TITLE: Number formation and bridged graphs, connection or coincidence?
QUESTION [1 upvotes]: Bridged graphs sequence
$g(n) =$ "Number of simple connected bridged graphs on $n+2$ nodes".
We have $g(n)=1, 3, 10, 52, 351, 3714,\dots$ from A052446.

Number formation sequence
We also have $f(n) =$ "Largest $N$ such that all numbers $1,\dots, N$ can be made using basic operations $+,-,\times,\div$ and parentheses $(\space)$, given optimal set of $n$ digits (to be defined for each $n$) such that all digits are used exactly once."
The conjectured solutions are: $f(n)=1, 3, 10, 52, 351$ so far for $n=1,\dots,5$.
The first three values are true and not hard to prove. The second two values are given by:
$f(4)\ge 52$ since (optimal?) digits $d=\{2,3,4,22\}$ can make numbers $1,\dots,52$.
$f(5)\ge 351$ since (optimal?) digits $d=\{3,6,8,12,37\}$ can make numbers $1,\dots,351$, which can be seen by clicking here. Note that solutions per $N_0\in[1,N]$ are not necessarily unique.
I conjecture these two are equalities based on computations. That is, I couldn't find a better digit set that would make larger $N$ possible in either case.

Question
Notice that so far I have a conjecture pattern:
$$f(n)=g(n)$$

Is this a coincidence or is there a connection between these two problems?
That is, can we provide arguments for either case?

I don't see a clear connection and would think this is just a coincidence. Can we  prove this is just a coincidence? That is, find $n$ and $d$ such that $f(n)\gt g(n)$? Or prove for some $n$ that for every $d$ we have $f(n)\lt g(n)$?
Or is this not a coincidence? Can we provide arguments why could these two problems be connected? It would be amazing if we could translate number formation to graphs and find optimal $d$ for some $n$ with graph theory.
Can we perhaps find $d$ such that $f(6)=3714$, and continue this pattern? My simple brute force c++ code is too slow to find significant records for $n\ge 6$. The best I have by sampling random digit cases is $f(6)\ge 2200$, so far.

Lower bound on $f(n)$
It can be shown $f(n)\gt 2^n,n\gt 2$, using $d=\{2^0,2^1,\dots,2^{n-1}\}$.
But this is not the best lower bound for this digit set. We have:
$f(n)\ge f_d(n)=1,3,10,26,76,596,2472,\dots$.
Notice this is a weak lower bound and is not sharp. Example: $f(6)\ge2200\gt f_d(6)=596$.
Notice $f(n)=f_d(n)$ for $n\le 3$. Better digit sets might give better bounds for $n\ge 4$.

REPLY [4 votes]: Coincidence.
$f(n)$ is not larger than the number of expressions on $n$ numbers. The expressions are in bijection with $n$-leaf rooted binary trees, where its non-leaf vertices are labeled with $\{+,-,×,÷\}$ and its leaves are in bijection with the optimal set of digits. The number of such binary trees is $4^{n-1}\text{Catalan}(n-1)n!$, which is $O(\exp(c n\log n))$ for some constant $c$.
The binary tree $T$ corresponding to a expression $d$ is constructed as follows:
$d$ takes the form $d_1⊕d_2$, where $d_1$ and $d_2$ are expressions, and $⊕$ is a operator. Label the root of $T$ with $⊕$, construct two subtrees $T_1$ and $T_2$ corresponding to $d_1$ and $d_2$, respectively. Let $T_1$ and $T_2$ be the two children of $T$. This is a bijection.
The (labeled) binary tree can be determined by the following information:

The structure of the binary tree ($\text{Catalan}(n-1)$ types)
The labels ($4$ choices for each non-leaf node)
The leaves ($n!$ bijections from the optimal digit set)

So the number of trees is $4^{n-1}\text{Catalan}(n-1)n!$.
But the number of bridged graphs on $n$ vertices is at least the number of graphs on $n/2$ vertices: just connect two graphs with an edge. The number of graphs on $n$ vertices is roughly $2^{\frac{n^2}2}$, which is faster than $\exp(c n\log n)$, so the number of bridged graphs eventually dominates $f(n)$.


TITLE: What would prevent two black holes from hitting speed of light in the flowing scenario?
QUESTION [0 upvotes]: If two super massive black hole are 10 l.y away and completely still relative to one another... They will start moving towards each other due gravity but they will never reach each other due universal expansion, in this case they will accelerate for an infinite amount of time, what will prevent them from reaching the speed of light once they reach 99,999...%  light speed  (and still accelerating)?

REPLY [0 votes]: You are making a wrong assumption that the space expansion "pulls" matter apart with a force. It does not. When two heavy objects are attracted to each other with a gravitational force, it is the only force that acts on them. The space expansion does not apply a counteracting force. In other words, in the universe expanding without acceleration, the space expansion is manifested by the recession speed, but not by force or acceleration. Therefore, in your example, nothing would "pull apart" the black holes, other than their initial speed.
At a 10-ly distance, the recession speed due to the space expansion is approximately 2.4 cm/sec, not a significant amount. However, the gravitational acceleration of a black hole of 2 solar masses at this distance is 0.000,000,000,000,03 m/s^2. It would take approximately 25,000 years for the black holes to stop moving apart and start moving toward each other with no effect from the space expansion.
According to the recent data, our universe may be expanding with a small acceleration. In this case the scenario would be slightly different. However, the acceleration is so small, that it would not produce a significant effect. So the overall picture would remain the same.
You can find more details here: Why does space expansion not expand matter?


TITLE: 1D Time independent Schrodinger equation applied to a ring of Radius $R$
QUESTION [0 upvotes]: In the lectures we have been doing basic examples of Applying the TISE to determine the solutions of simple situations such as the finite and infinite square well and I understand how to find the solutions (eigenstates) in the form u(x).
However if we consider a free particle of mass m which is constrained to move along a circumference of a ring of radius R (keeping it all in 1D) how would I begin to solve the 1D time independent schrodinger equation if V(x) = 0 along the ring and periodic boundary conditions were applied.
I am confident that the solution would be of the form $u(x) = Aexp(±ikx)$ but other than that I am stuck.
Any help would be appreciated.

REPLY [1 votes]: Let's start with the Schrodinger equation in 2D with $V = 0$.
$$
H \psi = - {\hbar^2 \nabla^2 \psi \over 2m}
= - {\hbar^2 \over 2m} \left({1 \over r} {\partial \over \partial r} \left( r {\partial \psi \over \partial r}\right) + {1 \over r^2} {\partial^2 \psi \over \partial \theta^2}\right).
$$
Now if we constrain the particle to move on a circle of radius R, the radial derivative drop out and we're left with
$$
H \psi = - {\hbar^2 \over 2m R^2}  {\partial^2 \psi \over \partial \theta^2}.
$$
The eigenfunctions of this operator are $e^{il \theta}$, where we require $l$ to be an integer in order to satisfy the periodicity condition.
Defining $\psi_l = {1 \over 2 \pi} \int_0^{2 \pi} d\theta\; e^{-il\theta} \psi(\theta)$, applying the azimuthal Fourier transform to both sides, and integrating by parts twice on the left-hand-side yields
$$
E_l \psi_l = {\hbar^2 l^2 \over 2 m R^2}\psi_l ~~~\Rightarrow~~~ E_l = {\hbar^2 l^2 \over 2 m R^2}.
$$
To find the time-dependent wavefunction, we simply tack on the exponential time-dependence for each of the eigenfunctions,
$$
\psi(\theta,t) = \sum_{-\infty}^{\infty} \psi_l(0) e^{il\theta - i E_l t/ \hbar},
$$
where $\psi_l(0) = {1 \over 2 \pi} \int_0^{2 \pi} d\theta\; e^{-il\theta} \psi(\theta,0)$. To get the time-independent version, just set $t = 0$.


TITLE: Entropy of floating number array
QUESTION [1 upvotes]: I am familiar with shanon's definition of entropy.
$$
H(P) = - \sum_{i=1}^n p_i \cdot \log_2(\mathcal p_i)
$$
I am today in the situation that I'd like to compute an entropy like function but for a set of (value/probability) where the value is a floating number.
The object i'm considering are point by point ratios betwen two discretized density fields which are basicaly functions $f: [a,b] \mapsto \mathbb R^+$.
The fact is there are a lot a floating point values ... a huge lot ... and computing the probability for each floating point value just doesn't sound right
Is there any standard methode to compute entropy/energy for real fields ?

REPLY [1 votes]: The first part of your question is a little confusing. $p_i$ cannot be an integer except 0 or 1. It's a probability and thus it's restricted to $0\leq p_i\leq 1$. The probabilities it represents are discrete.
Regarding the other part of your question, the entropy is completely independent of the values that the function takes (the "alphabet" in information theory terminology).

The fact is there are a lot a floating point values ... a huge lot ... and computing the probability for each floating point value just doesn't sound right

I don't understand what you mean here, but if there are a finite number of values then you can use the discrete entropy. However, if the function is continuous (has an uncountable infinite number of values), then you need to invoke a continuous version of the entropy function.


TITLE: Can this special case of a Theorem by Bonsall be generalised slightly?
QUESTION [0 upvotes]: When choosing $\Delta := \mathbb N$, $M_1 = M_2 = 1$ in Theorem 1 in this paper by F.F. Bonsall, we obtain the following variant:
Theorem 1 (Variant).
Let $(X, \| \cdot \|)$ be a Banach space, $(u_k)_{k \in \mathbb N} \subset X$ such that $\| u_k \| \le 1$ for all $k \in \mathbb N$ and
$$
\sup_{k \in \mathbb N} | \psi(u_k) | = \| \psi \|_{X^*}
$$
for all $\psi \in X^*$ and for $f \in X$
$$
L(f, (u_k)_{k \in \mathbb N})
:= \left\{ \lambda = (\lambda_k)_{k \in \mathbb N} \in \ell^1(\mathbb N): f = \sum_{k \in \mathbb N} \lambda_k u_k \right\}.
$$

Then $L(f, u) \ne \emptyset$ for all $f \in X$ and $\| f \| = \inf_{\lambda \in L(f, u)} \| \lambda \|_{\ell^1(\mathbb N)}$.

