# a note on squeezing function and its generalizations

Naveen Gupta  Department of Mathematics, University of Delhi, Delhi–110 007,
India<EMAIL_ADDRESS>and  Sanjay Kumar Pant  Department of
Mathematics, Deen Dayal Upadhyaya college, University of Delhi, Delhi–110 078,
India<EMAIL_ADDRESS>
###### Abstract.

This note investigates the relation between squeezing function and its
generalizations. Using the relation obtained, we present an alternate method
to find expression of generalized squeezing function of unit ball
corresponding to the generalized complex ellipsoids.

###### Key words and phrases:

squeezing function; holomorphic homogeneous regular domains; quasi balanced
domains.

###### 2010 Mathematics Subject Classification:

32F45, 32H02

##  1\. introduction

This note clarifies the relation between squeezing function and its
generalizations, namely, generalized squeezing function and $d$-balanced
squeezing function. The notion of squeezing function was first introduced by
Deng, Guan and Zhang [3]. Although its genesis can be traced back to the work
of Liu et al [8], [9] and of Yeung [13]. Rong and Yang[11] introduced the
notion of generalized squeezing function. The authors in [6] extended it to
$d$-balanced squeezing function. Let us see the definitions of all these terms
first.

$\mathbb{B}^{n}$ denotes unit ball in $\mathbb{C}^{n}$ and
$D\subseteq\mathbb{C}^{n}$ is used for bounded domain. The set of all
injective holomorphic maps from $D$ to a domain
$\Omega\subseteq\mathbb{C}^{n}$ is denoted by $\mathcal{O}_{u}(D,\Omega)$.

For $z\in{D}$ the squeezing function $S_{D}$ on $D$ is defined as

$S_{D}(z):=\sup_{f}\\{r:\mathbb{B}^{n}(0,r)\subseteq
f(D),f\in{\mathcal{O}_{u}(D,\mathbb{B}^{n})}\\},$

where $\mathbb{B}^{n}(0,r)$ denotes ball of radius $r$ centered at the origin.

A domain is said to be holomorphic homogeneous regular if its squeezing
function has a positive lower bound. Notion of holomorphic homogeneous regular
(HHR) manifolds was studied by Liu et al [8], [9]. In 2009 [13], Yeung renamed
HHR as uniform squeezing property. It was then formally defined in terms of
squeezing function by Deng et al in 2012[3]. Holomorphic homogeneous regular
domains have important significance due to several anlaytic and geometric
properties they ensure.

In [11], Rong and Yang introduced the concept of generalized squeezing
function $S^{\Omega}_{D}$ for bounded domains
$D,\Omega\subseteq\mathbb{C}^{n}$, where $\Omega$ is a balanced domain.

Let us quickly see the definitions of a balanced domain and Minkowski
function. A domain $\Omega\subseteq\mathbb{C}^{n}$ is said to be balanced if
$\lambda z\in{\Omega},$ for each $z\in{\Omega}$ and $|\lambda|\leq 1$. The
Minkowski function denoted by $h_{\Omega}$ on $\mathbb{C}^{n}$ is defined as

$h_{\Omega}(z):=\inf\\{t>0:\frac{z}{t}\in{\Omega}\\}.$

For $0<r\leq 1$, let $\Omega(r):=\\{z\in{\mathbb{C}^{n}:h_{\Omega}(z)<r\\}}$.
For a bounded domain $D\subseteq\mathbb{C}^{n}$ and a bounded, balanced,
convex domain $\Omega\subseteq\mathbb{C}^{n}$, Rong and Yang introduced the
notion of generalized squeezing function $S_{D}^{\Omega}$ on $D$ as

$S^{\Omega}_{D}(z):=\sup\\{r:\Omega(r)\subseteq
f(D),f\in{\mathcal{O}_{u}(D,\Omega)},f(z)=0\\}.$

Notion of balanced domains was introduced by Nikolov in his work [10]. Let
$d=(d_{1},d_{2},\ldots,d_{n})\in{\mathbb{Z}_{n}^{+}},n\geq 2$, we say that a
domain $\Omega\subseteq\mathbb{C}^{n}$ is $d$-balanced if for each
$z=(z_{1},z_{2},\ldots,z_{n})\in\Omega$ and
$\lambda\in{\overline{\mathbb{D}}}$,
$\left(\lambda^{d_{1}}z_{1},\lambda^{d_{2}}z_{2},\ldots,\lambda^{d_{n}}z_{n}\right)\in\Omega,$
where $\mathbb{D}$ denotes unit disk in $\mathbb{C}$. It is easy to see that
balanced domains are $(1,1,\ldots,1)$-balanced.

For a $d$-balanced domain $\Omega$, an analogue of Minkowski function, called
$d$-Minkowski function on $\mathbb{C}^{n}$, denoted by $h_{d,\Omega}$, is
defined as

$h_{d,\Omega}(z):=\inf\\{t>0:\left(\frac{z_{1}}{t^{d_{1}}},\frac{z_{2}}{t^{d_{2}}},\ldots,\frac{z_{n}}{t^{d_{n}}}\right)\in{\Omega}\\}.$

For each $0<r\leq 1$, we fix
$\Omega^{d}(r):=\\{z\in{\mathbb{C}^{n}:h_{d,\Omega}(z)<r\\}}$.

######  Definition 1.1.

For a bounded domain $D\subseteq\mathbb{C}^{n}$, and a bounded, convex,
$d=(d_{1},d_{2},\ldots,d_{n})$-balanced domain $\Omega$, _ $d$-balanced
squeezing function corresponding to $\Omega$_ of the domain $D$, denoted by
$S_{d,D}^{\Omega}$(also called the _ $d$-balanced squeezing function_ for
brevity) is given by:

$S_{d,D}^{\Omega}(z):=\sup\\{r:\Omega^{d}(r)\subseteq
f(D),f\in{\mathcal{O}_{u}(D,\Omega)},f(z)=0\\}.$

Similar to the case of squeezing function, holomorphic homogeneous regular
domains are the ones whose generalized squeezing function (or $d$-balanced
squeezing function) has a positive lower bound.

As we know that squeezing function was introduced in the context of uniform
squeezing property (HHR), so it is natural to expect that regularity remains
unaltered when one shifts from squeezing function to its generaliations. More
precisely, it is natural to expect that if a domain $D$ is regular in $S_{D}$,
then it should be regular in $S_{D}^{\Omega}$ or $S_{d,D}^{\Omega}$. To
address this we analyze relations between these objects in Theorems 2.1 and
3.2. We would like to point out that apart from addressing the question about
regularity of a domain, these results will also serve its purpose in
estimating squeezing function for different domains. Another similar question
related to regularity is behaviour of these generalizations corresponding to
different model domains. What we mean by this is: For a given domain $D$ and
two bounded, balanced, convex domains $\Omega_{1},\Omega_{2}$, if $D$ is
regular as per the setup of $S_{D}^{\Omega_{1}}$, will it be regular as per
$S_{D}^{\Omega_{2}}$? A similar question can be formed for $S_{d,D}^{\Omega}.$
One can not be unjust for expecting this and this is the content of our
Corollary 2.3 and Corollary 3.3.

##  2\. $S_{D}$ and $S_{D}^{\Omega}$

In [5, Lemma 1.2] a relation between the squeezing function and the squeezing
function corresponding to polydisk is presented. Such a relation immediately
deduces the equivalence between holomorphic homogeneous regular domains in
these two setups and is helpful in obtaining estimate for squeezing function
[See examples in [5]]. We present a similar relation between squeezing
function and generalized squeezing function, which in particular will imply
that the holomorphic homogeneous regular domains are the same for $S_{D}$ and
$S_{D}^{\Omega}$. We point out here that equivalence of holomorprhic
homogeneous regular domains is established in [11, Theorem 3.9], but the
inequality established here is more useful in estimating generalized squeezing
function. For a domain $D\subseteq\mathbb{C}^{n}$ and $z_{1},z_{2}\in D$, the
Carathéodory pseudodistance $c_{D}$ on $D$ is defined as

$c_{D}(z_{1},z_{2})=\sup_{f}\\{p(0,\mu):f\in{\mathcal{O}(\mathbb{D},D),f(z_{1})=0,f(z_{2})=\mu}\\},$

where $p$ denotes the Poincaré metric on unit disc $\mathbb{D}$. We take
$c_{D}^{*}=\tanh c_{D}$. Using [12, Lemma 3.2], for a bounded, balanced,
convex domain $D$, we have $c_{D}^{*}(0,z)=h_{D}(z)$ for $z\in D$.

