On the Radiation Reaction Force

T. Matolcsi111Department of Applied Analysis and Computational Mathematics,
Eötvös Loránd University, Budapest, Hungary

###### Abstract

The usual radiation self-force of a point charge is obtained in a
mathematically exact way and it is pointed out to that this does not call
forth that the spacetime motion of a point charge obeys the
Lorentz–Abraham–Dirac equation.

## Introduction

The equation of spacetime motion of a point charge under the action of an
external force is an old problem of electromagnetism. The original work of
Abraham (treated e.g. in [2], Ch. 16.3) and then Dirac’s formulation (treated
e.g. in [3], Ch III.3) suffers from a hardly acceptable mass renormalization
according to which the finite mass of a particle is the difference of the
positive infinite electric mass and the negative infinite mechanical mass.
Other, quick deductions based on Larmor’s formula and some other ‘natural’
requirements (see [2], Ch 16.2) are not convincing either. Moreover, the LAD
equation admits physically untenable run-away solutions. There are some
attmepts to exclude such non-physical solutions by imposing extra conditions.
(e.g. [11], [12]). There are a number of articles (e.g.[7], [5], [6], [9],
[4]) considering various continuum charge distributions instead of a point
charge and then taking or not the limit to a point. Though casting more and
more new light on the problem and its possible solution, neither way seems
satsifactory.

A recent article [8] – though its result rests on an erroneous basis, see [20]
– points out clearly to Dirac’s unjustified applications of Gauss–Stokes
theorem, Taylor expansion and the limit procedure when an extended charge is
shrunk to a point.

In some papers (e.g. [10], [13],[9]) special applications of Distribution
Theory can be found when treating point charges.

In the present paper, by a systematic use of Distribution Theory, it is shown
that not a mathematically incorrect derivation of the radiation reaction force
but a physical misapprehension is the source of why the LAD equation does not
work well.

The coordinate-free formulation of spacetime expounded in [1] is used which
makes formulae shorter and more easily comprehensible; find enclosed a brief
summary of the fundamental notions and some special notations.

Spacetime points and spacetime vectors (which are often confused in
coordinates) are distinguished:

– $\mathrm{M}$, mathematically an affine space, is the set of spacetime
points, its elements are denoted by normal letters, $x$, $y$ etc.,

– $\mathbf{M}$, mathematically a vector space, is the set of spacetime
vectors, its elements are denoted by boldface letters, $\mathbk x$, $\mathbk
y$ etc.,

– the set of time periods is $\mathbb{T}$,

– the exact treatment requires the tensorial quotients of $\mathbf{M}$ by
$\mathbb{T}$, here we do not refer to them explicitely,

– the Lorentz product of the vectors $\mathbk x$ and $\mathbk y$ is denoted by
$\mathbk x\cdot\mathbk y$ (which is $x_{k}y^{k}$ in coordinates); $\mathbk x$
is timelike if $\mathbk x\cdot\mathbk x<0$ (which means in coordinates the
signature $(-1,1,1,1)$ of the Lorentz form),

– an absolute velocity is a futurelike vector $\mathbk u$ for which $\mathbk
u\cdot\mathbk u=-1$,

– for an absolute velocity $\mathbk u$, $\mathbf{S}u:=\\{\mathbk
x\in\mathbf{M}\mid\mathbk u\cdot\mathbk x=0\\}$, the set of $\mathbk
u$-spacelike vectors, is a three dimensional Euclidean vector space,

– the action of linear and bilinear maps is denoted by a dot, too; e.g the
action of a linear map $\mathbk L$ on a vector $\mathbk x$ is $\mathbk
L\cdot\mathbk x$ (which is ${L^{i}}_{k}x^{k}$ in coordinates),

– the adjoint of a linear map $\mathbk L$ is the linear map $\mathbk L^{*}$
defined by $(\mathbk L^{*}\cdot\mathbk x)\cdot\mathbk y=\mathbk x\cdot(\mathbk
L\cdot\mathbk y)$ for all vectors $\mathbk x,\mathbk y$; shortly, $\mathbk
L^{*}\cdot\mathbk x=\mathbk x\cdot\mathbk L$,

– $\mathbk L$ is a Lorentz transformation if and only if $\mathbk
L^{*}=\mathbk L^{-1}$,

– the tensor product $\mathbk x\otimes\mathbk y$ of the vectors $\mathbk x$
and $\mathbk y$ is a linear map defined by $(\mathbk x\otimes\mathbk
y)\cdot\mathbk z:=\mathbk x(\mathbk y\cdot\mathbk z)$ (in coordinates it is
$x^{i}y^{k}$); $(x\otimes\mathbk y)^{*}=\mathbk y\otimes\mathbk x$; $\mathbk
x\land\mathbk y:=\mathbk x\otimes\mathbk y-(\mathbk x\otimes\mathbk y)^{*}$
and $\mathbk x\lor\mathbk y:=\mathbk x\otimes\mathbk y+(\mathbk
x\otimes\mathbk y)^{*}$ are the antisymmetric and symmetric tensor product,
respectively,

– the formulae of electromagnetism are written in the choice $c=1$, $\hbar=1$
and the electric charge is measured by real numbers.

– $\mathrm{D}$ denotes the differentiation in spacetime ($\partial_{k}$ in
coordinates). The antisymmetric derivative of a vector valued function
$\mathbk f$ is denoted by $\mathrm{D}\land\mathbk f$
($\partial_{k}f_{i}-\partial_{i}f_{k}$ in coordinates). The spacetime
divergence of $\mathbk f$ is denoted by $\mathrm{D}\cdot\mathbk f$
($\partial_{k}f^{k}$ in coordinates). Similarly, $\mathrm{D}\cdot\mathbk T$ is
the spacetime divergence of a tensor valued function $\mathbk T$
($\partial_{k}T^{ik}$ in coordinates).

– $\mathcal{D}$ denotes differentiation of other functions, e.g.
parameterizations.

The word ‘distribution’ will appear in two different senses: 1. having its
everyday meaning, 2. being a mathematical notion. To distinguish between them,
I write distribution for the first one and Distribution for the second one.

The usual setting of Distribution Theory is based on $\mathbb{R}^{n}$ ([16],
[17],[18], [19]). It is a quite simple generalization that spacetime and an
observer space are taken instead of $\mathbb{R}^{4}$ and $\mathbb{R}^{3}$.
Another simple generalization is that vector and tensor Distributions are
included, too. The present article can be understood without a thorough
knowledge of Distributions; besides the elementary notions the only non
trivial one is pole taming, described in the Appendix.

The guiding principle is that only quantities definable in Distribution Theory
can make sense. Distributions are denoted by calligraphic letters. Locally
integrable functions and the corresponding Distributions are distinguished in
notation; e.g. $\mathbk F$ is an electromagnetic field function,
$\boldsymbol{\mathcal{F}}$ is the corresponding electromagnetic field
Distribution.

## Electrodynamics of point charges

###  1 The retarded proper time

The existence of a material point in spacetime is described by a world line
function whose variable is the proper time; the range of a world line function
is a world line, a one dimensional submanifold in spacetime.

Let $r$ be a given continuously differentiable world line function. For a
world point $x$, $\mathbk s_{r}(x)$ will denote the retarded proper time
corresponding to $x$ i.e. the proper time point of $r$ for which $x-r(\mathbk
s_{r}(x))$ is future lightlike or zero; evidently, it is zero if and only if
$x$ is on the world line.

In other words, $\mathbk s_{r}(x)$ is defined implicitly by the relations

$(x-r(\mathbk s_{r}(x)))\cdot(x-r(\mathbk s_{r}(x)))=0,\qquad-\mathbk
u\cdot(x-r(\mathbk s_{r}(x)))\geq 0$

for an arbitrary absolute velocity $\mathbk u$.

According to the implicit function theorem, $\mathbk s_{r}$ is continuously
differentiable outside the world line; differentiating the first equality
above, we obtain that $2(x-r(\mathbk
s_{r}(x))\cdot\bigl{(}\boldsymbol{1}-\dot{r}(\mathbk
s_{r}(x))\otimes\mathrm{D}\mathbk s_{r}[x]\bigr{)}=0$ i.e.

$\mathrm{D}\mathbk s_{r}[x]=\frac{x-r(\mathbk s_{r}(x))}{\dot{r}(\mathbk
s_{r}(x))\cdot(x-r(\mathbk s_{r}(x)))}\qquad(x\notin\mathrm{Ran}r).$

Though $\mathbk s_{r}$ is not differentiable on the world line, it is
continuous there which can be seen as follows.

Let $\mathbk s_{-}<s_{0}<\mathbk s_{+}$ be arbitrary proper time points of the
world line. If $T^{\rightarrow}$ denotes the set of futurelike vectors then
$\bigl{(}r(\mathbk s_{-})+T^{\rightarrow}\bigr{)}\cap\bigl{(}r(\mathbk
s_{+})-T^{\rightarrow}\bigr{)}$ is a bounded open set containing $r(\mathbk
s_{0})$. If $x$ is in this set then $r(\mathbk s_{r}(x))$ is in it, too;
consequently the function $x\mapsto x-r(\mathbk s_{r}(x))$ is bounded.

The proper time passed between two points of the world line is less or equal
than the inertial time, thus

$\displaystyle|\mathbk s_{r}(x)-\mathbk s_{0}|^{2}$
$\displaystyle\leq-\bigl{(}r(\mathbk s_{r}(x)-r(\mathbk
s_{0}))\bigr{)}\cdot\bigl{(}r(\mathbk s_{r}(x)-r(\mathbk s_{0}))\bigr{)}=$
$\displaystyle=-\bigl{(}r(\mathbk s_{r}(x)-x)+(x-r(\mathbk
s_{0}))\bigr{)}\cdot\bigl{(}(r(\mathbk s_{r}(x)-x)+(x-r(\mathbk
s_{0}))\bigr{)}=$ $\displaystyle=2\bigl{(}x-r(\mathbk
s_{r}(x))\bigr{)}\cdot(x-r(\mathbk s_{0}))-(x-r(\mathbk
s_{0}))\cdot(x-r(\mathbk s_{0})).$

Because of the mentioned boundedness, the right hand side tends to zero as $x$
tends to $r(\mathbk s_{0})$.

###  2 The electromagnetic field produced
by a point charge

Let $r$ be a given twice differentiable world line function of a point charge
$e$.

