is the central extension of the Galilean algebra, known as the Bargmann algebra in which the commutator of boosts and momenta is proportional to the mass. The reasoning is given in the following arguments
1) The action on the configuration space: $Q = \{x^1, x^2, x^3, t\}$:
Here the translations and the boost operators act as vector fields and their commutator is zero:
Translation: $x^i \rightarrow x^i+c^i$, generating vector $P_i = \frac{\partial}{\partial x^i}$
Boost: $x^i \rightarrow x^i+v^i t$, generating vector $G_i = t \frac{\partial}{\partial x^i}$
This is a faithful action of the Galilean group: $[P_i, G_j] = 0$.
2) The lifted Galilean action to the phase space  $Q = \{x^1, x^2, x^3, p_1, p_2, p_3\}$
The meaning of lifting the action is to actually write the Lagrangian and finding the Noether charges of the above symmetry: The charges as  functions on the phase space will generate the centrally extended version of the group. An application of the Noether theorem, we obtain the following expressions of the Noether charges:
Translation: $P_i = p_i$
Boost: $ G_i = P_i t - m x^i$.
The canonical Poisson brackets at $t=0$ (because the phase space is the  space of initial data): $\{P_i, G_j\} = m \delta_{ij}$
The reason that the lifted action is a central extension lies in the fact that that the Poisson algebra of a manifold itself is a central extension of the space of Hamiltonian vector fields,
$$ 0\rightarrow \mathbb{R}\overset{i}{\rightarrow} C^{\infty}(M)\overset{X}{\rightarrow} \mathrm{Ham}(M)\rightarrow 0$$
Where the map $X$ generates a Hamiltonian vector field from a given
Hamiltonian:
$$X_H = \omega^{ij}\partial_{j}H$$
($\omega$ is the symplectic form. The exact sequence simply indicates that the Hamiltonian vector fields of constant functions are all zero).
Thus if the Lie algebra admits a nontrivial central extension, this  extension may materialize in the Poisson brackets (the result of a
Poisson bracket may be a constant function).
3) The reason that the action is also extended is that in quantum mechanics the wave functions are sections of a line bundle over the configuration manifold. A line bundle itself is a $\mathbb{C}$ bundle over the manifold:
$$ 0\rightarrow \mathbb{C}\overset{i}{\rightarrow} \mathcal{L}\overset{\pi}{\rightarrow} M\rightarrow 0$$
thus one would expect an extension in the lifted group action. Line bundles can acquire a nontrivial phases upon a given transformation. In the case of the boosts, the Schrödinger equation is not invariant under boosts unless the wave function transformation is of the form:
$$ \psi(x) \rightarrow  \psi'(x)  = e^{\frac{im}{\hbar}(vx+\frac{1}{2}v^2t)}\psi(x+vt)$$
The infinitesimal boost generators:
$$\hat{G}_i = im  x_i + \hbar t \frac{\partial}{\partial x_i}$$
Thus at $t=0$, we get: $[\hat{G}_i, \hat{P}_j] = -im \hbar\delta_{ij}$
Thus in summary, the Galilean group action on the free particle's configuration space is not extended, while the action on the phase space Poisson algebra and quantum line bundle is nontrivially central extended.
The classification of group actions on line bundles and central extensions may be
performed by means of  Lie group and Lie algebra cohomology.  A good reference
on this subject is the book by Azcárraga, and Izquierdo. This book contains a detailed treatment of the Galilean algebra cohomology. Also, there are two readable articles by van Holten: (first, second).
Group actions on line bundles (i.e. quantum mechanics)  is classified by the first Lie group cohomology group, while central extensions are classified by the second Lie  algebra cohomology group. The problem of finding central extensions to Lie algebras can be reduced to a manageable algebraic construction. One can form a BRST operator:
$$ Q = c^i  T_i + f_{ij}^k  c^i c^j b_k$$
Where $b$ abd $c$ are anticommuting conjugate variables: $\{b_i, c_j \} = \delta_{ij}$. $T_i$ are the Lie algebra generators.
It is not hard to verify that $Q^2 = 0$
If we can find a constant solution to the equation $Q \Phi = 0$ with $\Phi = \phi_{i j} c^i c^j$
which takes the following form in components, we have
$$ f_{[ij|}^k  \phi_{k|l]} = 0$$
(The brackets in the indices mean that the indices $i, j, l$ are anti-symmetrized. Then the following central extension closes:
$$ [\hat{T}_i, \hat{T}_j] = i f_{ij}^{k} \hat{T}_k + \phi_{ij}\mathbf{1}$$
The second Lie algebra cohomology group of the Poincaré group is vanishing, thus it does not have a nontrivial central extension. A hint for that can be found from the fact that the relativistic free particle action is invariant under Poincaré transformations. (However, this is not a full proof because it is for a specific realization). A general theorem in Lie algebra cohomology asserts that semisimple Lie algebras have a vanishing second cohomology group. Semidirect products of vector spaces and semisimple Lie algebras have also vanishing second cohomology provided that there are no invariant two forms on the vector space. This is the case of the Poincaré group. Of course, one can prove the special case of the Poincaré group by the BRST method described above.


