# Large convex sets in difference sets

Krishnendu Bhowmick  Johann Radon Institute for Computational and Applied
Mathematics
Linz, Austria<EMAIL_ADDRESS>,  Ben Lund  Institute for Basic
Science (IBS)
Daejeon, South Korea<EMAIL_ADDRESS>and  Oliver Roche-Newton  Institute
for Algebra, Johannes Kepler Universität
Linz, Austria<EMAIL_ADDRESS>
###### Abstract.

We give a construction of a convex set $A\subset\mathbb{R}$ with cardinality
$n$ such that $A-A$ contains a convex subset with cardinality $\Omega(n^{2})$.
We also consider the following variant of this problem: given a convex set
$A$, what is the size of the largest matching $M\subset A\times A$ such that
the set

$\\{a-b:(a,b)\in M\\}$

is convex? We prove that there always exists such an $M$ with
$|M|\geq\sqrt{n}$, and that this lower bound is best possible, up a
multiplicative constant.

##  1\. Introduction

This paper is concerned with the existence or non-existence of convex sets in
difference sets. A set $A\subset\mathbb{R}$ is said to be convex if its
consecutive differences are strictly increasing. That is, writing
$A=\\{a_{1}<a_{2}<\dots<a_{n}\\}$, $A$ is convex if

$a_{i+1}-a_{i}>a_{i}-a_{i-1}$

holds for all $i=2,\dots,n-1$. Most research on convex sets comes in the
context of sum-product theory, and one may think of the notion of a convex set
as a generalisation of a set with multiplicative structure. For instance, it
is known than convex sets determine many distinct sums and differences. In
particular, it was proven in [8] that the bound111Here and throughout this
paper, the notation $\gg$ is used to absorb a multiplicative constant. That is
$X\gg Y$ denotes that there exists an absolute constant $C>0$ such that $X\geq
CY$. The notation $Y\ll X$ has the same meaning.

$|A-A|\gg|A|^{8/5-o(1)}$

holds for any convex set $A$. Here, $A-A:=\\{a-b:a,b\in A\\}$ denotes the
difference set determined by $A$. This results captures the vague notion that
convex sets cannot be additively structured, and there has been considerable
effort expended to quantify and apply this idea in various way, see for
instance [1], [3] and [9].

Given a finite set $B\subset\mathbb{R}$, define

$\mathcal{C}(B):=\max_{A\subset B:A\text{ is convex}}|A|.$

That is, $\mathcal{C}(B)$ denotes the size of the largest convex subset of
$B$. The first question that we consider is the following: given a convex set
$A\subset\mathbb{R}$, what can we say about the possible value of
$\mathcal{C}(A-A)$? A first observation is that

(1) $|\mathcal{C}(A-A)|\geq|A|,$

as can be seen by considering the convex set $A-a\subset A-A$, where $a$ is an
arbitrary element of $A$. There are some simple constructions showing that the
lower bound (1) is optimal up to a multiplicative constant; for instance, we
can take $A=\\{i^{2}:1\leq i\leq n\\}$.

In this paper, we give a construction of a convex set $A$ whose difference set
contains a very large convex set.

######  Theorem 1.1.

For all $n\in\mathbb{N}$, there exists a convex set $A\subset\mathbb{R}$ with
$|A|=n$ such that $A-A$ contains a convex subset $S$ with cardinality $|S|\gg
n^{2}$.

Using the notation introduced earlier, Theorem 1.1 states that there exists a
convex set $A$ such that $\mathcal{C}(A-A)\gg|A|^{2}$. This result shares some
similarities with main result of [6], where it was established that there
exists a set $A\subset\mathbb{R}$ such that $A+A$ contains a convex subset
with cardinality $\Omega(|A|^{2})$. The main qualitative difference is that we
have the additional restriction that the set $A$ is also assumed to be convex.
Also, Theorem 1.1 provides a convex subset of the difference set, rather than
the sum set.

The simple construction giving rise to the lower bound (1) feels like
something of a cheat, and so we consider a variant of this problem where we
make a further restriction concerning the origin of the convex subset of a
difference set. A set $M\subset A\times A$ is a matching if the elements of
$M$ are pairwise disjoint. Given a matching $M\subset A\times A$, define the
restricted difference set

$A-_{M}A:=\\{a-b:(a,b)\in M\\}.$

Define

$\mathcal{CM}(A):=\max_{M\subset A\times A:M\text{ is a matching and
}A-_{M}A\text{ is convex.}}|M|.$

That is, $\mathcal{CM}(A)$ denotes the size of the largest matching on $A$
which gives rise to a convex subset of $A-A$.

Now, we ask a similar question for this quantity: given a convex set
$A\subset\mathbb{R}$, what can we say about the size of $\mathcal{CM}(A)$? In
particular, how small can this quantity be? Should we expect an analogue of
the bound (1) if we rule out this simple construction? In this paper we answer
this question by giving the following two complimentary results, showing that
$\mathcal{CM}(A)\geq\sqrt{|A|}$ and that this bound is optimal up to a
multiplicative constant.

######  Theorem 1.2.

Let $n\in\mathbb{N}$ be sufficiently large and suppose that
$A\subset\mathbb{R}$ is a convex set with cardinality $n$. Then there exists a
matching $M\subset A\times A$ such that $|M|\geq\sqrt{n}$ and $A-_{M}A$ is
convex.

######  Theorem 1.3.

For all $n\in\mathbb{N}$, there exists a convex set $A\subset\mathbb{R}$ with
cardinality $n$ such that, if $M\subset A\times A$ is a matching and $A-_{M}A$
is convex then $|M|\ll\sqrt{n}$.

##  2\. Proof of Theorem 1.1

Define

$a_{i}=i+c_{1}i^{2}+c_{2}i^{3},$

where

$c_{1}=\frac{75}{n^{2}}\,\,\,\,\text{ and }\,\,\,\,c_{2}=\frac{1}{n^{5}}.$

Assume that $n$ is a sufficiently large multiple of $100$. This assumption is
made only to simplify the notation slightly, and can be easily removed at the
price of introducing some floor and ceiling functions to the calculations.
Define $A=\\{a_{1}<\dots<a_{n}\\}$. Observe that $A$ is convex. Indeed, the
sequence $a_{i+1}-a_{i}$ is increasing, as can be seen by calculating that

$a_{i+1}-a_{i}=1+c_{1}(2i+1)+c_{2}(3i^{2}+3i+1).$

For each integer $k\in[0.009n,0.01n]$ define $D_{k}$ to be the set of $k$th
differences

$D_{k}:=\\{a_{i+k}-a_{i}:1\leq i\leq 0.99n\\}.$

The set $D_{k}$ is also convex. Indeed, let $d_{i}^{(k)}:=a_{i+k}-a_{i}$
denote the $i$th element of $D_{k}$. Then the sequence
$d_{i+1}^{(k)}-d_{i}^{(k)}$ increases with $i$. This can be seen by observing
that

(2)
$d_{i}^{(k)}=a_{i+k}-a_{i}=k+c_{1}(2ki+k^{2})+c_{2}(3i^{2}k+3ik^{2}+k^{3}),$

and hence

$d_{i+1}^{(k)}-d_{i}^{(k)}=2c_{1}k+3c_{2}k^{2}+3c_{2}k+6c_{2}ki$

increases with $i$.

We will find a large convex subset of $A-A$ by efficiently gluing together
consecutive convex sets $D_{k}$. We will make use of the following observation
from [6].

######  Lemma 2.1.

Suppose that $A=\\{a_{1}<\dots<a_{n}\\}$ and $B=\\{b_{1}<\dots<b_{m}\\}$ are
convex sets. Suppose that there exist $1\leq i\leq m$ and $1\leq j\leq n$ such
that

$b_{i}\leq a_{j}<a_{j+1}\leq b_{i+1}.$

Then the set

$\\{a_{1}<\dots<a_{j}<b_{i+1}<b_{i+2}<\dots<b_{m}\\}$

is convex.

Before we get into the details of the proof of Theorem 1.1, which involves
some rather tedious calculations, let us take a moment to try and explain the
idea behind it, with the help of some pictures.

Firstly, we note that, although the sets $D_{k}$ are convex, they are only
slightly convex, in the sense that, if we zoom out and take a look at $D_{k}$,
it appears to resemble an arithmetic progression with common difference
$2c_{1}k$. Note also that this common difference increases slightly as $k$
increases.

$d_{1}^{(10)}$$d_{2}^{(10)}$$d_{3}^{(10)}$ Figure 1. This picture shows the
first three elements of $D_{10}$ after setting $n=10000$. The three elements
form a convex set, but to the naked eye they appear to be arranged in an
arithmetic progression.

The other important feature of this construction is that we have chosen the
parameters in such a way that the $D_{k}$ have convenient overlapping
properties. In particular, each $D_{k}$ has diameter approximately
$\frac{3}{2}$, and starts at $k$. In particular, this means that neighbouring
$D_{k}$ have a significant overlap, but also that each $D_{k}$ takes sole
ownership of a section of the real line.

$D_{k}$$D_{k+1}$$D_{k+2}$$D_{k+3}$ Figure 2. This diagram illustrates the
intersection pattern of the sets $D_{k}$.

We can use this setup to form a convex set by gluing together consecutive
$D_{k}$. In the region where $D_{k}$ and $D_{k+1}$ overlap, the elements of
$D_{k}$ are slightly more dense (because the common difference of the
approximate arithmetic progression is smaller). This ensures that there exist
two consecutive elements of $D_{k}$ in this region, which allows for an
application of Lemma 2.1. Meanwhile, the existence of the non-overlapping
region ensures that this glued set contains many elements of $D_{k}$ for each
$k$.

Now we come to the formal details. We use the notation $d_{min}^{(k)}$ for the
smallest element of $D_{k}$ and $d_{max}^{(k)}$ for the largest element of
$D_{k}$. Note that

(3) $\displaystyle d_{min}^{(k)}$
$\displaystyle=d_{1}^{(k)}=k+c_{1}(2k+k^{2})+c_{2}(3k+3k^{2}+k^{3}),\text{
and}$ (4) $\displaystyle d_{max}^{(k)}$
$\displaystyle=d_{0.99n}^{(k+1)}=k+c_{1}\left(2k\frac{99n}{100}+k^{2}\right)+c_{2}\left(3\left(\frac{99n}{100}\right)^{2}k+3\frac{99n}{100}k^{2}+k^{3}\right).$

Observe that each $D_{k}$ is contained in the closed interval
$[b_{min}^{(k)},b_{max}^{(k)}]$. We will prove the following two facts about
the intersection properties of these intervals.