######  Proposition 2.1.

Let $D\subseteq\mathbb{C}^{n}$ be bounded and $\Omega\subseteq\mathbb{C}^{n}$
be bounded, balanced, convex domain. Then for $a\in D$ the following holds:

  1. (1)

$S_{D}(a)\geq\dfrac{\alpha}{R}S_{D}^{\Omega}(a),$

  2. (2)

$S_{D}^{\Omega}(a)\geq\dfrac{\alpha}{R}S_{D}(a),$

where $\alpha=dist(0,\partial\Omega)$ and $R=\dfrac{1}{2}diam(\Omega)$.

###### Proof.

  1. (1)

For $a\in D$, let $f:D\to\mathbb{B}^{n}$ be an injective holomorphic map with
$f(a)=0$. Let $r>0$ be such that $B^{n}(0,r)\subseteq f(D)$. Consider
$g:D\to\Omega$ defined as

$g(z):=\alpha f(z).$

Note that $g$ is injective holomorphic with $g(a)=0$ and $g(z)\in{\Omega}$ for
all $z\in D$. Also note that $B^{n}(0,\alpha r)\subseteq g(D)$. We claim that
$\Omega\left(\dfrac{\alpha r}{R}\right)\subseteq g(D).$ Before proving our
claim, we make an observation that $\Omega\subseteq B^{n}(0,R)$ and therefore

(2.1) $c^{*}_{B^{n}(0,R)}(0,z)\leq c^{*}_{\Omega}(0,z)=h_{\Omega}(z)\
\mbox{for all}\ z\in\Omega.$

In order to prove our claim, see that for $z\in{\Omega\left(\dfrac{\alpha
r}{R}\right)},$ using equation (2.1), we have

$c^{*}_{B^{n}(0,R)}(0,z)\leq c^{*}_{\Omega}(0,z)=h_{\Omega}(z)<\dfrac{\alpha
r}{R}.$

This gives us $z\in{\mathbb{B}^{n}(0,\alpha r)}$ and thus $z\in g(D).$
Therefore, we get our claim and hence $S_{D}^{\Omega}(a)\geq\dfrac{\alpha
r}{R}$, which implies $r\leq\dfrac{R}{\alpha}S_{D}^{\Omega}(a)$. This
establishes (1).

  2. (2)

Let $f:D\to\Omega$ be an injective holomorphic map with $f(a)=0$. Let $r>0$ be
such that $\Omega(r)\subseteq f(D)$. Consider $g:D\to\mathbb{B}^{n}$ defined
as

$g(z):=\dfrac{f(z)}{R}.$

Observe that $g$ is injective holomorphic with $g(a)=0$. Also,
$\Omega\left(\dfrac{r}{R}\right)=\dfrac{1}{R}\Omega(r)\subseteq
g(D)\subseteq\mathbb{B}^{n}.$ Since $\mathbb{B}^{n}(0,\alpha)\subseteq\Omega$,
therefore

(2.2) $h_{\Omega}(z)=c^{*}_{\Omega}(0,z)\leq c^{*}_{B^{n}(0,\alpha)}(0,z)\
\mbox{for all}\ z\in\mathbb{B}^{n}(0,\alpha).$

We claim that $\mathbb{B}^{n}\left(0,\dfrac{\alpha r}{R}\right)\subseteq
g(D).$ For $z\in\mathbb{B}^{n}\left(0,\dfrac{\alpha r}{R}\right),$ using
equation (2.2) (note that $r/R<1$), we have

$h_{\Omega}(z)=c^{*}_{\Omega}(0,z)\leq
c^{*}_{B^{n}(0,\alpha)}(0,z)=\dfrac{1}{\alpha}|z|<\dfrac{r}{R}.$

Thus we get $z\in\Omega\left(\dfrac{r}{R}\right)\subseteq g(D)$, which proves
our claim and therefore $S_{D}(a)\geq\dfrac{\alpha r}{R}$. This establishes
(2).

∎

######  Remark 2.2.

The two inequalities in the above theorem can not be attained simultaneously
unless $\Omega=\mathbb{B}^{n}$. To see this, note that the two inequalities
are attained simultaneously only when $\alpha=R$.

Note that previous theorem in particular implies that holomorphic homogeneous
regular domains in the set up of $S_{D}$ and $S_{D}^{\Omega}$ are the same.

The interaction between generalized squeezing function for different model
domains $\Omega$ is presented in the following easy corollary to the above
proposition. For relation between $S_{D}^{\Omega}$ for different model
domains, where $\Omega$ is _h_ omogeneous as well, one can see [11, Theorem
4.4]. We will see analogue of this result for $d$-balanced squeezing function
in Theorem 3.4.

######  Corollary 2.3.

Let $D\subseteq\mathbb{C}^{n}$ be a bounded domain and
$\Omega_{1},\Omega_{2}\subseteq\mathbb{C}^{n}$ be bounded, balanced, covex
domains, then the following holds:

  1. (1)

$S_{D}^{\Omega_{2}}(a)\geq\dfrac{\alpha_{1}\alpha_{2}}{R_{1}R_{2}}S_{D}^{\Omega_{1}}(a),$

  2. (2)

$S_{D}^{\Omega_{1}}(a)\geq\dfrac{\alpha_{1}\alpha_{2}}{R_{1}R_{2}}S_{D}^{\Omega_{2}}(a),$

where $\alpha_{i}=dist(0,\partial\Omega_{i})$ and
$R_{i}=\dfrac{1}{2}diam(\Omega_{i}),\ i=1,2$.

###### Proof.

To prove (1), use part (2) of Proposition 2.1 for $\Omega_{2}$ and then use
part (1) for $\Omega_{1}$. One can prove (2) in similar way. ∎

A simple consequence of the above stated corollary is: $S_{D}^{\Omega_{1}}$
has a positive lower bound if and only if $S_{D}^{\Omega_{2}}$ has positive
lower bound. We would like to mention here, this statement has been proved in
[11, Theorem 3.9 ] separately, but the bounds obtained here are the simplest
possible and are useful in its own right.

##  3\. $S_{D}$ and $S_{d,D}^{\Omega}$

Let us recall few properties of d-Minkowski function and $d$-balanced domains.

######  Result 3.1.

For a $d$-balanced domain $\Omega\subseteq\mathbb{C}^{n}$, the following
holds: ( see[7, Remark 2.2.14])

  1. (1)

$\Omega=\\{z\in{\mathbb{C}^{n}}:h_{d,\Omega}(z)<1\\}$.

  2. (2)

$h_{d,\Omega}\left(\lambda^{d_{1}}z_{1},\lambda^{d_{2}}z_{2},\ldots,\lambda^{d_{n}}z_{n}\right)=|\lambda|h_{d,\Omega}(z)$
for each $z=(z_{1},z_{2},\ldots,z_{n})\in{\mathbb{C}^{n}}$ and
$\lambda\in{\mathbb{C}}.$

  3. (3)

$h_{d,\Omega}$ is upper semicontinuous.

For a bounded, $d$-balanced, convex domain $\Omega$, using [2, Theorem 1.6],
we have

$\tanh^{-1}h_{d,\Omega}(z)^{L}\leq
c_{\Omega}(0,z)=k_{\Omega}(0,z)\leq\tanh^{-1}h_{d,\Omega}(z),$

where $L=\max_{1\leq i\leq n}d_{i}.$ Similar to the case of generalized
squeezing function, the following theorem for $d$-balanced squeezing function
holds.

######  Proposition 3.2.

Let $D\subseteq\mathbb{C}^{n}$ be bounded and $\Omega\subseteq\mathbb{C}^{n}$
be bounded, $d$-balanced, convex domain. Then for $a\in D$ the following
holds:

  1. (1)

$S_{D}(a)\geq\dfrac{\alpha}{P}S_{d,D}^{\Omega}(a)^{L},$

  2. (2)

$S_{d,D}^{\Omega}(a)\geq\dfrac{\alpha}{R}S_{D}(a),$

where $\alpha=dist(0,\partial\Omega)$ and
$R=\dfrac{1}{2}diam(\Omega),P=R+1,L=\max_{1\leq i\leq n}d_{i}.$

###### Proof.

Let $a\in D$ and $f:D\to\Omega$ be injective holomorphic map with $f(a)=0$.
Let $r>0$ be such that $\Omega^{d}(r)\subseteq
f(D)\subseteq\Omega=\Omega^{d}(1).$ Consider $g:D\to\mathbb{C}^{n}$ defined as

$g(z):=\left(\dfrac{f_{1}(z)}{P^{d_{1}}},\dfrac{f_{2}(z)}{P^{d_{2}}},\ldots,\dfrac{f_{n}(z)}{P^{d_{n}}}\right).$