TITLE: Prove that $|\sum\limits_{k=1}^{n} a_{k}| \ge |a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$.
QUESTION [1 upvotes]: I am abjectly disappointed that I could not prove this statement on my own.  I have tried it directly and by contradiction but hit a wall.  Here is the statement (again) and my proof (thus far):
Prove that $|\sum\limits_{k=1}^{n} a_{k}| \ge |a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$.
Proof:
Let $a_{1}, ..., a_{n}$ be real numbers. Consider $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|$. Then $|a_{1}| - \sum\limits_{k=2}^{n} |a_{k}| \leq ||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}||$ holds by a property of the Absolute Value function $\forall a_{1},...,a_{n} \in \mathbb{R}$. Now, by the Reverse Triangle Inequality, we have $||a_{1}| - \sum\limits_{k=2}^{n} |a_{k}|| \leq |a_{1} - \sum\limits_{k=2}^{n} a_{k}|$.
I then proceeded to show that $|a_{1} - \sum\limits_{k=2}^{n} a_{k}| \leq \sum\limits_{k=1}^{n} a_{k} \leq |\sum\limits_{k=1}^{n} a_{k}|$ by considering different cases for $a_{1}$ and $\sum\limits_{k=1}^{n} a_{k}$ (e.g. $a_{1}, \sum\limits_{k=1}^{n} a_{k} > 0$ as one case out of four), and was able to show one case to hold but am struggling with the others.  Is there a more simplified way to write this proof?
Any help would be greatly appreciated! This has hurt my brain for weeks now :(

REPLY [2 votes]: First observe $$\begin{align} -\left|\sum_{k = 2}^{n}a_k \right| &\le 0\\
\left|\sum_{k = 1}^{n}a_k\right| - |a_1| &\ge0 \end{align}$$
And the second inequality follows from triangle inequality


TITLE: show that the set is compact
QUESTION [0 upvotes]: Suppose $f$ is a reimann integrable function on [a,b]. Let V={$x\in[a,b]$ : $\int\limits_{x}^{b} f(t)dt$ is continuous}.Then show that V is compact
Clearly V is bounded.So we only have to prove that V is closed.For that I take a sequence $\{x_n\}$ in V which converges to x.How to prove that x is in V