######  Claim 2.2.

For each $k\in[0.009n,0.01n]$, there exist $1\leq i,j\leq 0.99n$ such that

$d_{i}^{(k+1)}\leq d_{j}^{(k)}<d_{j+1}^{(k)}\leq d_{i+1}^{(k+1)}.$

######  Claim 2.3.

For each $k\in[0.009n,0.01n]$, there are at least $Cn$ elements of $D_{k}$ in
the interval $(d_{max}^{(k-1)},d_{min}^{(k+1)})$, where $C>0$ is an absolute
constant.

Once we have proved these two claims, the proof will be finished. Indeed, we
can use Claim 2.2 together with Lemma 2.1 to glue together consecutive convex
sets $D_{k}$ to form a set

$S\subset\bigcup_{k=0.009n}^{0.01n}D_{k}\subset A-A.$

Claim 2.3 guarantees that, for each $k\in[0.009n,0.01n]$, there at least $Cn$
elements in $D_{k}\cap S$ that do not appear in $D_{j}\cap S$ for any $j\neq
k$. This implies that

$|S|\geq(Cn)\cdot(0.001n)\gg n^{2}.$

It remains to prove the two claims.

###### Proof of Claim 2.2.

We will show that the interval $I=[d_{min}^{(k+1)},d_{max}^{(k)}]$ contains at
least two more elements of $D_{k}$ than it does of $D_{k+1}$. It then follows
that there must exist two consecutive elements of $D_{k}$ in this interval,
which then implies the existence of the claimed configuration

$d_{i}^{(k+1)}\leq d_{j}^{(k)}<d_{j+1}^{(k)}\leq d_{i+1}^{(k+1)}.$

We begin by establishing a lower bound for $|D_{k}\cap I|$, namely

$\displaystyle|D_{k}\cap I|$ $\displaystyle=|\\{i\in[0.99n]:d_{i}^{(k)}\geq
d_{min}^{(k+1)}\\}|$ $\displaystyle\geq
0.99n-|\\{i\in\mathbb{N}:d_{i}^{(k)}<d_{min}^{(k+1)}\\}|.$

We need an upper bound for
$|\\{i\in\mathbb{N}:d_{i}^{(k)}<d_{min}^{(k+1)}\\}|$. To this end,

$\displaystyle|\\{i\in\mathbb{N}:d_{i}^{(k)}<d_{min}^{(k+1)}\\}|$
$\displaystyle\leq 2+|\\{i>2:d_{i}^{(k)}<d_{min}^{(k+1)}\\}|$ (5)
$\displaystyle=2+|\\{i>2:2c_{1}ki+c_{2}(3i^{2}k+3ik^{2})<1+c_{1}(4k+3)+c_{2}(6k^{2}+12k+7)\\}|$
$\displaystyle\leq 2+|\\{i>2:2c_{1}ki<1+c_{1}(4k+3)\\}|$ $\displaystyle\leq
2+\frac{1+c_{1}(4k+3)}{2c_{1}k}=\frac{1+c_{1}(8k+3)}{2c_{1}k}.$

Therefore,

(6) $|D_{k}\cap I|\geq\frac{1.98n\cdot c_{1}k-8c_{1}k-1-3c_{1}}{2c_{1}k}.$

We will compare the bound in (6) with an upper bound for $|D_{k+1}\cap I|$,
which we deduce now. Observe that

$\displaystyle|D_{k+1}\cap I|$
$\displaystyle=|\\{i\in[0.99n]:d_{i}^{(k+1)}\leq d_{max}^{(k)}\\}|$
$\displaystyle=|\\{i\in\mathbb{N}:1+c_{1}(2(k+1)i+2k+1)+c_{2}(3i^{2}(k+1)+3i(k+1)^{2}+3k^{2}+3k+1)$
$\displaystyle\leq
c_{1}2k\frac{99n}{100}+c_{2}\left(3\left(\frac{99n}{100}\right)^{2}k+3\frac{99n}{100}k^{2}\right)\\}|$
$\displaystyle\leq\left|\left\\{i\in\mathbb{N}:1+c_{1}(2(k+1)i+2k+1)\leq
2c_{1}k\frac{99n}{100}+c_{2}\left(3\left(\frac{99n}{100}\right)^{2}k+3\frac{99n}{100}k^{2}\right)\right\\}\right|.$

Note that the term which involves the multiple of $c_{2}$ in the previous step
is at most $1/n^{2}$. Therefore, this term is less than $c_{1}$, which allows
us to write

$\displaystyle|D_{k+1}\cap I|$
$\displaystyle\leq\left|\left\\{i\in\mathbb{N}:1+2c_{1}(k+1)i+2c_{1}k\leq
2c_{1}k\frac{99n}{100}\right\\}\right|$
$\displaystyle=\left|\left\\{i\in\mathbb{N}:i\leq\frac{2c_{1}k\left(\frac{99n}{100}-1\right)-1}{2c_{1}(k+1)}\right\\}\right|$
(7)
$\displaystyle\leq\frac{2c_{1}k\left(\frac{99n}{100}-1\right)-1}{2c_{1}(k+1)}.$

It remains to show that the lower bound given in (6) is at least as big as the
upper bound given in (7), plus two. That is, we need to show that

$\displaystyle\frac{1.98n\cdot c_{1}k-8c_{1}k-1-3c_{1}}{2c_{1}k}$
$\displaystyle\geq
2+\frac{2c_{1}k\left(\frac{99n}{100}-1\right)-1}{2c_{1}(k+1)}$
$\displaystyle=\frac{4c_{1}(k+1)+2c_{1}k\left(\frac{99n}{100}-1\right)-1}{2c_{1}(k+1)},$

which simplifies to give

$\frac{1.98n\cdot
c_{1}k-8c_{1}k-1-3c_{1}}{k}\geq\frac{4c_{1}(k+1)+2c_{1}k\left(\frac{99n}{100}-1\right)-1}{k+1},$

and eventually

$1.98nc_{1}k\geq 10c_{1}k^{2}+15c_{1}k+3c_{1}+1.$

Since we have $0.009n\leq k\leq 0.01n$, it would be sufficient to prove that

$1.98n\cdot c_{1}\cdot 0.009n\geq
10c_{1}\cdot(0.01n)^{2}+15c_{1}\cdot(0.01n)+3c_{1}+1.$

Substituting in the definition of $c_{1}$, the previous inequality becomes

$\frac{198\cdot 75\cdot 9}{100\cdot 1000}\geq\frac{75}{1000}+\frac{15\cdot
75}{100n}+\frac{3\cdot 75}{n^{2}}+1.$

This is equivalent to the inequality

$0.2615\geq\frac{15\cdot 75}{100n}+\frac{3\cdot 75}{n^{2}},$

which is valid for $n$ sufficiently large.

∎

###### Proof of Claim 2.3.

We will give an upper bound for $|D_{k}\cap(-\infty,d_{max}^{(k-1)}]|$ and a
lower bound for $|D_{k}\cap(-\infty,d_{min}^{(k+1)})|$, and combine these
bounds to deduce the claimed lower bound for
$|D_{k}\cap(d_{max}^{(k-1)},d_{min}^{(k+1)})|$. To upper bound
$|D_{k}\cap(-\infty,d_{max}^{(k-1)}]|$, we make use of (7) and deduce that

$\displaystyle|D_{k}\cap(-\infty,d_{max}^{(k-1)}]|$
$\displaystyle=|D_{k}\cap[d_{min}^{(k)},d_{max}^{(k-1)}]|$
$\displaystyle\leq\frac{2c_{1}(k-1)\left(\frac{99n}{100}-1\right)-1}{2c_{1}k}$
$\displaystyle\leq\frac{1}{2}\left(\frac{150}{n^{2}}\cdot(0.01n)\cdot
n-1\right)\cdot\left(\frac{1000}{9n}\cdot\frac{n^{2}}{75}\right)=\frac{10n}{27}.$

Next, we present a lower bound for
$|D_{k}\cap(-\infty,d_{min}^{(k+1)})|=|D_{k}\cap[d_{min}^{(k)},d_{min}^{(k+1)})|$.
Indeed,

$\displaystyle|D_{k}\cap[d_{min}^{(k)},d_{min}^{(k+1)})|$
$\displaystyle=|\\{i\in[0.99n]:d_{i}^{(k)}<d_{min}^{(k+1)})\\}|$
$\displaystyle=|\\{i\in[0.99n]:2c_{1}ki+c_{2}(3i^{2}k+3ik^{2})<1+c_{1}(4k+3)+c_{2}(6k^{2}+12k+7)\\}|$
$\displaystyle\geq|\\{i\in[0.99n]:2c_{1}ki+c_{2}(3i^{2}k+3ik^{2})<1+c_{1}(4k+3)\\}|$
(8) $\displaystyle\geq|\\{i\in[0.99n]:2c_{1}ki<0.99+c_{1}(4k+3)\\}|.$

The last inequality above uses the fact that the term $c_{2}(3i^{2}k+3ik^{2})$
is at most $0.01$, provided that $n$ is sufficiently large. Therefore,

$\displaystyle|D_{k}\cap[d_{min}^{(k)},d_{min}^{(k+1)})|$
$\displaystyle\geq|\\{i\in\mathbb{N}:2c_{1}ki<0.99+c_{1}(4k+3)\\}|$
$\displaystyle\geq\frac{0.99+c_{1}(4k+3)}{2c_{1}k}-1$
$\displaystyle\geq\frac{0.99}{2c_{1}k}-1$
$\displaystyle\geq\frac{99}{100}\cdot\frac{2n}{3}-1=\frac{66}{100}n-1\geq\frac{65}{100}n=\frac{13}{20}n.$

By combining this inequality with (2), it follows that

$|D_{k}\cap(d_{max}^{(k-1)},d_{min}^{(k+1)})|\geq\frac{13}{20}n-\frac{10}{27}n=\frac{151}{540}n,$

as required. ∎

##  3\. Matchings

###### Proof of Theorem 1.2.

Again, write $A=\\{a_{1}<a_{2}<\dots<a_{n}\\}$ and $d_{i}=a_{i+1}-a_{i}$.
Since $A$ is convex, we have

$d_{1}<d_{2}<\dots<d_{n-1}.$

For convenience, we the shorthand use $k:=\lceil\sqrt{n}\rceil$. The matching
$M$ is given by

$\displaystyle M$
$\displaystyle=\left\\{(a_{k+1},a_{k+2}),(a_{k},a_{k+4}),(a_{k-1},a_{k+7}),(a_{k-2},a_{k+11}),\dots,\left(a_{1},a_{k+1+\frac{k(k+1)}{2}}\right)\right\\}$
$\displaystyle=\left\\{\left(a_{k+1-i},a_{k+1+\frac{i(i+1)}{2}}\right):1\leq
i\leq k\right\\}.$

The matching $M$ has cardinality $k\geq\sqrt{n}$, and the choice of parameters
ensures that it is indeed a well-defined subset of $A\times A$, provided that
$n$ is sufficiently large. To verify this, we just need to check that
$k+1+\frac{k(k+1)}{2}\leq n$, which holds comfortably for $n\geq 36$.