Let us introduce the following functions:

$\mathbk R_{r}(x):=x-r(\mathbk s_{r}(x)),\qquad\mathbk
u_{r}(x):=\dot{r}(\mathbk s_{r}(x)),\qquad\mathbk a_{r}(x):=\ddot{r}(\mathbk
s_{r}(x)),$ (1)

$\mathbk L_{r}:=\frac{\mathbk R_{r}}{-\mathbk u_{r}\cdot\mathbk
R_{r}}=-\mathrm{D}\mathbk s_{r},\qquad\mathbk d_{r}:=\mathbk a_{r}+(\mathbk
a_{r}\cdot\mathbk L_{r})\mathbk u_{r}.$ (2)

Then it is known (see [2]) that the electromagnetic field produced by a point
charge $e$ with a given world line function $r$ is the regular Distribution
$\boldsymbol{\mathcal{F}}[r]$ defined by the locally integrable function

$\mathbk F[r]:=\frac{e}{4\pi}\frac{\mathbk u_{r}\land\mathbk L_{r}}{(-\mathbk
u_{r}\cdot\mathbk R_{r})^{2}}+\frac{e}{4\pi}\frac{\mathbk d_{r}\land\mathbk
L_{r}}{-\mathbk u_{r}\cdot\mathbk R_{r}};$ (3)

the first term is the tied field $\mathbk F[r]^{td}$ which is never zero, the
second term is the radiated field $\mathbk F[r]^{rd}$ which is zero if and
only if the charge is not accelerated.

It is emphasized by the notation $[r]$ that the formulae above make sense only
if the world line function is given because the retarded proper time is
defined only in that case.

###  3 ‘Energy-momentum tensor’ of a point charge

As usual, for an electromagnetic field function $\mathbk F$,

$\mathbk T:=-\mathbk F\cdot\mathbk F-\frac{1}{4}\mathrm{Tr}\mathbk
F\cdot\mathbk F$ (4)

is considered the energy-momentum tensor whose negative spacetime divergence
$-\mathrm{D}\cdot\mathbk T$ is the spacetime force density.

The problem is that $\mathbk T$, in general, is not locally integrable, thus
$\mathbk T$ does not define a Distribution a priori, so it has no physical
meaning. In particular, this occurs for a point charge; that is why the
quotation mark appeared in the section title. According to (3), the ‘energy-
momentum tensor’ is

$\mathbk T[r]=\mathbk T[r]^{td}+\mathbk T[r]^{rtd}+\mathbk T[r]^{rd}$ (5)

where

$\mathbk T[r]^{td}:=-\mathbk F[r]^{td}\cdot\mathbk
F[r]^{td}-\frac{1}{4}\mathrm{Tr}\mathbk F[r]^{td}\cdot\mathbk F[r]^{td},$
$\mathbk T[r]^{rtd}:=-(\mathbk F[r]^{td}\cdot\mathbk F[r]^{rd}+\mathbk
F[r]^{rd}\cdot\mathbk F[r]^{td})-\frac{1}{4}\mathrm{Tr}(\mathbk
F[r]^{td}\cdot\mathbk F[r]^{rd}+\mathbk F[r]^{rd}\cdot\mathbk F[r]^{td}),$
$\mathbk T[r]^{rd}:=-\mathbk F[r]^{rd}\cdot\mathbk
F[r]^{rd}-\frac{1}{4}\mathrm{Tr}\mathbk F[r]^{rd}\cdot\mathbk F[r]^{rd},$

for which we have the following equalities:

$-\mathbk F^{td}[r]\cdot\mathbk F^{td}[r]=\frac{e^{2}}{16\pi^{2}}\frac{\mathbk
u_{r}\otimes\mathbk L_{r}+\mathbk L_{r}\otimes\mathbk u_{r}-\mathbk
L_{r}\otimes\mathbk L_{r}}{(\mathbk u_{r}\cdot\mathbk R_{r})^{4}},$ (6)

$-\mathrm{Tr}(\mathbk F^{td}[r]\cdot\mathbk
F^{td}[r])=\frac{e^{2}}{16\pi^{2}}\frac{2}{(\mathbk u_{r}\cdot\mathbk
R_{r})^{4}},$ (7)

$-\bigl{(}\mathbk F^{td}[r]\cdot\mathbk F^{rd}[r]+\mathbk
F^{rd}[r]\cdot\mathbk F^{td}[r]\bigr{)}=\frac{e^{2}}{16\pi^{2}}\frac{\mathbk
d_{r}\otimes\mathbk L_{r}+\mathbk L_{r}\otimes\mathbk d_{r}+2(\mathbk
u_{r}\cdot\mathbk d_{r})\mathbk L_{r}\otimes\mathbk L_{r}}{(-\mathbk
u_{r}\cdot\mathbk R_{r})^{3}},$ (8)

$\mathrm{Tr}(\mathbk F^{td}[r]\cdot\mathbk F^{rd}[r])=0,$ (9)

$-\mathbk F^{rd}[r]\cdot\mathbk
F^{rd}[r]=\frac{e^{2}}{16\pi^{2}}\frac{|\mathbk d_{r}|^{2}\mathbk
L_{r}\otimes\mathbk L_{r}}{(\mathbk u_{r}\cdot\mathbk R_{r})^{2}},$ (10)

$\mathrm{Tr}(\mathbk F^{rd}[r]\cdot\mathbk F^{rd}[r])=0.$ (11)

Evidently, $\mathbk T[r]$ is differentiable outside the world line; its
properties near the world line will be examined in the forthcoming sections.

###  4 Radial expansion of the retarded proper time

The retarded proper time is not differentiable on the world line, so it has
not a Taylor polynomial around the world line. Nevertheless, it can be
expanded in ‘radial directions’ in powers of the ‘radial distance’ as follows.

We take the parameterization (60) of spacetime around a given world line
$\mathrm{Ran}r$; in this parameterization, at every proper time point $\mathbk
s$ of $r$, the spacelike vectors Lorentz orthogonal to $\dot{r}(\mathbk s)$
are taken into account; then according to the notations introduced in (62),
$\rho$ and $\mathbk n$ describe distances and directions, respectively, in
that Euclidean space. The difference between the parameter proper time and the
retarded proper time is

$\Theta(\mathbk s,\mathbk q):=\mathbk s-\mathbk s_{r}\bigl{(}r(s)+\mathbk
L(\mathbk s)\mathbk q\bigr{)}.$

$\Theta$ is a continuous function, non-negative and $\Theta(\mathbk s,\mathbk
q)=0$ if and only if $\mathbk q=0$. Thus,

$\lim_{\mathbk q\to\boldsymbol{0}}\Theta(\mathbk s,\mathbk q)=0$ (12)

holds as well.

In the sequel, $r$ is supposed to be three times continuously differentiable,
the variables are omitted for the sake of brevity and $\mathrm{Ordo}$ denotes
a function whose limit at zero is zero.

First, let us observe that according to (12)

$\Theta=\mathrm{Ordo}(\rho).$

Then starting with $\mathbk s_{r}=\mathbk s-\Theta$, we have

$r(\mathbk s-\Theta)=r(\mathbk s)-\dot{r}(\mathbk
s)\Theta+\frac{1}{2}\ddot{r}(\mathbk s)\Theta^{2}-\frac{1}{6}\dddot{r}(\mathbk
s)\Theta^{3}+\Theta^{3}\mathrm{Ordo}(\Theta).$ (13)

Subtracting both sides from the parameterization (63), we get (see (1))

$\mathbk R_{r}=\rho\mathbk
n+\dot{r}\Theta-\frac{1}{2}\ddot{r}\Theta^{2}+\frac{1}{6}\dddot{r}\Theta^{3}+\Theta^{3}\mathrm{Ordo}(\Theta).$
(14)

The left hand side is a lightlike vector or zero. The Lorentz square of both
sides, with the equalities

$\dot{r}\cdot\dot{r}=-1,\ \ \mathbk n\cdot\mathbk n=1,\ \ \dot{r}\cdot\mathbk
n=0,\ \ \dot{r}\cdot\ddot{r}=0,\ \
\dot{r}\cdot\dddot{r}=-\ddot{r}\cdot\ddot{r},$

results in

$0=\rho^{2}-(1+\rho\mathbk n\cdot\ddot{r})\Theta^{2}+\frac{1}{3}\rho\mathbk
n\cdot\dddot{r}\Theta^{3}-\frac{1}{12}\ddot{r}\cdot\ddot{r}\Theta^{4}+(\rho\Theta^{3}+\Theta^{4})\mathrm{Ordo}(\Theta).$

In another form,

$1=\frac{\Theta^{2}}{\rho^{2}}\left(1+\rho\mathbk
n\cdot\ddot{r}-\frac{1}{3}\rho\mathbk
n\cdot\dddot{r}\Theta+\frac{1}{12}|\ddot{r}|^{2}\Theta^{2}-(\rho\Theta+\Theta^{2})\mathrm{Ordo}(\Theta)\right),$
(15)

which says that $\lim_{\rho\to 0}\frac{\Theta^{2}}{\rho^{2}}=1$. Since
$\Theta$ is non-negative, $\lim_{\rho\to 0}\frac{\Theta}{\rho}=1$ holds as
well. Then

$\Theta=\rho(1+A)\qquad\text{where}\ A=\mathrm{Ordo}(\rho)$ (16)

and (15) can be rewritten in the form

$1=\frac{\Theta^{2}}{\rho^{2}}+\frac{\Theta^{2}}{\rho^{2}}(\rho\mathbk
n\cdot\ddot{r}+\rho\mathrm{Ordo}(\rho)).$

Putting (16) in the first term on the right hand side, we get

$\frac{2A+A^{2}}{\rho}=-\frac{\Theta^{2}}{\rho^{2}}(\mathbk
n\cdot\ddot{r}+\mathrm{Ordo}(\rho)).$

Let $\rho$ tend to zero. Since $A=\mathrm{Ordo}(\rho)$, we get

$\lim_{\rho\to}\frac{2A}{\rho}=-\mathbk n\cdot\ddot{r},$

in other words, $A=\rho(-\frac{1}{2}\mathbk
n\cdot\ddot{r}+\mathrm{Ordo}(\rho))$; thus,

$\Theta=\rho-\frac{\rho^{2}}{2}(\mathbk n\cdot\ddot{r}+B)\qquad\text{where}\
B=\mathrm{Ordo}(\rho).$ (17)

Then (15) can be written in the form:

$1=\frac{\Theta^{2}}{\rho^{2}}(1+\rho\mathbk
n\cdot\ddot{r})-\frac{\Theta^{2}}{\rho^{2}}\left(\frac{1}{3}\rho\mathbk
n\cdot\dddot{r}\Theta-\frac{1}{12}|\ddot{r}|^{2}\Theta^{2}+\rho^{2}\mathrm{Ordo}(\rho)\right).$
(18)