It is obvious that $g$ is injective holomorphic with $g(a)=0$. Also using
Result 3.1(2) and the fact that $\Omega\subseteq\mathbb{B}^{n}(0,P)$ it is
easy to infer that $g(D)\subseteq\mathbb{B}^{n}$. Therefore,
$g:D\to\mathbb{B}^{n}$. Next we claim that $\Omega^{d}\left(\dfrac{\alpha
r^{L}}{P}\right)\subseteq g(D)$. Let $z\in\Omega^{d}\left(\dfrac{\alpha
r^{L}}{P}\right)$, that is, $h_{d,\Omega}(z)<\dfrac{\alpha r^{L}}{P}<1$.
Therefore, we get $z\in\Omega$. Also using the decreasing property of
Caratheodory metric, we have $c^{*}_{\mathbb{B}^{n}(0,P)}(0,z)\leq
c^{*}_{\Omega}(0,z)$ for $z\in{\Omega}$. Thus we have $\dfrac{1}{P}|z|\leq
c^{*}_{\Omega}(0,z)\leq h_{d,\Omega}(z)<\dfrac{\alpha r^{L}}{P}$. Thus,
$|z|<\alpha r^{L}\leq\alpha$ and for $z\in{\mathbb{B}^{n}(0,\alpha)}$, using
the decreasing property, we have $h_{d,\Omega}(z)^{L}\leq
c^{*}_{\Omega}(0,z)\leq
c^{*}_{\mathbb{B}^{n}(0,\alpha)}(0,z)=\dfrac{1}{\alpha}|z|.$ Therefore, we
obtain $h_{d,\Omega}(z)<\dfrac{\alpha r}{P}$. Now using Result 3.1(2), we get
$z\in g(D)$. Therefore, $\Omega^{d}\left(\dfrac{\alpha
r^{L}}{P}\right)\subseteq g(D)$ and this finally yeilds (1). Proceding on the
similar lines, one can prove (2). ∎

The following corollary ensures that the holomorohic homogeneous regular
domains are same corresponding to different $d$-balanced domains.

######  Corollary 3.3.

Let $D\subseteq\mathbb{C}^{n}$ be bounded and
$\Omega_{1},\Omega_{2}\subseteq\mathbb{C}^{n}$ be bounded, convex domains.
Assume that $\Omega_{1}$ is $d$-balanced, and $\Omega_{2}$ is
$d^{\prime}$-balanced, then for $a\in D$, the following holds:

  1. (1)

$S_{d^{\prime},D}^{\Omega_{2}}(a)\geq\dfrac{\alpha_{1}\alpha_{2}S_{d,D}^{\Omega_{1}}(a)^{L}}{P_{1}R_{2}},$

  2. (2)

$S_{d,D}^{\Omega_{1}}(a)\geq\dfrac{\alpha_{1}\alpha_{2}S_{d^{\prime},D}^{\Omega_{2}}(a)^{L^{\prime}}}{R_{1}P_{2}},$

where $P_{i}=R_{i}+1,\alpha_{i}=dist(0,\partial\Omega_{i})$,
$R_{i}=\dfrac{1}{2}diam(\Omega_{i}),\ \ i=1,2$; $L=\max_{1\leq i\leq n}d_{i}$
and $L^{\prime}=\max_{1\leq i\leq n}d_{i}^{\prime}$.

###### Proof.

One can prove this by proceeding on the same lines as in the Corollary 2.3. ∎

Similarly to the Theorem [12, Theorem 4.4] the following theorem gives
relation between $d$-balanced squeezing function for different model domains,
when domains in consideration are bounded, $d$-balanced, convex and
_homogeneous_.

######  Theorem 3.4.

let $\Omega_{1},\Omega_{2}\subseteq\mathbb{C}^{n}$ be bounded, convex,
homogeneous domains and $D\subseteq\mathbb{C}^{n}$ be a bounded domain. Assume
that $\Omega_{1}$ is $d$-balanced and $\Omega_{2}$ is $d^{\prime}$-balanced.
Then

  1. (1)

$S_{d^{\prime},D}^{\Omega_{2}}(z)\geq
S_{d,\Omega_{1}}^{\Omega_{2}}(0)S_{d,D}^{\Omega_{1}}(z)^{L},$

  2. (2)

$S_{d,D}^{\Omega_{1}}(z)\geq
S_{d^{\prime},\Omega_{2}}^{\Omega_{1}}(0)S_{d^{\prime},D}^{\Omega_{2}}(z)^{L^{\prime}},$

where $L=\max_{1\leq i\leq n}d_{i}$ and $L^{\prime}=\max_{1\leq i\leq
n}d_{i}^{\prime}$.

We will need the following lemma to prove this theorem.

######  Lemma 3.5.

Let $\Omega\subseteq\mathbb{C}^{n}$ be a $d$-balanced domain and
$\Omega_{1}=\Omega^{d}(r),$ where $0<r\leq 1$. Then

$h_{d,\Omega_{1}}=\dfrac{1}{r}h_{d,\Omega}.$

###### Proof.

First of all, observe that $\Omega_{1}$ is bounded, $d$-balanced and
$\Omega_{1}\subseteq\Omega^{\prime}$. Let
$A=\\{t>0:\left(\frac{z_{1}}{t^{d_{1}}},\frac{z_{2}}{t^{d_{2}}},\ldots,\frac{z_{n}}{t^{d_{n}}}\right)\in{\Omega}\\}$
and
$A_{1}=\\{t>0:\left(\frac{z_{1}}{t^{d_{1}}},\frac{z_{2}}{t^{d_{2}}},\ldots,\frac{z_{n}}{t^{d_{n}}}\right)\in{\Omega^{\prime}}\\}$.
It is easy to see that $t\in A$ if and only if $\dfrac{t}{r}\in A_{1}$.
Therefore we get $A_{1}=\dfrac{1}{r}A$ and hence the result follows. ∎

###### Proof of theorem 3.4.

Let $z\in D$ and $f:D\to\Omega_{1}$ be injective holomorpphic with $f(z)=0$.
Let $r>0$ be such that

(3.1) $\Omega_{1}^{d}(r)\subseteq f(D).$

Let $\alpha=S_{d,\Omega_{1}}^{\Omega_{2}}(0)$, then using Theorem [6, Theorem
4.6 ], there exists an injective holomorphic map $g:\Omega_{1}\to\Omega_{2}$
with $g(0)=0$ such that

(3.2) $\Omega_{2}^{d^{\prime}}(\alpha)\subseteq g(\Omega_{1}).$

Consider $F:D\to\Omega_{2}$ defined as $F=g\circ f$, then $F$ is injective
holomorphic with $F(z)=0$. Let us fix
$\Omega=\Omega_{2}^{d^{\prime}}(\alpha)$. We will first prove that
$B_{\Omega}^{l}\left(0,\beta\right)\subseteq g\left(\Omega_{1}^{d}(r)\right),$
where $\beta=\tanh^{-1}r^{L}$ and $l$ denotes Lempert’s function.

Let $z\in B_{\Omega}^{l}\left(0,\beta\right)$, that is,
$l_{\Omega}(0,z)<\beta$. Since $\Omega\subseteq g(\Omega_{1})$, therefore
$z=g(w)$ for some $w\in\Omega$ and using decreasing property of Lempert’s
function, we get $l_{g(\Omega_{1})}(0,g(w))<\beta$, thus we get
$l_{\Omega_{1}}(0,w)<\beta$. Also using [2, Theorem 1.6], we have
$\tanh^{-1}h_{d,\Omega_{1}}(z)\leq l_{\Omega_{1}}(0,w)$. Thus we obtain
$h_{d,\Omega_{1}}(w)<r$ and therefore
$z=g(w)\in{g\left(\Omega_{1}^{d}(r)\right)}.$

So we have obtained $B_{\Omega}^{l}\left(0,\beta\right)\subseteq
g\left(\Omega_{1}^{d}(r)\right)$ and therefore using Equation (3.1), we have

(3.3) $B_{\Omega}^{l}\left(0,\beta\right)\subseteq g(f(D))=F(D).$

Next we claim to prove $\Omega_{2}^{d^{\prime}}(\alpha r^{L})\subseteq
B_{\Omega}^{l}\left(0,\beta\right)$. For $z\in{\Omega_{2}^{d^{\prime}}(\alpha
r^{L})}$, $h_{d^{\prime},\Omega_{2}}(z)<\alpha r^{L}$. Using Lemma 3.5, we get
$h_{d^{\prime},\Omega}(z)<r^{L}$. Now observing that the right hand side of
inequality in [2, Theorem 1.6] is true without pseudoconvexity assumption of
the domain, we get $\tanh l_{\Omega}(0,z)\leq h_{d^{\prime},\Omega}(z)<r^{L}$.
This implies $z\in{B_{\Omega}^{l}\left(0,\beta\right)}$. This proves our claim
and thus using Equation (3.3), we get

$\Omega_{2}^{d^{\prime}}(\alpha r^{L})\subseteq F(D).$

Hence we get $S_{d^{\prime},D}^{\Omega_{2}}(z)\geq\alpha r^{L}$ and this
completes the proof of (1). Similarly one can prove (2). ∎

##  4\. generalized squeezing function for complex ellipsoid

A _complex ellipsoid_ in $\mathbb{C}^{n}$ is a domain of the form

$E(m)=\\{z\in\mathbb{C}^{n}:\sum_{j\in{I_{1}}}|z_{j}|^{2m_{j}}<1,\,|z_{k}|<1\,\forall
k\in{I_{2}}\\},$

where $1\leq m_{i}\leq\infty$, $I_{1}=\\{1\leq j\leq n:m_{j}\neq\infty\\}$ and
$I_{2}=\\{1,2,\ldots,n\\}\setminus I_{1}$. $E(m)$ is convex if and only if
$m_{1},m_{2},\ldots,m_{n}\geq\frac{1}{2}$. $E(m)=\mathbb{B}^{n},$ when
$m=(1,1,\ldots,1)$ and $E(m)=\mathbb{D}^{n}$, when $m_{j}=\infty$ for all
$1\leq j\leq n$.

