TITLE: Calculating the surface area of sphere above a plane
QUESTION [2 upvotes]: How do I calculate the surface area of the unit sphere above the plane $z=\frac12$?
EDIT: I have been attempting things and I am thinking about parameterizing this... While I know that surface area is given by the double integral of the cross products of partial derivatives of the new parameters, I don't know what to set them to.. (sorry I'm not good with the fancy notation)

REPLY [2 votes]: Surface area is given by
$$
\iint_R \left| \vec r_u \times \vec r_v \right| \ dA
$$
where $\vec r(u,v)$ is the parametrization of the surface. We can rewrite this as (derivation shown here: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx):
$$
\iint_D \sqrt{ \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 + 1} \ dA
$$
for a function $z = f(x,y)$ where $D$ is the projection of the surface onto the xy-plane.
Since we are only concerned with the portion of the unit sphere above $z = 0$, we can write it as
$$
z = \sqrt{1-x^2-y^2}
$$
Computing the partial derivatives with respect to $x$ and $y$,
$$
\frac{\partial z}{\partial x} = \frac{-x}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial x}\right)^2 = \frac{x^2}{1-x^2-y^2}
$$
$$
\frac{\partial z}{\partial y} = \frac{-y}{\sqrt{1-x^2-y^2}} \rightarrow \left(\frac{\partial z}{\partial y}\right)^2 = \frac{y^2}{1-x^2-y^2}
$$
Substituting these into our expression for surface area,
$$
\iint_D \sqrt{ \frac{x^2}{1-x^2-y^2} + \frac{y^2}{1-x^2-y^2} + 1} \ dA
$$
which simplifies to (omitting a bit of algebra)
$$
\iint_D \frac{1}{\sqrt{1-x^2-y^2}} \ dA
$$
Observe that $D$ (the projection of our surface into the xy-plane) is given by
$$
z = \sqrt{1-x^2-y^2}
$$
$$
\frac{1}{2} = \sqrt{1-x^2-y^2}
$$
$$
\frac{1}{4} = 1-x^2-y^2
$$
$$
x^2+y^2 = \frac{3}{4}
$$
which is a circle of radius $\frac{\sqrt{3}}{2}$. The integral over $D$ is easiest done in polar coordinates. I'll assume you know how to do that and omit the computation.
$$
\int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \frac{1}{\sqrt{1-r^2}} \ r \ dr \ d\theta
$$
$$
= \pi
$$


TITLE: Compactly supported sections of coherent sheaves and the dualizing complex
QUESTION [8 upvotes]: Suppose $U$ is a (possibly singular) scheme and $X$ is a compactification (potentially unnecessary at least in characteristic $0$). Let $\pi:X\to *$ be the map to the point (though one can consider more general maps as well). There is a classically known pro-algebraic "compactly supported global sections" functor defined by Deligne in the appendix to Hartshorne's "Residues and Duality". Namely, one has a functor $$\pi_!: \mathrm{Coh}_U\to D^b \mathrm{ProFinVect},$$ given informally by taking a sheaf $F$ to the fiber of the map $$\Gamma(U, F)\to \Gamma(\mathring{\delta}_X, F),$$
where $\mathring{\delta}_X: = \widehat{X\setminus U}\cap U$ is the punctured formal boundary. (More explicitly, one defines a functor on coherent sheaves on $X$ and applies it to any continuation of $F$, or equivalently, to $j_*F$ as an ind-object. It's easy to see that this can be done fully inside the $\infty$-category of complexes of pro-coherent sheaves on $X$.)
This functor is described in the condensed language in Lecture 11 of Scholze and Clausen's Lectures, though here my understanding is limited (and the lecture works under a smoothness condition, which is surely unnecessary in defining $\pi_!$).
Since $\pi_!$ is valued in complexes of pro-vector spaces, one can define the dual contravariant functor $(\pi_!)^*:D^b Coh(U)\to D^b Vect$ (as a functor of $\infty$-categories). As $U$ varies, this forms a presheaf of "distributions on $F$" (which I think is a sheaf in general) which, for $X$ smooth, agrees with the Serre dualizing sheaf. Write $S_X$ for the sheaf of distributions on the constant sheaf, i.e. (the sheafification of) $$S_X:U\mapsto (\pi_!^U)^*(\mathcal{O}_X).$$
I have three questions about this construction.