My Question.
If I take a set $E \subset X$ such that $\| u \| \le 1$ for all $u \in E$ and $\sup_{u \in E} | \psi(u) | = \| \psi \|_{X^*}$ for all $\psi \in X^*$ instead of the sequence $(u_k)_{k \in \mathbb N} \subset X$ with those properties, can I still conclude that the set
$$
L(f, E) := \left\{ \lambda = (\lambda_k)_{k \in \mathbb N} \in \ell^1(\mathbb N): f = \sum_{k \in \mathbb N} \lambda_k u_k, \ (u_k)_{k \in \mathbb N} \subset E \right\}
$$
is nonempty for each $f \in X$ and that $\| f \| = \inf_{c \in L(f, E)} \| c \|_{\ell^1(\mathbb N)}$?
I hope this is true as one can write $E$ as a subset of the union of all sequences $u$ considered in the variant of theorem 1.
My try.
Nonemptyness:
By the properties of $E$, there is a sequence $(u_k)_{k \in \mathbb N} \subset E$ fulfilling the conditions in the variant of Theorem 1.
Let $f \in X$.
There there exists a $\tilde{\lambda} \in L(f, (u_k)_{k \in \mathbb N}) \subset L(f, E)$ by the variant of Theorem 1, so $L(f, E) \ne \emptyset$.
Is the norm condition true, as well?
Note.
I am aware that the result I seek is true, as it is proven in this later paper by the same author. But that proof is rather cumbersome and unintuitive, as it strictly avoids the use of Banach's closed range theorem (because later in the paper the author demonstrates an easy proof of the closed range theorem with the help of the above result). The proof in the first paper uses the closed range theorem and thus is much shorter.

REPLY [1 votes]: From
$$
|\psi(u)| \le \|u\|_X \, \|\psi\|_{X^*}
$$
you still have $\sup_{u \in E} |\psi(u)| = \|\psi\|_{X^*}$,
i.e. there is no difference in this assumption.
As you said, there is a sequence $(u_k) \subset E$ such that theorem 1 can be applied. Hence,
$$
\inf_{c \in L(f,E)} \|c\| \le \inf_{c \in L(f, (u_k))} \|c\| = \|f\|.$$
To the contrary,
for every $c \in L(f,E)$, there is a sequence $(u_k^{(c)}) \subset E$ (take the above sequence amended by the terms appearing in the statement of $c \in L(f,E)$), such that
it satisfies the assumptions of theorem 1 and
$c \in L(f,(u_k^{(c)}))$.
Hence,
$$\|c\| \ge \inf_{d \in L(f,(u_k^{(c)}))}\|d\| = \|f\|.$$
This shows the norm statement.


TITLE: Running backwards on a carousel
QUESTION [3 upvotes]: Let's suppose I'm on a carousel rotating with a speed $\omega$ and start running on it in the opposite direction with a speed $v$ such that $\omega r=-v$ where $r$ is the distance between me and the center of the carousel.
What are the forces that act on me? Do I feel Coriolis force?
Intuitively I'm "still" so no force should act on me, but also intuitively this explanation seems too easy to be the real one.

REPLY [2 votes]: As viewed in an inertial frame, no force acts on you.  This makes sense, because you are at rest in that frame.
As viewed in the rotating reference frame attached to the carousel, two forces are acting on you:  the Coriolis force and the centrifugal force.  The Coriolis force has magnitude $2 m \omega v = 2 m \omega^2 r$ inwards.  The centrifugal force has magnitude $m \omega^2 r$ outwards.  Thus, in the rotating reference frame there is a net force of $m \omega^2 r$ inwards.  This is completely in line with your motion in the rotating reference frame, since in that frame you are executing uniform circular motion about the origin with radius $r$ and angular frequency $\omega$.


TITLE: sum of generalized harmonic numbers
QUESTION [0 upvotes]: For $\alpha>0$, let
\begin{equation*}
H_m^{(\alpha)}=\sum_{k=1}^m\frac{1}{k^{\alpha}}
\end{equation*}
What is
$
\sum_{m=1}^n H_m^{(\alpha)}  ?
$
Does simply identity exist?

REPLY [5 votes]: It's $$\sum_{m=1}^{n} H_{m}^{(\alpha)}=\sum_{k=1}^{n}\frac{n-k+1}{k^\alpha} = (n+1)H_n^{(\alpha)}-H_{n}^{(\alpha-1)}$$


TITLE: Geometrically describe these Cosets and form a bijection with the Orbit-Stabilizer relation.
QUESTION [1 upvotes]: I am beginning to study abstract algebra/group theory and I have some seemingly simple practice questions here. I just want make sure I am understanding the concepts correctly. Here are the questions:

Let $H = \{  \begin{pmatrix}   c & d  \\   0 & 1  \end{pmatrix} |
\space c,d \in \mathbb{R} , c > 0   \} $ and $ J = \{  \begin{pmatrix}
a & 0  \\   0 & 1  \end{pmatrix} | \space a \in \mathbb{R} , a > 0  \} $
Identify the element $\begin{pmatrix}   c & d  \\   0 & 1 \end{pmatrix}$ of $H$ with the point $(c,d) \in \mathbb{R}^2$ This
  identifies $H$ with a half plane in $\mathbb{R}^2$ .
(a) Describe the left and right cosets of $J$ geometrically in this
  half plane.
(b) The orbit-stabilizer relation states that there is a bijection
  between the orbit of a point and the left-cosets of its stabilizer.
  Describe this bijection explicitly for the orbit of $\begin{pmatrix}
 0  \\   1  \end{pmatrix}$ under $H$ and the left cosets of it's
  stabilizer.

For (a), I have that the left cosets are of the form (for a fixed $h \in H$):
$$ hJ = \begin{pmatrix}   c & d  \\   0 & 1 \end{pmatrix} \begin{pmatrix}   a & 0  \\   0 & 1 \end{pmatrix} = \begin{pmatrix}   ac & 0  \\   0 & 1 \end{pmatrix}$$ with $c>0$ fixed, $ac > 0$ and $a$ varying over $(0, \infty) $ for every $j \in J$. Geometrically, this would then be all the points on the $x$ axis $\in (0,\infty) $.
The right cosets are then:
$$ Jh = \begin{pmatrix}   a & 0  \\   0 & 1 \end{pmatrix} \begin{pmatrix}   c & d  \\   0 & 1 \end{pmatrix} = \begin{pmatrix}   ac & ad  \\   0 & 1 \end{pmatrix}$$ with $c>0$ fixed, $ac > 0$ and $a$ varying over $(0, \infty) $ for every $j \in J$, $d$ fixed and $a$ varying over $(0, \infty) $.
Geometrically, if $d < 0$ then it is the bottom right quadrant (not including the axes) as $a$ varies over $\in (0,\infty) $. If $d = 0$ then it is all the points on the $x$ axis $\in (0,\infty) $, and if $d>0$ it is the top right quadrant (not including the axes) as $a$ varies over $\in (0,\infty) $.
Do I have the right idea here?
For (b),
I'm not quite sure if how I have formed the bijection is correct. I have that the orbit of $\begin{pmatrix}
 0  \\   1  \end{pmatrix}$ under $G$ are the points given by the vectors of the form $\begin{pmatrix} d  \\   1  \end{pmatrix}$ as $d$ varies over $(-\infty,\infty) $ and the stabilizer are matrices of the form $\begin{pmatrix}   c & 0  \\   0 & 1 \end{pmatrix}$ where $c$ varies over $(0,\infty) $.
Therefore, the left-cosets of the stabiliser of $\begin{pmatrix} 0  \\   1  \end{pmatrix}$ are of the form:
$$ \begin{pmatrix}   c & d  \\   0 & 1 \end{pmatrix} \begin{pmatrix}   c & 0  \\   0 & 1 \end{pmatrix} = \begin{pmatrix}   c^2 & d  \\   0 & 1 \end{pmatrix}$$ with $c^2$ varying over $(0,\infty)$ and $d$ varying over $(-\infty, \infty)$. Thus, I need to form a bijection between:
$$\begin{pmatrix}   c^2 & d  \\   0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} d  \\   1  \end{pmatrix}$$
It seems natural to simply map $d$ explicitly to itself for every $d \in \mathbb{R} $ as it varies over $(-\infty, \infty)$. Since $c$ and $d$ vary over infinite intervals, every $c^2$ which occurs in the left matrix will be paired with a $d$.
Is this a correct bijection?
I really appreciate any help you can give me. Thanks.

REPLY [1 votes]: Regarding part (a)
You are close, but there are some issues. First of all, the left cosets will be of the form
$$
hJ = \begin{pmatrix}   c & d  \\   0 & 1 \end{pmatrix} \begin{pmatrix}   a & 0  \\   0 & 1 \end{pmatrix} = \begin{pmatrix}   ac & d  \\   0 & 1 \end{pmatrix}
$$
Then to the conclusions:

$hJ$ constitutes a half line $(x,d)$, where $d$ is some fixed $y$-coordinate and $x=ac\in(0,\infty)$
For $Jh$ we have half lines of the form $(ac,ad)$ so in effect they can be described via $x=ac\in(0,\infty)$ and $y=ad$ which can be combined to $y=\frac{d}{c}\cdot x$ where $x>0$. For $d>0$ this is increasing, for $d=0$ it is constant and is one half of the x axis as you already stated, and for $d<0$ it is decreasing.