Putting (17) in the first term on the right hand side, we get

$\left(1-\rho\mathbk n\cdot\ddot{r}-\rho B+\frac{\rho^{2}}{4}\bigl{(}(\mathbk
n\cdot\ddot{r})^{2}+2\mathbk
n\cdot\ddot{r}B+B^{2}\bigr{)}\right)(1+\rho\mathbk n\cdot\ddot{r})=\\\
=1-\rho^{2}(\mathbk n\cdot\ddot{r})^{2}-\rho
B+\frac{\rho^{2}}{4}\bigl{(}(\mathbk
n\cdot\ddot{r})^{2}+\mathrm{Ordo}(\rho)\bigr{)}.$

Then (18) divided by $\rho^{2}$ becomes

$0=-(\mathbk n\cdot\ddot{r})^{2}-\frac{B}{\rho}+\frac{1}{4}\bigl{(}(\mathbk
n\cdot\ddot{r})^{2}+\mathrm{Ordo}(\rho)\bigr{)}-\frac{\Theta^{2}}{\rho^{2}}\left(\frac{1}{3}\mathbk
n\cdot\dddot{r}\frac{\Theta}{\rho}-\frac{1}{12}|\ddot{r}|^{2}\frac{\Theta^{2}}{\rho^{2}}+\mathrm{Ordo}(\rho)\right)$

from which

$\lim_{\rho\to 0}\frac{B}{\rho}=-\frac{3}{4}(\mathbk
n\cdot\ddot{r})^{2}-\frac{1}{3}\mathbk
n\cdot\dddot{r}+\frac{1}{12}|\ddot{r}|^{2}.$

Then

$B=\rho\left(-\frac{3}{4}(\mathbk n\cdot\ddot{r})^{2}-\frac{1}{3}\mathbk
n\cdot\dddot{r}+\frac{1}{12}|\ddot{r}|^{2}+\mathrm{Ordo}(\rho))\right).$

Then we have the final result

$\Theta=\rho\left(1-\rho\frac{\mathbk
n\cdot\ddot{r}}{2}+\frac{\rho^{2}}{2}\left(\frac{9(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{12}\ +\frac{\mathbk
n\cdot\dddot{r}}{3}\right)+\rho^{2}\mathrm{Ordo}(\rho)\right).$ (19)

For the sake of simplicity, without the danger of confusion, we shall omit the
terms $\mathrm{Ordo}(\rho)$, writing approximative equalities:

$\Theta\approx\rho\left(1-\rho\frac{\mathbk
n\cdot\ddot{r}}{2}+\frac{\rho^{2}}{2}\left(\frac{9(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{12}\ +\frac{\mathbk
n\cdot\dddot{r}}{3}\right)\right),$ (20)

$\Theta^{2}\approx\rho^{2}(1-\rho(\mathbk
n\cdot\ddot{r})),\qquad\Theta^{3}\approx\rho^{3}.$ (21)

It is emphasized again that these formulae do not come from a Taylor expansion
because of the non differentiability of the retarded proper time on the world
line.

###  5 Further radial expansions

Using the previous results, we give the radial expansion of the functions
occurring in the formulae of the electromagnetic field (3).

According to (20) and (21), the retarded velocity and acceleration introduced
in (1) have radial expansion

$\mathbk
u_{r}\approx\dot{r}-\ddot{r}\Theta+\frac{1}{2}\dddot{r}\Theta^{2}\approx\dot{r}-\rho\ddot{r}+\frac{\rho^{2}}{2}\bigl{(}(\mathbk
n\cdot\ddot{r})\dot{r}+\dddot{r}\bigr{)},$ (22) $\mathbk
a_{r}\approx\ddot{r}-\dddot{r}\Theta\approx\ddot{r}-\rho\dddot{r}.$ (23)

Further, (14) gives us

$\displaystyle\mathbk R_{r}\approx$ $\displaystyle\rho\mathbk
n+\dot{r}\Theta-\frac{1}{2}\ddot{r}\Theta^{2}+\frac{1}{6}\dddot{r}\Theta^{3}\approx$
$\displaystyle\approx$ $\displaystyle\rho\Biggl{(}\mathbk
n+\ddot{r}-\rho\frac{\ddot{r}+(\mathbk n\cdot\ddot{r})\dot{r}}{2}+$
$\displaystyle\qquad\ +\frac{\rho^{2}}{2}\left(\left(\frac{9(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{12}+\frac{\mathbk
n\cdot\dddot{r}}{3}\right)\dot{r}+(\mathbk
n\cdot\dot{r})\ddot{r}+\frac{\dddot{r}}{3}\right)\Biggl{)},$

$(-\mathbk u_{r}\cdot\mathbk R_{r})\approx\rho\left(1+\rho\frac{(\mathbk
n\cdot\ddot{r})}{2}-\frac{\rho^{2}}{2}\left(\frac{(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{4}+\frac{2\mathbk
n\cdot\dddot{r}}{3}\right)\right),$

$\frac{1}{-\mathbk u_{r}\cdot\mathbk
R_{r}}\approx\frac{1}{\rho}\Biggl{(}1-\rho\frac{\mathbk
n\cdot\ddot{r}}{2}+\frac{\rho^{2}}{2}\left(\frac{3(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{4}+\frac{2\mathbk
n\cdot\dddot{r}}{3}\right)\Biggr{)},$

$\frac{1}{(\mathbk u_{r}\cdot\mathbk
R_{r})^{2}}\approx\frac{1}{\rho^{2}}\Biggl{(}1-\rho\mathbk
n\cdot\ddot{r}+\frac{\rho^{2}}{2}\left(\frac{4(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{2}+\frac{4\mathbk
n\cdot\dddot{r}}{3}\right)\Biggr{)},$

$\frac{1}{(-\mathbk u_{r}\cdot\mathbk
R_{r})^{3}}\approx\frac{1}{\rho^{3}}\Biggl{(}1-\rho\frac{3\mathbk
n\cdot\ddot{r}}{2}+\frac{\rho^{2}}{2}\left(\frac{15(\mathbk
n\cdot\ddot{r})^{2}-3|\ddot{r}|^{2}}{4}+2\mathbk
n\cdot\dddot{r}\right)\Biggr{)},$

$\frac{1}{(\mathbk u_{r}\cdot\mathbk
R_{r})^{4}}\approx\frac{1}{\rho^{4}}\Biggl{(}1-\rho 2\mathbk
n\cdot\ddot{r}+\frac{\rho^{2}}{2}\left(6(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}+\frac{8}{3}\mathbk
n\cdot\dddot{r}\right)\Biggr{)}.$

Thus, for the quantities introduced in (2),

$\begin{split}\mathbk L_{r}\approx\mathbk
n+\dot{r}-&\rho\left(\frac{\ddot{r}}{2}+(\mathbk
n\cdot\ddot{r})\left(\dot{r}+\frac{\mathbk n}{2}\right)\right)+\\\
&+\frac{\rho^{2}}{2}\left(\frac{6(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{3}+\mathbk
n\cdot\dddot{r}\right)\dot{r}+\\\ &+\frac{\rho^{2}}{2}\left(\frac{3(\mathbk
n\cdot\ddot{r})^{2}-|\ddot{r}|^{2}}{4}+\frac{2\mathbk
n\cdot\dddot{r}}{3}\right)\mathbk n+\\\
&+\frac{\rho^{2}}{2}\left(\frac{3(\mathbk
n\cdot\ddot{r})\ddot{r}}{2}+\frac{\dddot{r}}{3}\right),\end{split}$ (24)

$\mathbk d_{r}\approx\ddot{r}+(\mathbk
n\cdot\ddot{r})\dot{r}+\rho\left(\left(\frac{|\ddot{r}|^{2}-(\mathbk
n\cdot\ddot{r})^{2}}{2}-\mathbk n\cdot\dddot{r}\right)\dot{r}-(\mathbk
n\cdot\ddot{r})\ddot{r}-\dddot{r}\right)$ (25)

###  6 Radial expansion of the
‘energy-momentum tensor’

Let us take (6)-(11) and, for the sake of simplicity, let us rewrite (24) in
the form

$\mathbk L_{r}\approx\mathbk n+\dot{r}-\rho\mathbk A+\frac{\rho^{2}}{2}\mathbk
B.$

Then – $\lor$ denotes the symmetric tensor product –

$\displaystyle(\mathbk u_{r}\lor\mathbk L_{r})$ $\displaystyle\approx\mathbk
u\lor(\mathbk n+\mathbk u)-\rho\bigl{(}\mathbk u\lor\mathbk
A+\ddot{r}\lor(\mathbk n+\mathbk u)\bigr{)}+$
$\displaystyle+\frac{\rho^{2}}{2}\Bigl{(}\dot{r}\lor\mathbk B+((\mathbk
n\cdot\ddot{r})\ddot{r}+\dddot{r})\lor(\mathbk n+\dot{r})+2\ddot{r}\lor\mathbk
A\Bigr{)},$ $\displaystyle(\mathbk L_{r}\otimes\mathbk L_{r})$
$\displaystyle\approx(\mathbk n+\dot{r})\otimes(\mathbk
n+\dot{r})-\rho\big{(}(\mathbk n+\dot{r})\lor\ddot{r}\bigr{)}+$
$\displaystyle+\frac{\rho^{2}}{2}\Bigl{(}(\mathbk n+\dot{r})\lor\mathbk
B+\mathbk A\lor\mathbk A\Bigr{)}.$

Further, rewriting (25) in the form

$\mathbk d_{r}\approx\ddot{r}+(\mathbk
n\cdot\ddot{r})\dot{r}+\rho(\beta\dot{r}-(\mathbk
n\cdot\ddot{r})\ddot{r}-\dddot{r})$

we have

$\displaystyle(\mathbk d_{r}\lor\mathbk L_{r})$
$\displaystyle\approx(\ddot{r}+(\mathbk n\cdot\ddot{r})\dot{r})\lor(\mathbk
n+\dot{r})+$ $\displaystyle+\rho\Bigl{(}-\bigl{(}\ddot{r}+(\mathbk
n\cdot\ddot{r})\dot{r}\bigr{)}\lor\mathbk A+(\beta\dot{r}-(\mathbk
n\cdot\ddot{r})\ddot{r}-\dddot{r})\lor(\mathbk n+\dot{r})\Bigr{)},$ $(\mathbk
u_{r}\cdot\mathbk d_{r})\approx-(\mathbk n\cdot\ddot{r}+\rho\beta),$ (26)

and

$\displaystyle(\mathbk u_{r}\cdot\mathbk d_{r})(\mathbk L_{r}\otimes\mathbk
L_{r})$ $\displaystyle\approx-(\mathbk n\cdot\ddot{r})\bigl{(}(\mathbk
n+\dot{r})\otimes(\mathbk n+\dot{r})\bigr{)}+$
$\displaystyle+\rho\bigl{(}(\mathbk n\cdot\ddot{r})(\mathbk
n+\dot{r})\lor\mathbk A-\beta(\mathbk n+\dot{r})\otimes(\mathbk
n+\dot{r})\bigr{)};$

finally,

$|\mathbk d_{r}|^{2}\approx|\ddot{r}|^{2}-(\mathbk n\cdot\ddot{r})^{2}.$

All these together give that the ‘energy-momentum tensor’ (5) has the radial
expansion