REPLY [1 votes]: We can prove a stronger result, namely that $V=[a,b]$. I will do this without using sequences.
It was given that $f$ is Riemann integrable on $[a,b]$, so for all $x\in[a,b]$, $\int_{x}^{b}f(t)dt$ exists.
We will now prove that the function $F(x)=\int_{x}^{b}f(t)dt$ is continuous on $[a,b]$.
Given an arbitrary $c\in[a,b]$, notice that $F(c)-F(x)=\int_{c}^{b}f(t)dt-\int_{x}^{b}f(t)dt=\int_{c}^{x}f(t)dt$ holds for all $x\in[a,b]$. It was assumed that $f$ is Riemann integrable on $[a,b]$, so it must be bounded there, that is $m\leq f(t)\leq M$ for some $m,M\in\mathbb{R}$ and for all $t\in\mathbb{R}$. Basic integral inequalities then establish that
$$m(x-c)\leq \int_{c}^{x}f(t)dt\leq M(x-c)$$
so $m(x-c)\leq F(c)-F(x)\leq M(x-c)$. If $a<c<b$, then as $x\to c$, both $m(x-c)$ and $M(x-c)$ approach $0$, so the squeeze theorem gives
$$\lim_{x\to c}[F(c)-F(x)]=0$$
which implies that $\lim_{x\to c}F(x)=F(c)$. This shows that $F$ is continuous on $(a,b)$. If $c=a$ or $c=b$, a similar argument applies with right-hand and left-hand limits, respectively. It follows that $F(x)=\int_{x}^{b}f(t)dt$ is continuous on $[a,b]$. Thus, $V=[a,b]$.


TITLE: how to interpret the solution of a linear system?
QUESTION [1 upvotes]: i have the following reduced echelon form matrix
[(1,0,1,)(0,1,0)(0,0,0)] and the solutions are (2,1,0)
EDIT This should be the system of linear equations
$$
\left\{
\begin{matrix}
x&+&&&z&=&2\\
&&y&&&=&1\\
\end{matrix}
\right.
$$
now I am asked which one of the following options describes the solution of the system:
a) single point at (2,1,0)  b) there are no solutions  c) a line through (0,0,0) and (2,1,0) d) a line through (2,1,0) and (0,1,2)
I know that it has infinite many solutions, as as the number of unknowns#equations and it doesnt have any inconsistency , so a and b are not the right answer.
ill give z the value of t, so I  have (x,y,z)=(2-t, 1, t).... and from now on, in case I have done everything  well I thought that I could get the general equation of that line , and plug the values inside and see if it works...but not sure..the two equations of the line I got are: x+z-2=0 and y-1=0 and then I plug them in and I got d as a result, as it fulfills the equation....is that right???
then they ask me that suppose that z=t for all real numbers is taken as a free variable in that linear system, what statement is true?
a) x doesnt depend on t b) z does not depend on t, c)y doesnt depend on t  d0 all depend on t.   I think is c.
please correct me if I am wrong, and pretty newbie with that and I have an exam coming up on tuesday....:S

REPLY [2 votes]: I have taken the liberty of editing your post, writing the system of linear equations explicitly. Please check to see whether I did it right, according to your intentions.
I believe you are right on both counts, the set $(x,y,z)=(2-t, 1, t) = (2,1,0) + t (-1,0,1)$ describes indeed a line through $(2,1,0)$ and $(0,1,2)$. (The first point is obtained for $t = 0$, the second for $t = 2$). And $y = 1$ shows that $y$ does not depend on $t$.
For the first part, it is useful to remember that the set of solutions of a system of linear homogeneous equations will be a subspace of the appropriate vector space. Whereas if the system is non-homogeneous (as in your case) it will be an affine subspace, that is, a translate of a subspace. In your case it is a line. Since $(2,1,0)$ and $(0,1,2)$ are distinct solutions, it will be indeed the line through them.


TITLE: Showing that $S^2 \vee S^4$ is not homotopy equivalent to any closed 4-manifolds.
QUESTION [6 upvotes]: I am working on a problem that asks to show that $S^2 \vee S^4$ is not homotopy equivalent to any closed $4$-manifold. My only understanding of manifolds is from Hatcher's chapter on Cohomology (chapter 3) so all manifolds in question are general topological manifolds.
From what I was told, I should consider working with $F_2$ coefficients and using poincare duality. However, I do not see what to do.