Regarding part (b)
You are close again, but again there are issues. First of all we have the stabilizer of $v=\begin{pmatrix} 0\\1\end{pmatrix}$ as
$$
H_v=\{\begin{pmatrix}a&0\\0&1\end{pmatrix}|\ a\in\mathbb R, a>0\}=J
$$
So in fact the stabilizer equals $J$ from part (a). There we already identified each left coset to be defined by $d$ being a line of the form $\{(x,d)|\ x>0\}$. Thus each left coset $hJ$ depends only on $d$ as does each element $\begin{pmatrix} d  \\   1  \end{pmatrix}$ in the orbit. For each fixed pair $c,d$ and the corresponding element $h=\begin{pmatrix}c&d\\0&1\end{pmatrix}\in H$ we have the bijections
$$
hJ\longmapsto\{(x,d)|\ x>0\}\longmapsto d\longmapsto\begin{pmatrix}d\\1\end{pmatrix}
$$


TITLE: What is the most important notion of sparsity for the design of efficient graph algorithms?
QUESTION [13 upvotes]: There are several competing notions of a "sparse graph".  For instance, a surface-embeddable graph could be considered sparse.  Or a graph with bounded edge density.  Or a graph with high girth.  A graph with large expansion.  A graph with bounded treewidth.  (Even within the subfield of random graphs, it is slightly ambiguous as to what could be called sparse.)  Et cetera.
What notion of "sparse graph" has had the most impact on the design of efficient graph algorithms, and why?  Similarly, what notion of "dense graph" ... ?  (NB: Karpinski has worked a great deal on approximation results for one standard model of dense graphs.)
I have just seen a talk by J. Nesetril on a program of his (together with P. Ossona de Mendez) to capture measures of sparsity in graphs within a unified (asymptotic) framework.  My question -- yes, maybe quite subjective and I expect different camps -- is motivated by the desire to catch a multi-faceted perspective on the use of sparsity in algorithms (and plug any gaps in my own understanding of the issue).

REPLY [1 votes]: One can think of a graph as an adjacency matrix - there are several definitions for matrix sparsity (% of zero entries, for example) which could translate back to the graph itself. Other than % of zero entries, matrix bandwidth under a reordering could be a good proxy for graph sparsity (it looks like bandwidth is related to degeneracy).


TITLE: complex integrals springing from IMO question?
QUESTION [0 upvotes]: I was looking at problem $A2$ here: https://www.imo-official.org/problems/IMO2006SL.pdf
The comment following $A2$ suggests that complex contours lead to a nice expression, but I don’t see the connection. It may have to do with the residue theorem.

REPLY [0 votes]: By the comment following A2 we know
$$\sum_{k=0}^\infty a_k z^k=\frac{z}{\log (1-z)},\quad (|z|<1)$$
where the branch of $\log(1-z)$ is chosen so that $\log 1 =0$ of course.
We can rewrite this as
$$\sum_{k=0}^\infty a_k z^k=\frac{z}{\log (z-1)-i\pi},$$
where the branch of $\log(z-1)$ is chosen so that $0\le \operatorname{Im}\,\log (z-1)<2\pi$.
Dividing both sides by $z^{n+1}$ we have $$
\sum_{k=0}^\infty a_k z^{k-n-1}=\frac{1}{z^n(\log (z-1)-i\pi)}$$
and hence the residue therem yields
$$
a_n=\frac{1}{2\pi i}\int_{|z|=r}  \frac{dz}{z^n(\log (z-1)-i\pi)} \quad (r<1).$$
Now we consider the keyhole-shaped contour consisting of
1. the horizontal line segment $I_1$ from $1+i\varepsilon $ to $\sqrt{R^2-\varepsilon ^2}+i\varepsilon $;
2. the circular arc $C_R$ of radius $R$ traced counterclockwise from $\sqrt{R^2-\varepsilon ^2 } +i\varepsilon $ to $\sqrt{R^2-\varepsilon ^2 } -i\varepsilon$;
3. the horizontal line segment $I_2$ from $\sqrt{R^2-\varepsilon ^2}-i\varepsilon $ to $1-i\varepsilon $;
4. the semi-circle $C_\varepsilon $ of radius $\varepsilon $ traced clockwise from $1-i\varepsilon $ to $1+i\varepsilon $.
Then we have \begin{align}
a_n=\frac{1}{2\pi i}\left(\int_{I_1}+\int_{C_R}+\int_{I_2}+\int_{C_\varepsilon }\right)  \frac{dz}{z^n(\log (z-1)-i\pi)} ,\tag{1}\end{align}
since the integrand is analytic in $\mathbb{C}\setminus (\{0\}\cup [1,\infty))$.
Note that \begin{align}
\lim_{\varepsilon \to 0} \int_{I_2} \frac{dz}{z^n(\log (z-1)-i\pi)}&=\int_R^1 \frac{dx}{x^n(\log (x-1)+2i\pi-i\pi)}\\
&=\int_R^1 \frac{dx}{x^n(\log (x-1)+i\pi)}.\end{align}
Therefore letting $R \to \infty$, $\varepsilon \to 0$ in $(1)$, we have
\begin{align}
a_n&=\frac{1}{2\pi i}\left(\int_1^\infty \frac{dx}{x^n\left(\log (x-1)-i\pi\right)}+\int_\infty^1 \frac{dx}{x^n\left(\log (x-1)+i\pi\right)}\right)\\
&=\int_1^\infty \frac{dx}{x^n\left(\log^2(x-1)+\pi^2\right)},
\end{align}
since integrals along $C_\varepsilon $ and $C_R$ tend to $0$.


TITLE: Minimising the integral of two functions
QUESTION [0 upvotes]: I've been given two functions: $f,g: [a,b] \to \mathbb R$. I have to minimize
$$t\mapsto \int_a^b (f(x)-tg(x))^2 dx.$$
I have been asked to do this in order to show
$$\int_a^b  (f(x)g(x))^2 dx \le \int_a^b f(x)^2 dx\int_a^b g(x)^2dx.$$
I haven't been able to get very far, as I keep getting confused. I assume I have to find the derivative of the original integral, but I don't really know how. Any help would be greatly appreciated. Thank you

REPLY [3 votes]: If the original integral is $$I=\int_a^b (f(x)-tg(x))^2 dx$$
then you can rewrite it as
$$\begin{align}I&=\int_a^b (f^2(x) - 2f(x)g(x)\cdot t + g^2(x)\cdot t^2)dx\\
&= \int_a^b f^2(x) - t\cdot 2\int_a^b f(x)g(x)dx + t^2\cdot \int_a^b g^2(x)dx\end{align}$$
so basically, your function is of the form $A t^2 + Bt + C$ for some constants $A,B,C\in\mathbb R$. You probably know how to minimise this function.


TITLE: Reference Text that develops Linear Algebra with Knowledge of Abstract Algebra
QUESTION [5 upvotes]: Background: Due to some unfortunate sequencing, I have developed my abstract algebra skills before most of my linear algebra skills. I've worked through Topics in Algebra by Herstein and generally liked his approach to vector spaces and modules. Besides a very elementary course in linear algebra (where most of the time went towards matrix multiplication), I have not developed any other linear algebra skills.
But it seems that it is now necessary for me to do so. Most of the topics that I am looking at now require some background of linear algebra and I still lack understanding of ideas like: bilinear forms, invariant subspaces, eigenvalues, requirements for diagonalization of a matrix and so forth.
This brings me to my question (provided there are any)

Question: What are some good texts that develop the theory of linear algebra from a perspective of general algebra?

Are there any texts that develop the key (elementary) ideas of linear algebra in an abstract setting? Thank you for the help!

REPLY [1 votes]: Advanced Linear Algebra by Roman certainly develops the key ideas of linear algebra in an abstract setting. The preliminaries section contains a quick introduction to some concepts and definitions from abstract algebra, but the book is more appropriate for someone, like yourself, who has already studied some abstract algebra. I'd also recommend having an introductory linear algebra book on hand for additional examples.


TITLE: compactness and fundamental group of $SL(n,\mathbb{C})$
QUESTION [2 upvotes]: could any one help me to prove the heading?
$SL(n,\mathbb{C})$ is closed I can prove only, what are the other tools I need?
$SL(n,\mathbb{C})$ connected?simply connected?

REPLY [1 votes]: The case $n=1$ is trivial, so assume $n\ge 2$. It is easy to see that $SL(n,\mathbb{C})$ is not compact, just look at diagonal matrices with entries $\lambda, \lambda^{-1}, 1, \ldots,1$, and let $\lambda \to \infty$. It is equally easy to see that it is connected, just deform the Jordan normal form to the identity matrix. For the fundamental group you need a little more sophisticated tools, see the other answer.


TITLE: Characterized by measurable function
QUESTION [1 upvotes]: Is there a function that inverse function take a borel set to measurable set, but don’t take a measurable set to measurable set?

REPLY [0 votes]: Let $A\subseteq \Bbb R$ be a set which is measurable but is not borel __ I think only if you accept Axiom of Choice there's such a set __ and choose $c\in A^c$. Then I think the function
$$f:\Bbb R\to \Bbb R$$
$$(\forall x\in A^c)(f(x)=x)$$
$$f(A)=\{c\}$$
is probably what you are looking for.


TITLE: Complex Integral
QUESTION [0 upvotes]: I have a question about a task, that I am supposed to solve in the Complex analysis course.
Let  $g(z)=\frac {\sin z}{z^2+i}$, and $K$ is a curve running from $-1$ to $1$, inside the domain of $g$ . What values can the integral take:
$$\int_{K}g(z) \space dz \quad ?$$
I have tried completing the curve into a closed curve and have then used the residue theorem, I did also find all the singularities of g, but I just can't find a logicall answer.

REPLY [1 votes]: So... $g(z)$ has two singuar points : $a=-\frac {\sqrt2}{2}+\frac {\sqrt2}{2}i$ and $\frac {\sqrt2}{2}-\frac {\sqrt2}{2}i$ . If I complete the $K$ curve to a full curve, the closed curve now consists of two parts, $K$ and $G$, if $G$ does not enclose any of the singular points $\int _{K\bigcup G} g(z)dz =0 $  which means:
$\int _{K\bigcup G} g(z)dz =\int _{K} g(z)dz \space +\space \int _{G} g(z)dz = 0$  so $\int _{K} g(z)dz = -\int _{G} g(z)dz $.  So the integral takes the value of   $\int _{K} g(z)dz\space =-(g(1)-g(-1)). $ But if on the other hand the integral encloses one or both singular points then it should take the value of $\int _{K} g(z)=k\space Res(g,a)+h\space Res(g,b) $, where $Res(g,a)$ means the residue of function $g$ in $a$, and $k$ and $h$ are both integers,via the residue theorem


TITLE: What is $3^{99} \pmod{100}$?
QUESTION [1 upvotes]: I saw this post on how to solve $3^{123}\pmod{100}$ using Euler's Totient Theorem.
How about for $3^{99}\pmod{100}$? It seems more complicated because applying Euler's Totient Theorem gets us $3^{40}\equiv 1\pmod{100}$. This means $3^{80}\equiv 1 \pmod{100}$, which isn't enough, because we still need to find $3^{19}\pmod{100}$.
Now, when the terms are listed, a pattern does emerge.
$$\begin{array}{|c|c|c|c|}
\hline
3 & 9 & 27 & 81 \\ \hline
43 & 29 & 87 & 61 \\ \hline
83 & 49 & 47 & 41\\ \hline
23 & 69 & 07 & 21 \\ \hline
03 & 89 & 67 & 01 \\ \hline
\end{array}$$
And $3^{19}$ ends in $67$.
But how can I find this in other methods, besides bashing? Is there some sort of theorem that I can use?