$\begin{split}\frac{16\pi^{2}}{e^{2}}\mathbk
T[r]&=\frac{\dot{r}\otimes\dot{r}-\mathbk n\otimes\mathbk
n+\frac{\boldsymbol{1}}{2}}{\rho^{4}}-\\\ &+\frac{\frac{\mathbk
n\otimes\ddot{r}+\ddot{r}\otimes\mathbk n}{2}-(\mathbk
n\cdot\ddot{r})(2\dot{r}\otimes\dot{r}-\mathbk n\otimes\mathbk
n+\boldsymbol{1})}{\rho^{3}}-\\\
&-\frac{1}{\rho^{2}}\left(\frac{\ddot{r}\otimes\ddot{r}}{4}+\frac{\dot{r}\otimes\dddot{r}+\dddot{r}\otimes\dot{r}}{2}\right)+\\\
&-\frac{1}{\rho^{2}}\left(\frac{2(\mathbk
n\otimes\dddot{r}+\dddot{r}\otimes\mathbk n)}{3}+(\mathbk
n\cdot\ddot{r})(\mathbk n\otimes\ddot{r}+\ddot{r}\otimes\mathbk n)\right)+\\\
&+\frac{1}{\rho^{2}}\left(|\ddot{r}|^{2}+3(\mathbk
n\cdot\ddot{r})^{2}+\frac{4\mathbk
n\cdot\dddot{r}}{3}\right)\dot{r}\otimes\dot{r}+\\\
&+\frac{1}{\rho^{2}}\left(\frac{3|\ddot{r}|^{2}}{4}-(\mathbk
n\cdot\ddot{r})^{2}\right)\mathbk n\otimes\mathbk n+\\\
&+\frac{1}{\rho^{2}}\left(\frac{2|\ddot{r}|^{2}}{3}+\frac{\mathbk
n\cdot\dddot{r}}{2}\right)(\dot{r}\otimes\mathbk n+\mathbk
n\otimes\dot{r})+\\\
&+\frac{1}{\rho^{2}}\left(\frac{|\ddot{r}|^{2}}{4}+\frac{3(\mathbk
n\cdot\ddot{r})^{2}}{2}+\frac{2\mathbk
n\cdot\dddot{r}}{3}\right)\boldsymbol{1}+\frac{\mathrm{Ordo}(\rho)}{\rho^{2}}.\end{split}$
(27)

###  7 Distribution of energy-momentum

The radial expansion (27) shows that the ‘energy-momentum tensor’ $\mathbk
T[r]$, for every proper time value $\mathbk s$ of the given world line
function $r$, has a pole of fourth order and a pole of third order in the
three dimensional Euclidean space Lorentz orthogonal to $\dot{r}(\mathbk s)$.
We can tame its poles, obtaining $\mathrm{tm}\mathbk T[r]$ as follows. For the
sake of avoiding misunderstandings, we shall write simply $\mathbk T$ instead
of $\mathbk T[r]$ when integrating functions.

For an $x$ in the neighbourhood of the world line in question let
$\hat{\mathbk}s(x)$ denote the proper time value for which

$\dot{r}(\hat{\mathbk}s(x))\cdot\bigl{(}x-r(\hat{\mathbk}s(x)\bigr{)}=0,$ (28)

holds, i.e. $\hat{\mathbk}s$ is the first component of the inverse of the
parameterization (60).

######  Proposition 7.1.

If the support of the test function $\phi$ is disjoint from the range of $r$
then

$\bigl{(}\mathrm{tm}\mathbk
T[r]\mid\phi\bigr{)}:=\int\limits_{\mathrm{M}}\mathbk T(x)\phi(x)\ dx;$

if the support of the test function $\phi$ is in a world roll $H_{I,R_{I}}$
then

$\bigl{(}\mathrm{tm}\mathbk T[r]\mid\phi\bigr{)}:=\\\
:=\int\limits_{\mathrm{M}}\Biggl{(}\mathbk
T(x)\phi(x)-\frac{e^{2}}{16\pi^{2}}\frac{4\dot{r}(\hat{\mathbk}s(x))\otimes\dot{r}(\hat{\mathbk}s(x))+\boldsymbol{1}}{6|x-r(\hat{\mathbk}s(x))|^{4}}\phi(r(\hat{\mathbk}s(x)))\Biggr{)}\
dx\ !!$ (29)

where the double exclamation mark says the integral must be taken in the
radial parameterization of the world roll $H_{I,R_{I}}$ and in a given order.

Proof The integral (29) in the radial parameterization reads

$\displaystyle\int\limits_{\mathbb{T}}\Biggl{(}\int\limits_{0}^{\infty}\Biggl{(}\int\limits_{S_{1}(0)}$
$\displaystyle\Bigl{(}\mathbk T(r(\mathbk s)+\rho\mathbk
n)\phi(r(s)+\rho\mathbk n)-$
$\displaystyle-\frac{e^{2}}{16\pi^{2}}\frac{4\dot{r}(\mathbk
s)\otimes\dot{r}(\mathbk s)+\boldsymbol{1}}{6\rho^{4}}\phi(r(\mathbk
s))\Bigr{)}(1+\rho(\mathbk n\cdot\ddot{r}(\mathbk s))\ d\mathbk
n_{0}\Biggr{)}\ \rho^{2}d\rho\Biggr{)}\ d\mathbk s$

with $\mathbk n=\mathbk L(\mathbk s)\mathbk n_{0}$.

Let us write the radial expansion of the ‘energy-momentum tensor’ in the form

$\frac{A(\mathbk n)}{\rho^{4}}+\frac{B(\mathbk n)}{\rho^{3}}+\frac{C(\mathbk
n)+\mathrm{Ordo}(\rho)}{\rho^{2}}.$ (30)

Then the function

$\displaystyle\left(\frac{A(\mathbk n)}{\rho^{4}}+\frac{B(\mathbk
n)}{\rho^{3}}+\frac{C(\mathbk
n)+\mathrm{Ordo}(\rho)}{\rho^{2}}\right)(1+\rho\mathbk n\cdot\ddot{r})=$ (31)
$\displaystyle=\frac{A(\mathbk n)}{\rho^{4}}+\frac{A(\mathbk n)(\mathbk
n\cdot\ddot{r})+B(\mathbk n)}{\rho^{3}}+\frac{B(\mathbk n)(\mathbk
n\cdot\ddot{r})+C(\mathbk n)+\mathrm{Ordo}(\rho)}{\rho^{2}}$ (32)

is to be pole tamed for each fixed $\mathbk s$.

$A(\mathbk n)$ is an even tensor product of $\mathbk n$ (linear function of an
even tensor muliple of $\mathbk n_{0}$), so the first term of (32) can be
tamed: the function

$\frac{A(\mathbk n)\bigl{(}\phi(r(\mathbk s)+\rho\mathbk n)-\phi(r(\mathbk
s))\bigr{)}}{\rho^{4}}$

is to be integrated in the radial variables (because the derivative of $\phi$
drops out from the Taylor polynomial).

The part containing $\phi(r(\mathbk s)+\rho\mathbk n)$ remains as it is,

$\frac{A(\mathbk n)}{\rho^{4}}\phi(r(\mathbk s)+\rho\mathbk n);$ (33)

the part containing $\phi(r(\mathbk s))$ becomes

$\frac{e^{2}}{16\pi^{2}}\left(\frac{4\dot{r}(s)\otimes\dot{r}(\mathbk
s)+\boldsymbol{1}}{6\rho^{4}}\right)\phi(r(\mathbk s))$ (34)

which is obtained from the integral

$\frac{1}{4\pi}\int_{S_{1}(0)}\left(\dot{r}(\mathbk s)\otimes\dot{r}(\mathbk
s)-\mathbk L(\mathbk s)\mathbk n_{0}\otimes\mathbk L(\mathbk s)\mathbk
n_{0}+\frac{\boldsymbol{1}}{2}\right)\ d\mathbk n_{0}$

by the use of

$\int\limits_{S_{1}(\boldsymbol{0})}\mathbk L(\mathbk s)\mathbk
n_{0}\otimes\mathbk L(\mathbk s)\mathbk n_{0}\ d\mathbk
n_{0}=\frac{4\pi}{3}\bigl{(}\boldsymbol{1}+\mathbk L(\mathbk
s)\dot{r}(0)\otimes\mathbk L(\mathbk
s)\dot{r}(0)\bigr{)}=\frac{4\pi}{3}\bigl{(}\boldsymbol{1}+\dot{r}(\mathbk
s)\otimes\dot{r}(\mathbk s)\bigr{)}.$

$A(\mathbk n)\mathbk n\cdot\ddot{r}+B(\mathbk n)$ is a linear function of an
odd tensor product of $\mathbk n$, so the second term in (32) can be tamed:
the function

$\frac{A(\mathbk n)\mathbk n\cdot\ddot{r}+B(\mathbk
n)}{\rho^{3}}\phi(r(s)+\rho\mathbk n)$ (35)

is to be integrated in the radial variables (because $\phi(r(\mathbk s))$
drops out from the Taylor polynomial).

The third term in (32) defines a regular Distribution, so the function

$\frac{B(\mathbk n)\mathbk n\cdot\ddot{r}+C(\mathbk
n)+\mathrm{Ordo}(\rho)}{\rho^{2}}\phi(r(\mathbk s)+\rho\mathbk n)$ (36)

is to be integrated.

Summarizing: the sum of (33), (35) and (36) gives the first term in (29) and
(34) gives the second term because in the radial parameterization
$\hat{\mathbk}s(x)=\mathbk s$ and $|x-r(\hat{\mathbk}s(x))|=\rho$.
$\blacksquare$

$\mathrm{tm}\mathbk T[r]$ will be called the Distribution of energy-momentum
(which is not an energy-momentum distribution!) of the point charge with a
given world line function $r$.

###  8 The radiation reaction force

The Distribution of energy-momentum of a point charge is a mathematical
object. Has it some physical meaning? The negative spacetime divergence of the
energy-momentum tensor of a continuous charge-current density is the force
density; that is why we can hope here a similar physical meaning.

Recall that $\lambda_{\mathrm{Ran}r}$ is the Lebesgue measure of the world
line $\mathrm{Ran}r$ (see (53)).