A _generalized complex ellipsoid_ denoted by $E(p,m)$ with
$p=(p_{1},p_{2},\ldots,p_{k}),\,m=(m_{1},m_{2},\ldots,m_{k})$,
$p_{1}+p_{2}+\ldots+p_{k}=n$ is given by

$E(p,m)=\\{z\in\mathbb{C}^{p_{1}}\times\mathbb{C}^{p_{2}}\times\ldots\times\mathbb{C}^{p_{k}}:\sum_{j=1}^{k}\|z_{j}\|^{2m_{j}}<1\\}.$

If any of $m_{j}=\infty$, then $E(p,m)$ is understood as in the case of
$E(m).$ In particular, when $m_{j}=\infty$ for each $j=1,2,\ldots,k$,

$E(p,\infty)=\mathbb{B}^{p_{1}}\times\mathbb{B}^{p_{2}}\times\ldots\times\mathbb{B}^{p_{k}}.$

Note that $E(p,\infty)$ is bounded, balanced and convex. Also, it is easy to
see that $\Omega$ is homoegeneous.

Let us denote by $\Omega_{i}=\mathbb{B}^{p_{i}}\,,i=1,2,\ldots,k$, and
$D=\mathbb{B}^{n},$ then
$\Omega=E(p,\infty)=\Omega_{1}\times\Omega_{2}\times\ldots\times\Omega_{k}$.
Therefore, using [12, Corollary 4.3 ], we get that

$S_{D}^{\Omega}(z)=S^{D}_{\Omega}(w),$

for each $z\in D$ and $w\in\Omega$. It is readily seen using [3, Theorem 7.5]
that

$S^{D}_{\Omega}(w)=\left(S_{\Omega_{1}}(w_{1})^{-2}+S_{\Omega_{2}}(w_{2})^{-2}+\ldots+S_{\Omega_{k}}(w_{k})^{-2}\right)^{-1/2}=\dfrac{1}{\sqrt{k}}$

and therefore $S_{D}^{\Omega}(z)=\dfrac{1}{\sqrt{k}}$. We will present here an
alternate method to obtain this expression using the inequality obtained in
Proposition 2.1. We will require the following observation for this.

It is easy to see that $\alpha=dist(0,\partial\Omega)=1$ and
$R=\frac{1}{2}diam(\Omega)=\sqrt{k}$. We also need the following lemma, whose
proof is based on the technique used by H. Alexander in the proof of [1,
Proposition 2].

######  Lemma 4.1.

Let $f:D\to\Omega$ be injective holomorphic with $f(0)=0$. Let $r>0$ be such
that $\Omega(r)\subseteq f(D)$, then $r\leq\dfrac{1}{\sqrt{k}}.$

###### Proof.

Let us consider a function $g:\Omega(r)\to D$ defined as

$g(z):=f^{-1}(z).$

For any $\epsilon>0$, consider
$\mathbb{B}^{n}\left(0,\dfrac{1+\epsilon}{\sqrt{k}}\right)$. Note that using
[3, Theorem 7.5], it follows that $S_{\Omega(r)}^{D}=\dfrac{1}{\sqrt{k}}$,
therefore $g(\Omega(r))$ does not contain
$\mathbb{B}^{n}\left(0,\dfrac{1+\epsilon}{\sqrt{k}}\right)$. Let
$a\in{\mathbb{B}^{n}\left(0,\dfrac{1+\epsilon}{\sqrt{k}}\right)}$ be such that
$b=f(a)\notin{\Omega(r)}=\mathbb{B}^{p_{1}}(r)\times\mathbb{B}^{p_{2}}(r)\times\ldots\times\mathbb{B}^{p_{k}}(r)$.
This yields one of $b_{1},b_{2},\ldots,b_{k}$, say $b_{1}$, has modulus $\geq
r$. Therefore, using Schwarz’s lemma, we get

$r\leq\|b_{1}\|=\|f_{1}(a)\|\leq\|a\|\leq\dfrac{1+\epsilon}{\sqrt{k}},$

which completes the proof.

∎

######  Example 4.2.

For $D=\mathbb{B}^{n}$, $\Omega=E(p,\infty)$,

$S_{D}^{\Omega}(z)=\frac{1}{\sqrt{k}},$

for $z\in D.$

Before we prove this, observe that when $p_{i}=1$ for all $i=1,2,\ldots,k$,
then $\Omega=\mathbb{D}^{n}$. For this case, we have
$S_{D}^{\Omega}(z)=\dfrac{1}{\sqrt{n}},$ [5, Example 1].

###### Proof.

Using Proposition 2.1, we know that

$S_{D}^{\Omega}(a)\geq\dfrac{\alpha}{R}S_{D}(a)=\dfrac{\alpha}{R},$

where $\alpha$ and $R$ as in the proposition. Therefore, we get

$S_{D}^{\Omega}(z)\geq\dfrac{1}{\sqrt{k}}.$

On the other hand, using Lemma 4.1, we get

$S_{D}^{\Omega}(z)\geq\dfrac{1}{\sqrt{k}}.$

∎

## Acknowledgement

The first author is thankful to Peter Pflug for fruitful discussions for some
parts of this note.

## References

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# Exact form of the generalized Lorentz force in Fock’s nonlinear relativity

N. Takka 111E-mail<EMAIL_ADDRESS>
Laboratoire de Physique Théorique, Faculté des Sciences Exactes,
Université de Bejaia, 06000 Bejaia, Algeria

(August 27, 2024)

###### Abstract

This work completes a serie of two papers devoted to the extension of the
fundamental laws of electrodynamics in the context of Fock’s nonlinear
relativity (FNLR). Indeed, after having established in the previous study the
exact generalizations of both Maxwell’s equations and Dirac magnetic monopole,
we present here the remaining exact Lorentz force. As in $\kappa$-Minkowski
spacetime, two different nature contributions appear in the corresponding
equation of motion where the new effect is interpreted as the gravitational-
type Lorentz force. This common point separately induced by the radius of the
universe in our case or Planck energy in other works, reinforces once more the
analogy between electromagnetism and gravity in two different scientific
approaches. As a relative difference, it is very important to highlight that
more homogeneity characterizes our results where each effect is exclusively
generated by mass or charge but not both at the same time. Even more, the new
effect emerges as the result of the triple effect of $R$-deformation, mass and
the square of velocities but completely independent of electromagnetic field.