REPLY [0 votes]: Just to give a different approach, observe that
$$3^{99}=3^{1+2\cdot49}=3\cdot9^{49}=-3(1-10)^{49}\equiv-3(1-49\cdot10)\mod100$$
(where in the binomial expansion $(1-10)^n=1-10n+{n\choose2}100+\cdots$ we drop everything after the first two terms), and
$$-3(1-490)=3\cdot489\equiv3\cdot89\equiv-3\cdot11=-33\equiv67\mod100$$


text:
\begin{document}

\title[{\rm J A Heras and R Heras}]{{\fontfamily{qag}\selectfont {\LARGE \textcolor[rgb]{0.00,0.00,0.49}{Helmholtz's theorem for two retarded fields and its application to Maxwell's equations}}}}

\vskip 30pt

\author{Jos\'e A. Heras$^1$ and Ricardo Heras$^2$}
\address{$^1$Instituto de Geof\'isica, Universidad Nacional Aut\'onoma de M\'exico, Ciudad de M\'exico 04510, M\'exico.
E-mail: herasgomez@gmail.com\\
$^2$Department of Physics and Astronomy, University College London, London WC1E 6BT, UK. E-mail: ricardo.heras.13@ucl.ac.uk }
\begin{abstract}
\noindent An extension of the Helmholtz theorem is proved, which states that two retarded vector fields ${\bf F}_1$ and ${\bf F}_2$
satisfying appropriate initial and boundary conditions are uniquely determined by specifying their divergences $\nabla\cdot{\bf F}_{1}$ and $\nabla\cdot{\bf F}_{2}$ and their coupled curls $-\nabla\times{\bf F}_{1}-\partial {\bf F}_{2}/\partial t$ and $\nabla\times{\bf F}_{2}-(1/c^2)\partial {\bf F}_{1}/\partial t$, where $c$ is the propagation speed of the fields. When a corollary of this theorem is applied to Maxwell's equations, the retarded electric and magnetic fields are directly obtained. The proof of the theorem relies on a novel demonstration of the uniqueness of the solutions of the vector wave equation.
\end{abstract}
\vskip 30pt

\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{1. Introduction}}}}}

 The mathematical foundations of electrostatics and magnetostatics relies on the Helmholtz theorem of vector analysis.
Among the several equivalent formulations of this theorem presented in standard textbooks \cite{1,2,3,4}, let us consider the formulation given by Griffiths \cite{1}. The theorem states that if the divergence $\nabla\cdot\v F = D(\v r)$ and the curl $\nabla\times\v F=\v C(\v r)$ of a vector function $\v F(\v r)$ are specified, and if they both go to zero faster than $1/r^2$ as $r\rightarrow\infty,$ and if $\v F(\v r)$ goes to zero as $r\rightarrow\infty,$ then  $\v F$ is uniquely given by $\v F = - \nabla U  + \nabla\times \v W,$ where
\begin{eqnarray}
 U(\v r)\equiv \frac{1}{4\pi}\int \frac{D ( \v r') }{R} \, d^3r',\;\; W(\v r)\equiv \frac{1}{4\pi}\int \frac{\v C ( \v r') }{R}\, d^3r'.
\end{eqnarray}
Here $\v r$ is the field point and $r=|\v r|$ is its magnitude, $\v r'$ is the source point and $R=\!|\v r-\v r'|$. The integrals are over all space and
 $d^3r'$ is the volume element. This theorem has a useful corollary: Any vector function $\v F(\v r)$ that goes to zero faster than $1/r$ as $r\rightarrow\infty$ can be expressed as
\begin{eqnarray}
\v F (\v r)= -\nabla\bigg(\frac{1}{4\pi}\int \frac{\nabla' \cdot \v F ( \v r') }{R} \, d^3r'\bigg)  + \nabla\times \bigg(\frac{1}{4\pi}
\int \!\frac{\nabla' \times \v F ( \v r') }{R}\, d^3r'\bigg).
\end{eqnarray}
In a first application, we write $\v F=\v E$ and use the electrostatic equations: $\nabla \cdot \v E=\rho/\epsilon_0$ and $\nabla \times \v E=0$ to obtain the electrostatic field $\v E(\v r)$ produced by the charge density $\rho(\v r)$:
\begin{eqnarray}
\v E (\v r)= -\nabla\bigg(\frac{1}{4\pi\epsilon_0}\int \frac{\rho(\v r') }{R}\, d^3r'\bigg),
\end{eqnarray}
or equivalently $\v E=-\nabla \Phi,$ where $\Phi$ is the scalar potential:
\begin{eqnarray}
\Phi(\v r)= \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\v r') }{R}\, d^3r'.
\end{eqnarray}
 In a second application we make $\v F=\v B$ and use the magnetostatic  equations: $\nabla\cdot \v B=0$ and $\nabla \times \v B=\mu_0\v J$ to obtain the magnetostatic field $\v B(\v r)$ produced by the current density $\v J(\v r)$:
\begin{eqnarray}
\v B (\v r)= \nabla \times\bigg(\frac{\mu_0}{4\pi}\int \frac{\v J(\v r')}{R}\, d^3r'\bigg),
\end{eqnarray}
or equivalently $\v B=\nabla\times \v A,$ where $\v A$ is the vector potential:
\begin{eqnarray}
\v A (\v r)= \frac{\mu_0}{4\pi}\int \frac{\v J(\v r')}{R}\, d^3r'.
\end{eqnarray}
Here $\epsilon_0$ and $\mu_0$ are the permittivity and permeability of vacuum which satisfy $\epsilon_0\mu_0=1/c^2$, with $c$ being the speed of light in vacuum.

The Helmholtz theorem is also applicable to the time-dependent regime of Maxwell's equations \cite{5,6}. The reason is simple: the derivation of (2) does not involve time and therefore it can be applied to a time-dependent vector field $\v F(\v r,t)$. We simply make the replacements $\v F(\v r)\!\to\! \v F(\v r,t)$ and $\v F(\v r')\!\to\! \v F(\v r',t)$ in (2) and obtain an {\it instantaneous} form of the theorem \cite{6}:
\begin{eqnarray}
\v F (\v r, t)\!= -\nabla\bigg(\frac{1}{4\pi}\int \frac{\nabla' \cdot \v F ( \v r',t) }{R} \,d^3r'\bigg)  + \nabla\times \bigg(\frac{1}{4\pi}
\int \!\frac{\nabla' \times \v F ( \v r',t) }{R}\, d^3r'\bigg).
\end{eqnarray}
 However, this form of the theorem is of {\it limited} practical utility in the time-dependent regime of Maxwell's equations because  the Faraday law
and the Amp$\grave{\rm e}$re-Maxwell law in this regime do not specify the curls of the fields $\v E$ and $\v B$ as such in terms of sources, rather they specify the {\it hybrid} quantities $\nabla\times \v E+\partial \v B/\partial t$ and $\nabla\times \v B-(1/c^2)\partial \v E/\partial t$, which couple space and time variations of both fields. In the more general case in which there are magnetic monopoles, the Faraday
and Amp$\grave{\rm e}$re-Maxwell laws read $-\nabla\times \v E-\partial \v B/\partial t=\mu_0\v J_g$ and $\nabla\times \v B-(1/c^2)\partial \v E/\partial t=\mu_0 \textbf{J}_e$, where $\textbf{J}_g$ is  the magnetic current density and $\textbf{J}_e$ is  the electric current density.  As may be seen, these laws connect the hybrid quantities $-\nabla\times \v E-\partial \v B/\partial t$ and $\nabla\times \v B-(1/c^2)\partial \v E/\partial t$ with their respective electric and magnetic currents. It is clear that the time derivatives of the fields couple their corresponding curls. Let us call the first hybrid quantity the \emph{coupled curl} of $\v E$ and the second quantity the \emph{coupled curl} of $\v B$.

It would be desirable to have an extension of the Helmholtz theorem which allows us to directly find the retarded electric and magnetic fields in terms of the retarded scalar and vector  potentials, in the same form that the standard Helmholtz theorem allows us to directly find the electrostatic and magnetostatic fields in terms of their respective static scalar and vector potentials. Expectably, this appropriate generalisation of the theorem should be formulated for two retarded vector fields and in terms of their respective divergences and coupled curls.

In this paper we formulate a generalisation of the Helmholtz theorem for two retarded vector functions. We show how a corollary of this theorem allows us to find the retarded electric and magnetic fields and introduce the corresponding retarded potentials in three specific cases: for Maxwell equations with the electric charge and current densities (the standard case), when these equations additionally have polarisation and magnetisation densities and finally when they additionally have magnetic charge and current densities. The proof of the theorem relies on the uniqueness of the solutions of the vector wave equation. A novel demonstration of this uniqueness is presented in the Appendix A. Although more complicated than the standard Helmholtz theorem, the extension of this theorem for two retarded vector fields formulated here may be presented in an advanced undergraduate course of electrodynamics.