######  Proposition 8.1.

$\displaystyle-\mathrm{D}\cdot\mathrm{tm}\mathbk T[r]$
$\displaystyle=\frac{1}{4\pi}\frac{2e^{2}}{3}\bigl{(}\dddot{r}+(\dot{r}\cdot\dddot{r})\dot{r}\bigr{)}\lambda_{\mathrm{Ran}r}=$
(37)
$\displaystyle=\frac{1}{4\pi}\frac{2e^{2}}{3}(\dot{r}\land\dddot{r})\cdot\dot{r}\lambda_{\mathrm{Ran}r}.$
(38)

Proof The Distribution $\mathrm{tm}\mathbk T[r]$ restricted to an open subset
disjoint from the world line equals the regular Distribution corresponding to
the restriction of $\mathbk T[r]$. Since $\mathrm{D}\cdot\mathbk T[r]=0$
except the world line, we have to take test functions $\phi$ whose support is
in a world roll $H_{I,R_{I}}$ around the world line. For them

$\displaystyle(-\mathrm{D}\cdot\mathrm{tm}\mathbk
T[r])\mid\phi)=(\mathrm{tm}\mathbk T[r]|\cdot\mathrm{D}\phi)=$
$\displaystyle=\int\limits_{\mathrm{M}}\Biggl{(}\mathbk
T(x)\cdot\mathrm{D}\phi[x]-\frac{e^{2}}{16\pi^{2}}\frac{4\dot{r}(\hat{\mathbk}s(x))\otimes\dot{r}(\hat{\mathbk}s(x))+\boldsymbol{1}}{6|x-r(\hat{\mathbk}s(x)|^{4}}\cdot\mathrm{D}\phi[r(\hat{\mathbk}s(x))]\Biggr{)}\
dx\ !!=$ $\displaystyle\qquad\qquad=\lim_{R\to
0}\int\limits_{\mathrm{M}\setminus H_{I,R}}\dots\ !!$

Because of $\mathrm{D}\cdot\mathbk T=0$ in $\mathrm{M}\setminus H_{I,R}$,, the
first term in the integrand equals $\mathrm{D}\cdot(\mathbk T\phi)$.

For examining the second term let us consider $\rho$ and $\mathbk n$ as
functions of the spacetime points $x$ instead of the radial parameters
$(\mathbk s,\mathbk q)$, i.e. let us introduce

$\hat{\rho}(x):=|x-r(\hat{\mathbk}s(x))|,\quad\hat{\mathbk}n(x):=\frac{x-r(\hat{\mathbk}s(x))}{|x-r(\hat{\mathbk}s(x))|};$
(39)

moreover, let

$\hat{\mathbk}u(x):=\dot{r}(\hat{\mathbk}s(x)),\qquad\hat{\mathbk}a(x):=\ddot{r}(\hat{\mathbk}s(x)).$

Then

$(\hat{\rho}\hat{\mathbk}n)(x)=x-r(\hat{\mathbk}s(x))$

and

$\hat{\mathbk}n\cdot\hat{\mathbk}n=1,\qquad\hat{\mathbk}u\cdot\hat{\mathbk}n=0.$
(40)

By differentiation we get

$\boldsymbol{1}-\hat{\mathbk}u\otimes\mathrm{D}\hat{\mathbk}s=\mathrm{D}(\hat{\rho}\hat{\mathbk}n)=\hat{\mathbk}n\otimes\mathrm{D}\hat{\rho}+\hat{\rho}\mathrm{D}\hat{\mathbk}n,$
(41)

from which, by (40) we deduce

$\mathrm{D}\hat{\rho}=\hat{\mathbk}n.$ (42)

Further,

$\mathrm{D}\hat{\mathbk}u=\hat{\mathbk}a\otimes\mathrm{D}\hat{\mathbk}s,$

and

$0=\mathrm{D}(\hat{\mathbk}u\cdot\hat{\rho}\hat{\mathbk}n)=\hat{\rho}\hat{\mathbk}n\cdot(\hat{\mathbk}a\otimes\mathrm{D}\hat{\mathbk}s)+\hat{\mathbk}u\cdot(\boldsymbol{1}-\hat{\mathbk}u\otimes\mathrm{D}\hat{\mathbk}s),$

therefore

$\mathrm{D}\hat{\mathbk}s=-\frac{\hat{\mathbk}u}{1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a}$
(43)

and as a consequence,

$\mathrm{D}\cdot(\hat{\rho}\hat{\mathbk}n)=4-\frac{1}{1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a},$
(44)

Further, let

$\mathbk Z(\mathbk s):=\frac{(4\dot{r}(\mathbk s)\otimes\dot{r}(\mathbk
s)+\boldsymbol{1})\cdot\mathrm{D}\phi[r(\mathbk
s)]}{6},\qquad\hat{\mathbk}Z(x):=\mathbk Z(\hat{\mathbk}s(x)).$

Then we find that
$\mathrm{D}\hat{\mathbk}Z=\dot{\mathbk}Z(\hat{\mathbk}s)\otimes\mathrm{D}\hat{\mathbk}s$
from which by (43) $(\mathrm{D}\hat{\mathbk}Z)\cdot\hat{\mathbk}n=0$ follows.
Then taking into account (44) and (42),

$\displaystyle\mathrm{D}\cdot\frac{\hat{\mathbk}Z\otimes\hat{\rho}\hat{\mathbk}n}{\hat{\rho}^{4}}=$
$\displaystyle\hat{\mathbk}Z\left(\left(4-\frac{1}{1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a}\right)\frac{1}{\hat{\rho}^{4}}-\frac{\hat{\rho}\hat{\mathbk}n\cdot
4\hat{\mathbk}n}{\hat{\rho}^{5}}\right)=$ $\displaystyle=$
$\displaystyle-\frac{\hat{\mathbk}Z}{(1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a)\hat{\rho}^{4}}=-\frac{\hat{\mathbk}Z}{\hat{\rho}^{4}}+\frac{\hat{\mathbk}Z(\hat{\mathbk}n\cdot\hat{\mathbk}a)}{(1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a)\hat{\rho}^{3}},$

in other words,

$-\frac{\hat{\mathbk}Z}{\hat{\rho}^{4}}=\mathrm{D}\cdot\frac{\hat{\mathbk}Z\otimes\hat{\mathbk}n}{\hat{\rho}^{3}}-\frac{\hat{\mathbk}Z(\hat{\mathbk}n\cdot\hat{\mathbk}a)}{(1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a)\hat{\rho}^{3}}.$

Thus, we arrive at

$(-\mathrm{D}\cdot\mathrm{tm}\mathbk T[r]\mid\phi)=\\\ =\lim_{R\to
0}\int\limits_{\mathrm{M}\setminus
H_{I,R}}\Biggl{(}\mathrm{D}\cdot\Biggl{(}\mathbk
T\phi+\frac{e^{2}}{16\pi^{2}}\frac{\hat{\mathbk
Z}\otimes\hat{\mathbk}n}{\hat{\rho}^{3}}\Biggr{)}-\frac{e^{2}}{16\pi^{2}}\frac{\hat{\mathbk
Z}(\mathbk
n\cdot\hat{\mathbk}a)}{(1+\hat{\rho}\hat{\mathbk}n\cdot\hat{\mathbk}a)\hat{\rho}^{3}}\Biggr{)}.\
!!$

The integral of the second term in the parameterization around the world line
becomes a multiple of

$\int\limits_{I}\int\limits_{R}^{R_{I}}\rho^{2}\int\limits_{S_{1}(\boldsymbol{0})}\frac{\mathbk
Z(\mathbk s)(\mathbk L(\mathbk s)\cdot\mathbk n_{0})\cdot\ddot{r}(\mathbk
s)}{\rho^{3}}\ d\mathbk n_{0}\ d\rho\ d\mathbk s$

which is zero because of (54).

By Gauss’ theorem, the integral of the first term can be transformed to an
integral on the corresponding world tube in such a way that the tube is
directed ‘inwards’ i.e. the normal vector is $-\hat{\mathbk}n$:

$\begin{split}(-\mathrm{D}\cdot\mathrm{tm}\mathbk T[r]\mid\phi)&=\lim_{R\to
0}\int\limits_{\mathrm{M}\setminus H_{I,R}}\mathrm{D}\cdot\Biggl{(}\mathbk
T\phi+\frac{e^{2}}{16\pi^{2}}\frac{\hat{\mathbk
Z}\otimes\hat{\mathbk}n}{\hat{\rho}^{3}}\Biggr{)}\ !!=\\\ &=-\lim_{R\to
0}\int\limits_{P_{I,R}}\Biggl{(}\mathbk
T\phi+\frac{e^{2}}{16\pi^{2}}\frac{\hat{\mathbk
Z}\otimes\hat{\mathbk}n}{\hat{\rho}^{3}}\Biggr{)}\cdot\hat{\mathbk}n\
d\lambda_{P_{I,R}}\ !!=\\\ &=-\lim_{R\to
0}\int\limits_{I}R^{2}\int\limits_{S_{1}(0)}...(\mathbk n_{0},\mathbk s)\
d\mathbk n_{0}\ d\mathbk s\ !\end{split}$ (45)

where the integrand is

$\Biggl{(}(\mathbk T\phi)\bigl{(}r(\mathbk s)+R\mathbk n(\mathbk
s)\bigr{)}+\frac{e^{2}}{16\pi^{2}}\frac{\mathbk Z(\mathbk s)\otimes\mathbk
n(\mathbk s)}{R^{3}}\Biggr{)}\cdot\mathbk n(\mathbk s)(1+R\mathbk n(\mathbk
s)\cdot\ddot{r}(\mathbk s))$ (46)

whith $\mathbk n(\mathbk s):=\mathbk L(\mathbk s)\mathbk n_{0}$.

Then

$\frac{\mathbk Z\otimes\mathbk n}{\mathbk R^{3}}\cdot\mathbk n(1+R\mathbk
n\cdot\ddot{r})=\frac{(4\dot{r}\otimes\dot{r}+\boldsymbol{1})\cdot\mathrm{D}\phi[r]}{6}\left(\frac{1}{R^{3}}+\frac{\mathbk
n\cdot\ddot{r}}{R^{2}}\right),\ $

and on the base of the radial expansion (27)

$\displaystyle(\mathbk T(r+R\mathbk n)$ $\displaystyle\cdot\mathbk
n(1+R\mathbk n\cdot\ddot{r})=$
$\displaystyle=\frac{e^{2}}{16\pi^{2}}\Bigl{(}-\frac{\mathbk
n}{2R^{4}}+\frac{\dot{r}}{2R^{3}}+$ $\displaystyle+\frac{|\dot{r}|^{2}\mathbk
n+\frac{2}{3}|\dot{r}|^{2}\dot{r}-\frac{3}{4}(\mathbk
n\cdot\dot{r})\dot{r}-\frac{2}{3}\dddot{r}}{R^{2}}+$
$\displaystyle+\frac{\mathrm{Ordo}(R)}{R^{2}}\Bigr{)}.$

This, together with the expansion

$\phi(r+R\mathbk n)=\phi(r)+R\mathbk
n\cdot\mathrm{D}\phi[r]+R^{2}\mathrm{Ordo}(R),$

give that the integral (45) of the Ordo terms in (46), because of the three
times differentiability of $r$, is less than a multiple of $R$, so their limit
is zero when $R$ tends to zero; moreover, the integrals of the terms linear
and trilinear in $\mathbk n$ in (46) are zero.