PACS numbers : 03.30.+p, 03.50.De

Keywords : Fock-Lorentz transformation, $R$-Minkowski spacetime,
Electromagnetism

##  1 Introduction

Among many approaches to present the laws of electrodynamics, Feynman’s post-
humous paper, published by Freeman J. Dyson in $1990$, constitutes a very
elegant mathematical way for such derivation[1]. The initial motivation behind
that curiosity was to go as far as possible beyond the framework of
conventional physics in order to develop a quantization procedure without
resorting to a Lagrangian or Hamiltonian. In its original version, this
approach is based on the commutation relations between position and velocity
for a nonrelativistic particle and Newton’s law of motion valid in three-
dimensional Euclidean space. To reconcile the aforementioned proof with some
conflicts between Lorentz covariance and the Euclidean assumptions raised in
Refs. [2]-[7], Tanimura formulated the corresponding special and general
relativistic versions [8]. In this latter, it has also shown that the only
possible fields that can consistently act on a quantum particle are scalar,
gauge and gravitational fields. Some years later, the use of Hodge duality
made possible the derivation of the two groups of Maxwells equations with a
magnetic monopole in flat and curved spaces [9]. In many other works, Bérard
et al. also established a connection between Lorentz algebra symmetry and
Dirac’s magnetic monopole, e.g. Refs. [10] and [11]. Thinking in terms of
analogy, the equivalence between Feynman’s proof and minimal coupling has been
discussed in commutative space–time [14]. Moving forward to noncommutative
space–time, it is turned out that the extended first approximation of the laws
of electrodynamics in $\kappa$-Minkowski space–time depends on the particle
mass [12] and [13]. Here of course, $\kappa$ represents Planck energy divided
by the speed of light in a vacuum. In Ref. [13] this is done by power series
expansion of noncommutative coordinates, momenta and some derived functions,
in terms of commutative coordinates, momenta and the deformation parameter
$\kappa$. Finally, we have suggested in our turn a new version of Feynman’s
proof within which the electromagnetism can be studied formally [15]. With
such reasoning, we have then found the first-order approximation of Maxwell’s
equations, Lorentz force and Dirac’s magnetic monopole in FNLR [15]. A year
later and except for the equation of motion, we have succeeded to go further
than our first paper by establishing the corresponding exact formulations in
the second one [16]. Calculating the flux of a magnetostatic field generated
by the hypothetical magnetic charge across a closed surface, it is turned out
that the universe could locally contain the magnetic charge but in its
totality it is still neutral. Now, to talk briefly about FNLR, let us take as
origin [17] where the general form of the coordinates transformation between
two inertial frames, based only on the first principle of relativity, is given
by the formula

$t^{\prime}=\frac{\gamma\big{(}t-ux/c^{2}\big{)}}{\alpha_{R}},\qquad
x^{\prime}=\frac{\gamma(x-ut)}{\alpha_{R}},\qquad
y^{\prime}=\frac{y}{\alpha_{R}},\qquad z^{\prime}=\frac{z}{\alpha_{R}},$ (1)

here obviously $\gamma=\big{(}1-u^{2}/c^{2}\big{)}^{-1/2}$ and
${\alpha_{R}}=1+\left[(\gamma-1)ct-\gamma xu/c\right]/R$ where $R$ designates
the radius of the universe. To obtain a coherent theory, this latter has
undergone several steps. In fact, unlike deformed (or doubly) special
relativity (DSR), whose construction is based on three postulates by imposing
a maximal energy to special relativity [18]-[20], FNLR takes the diametrically
opposed direction by studying the implications of the non-constancy of the
speed of light in vacuum. Subsequently, this trend is greatly reinforced by
the desire to resolve some problems and paradoxes in cosmology [21] and in
quantum gravity [22]. A few years later, a further step in that perspective
has been realized by inspiring from one of the last versions of DSR [23]. In
other words, the usual Fock-Lorentz transformation (1) has been reproduced and
a new momentum transformation suggested as a result of the definition of the
appropriate deformed Poisson brackets [24]. The main novelty of such rewriting
is that the contraction $x_{\mu}p^{\mu}$ becomes an invariant which made
possible the coherent description of free particles in this context. Focusing
on the generalizations of both Klein-Gordon and Dirac equations in [25] and
[26] respectively, the correspondence between $R$-Minkowski and de Sitter
spacetime has been highlighted.

This paper is organized as follows. In section $2$, we briefly recall some
needed results found in [16] in order to find the exact Lorentz force in FNLR.
To this end, we have used one of the common points between our version of
Feynman’s proof and that developed by Tanimura [8], namely, the relativistic
Newton’s law. As consequence, a partial dependance on particle mass emerges
naturally in the corresponding equation of motion. Section $3$ is devoted to
conclusion where a special case is discussed.

##  2 Exact $R$-Lorentz force

To begin, let us recall that the quantized $R$-deformed phase space algebra is
described by the following three commutators [25]

$\displaystyle\left[x^{\mu},x^{\nu}\right]$ $\displaystyle=0,$ (2)
$\displaystyle\left[x^{\mu},p^{\nu}\right]$
$\displaystyle=-i\hbar\eta^{\mu\nu}+\frac{i\hbar}{R}\eta^{0\nu}x^{\mu},$ (3)
$\displaystyle\left[p^{\mu},p^{\nu}\right]$
$\displaystyle=-\frac{i\hbar}{R}\left(p^{\mu}\eta^{0\nu}-p^{\nu}\eta^{\mu
0}\right),$ (4)

where $\eta^{\mu\nu}=(+1,-1,-1,-1)$ and $\mu,\nu=0,1,2,3$. To derive the
generalized fundamental laws of electrodynamics in the context of FNLR, we
have developed in [15] a new relativistic version of Feynman’s proof. This
latter is based on the knowledge of phase space algebra and the explicit form
of the four-momentum valid in the absence of electromagnetic field. To this
end, we were inspired by the equivalence between the minimal coupling
prescription and Feynman’s approach highlighted in the first paper [1] and
confirmed in further works, e.g. [14] and [13]. On a quest for new
generalizations even more satisfying, we set up a symmetrization mechanism
giving the following exact formulations of momentum and commutation relations
respectively [16]

$p^{\mu}=m\Big{(}1-\frac{x^{0}}{R}\Big{)}^{-3}\left\\{\dot{x}^{\mu}+\frac{1}{R}\big{(}x^{\mu}\dot{x}^{0}-x^{0}\dot{x}^{\mu}\big{)}\right\\}+\frac{3i\hbar}{2R}\eta^{\mu
0},$ (5)

$[x^{\mu},\dot{x}^{\nu}]=\frac{i\hbar}{m}\Big{(}1-\frac{x^{0}}{R}\Big{)}^{2}\left\\{-\eta^{\mu\nu}+\frac{1}{R}\big{(}\eta^{0\nu}x^{\mu}+\eta^{\mu
0}x^{\nu}\big{)}-\frac{x^{\mu}x^{\nu}}{R^{2}}\right\\},$ (6)

$[\dot{x}^{\mu},\dot{x}^{\nu}]=-\frac{i\hbar
q}{m^{2}}F^{\mu\nu}(x)-\frac{i\hbar}{mR}\Big{(}1-\frac{x^{0}}{R}\Big{)}^{2}\Big{\\{}\big{(}\eta^{\mu
0}\dot{x}^{\nu}-\eta^{0\nu}\dot{x}^{\mu}\big{)}-\frac{1}{R}\big{(}x^{\mu}\dot{x}^{\nu}-x^{\nu}\dot{x}^{\mu}\big{)}\Big{\\}},$
(7)

obviously here $F^{\mu\nu}(x)$ describes the generalized electromagnetic
tensor. The unique dependence of this latter on position has been deduced from
the following result $\big{[}x^{\lambda},F^{\mu\nu}\big{]}=0$. Now, by
differentiating the commutator involving one coordinate and one velocity with
respect to the time parameter $\tau$, the equation of motion describing a
moving charged particle under the influence of electromagnetic field is given
by [8]

$\left[x^{\mu},F^{\nu}\right]=m\left[x^{\mu},\ddot{x}^{\nu}\right]=m\frac{d}{d\tau}\left[x^{\mu},\dot{x}^{\nu}\right]-m\left[\dot{x}^{\mu},\dot{x}^{\nu}\right].$
(8)

Above, the relativistic Newton’s law is used as an assumption allowing the
recovery of the usual results of special relativity. Given that the derivation
of $F^{\nu}$ is conditioned by the knowledge of the two intervening
commutators, to determine $\frac{d}{d\tau}\left[x^{\mu},\dot{x}^{\nu}\right]$,
we propose below a mechanism giving such calculation. First of all, we begin
by taking the time derivative of (6), to get

$\displaystyle m\frac{d}{d\tau}\left[x^{\mu},\dot{x}^{\nu}\right]$
$\displaystyle=$
$\displaystyle\frac{2i\hbar}{R}\eta^{\mu\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)\dot{x}^{0}\Big{>}-\frac{2i\hbar}{R^{2}}\left\\{\eta^{0\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}\dot{x}^{0}\Big{>}\right.$
(9) $\displaystyle+$ $\displaystyle\left.\eta^{\mu
0}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\nu}\dot{x}^{0}\Big{>}\right\\}+\frac{2i\hbar}{R^{3}}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}x^{\nu}\dot{x}^{0}\Big{>}$
$\displaystyle+$ $\displaystyle\frac{i\hbar}{R}\left\\{\eta^{\mu
0}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\nu}\Big{>}+\eta^{0\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\mu}\Big{>}\right\\}$
$\displaystyle-$
$\displaystyle\frac{i\hbar}{R^{2}}\left\\{\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\mu}\dot{x}^{\nu}\Big{>}+\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\nu}\dot{x}^{\mu}\Big{>}\right\\},$

where $<.>$ refers to the symmetrization operator [8]. To explicit each of the
above terms, we proceed one by one. First, starting from the definition of the
commutator