\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{2. The Helmholtz theorem for two retarded fields}}}}}

\noindent We begin by defining a {\it retarded} function as one function of space and time whose sources are evaluated at the retarded time.
For example, the vector $\cF(\v r,t)= \int \bfg( \v r',t-R/c )d^3r'$
is a retarded field because its vector source $\bfg$ is evaluated at the retarded time $t-R/c$. In order to simplify the notation, we will use the retardation brackets $[\quad]$ to indicate that the enclosed quantity is to be evaluated at the source point $\v r'$ and at the retarded time $t-R/c,$ that is, $[\cF]=\cF(\v r',t-R/c)$. For example, $\cF= \int [\bfg]d^3r'$ denotes a retarded quantity.

Suppose we are told that the divergences $\nabla\cdot\v F_1$ and $\nabla\cdot\v F_2$ of two retarded vector functions $\v F_1=\v F_1(\v r,t)$ and $\v F_2=\v F_2(\v r,t)$ are  specified by the time-dependent scalar functions $D_1=D_1(\v r,t)$ and  $D_2=D_2(\v r,t)$:
\begin{eqnarray}
\nabla\cdot\v F_1=D_1, \qquad \nabla\cdot\v F_2=D_2,
\end{eqnarray}
and that the coupled curls  $-\nabla\times\v F_1-\partial \v F_2/\partial t$ and $\nabla\times\v F_2-(1/c^2)\partial \v F_1/\partial t$ are specified by the time-dependent vector functions $\v C_1=\v C_1(\v r,t)$ and $\v C_2=\v C_2(\v r,t)$:
\begin{eqnarray}
-\nabla\times\v F_1-\frac{\partial \v F_2}{\partial t}=\v C_2, \qquad \nabla\times\v F_2-\frac{1}{c^2}\frac{\partial \v F_1}{\partial t}=\v C_1,
\end{eqnarray}
where $c$ is the propagation speed of the fields $\v F_1$ and $\v F_2$.\footnote[1]{Here the speed $c$ does not necessarily have to represent the speed of light. However, if $c$ is identified with the speed of light in vacuum and the following identifications: $\v F_1=\v E,\v F_2=\v B, D_1=\rho/\epsilon_0,  D_2=0, \v C_1=\mu_0\v J$ and $\v C_2=0$ with $\epsilon_0\mu_0=1/c^2$ are made, then (8) and (9) become Maxwell's equations in SI units. Analogously, if the identification $\v F_1=c\v E,\v F_2=\v B, D_1=4\pi c \rho,  D_2=0, \v C_1=(4\pi/c)\v J$ and $\v C_2=0$ are made then (9) and (10) become Maxwell's equations in Gaussian units.} For consistence, $\v C_1, \v C_2, D_1$ and $D_2$ must satisfy
the ``continuity'' equations:
\begin{eqnarray}
\nabla\cdot\v C_1+\frac{1}{c^2}\frac{\partial D_1}{\partial t}=0, \qquad \nabla\cdot\v C_2+\frac{\partial D_2}{\partial t}=0,
\end{eqnarray}
which are implied by (8) and (9). {\it Question}: Can we, on the basis of this information, uniquely determine $ \v F_1$ and $\v F_2$? If $D_1,D_2,\v C_1$ and $\v C_2$ go to zero sufficiently rapidly at infinity, the answer is $yes$, as we will show by explicit
construction.\footnote[2]{The answer is also $yes$ for the case in which the quantities $D_1,D_2,\v C_1$ and $\v C_2$ are localised in space, that is, when they are zero outside a finite region of space. This frequently occurs  in practical applications because these quantities usually play the role of sources, which are physically localised in space.
}

We claim that the solution of (8) and (9) is
\begin{eqnarray}
\v F_1 = - \nabla U_1  - \nabla\times \v W_1 - \frac{\partial \v W_2}{\partial t},\;\;\;  \v F_2 = - \nabla U_2  + \nabla\times \v W_2 - \frac{1}{c^2}\frac{\partial \v W_1}{\partial t} ,
\end{eqnarray}
where
\begin{eqnarray}
U_1= {1\over 4\pi}\int \frac{[D_1]}{R} \,d^3r',\qquad U_2= {1\over 4\pi}\int {[D_2]\over R} ]\,d^3r',
\end{eqnarray}
and
\begin{eqnarray}
\v W_1= {1\over 4\pi}\int {[\v C_2]\over R} \,d^3r',\qquad \v W_2= {1\over 4\pi}\int {[\v C_1]\over R} \,d^3r'.
\end{eqnarray}
The integrals are over all space. Our demonstration requires two {\it additional} sets of equations. The first set is formed by the  ``Lorenz'' conditions:
\begin{eqnarray}
\nabla\cdot\v W_1+\frac{\partial U_2}{\partial t}=0,\qquad \nabla\cdot\v W_2+\frac{1}{c^2}\frac{\partial U_1}{\partial t}=0.
\end{eqnarray}
We will prove the first condition using the result \cite{2}: $\nabla\cdot\big([\cF]/R\big)\!+\!\nabla'\cdot\big([\cF]/R \big)\!=\![\nabla'\cdot\cF]/R$,
\begin{eqnarray}
\nabla\cdot\v W_1&=\frac{1}{4\pi}\int\nabla\cdot\bigg(\frac{[\v C_2]}{R}\,d^3r'\bigg)\nonumber\\
                 &= \frac{1}{4\pi}\int \frac{[\nabla'\cdot\v C_2]}{R}\,d^3r'-\frac{1}{4\pi}\int\nabla'\cdot\bigg(\frac{[\v C_2]}{R}\bigg)\,d^3r'\nonumber\\
                 &=-\frac{\partial }{\partial t}\bigg(\frac{1}{4\pi}\int\frac{[D_2]}{R}\,d^3r'\bigg) -\frac{1}{4\pi}\oint\frac{\hat{\v n}\cdot[\v C_2]}{R}\,dS=-\frac{\partial U_2 }{\partial t},
\end{eqnarray}
where we have used the second equation given in (10), the property \cite{7}:  $[\partial\cF/\partial t]=\partial[\cF]/\partial t$, the Gauss theorem to transform the second volume integral of the second line into a surface integral ($dS$ is the surface element), which is seen to vanish at infinity by assuming the boundary condition that $\v C_2$  goes to zero faster than $1/r^2$ as $r\rightarrow\infty$, and finally considering the second equation in (12).\footnote[3]{In physical applications, the vector $\v C_2$ usually represents a localised quantity. This means that $\v C_2$ is different from zero only within a limited region of space. However, the surface of integration $S$ in the surface integral on the third line of (15) encloses all space, and therefore $S$ is outside the region where $\v C_2$ is different from zero. It follows that $\v C_2$ is zero everywhere on $S$ and then the surface integral vanishes.} Following a similar procedure, we can prove the second ``Lorenz'' condition given in (14). The required second set of equations is formed by the wave equations
\begin{eqnarray}
\Box^2U_1=-D_1,\quad \square^2U_2=-D_2,\quad \square^2\v W_1=-\v C_2,\quad \square^2\v W_2=-\v C_1,
\end{eqnarray}
where $\Box^{2}\equiv\nabla^2-(1/c^2)\partial^2/\partial t^2$ is the d'Alembertian operator. Using the result \cite{7}:
\begin{eqnarray}
\Box^2\bigg(\frac{[{\cX}]}{R}\bigg)=-4\pi[{\cX}]\delta(\v r-\v r'),
\end{eqnarray}
where $\cX$ represents a scalar or vector function,\footnote[4]{This identity is true for functions ${\cX}$ such that the quantities $[{\cX}]/R$ have not the form  $[{\cX}]/R=f(R)[\v F]$ with $f(R)$ being a polynomial function. If for example $f(R)=R$ then
$[{\cX}]/R=R[\v F]$. It follows that $\Box^2 (R[\v F])=-4\pi R[\v F]\delta(\v r-\v r')=0$ since the factor $R\delta(\v r-\v r')$
vanishes for $\v r\not=\v r'$ because of the delta function and also for $\v r=\v r'$  because this equality implies $R=0$.} we can show the first wave equation in (16),
\begin{eqnarray}
\square^2 U_1 =\frac{1}{4\pi}\int\! \square^2\bigg(\frac{[D_1]}{R}\bigg)\,d^3r'=-\int\![D_1]\delta(\v r-\v r')\,d^3r'=-D_1.
\end{eqnarray}
By an entirely similar procedure, we can show the remaining wave equations given  in (16). Considering (11), (14) and (16), we can now prove that $\v F_1$ and $\v F_2$ satisfy (8) and (9). We take the divergences to $\v F_1$ and $\v F_2$ and obtain
\begin{eqnarray}
\nabla\cdot\v F_1&=-\nabla^2U_1-\frac{\partial}{\partial t}\nabla\cdot\v W_2=-\nabla^2U_1+\frac{1}{c^2}\frac{\partial^2 U_1}{\partial t^2}=-\Box^2U_1=D_1,\\
\nabla\cdot\v F_2&=-\nabla^2U_2-\frac{1}{c^2}\frac{\partial}{\partial t}\nabla\cdot\v W_1=-\nabla^2U_1+\frac{1}{c^2}\frac{\partial^2 U_2}{\partial t^2}=-\Box^2U_2=D_2.
\end{eqnarray}
We now calculate the coupled curls of $\v F_1$ and $\v F_2$ and obtain
\begin{eqnarray}
-\nabla\times\v F_1-\frac{\partial \v F_2}{\partial t} =\;\nabla(\nabla\cdot\v W_1)-\nabla^2\v W_1+\frac{\partial}{\partial t}\nabla\times \v W_2
\nonumber\\
\qquad\qquad\qquad+\nabla\bigg(\frac{\partial U_2}{\partial t }\bigg)-\nabla\times \frac{\partial \v W_2}{\partial t }+\frac{1}{c^2}\frac{\partial^2 \v W_1}{\partial t^2}\nonumber \\
\qquad\qquad\quad\;\;=\;\nabla\bigg(\nabla\cdot\v W_1+\frac{\partial U_2}{\partial t}\bigg)-\square^2\v W_1=-\square^2\v W_1=\v C_2,\\
\;\;\nabla\times\v F_2-\frac{\partial \v F_1}{\partial t} =\;\nabla(\nabla\cdot\v W_2)-\nabla^2\v W_2+\frac{\partial}{\partial t}\nabla\times \v W_1
\nonumber\\
\qquad\qquad\qquad+\nabla\bigg(\frac{\partial U_1}{\partial t }\bigg)-\nabla\times \frac{\partial \v W_1}{\partial t }+\frac{1}{c^2}\frac{\partial^2 \v W_2}{\partial t^2}\nonumber \\
\qquad\qquad\quad\;\;=\;\nabla\bigg(\nabla\cdot\v W_2+\frac{\partial U_1}{\partial t}\bigg)-\square^2\v W_2=-\square^2\v W_2=\v C_1.
\end{eqnarray}
Therefore the functions $\v F_1$ and $\v F_2$ given in (11) constitute a solution of (8) and (9).