The integral of the terms independent of $\mathbk n$ results in a
multiplication by $4\pi$, the integral of the terms bilinear in $\mathbk n$
results in a multiplication by
$\frac{4\pi}{3}(\boldsymbol{1}+\dot{r}\otimes\dot{r})$.

Thus, $\mathrm{D}\phi[r]$ will be multiplied by

$-\frac{4\pi}{3}\frac{\boldsymbol{1}+\dot{r}\otimes\dot{r}}{2R}+4\pi\frac{4\dot{r}\otimes\dot{r}+\boldsymbol{1}}{6R}=\frac{4\pi}{2R}\dot{r}\otimes\dot{r}$

and $\phi(r)$ will be multiplied by
$\frac{4\pi}{2R}\ddot{r}+\frac{2}{3}\left(|\ddot{r}|^{2}\dot{r}-\dddot{r}\right)$.
Since

$\frac{4\pi}{2R}\Bigl{(}\dot{r}(\mathbk s)(\dot{r}(\mathbk
s)\cdot\mathrm{D}\phi[r(\mathbk s)])+\frac{4\pi}{2R}\ddot{r}(\mathbk
s)\Bigr{)}=\frac{4\pi}{2R}\frac{d}{d\mathbk s}\bigl{(}\dot{r}(\mathbk
s)\phi(r(\mathbk s))\bigr{)}$ (47)

and its integral is zero because the support of $\phi$ is in $H_{I,R_{I}}$, it
remains only

$-\frac{e^{2}}{16\pi^{2}}\int\limits_{I}4\pi\frac{2}{3}(|\ddot{r}(\mathbk
s)|^{2}\dot{r}(\mathbk s)-\dddot{r}(\mathbk s))\phi(r(\mathbk s))\ d\mathbk s$

which gives the desired result by $|\ddot{r}|^{2}=-\dot{r}\cdot\dddot{r}$.
$\blacksquare$

###  9 Discussion

####  9.1 On the self-force

The usual formula (38) of the radiation reaction force is obtained in a
mathematically exact way, without unjustified application of Gauss–Stokes
theorem, of Taylor expansion and without a doubtful limit to zero.

It is an important fact that the self-force is obtained for a given world line
function $r$. Consequently, its physical meaning is the following:

If the world line function is $r$ then the self-force is
$\frac{1}{4\pi}\frac{2e^{2}}{3}(\dot{r}\land\dddot{r})\cdot\dot{r}$.

Accordingly, if the force $\mathbk f$ is necessary to get a given $r$ without
radiation i.e. $m\ddot{r}=\mathbk f(r,\dot{r})$ would valid without radiation,
then besides $\mathbk f$ the force opposite to the self-force must be applied
for getting the desired world line function $r$.

An actual example is when an elementary particle is revolved in a cyclotron by
a homogeneous static magnetic field. In spacetime formulation: there is given
a constant electromagnetic field $\mathbk F$ and the world line without
radiation would satisfy $m\ddot{r}=e\mathbk F\dot{r}$. Then

$\dddot{r}=\frac{e^{2}}{m^{2}}\mathbk F\cdot\mathbk
F\cdot\dot{r},\qquad\dot{r}\cdot\dddot{r}=\frac{e^{2}}{m^{2}}(\dot{r}\cdot\mathbk
F)\cdot(\mathbk F\cdot\dot{r})=-\frac{e^{2}}{m^{2}}(\mathbk
F\cdot\dot{r})\cdot(\mathbk F\cdot\dot{r}),$

thus the self-force is

$\frac{1}{4\pi}\frac{2e^{4}}{3m^{2}}\bigl{(}\mathbk F\cdot\mathbk
F\cdot\dot{r}-|\mathbk F\cdot\dot{r}|^{2}\dot{r}\bigr{)};$ (48)

to keep the particle on a prescribed orbit, the opposite of the above force
must be applied besides the magnetic field,

For better seeing, let us take the standard inertial frame $\mathbk u$ in
which the cyclotron is at rest. Then the $\mathbk u$-spacelike component of
$\mathbk F$, for which $\mathbk F\cdot\mathbk u=0$, is the magnetic field
$\underline{\mathbk B}$, a three dimensional antisymmetric tensor.

Using the relative velocity $\mathbk v:=\frac{\dot{r}}{-\mathbk
u\cdot\dot{r}}-\mathbk u$ with respect to the inertial frame, we have

$\mathbk F\cdot\dot{r}=\mathbk F\cdot\left(\frac{\dot{r}}{-\mathbk
u\cdot\dot{r}}-\mathbk u\right)(-\mathbk
u\cdot\dot{r})=\frac{\underline{\mathbk B}\cdot\mathbk v}{\sqrt{1-|\mathbk
v|^{2}}}$

and the self-force relative to the inertial frame is

$\frac{1}{4\pi}\frac{2e^{4}}{3m^{2}}\frac{(\boldsymbol{1}+\mathbk
u\otimes\mathbk u)}{-\mathbk u\cdot\dot{r}}\bigl{(}\mathbk F\cdot\mathbk
F\cdot r-|\mathbk
F\cdot\dot{r}|^{2}\dot{r}\bigr{)}=\frac{1}{4\pi}\frac{2e^{4}}{3m^{2}}\left(\underline{\mathbk
B}\cdot\underline{\mathbk B}\cdot\mathbk v-\frac{|\underline{\mathbk
B}\cdot\mathbk v|^{2}}{1-|\mathbk v|^{2}}\mathbk v\right).$

Introducing the customary axial vector $\mathbk B$ corresponding to
$\underline{\mathbk B}$ we have $|\mathbk B|=|\underline{\mathbk B}|$,
$\underline{\mathbk B}\cdot\mathbk v=\mathbk v\times\mathbk B$,
$\underline{\mathbk B}\cdot\underline{\mathbk B}\cdot\mathbk v=(\mathbk
v\times\mathbk B)\times\mathbk B$. If the relative velocity is orthogonal to
$\mathbk B$ then $\underline{\mathbk B}\cdot\underline{\mathbk B}\cdot\mathbk
v=-|\mathbk B|^{2}\mathbk v$, $|\underline{\mathbk B}\cdot\mathbk
v|^{2}=|\mathbk B|^{2}|\mathbk v|^{2}$ and the self-force relative to the
inertial frame

$-\frac{1}{4\pi}\frac{2e^{4}}{3m^{2}}\frac{|\mathbk B|^{2}\mathbk
v}{1-|\mathbk v|^{2}}$ (49)

brakes the motion; the same opposite force must be applied additionally to
avoid braking and to ensure that the particle moves on the prescribed orbit.

####  9.2 On the LAD equation

Besides the unjustified application of Gauss–Stokes theorem, of Taylor
expansion and a doubtful limit to zero there is another, a fundamental error
in usual derivations of the LAD equation which is clearly seen now.

The self-force
$\frac{1}{4\pi}\frac{2e^{2}}{3}(\dot{r}\land\dddot{r})\cdot\dot{r}$ is valid
for a given world line function $r$; therefore it is unjustified to put it in
a Newtonian-like equation to obtain the LAD equation which would serve to
determine $r$. Thus, the LAD equation is a misconception and its pathological
properties are not surprising.

The well known fundamental problem ([24]) is that both electrodynamics and
mechanics in their known forms are theories of action:

– the Maxwell equations define the electromagnetic field
$\boldsymbol{\mathcal{F}}$ produced by a given world line function $r$ of a
particle,

– the Newtonian equation defines the world line function $r$ of a particle in
a given force (e.g. the Lorentz force in an extraneous electromagnetic field
$\mathbk F$),

and at present we have not a well working theory of interaction which would
define both the electromagnetic field $\boldsymbol{\mathcal{F}}$ and the world
line function $r$ together.

To see the problem more closely, let us consider a point charge under the
action of a given force $\mathbk f$. Then according to the general conception
(see e.g. [9]) – formulated in our language – the Newtonian equation is
replaced with the balance equation

$-\mathrm{D}\cdot(\boldsymbol{\mathcal{T}}_{m}[r]+\boldsymbol{\mathcal{T}}_{e}[\boldsymbol{\mathcal{F}}])+\mathbk
f(r,\dot{r})\lambda_{\mathrm{Ran}r}=0$ (50)

where $\boldsymbol{\mathcal{T}}_{m}[r]$ is the tensor Distribution of
mechanical energy-momentum constructed somehow of $r$ and
$\boldsymbol{\mathcal{T}}_{e}[\boldsymbol{\mathcal{F}}]$ is the tensor
Distribution of electromagnetic energy-momentum constructed somehow of
$\boldsymbol{\mathcal{F}}$.

It seems evident that
$\boldsymbol{\mathcal{T}}_{m}[r]=m\dot{r}\otimes\dot{r}\lambda_{\mathrm{Ran}r}$
and then the balance equation

$m\ddot{r}\lambda_{\mathrm{Ran}r}=\mathbk
f(r,\dot{r})\lambda_{\mathrm{Ran}r}-\mathrm{D}\cdot\boldsymbol{\mathcal{T}}_{e}[\boldsymbol{\mathcal{F}}]$
(51)

together with the Maxwell equations

$\mathrm{D}\cdot\boldsymbol{\mathcal{F}}=\dot{r}\lambda_{\mathrm{Ran}r},\qquad\mathrm{D}\land\boldsymbol{\mathcal{F}}=0$
(52)

would describe that $r$ and $\boldsymbol{\mathcal{F}}$ determine each other
mutually.

It is not sure that there is a conveniently defined
$\boldsymbol{\mathcal{T}}_{e}[\boldsymbol{\mathcal{F}}]$ with which (51) and
(52) form a consistent system of equations.

It is sure, however, that
$\boldsymbol{\mathcal{T}}_{e}[\boldsymbol{\mathcal{F}}]$ cannot be replaced
with $\mathrm{tm}\mathbk T[r]$ which is valid for a given $r$; the LAD
‘equation’ is obtained by committing this replacement.

####  9.3 Equation, equality

At last, let the LAD ‘equation’ be looked at from another point of view.
Namely, we have to distinguish clearly between an equation and an equality,
both denoted by the same symbol $=$.

An equation is a definition: it defines a set (the set of its solutions).

An equality is a statement: it states that two sets are the same.

Let $S(e,m,\mathbk f)$ be the set of processes of a point charge $e$ with mass
$m$ under the action of a force $\mathbk f$.