REPLY [4 votes]: The key point is that Poincare duality is given by capping against the fundamental class.
You can write down the cohomology ring of S2∨S4 (wedge sum of spaces), where it should be clear that all products are zero, and conclude that when you cap the fundamental class against a generator for the cohomology in degree 2, you get zero, which is not what would happen under Poincare duality...
Assume for contradiction that Poincare duality holds for $S^2 \vee S^4$. In particular, we let $\phi$ generate the second cohomology, $[M]$ be the fundamental class (generator of top homology), and let $\sigma$ be a generator of second homology.
Assuming it has Poincare duality implies that $\phi \cap [M] = \sigma$ (we're using $F_2$ coefficients).
However, because the general formula $(\alpha \cap \phi, \psi) = (\alpha, \phi \cup \psi)$ holds ( $(,)$ denotes the pairing betwee homology and cohomology... see the Hatcher document below), we have $1 = (\sigma, \phi) = ([M] \cap \phi, \phi) = ([M], \phi \cup \phi) = ([M], 0) = 0$, which is a contradiction.
math.cornell.edu/~hatcher/AT/ATcapprod.pdf gives a helpful discussion


TITLE: Why is it that $ V_{+}(I) \cup V(J) $ = $ V_{+}(I \cap J), $ where $ I $ and $ J $ are homogeneous ideals of a graded ring $ S$?
QUESTION [0 upvotes]: Define $ V_{+}(I) = \lbrace \mathfrak{p} \in \text{Proj}(S) \;|\; \mathfrak{p} \supset I \rbrace. $
I can see that $ V_{+}(I) \cup V_{+}(J) \subset V_{+}(I \cap J) $ since if $ \mathfrak{p} \in V_{+}(I) \cup V_{+}(J), $ then $ \mathfrak{p} $ contains $ I $ or $ J $ and therefore contains $ I \cap J $ either way.
But I am struggling to show that $ V_{+}(I \cap J) \subset V_{+}(I) \cup V_{+}(J). $

REPLY [1 votes]: Let $\mathfrak{p} \in V_+(I \cap J).$ Suppose for a contradiction that $\mathfrak{p} \notin V_+(I)$ and $\mathfrak{p} \notin V_+(J).$ Then there exists $i \in I$ and there exists $j \in J$ for which we have $i \notin \mathfrak{p}$ and $j \notin \mathfrak{p}.$ However, we have $ij \in (I \cap J) \subseteq \mathfrak{p}.$ Since $\mathfrak{p}$ is prime, it follows $i \in \mathfrak{p}$ or $j \in \mathfrak{p}.$ This is a contradiction.


TITLE: How the quotients of central series forms a Lie algebra
QUESTION [1 upvotes]: Let $G$ be a group. Then the subgroups of the central series are $\gamma_1(G)=G$, $\gamma_2(G)=[\gamma_1(G),G],..., \gamma_n(G)=[\gamma_{n-1}(G),G]$. Then in the reference paper given below it is given that the direct sum $\bigoplus_{n\geq1}\frac{\gamma_n(G)}{\gamma_{n+1}(G)}$ forms a Lie algebra over $\mathbb{Z}$. (http://emis.impa.br/EMIS/journals/GMJ/vol9/v9n4-9.pdf).
My first question is that if the notion of Lie algebra is defined over some field, then how we can use $\mathbb{Z}$ (set of integers)?
Kindly explain how this direct sum can become a Lie algebra. In the reference paper above it is given that the commutator map $\gamma_n \times \gamma_m\rightarrow \gamma_{m+n}$ is given by $(x,y)\longmapsto [x,y]$ but I can't understand this mapping.
One more thing: it is given that there is a canonical map from $G_{ab}$ to $\bigoplus_{n\geq1}\frac{\gamma_n(G)}{\gamma_{n+1}(G)}$ inducing a surjective homomorphism from $L(G_{ab})$ to $\bigoplus_{n\geq1}\frac{\gamma_n(G)}{\gamma_{n+1}(G)}$, where $L(G_{ab})$ is the free Lie algebra over $G_{ab}$. How we can define this when $\gamma_2$ is present in the first summand of the direct sum?
I know these are very silly questions but I tried a lot to understand this but was not able to understand. Please help me out to know about this as I am a newcomer to the field of Lie algebras.