What is the name of the resulting complex for a general proper (and arbitrarily singular) $X$? I want to call it the dualizing complex, but have only seen that defined under some homological singularity restrictions on $X$ like the Gorenstein property.
Is there a way to interpret this construction in the condensed language (e.g., is this what would be called $\pi^!\mathcal{O}$ in the Clausen-Scholze lectures? Is it clear that their construction works in the non-smooth context?)
The construction in the Deligne appendix only defines $\pi_! F$ as a complex of pro-vector spaces, not pro-finite vector spaces. It is very easy to make it pro-finite-dimensional (and thus with an ind-finite dimensional dual) by rewriting his construction $\infty$-categorically, but I have never seen this done. Is there a reference to this?

REPLY [4 votes]: Isn't the dualizing complex defined in general in the proper case by taking applying the right adjoint of $\pi_\ast$ to $k$? That's what I'll take as the definition anyways. The Gorenstein property just means that the dualizing complex is invertible.
It is true that this can be computed by the formula you write down, and one way to see this is as in our lecture notes. (We do define compactly supported cohomology without the smoothness assumption.) So yes, this can be interpreted in our setting.
I know only few references on those compactly supported cohomology groups, and I'm not aware of any that work $\infty$-categorically.
And as you guessed, you actually don't have to sheafify this construction, it's already a sheaf.


TITLE: Why is the spectrometer so big in KATRIN experiment?
QUESTION [1 upvotes]: I've been studying KATRIN experiment for a popularizing science project and I was wondering, why is the spectrometer so big ? I mean the thing is so huge that they had to carry it over 8000km on the sea, it is 24meters long and had a diameter of 10m. But in all the publications related I didn't find any hint, only that the precision of the filter is given by:
$$\frac{\Delta E}{E} = \frac{B_{\text{min}}}{B_{\text{max}}}$$
which involves only magnitude of magnetic field. They also want the magnetic field to vary adiabatically which involves:
$$\nabla B/ B << 1$$
That can explain the length, but why this ovoid shape ?
I'm not an expert of this field if someone can give a clear and precise answer I'll be very grateful !
Thank you,

REPLY [1 votes]: I'm assumming you obtained your information from the page on the katrin.kit.edu website where the measurement principle of a MAC-E-filter is discussed
The objective is to obtain the highest resolution that current technology can obtain.
The electrons entering the main filter have already undergone multiple stages of pre-filtering, second to last in a smaller (but still very large) MAC-E-filter, so the electrons entering the main filter already have an energy very close to each other.
I don't see explicit mention anywhere of why the main chamber is so large.
I assume that if the "main" chamber would be of similar size to the chamber of the pre-filter the "main" chamber wouldn't add all that much.
I assume that the larger the chamber of the main filter the higher the resolution of the cut-off that the setup is designed to obtain.

From the 2004 KATRIN design report:
Section 2.1.1 (page 36) describes that it is also possible to build the setup in such a way that it also support a time-of-flight mode.
"By measuring the time-of-flight of the electrons those of interest can be distinguished fromthe high energy electrons and the MAC-E-Filter transforms from an energy high-passfilter into an energy band-pass filter of the same energy resolution and nearly the sameluminosity as the MAC-E-Filter without time-of-flight measurement"
When set up to operate in time-of-flight mode the separation is based on duration of the flight. The bigger the chamber, the longer the total duration of the flight, the more time available for separation.

As you know, the researchers are interested only in the electrons that carry away so much energy from the decay event that the neutrino formed in that event received only the energy that is the minimum it needs to be formed at all. All electrons that carry away less than that maximum need to be washed away. The electrons that even make it to the main chamber will have energies very, very close to each other. That of course makes separation very, very hard to obtain.


text:
\section{Reductio ad Absurdum/Variant 1}
Tags: Reductio ad Absurdum

\begin{theorem}
: $\neg p \implies \bot \vdash p$
\end{theorem}

config: wiki


TITLE: A question related to Pigeonhole Principle
QUESTION [22 upvotes]: In a room there are $10$ people, none of whom are older than $60$, but each of whom is at least $1$ year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same.