Because of the non-physical run-away solutions, the LAD ‘equation’ does not
define $S(e,m,\mathbk f)$, thus it is not an equation from a physical point of
view.

The various attempts to find satisfactory conditions to exclude non-physical
solutions suggest in this context that, in fact, the LAD ‘equation’ would be
an equality, stating that $S(e,m,\mathbk f)$ is a subset of all the solutions.

Unfortunately, this is not true, either, as it is shown in the excellent paper
[21]: $\mathbk f$ being the Coulomb force of a point charge, no physical
motion in radial direction is a solution of the LAD ‘equation’.

####  9.4 An illustrative example

The following example can illustrate the problem of describing the
interaction. Let two particles with the same charge and mass rest in the space
of a standard inertial frame. They repulse each other by the Coulomb forces
which are compensated by some constraint to attain the rest. If the constraint
is left off at an instant then the particles start to move. Figure 1 shows
what happens (time passes from the left to the right).

Figure 1:  Interaction of two point charges

While resting, the particles send Coulomb-actions to each other in the
positive light cone, represented by sparsely dotted lines. At the instant when
the constraint is left off, the acceleration of each particles is determined
by the Coulomb force of the other. Then for some time the acceleration of a
particle is determined by the Coulomb force of the other particle and the
radiation self-force, and then it sends some action, different from the
Coulomb force to the other particle; this is represented by the densely dotted
lines. We have no formula for this acceleration and action. Later the
situation becomes more complicated: it is not known which action and self-
force give rise to an acceleration.

####  9.5 Final remarks

It is emphasized that the retarded field is meaningful only for a given world
line, therefore it cannot be used for determining the world line.

The various attempts to obtain a physically acceptable equation with continuum
charge distributions instead of a point charge suffer from the same fault of
using the retarded field (and sometimes the advanced one, too).

## Appendix

###  10 Measures

Elementary notions of (vector) measures and integration by them can be found
in [25].

The Lebesgue measure of an affine space is the measure given by the usual
integration (customary for $\mathbb{R}^{n}$).

The Lebesgue measure $\lambda_{H}$ of a submanifold $H$ in spacetime is given
by the formula

$\int\limits_{H}f\
d\lambda_{H}:=\int\limits_{\mathrm{Dom}p}f(p(\xi))\sqrt{|\det(\mathcal{D}p[\xi]^{*}\mathcal{D}p[\xi])|}\
d\xi$

where $p$ is an arbitrary parameterization of the submanifold i.e. a
continuously differentiable injective map defined in an affine space, having
$H$ as its range and $\mathcal{D}p[\xi]$, the derivative of $p$ at $\xi$ is
injective for all $\xi$.

In particular, for a world line function $r$, the Lebesgue measure of the
corresponding world line is $\lambda_{\mathrm{Ran}r}$ which gives the
integration along the world line. In particular, it is considered a
Distribution in spacetime,

$\left(\lambda_{\mathrm{Ran}r}\mid\phi\right)=\int\limits_{\mathbb{T}}\phi(r(\mathbk
s))\ d\mathbk s$ (53)

for arbitrary test functions $\phi$.

###  11 Three dimensional integration

Let $\mathbk u_{0}$ be an absolute velocity in spacetime and $\mathbf{S}_{0}$
the three dimensional Euclidean vector space Lorentz orthogonal to $\mathbk
u_{0}$; $S_{1}(\boldsymbol{0})$ is the unit sphere around zero, its elements
are denoted by $\mathbk n_{0}$.

In the usual parameterization given by the angles $\vartheta,\varphi$ relative
to an orthonormal bases, the Lebesgue measure of $S_{1}(\boldsymbol{0})$ is
obtained by the symbolic formula $d\mathbk n_{0}\sim\sin\vartheta\ d\vartheta\
d\varphi$ with which we easily find the equalities

$\int\limits_{S_{1}(\boldsymbol{0})}\mathbk n_{0}\ d\mathbk
n_{0}=0,\qquad\int\limits_{S_{1}(\boldsymbol{0})}\mathbk n_{0}\otimes\mathbk
n_{0}\otimes\mathbk n_{0}\ d\mathbk n_{0}=0,$ (54)
$\int\limits_{S_{1}(\boldsymbol{0})}\mathbk n_{0}\otimes\mathbk n_{0}\
d\mathbk n_{0}=\frac{4\pi}{3}(\boldsymbol{1}+\mathbk u_{0}\otimes\mathbk
u_{0})$ (55)

where $\boldsymbol{1}$ is the identity map of spacetime vectors; and so on,
the integral of an even tensor products is a non-zero tensor, the integral of
an odd tensor products is zero.

The radial coordinates of $\mathbk q\in\mathbf{S}_{0}$ are

$\rho(\mathbk q):=|\mathbk q|\geq 0,\quad\mathbk n_{0}(\mathbk
q):=\frac{\mathbk q}{|\mathbk q|}\in S_{1}(\boldsymbol{0}).$

Then the radial parameterization of $\mathbf{S}_{0}$ is

$(\rho,\mathbk n_{0})\mapsto\rho\mathbk n_{0}.$

We applied the usual ambiguity that $\rho$ and $\mathbk n_{0}$ denote
functions in some respect and variables in an other respect.

The other well known integral formula

$\int\limits_{\mathbf{S}_{0}}f(\mathbk q)\ d\mathbk
q=\int\limits_{0}^{\infty}\rho^{2}\int\limits_{S_{1}(\boldsymbol{0})}f(\rho\mathbk
n_{0})\ d\mathbk n_{0}\ d\rho$ (56)

will be often used, too.

###  12 Pole taming

There is a well defined method to attach a Distribution to a locally non-
integrable function, having a pole singularity in a Euclidean space.
Originally the method was called regularization ([16]). Since the Distribution
obtained in this way is not regular, I call the method pole taming222Áron
Szabó proposed this name.

Only a special case is necessary for us.

For the function $\rho(\mathbk q):=|\mathbk q|$ $(\mathbk
q\in\mathbf{S}_{0})$, $\frac{1}{\rho^{2+m}}$ is not locally integrable if
$m>0$. For $m$ being an even positive integer, a Distribution can be attached
to it by pole taming defined as follows.

For a test function $\psi$ let $\mathbk q\mapsto T_{\psi}^{(m-1)}(\mathbk q)$
denote the Taylor polynomial of order $m-1$ at zero. In a neighbourhood of
zero, $\psi-T_{\psi}^{(m-1)}$ is a continuous function of order $\rho^{m}$,
therefore $\frac{\psi-T_{\psi}^{(m-1)}}{\rho^{2+m}}$ is integrable there.
Outside the support of $\psi$, however, the term containing the $(m-1)$-th
derivative is not integrable.

$m$ is even; in the radial parameterization (11),
$\mathcal{D}^{(m-1)}\psi(\boldsymbol{0})(\rho\mathbk n_{0},\dots,\rho\mathbk
n_{0})$ is a linear function of an odd tensor product of $\mathbk n_{0}$,
thus, integrating in the order given by (56), this term will drop out and an
integrable function remains.

We repeat for avoiding misundertanding: it is known that, for an integrable
function, the order of integration by $\mathbk n_{0}$ and $\rho$ can be
interchanged in (56). If a function is not integrable, it can occur that in
one of the orders the integral exists (but in the other order not). Here we
take the advantage that the integral exists in the given order. In this way
the pole taming of $\frac{1}{\rho^{2+m}}$ results in the Distribution defined
by

$\left(\mathrm{tm}\frac{1}{\rho^{2+m}}\Bigm{|}\psi\right):=\int\limits_{0}^{\infty}\left(\rho^{2}\int\limits_{S_{1}(\boldsymbol{0})}\frac{\psi(\rho\mathbk
n_{0})-T_{\psi}^{(m-1)}(\rho\mathbk n_{0})}{\rho^{2+m}}\ d\mathbk
n_{0}\right)\ d\rho\ !$ (57)

where the exclamation mark calls attention to that the order of integration
cannot be interchanged.

For $m=0$ we mean the pole taming formally by taking the corresponding regular
Distribution.

Later, for the sake of brevity, we also use the formula

$\left(\mathrm{tm}\frac{1}{\rho^{2+m}}\Bigm{|}\psi\right):=\int\limits_{\S_{0}}\frac{\psi(\mathbk
q)-T_{\psi}^{(m-1)}(\mathbk q)}{|\mathbk q|^{2+m}}\ d\mathbk q\ !!$ (58)

where the double exclamation mark calls attention to that the integral must be
taken in radial parameterization and in the given order.

We have to extend the notion of pole taming as follows. For a $k$-th tensor
power $\mathbk n_{0}^{\otimes k}:=\mathbk n_{0}\otimes\mathbk
n_{0}\otimes\dots\mathbk n_{0}$ we put

$\left(\mathrm{tm}\frac{\mathbk n_{0}^{\otimes
k}}{\rho^{2+m}}\Bigm{|}\psi\right):=\int\limits_{0}^{\infty}\rho^{2}\left(\int\limits_{S_{1}(\boldsymbol{0})}\frac{\mathbk
n_{0}^{\otimes k}\bigl{(}\psi(\rho\mathbk n_{0})-T_{\psi}^{(m-1)}(\rho\mathbk
n_{0}\bigr{)}}{\rho^{2+m}}\ d\mathbk n_{0}\right)\ d\rho\ !$ (59)

which makes sense if $m-1+k$ is odd; therefore

– if $m$ is even then $k$ must be even, too,

– if $m$ is odd then $k$ must be odd, too.

###  13 World rolls and world tubes
around a world line

Let $r$ be a twice continuously differentiable world line function. The
Lorentz boost (see [1])

$\mathbk L(\mathbk s):=\boldsymbol{1}+\frac{(\dot{r}(0)+\dot{r}(\mathbk
s))\otimes(\dot{r}(0)+\dot{r}(\mathbk s))}{1-\dot{r}(0)\cdot\dot{r}(\mathbk
s)}-2\dot{r}(\mathbk s)\otimes\dot{r}(0)$

maps $\dot{r}(0)$ into $\dot{r}(\mathbk s)$.

Further, we use the notations $\mathbk u_{0}:=\dot{r}(0)$ and $\mathbf{S}_{0}$
for the linear subspace consisting of vectors Lorentz orthogonal to $\mathbk
u_{0}$; then $\\{\mathbk L(s)\cdot\mathbk q\mid\mathbk q\in\mathbf{S}_{0}\\}$
is the subspace consisting of vectors Lorentz orthogonal to $\dot{r}(\mathbk
s)$. Then

$p:\mathbb{T}\times\mathbf{S}_{0}\to\mathrm{M},\qquad(\mathbk s,\mathbk
q)\mapsto r(\mathbk s)+\mathbk L(\mathbk s)\cdot\mathbk q$ (60)

is a parameterization of a neighbourhood of the world line.