$\dot{x}^{\mu}x^{0}=x^{0}\dot{x}^{\mu}-[x^{0},\dot{x}^{\mu}],$ (10)

where

$[x^{\mu},[x^{\nu},\dot{x}^{\lambda}]]=0$ (11)

and

$[x^{\mu},\dot{x}^{\nu}]=[x^{\nu},\dot{x}^{\mu}],$ (12)

we get the following generalization [16]

$\big{<}f(x^{0})\dot{x}^{\mu}\big{>}=f(x^{0})\dot{x}^{\mu}-\frac{1}{2}\frac{df(x^{0})}{dx^{0}}[x^{0},\dot{x}^{\mu}],$
(13)

which, by means of (6), gives

$\frac{2i\hbar}{R}\eta^{\mu\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)\dot{x}^{0}\Big{>}=\frac{2i\hbar}{R}\eta^{\mu\nu}\left(1-\frac{x^{0}}{R}\right)\dot{x}^{0}+\frac{\hbar^{2}}{mR^{2}}\eta^{\mu\nu}\left(1-\frac{x^{0}}{R}\right)^{4}.$
(14)

To calculate the other two terms appearing in the same order in (9), let us
recall that for three operators $A$, $B$, and $C$ satisfying the following
commutation relations

$[A,B]=0\hskip 5.69046pt\text{and}\hskip 5.69046pt[A,[B,C]]=0,$ (15)

we have [16]

$\big{<}ABC\big{>}=\frac{1}{2}A\big{<}BC\big{>}+\frac{1}{2}\big{<}BC\big{>}A$
(16)

and then the second term can be written as

$\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}\dot{x}^{0}\Big{>}=\frac{1}{2}\left(1-\frac{x^{0}}{R}\right)\left<x^{\mu}\dot{x}^{0}\right>+\frac{1}{2}\left<x^{\mu}\dot{x}^{0}\right>\left(1-\frac{x^{0}}{R}\right).$
(17)

Using Eq. (6), we can make sure that

$\left<x^{\mu}\dot{x}^{0}\right>=x^{\mu}\dot{x}^{0}+\frac{1}{2}\left[\dot{x}^{0},x^{\mu}\right]=x^{\mu}\dot{x}^{0}+\frac{i\hbar}{2m}\left(1-\frac{x^{0}}{R}\right)^{3}\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right)$ (18)

and similarly

$\dot{x}^{0}\left(1-\frac{x^{0}}{R}\right)=\left(1-\frac{x^{0}}{R}\right)\dot{x}^{0}-\frac{i\hbar}{mR}\left(1-\frac{x^{0}}{R}\right)^{4},$
(19)

which, after substitution into (17), yields

$\displaystyle-\frac{2i\hbar}{R^{2}}\eta^{0\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}\dot{x}^{0}\Big{>}=$
$\displaystyle-$
$\displaystyle\frac{2i\hbar}{R^{2}}\eta^{0\nu}\left(1-\frac{x^{0}}{R}\right)x^{\mu}\dot{x}^{0}-\frac{2\hbar^{2}}{mR^{3}}\eta^{0\nu}\left(1-\frac{x^{0}}{R}\right)^{4}x^{\mu}$
(20) $\displaystyle+$
$\displaystyle\frac{\hbar^{2}}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\eta^{\mu
0}\eta^{0\nu}.$

After interchanging $\mu$ with $\nu$ in the last result, the third term is
given by

$\displaystyle-\frac{2i\hbar}{R^{2}}\eta^{\mu
0}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\nu}\dot{x}^{0}\Big{>}=$
$\displaystyle-$ $\displaystyle\frac{2i\hbar}{R^{2}}\eta^{\mu
0}\left(1-\frac{x^{0}}{R}\right)x^{\nu}\dot{x}^{0}-\frac{2\hbar^{2}}{mR^{3}}\eta^{\mu
0}\left(1-\frac{x^{0}}{R}\right)^{4}x^{\nu}$ (21) $\displaystyle+$
$\displaystyle\frac{\hbar^{2}}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\eta^{\mu
0}\eta^{0\nu}.$

To continue, we generalize in its turn Eq. (16). Indeed, for four operators
$A$, $B$ $C$ and $D$ satisfying the following commutation relations

$[A,B]=[B,C]=0\hskip 5.69046pt\text{and}\hskip
5.69046pt[A,[B,D]]=[A,[C,D]]=0,$ (22)

we get

$\big{<}ABCD\big{>}=\frac{1}{4}\left[AB\big{<}CD\big{>}+B\big{<}CD\big{>}A+A\big{<}CD\big{>}B+\big{<}CD\big{>}AB\right].$
(23)

Identifying the last relation with the four term listed into (9), we obtain

$\displaystyle\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}x^{\nu}\dot{x}^{0}\Big{>}$
$\displaystyle=$
$\displaystyle\frac{1}{4}\left[\left(1-\frac{x^{0}}{R}\right)x^{\mu}\left<x^{\nu}\dot{x}^{0}\right>+x^{\mu}\left<x^{\nu}\dot{x}^{0}\right>\left(1-\frac{x^{0}}{R}\right)\right.$
(24) $\displaystyle\hskip
5.69046pt\left.+\left(1-\frac{x^{0}}{R}\right)\left<x^{\nu}\dot{x}^{0}\right>x^{\mu}+\left<x^{\nu}\dot{x}^{0}\right>\left(1-\frac{x^{0}}{R}\right)x^{\mu}\right].$

With the same reasoning, we can check that the use of Eqs. (6), (10), (18) and
(19) transforms the last relation as follows

$\displaystyle\frac{2i\hbar}{R^{3}}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}x^{\nu}\dot{x}^{0}\Big{>}$
$\displaystyle=$
$\displaystyle\frac{2i\hbar}{R^{3}}\left(1-\frac{x^{0}}{R}\right)x^{\mu}x^{\nu}\dot{x}^{0}-\frac{\hbar^{2}}{mR^{3}}\left(1-\frac{x^{0}}{R}\right)^{4}\left[\left(\eta^{\mu
0}x^{\nu}+\eta^{0\nu}x^{\mu}\right)\right.$ (25) $\displaystyle-$
$\displaystyle\left.\frac{3}{2}x^{\mu}x^{\nu}\right].$

After that, taking into account Eqs. (6) and (13), we can make sure that the
fifth and sixth terms are respectively written as

$\frac{i\hbar}{R}\eta^{\mu
0}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\nu}\Big{>}=\frac{i\hbar}{R}\eta^{\mu
0}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\nu}+\frac{\hbar^{2}}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\eta^{\mu
0}\left(\eta^{0\nu}-\frac{x^{\nu}}{R}\right)$ (26)

and

$\frac{i\hbar}{R}\eta^{0\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\mu}\Big{>}=\frac{i\hbar}{R}\eta^{0\nu}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\mu}+\frac{\hbar^{2}}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\eta^{0\nu}\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right).$ (27)

To find the seventh one, making use of Eq. (16) to get

$\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\mu}\dot{x}^{\nu}\Big{>}=\frac{1}{2}\left(1-\frac{x^{0}}{R}\right)^{2}\left<x^{\mu}\dot{x}^{\nu}\right>+\frac{1}{2}\left<x^{\mu}\dot{x}^{\nu}\right>\left(1-\frac{x^{0}}{R}\right)^{2}.$
(28)

As for (18), we can check that

$\left<x^{\mu}\dot{x}^{\nu}\right>=x^{\mu}\dot{x}^{\nu}+\frac{i\hbar}{2m}\Big{(}1-\frac{x^{0}}{R}\Big{)}^{2}\left\\{\eta^{\mu\nu}-\frac{1}{R}\big{(}\eta^{0\nu}x^{\mu}+\eta^{\mu
0}x^{\nu}\big{)}+\frac{x^{\mu}x^{\nu}}{R^{2}}\right\\}.$ (29)

Exploiting Eqs. (10) and (11), we can prove by mathematical induction that for
any regular function $f$ of $x^{0}$, we have

$\dot{x}^{\nu}f(x^{0})=f(x^{0})\dot{x}^{\nu}-\frac{df(x^{0})}{dx^{0}}[x^{0},\dot{x}^{\nu}]$
(30)

and then

$\dot{x}^{\nu}\left(1-\frac{x^{0}}{R}\right)^{2}=\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\nu}-\frac{2i\hbar}{mR}\left(1-\frac{x^{0}}{R}\right)^{4}\left(\eta^{0\nu}-\frac{x^{\nu}}{R}\right).$
(31)