It remains to check that $A-_{M}A$ is convex. Let

$e_{i}:=a_{k+1+\frac{i(i+1)}{2}}-a_{k+1-i}$

denote the $i$th element of $A-_{M}A$. We need to check that
$e_{i+1}-e_{i}>e_{i}-e_{i-1}$ holds for all $2\leq i\leq k-1$. A telescoping
argument gives

$\displaystyle e_{i}$
$\displaystyle=(a_{k+2-i}-a_{k+1-i})+(a_{k+3-i}-a_{k+2-i})+\dots+\left(a_{k+1+\frac{i(i+1)}{2}}-a_{k+\frac{i(i+1)}{2}}\right)$
(9) $\displaystyle=d_{k+1-i}+d_{k+2-i}+\dots+d_{k+\frac{i(i+1)}{2}},$

and therefore

$\displaystyle e_{i+1}-e_{i}$
$\displaystyle=d_{k+1-(i+1)}+d_{k+2-(i+1)}+\dots+d_{k+\frac{(i+1)(i+2)}{2}}$
$\displaystyle-\left(d_{k+1-i}+d_{k+2-i}+\dots+d_{k+\frac{i(i+1)}{2}}\right)$
$\displaystyle=d_{k-i}+d_{k+\frac{i(i+1)}{2}+1}+d_{k+\frac{i(i+1)}{2}+2}+\dots+d_{k+\frac{(i+1)(i+2)}{2}}.$

It then follows that

$\displaystyle(e_{i+1}-e_{i})-(e_{i}-e_{i-1})$
$\displaystyle=d_{k-i}+d_{k+\frac{i(i+1)}{2}+1}+d_{k+\frac{i(i+1)}{2}+2}+\dots+d_{k+\frac{(i+1)(i+2)}{2}}$
$\displaystyle-\left(d_{k-i+1}+d_{k+\frac{i(i-1)}{2}+1}+d_{k+\frac{i(i+1)}{2}+2}+\dots+d_{k+\frac{i(i+1)}{2}}\right)$

There are $i+2$ terms with a positive sign and $i+1$ with a negative sign. We
can pair off the $i+1$ largest positive terms with smaller (in absolute value)
negative terms to conclude the proof, as follows:

$\displaystyle(e_{i+1}-e_{i})-(e_{i}-e_{i-1})$
$\displaystyle=d_{k-i}+\left(d_{k+\frac{i(i+1)}{2}+1}-d_{k+\frac{i(i+1)}{2}}\right)+\left(d_{k+\frac{i(i+1)}{2}+2}-d_{k+\frac{i(i+1)}{2}-1}\right)$
$\displaystyle+\dots+\left(d_{k+\frac{(i+1)(i+2)}{2}-1}-d_{k+\frac{i(i-1)}{2}+1}\right)+\left(d_{k+\frac{(i+1)(i+2)}{2}}-d_{k-i+1}\right)$
$\displaystyle>0.$

∎

###### Proof of Theorem 1.3.

For each $j=1,\dots,n$, define

$a_{j}:=j(2n)^{n}+(j-1)(2n)^{n-1}+\dots+2(2n)^{n-(j-2)}+(2n)^{n-(j-1)}.$

The set $A=\\{a_{j}:1\leq j\leq n\\}$ is a convex set. Indeed, the consecutive
differences of $A$ are given by

$d_{j}=a_{j+1}-a_{j}=(2n)^{n}+(2n)^{n-1}+\dots+(2n)^{n-j},$

a sequence which is strictly increasing.

Let $M\subset A\times A$ be a matching such that $A-_{M}A$ is convex. Our goal
is to prove that $|M|\ll\sqrt{n}$.

Let $k\leq n-1$ be an integer. Repeating notation used earlier in the paper,
set $d_{j}^{(k)}:=a_{j+k}-a_{j}$ and

$D_{k}=\\{d_{j}^{(k)}:1\leq j\leq n-k\\}.$

We calculate that

$\displaystyle d_{j}^{(k)}$
$\displaystyle=k[(2n)^{n}+(2n)^{n-1}+\dots+(2n)^{n-j}]$
$\displaystyle+(k-1)(2n)^{n-(j+1)}+(k-2)(2n)^{n-(j+2)}+\dots+(2n)^{n-(j+k-1)}.$

An important feature of this construction is that the diameter of the
components $D_{k}$, which is approximately $(2n)^{n-2}$, is significantly
smaller than the gaps between consecutive components, which is approximately
$(2n)^{n}$.

$D_{1}$$D_{2}$$D_{3}$ Figure 3. This diagram illustrates how the gaps between
the consecutive $D_{i}$ are significantly larger than the diameters of the
individual $D_{i}$. This is the heuristic reason why Claims 3.1 and 3.2 are
valid.

This allows us to conclude that, with at most one exception, a convex set can
have at most one representative from each $D_{k}$. This is formalised in the
following claim.

######  Claim 3.1.

Suppose that $S\subset A-A$ is convex. Then there exists at most one
$k\in\mathbb{N}$ such that $|S\cap D_{k}|\geq 2$. Moreover, if $|S\cap
D_{k_{2}}|\geq 2$ then $S\cap D_{k_{1}}=\varnothing$ for all $k_{1}<k_{2}$.

###### Proof.

The first sentence of the claim follows from the second, and so it is
sufficient to prove only the second sentence. Suppose for a contradiction that
$k_{1}<k_{2}$, $|S\cap D_{k_{1}}|\geq 1$ and $|S\cap D_{k_{2}}|\geq 2$. Let
$d_{j}^{k_{2}}$ be the smallest element of $S\cap D_{k_{2}}$. Since $S\cap
D_{k_{1}}$ is non-empty, it follows that $d_{j}^{(k_{2})}$ is not the first
element of $S$, and since $S$ also contains a larger element of $D_{k_{2}}$,
we also know that $d_{j}^{(k_{2})}$ is not the last element of $S$. Let $x$ be
the element of $S$ preceding $d_{j}^{(k_{2})}$ and let $y$ be the element of
$S$ following $d_{j}^{(k_{2})}$. By the convexity of $S$,

(10) $d_{j}^{(k_{2})}-x<y-d_{j}^{(k_{2})}.$

On the other hand

$\displaystyle d_{j}^{(k_{2})}-x\geq
d_{1}^{(k_{2})}-d_{n-(k_{2}-1)}^{(k_{2}-1)}$
$\displaystyle>k_{2}[(2n)^{n}+(2n)^{n-1}]-(k_{2}-1)[(2n)^{n}+\dots+(2n)]$
$\displaystyle=(2n)^{n}+(2n)^{n-1}-(k_{2}-1)[(2n)^{n-2}+\dots+(2n)]$
$\displaystyle>(2n)^{n}+(2n)^{n-1}-n[(2n)^{n-2}+\dots+(2n)]$ (11)
$\displaystyle\geq(2n)^{n}+(2n)^{n-1}-n\cdot 2\cdot(2n)^{n-2}=(2n)^{n}.$

The second inequality uses the fact that $k_{2}\leq n$, while the third
inequality is an application of the inequality

(12) $(2n)^{j}+(2n)^{j-1}+\dots+(2n)\leq 2\cdot(2n)^{j},$

which is valid for all $j,n\in\mathbb{N}$.

Meanwhile, since $y\in D_{k_{2}}$, a similar calculation yields

(13) $y-d_{j}^{(k_{2})}\leq d_{n-k_{2}}^{(k_{2})}-d_{1}^{(k_{2})}\leq
k_{2}[(2n)^{n-2}+(2n)^{n-3}+\dots+(2n)]\leq(2n)^{n-1}.$

Comparing (3), (13) and (10), we observe a contradiction. ∎

Using the same basic fact about the blocks $D_{k}$ again, namely that the gap
between consecutive blocks is significantly larger than their individual
diameters, we now show that the blocks which contain elements of a convex set
must occur in a weakly convex form. A set
$S=\\{s_{1}<s_{2}<\dots<s_{n}\\}\subset\mathbb{R}$ is said to be weakly convex
if the inequality

$s_{i+1}-s_{i}\geq s_{i}-s_{i-1}$

holds for all $i=2,\dots,n-1$. For a given set $S\subset A-A$, we define

$K(S)=\\{k\in\mathbb{N}:|S\cap D_{k}|\geq 1\\}.$

######  Claim 3.2.

Suppose that $S\subset A-A$ is convex. Then the set $K(S)$ is weakly convex.

###### Proof.

Suppose for a contradiction that $K(S)$ is not weakly convex. Then there
exists three consecutive elements

$d_{j_{1}}^{(k_{1})},d_{j_{2}}^{(k_{2})},d_{j_{3}}^{(k_{3})}\in S$

such that $k_{2}-k_{1}>k_{3}-k_{2}$. Since the $k_{i}$ are integers, it
follows that

(14) $2k_{2}-k_{1}-k_{3}\geq 1.$

The difference between $d_{j_{2}}^{(k_{2})}$ and $d_{j_{1}}^{(k_{1})}$ is

$\displaystyle d_{j_{2}}^{(k_{2})}-d_{j_{1}}^{(k_{1})}$ $\displaystyle\geq
d_{1}^{(k_{2})}-d_{n-k_{1}}^{(k_{1})}$
$\displaystyle>k_{2}[(2n)^{n}+(2n)^{n-1}]-k_{1}[(2n)^{n}+(2n)^{n-1}+\dots+(2n)]$
$\displaystyle=(k_{2}-k_{1})[(2n)^{n}+(2n)^{n-1}]-k_{1}[(2n)^{n-2}+(2n)^{n-3}+\dots+(2n)]$

Meanwhile, the next difference can be bounded by

$\displaystyle d_{j_{3}}^{(k_{3})}-d_{j_{2}}^{(k_{2})}$ $\displaystyle\leq
d_{n-k_{3}}^{(k_{3})}-d_{1}^{(k_{2})}$
$\displaystyle<k_{3}[(2n)^{n}+(2n)^{n-1}+\dots+(2n)]-k_{2}[(2n)^{n}+(2n)^{n-1}]$
$\displaystyle=(k_{3}-k_{2})[(2n)^{n}+(2n)^{n-1}]+k_{3}[(2n)^{n-2}+(2n)^{n-3}+\dots+(2n)].$

However, by the convexity of $S$, we also have
$d_{j_{2}}^{(k_{2})}-d_{j_{1}}^{(k_{1})}<d_{j_{3}}^{(k_{3})}-d_{j_{2}}^{(k_{2})}$.
Combining this with the previous two inequalities and applying (14) yields

$(k_{3}+k_{1})[(2n)^{n-2}+(2n)^{n-3}+\dots+(2n)]>(2k_{2}-k_{3}-k_{1})[(2n)^{n}+(2n)^{n-1}]\geq[(2n)^{n}+(2n)^{n-1}].$

We then once again use inequality (12) to obtain

(15) $(k_{3}+k_{1})\cdot 2\cdot(2n)^{n-2}>(2n)^{n}.$

Finally, note that $k_{3}+k_{1}<2n$. This holds because $k_{3},k_{1}\leq n-1$,
as the sets $D_{k_{i}}$ are only defined within this range. Plugging this into
(15), we obtain the contradiction

$2\cdot(2n)^{n-1}>(2n)^{n}.$

∎

Another useful feature of this construction is that the consecutive
differences within the components $D_{k}$ shrink rapidly, which makes it
difficult to find a large convex sets in $A-A$.

