# On the Joint Evolution Problem
for a Scalar Field and its Singularity

Aditya Agashe<EMAIL_ADDRESS>Brown University  Ethan Lee
<EMAIL_ADDRESS>Rutgers University  Shadi Tahvildar-Zadeh
<EMAIL_ADDRESS>Rutgers University

(January 19, 2023)

###### Abstract

In the classical electrodynamics of point charges in vacuum, the
electromagnetic field, and therefore the Lorentz force, is ill-defined at the
locations of the charges. Kiessling resolved this problem by using the
momentum balance between the field and the particles, extracting an equation
for the force that is well-defined where the charges are located, so long as
the field momentum density is locally integrable in a neighborhood of the
charges.

In this paper, we examine the effects of such a force by analyzing a
simplified model in one space dimension. We study the joint evolution of a
massless scalar field together with its singularity, which we identify with
the trajectory of a particle. The static solution arises in the presence of no
incoming radiation, in which case the particle remains at rest forever. We
will prove the stability of the static solution for particles with positive
bare mass by showing that a pulse of incoming radiation that is compactly
supported away from the point charge will result in the particle eventually
coming back to rest. We will also prove the nonlinear instability of the
static solution for particles with negative bare mass by showing that an
incoming radiation with arbitrarily small amplitude will cause the particle to
reach the speed of light in finite time. We conclude by discussing
modifications to this simple model that could make it more realistic.

##  1 Introduction and Main Result

Classical electromagnetism has a fundamental problem: For a charged point-
particle in an electromagnetic field that is at least partly sourced by that
particle, the field is not defined at the location of the particle. Because
the Lorentz force that the field exerts on the particle depends on the values
of the field at the particle’s location, the force is also undefined where
it’s needed, i.e. for the particle’s equations of motion to make sense. This
is the infamous radiation-reaction problem. This problem has been the subject
of intense study by some of the world’s most renowned physicists and
mathematicians, including Poincaré [13] and Dirac [3], for more than a
century. An excellent account of this endeavor can be found in [14], where an
entire chapter is devoted to recounting its history111It is outside the scope
of this article to mention all the various directions taken by researchers to
resolve this issue. Interested readers are referred to [14] and its copious
bibliography..

An important breakthrough came in 2019 when, following up on some ideas of
Poincaré [13], Kiessling [7, 8] showed that if one postulated local
conservation laws to hold for the total (field + particle) energy density-
momentum density-stress tensor, i.e. (employing the Einstein summation
convention, where repeated indices are summed over their range)

$\partial^{\mu}T^{\mbox{\tiny total}}_{\mu\nu}=0,\quad\mbox{where}\quad
T_{\mu\nu}^{\mbox{\tiny total}}=T_{\mu\nu}^{\mbox{\tiny
field}}+T_{\mu\nu}^{\mbox{\tiny particle}},$ (1)

(once these expressions are properly defined) then the force can be determined
using the law of momentum balance, provided the field momentum density is
locally integrable in a neighborhood of the charge. This integrability
assumption rules out the classical electromagnetic vacuum law $E=D,B=H$
postulated by Maxwell, but admits others, such as the Bopp-Landé-Thomas-
Podolsky (BLTP for short) vacuum law [1, 2, 10, 11, 12].

It is of interest to study the effect of the Kiessling force on the motion of
an electromagnetic point-charge. In three space dimensions using the standard
electromagnetic vacuum laws, this is not possible. One can use other vacuum
laws in three space dimensions, such as BLTP, that do meet Kiessling’s
criterion, and for which one can prove local well-posedness of the joint field
and particle dynamics [7, 9] as well as global existence for the solution to
the scattering problem of a single particle by a smooth potential [6].
However, the expression for the force in the BLTP case is quite complicated,
which makes it hard to figure out what the particle trajectories actually look
like. On the other hand, by a simple scaling analysis, it is easy to see that
Kiessling’s criterion may be satisfied for Maxwell’s vacuum law, so long as
one works in one space dimension. Since there is however no viable
electromagnetism in one space dimension, we instead turn to the simpler model
of a scalar charge. Such a model has been proposed before by many authors, see
e.g. [4] and references therein. To simplify matters even further, we will be
focusing on the case of a single particle perturbed by scalar radiation.
Following Weyl’s ideas on singularity theories of matter [15], we will take
the evolving singularity of the scalar field to represent the path of the
particle in space-time. This is thus a joint evolution problem for a scalar
field $u(t,s)$, and the trajectory of its singularity, $s=q(t)$. The governing
equations are as follows (see [4], Eqs. 7–9): The field satisfies

$\begin{cases}\partial_{t}^{2}u-\partial_{s}^{2}u=a\delta(s-q(t))\\\
u(0,s)=-\frac{a}{2}\lvert s\rvert+V_{0}(s)\\\ \partial_{t}u(0,s)=V_{1}(s)\\\
\end{cases}$ (2)

($\delta$ is the Dirac delta-function) while the equations of motion for the
particle are

$\begin{cases}\dot{q}=\frac{p}{m\sqrt{1+\frac{p^{2}}{m^{2}}}}\\\
\dot{p}=f(t,q,\dot{q})\\\ q(0)=0\\\ \dot{q}(0)=0\end{cases}$ (3)

(The speed of light has been set equal to one.)