This is a reformulation of a usual setting: $(\mathbk s,\mathbk q)$ correspond
to the so called Fermi coordinates.

To be precise, the following details are necessary. $p$ is evidently
continuously differentiable and its derivative

$\mathcal{D}p[\mathbk s,\mathbk q]=\Bigl{(}\dot{r}(\mathbk
s)+\dot{\mathbk}L(\mathbk s)\cdot\mathbk q\ \ \mathbk L(\mathbk
s)|_{\mathbf{S}_{0}}\Bigr{)}$ (61)

($\mathbk L(\mathbk s)|_{\mathbf{S}_{0}}$ is the restriction of $\mathbk
L(\mathbk s)$ to $\mathbf{S}_{0}$) is injective for all $\mathbk s$ and for
$\mathbk q=\boldsymbol{0}$, thus, according to the inverse function theorem
and the fact that $p(\mathbk s,\boldsymbol{0})=r(\mathbk s)$, there is an open
subset in $\mathbb{T}\times\mathbf{S}_{0}$ which contains
$\mathbb{T}\times\\{\boldsymbol{0}\\}$ where both $p$ and all the values
$\mathcal{D}p[\ \cdot\ ]$ are injective.

Let $I$ be a bounded and closed time interval. Then $\\{r(\mathbk
s)\mid\mathbk s\in I\\}$ is a compact set and each of its points has a
neighbourhood in the range of the parameterization; thus, $\\{r(\mathbk
s)\mid\mathbk s\in I\\}$ can be covered by finite many such neighbourhoods. As
a consequence, there is an $R_{I}>0$ in such a way that the part $\\{r(\mathbk
s)\mid\mathbk s\in I\\}$ of the world line is contained in

$H_{I,R}:=\\{r(\mathbk s)+\mathbk L(\mathbk s)\cdot\mathbk q\mid\mathbk s\in
I,|\mathbk q|<R\\}$

for all $0<R\leq R_{I}$.

$H_{I,R}$ is called a world roll of length $I$ and radius $R$. With the
notations

$\rho(\mathbk q):=|\mathbk L(\mathbk s)\cdot\mathbk q|=|\mathbk
q|,\qquad\mathbk n(\mathbk s,\mathbk q):=\frac{\mathbk L(\mathbk
s)\cdot\mathbk q}{\rho(\mathbk q)},\qquad\mathbk n_{0}(\mathbk
q):=\frac{\mathbk q}{\rho(\mathbk q)},$ (62)

the parameterization of $H_{I,R}$ can be rewritten in the form

$r(\mathbk s)+\rho(\mathbk q)\mathbk n(\mathbk s,\mathbk q)=r(\mathbk
s)+\mathbk L(s)\cdot\mathbk n_{0}(\mathbk q).$ (63)

The cylinder around $H_{I,R}$,

$P_{I,R}:=\\{r(\mathbk s)+\mathbk L(\mathbk s)\cdot\mathbk q\mid\mathbk s\in
I,\ |\mathbk q|=R\\}$

is called the corresponding world tube.

The tangent space of $P_{I,R}$ at the point $r(\mathbk s)+\mathbk L(\mathbk
s)\cdot\mathbk q$ is

$\\{\bigl{(}\dot{r}(\mathbk s)+\dot{\mathbk}L(\mathbk s)\cdot\mathbk
q\bigr{)}\mathbk t+\mathbk L(\mathbk s)\cdot\mathbk h\mid\mathbk
t\in\mathbb{T},\ \mathbk h\in\mathbf{S}_{0},\ \mathbk h\cdot\mathbk q=0\\}.$

For $\mathbk q\in\mathbf{S}_{0}$ we have $(\mathbk L(\mathbk s)\cdot\mathbk
q)\cdot\dot{r}(\mathbk s)=\boldsymbol{0}$, $(\mathbk L(\mathbk s)\cdot\mathbk
q)\cdot(\mathbk L(\mathbk s)\cdot\mathbk h)=\mathbk q\cdot\mathbk h=0$ and
$(\dot{\mathbk}L(\mathbk s)\cdot\mathbk q)\cdot(\mathbk L(\mathbk
s)\cdot\mathbk q)=\frac{1}{2}\frac{\partial}{\partial\mathbk s}|\mathbk
L(\mathbk s)\cdot\mathbk q|^{2}=0$; thus,

$\Bigl{(}\bigl{(}\dot{r}(\mathbk s)+\dot{\mathbk}L(\mathbk s)\cdot\mathbk
q\bigr{)}\mathbk t+\mathbk L(\mathbk s)\cdot\mathbk h\Bigr{)}\cdot(\mathbk
L(\mathbk s)\cdot\mathbk q)=0$

and we find that the outward normal vector of $P_{I,R}$ at the point
$r(\mathbk s)+\rho(\mathbk q)\mathbk n(\mathbk s,\mathbk q)$ is

$\mathbk n(\mathbk s,\mathbk q).$ (64)

###  14 Integration in world rolls

Integrals in a world roll will be computed in the parameterization (60) for
which we have

$\sqrt{|\det\mathcal{D}p[\mathbk s,\mathbk q]^{*}\mathcal{D}p[\mathbk
s,\mathbk q]|}=1+\rho(\mathbk q)\mathbk n(\mathbk s,\mathbk
q)\cdot\ddot{r}(\mathbk s).$ (65)

To show it, for the sake of brevity, we introduce the notation

$\mathbk z(\mathbk s,\mathbk q):=\dot{r}(\mathbk s)+\dot{\mathbk}L(\mathbk
s)\cdot\mathbk q.$

Then from (61) we get

$\displaystyle\mathcal{D}p[\mathbk s,\mathbk q]^{*}\mathcal{D}p[\mathbk
s,\mathbk q]$ $\displaystyle=\begin{pmatrix}\mathbk z(\mathbk s,\mathbk q)\\\
\bigl{(}\mathbk L(\mathbk
s)|_{\mathbf{S}_{0}}\bigr{)}^{*}\end{pmatrix}\begin{matrix}\cdot\begin{pmatrix}\mathbk
z(\mathbk s,\mathbk q)&\mathbk L(\mathbk s)|_{\mathbf{S}_{0}}\end{pmatrix}\\\
&\end{matrix}=$ (66) $\displaystyle=\begin{pmatrix}\mathbk z(\mathbk s,\mathbk
q)\cdot\mathbk z(\mathbk s,\mathbk q)&\mathbk z(\mathbk s,\mathbk
q)\cdot\mathbk L|(\mathbk s)_{\mathbf{S}_{0}}\\\ (\mathbk L(\mathbk
s)|_{\mathbf{S}_{0}})^{*}\cdot\mathbk z(\mathbk s,\mathbk
q)&\boldsymbol{1}_{\mathbf{S}_{0}}\end{pmatrix}.$ (67)

Omitting the variables for the sake of perspicuity, we can write $\mathbk
L|_{\mathbf{S}_{0}}=\mathbk L\cdot(\boldsymbol{1}+\mathbk u_{0}\otimes\mathbk
u_{0})$, thus $(\mathbk L|_{\mathbf{S}_{0}})^{*}=(\boldsymbol{1}+\mathbk
u_{0}\otimes\mathbk u_{0})\cdot\mathbk L^{*}$; accordingly,
$(\boldsymbol{1}+\mathbk u_{0}\otimes\mathbk u_{0})\cdot\mathbk
L^{*}\cdot\mathbk z=\mathbk L^{*}\cdot\mathbk z+\mathbk u_{0}((\mathbk
u_{0}\cdot\mathbk L^{*})\cdot\mathbk z$ which equals $\mathbk z\cdot\mathbk
L+\mathbk u_{0}(\dot{r}\cdot\mathbk z)=\mathbk z\cdot(\mathbk
L\cdot(\boldsymbol{1}+\mathbk u_{0}\otimes\mathbk u_{0}))$.

The block matrix (67) is symmetric, its determinant is $\mathbk z\cdot\mathbk
z-(\mathbk z\cdot\mathbk L|_{\mathbf{S}_{0}})\cdot((\mathbk
L|_{\mathbf{S}_{0}})^{*}\cdot\mathbk z)$. The second term here equals
$(\mathbk z\cdot\mathbk L)\cdot(\mathbk z\cdot\mathbk L)+2(\mathbk
z\cdot\mathbk L)\cdot\mathbk u_{0}(\dot{r}\cdot\mathbk
u_{0})-(\dot{r}\cdot\mathbk u_{0})^{2}=\mathbk z\cdot\mathbk
z+(\dot{r}\cdot\mathbk u_{0})^{2},$ thus we have

$\det(\mathcal{D}p^{*}\mathcal{D}p)=-(\dot{r}\cdot\mathbk
z)^{2}=-(1+\rho\mathbk n\cdot\ddot{r})^{2};$

the last equality comes from

$\dot{r}(\mathbk s)\cdot\mathbk z(\mathbk s,\mathbk q)=-1+\dot{r}(\mathbk
s)\cdot\dot{\mathbk}L(\mathbk s)\cdot\mathbk
q=-1+\frac{\partial}{\partial\mathbk s}\bigl{(}\dot{r}(\mathbk s)\cdot\mathbk
L(s)\cdot\mathbk q\bigr{)}-\ddot{r}(\mathbk s)\cdot\mathbk L(\mathbk
s)\cdot\mathbk q$

and the quantity in the parenthesis in the middle term on the right hand side
is zero.

Finally, if $\rho(\mathbk q)$ is ‘sufficiently small’ then $1+\rho(\mathbk
q)\mathbk n(\mathbk s,\mathbk q)\cdot\ddot{r}(\mathbk s)$ is positive, thus it
is positive for all the possible $(\mathbk s,\mathbk q)$ because
$\mathcal{D}p[\mathbk s,\mathbk q]$ is injective, the determinant cannot be
zero.

Further, because of $\mathbk n(\mathbk s,\mathbk q)=\mathbk L(\mathbk
s)\mathbk n_{0}(\mathbk q)$, using the radial parameterization (11) of
$\mathbf{S}_{0}$ we have

$\int\limits_{H_{I,R}}\dots(x)\
dx=\int\limits_{0}^{R}\rho^{2}\int\limits_{S_{1}(0)}\dots\bigl{(}r(\mathbk
s)+\rho\mathbk L(\mathbk s)\mathbk n_{0}\bigr{)}(1+\rho\ddot{r}(\mathbk
s)\cdot\mathbk L(\mathbk s))\ d\mathbk n_{0}\ d\rho.$ (68)

###  15 Acknowledgement

I am grateful to T.Gruber for checking all the not too simple formulas and to
P.Ván for helping me to compose this article.

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