$d_{1}^{(1)}$$d_{2}^{(1)}$$d_{3}^{(1)}$$d_{4}^{(1)}$ Figure 4. In this
picture, we zoom in to take a closer look at the way the elements of $D_{k}$
are distributed (here we consider the set $D_{1}$ with $n=5$). Crucially, the
gaps between consecutive elements of $D_{k}$ shrink rapidly, with the
conseutive differences resembling a geometric progression with a small common
ratio. This picture can be used for a sketchy justification of Claims 3.3 and
3.4

We use this fact in the following claim to establish that a convex set cannot
contain more than two elements from any $D_{k}$.

######  Claim 3.3.

Suppose that $S\subset A-A$ is convex. Then

$|S\cap D_{k}|\leq 2$

holds for any $k=1,\dots,n-1$.

###### Proof.

Suppose for a contradiction that there exist three consecutive elements of $S$
belonging to the same block $D_{k}$. In particular, we have
$d_{i_{1}}^{(k)}<d_{i_{2}}^{(k)}<d_{i_{3}}^{(k)}$ satisfying

(16) $d_{i_{3}}^{(k)}-d_{i_{2}}^{(k)}>d_{i_{2}}^{(k)}-d_{i_{1}}^{(k)}.$

Then, since $i_{2}\geq i_{1}+1$, we have

$d_{i_{2}}^{(k)}-d_{i_{1}}^{(k)}\geq
d_{i_{2}}^{(k)}-d_{i_{2}-1}^{(k)}=(2n)^{n-i_{2}}+(2n)^{n-(i_{2}+1)}+\dots+(2n)^{n-(i_{2}+k-1)}>(2n)^{n-i_{2}}.$

On the other hand,

$\displaystyle d_{i_{3}}^{(k)}-d_{i_{2}}^{(k)}\leq
d_{n-k}^{(k)}-d_{i_{2}}^{(k)}$
$\displaystyle<k[(2n)^{n-(i_{2}+1)}+\dots+(2n)]$ $\displaystyle\leq k\cdot
2\cdot(2n)^{n-(i_{2}+1)}$ $\displaystyle<n\cdot
2\cdot(2n)^{n-(i_{2}+1)}=(2n)^{n-i_{2}}.$

Combining the previous two inequalities with (16), we obtain the intended
contradiction

∎

By proving Claims 3.1 and 3.3, we have essentially proved that each block of
$D_{k}$ in $A-A$ can contain at most one element of a convex set $S\subset
A-A$. We can be a little more precise; taking the potential exceptional block
into account, we have the bound

(17) $|S|\leq|K(S)|+1.$

It remains to upper bound the size of the indexing set $K(S)$. We need one
more claim to allow us to achieve this goal. Note that the following claim
represents the first time in the proof where we use the fact that the convex
set $S$ is derived from a matching.

######  Claim 3.4.

Suppose that $M\subset A\times A$ is a matching and that $S=A-_{M}A$ is a
convex set. Then the indexing set $K(S)$ does not contain four consecutive
elements which form an arithmetic progression.

###### Proof.

Suppose for a contradiction that four consecutive elements of $K(S)$ form an
arithmetic progression. It then follows from Claim 3.1 that there exist four
consecutive elements $d_{j_{1}}^{(k)},d_{j_{2}}^{(k+t)},d_{j_{3}}^{(k+2t)}$
and $d_{j_{4}}^{(k+3t)}$ in $S$, for some positive integers $k,t$ such that
$k+3t\leq n-1$. Since $S$ is derived from a matching, it must be the case that
the $j_{i}$ are pairwise distinct. Write

$\displaystyle e_{1}$ $\displaystyle=d_{j_{2}}^{(k+t)}-d_{j_{1}}^{(k)}$
$\displaystyle e_{2}$ $\displaystyle=d_{j_{3}}^{(k+2t)}-d_{j_{2}}^{(k+t)}$
$\displaystyle e_{3}$ $\displaystyle=d_{j_{4}}^{(k+3t)}-d_{j_{2}}^{(k+2t)}.$

Since $S$ is convex, we have $e_{1}<e_{2}<e_{3}$.

We will now show that it must be the case that $j_{2}<j_{1}$. Suppose for a
contradiction that this is not true, and so $j_{2}>j_{1}$. Then

$e_{1}=d_{j_{2}}^{(k+t)}-d_{j_{1}}^{(k)}>t[(2n)^{n}+\dots+(2n)^{n-j_{1}}]+(t+1)(2n)^{n-(j_{1}+1)}.$

On the other hand

$\displaystyle e_{2}=d_{j_{3}}^{(k+2t)}-d_{j_{2}}^{(k+t)}$ $\displaystyle\leq
d_{n-(k+2t)}^{(k+2t)}-d_{j_{2}}^{(k+t)}$
$\displaystyle<t[(2n)^{n}+\dots+(2n)^{n-j_{2}}]+(k+2t)[(2n)^{n-(j_{2}+1)}+\dots+(2n)].$

It follows from the previous two bounds that

$\displaystyle e_{1}-e_{2}$
$\displaystyle>(2n)^{n-(j_{1}+1)}-(k+2t)[(2n)^{n-(j_{1}+2)}+\dots+(2n)]$
$\displaystyle\geq(2n)^{n-(j_{1}+1)}-(k+2t)\cdot 2\cdot(2n)^{n-(j_{1}+2)}$
$\displaystyle>(2n)^{n-(j_{1}+1)}-n\cdot 2\cdot(2n)^{n-(j_{1}+2)}=0$

This is a contradiction, and we have thus established that $j_{2}<j_{1}$. The
exact same argument implies that $j_{3}<j_{2}$.

Now, since $j_{2}<j_{1}$, we have

$\ e_{1}=d_{j_{2}}^{(k+t)}-d_{j_{1}}^{(k)}\geq
t[(2n)^{n}+\dots+(2n)^{n-j_{2}}]-k[(2n)^{n-(j_{2}+1)}+\dots+(2n)],$

Similarly, since $j_{3}<j_{2}$, it follows that

$e_{2}=d_{j_{3}}^{(k+2t)}-d_{j_{2}}^{(k+t)}\leq
t[(2n)^{n}+\dots+(2n)^{n-j_{3}}]+(t-1)[(2n)^{n-(j_{3}+1)}+\dots+(2n)].$

Combining the previous two inequalities, and again making use of the fact that
$j_{3}<j_{2}$, we have

$\displaystyle e_{1}-e_{2}$
$\displaystyle\geq(2n)^{n-(j_{3}+1)}-(k+t-1)[(2n)^{n-(j_{3}+2)}+\dots+(2n)]$
$\displaystyle\geq(2n)^{n-(j_{3}+1)}-(k+t-1)\cdot 2\cdot(2n)^{n-(j_{3}+2)}$
$\displaystyle>(2n)^{n-(j_{3}+1)}-n\cdot 2\cdot(2n)^{n-(j_{3}+2)}=0.$

This contradicts the fact that $e_{1}<e_{2}$ and completes the proof of the
claim. ∎

When we combine Claim 3.2 and Claim 3.4, we see that the set $K$ is a weakly
convex subset of $\\{1,\dots,n\\}$ which does not contain four consecutive
terms in an arithmetic progression. It follows that

$|K|\ll\sqrt{n}.$

Combining this with (17), the proof is complete.

∎

##  4\. Concluding remarks; sums instead of differences

The problems considered in this paper were partly motivated by a potential
application to a problem in discrete geometry concerning the minimum number of
angles determined by a set of points in the plane in general position. This
problem was considered recently in [2], and similar problems can be traced
back to the work of Pach and Sharir [5]. We found that progress on this
question could be given by a solution to the following problem: given a convex
set $A\subset\mathbb{R}$ estimate the size of the largest matching on $A$
which gives rise to a convex set in the image set $f(A,A)$, where
$f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ is a specific bivariate
function whose rather complicated formula is omitted here. Theorems 1.2 and
1.3 of this paper solve this problem for this simplified case when
$f(x,y)=x-y$.

With the potential application to the distinct angles problem in mind, an
interesting future research direction could be to generalise the problems
considered in this paper by considering an arbitrary
$f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ in place of the function
$f(x,y)=x-y$. We conclude this paper with some remarks about the most natural
case, whereby $f(x,y)=x+y$.

It is interesting to see that we can quite easily obtain an optimal result,
giving a significant quantitative improvement to Theorem 1.2, if we consider
sums instead of differences, as follows.

######  Theorem 4.1.

Let $n\in\mathbb{N}$ and suppose that $A\subset\mathbb{R}$ is a convex set
with cardinality $n$. Then there exists a matching $M\subset A\times A$ such
that $|M|\geq\lfloor\frac{n}{2}\rfloor$ and $A+_{M}A$ is convex.

###### Proof.

Suppose that $n$ is even and consider the matching

$M=\\{(a_{1},a_{n/2+1}),(a_{2},a_{n/2+2}),...,(a_{n/2},a_{n})\\}.$

Then the set

$A+_{M}A=\\{a_{n/2+k}+a_{k}:k\in\\{1,\dots,n/2\\}\\}$

is convex. If $n$ is odd then we omit $a_{n}$ and use the same argument as
above. ∎

In particular, it follows from Theorem 4.1 that an analogue of the
construction in the proof of Theorem 1.3 is not possible if we take sums
instead of differences. There are other cases in which problems concerning
additive properties of convex sets are sensitive to sums and differences. For
instance, a construction in [4] (see also [7]) shows that there exists a
convex set $A\subset\mathbb{R}$ and $x\in A-A$ with

$|\\{(a,b)\in A:a-b=x\\}|\gg|A|.$

On the other hand, incidence geometry can be used to give the upper bound

$|\\{(a,b)\in A:a+b=x\\}|\ll|A|^{2/3}$

for any convex $A\subset\mathbb{R}$ and $x\in A+A$.