My approach:  There are $2^{10}=1024$ subsets, $1023$ non-empty subsets. Therefore there are $1023$ sums of ages and each sum is between $1$ and $600$. Then there are $600$ possible values, but $1023$ sums. Therefore at least two of them must be equal, i.e. there exist different subsets $\{P_{i1}, \ldots, P_{in}\}$ and $\{P_{j1}, \ldots, P_{jn}\}$ such that the sum of the ages agree. Now take out the people present in both subsets.

Can $10$ people be replaced by a smaller number?

I guess, it cannot. For example if there were to be $9$ people, then I would have $2^9-1 = 511$ proper subsets and since now I have $9\cdot 60=540$ possible totals, it is not guaranteed that there exists two disjoint groups of people such that the sum of whose ages are the same.
Am I right?

REPLY [0 votes]: In your solution, you have mentioned that the maximum sum is 60. But notice that if there were two people of the same age then we are done as we can easily choose the two persons as a different group. So it should be distinct, i.e., max sum= $\displaystyle\sum_{i=0}^9 (51+i)=\dfrac{10}{2}(51+60)=555$. But here comes the surprise, the minimum sum = $\displaystyle\sum_{i=0}^9 (1+i)=\dfrac{10}{2}(1+10)=55$, and the total number of sums using inclusion and exclusion principle we get, $555-55+1=501$. Here see that $2^9-1=511$, hence  @Ilmari Karonen did not encounter any problem with 9 number set with his brute force algorithm.
But we are not done here, notice that $2^8-1=255<501$. It simply implies that we can always find a subset such that no 2 groups of people will have the same age. Here we have reduced the number of possible sums, which is very important as no we get the reduced possibilities ignoring all that impossible cases. You can try to prove that 8- number set is not possible to obtain.


TITLE: What does $\int_0^1 x^n d\alpha(x)$ converge to?
QUESTION [2 upvotes]: Let $\alpha$ be a monotonically increasing function on $[0,1]$. What does the sequence of Riemann-Stieltjes integrals
\begin{equation*}
\int_0^1 x^n d\alpha(x)
\end{equation*}
converge to as $n \to \infty$?
I know that if $\alpha(x) = x$ then the sequence converges to $0$. More generally, if $\alpha(x)$ is continuous, then the sequence converges to zero. But what about for general monotone functions $\alpha(x)$?

REPLY [4 votes]: Write this integral (with $0 < \epsilon < 1$) as
$$\int_0^1 x^n \, d\alpha =\int_0^{1-\epsilon}x^n \, d\alpha + \int_{1-\epsilon}^{1}x^n \, d\alpha $$
Considering the first integral on the RHS, we have
$$0 \leqslant  \int_0^{1-\epsilon}x^n \, d\alpha \leqslant (1-\epsilon)^n[\alpha(1) - \alpha(0)] \\ \implies \lim_{n \to \infty}\int_0^{1-\epsilon}x^n \, d\alpha = 0 $$
Considering the second integral, we have for all $n > 1/\sqrt{\epsilon}$
$$\int_{1-\epsilon}^{1}x^n \, d\alpha \geqslant \int_{1-1/n^2}^{1}x^n \, d\alpha \geqslant \left(1 - \frac{1}{n^2} \right)^n[\alpha(1) - \alpha(1- 1/n^2)] \\ \implies \liminf_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha \geqslant \alpha(1) - \alpha(1-),$$
since $1 - 1/n \leqslant (1- 1/n^2)^n \leqslant 1$ using the Bernoulli inequality, and, hence, $(1-1/n^2)^n \to 1$.
We also have
$$\int_{1-\epsilon}^{1}x^n \, d\alpha \leqslant \alpha(1) - \alpha(1-\epsilon) \leqslant \alpha(1) - \alpha(1-)$$
Thus,
$$ \alpha(1) - \alpha(1-) \leqslant \liminf_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha  \leqslant \limsup_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha \leqslant  \alpha(1) - \alpha(1-),$$
and it follows that
$$  \lim_{n \to \infty}\int_{0}^{1}x^n \, d\alpha =\lim_{n \to \infty}\int_{1-\epsilon}^{1}x^n \, d\alpha  = \alpha(1) - \alpha(1-)$$