REPLY [2 votes]: Re first question: Although very often Lie algebras are defined over fields, one can just extend the definition to any (e.g.) commutative ring $R$:

A Lie algebra over $R$ is an $R$-module $L$ with a Lie bracket $[\cdot, \cdot]: L \times L \rightarrow L$ which is bilinear, alternating ($[x,x]=0$ for all $x\in L$) and satisfies  the Jacobi identity.

E.g. Bourbaki only restricts to $R$ being a field after paragraph 3 in the first chapter of their Lie Groups and Lie Algebras book. Also, Lie algebras over the $p$-adic integers $\mathbb Z_p$ are quite common in the theory of $p$-adic analytic groups (Lazard etc.). As mentioned in a comment, sometimes this more general object is called a "Lie ring". Of course one has to be careful about what "standard" results still hold true for this more general definition.
Re second question: First of all, I take it that for elements $x,y \in G$, the (group) commutator $[x,y]$ is defined as $xyx^{-1}y^{-1}$ (sometimes one defines it as $x^{-1}y^{-1}xy$; insignificant changes apply).
Now let's check the above definition.
A. $\bigoplus \gamma_n(G)/\gamma_{n+1}(G)$ is a $\mathbb Z$-module. Clear, because every summand, being an abelian group, is a $\mathbb Z$-module (write the group operation additively as $\bar x +\bar y := \overline{xy}$).
B. Defining the Lie bracket: First, let's define the Lie bracket on a pair of homogeneous elements (i.e. elements which are contained in one of the summands). This is done as you say: if $\bar x \in \gamma_m(G)/\gamma_{m+1}(G), \bar y \in \gamma_n(G)/\gamma_{n+1}(G)$, then we define the Lie bracket $[x,y]$ as the residue class, in $\gamma_{m+n}(G)/\gamma_{m+n+1}(G)$, of the group commutator $[x,y]$. One checks that this is well-defined, $\mathbb Z$-bilinear, alternating, and satisfies the Jacobi identity.
Now, since every element of the direct sum is a finite sum of homogeneous elements, there is only one way to extend the definition bilinearly to all elements, namely
$$[\sum \bar x_i, \sum \bar y_j] = \sum_{i,j} [\bar x_i, \bar y_j]$$
($\bar x_i \in \gamma_i(G)/\gamma_{i+1}(G)$ etc.) and this is easily checked to still be alternating and satisfy the Jacobi identity.
Re third question: The abelianisation $G_{ab}:= G/[G,G]$ literally is $\gamma_1(G)/\gamma_2(G)$, the first summand in the direct sum, so the "canonical map" is indeed just the inclusion of that first summand into the entire sum. That this induces (or: "lifts to") a homomorphism from the free Lie algebra $L(G_{ab})$ to the sum is part of the universal property of a free object (or whatever definition of free Lie algebra you use, it should follow immediately from that). That it is surjective comes from the fact that additively, the direct sum is generated by all iterated group commutators $[g_1,[g_2,[g_3, .... ]]]$, and these are in the image of the free Lie algebra $L(G_{ab})$. Notice that for example there are generators $h$ of $\gamma_{2}(G)/\gamma_3(G)$ for which there are $x,y \in G$ such that $[x,y] + \gamma_3(G) = h$, and if $x_{ab}, y_{ab} \in G_{ab}$ are the respective residue classes in $G_{ab}$, then in $L(G_{ab})$ there is the element $[x_{ab},y_{ab}]_{L(G_{ab})}$ which via the univeral homomorphism gets mapped to $h$.