We would also be interested to know whether Theorem 1.1 is still valid when
$A-A$ is replaced by $A+A$. We were unable to prove anything non-trivial for
this question.

## Acknowledgements

Krishnendu Bhowmick and Oliver Roche-Newton were supported by the Austrian
Science Fund FWF Project P 34180. Ben Lund was supported by the Institute for
Basic Science (IBS-R029-C1). Part of this work was carried out at Vietnam
Institute for Advanced Study in Mathematics (VIASM) in Hanoi, and we thank
VIASM for their hospitality and the excellent working conditions. We also
thank Eyvindur Ari Palsson, Steven Senger and Audie Warren for helpful
discussions.

## References

  * [1] G. Elekes, M. Nathanson and I. Ruzsa, ‘Convexity and sumsets’, J. Number Theory. 83 (1999), 194-201.
  * [2] H. L. Fleischmann, S. V. Konyagin, S. J. Miller, E. A. Palsson, E. Pesikoff and C. Wolf ‘Distinct angles in general position’, Discrete Math. 346 (2023), no. 4, Paper No. 113282, 4pp.
  * [3] B. Hanson, O. Roche-Newton and M. Rudnev, ‘Higher convexity and iterated sum sets’, Combinatorica 42 (2022), no. 1, 71-85.
  * [4] O. Roche-Newton and A. Warren, ‘A convex set with a rich difference’, Acta Math. Hungar. 168 (2022), no. 2, 587-592.
  * [5] J. Pach and M. Sharir, ‘Repeated angles in the plane and related problems’, J. Combin. Theory Ser. A 59 (1992), no. 1, 12-22.
  * [6] I. Z. Ruzsa and D. Zhelezov, ‘Convex sequences may have thin additive bases’, Mosc. J. Comb. Number Theory 8 (2019), no. 1, 43-46.
  * [7] T. Schoen, ‘On Convolutions of Convex Sets and Related Problems’, Canad. Math. Bull. 57 (2014), no. 4, 877-883.
  * [8] T. Schoen and I. Shkredov, ‘On sumsets of convex sets’, Combin. Probab. Comput. 20 (2011), no. 5, 793-798.
  * [9] S. Stevens and A. Warren, ‘On sum sets and convex functions’, Electron. J. Combin. 29 (2022), no. 2, Paper No. 2.18, 19 pp.


# Energy Loss of Monopolium in a Medium

Huner Fanchiotti  Carlos A. García Canal  IFLP(CONICET) and Department of
Physics, University of La Plata, C.C. 67 1900, La Plata, Argentina  Vicente
Vento  Departamento de Física Teórica and IFIC, Universidad de Valencia -
CSIC, E-46100 Burjassot (Valencia), Spain.

###### Abstract

We study the energy loss of excited monopolium in an atomic medium. We perform
a classical calculation in line with a similar calculation performed for
charged particles which leads in the non relativistic limit to the Bethe-Bloch
formula except for the density dependence of the medium, which we do not
consider in this paper. Our result shows that for maximally deformed Rydberg
states the ionization of monopolium in a light atomic medium is similar to
that of light ions.

##  I Introduction

Inspired by the old idea of Dirac and Zeldovich Dirac:1931qs ;
Zeldovich:1978wj  that monopoles are not seen free because they are confined
by their strong magnetic forces, we have studied a monopole-antimonopole bound
state that we have called monopolium Vento:2007vy . This state is the strongly
coupled dual analog of positronium and decays into photons Epele:2012jn . The
discovery of monopole and dipole solutions in Kaluza Klein theories
Kaluza:1921tu ; Klein:1926tv  made these theories very exciting from a
theoretical point of view Gross:1983hb . In particular the dipole solution,
which is classically stable and therefore very long lived, forms a very
interesting state which we have also called monopolium in analogy with its
decaying cousin in gauge theories. The Kaluza Klein monopolium is extremely
massive with a mass of the order of the Planck mass and therefore impossible
to produce in laboratories. However, there might be clouds of them in the
cosmos which might enter our detectors Vento:2020vsq .

Much experimental research has been carried out into the search for monopoles
Cabrera:1982gz ; Milton:2006cp ; MoEDAL:2014ttp ; Acharya:2014nyr ;
Patrizii:2015uea . However, our interest lies in the detection of monopolium,
which is chargeless but can manifest a magnetic moment in the presence of
magnetic fields (dual Stark effect) in excited deformed states Vento:2020vsq ;
Vento:2019auh ; Fanchiotti:2022xvx . In this work we want to analyze the
capacity of monopolium to produce ionization when traversing a medium by
calculating its energy loss when colliding with atomic electrons. The idea
behind this investigation is to find out if TPCs Nygren:2018sjx  could in
principle be used for detecting monopolium.

We start by reviewing in section II a classical calculation of the energy loss
by a heavy charged particle, a well known problem, that will serve to present
the formalism that will be later used in the case of monopolium. This
classical result provides some of the dependences that describe the stopping
power of a heavy charged particle in a medium. Thereafter, in section III we
follow the same procedure to calculate the stopping power for a heavy particle
with no charge but with magnetic moment. We next compare the result for the
charged particle and neutral magnetic by establishing a ratio of energy
losses. In section IV we apply the described analysis to monopolium. We end up
by some concluding remarks.

##  II Energy Loss by a heavy charged particle traversing a medium

A heavy charged particle traversing a medium interacts electromagnetically
mainly with the atomic electrons in the medium. In the process it transfers
energy to them ionizing the atoms of the medium. The quantum mechanical
calculation is quite complicated and leads to the famous Bethe-Bloch formula
Bethe:1930ku ; Bloch:1933zza . However, since we would like to compare this
case with the case of monopolium we will resorts to the much simpler classical
calculation which shows the main dependencies of the process Foudas:2007cu ;
Ahlen:1976jw ; Ahlen:1982mx .

Let us solve the problem of calculating the energy transfer from a heavy
charged particle that moves with velocity $v<<c$ towards a stable electron
(see Fig. 1). Since we want to calculate the energy loss $dE/dz$ the distance
traveled $dz$ is small, thus to keep $v$ constant is a good approximation.

Figure 1: Heavy charged particle traveling with velocity $v$ along the $z$
direction passing at impact parameter $b$ from an atomic electron.

The force acting on the electron is

$\vec{F}=\frac{Qe\hat{r}}{|\vec{r}(t)|^{2}}=F_{x}\hat{x}+F_{z}\hat{z}.$ (1)

Due to the symmetry of the problem $F_{z}$ cancels out as shown in Fig. 1 and
$F_{x}$ is

$F_{x}=\frac{Qe\sin{\theta}}{|\vec{r}(t)|^{2}}=\frac{Qeb}{|\vec{r}(t)|^{3}}=\frac{Qeb}{(b^{2}+(vt)^{2})^{\nicefrac{{3}}{{2}}}},$
(2)

where $\theta$ is the angle between $\vec{F}$ and the $z$ axis. From $F_{x}$
we get the velocity

$v_{x}=\frac{Qe}{m}\int_{-\infty}^{-\infty}\frac{b}{(b^{2}+(vt)^{2})^{\nicefrac{{3}}{{2}}}}dt=\frac{2Qe}{mvb},$
(3)

where $v=|\vec{v}|$.

The energy gained by the electron is

$\Delta E=\frac{1}{2}mv_{x}^{2}=\frac{2Q^{2}e^{2}}{mv^{2}b^{2}}.$ (4)

In order to obtain the contribution of all the electrons interacting with the
heavy particle we need to consider the cylindrical volume in Fig. 2.

$dV=2\pi bdbdz.$ (5)

Note that although we have calculated the interaction in the plane $x-z$ the
interaction has cylindrical symmetry and the active force goes into the radial
direction in all planes. In order to calculate the total energy loss we have
to take into account all the electrons per unit volume and multiply by N the
electron number density.

Figure 2: The elementary Integration volume to obtain the interaction with all
the electrons

Thus the total energy loss is this given by

$\frac{dE}{dz}=2\pi\frac{Q^{2}e^{2}N}{mv^{2}}\log{\frac{b_{max}}{b_{min}}}$
(6)

Given the relation between the impact parameter and the energy, Eq.(4), we
obtain for the energy loss

$\frac{dE}{dz}=2\pi\frac{Q^{2}e^{2}N}{mv^{2}}\log{\frac{E_{max}}{E_{min}}}.$
(7)

The maximum energy to leading order in $(m/M)$ is given by Ahlen:1976jw

$E_{max}=\frac{2mc^{2}\beta^{2}}{1-\beta^{2}}$ (8)

where $\beta=v/c$, and therefore

$\frac{dE}{dz}=2\pi\frac{Q^{2}e^{2}N}{mc^{2}\beta^{2}}\log{\frac{2m\beta^{2}}{(1-\beta^{2})E_{min}}}.$
(9)

which agrees with the more sophisticated quantum calculation by Ahlen
Ahlen:1976jw  for small velocity $Z/\beta>137$. This result is an
approximation to the more complete Bethe-Bloch Bethe:1930ku ; Bloch:1933zza
relation which includes the effects of atomic or molecular binding and
relativistic corrections,

$\frac{dE}{dz}=2\pi\frac{Q^{2}e^{2}N}{mc^{2}\beta^{2}}\left(\log{\frac{2m\beta^{2}}{(1-\beta^{2})E_{min}}}-\beta^{2}-\delta_{e}/2\right),$
(10)

where $\delta_{e}$ characterizes the density effect and $E_{min}$ is the mean
ionization potential which can be approximated by $10Z$ eV Foudas:2007cu ;
Ahlen:1976jw , where $Z$ is the atomic number of the atom in the medium . In
this initial attempt our aim is to look at the behavior of the first term
which is dominant for very small velocity and disregard the density
dependence. Our comparison will be valid therefore only in the kinematical
region where the first term of Eq.(10) is dominant.

Note that we use $4\pi\epsilon_{0}=1$, $\mu_{0}/4\pi=1$ and therefore $c=1$.
Sometimes though we shall keep explicitly $c$ during the calculation process,
to respect the notation of the classical equations, although at the end we
will obtain all our equations in natural units.

##  III Energy Loss by heavy neutral particles with magnetic moment in a
medium

Our interest lies in the possibility of detecting monopolium Vento:2007vy , a
bound monopole anti-monopole state, by its ionization. Monopolium is neutral
and is heavy, but in its excited states it has a magnetic moment in the
presence of even weak magnetic fields Vento:2020vsq ; Fanchiotti:2022xvx .
Heavy neutral particles with magnetic moment (this work) or electric dipole
moment PhysRevA.65.022902  interact electromagnetically mainly with the atomic
electrons in the medium and transfer energy to them in a process which results
in ionizing the atoms of the medium. Calculating in this case the energy loss
in a medium using quantum theory is quite complicated, however we have just
shown that for small velocities the classical calculation agrees well with the
quantum mechanical one. The classical calculation demonstrates the origin of
the main features and dependencies of the energy loss. Let us proceed
therefore by developing the classical dynamics of the process as shown in Fig.
1 changing $(Q,M)$ by $(\overrightarrow{\mathcal{M}},M)$ and $(-e,m)$ by
$(-e,\vec{\mu},m)$.