By means of the last result, the substitution of Eq. (29) into (28) yields

$\displaystyle-\frac{i\hbar}{R^{2}}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\mu}\dot{x}^{\nu}\Big{>}=$
$\displaystyle-$
$\displaystyle\frac{i\hbar}{R^{2}}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\mu}\dot{x}^{\nu}-\frac{\hbar^{2}}{mR^{3}}\left(1-\frac{x^{0}}{R}\right)^{4}\left(\eta^{0\nu}-\frac{x^{\nu}}{R}\right)x^{\mu}$
(32) $\displaystyle+$
$\displaystyle\frac{i\hbar}{2R^{2}}\left(1-\frac{x^{0}}{R}\right)^{2}[x^{\mu},\dot{x}^{\nu}],$

which, after interchanging $\mu$ with $\nu$, gives the following last term

$\displaystyle-\frac{i\hbar}{R^{2}}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\nu}\dot{x}^{\mu}\Big{>}=$
$\displaystyle-$
$\displaystyle\frac{i\hbar}{R^{2}}\left(1-\frac{x^{0}}{R}\right)^{2}x^{\nu}\dot{x}^{\mu}-\frac{\hbar^{2}}{mR^{3}}\left(1-\frac{x^{0}}{R}\right)^{4}\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right)x^{\nu}$ (33) $\displaystyle+$
$\displaystyle\frac{i\hbar}{2R^{2}}\left(1-\frac{x^{0}}{R}\right)^{2}[x^{\mu},\dot{x}^{\nu}].$

Finally, by substituting all intervening terms expressed by (14), (20), (21),
(25), (26), (27), (32) and (33) into (9), we get after simplification

$\displaystyle m\frac{d}{d\tau}\left[x^{\mu},\dot{x}^{\nu}\right]$
$\displaystyle=$
$\displaystyle-\frac{2m}{R}\left(1-\frac{x^{0}}{R}\right)^{-1}[x^{\mu},\dot{x}^{\nu}]\dot{x}^{0}+\frac{i\hbar}{R}\left(1-\frac{x^{0}}{R}\right)^{2}\left\\{\left(\eta^{0\nu}\dot{x}^{\mu}+\eta^{\mu
0}\dot{x}^{\nu}\right)\right.$ (34)
$\displaystyle\left.-\frac{1}{R}\left(x^{\nu}\dot{x}^{\mu}+x^{\mu}\dot{x}^{\nu}\right)+\frac{1}{R}[x^{\mu},\dot{x}^{\nu}]\right\\}+\frac{\hbar^{2}}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\left\\{\left(\eta^{\mu\nu}+4\eta^{\mu
0}\eta^{0\nu}\right)\right.$
$\displaystyle\left.-\frac{5}{R}\left(\eta^{0\nu}x^{\mu}+\eta^{\mu
0}x^{\nu}\right)+\frac{5}{R^{2}}x^{\mu}x^{\nu}\right\\}.$

Taking into account (7) and (34), Eq.(8) gives

$\displaystyle\left[x^{\mu},F^{\nu}\right]$ $\displaystyle=$
$\displaystyle\frac{i\hbar
q}{m}F^{\mu\nu}(x)-\frac{2m}{R}\left(1-\frac{x^{0}}{R}\right)^{-1}\left\\{[x^{\mu},\dot{x}^{\nu}]\dot{x}^{0}+[x^{\mu},\dot{x}^{0}]\dot{x}^{\nu}\right\\}$
(35) $\displaystyle+$
$\displaystyle\frac{i\hbar}{R^{2}}\left(1-\frac{x^{0}}{R}\right)^{2}[x^{\mu},\dot{x}^{\nu}]+\frac{\hbar^{2}}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\left\\{\left(\eta^{\mu\nu}+4\eta^{\mu
0}\eta^{0\nu}\right)\right.$ $\displaystyle-$
$\displaystyle\left.\frac{5}{R}\left(\eta^{0\nu}x^{\mu}+\eta^{\mu
0}x^{\nu}\right)+\frac{5}{R^{2}}x^{\mu}x^{\nu}\right\\}.$

Observing the last result where
$[x^{\mu},\dot{x}^{\nu}]=[x^{\mu},\dot{x}^{\nu}](x)$, we remark that
$\left[x^{\mu},F^{\nu}\right]$ contains two kind of terms. One depends only on
position and another is proportional to the velocity. As consequence, Eqs. (2)
and (11) indicate that the resulting force can be written as

$F^{\nu}=q\big{<}\tilde{F}^{\nu\alpha}(x)\dot{x}_{\alpha}\big{>}+m\big{<}\Gamma^{\nu\alpha\beta}(x)\dot{x}_{\alpha}\dot{x}_{\beta}\big{>}+G^{\nu}(x)$
(36)

and then

$\left[x^{\mu},F^{\nu}\right]=q\tilde{F}^{\nu\alpha}(x)[x^{\mu},\dot{x}_{\alpha}]+m\big{<}\Gamma^{\nu\alpha\beta}(x)\left\\{\left[x^{\mu},\dot{x}_{\alpha}\right]\dot{x}_{\beta}+\dot{x}_{\alpha}\left[x^{\mu},\dot{x}_{\beta}\right]\right\\}\big{>},$
(37)

which, after identification with Eq. (35), yields

$\Gamma^{\nu\alpha\beta}(x)=-\frac{2}{R}\left(1-\frac{x^{0}}{R}\right)^{-1}\eta^{0\alpha}\eta^{\nu\beta}.$
(38)

Above $\Gamma^{\nu\alpha\beta}(x)$ represents the electromagnetic-type
Christoffel symbols and $G^{\nu}(x)$ is just an integration function depending
only on position. To rigorously confirm the correctness of such mathematical
deduction, we seek the appropriate $\tilde{F}^{\nu\alpha}(x)$ by the
identification reasoning. To do it, we exploit (38) in order to calculate the
two last terms appearing on the right side of (37). First, using the last
result, it follows that Eqs. (6) and (16) give

$\displaystyle\big{<}\Gamma^{\nu\alpha\beta}(x)\left[x^{\mu},\dot{x}_{\alpha}\right]\dot{x}_{\beta}\big{>}$
$\displaystyle=$ $\displaystyle\frac{i\hbar}{mR}\left\\{\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right)\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\nu}\Big{>}\right.$
(39) $\displaystyle+$
$\displaystyle\left.\Big{<}\left(1-\frac{x^{0}}{R}\right)^{2}\dot{x}^{\nu}\Big{>}\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right)\right\\},$

which, by using (26) where

$\dot{x}^{\nu}\left(\eta^{\mu 0}-\frac{x^{\mu}}{R}\right)=\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right)\dot{x}^{\nu}-\frac{1}{R}\left[\dot{x}^{\nu},x^{\mu}\right],$
(40)

reduces to

$\displaystyle\big{<}\Gamma^{\nu\alpha\beta}(x)\left[x^{\mu},\dot{x}_{\alpha}\right]\dot{x}_{\beta}\big{>}$
$\displaystyle=$
$\displaystyle-\frac{2}{R}\left(1-\frac{x^{0}}{R}\right)^{-1}\left[x^{\mu},\dot{x}^{0}\right]\dot{x}^{\nu}+\frac{i\hbar}{mR^{2}}\left(1-\frac{x^{0}}{R}\right)^{2}\left[x^{\mu},\dot{x}^{\nu}\right]$
(41)
$\displaystyle+\frac{2\hbar^{2}}{m^{2}R^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\left\\{\eta^{\mu
0}\eta^{0\nu}-\frac{1}{R}\big{(}\eta^{0\nu}x^{\mu}+\eta^{\mu
0}x^{\nu}\big{)}+\frac{x^{\mu}x^{\nu}}{R^{2}}\right\\}.$

Next, taking into account (6) and (38), we get

$\displaystyle\big{<}\Gamma^{\nu\alpha\beta}(x)\left[x^{\mu},\dot{x}_{\beta}\right]\dot{x}_{\alpha}\big{>}$
$\displaystyle=$
$\displaystyle\frac{2i\hbar}{mR}\eta^{\mu\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)\dot{x}^{0}\Big{>}-\frac{2i\hbar}{mR^{2}}\left\\{\eta^{\mu
0}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\nu}\dot{x}^{0}\Big{>}\right.$ (42)
$\displaystyle+$
$\displaystyle\left.\eta^{0\nu}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}\dot{x}^{0}\Big{>}\right\\}+\frac{2i\hbar}{mR^{3}}\Big{<}\left(1-\frac{x^{0}}{R}\right)x^{\mu}x^{\nu}\dot{x}^{0}\Big{>}$

which, after using (14), (20), (21) and (25), yields

$\displaystyle\big{<}\Gamma^{\nu\alpha\beta}(x)\left[x^{\mu},\dot{x}_{\beta}\right]\dot{x}_{\alpha}\big{>}$
$\displaystyle=$
$\displaystyle-\frac{2}{R}\left(1-\frac{x^{0}}{R}\right)^{-1}\left[x^{\mu},\dot{x}^{\nu}\right]\dot{x}^{0}+\frac{\hbar^{2}}{m^{2}R^{2}}\left(1-\frac{x^{0}}{R}\right)^{4}\left\\{\left(\eta^{\mu\nu}+2\eta^{\mu
0}\eta^{0\nu}\right)\right.$ (43)
$\displaystyle\left.-\frac{3}{R}\big{(}\eta^{0\nu}x^{\mu}+\eta^{\mu
0}x^{\nu}\big{)}+\frac{3}{R^{2}}x^{\mu}x^{\nu}\right\\}.$