TITLE: Categorical Kähler differentials and the Leibniz rule
QUESTION [4 upvotes]: From nlab, the module of Kähler differentials over some category $\mathcal{C}$ is the free functor:
$$\Omega: \mathcal{C} \to \mathsf{Mod_{\mathcal{C}}}$$
left-adjoint to the (forgetful) embedding:
$$u: \mathsf{Mod}_{\mathcal{C}} \cong \mathsf{Ab}(\mathcal{C}) \hookrightarrow \mathcal{C}$$
with $\mathsf{Ab}(\mathcal{C})$ denoting abelian group objects of $\mathcal{C}$, and $\mathsf{Mod}_{\mathcal{C}}$ denoting the category of modules of $\mathcal{C}$. By this definition, we automatically get a bijection of hom-sets:
$$\mathsf{Mod_{\mathcal{C}}}(\Omega(R), M) \cong \mathcal{C}(R, u(M))$$
Also, according to the same nlab page as above, $R$-derivations taking values in some $R$-module $M$ (with $R$ some object of $\mathcal{C}$) are morphisms:
$$d: \Omega(R) \to M$$
in $\mathsf{Mod}_R \cong \mathsf{Mod}_{R/\mathcal{C}}$. Thus, they could be identified by $\mathcal{C}$-morphisms:
$$X: R \to u(M)$$
because of the adjunction $\Omega \dashv u$.
At this point, I have two questions:

When $\mathcal{C} = \mathsf{CRing}$, the category of commutative and unital rings, do we automatically get the Leibniz/product rule ? Why or why not ?
If we do automatically get the Leibniz rule, then is it also the case in categories more general than $\mathsf{CRing}$ ?

Thank you.

REPLY [2 votes]: 1. The Leibniz rule follows immediately from the last description
of derivations as morphisms of commutative rings X:R→u(M).
Indeed, u(M) is the square-zero extension of some R-module M'
(in the traditional sense), i.e., u(M)=R⊕M'.
Now a morphism of commutative rings f:R→R⊕M' in the slice category C/R
(not in C, as is claimed in the main post) necessarily has the form
r↦(r,φ(r)), for some linear map φ.
Since f is a homomorphism, we have f(1)=(1,φ(1))=(1,0), so φ(1)=0.
Also $$(rr',φ(rm'+r'm))=f(rr',rm'+r'm)=f((r,m)(r',m'))=f(r,m)f(r',m')=(r,φ(m))(r',φ(m'))=(rr',rφ(m')+r'φ(m)),$$
so $$φ(rm'+r'm)=rφ(m')+r'φ(m).$$
This is precisely the Leibniz rule.
2. Yes, for instance, this is true for algebras over Fermat theories.
See Carchedi and Roytenberg's Homological Algebra for Superalgebras of Differentiable Functions.


TITLE: Conditional Statements: "only if"
QUESTION [23 upvotes]: For some reason, be it some bad habit or something else, I can not understand why the statement "p only if q" would translate into p implies q. For instance, I have the statement "Samir will attend the party only if Kanti will be there." The way I interpret this is, "It is true that Samir will attend the party only if it is true that Kanti will be at the party;" which, in my mind, becomes "If Kanti will be at the party, then Samir will be there."
Can someone convince me of the right way?
EDIT:
I have read them carefully, and probably have done so for over a year. I understand what sufficient conditions and necessary conditions are. I understand the conditional relationship in almost all of its forms, except the form "q only if p." What I do not understand is, why is p the necessary condition and q the sufficient condition. I am not asking, what are the sufficient and necessary conditions, rather, I am asking why.

REPLY [5 votes]: The mathematician R.L. Moore, who was very careful with his language, interpreted "only if" to mean "if and only if".  In his mind, "A only if B" was a stronger statement than "A if B".  In other words, "A only if B" tells us that "A if B", but also gives us a little extra information: "A only if B".  He would have insisted that any other interpretation of "only if" is inconsistent with the standard use of the English language.
(However, Moore did frequently say "if and only if" for the sake of clarity.)
Therefore, if you think that "only if" should mean "if and only if", you are not alone.  But, mathematicians have established the convention that "A only if B" means "if A then B", and so now we must follow this convention.