We begin by describing the dynamics of the collision between a moving neutral
heavy particle with magnetic moment and an atomic electron. We assume a
neutral heavy particle with magnetic moment $\overrightarrow{\mathcal{M}}$
moving with velocity $\vec{v}$ along the $z$ direction. In its own frame of
reference it will create a magnetic field

$\displaystyle\overrightarrow{B_{M}}=$
$\displaystyle\left(\frac{3\vec{r}\cdot\overrightarrow{\mathcal{M}}\,\vec{r}}{r^{5}}-\frac{\overrightarrow{\mathcal{M}}}{r^{3}}\right),$
(11)

The electron has charge $-e$ and magnetic moment $\vec{\mu}$ therefore in its
own frame creates an electric field

$\overrightarrow{E_{e}}=-\frac{e}{r^{2}}\,\hat{r}$ (12)

and a magnetic field

$\displaystyle\overrightarrow{B_{e}}=$
$\displaystyle\left(\frac{3\vec{r}\cdot\vec{\mu}\,\vec{r}}{r^{5}}-\frac{\vec{\mu}}{r^{3}}\right).$
(13)

In the frame of the electron the heavy particle has both electric and magnetic
fields as dictated by the Lorentz transformation from the heavy particle
reference system $S$ to the electron reference system $S^{\prime}$, which is
given by

$\displaystyle E^{\prime}_{Mz}$ $\displaystyle=0$ $\displaystyle
E^{\prime}_{Mx}$ $\displaystyle=-\gamma\beta B_{My}$ $\displaystyle
E^{\prime}_{My}$ $\displaystyle=\gamma\beta B_{Mx},$ (14)

and a magnetic field

$\displaystyle B^{\prime}_{Mz}$ $\displaystyle=B_{Mz}$ $\displaystyle
B^{\prime}_{Mx}$ $\displaystyle=\gamma B_{Mx}$ $\displaystyle B^{\prime}_{My}$
$\displaystyle=\gamma B_{My},$ (15)

where $\gamma=1/\sqrt{1-\beta^{2}}$. In the following subsections we calculate
the contribution to the energy loss of a heavy moving particle with magnetic
moment taking into account the interaction of the electron charge with this
electric field and that of the electron magnetic moment with the magnetic
field.

We expect slow particles to ionize more like in the charge case thus a non-
relativistic limit in the case of a very heavy particle is reasonable. From
Eqs.(14) and (15) it would seem that for low velocities the interaction of the
magnetic moment of the electron $\vec{\mu}$ with the magnetic field of the
heavy particle Eq.(15), $\overrightarrow{B_{M}}$ is $\beta$ dominant. We will
see that this only happens for extremely low velocities due to the dependence
on the distance of the interactions.

###  III.1 Energy Loss produced by the Magnetic field interaction

Let us start by calculating the energy loss of the magnetic field interaction.
The interaction in this case is given by

$H=-\vec{\mu}\cdot\overrightarrow{B_{M}}$ (16)

and following the classical description of the previous section we have a
force acting on the electron,

$\overrightarrow{F_{e}}=-\vec{\nabla}H=\vec{\nabla}\left(\frac{3\vec{r}\cdot\overrightarrow{{\mathcal{M}}}\vec{r}\cdot\vec{\mu}}{r^{5}}-\frac{\overrightarrow{{\mathcal{M}}}\cdot\vec{\mu}}{r^{3}}\right).$
(17)

Thus

$F_{ei}=3\frac{{\mathcal{M}}_{i}\vec{r}\cdot\vec{\mu}+\mu_{i}\vec{r}\cdot\overrightarrow{{\mathcal{M}}}}{r^{5}}-15\frac{r_{i}\vec{r}\cdot\overrightarrow{{\mathcal{M}}}\vec{r}\cdot\vec{\mu}}{r^{7}}+5\frac{r_{i}\overrightarrow{{\mathcal{M}}}\cdot\vec{\mu}}{r^{5}}.$
(18)

where $i=x,y,z$ and $r_{x}=x,r_{y}=y$ and $r_{z}=z$.

Fig.1 shows also now that $F_{ez}$ will vanish and therefore we are left with
$F_{ex}$ and $F_{ey}$, and with the corresponding velocities $v_{ex}$ and
$v_{ey}$.

Recall that $r=\sqrt{b^{2}+(vt)^{2}}$ and
$\vec{r}=vt\hat{k}+b\hat{\rho}=vt\hat{k}+b\cos{\phi}\hat{i}+b\sin{\phi}\hat{j}$,
where $\hat{i},\hat{j},\hat{k}$ form the cartesian unit basis, $\phi$ is the
polar angle in the $x,y$ plane and we use $c\neq 1$ until further notice. Let
us calculate the velocity in an arbitrary point in the circle of radius $b$ in
the $x-y$ plane $(b\cos{\phi},b\sin{\phi})$, whose components in the plane are
given by

$\displaystyle mv_{ex}=$ $\displaystyle\int_{\infty}^{\infty}F_{ex}dt$
$\displaystyle mv_{ey}=$ $\displaystyle\int_{\infty}^{\infty}F_{ey}dt$ (19)

All the integrals can be easily performed and the results is

$\displaystyle v_{ex}$ $\displaystyle=$
$\displaystyle\frac{1}{4mvb^{3}}((2{\mathcal{M}}_{x}\mu_{x}-{\mathcal{M}}_{z}\mu_{z}+\frac{5}{3}\overrightarrow{{\mathcal{M}}}\cdot\vec{\mu})\cos{\phi}+({\mathcal{M}}_{x}\mu_{y}+{\mathcal{M}}_{y}\mu_{x})\sin{\phi}$
(20)
$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;-4({\mathcal{M}}_{y}\mu_{y}\cos{\phi}\sin^{2}{\phi}+({\mathcal{M}}_{x}\mu_{y}+{\mathcal{M}}_{y}\mu_{x})\cos^{2}{\phi}\sin{\phi}+{\mathcal{M}}_{x}\mu_{x}\cos^{3}{\phi}))=\frac{4{\mathcal{M}}\mu_{b}}{mvb^{3}}\Phi_{\mu
x},$ $\displaystyle v_{ey}$ $\displaystyle=$
$\displaystyle\frac{1}{4mvb^{3}}((2{\mathcal{M}}_{y}\mu_{y}-{\mathcal{M}}_{z}\mu_{z}+\frac{5}{3}\overrightarrow{{\mathcal{M}}}\cdot\vec{\mu})\sin{\phi}+({\mathcal{M}}_{x}\mu_{y}+{\mathcal{M}}_{y}\mu_{x})\cos{\phi}$
(21)
$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;-4({\mathcal{M}}_{y}\mu_{y}\sin^{3}{\phi}+({\mathcal{M}}_{x}\mu_{y}+{\mathcal{M}}_{y}\mu_{x})\cos{\phi}\sin^{2}{\phi}+{\mathcal{M}}_{x}\mu_{x}\cos^{2}{\phi}\sin{\phi}))=\frac{4{\mathcal{M}}\mu_{b}}{mvb^{3}}\Phi_{\mu
y}.$

These equations define $\Phi_{\mu x}$ and $\Phi_{\mu y}$ which enter the
deposited energy as

$E=\frac{8{\mathcal{M}}^{2}\mu_{B}^{2}}{mv^{2}b^{6}}(\Phi_{\mu
x}^{2}+\Phi_{\mu y}^{2})$ (22)

where $\mu_{B}$ is Bohr’s magneton and $\Phi_{\mu i}^{2},i=x,y$ are the
angular dependences determined from $v_{ei}$ by substituting
$\overrightarrow{{\mathcal{M}}}={\mathcal{M}}(\sin{\theta_{{\mathcal{M}}}}\cos{\phi_{{\mathcal{M}}}},\sin{\theta_{{\mathcal{M}}}}\sin{\phi_{{\mathcal{M}}}},\cos{\theta_{{\mathcal{M}}}})$
and
$\vec{\mu}=\mu_{B}(\sin{\theta_{\mu}}\cos{\phi_{\mu}},\sin{\theta_{\mu}}\sin{\phi_{\mu}},\cos{\theta_{\mu}})$.
Note that we are taking the maximum component of the magnetic moments in
complete arbitrary directions.

We have to add now the contribution from all electrons. Given the $\phi$
dependence our elementary volume, shown in Fig.2 is now

$dV=bdbd\phi dz$ (23)

and we have to integrate over $0<\phi\leq 2\pi$ to obtain the contribution of
the full volume.

Since the calculated energy loss will be an average value, we also average
over the electron magnetic moment angles. Let us call the average magnetic
polarization function in terms of the magnetic polarization of the incoming
heavy particle,

$\Phi_{\mu}^{2}(\theta_{{\mathcal{M}}},\phi_{{\mathcal{M}}})=\frac{1}{2\pi^{2}}\int_{0}^{\pi}d\theta_{e}\int_{0}^{2\pi}d\phi_{e}\int_{0}^{2\pi}d\phi(\Phi_{\mu
x}^{2}+\Phi_{\mu y}^{2}),$ (24)

which is shown in Fig.3.

We have to take into account the number of electrons. Recall discussion around
Eqs.(5) to 6. Thus, our estimate for the deposited energy as a function of the
incoming magnetic polarization is

$\frac{dE}{dz}(\theta_{{\mathcal{M}}},\phi_{{\mathcal{M}}})=\frac{{\mathcal{M}}^{2}\mu_{B}^{2}N}{2mv^{2}}\Phi_{\mu}^{2}(\theta_{{\mathcal{M}}},\phi_{{\mathcal{M}}})\int_{b_{min}}^{b_{max}}\frac{db}{b^{5}}$
(25)

where $N$ is the electron number density.

Figure 3: The averaged magnetic polarization function in terms of the magnetic
polarization orientation of the incoming heavy particle.