Now, assuming that the conditions on $D_1,D_2,\v C_1$ and $\v C_2$ are met, is the solution in (11) $unique$? The answer is clearly $no$, for we can add to $\v F_1$ the function $\v H_1$ and $\v F_2$ the function $\v H_2$, with the divergences and coupled curls of $\v H_1$ and $\v H_2$ being zero, and the result still has divergences $D_1$ and $D_2$ and coupled curls $\v C_1$ and $\v C_2$. However, it so happens that if $\v F_1, \v F_2, \partial\v F_1/\partial t$ and
$ \partial\v F_2/\partial t$ vanish at $t=0$, and if  $\nabla \v F_1,\nabla \v F_2, \partial\v F_1/\partial t$ and
$ \partial\v F_2/\partial t$ go to zero faster than $1/r^2$ as $r\rightarrow\infty$, and if $\v F_1$ and $\v F_2$ themselves go to zero as $r\rightarrow\infty$,
then $\v H_1$ and $\v H_1$ are zero and therefore the solution (11) is unique. In the notation we are using a generic  second-order tensor $\nabla\cF$ denotes the gradient of the vector $\cF$. The explicit proof that $\v H_1$ and $\v H_1$ are zero under their specified initial and boundary conditions is presented in Appendix A. Notice that the conditions that $\nabla \v F_1,\nabla \v F_2, \partial\v F_1/\partial t$ and
$ \partial\v F_2/\partial t$ go to zero faster than $1/r^2$ as $r\rightarrow\infty$ {\it automatically} imply that $D_1,D_2,\v C_1$ and $\v C_2$ go to zero faster than $1/r^2$ as $r\rightarrow\infty$.
\vskip 5pt
Now we can state the Helmholtz theorem for two retarded fields more rigorously:
\vskip 5pt
\noindent \textbf{Theorem.} If the divergences $D_1$ and $D_2$ and the coupled curls $\v C_1$ and $\v C_2$ of two retarded vector functions $\v F_1$ and $\v F_2$ are specified, and if $\v F_1,\v F_2, \partial\v F_1/\partial t$ and
$ \partial\v F_2/\partial t$ vanish at $t=0$, and if $\nabla \v F_1,\nabla \v F_2, \partial\v F_1/\partial t$ and
$ \partial\v F_2/\partial t$ go to zero faster than $1/r^2$ as $r\rightarrow\infty$, and if $\v F_1$ and $\v F_2$ themselves go to zero as $r\rightarrow\infty$, then $\v F_1$ and $\v F_2$ are uniquely given by (11).
\vskip 5pt
This theorem has the following corollary:
\noindent The fields $\v F_1$ and $\v F_2$ can be expressed as
\begin{eqnarray}
\v F_1 =& - \nabla \bigg\{\int\!\frac{[\nabla'\cdot\v F_1]}{4\pi R} \,d^3r'\bigg\}
- \nabla \times \bigg\{\int\!\frac{[-\nabla'\!\times\v F_1-\partial \v F_2/\partial t]}{4\pi R} \,d^3r'\bigg\}\nonumber\\
& - \frac{\partial}{\partial t}\bigg\{\int\!\frac{[\nabla'\!\times\v F_2-(1/c^2)\partial \v F_1/\partial t]}{4\pi R} \,d^3r'\bigg\},\\
\v F_2 =& - \nabla \bigg\{\int\!\frac{[\nabla'\cdot\v F_2]}{4\pi R} \,d^3r'\bigg\}
+\nabla \times\! \bigg\{\int\!\frac{[\nabla'\!\times\v F_2-(1/c^2)\partial \v F_1/\partial t]}{4\pi R} \,d^3r'\bigg\}\nonumber\\
& - \frac{\partial}{\partial t}\bigg\{\int\!\frac{[-\nabla'\!\times\v F_1-\partial \v F_2/\partial t]}{4\pi R} \,d^3r'\bigg\}.
 \end{eqnarray}
These expressions of the Helmholtz theorem for two retarded vector functions are considerably more complicated than the expression of the Helmholtz theorem (2) for a static vector function. However, the practical advantage (23) and (24) is that they allows us to directly find the retarded fields of Maxwell's equations, as we will see in the next section.

\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{3. Applications }}}}}
 \noindent In a first application of the corollary expressed in (23) and (24), we make the identifications:
$\v F_1=\v E$ and $\v F_2=\v B$ in (23) and (24) and subsequently use the Maxwell equations in SI units
\begin{eqnarray}
\nabla\cdot\v E=\,\frac{1}{\epsilon_0}\rho,\qquad \qquad\quad\;\;\nabla\cdot\v B= 0, \\
\nabla\times \v E+\frac{\partial \v B}{\partial t}=0,\quad
\nabla\times \v B-\frac{1}{c^2}\frac{\partial \v E}{\partial t}= \mu_0\v J,
\end{eqnarray}
to directly obtain the familiar retarded fields
\begin{eqnarray}
\v E =& - \nabla \bigg({1\over 4\pi\epsilon_0}\int\!\frac{[\rho]}{R} \,d^3r'\bigg)
- \frac{\partial}{\partial t}\bigg (\frac{\mu_0}{4\pi}\int\frac{[\v J]}{R}\,d^3r'\bigg),\\
\v B =& \nabla \times \bigg(\frac{\mu_0}{4\pi}\int\frac{[\v J]}{R}\,d^3r'\bigg),
 \end{eqnarray}
or more compactly,
\begin{eqnarray}
\v E = - \nabla \Phi - \frac{\partial \v A}{\partial t},\quad\v B =\nabla \times \v A,
 \end{eqnarray}
 where we have defined the retarded scalar and vector potentials as
 \begin{eqnarray}
\Phi=& {1\over 4\pi\epsilon_0}\int\!\frac{[\rho]}{R}\,d^3r',\quad \v A=\frac{\mu_0}{4\pi}\int\frac{[\v J]}{R}\,d^3r'.
 \end{eqnarray}
Here $c$ is the speed of light in vacuum and is defined by $c=1/\sqrt{\epsilon_0\mu_0}$.\footnote[5]{It is interesting to note that (23) and (24) can be used to find the retarded fields of the gravitational theory suggested by Heaviside \cite{8}, which is similar in form to that of Maxwell.
If we write
$\v F_1=\v g$ and $\v F_2=\v k$, where the time-dependent vector fields $\v g$ and $\v k$ are respectively the gravitoelectric and gravitomagnetic fields, use equations (23) and (24), and consider the gravitational equations in SI units: $\nabla\cdot\v g=-\rho_{\textsc{g}}/\tilde{\epsilon_0},\nabla\cdot\v k= 0,
\nabla\times \v g+\partial \v k/\partial t=0, \nabla\times \v k-(1/c^2)\partial \v g/\partial t= -\tilde{\mu}_0\v J_{\textsc{g}},$
where $\rho_{\textsc{g}}$ and $\v J_{\textsc{g}}$ are the mass density and the mass current density, which satisfy the continuity equation $\nabla\cdot\v J_{\textsc{g}}+\partial\rho_{\textsc{g}}/\partial t=0$ (which represents the mass conservation), then we get the retarded fields $\v g = -\nabla V_{\textsc{g}} - \partial \v A_{\textsc{g}}/\partial t$ and $\v k =\nabla \times \v A_{\textsc{g}}$,
 where the retarded scalar and vector potentials are
 \[
V_{\textsc{g}}=- {1\over 4\pi\tilde{\epsilon}_0}\int\!\frac{[\rho_{\textsc{g}}]}{R} \,d^3r',\quad \v A_{\textsc{g}}=-\frac{\tilde{\mu}_0}{4\pi}\int\frac{[\v J_{\textsc{g}}]}{R}\,d^3r'.
 \]
Here the gravitational permittivity constant of vacuum is defined as $\tilde{\epsilon}_0=1/(4\pi \textsc{G})$, where $\textsc{G}$ is the universal gravitational constant. The gravitational permeability constant of vacuum is defined as $\tilde{\mu}_0=4\pi\mathfrak{S}/c^2$. It follows that $\tilde{\epsilon}_0\tilde{\mu}_0=1/c^2$, where $c$ is the speed of light in vacuum. Jefimenko \cite{9}, McDonald \cite{10} and recently,  Vieira and Brentan \cite{11} have discussed this gravitational theory, whose equations have also been obtained using an alternative approach \cite{12}.}

 In a second application of (23) and (24) , we make the identifications:
$\v F_\texttt{1}=\v E$ and $ \v F_{\texttt{2}}=\v B$ in (23) and (24) and subsequently use the Maxwell equations with material sources (SI units):
\begin{eqnarray}
\nabla\cdot\v E=\,\frac{1}{\epsilon_0}(\rho-\nabla\cdot \v P),\qquad\qquad\;\nabla\cdot\v B=0, \\
\nabla\times \v E+\frac{\partial \v B}{\partial t}=0,\qquad\quad
\nabla\times \v B-\frac{1}{c^2}\frac{\partial \v E}{\partial t}= \mu_0\bigg(\v J+\nabla\times \v M+\frac{\partial \v P}{\partial t}\bigg),
\end{eqnarray}
where $\v P$ and $\v M$ are the polarisation and magnetisation vectors, to get the retarded fields \cite{12}:
\begin{eqnarray}
\v E =&\! - \nabla \bigg({1\over 4\pi\epsilon_0}\int\!\frac{[\rho-\nabla'\cdot\v P]}{R}\,d^3r' \bigg)
- \frac{\partial}{\partial t}\bigg (\frac{\mu_0}{4\pi}\int\frac{[\v J+\nabla'\times\v M+\partial\v P/\partial t]}{R}d^3r'\bigg),\nonumber\\
\\
\v B =& \nabla \times \bigg(\frac{\mu_0}{4\pi}\int\frac{[\v J +\nabla'\times\v M+\partial\v P/\partial t]}{R}\,d^3r'\bigg),
 \end{eqnarray}
or more compactly,
\begin{eqnarray}
\v E = - \nabla \Phi - \frac{\partial \v A}{\partial t},\quad\v B =\nabla \times \v A,
 \end{eqnarray}
 where we have defined the retarded scalar and vector potentials as
 \begin{eqnarray}
\Phi= {1\over 4\pi\epsilon_0}\!\int\!\frac{[\rho-\nabla'\cdot\v P]}{R}\,d^3r',\;\v A=\frac{\mu_0}{4\pi}\!\int\!\frac{[\v J +\nabla'\times\v M+\partial\v P/\partial t]}{R}\,d^3r'.
 \end{eqnarray}
Notice that if the vectors $\textbf{P}$ and $\textbf{M}$ are vanished in (33) and (34) then we recover (27) and (28).