TITLE: Surface integral of a sphere inside a cylinder
QUESTION [1 upvotes]: Find the surface area of the portion of the origin-centred sphere of radius $R=4$ that lies inside the cylinder $x^2 +y^2=12$ and above the $xy$ plane.

Does this question make sense? How can surface area lie inside the cylinder given that the radius of sphere is greater than radius of cylinder?

REPLY [1 votes]: When the sphere is cut by the cylinder you have two spherical caps remaining. The figure below shows the 2D representation.
Now, the basics are
$$
\theta=\sin^{-1}\frac{r}{R}\\
d=R\cos\theta\\
h=R-d\\
$$
The surface area of each spherical cap is
$$S=2\pi Rh$$
For your case I find that the surface area above the $x-y$ plane is $S=16\pi$.
I have verified this calculation numerically by calculating the surface area of of surface of revolution (of the red line).


TITLE: Isotropy of Apollonian disk-packing
QUESTION [6 upvotes]: Is there any sense in which the "epsilon-tail" of an Apollonian disk-packing (by which I mean the union of the disks of radius less than epsilon) exhibits more and more statistical isotropy as epsilon goes to zero?
Here's one example of the kind of thing I mean. Label the disks $D_1,D_2,\dots$ in order of weakly decreasing radius. It seems likely that for all $n$ in a set of density 1, there is a unique largest disk $D'_n$ tangent to disk $D_n$, and the ray from the center of $D_n$ to the center of $D'_n$ determines a unit vector $v_n$.  Now we can ask whether the end-points of the unit vectors $v_1,v_2,\dots$ are uniformly distributed on the circle.
I'm interested in finding out what's known about questions like this.

REPLY [3 votes]: Hee Oh has replied to my question and given me permission to post her reply here.  What follows are her words (so "I" means "Hee") with my reformatting.

What I proved with Shah (Inventiones, 2012) says that for a given Apollonian circle packing $P$, considered as a countable union of circles in the plane, "small circles in $P$ are uniformly distributed with respect to the $\alpha$-dimensional Hausdorff measure of the residual set of $P$, which is the closure of $P$", where $\alpha=1.305...$. For any region $E$ with piecewise smooth boundary the number of circles intersecting $E$ of radius at least $t$ is asymptotic to $c t^{-\alpha} H_{\alpha}(E)$.
Chapter 8 of the following article can be useful: http://gauss.math.yale.edu/~ho2/newMSRI_Oh.pdf
So it seems that your question amounts to asking how the $\alpha$-dimensional Haudorff measure of the residual set of $P$ (${\rm Res}(P$)) behaves. Since $\alpha$ is bigger than 1, a theorem of Marstrand says that a typical point $x$ in ${\rm Res}(P$) -- typical in the sense of the $\alpha$-dimensional Hausdorff measure -- is a condensation point for ${\rm Res}(P$), meaning that for almost all directions $\theta$, $x$ is a limit point of the intersection of $(x, \theta)$ and Res($P$) where $(x, \theta)$ is the ray from $x$ in the direction of $\theta$.
I am referring to Theorem 6.1 of the following paper http://gauss.math.yale.edu/~ho2/BR_JAMS_Final.pdf where we copied a theorem of Marstrand in 1954.
So, you do see circles in almost all directions in the above sense.

Link to Marstrand's paper: http://dx.doi.org/10.1112/plms/s3-4.1.257


TITLE: Applications of idempotent ultrafilters
QUESTION [12 upvotes]: Recently Justin Moore has posted a solution to the amenability of Thompson's group F. A key(?) step exploits the existence of idempotent ultrafilters on $\mathbb N$ to construct an idempotent measure on the free non-associative semigroup on one-generator.
Previously the main applications I knew of idempotent ultrafilters involved Ramsey theory, most specifically Hindman's theorem.