In order to get a feasibility study of detection of magnetic particles it is
more convenient instead of calculating the deposited energy, to calculate the
ratio of the deposited energy of a heavy neutral magnetic particle to that of
a heavy charged particle, namely

$Ratio=\frac{\mbox{magnetic
moment}}{\mbox{charge}}=\frac{{\mathcal{M}}^{2}\mu_{B}^{2}\Phi_{\mu}^{2}}{2\pi
Q^{2}e^{2}}\;\frac{\frac{1}{b^{4}_{max}}-\frac{1}{b^{4}_{min}}}{\log{\frac{E_{max}}{E_{min}}}}.$
(26)

where $Q=Z_{H}e$ is the charge of the heavy particle. The impact parameter can
be calculated in terms of the lab energy for one electron from Eq.(22)

$b=\left(\frac{{8\mathcal{M}}^{2}\mu_{B}^{2}\left(\frac{\Phi_{\mu}^{2}}{2\pi}\right)}{mv^{2}E}\right)^{\nicefrac{{1}}{{6}}}.$
(27)

We average over $\phi$ to get an estimate in the equation of one electron
which leads to $\frac{\Phi_{\mu}^{2}}{2\pi}$ as written in the previous
equation explicitly. The maximum energy to leading order in $m/M$ is
$E_{max}=2mv^{2}/(1-v^{2}/c^{2})$. For $E_{min}$ it is customary to choose
$10Z$ eV Foudas:2007cu ; Ahlen:1976jw , where $Z$ is the atomic number of the
medium. In all shown results we shall use as a medium liquid Argon whose
$Z=18$. Except for clarification purposes we use $\beta=v/c$ for the velocity
and $mc^{2}$ for the electron mass.

###  III.2 Energy Loss produced by the Electric field interaction

In this case the force $\vec{F}_{E}=q\vec{E}^{\prime}$ becomes the
conventional Lorentz force

$\vec{F}_{E}=e\beta\gamma B_{y}\hat{i}-e\beta\gamma
B_{x}\hat{j}\rightarrow\frac{e}{c}\vec{v}\times\vec{B},$ (28)

in the low velocity limit.

Explicitly the forces become

$\displaystyle F_{Ex}$ $\displaystyle=$ $\displaystyle
e\beta\gamma\left(3\frac{\vec{r}\cdot{\overrightarrow{\mathcal{M}}r_{y}}}{r^{5}}-\frac{{\mathcal{M}}_{y}}{r^{3}}\right),$
(29) $\displaystyle F_{Ex}$ $\displaystyle=$
$\displaystyle-e\beta\gamma\left(3\frac{\vec{r}\cdot{\overrightarrow{\mathcal{M}}r_{x}}}{r^{5}}-\frac{{\mathcal{M}}_{x}}{r^{3}}\right).$
(30)

The velocities are obtained by integrating the accelerations and lead to

$\displaystyle v_{ex}$ $\displaystyle=$
$\displaystyle\frac{2e\gamma}{mb^{2}}(2({\mathcal{M}}_{x}\cos\phi-{\mathcal{M}}_{y}\sin\phi)\sin\phi-{\mathcal{M}}_{y})=\frac{2e\gamma{\mathcal{M}}}{mb^{2}}\Phi_{ex},$
(31) $\displaystyle v_{ey}$ $\displaystyle=$
$\displaystyle\frac{2e\gamma}{mb^{2}}(2({\mathcal{M}}_{x}\cos\phi-{\mathcal{M}}_{y}\sin\phi)\cos\phi-{\mathcal{M}}_{x})=-\frac{2e\gamma{\mathcal{M}}}{mb^{2}}\Phi_{ey}.$
(32)

Therefore

$E=\frac{2e^{2}\gamma^{2}{\mathcal{M}}^{2}}{mb^{4}}(\Phi_{ex}^{2}+\Phi_{ey}^{2}))=\frac{2e^{2}\gamma^{2}{\mathcal{M}}^{2}}{mb^{4}}\Phi_{e}^{2}.$
(33)

This equation defines the as before the impact parameter $b$ in terms of $E$.
Integrating over $\phi$ and $bdb$ we get

$Ratio=\frac{\mbox{magnetic
moment}}{\mbox{charge}}=\frac{e^{2}{\mathcal{M}}^{2}\gamma^{2}\beta^{2}\Phi_{e}^{2}}{4\pi
Q^{2}e^{2}}\;\frac{\frac{1}{b^{2}_{max}}-\frac{1}{b^{2}_{min}}}{\log{\frac{E_{max}}{E_{min}}}},$
(34)

and

$b=\left(\frac{2e^{2}\gamma^{2}{\mathcal{M}}^{2}\left(\frac{\Phi_{e}^{2}}{2\pi}\right)}{mE}\right)^{\nicefrac{{1}}{{4}}}$
(35)

The integrated $\Phi_{e}(\phi_{\mathcal{M}},\theta_{\mathcal{M}})^{2}$ is
shown in Fig. 4.

Figure 4: The averaged polarization function in terms of the polarization
orientation of the incoming heavy particle.

We next study the two interactions as a functions of $\beta$. We take the
actual values for the electron parameters $\mu_{B}=0.0836$ MeV-1,
$m=0.511$MeV, $q=-e=-\sqrt{1/137}$ and plot the ratio for three magnetic
moments of the heavy particles. We average over $\Phi_{\mathcal{M}}$ and
$\theta_{\mathcal{M}}$ getting for the average values of
${\Phi_{\mu}}^{2}=2.618$ and of ${\Phi_{e}}^{2}=\pi$. The result is shown in
Fig. 5. We see that for the higher magnetic moments the charge magnetic moment
interaction dominates due to its softer radial dependence for almost all the
range of $\beta$ studied. For the lower magnetic moments, at the level of
nuclear magnetons, the magnetic moments interaction dominates due to the
velocity dependence. The $Z_{H}$ dependence is easy to understand it simply
divides the Ratio by $Z_{H}^{2}$.

Figure 5: We plot the ratio of the two magnetic moment interactions with
respect to a conventional charge interaction, for the proton, as a function of
velocity $\beta$ for three magnetic moments of the heavy particle. The upper
curves correspond to $2000\mu_{B}$, the middle curves for $\mu_{B}$ and the
lower curves for $\mu_{B}/2000$, which is approximately the value of the
nuclear magneton. The solid line signals the magnetic moments interaction and
the dotted line the charge magnetic moment interaction. Figure 6: We plot the
ratio of the two magnetic moment interactions for the neutron with respect to
a conventional charge interaction, for the alpha particle, as a function of
velocity $\beta$. The solid line corresponds to the magnetic moments
interaction and the dotted line to the charge magnetic moment interaction.

As an application of our calculation to a well known systems we plot the
comparison between the ratio of the energy loss of a neutron to that of an
alpha particle in Fig.6. In this case the magnetic moment interaction
dominates over the whole range of velocities studied due to the smallness of
the nuclear magneton .

###  III.3 Energy Loss associated with the full interaction

The aim of this work is to analyze the energy loss of monopolium and for that
purpose we are going to implement the full interaction. The procedure consists
in adding all forces acting of the heavy particle and constructing from them
the velocity which leads immediately to the lost energy.

The total velocity requires the addition of Eqs.(21) and (32) and can be
written as

$\displaystyle v_{ex}$ $\displaystyle=$
$\displaystyle\frac{4{\mathcal{M}}\mu_{b}}{\beta
b^{3}m}\Phi^{\mu}_{x}+\frac{2e\gamma{\mathcal{M}}}{mb^{2}}\Phi^{e}_{x}$ (36)
$\displaystyle v_{ex}$ $\displaystyle=-$
$\displaystyle\frac{4{\mathcal{M}}\mu_{b}}{\beta
b^{3}m}\Phi^{\mu}_{y}+\frac{2e\gamma{\mathcal{M}}}{mb^{2}}\Phi^{e}_{y}$ (37)

where we use the natural system of units and $\Phi^{\mu}_{i}$ and
$\Phi^{e}_{i}$ are the polarization functions appearing in Eqs.(21) and (32)
when substituting
$\overrightarrow{{\mathcal{M}}}={\mathcal{M}}(\sin{\theta_{{\mathcal{M}}}}\cos{\phi_{{\mathcal{M}}}},\sin{\theta_{{\mathcal{M}}}}\sin{\phi_{{\mathcal{M}}}},\cos{\theta_{{\mathcal{M}}}})$,
$\vec{\mu}=\mu_{B}(\sin{\theta_{\mu}}\cos{\phi_{\mu}},\sin{\theta_{\mu}}\sin{\phi_{\mu}},\cos{\theta_{\mu}})$.

The energy transmitted to the electron for fixed
$\theta_{\mu},\phi_{\mu},\theta_{\mathcal{M}},\phi_{\mathcal{M}},\phi$ and $b$
becomes

$\frac{2{\mathcal{M}}^{2}}{mb^{4}}\sum_{i=x,y}\left(\frac{2\mu_{B}}{\beta
b}\Phi^{\mu}_{i}+e\gamma\Phi^{e}_{i})^{2}\right).$ (38)

We next integrate this expression $\frac{1}{2\pi^{2}}\int
d\theta_{\mu}d\phi_{\mu}d\phi$. The cross terms
$\frac{1}{2\pi^{2}}\int\Phi^{\mu}_{i}\Phi^{e}_{i}d\theta_{\mu}d\phi_{\mu}d\phi$
vanish, the energy transferred to all electrons with fixed
$\theta_{\mathcal{M}},\phi_{\mathcal{M}}$ and $b$ is

$E=\frac{8{\mathcal{M}}^{2}\mu_{B}^{2}}{m\beta^{2}}\left(\frac{\frac{\Phi_{\mu}^{2}}{2\pi}}{b^{6}}+\frac{e^{2}\beta^{2}\frac{\Phi_{e}^{2}}{2\pi}}{4\mu_{B}^{2}}{b^{4}}\right),$
(39)

where
$\Phi^{2}\sim\frac{1}{2\pi^{2}}\int\sum_{i}\Phi_{i}^{2}d\phi_{\mu}d\theta_{\mu}d\phi$,
with $i=x,y$. Figs. 3 and 4 show the structure of the polarization functions
after this integration.