By summing (41) with (43) and identifying the final result with (35), we find

$\displaystyle\left[x^{\mu},F^{\nu}\right]$ $\displaystyle=$
$\displaystyle\frac{i\hbar
q}{m}F^{\mu\nu}(x)+m\big{<}\Gamma^{\nu\alpha\beta}(x)\left(\left[x^{\mu},\dot{x}_{\alpha}\right]\dot{x}_{\beta}+\dot{x}_{\alpha}\left[x^{\mu},\dot{x}_{\beta}\right]\right)\big{>},$
(44)

which, by comparing it with (37), gives

$\tilde{F}^{\nu\alpha}(x)[x^{\mu},\dot{x}_{\alpha}]=\frac{i\hbar}{m}F^{\mu\nu}(x).$
(45)

To go further, we put $\mu=0$ in the last relation in order to determine the
final expression of $\tilde{F}^{\nu\alpha}(x)$. Indeed, using Eq. (6), it
results

$\frac{1}{R}\tilde{F}^{\nu}_{\alpha}x^{\alpha}=-\left(1-\frac{x^{0}}{R}\right)^{-3}F^{\nu
0}+\tilde{F}^{\nu 0}.$ (46)

which, by substituting it in (45), yields

$\tilde{F}^{\nu\mu}=\left(1-\frac{x^{0}}{R}\right)^{-2}F^{\nu\mu}-\left(1-\frac{x^{0}}{R}\right)^{-3}\left(\eta^{\mu
0}-\frac{x^{\mu}}{R}\right)F^{\nu 0}+\eta^{\mu 0}\tilde{F}^{\nu 0}.$ (47)

By putting above $\nu=0$, it follows

$\tilde{F}^{0\mu}=\left(1-\frac{x^{0}}{R}\right)^{-2}F^{0\mu}$ (48)

and then (47) takes the following final form

$\tilde{F}^{\nu\mu}=\left(1-\frac{x^{0}}{R}\right)^{-2}F^{\nu\mu}+\frac{1}{R}\left(1-\frac{x^{0}}{R}\right)^{-3}\left(x^{\mu}-\eta^{\mu
0}x^{0}\right)F^{\nu 0}.$ (49)

To zeroth order in $1/R$, we get $\tilde{F}^{\nu\mu}=F^{\nu\mu}$. Finally, the
substitution of (38) and (49) into (36) gives the following exact form of
Lorentz force in Fick’s nonlinear relativity

$\displaystyle F^{\nu}$ $\displaystyle=$ $\displaystyle
q\Big{<}\left\\{\left(1-\frac{x^{0}}{R}\right)^{-2}F^{\nu\mu}+\frac{1}{R}\left(1-\frac{x^{0}}{R}\right)^{-3}\left(x^{\mu}-\eta^{\mu
0}x^{0}\right)F^{\nu 0}\right\\}\dot{x}_{\mu}\Big{>}$ (50) $\displaystyle-$
$\displaystyle\frac{2m}{R}\Big{<}\left(1-\frac{x^{0}}{R}\right)^{-1}\dot{x}^{0}\dot{x}^{\nu}\Big{>}+G^{\nu}(x).$

Analysing the above result, we remark that the particle mass acts so that for
any two particles with the same charge, submitted to the same electromagnetic
field, will not necessarily feel the same force if their masses are different.
Surprisingly, it is notable that mass and charge emerge independently of each
other in two distinct terms where the new effect is completely independent of
electromagnetic field. As one consequence, even for a neutral particle, moving
without the influence of electromagnetic field, a particular kind of force
acts on it (gravitational-type Lorentz force). Using Eqs. (46) and (48), we
can check that Eq. (50) reproduces the first order approximation previously
found in [15]. In the temporal limit where $x^{0}=R$, it is obvious that no
force is generated $(F^{0}=0,\hskip 2.84544pt\vec{F}=\vec{0})$.

##  3 Conclusion

This work constitutes another effort to better understand the impact of the
variation of the fundamental physical constants on the basic laws of
electrodynamics. Thus, to find the first-order approximation of Lorentz force
in the context of Fock’s nonlinear relativity, we have developed in Ref. [15]
an adequate procedure for such derivation. Building on this latter, we have
highlighted in Ref. [16] that an iterative method could be used in order to
move towards the higher-order terms. However, it is turned out that this
reasoning quickly becomes obsolete as we seek to increase the precision (even
for the second order, the calculation is already too complicated). To overcome
this difficulty, we have succeeded here not only to simplify the first
procedure but we also went further by proposing an elegant exact study. In
doing so, we have shown that the aforementioned force is decomposed into two
different contributions where each effect is reinforced in its proper nature
by a separability of variables. Indeed, in comparison with the generalized
Lorentz force found in Ref. [13], valid up to first order in the deformation
parameter $\kappa$, we emphasize that mass and charge intervene linearly and
independently of each other in our work but as a product in the
$\kappa$-deformed one. Interpreting the two effects appearing in the final
result, the first term linear in velocity, depending on charge but not on
mass, is named the electric-type Lorentz force, the second one proportional to
the square of the velocity, depending on mass but not on charge, is
interpreted as the gravitational-type Lorentz force. Consequently, more
homogeneity regarding the cause and effect relationship takes place as a new
important particularity in our context. More specifically, it thus seems that
the similarity between electromagnetism and gravity is more visible in FNLR
than in $\kappa$-Minkowski space–time.

To summarize a little bit better what has been shown until now, it is very
useful to shed light on some important points. First of all, the reasoning
developed above is nothing else than a logical continuation of the two last
papers where the theoretical background is constructed in a coherent way.
Indeed, our approach is based on our version of Feynman’s proof, itself based
on both the relativistic version of this latter and the use of the momentum
valid in the absence of the electromagnetic field. Mathematically, all above
calculations can be viewed as a symmetrization mechanism allowing an
identification work. As one perspective, the ordinary Lagrangian formulation
could be envisaged as an approximate study since the calculations seem to be
complicated even for the first order, e.g. DSR [27]. Generally speaking, the
standardization of our resolution procedure is technically possible but only
conditioned by the mathematical development taking into account the
specificities of each new context. Secondly, in addition to the exact nature
of our predictions, some new particularities concerning both the content and
form can be illustrated. For example, the separability between mass and charge
in the R-deformed Lorentz force is in accordance with the independence of the
two physical characteristics in standard physics. From a practical point of
view, the relevance of our studies is the proposition of new theoretical
challenges to check the validity of the invariance of the speed of light, the
homogeneity of space and the uniformity of time in the cosmological context
[24]. In an equivalent way, the theoretical results highlighted in FNLR, [15],
[16], [25],[26] provide a new series of occasions to test one of the most
fundamental concepts of physics, namely, the conservation of the energy at the
scale of the universe [24]. To examine the scientific soundness of our
predictions, it is obvious that the empirical or observational studies must be
focused on the physical phenomena of the large scale in space and/or time. In
the same order of ideas, we emphasize that all our final results take a very
condensed form despite the fact that the calculations are very long. Also, the
non-use of any other assumption beside the relativistic Newton’s law, making
possible the reproduction of the usual Lorentz force, constitutes in itself
another technical and mathematical prowess. Before finishing, let us take the
most simplest case characterized by the classical static limit. In fact, by
considering two identical charged particles $q$ at rest and separated by a
distance $r$, it follows that $\dot{x}^{0}=1$ and $\dot{x}^{i}=0$ which
implies that Eq. (50) reduces to

$\vec{F}=q\left(1-\frac{x^{0}}{R}\right)^{-2}\vec{E}.$ (51)

In contrast to $\kappa$-Minkowski space–time, where the first extended form of
Coulomb’s law depends on mass and charge, [13] we note here an exclusive
dependence on charge in our exact result. This suggests that the
gravitational-type effect is mainly related to the movement which supports
even more the conclusion drawn just before. Based on these consequences, many
other analogies and more detailed analysis could be envisaged in other works.

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