Question: What are other applications of idempotent ultrafilters?

I have made this a big-list CW question, although a big list would pleasantly surprise me.

REPLY [5 votes]: There are numerous applications of ultrafilters within ergodic theory/combinatorial number theory. Vitaly Bergelson's article gives a very nice review of these. Among the results presented there are the following:

Strengthened Hindman: In a finite partition, one can find a cell that both and additive IP-set and a multiplicative IP-set.
Partition regularity of $a + b = cd $:  In a finite partition, one can find a cell that contains $a,b,c,d$ with $a + b = cd$.
Integer approximation of polynomials: If $f:\ \mathbb{R} \to \mathbb{R}$ is a polynomial with $f(0) = 0$, then for fixed $\varepsilon > 0$, the distance from $f(n)$ to a closest integer is less than $\varepsilon$ for $IP^*$-many $n \in \mathbb{Z}$.
Combinatorial richness of return times: If $f:\ \mathbb{Z} \to \mathbb{Z}$ is a polynomial with $f(0) = 0$, and $(X,T,\mu)$ is an invertible measure preserving system, then for fixed $\varepsilon > 0$, it holds that $\mu(T^{-f(n)}A \cap A) > \mu(A)^2-\varepsilon$ for $IP^*$-many $n \in \mathbb{Z}$. Under Furstenberg correspondence, this turns into the statement that if $E \subset \mathbb{N}$ has positive density $d^*(E) > 0$, then $d^*(E \cap (E − f(n))) > d^∗(E)^2 − \varepsilon$ for $IP^*$-many $n \in \mathbb{Z}$.

In fact, the last two results can be substantially extended, for instance $f$ can be a generalised polynomial, if we take care of some additional assumptions.
A more sophisticated application can be found in another article by Bergelson and McCutcheon, where a version of Szemeredi's theorem is proved for generalised polynomials. They prove a number of amusing auxiliary results, such as characterisation of weak mixing involving ultrafilters, or some convergence results for ultrafilter convergence in weakly mixing systems. There are also nice results for multiple operators re-proved in an article by Schnell.
I read somewhere a slogan saying approximately that ultrafilters allow one to do ergodic theory without ergodic averages. (So, instead of $\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N T^n $, look at $p\!-\!\lim_{n } T^n $, where $p$ is an ultrafilter (normally: idempotent or even minimal)).
There is a MSc thesis by J.G. Zirnstein, which covers some more applications, including a variety of Ramsey theoretical results (which go far beyond Hindman, but include also van der Waerden and Jin theorems), and is written in a very nice and accessible tone.


TITLE: Exactly how does an oscillating electric field produce an oscillating magnetic field?
QUESTION [1 upvotes]: Let's say we have a capacitor which is connected to a sinusoidal voltage source, that means that the electric  field within the capacitor is a sinusoidal function(assuming that the capacitor is a parallel plate capacitor ),  but how does this model generate any magnetic field? If it does,  then doesn't that mean that light is generated within the capacitor?

REPLY [1 votes]: According to the Maxwell equation $$ curl{\vec H} = \vec j + \epsilon \frac{∂\vec E}{∂t}$$ An oscillating electric $\vec E∝ exp{(i\omega t)}$ field creates a displacement current density $j_D=\epsilon \frac{∂\vec E}{∂t}=i\omega\vec E$ that is equivalent to a current in creating a magnetic field that encircles the sum of convection and displacement current. Therefore, in a capacitor with an oscillating electric field producing a displacement current $j_D$, a magnetic field encircles the displacement current in the capacitor just like the corresponding conduction current in the wire leads to the capacitors. In a normal capacitor it is practically impossible to have field oscillations in the light frequency range. Therefore you will have no light generation. But you could have radio frequency wave generation. This is, e.g., the case in an oscillating dipole antenna which is similar to an open capacitance-inductance electric oscillator.