This equation relates $E$ to $b$ and will be used to eliminate $b$ in terms of
$E$ in the final calculation. However for one electron we average the
polarization function over $\phi$, thus the contribution will be in that case
$\frac{\Phi^{2}}{2\pi}$. Following the procedure outlined before we have next
to perform the integration over $\int bdb$ to obtain the energy for all the
electrons in the $b,b+db$ cylinder. Thus the energy lost by the particle as it
travels an infinitesimal $dz$ is

$\frac{dE}{dz}=\frac{2{\mathcal{M}}^{2}\mu_{B}^{2}}{m\beta^{2}}\left(\Phi_{\mu}^{2}\left(\frac{1}{b_{max}^{4}}-\frac{1}{b_{min}^{4}}\right)+\frac{e^{2}\beta^{2}\Phi_{e}^{2}}{2\mu_{B}^{2}}\left(\frac{1}{b_{max}^{2}}-\frac{1}{b_{min}^{2}}\right)\right).$
(40)

The next step is to use Eq.(38) to relate $b$ to the energy. In order to get
estimates we shall average over polarization angles and $\phi$. The result
becomes

$\sigma^{3}-P\sigma-Q=0$ (41)

where

$\displaystyle b$ $\displaystyle=$ $\displaystyle\frac{\sqrt{\sigma}}{m},$
$\displaystyle P$ $\displaystyle=$
$\displaystyle\frac{2{\mathcal{M}}^{2}e^{2}\frac{\Phi_{e}^{2}}{2\pi}m^{3}}{E},$
$\displaystyle Q$ $\displaystyle=$
$\displaystyle\frac{8{\mathcal{M}}^{2}\mu_{B}^{2}\frac{\Phi_{\mu}^{2}}{2\pi}m^{5}}{\beta^{2}E}.$
(42)

The coefficients and the variables in the Cardano polynomial have been
multiplied by appropriate powers of $m$ to have all of them adimensional. This
equation defines $b_{max},b_{min}$ as a function of $E_{max},E_{min}$.

$Ratio=\frac{2{\mathcal{M}}^{2}\mu_{B}^{2}\left(\Phi_{\mu}^{2}\left(\frac{1}{b_{max}^{4}}-\frac{1}{b_{min}^{4}}\right)+\frac{e^{2}\beta^{2}\Phi_{e}^{2}}{2\mu_{B}^{2}}\left(\frac{1}{b_{max}^{2}}-\frac{1}{b_{min}^{2}}\right)\right)}{4\pi
Q^{2}e^{2}\log{\frac{E_{max}}{E_{min}}}}$ (43)

As an application of the procedure we show in Fig. 7 the full calculation
(thick) for the neutron and we compare it with the results of the magnetic
moment interaction (dashed) and electric charge magnetic moment interaction
(dotted). The magnetic moments interaction is barely noticed, because it is
dominant for the neutron magnetic moment and the range of velocities studied,
and therefore almost coincides with the full calculation as can be confirmed
by comparing this figure with Fig. 6. We have extended in this case $\beta$ to
$0.3$ simply to make manifest the difference between the full and the magnetic
moments interaction.

Figure 7: We plot the ratio of the full interaction for the neutron with
respect to a conventional charge interaction for the alpha particle, as a
function of velocity $\beta$. The dashed line shows the result of only the
magnetic moments interaction and the dotted line that of the charge magnetic
moment interaction.

Having finished the theoretical developments leading to the full interaction
ratio we proceed to apply our result to monopolium.

##  IV Analysis for Monopolium

In this section we apply the previous developments to the excited high lying
Rydberg maximally deformed states of monopolium. In the conventional studies
of energy loss the value of $E_{min}$ used is $\sim 10Z$ eV, where $Z$ is the
atomic number of the atom in the medium and as mentioned before we use liquid
Argon, $Z=19$.

Figure 8: Left:We plot the relation between the corresponding magnetic moment
in Bohr magnetons and their corresponding absolute value of the monopolium’s
$n$th state binding energy. Right: we show the relation between the binding
energy and the principal quantum number of the state for two monopole masses
$M=1$ TeV (solid) and $M=10$ TeV (dotted).

In monopolium the high $n$ states are those of a magnetic Coulomb potential
$\frac{g}{r}$ and therefore for each $n$ there is a degeneracy in $l$ in the
absence of external fields. Consequently even with a weak magnetic or electric
field, which will always be present in our experiments, there will be a
complete $l$ mixing. Calculating the dual Stark effect in parabolic
coordinates, and comparing with the energy shift of a magnetic dipole field,
we obtain the magnetic moment of a Rydberg state in the presence of a magnetic
field. For high $n$ the maximal magnetic moment is Damburg_1979 ;
Gallagher_1988 ; Hogan:2013qla ; Alonso:2018sun ; Fanchiotti:2022xvx ,

${\mathcal{M}_{n}}\sim\frac{3}{2}n^{2}g\;r_{Bohr}\sim
g<n0|r|0n>\sim\frac{3}{32\,\alpha^{\nicefrac{{3}}{{2}}}\,E_{n}},$ (44)

where $E_{n}$ is the absolute value of the $n$th state binding energy. These
expressions correspond to the large $n$ limit.

We show in Fig. 9 the largest possible ratio of the energy loss of monopolium
by comparing the monopolium energy loss with that of a proton $Z_{H}=1$. The
ratio for any other ion is obtained dividing by its $Z_{H}^{2}$. The remaining
inputs are Bohr’s magneton given by $\mu_{B}\sim 0.0837$MeV-1, the electron
mass $m\sim 0.511$ MeV, and the unit charge $e\sim\sqrt{1/137}=0.0854$

Figure 9: Left: Ratio of energy losses between monopolium and the proton for
$E_{n}=1000$ eV as a function of $\beta$. Right: Ratio of energy loses between
monopolium and the proton for $\beta=0.001$ as a function of binding energy
$E_{n}$ in eV. The solid curve represents the full calculation, the dashed
curve the ratio taking into account only the magnetic moment interaction, and
the dotted curve the ratio taking into account only the charge magnetic moment
interaction.

We express our results as a function of the absolute value of the binding
energy $E_{n}$. For physical purposes we show in Fig. 8 the relation between
$E_{n}$ in eV and the maximal magnetic moment ${\mathcal{M}}$ of that $n$th
level. In the right plot of the same figure we show the correspondence between
the binding energy and the principal quantum number for monopole masses of $1$
TeV (dashed) and $10$ TeV (dotted) as given by Eq. (45),

$E_{n}=\frac{3M}{48\alpha^{2}n^{2}},$ (45)

where $M$ is the mass of the monopole.

In Fig. 9 we show (left) the ratio of energy losses between a high lying
monopolium Rydberg state of $E_{n}=1000$ eV and a proton The right figure
shows the same ratio of energy losses as a function of binding energy for
$\beta=0.001$. For larger $\beta$s the charge magnetic moment interaction will
dominate and for smaller the magnetic moments interaction will be dominant.
These results point out that if highly deformed monopolia would reach a liquid
$Ar$ detector they would produce a ionization of the same order of magnitude
as that of light-medium ions.

One might be surprised by the extremely large magnetic moments required for
the observation corresponding to huge principal quantum numbers. Will these
very high $n$ monopolium Rydberg states reach detectors on Earth? The Rydberg
states with long lifetimes and large magnetic moments are lightly bound
monopole-antimonopole pairs with binding energies of the order of eVs. The
small Parker bound, an upper bound on the density of magnetic monopoles
Turner:1982ag , can be obviated because we are dealing with bound states
Dicus:1983ry . The Parker bound is based on the monopoles taking energy from
the intergalactic magnetic field. However, monopolia below the dissociation
energy do not take energy from the magnetic field since magnetic moment
interactions at large distances are much weaker than magnetic charge
interactions. Moreover, the highly deformed Rydberg states which are in high
$l$ states, therefore with vanishing wave function at the origin, have the
monopole and antimonopole far away from each other and therefore do not
annihilate easily as it is the case of the low lying states. They will de-
excite by spontaneous emission of photons to the lower levels where they will
annihilate either as S states or to non leading order as $l\neq 0$ state
decays. In order to get an estimate for spontaneous emission as a function of
$n$ using the naive Bohr model one gets Dicus:1983ry ,

$\Gamma^{se}_{n}(E)=4096\,\beta_{n}^{4}\,\alpha\,m\left(\frac{E_{n}}{m}\right)^{3}.$
(46)

This width’s $n$ dependence dependence goes like $1/n^{6}$, therefore
lifetimes of these Rydberg states increase notably with $n$ as seen in Fig.
10. By astronomical standards the distances they can travel are small and
therefore they will have to be produced by high energy photons in our
vecinity.

Similar calculations can be extended easily to Kaluza-Klein (KK) monopolium as
was shown in ref. Vento:2007vy , but in this case since the monopoles do not
decay classically their lifetime is enormous and therefore they could reach us
from all parts of the universe, most probably quite cold and therefore our non
relativistic result applies to them very nicely. To translate to that
calculation we have simply to transfer the results to magnetic moments and
apply the formula for magnetic moments to the magnetic moment of that of the
KK monopolia in terms of the distance between the poles Gross:1983hb ;
Vento:2007vy .

Figure 10: Lifetime of monopolium Rydberg states in years as a function of
binding energies.

##  V Concluding remarks

Monopolium, a bound state of monopole-antimonopole, has no magnetic charge and
in its ground state no magnetic moment, being therefore very difficult to
detect directly. In conventional gauge theories it has the quantum numbers of
the vacuum and annihilates into photons, and these disintegrations have been
intensively studied Epele:2007ic ; Epele:2008un ; Epele:2012jn ;
Barrie:2016wxf ; Fanchiotti:2017nkk ; Barrie:2021rqa . In Kaluza-Klein
theories monopolium is classically stable and therefore very long lived
Gross:1983hb ; Vento:2020vsq  but its detection problems still persist since
it only interacts gravitationally. What happens with excited monopolium
states? This paper deals with the study of the excited states of monopolium.
Excited states have in the presence of even weak magnetic fields magnetic
moments and their interaction with magnetic and electric fields is known.
Recently we used a theoretical analysis of magnetic dipole detection to learn
about excited monopolium states Fanchiotti:2022xvx . In this work we have
analyzed the capacity of monopolium to produce ionization when traversing a
medium by calculating its energy loss when colliding with atomic electrons.
The idea behind this investigation is to find out if TPCs Nygren:2018sjx
could be used for detecting monopolium. We have calculated the energy loss of
monopolium in an $Ar$ medium by a classical analog of the Bethe-Bloch formula.
The analogous calculation for a charge particle and for a monopole agrees well
with the Bethe-Bloch equation Bethe:1930ku ; Bloch:1933zza ; Ahlen:1976jw  in
the non relativistic limit, and thus we expect that the same will happen for
monopolium. We have seen that in those kinematical regions where the density
dependence is not important and for non relativistic particles the ionization
of monopolium excited Rydberg states is of the same magnitude as that to
medium-light ions, the result being independent of the heavy particle mass
(recall Fig. 9). The Bethe-Bloch formulation is an average description and
therefore our result does not determine the precise ionization of one
monopolium event but certainly determines the order of magnitude of the effect
from the perspective of the sensitivity of the detectors.

## Acknolwedgements

HF and CAGC were partially supported by ANPCyT, Argentina. VV was supported by
MCIN/AEI/10.13039/501100011033, European Regional Development Fund Grant No.
PID2019-105439 GB-C21 and by GVA PROMETEO/2021/083. VV acknowledges
correspondence with Alan Chodos and David Nygren which motivated this
investigation, and useful discussions on monopole ionization with Laura
Patrizii.

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