In the last application  we write $\v F_1=c\v E$ and $\v F_2=\v B$ in (23) and (24), and subsequently use Maxwell's equations with magnetic monopoles expressed in Gaussian units as
\begin{eqnarray}
\nabla\cdot c\v E=\,4\pi c\rho_e,\qquad \qquad\;\nabla\cdot\v B=4\pi\rho_m, \\
-\nabla\times c\v E-\frac{\partial \v B}{\partial t}=4\pi\v J_m,\quad
\nabla\times \v B-\frac{1}{c}\frac{\partial \v E}{\partial t}= \frac{4\pi}{c}\v J_e,
\end{eqnarray}
where $\rho_e$ and $\v J_e$ are the electric charge and current densities and $\rho_m$ and $\v J_m$ are the magnetic charge and current densities,
to directly obtain the retarded fields with magnetic monopoles
\begin{eqnarray}
\v E = - \nabla\bigg( \int\!\frac{[\rho_e]}{R} \,d^3r'\bigg)
- \nabla \times \bigg(\int\frac{[\v J_m]}{R c}\,d^3r'\bigg)-\frac{1}{c} \frac{\partial}{\partial t}\bigg (\int\frac{\v [\v J_e]}{R c}\,d^3r'\bigg),\\
\v B = - \nabla \bigg(\int\!\frac{[\rho_m]}{R}\,d^3r'\bigg)
 +\nabla \times \bigg(\int\frac{[\v J_e]}{R c}\,d^3r'\bigg)- \frac{1}{c}\frac{\partial}{\partial t}\bigg (\int\frac{[\v J_m]}{R c}\,d^3r'\bigg),
 \end{eqnarray}
or more compactly:
\begin{eqnarray}
\v E = - \nabla \Phi_e  - \nabla\times \v A_m -\frac{1}{c} \frac{\partial \v A_e}{\partial t},\;\;\; \v B = - \nabla \Phi_m  + \nabla\times \v A_e - \frac{1}{c}\frac{\partial \v A_m}{\partial t} ,
\end{eqnarray}
where we have defined the retarded electric and magnetic scalar potentials as
 \begin{eqnarray}
\Phi_e=& \int\!\frac{[\rho_e]}{R}\,d^3r',\;\;
\Phi_m=\int\!\frac{[\rho_m}{R} \,d^3r',
 \end{eqnarray}
 and the retarded electric and magnetic vector potentials as
 \begin{eqnarray}
\v A_e=\int\frac{[\v J_e]}{R c}\,d^3r', \;\;\v A_m=\int\frac{[\v J_m]}{R c}\,d^3r'.
 \end{eqnarray}
Notice that if the magnetic densities $\rho_m$ and $\v J_m$ are vanished in (39) and (40) then we recover (27) and (28).

\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{4. Pedagogical comment}}}}}
Standard textbook presentations of electromagnetism follow a different route to the electric and magnetic fields in the time-independent regime of Maxwell's equations than in the time-dependent regime of these equations. While in the time-independent regime, the Helmholtz theorem of the vector analysis is commonly applied to  find the electrostatic and magnetostatic fields in terms of their respective scalar and vector potentials, in the time-dependent regime Maxwell's sourceless equations are commonly used to introduce the scalar and vector potentials which are then inserted into Maxwell's  source equations, obtaining two coupled second-order equations involving potentials, which are shown to be  gauge invariant. The Lorenz-gauge condition is usually adopted to decouple these second-order equations and as a result, we arrive at two wave equations, which are solved to obtain the retarded scalar and vector potentials and by a subsequent differentiation of them we finally obtain the retarded electric and magnetic fields. It is evident that the usual method followed in the  time-dependent regime is somewhat more complicated than the usual method used in the time-independent regime which is based in the Helmholtz theorem. It is clear that for pedagogical reasons it is worth formulating an extension of the Helmholtz theorem, which may be useful in the time-dependent regime of Maxwell's equations. This has been made in the past and different extensions of the Helmholtz theorem to include the time-dependence of the fields have been formulated \cite{13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28}. But, as far as we are aware, none of these extensions of the Helmholtz theorem has yet been included in standard textbooks.

In our opinion, equations (23) and (24) for the retarded fields $\mathbf{F}_1$ and $\mathbf{F}_2$ could be included in a undergraduate presentation of Maxwell's equations. Similarly, the analogous equation but for a single retarded field \cite{18}:
\begin{eqnarray}
\v F= - \nabla \int\!\frac{[\nabla'\cdot\v F]}{4\pi R} d^3r'\! +\nabla\!\times \int\!\frac{[\!\nabla'\!\times\v F]}{4\pi R} d^3r'\! + \frac{1}{c^2}\frac{\partial}{\partial t}\int\!\frac{[\partial \v F/\partial t]}{4\pi R} d^3r',
 \end{eqnarray}
could alternatively be included in such an undergraduate presentation. We notice that if any of the fields $\mathbf{F}_1$ and $\mathbf{F}_2$ defined by  (23) and (24) is vanished then we obtain (44). The disadvantage of (23) and (24) is that they are somewhat complicated but their
advantage is that they \emph{directly} yield the retarded fields. In contrast, (44) has the advantage of being simpler but
the disadvantage is that it yields the retarded fields in an \emph{indirect} way.

\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{5. Conclusion}}}}}

\noindent How to solve the time-independent Maxwell's equations? Answer: using the Helmhotz theorem of the vector analysis. How to solve the time-dependent Maxwell's equations? Answer: using the Helmholtz theorem for two retarded fields, which was formulated in this paper. The {\it key} to formulate this generalised theorem was the observation that the curls in the first regime are \emph{decoupled} quantities: $\nabla\times \v E=0$ and $\nabla\times \v B=\mu_0\v J$, while the curls in the second regime are \emph{coupled} quantities: $\nabla\times \v E+\partial \v B/\partial t=0$ and
 $\nabla\times \v B-(1/c^2)\partial \v E/\partial t=\mu_0\v J$. Accordingly, we formulated the Helmholtz theorem for two retarded vectors $\v F_1$ and $\v F_1$ in terms of their divergences: $\nabla\cdot\v F_\texttt{1}$ and $\nabla\cdot\v F_2$ and coupled curls: $-\nabla\times\v F_1-\partial \v F_2/\partial t$ and $\nabla\times\v F_2-(1/c^2)\partial \v F_1/\partial t$. The proof of the theorem required of the uniqueness of the solutions of the homogeneous wave equation, which was explicitly demonstrated in Appendix A. As applications, we applied the theorem to Maxwell's equations when they have electric charge and current densities, when they additionally have polarisation and magnetisation densities and when they additionally have magnetic charge and current densities. For each case we obtained the retarded fields in terms of their retarded potentials.  Standard pedagogy generally considers the Helmholtz theorem for a static field to {\it justify} the mathematical form of Maxwell's equations in the time-independent regime. Standard pedagogy might also consider the Helmholtz theorem for two retarded vector fields to {\it justify} the mathematical form of
 Maxwell's equations in the time-dependent regime.

 \appendix
\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{Appendix A. Uniqueness of the solutions of the wave equation}}}}}

\noindent We have shown that the set of functions $\v F_1$ and $\v F_2$ given (11) satisfies (8) and (9). We have also pointed out that this set of functions is not generally unique and that the set formed by $\v F_1 +\v H_1$ and $\v F_2 +\v H_2$ also satisfies  Eqs.~(8) and (9) provided the functions $\v H_1$ and $\v H_2$ satisfy
\begin{eqnarray}
\qquad\;\;\;\;\;\nabla\cdot\v H_1=0, \qquad\qquad\;\; \,\,\nabla\cdot\v H_2=0,\\
-\nabla\times\v H_1-\frac{\partial \v H_2}{\partial t}=0, \quad \nabla\times\v H_2-\frac{1}{c^2}\frac{\partial \v H_1}{\partial t}=0.
\end{eqnarray}
The uniqueness of  $\v F_1$ and $\v F_2$ will be guaranteed if we are able to show that  $\v H_1$ and $\v H_2$ vanish everywhere. From (A.1) and (A.2) we obtain the homogeneous wave equations $\Box^2\v H_1=0$ and $\Box^2\v H_2=0.$ Our strategy is now to construct a relation that allows us to discover those initial and boundary conditions that guarantee that the only solutions of these wave  equations are the trivial ones:
$\v H_1=0$ and $\v H_2=0.$ We will see that these conditions are precisely those given in the formulation of the Helmholtz theorem for two retarded vector fields.

In order to proceed in a more rigorous way, we adopt the following notation: the Cartesian components of $\nabla\cF$ are  given by  $(\nabla\cF)^{ij}=\partial^i{\cal F}^{j}$. Latin indices $i,j,k...$ run from 1 to 3 and the summation convention $a_i a^i$  on repeated indices (one covariant and the other one contra-variant) is adopted. The normal derivative of $\cF$ at the surface $S$ (directed outwards from inside the volume $V$)
is denoted by the vector $\v n\!\cdot\!\nabla\cF$ and defined by its components as $(\v n\!\cdot\!\nabla\cF)^i=n_j\partial^j{\cal F}^i$ where $\v n$ is a unit vector outward to the surface  with $(\v n)^j=n^j.$  Now we can state the following uniqueness theorem for the solutions of the homogeneous vector wave equation:
\vskip 3pt
\noindent \textbf{Theorem.} If the vector function $\cF(\v r,t)$ satisfies the homogeneous wave equation $\Box^2\cF=0$,
the initial conditions that $\cF$ and $\partial\cF/\partial t$ vanish at $t=0$, and the boundary condition that
$\v n\!\cdot\!\nabla \cF$ is zero at the surface $S$ of the volume $V$, then $\cF(\v r,t)$ is identically zero.
\vskip 3pt
Proof. We write the wave equation $\Box^2\cF=0$ in index notation
\begin{eqnarray}
\partial_j\partial^j{\cal F}^i -\frac{1}{c^2}\frac{\partial^2 {\cal F}^i}{\partial t^2}=0.
\end{eqnarray}
We multiply this equation by $\partial {\cal F}_i/\partial t$ and obtain
\begin{eqnarray}
\frac{\partial {\cal F}_i}{\partial t}\bigg(\partial_j\partial^j{\cal F}^i -\frac{1}{c^2}\frac{\partial^2 {\cal F}^i}{\partial t^2}\bigg)=0.
\end{eqnarray}
The left-hand side can be re-written as
\begin{eqnarray}
\partial_j\bigg(\frac{\partial {\cal F}_i}{\partial t}\partial^j{\cal F}^i\bigg)-\frac{\partial}{\partial t}\bigg\{
\frac{\partial_j{\cal F}_{i}\partial^j{\cal F}^{i}}{2}+\frac{1}{2c^2}\frac{\partial{\cal F}_{i} }{\partial t}\frac{\partial{\cal F}^{i} }{\partial t}
\bigg\}=0.
\end{eqnarray}
The volume integration of this equation implies
\begin{eqnarray}
\frac{\partial}{\partial t}\int_V\bigg(\frac{\partial_j{\cal F}_{i}\partial^j{\cal F}^{i}}{2}+\frac{1}{2c^2}\frac{\partial{\cal F}_{i} }{\partial t}\frac{\partial{\cal F}^{i} }{\partial t}\bigg)\,d^3r=\int_V \partial_j\bigg(\frac{\partial {\cal F}_i}{\partial t}\partial^j{\cal F}^i\bigg)\, d^3r.
\end{eqnarray}
The volume integral on the right-side can be transformed into a surface integral,
\begin{eqnarray}
\frac{\partial}{\partial t}\int_V\bigg(\frac{\partial_j{\cal F}_{i}\partial^j{\cal F}^{i}}{2}+\frac{1}{2c^2}\frac{\partial{\cal F}_{i} }{\partial t}\frac{\partial{\cal F}^{i} }{\partial t}\bigg)\,d^3r=\oint_S\bigg(\frac{\partial {\cal F}_i}{\partial t}\bigg)\Big( n_j\partial^j{\cal F}^i\Big)\, dS.
\end{eqnarray}
From the boundary condition that $n_j\partial^j{\cal F}^i$ is zero at $S$ it follows that
 \begin{eqnarray}
\frac{\partial}{\partial t}\int_V\bigg(\frac{\partial_j{\cal F}_{i}\partial^j{\cal F}^{i}}{2}+\frac{1}{2c^2}\frac{\partial{\cal F}_{i} }{\partial t}\frac{\partial{\cal F}^{i} }{\partial t}\bigg)\,d^3r=0,
\end{eqnarray}
and therefore the integral is at most a function of space $g(x^i)$. The initial condition that
${\cal F}^{i}$ is zero at  $t=0$ implies the condition that $\partial^j{\cal F}^{i}$ is zero at $t=0$, which is used together with  the condition that $\partial{\cal F}^{i}/\partial t$ is zero at $t=0$ to show that $g(x^i)=0$. Accordingly,
\begin{eqnarray}
\int_V\bigg(\frac{\partial_j{\cal F}_i\partial^j{\cal F}^i}{2}+\frac{1}{2c^2}\frac{\partial_i{\cal F}_i }{\partial t}\frac{\partial{\cal F}^{i} }{\partial t}\bigg)\,d^3r =0,
\end{eqnarray}
or in the more familiar vector notation
  \begin{eqnarray}
\int_V\bigg(\frac{(\nabla\cF)^2}{2}+\frac{1}{c^2}\frac{(\partial\cF/\partial t)^2}{2}\bigg)\,d^3r =0.
\end{eqnarray}
Since the volume $V$ is arbitrary the integrand must vanish and considering that $\cF$ is a real function it follows that $\nabla\cF=0$ and $\partial\cF/\partial t=0$, which imply that $\cF=\;$ constant. Using the condition that $\cF$ is zero at $t=0$, it follows that this constant vanishes and then $\cF=0$, which proves the theorem.

The above theorem clearly describes a uniqueness theorem for the solutions of the homogeneous wave equation. Suppose we have two vector fields
$\cF_1$ and $\cF_2$ that satisfy the {\it same} wave equation $\Box^2\cF_1=0$ and  $\Box^2\cF_2=0$, the {\it same} initial conditions that $ \cF_1, \cF_2, \partial\cF_1/\partial t$ and $\partial\cF_2/\partial t$ vanish at $t=0$, and the {\it same} boundary conditions that
$\v n\!\cdot\!\nabla \cF_1$ and $\v n\!\cdot\!\nabla \cF_2$ are zero at the surface $S$ of the volume $V$. Now, let us write $\cF=\cF_2-\cF_1$. It follows that $\cF$ satisfies $\Box^2\cF=0$, the initial conditions that $\cF$ and $\partial\cF/\partial t$ vanish at $t=0$, and the boundary condition that $\v n\!\cdot\!\nabla \cF$ zero at the surface $S$. By the uniqueness theorem we have $\cF=0$ and therefore $\cF_1=\cF_2$, i.e., the solution is unique.

In particular, if the surface goes to infinity then we can assume the boundary conditions that $\nabla \cF$ and $\partial\cF/\partial t$
go to zero faster than $1/r^2$ as $r\rightarrow\infty$. Under these conditions, the surface integral in (A.7) vanishes. In this case we also arrive at (A.8) but with the volume $V$ extended over all space. Using the initial conditions that $\cF$ and $\partial\cF/\partial t$ vanish at $t=0$ we again imply (A.10) with the volume $V$ extended over all space and therefore  ${\cF}=0$ because $\nabla \cF$ and $\partial\cF/\partial t$ are zero in the limit
$r\rightarrow\infty$. But  ${\cF}=0$ means  $\cF_1=\cF_2$, i.e., the solution is unique.

We now return to the solutions $\v F_1 +\v H_1$ and $\v F_2 +\v H_2$ of (8) and (9). As previously noted: $\square^2\v H_1=0$ and $\square^2 \v H_2=0$. Therefore we can apply the uniqueness theorem to both $\v H_1$ and $\v H_2$ by assuming the initial conditions that $\v H_1, \v H_2, \partial\v H_1/\partial t, \partial\v H_2/\partial t$ vanish at $t=0$, the boundary conditions that
$\nabla \v H_1, \nabla \v H_2, \partial \textbf{H}_1/\partial t$ and $\partial \textbf{H}_2/\partial t$ go to zero faster than $1/r^2$ as $r\rightarrow\infty$, and the boundary conditions that $\v H_1$ and $\v H_2$ go to zero as $r\rightarrow\infty$. Under these conditions, the uniqueness theorem states that $\v H_1=0$ and $\v H_2=0.$

Therefore, the uniqueness of the vector functions $\v F_1$ and $\v F_2$ of the Helmholtz theorem for two retarded fields is guaranteed by assuming the initial conditions that $ \v F_1, \v F_2,
\partial\v F_1/\partial t$ and $\partial\v F_2/\partial t$  vanish at $t=0$, the boundary conditions that $\nabla \v F_1,\nabla \v F_2,
\partial\v F_1/\partial t$ and $\partial\v F_2/\partial t$ go to zero faster than $1/r^2$ as $r\rightarrow\infty$, and the boundary conditions that $\v F_1$ and $\v F_2$ go to zero as $r\rightarrow\infty$. These boundary conditions {\it imply} that the quantities $D_1, D_2, \v C_1$ and $\v C_2$, considered in the Helmholtz theorem for two retarded fields, go to zero faster than $1/r^2$ as $r\rightarrow\infty$.

\section*{{{\fontfamily{qag}\selectfont {\large \textcolor[rgb]{0.00,0.00,0.49}{References}}}}}

config: arxiv


TITLE: Closed points of field extension of k-scheme under projection
QUESTION [4 upvotes]: I really couldn't figure out the answer to the following question: Let $X$ be a scheme of finite type over a field $k$ and let $K$ be an extension field of $k$. Let $X_K := K \times_k X$ be the base extension and let $p: X_K \rightarrow X$ be the projection. Is it true that $p$ maps closed points to closed points?
This should be true if $K$ is of finite type over $k$ (so not necessarily finite!). Proof: $X$ is Jacobson, the projection $p$ is of finite type and now it follows for example from the Stacks Project Lemma 27.17.8 (see http://stacks.math.columbia.edu/tag/01TB) that $p$ maps closed points to closed points. But I would like to have the same for any $K$. Is this true?

REPLY [9 votes]: No, it is not ! For instance, if $k=\bar{\mathbb{Q}}$, $K=\mathbb{C}$ and $X$ has dimension $\geq 1$, there is exactly one closed point of $X_K$ above each closed point of $X$. But there is only countably many closed points of $X$, whereas $X_K$ has uncountably many closed points.
In general, a point of $X$ will be the image of a closed point of $X_K$ if and only if its residue field may be embedded in a finite extension of $K$ (in a way that respects $k$).
In particular, your statement is not true in general, even when $K$ is of finite type over $k$ : for instance, if $K$ is the residue field of a generic point of $X$. The problem with your argument is that $K$ is of finite type over $k$ as a field, but not as an algebra.