Here, (2) is the Cauchy problem for a massless wave equation sourced by the
particle. We have added $V_{0}(s)$ and $V_{1}(s)$ to the initial data to
represent smooth incoming radiation that is compactly supported away from the
point charge. Thus $V_{0},V_{1}\in
C^{\infty}_{c}(\mathbb{R}\setminus\\{0\\})$. Real constants $a$ and $m$
represent the charge and the (bare) rest mass of the particle. Equations (3)
are simply Newton’s equations of motion with Einstein’s special-relativistic
relation between momentum and velocity instead of Newton’s. $f$ is the force
exerted by the field on the particle, and its precise expression needs to be
determined using another principle. Here, following Kiessling, we will use
momentum conservation to determine $f$.

######  Remark 1.

We note that the above system of equations is not fully Lorentz-covariant: the
right-hand side of the wave equation in (2) does not transform correctly under
a Lorentz transformation. This is a defect of the model (which was also
pointed out in [4]). This defect can be corrected, but the resulting system
becomes harder to analyze. Some of the results in this paper have also been
obtained for the fully relativistic version, and will appear elsewhere [5].

The initial conditions in (3) can always be satisfied by going into the
initial rest frame of the particle. We will use Kiessling’s prescription to
determine the force $f$ on the particle. This will depend on the field $u$,
which makes (2-3) a coupled system of equations for the joint evolution of the
field and its singularity.

Consider first the case of no incoming radiation, i.e. $V_{0}\equiv 0$ and
$V_{1}\equiv 0$. In that case, $u=-\frac{a}{2}|s|$ where $q(t)=0$ for all $t$
is clearly a time-independent solution to (2), i.e. the particle remains at
rest forever. We shall see that in this case, $f\equiv 0$. In this paper we
will prove:

######  Theorem 1.

  1. (a)

Suppose $m>0$. For all smooth initial data $(V_{0},V_{1})$ for (2) that are
compactly supported away from the origin, there exists a solution
$(u(t,s),q(t))$ to the joint field-particle evolution problem (2-3), with the
property that:

    1. (i)

the field $u$ is Lipschitz everywhere and the particle trajectory $q$ is
$C^{1}$,

    2. (ii)

$u$ is at least $C^{1}$ away from the particle path $s=q(t)$, and

    3. (iii)

for all $\epsilon>0$, $\exists T>0$ such that $t>T$ implies
$|\dot{q}(t)|<\epsilon$.

  2. (b)

Suppose $m<0$. For all $\epsilon>0$, there exists smooth, compactly supported
initial data $(V_{0},V_{1})$ for (2), with
$\|V_{0}\|_{C^{1}(\mathbb{R})}+\|V_{1}\|_{C^{0}(\mathbb{R})}<\epsilon$, and a
solution $(u(t,s),q(t))$ to the joint field-particle evolution problem (2-3),
with the property that:

    1. (i)

the field $u$ is Lipschitz everywhere and the particle trajectory $q$ is
$C^{1}$,

    2. (ii)

$u$ is at least $C^{1}$ away from the particle path $s=q(t)$, and

    3. (iii)

the particle reaches the speed of light in finite time, i.e. $\exists T>0$
s.t. $|\dot{q}(T)|=1$.

Outline of the proof: In Section 2 we solve the field equations (2) assuming
the trajectory $s=q(t)$ of the singularity is given. We do this by decomposing
the field into a smooth part depending only on the incoming radiation, and a
singular part sourced by the particle. In Section 3 we use this field to
compute the Kiessling force $f$ in (3), and show that it depends only on the
smooth part of the field. We can thus eliminate the field from (3) and have
$q(t)$ be the only unknown. In Section 4 we study (3) by turning it into a
dynamical system in the plane and analyzing its phase portrait, which will
allow us to prove the stability claim in Section 5 and the instability claim
in Section 6.

We conclude in Section 7 by speculating on the mechanism for instability, and
propose various modifications to our model that could perhaps avoid such
instabilities.

##  2 Solving the field equations

######  Proposition 1.

For any given trajectory $q(t)$ with $|\dot{q}|<1$, $q(0)=0$, and
$\dot{q}(0)=0$, the following initial value problem

$\begin{cases}\partial_{t}^{2}u-\partial_{s}^{2}u=a\delta(s-q(t))\\\
u(0,s)=-\frac{a}{2}\lvert s\rvert+V_{0}(s)\\\ \partial_{t}u(0,s)=V_{1}(s)\\\
\end{cases}$ (4)

has the unique solution:

$u(t,s)=\frac{a}{2}\begin{cases}s+V(t,s)&s<-t\\\
T_{+}(s+t)-t+V(t,s)&-t<s<q(t)\\\ T_{-}(s-t)-t+V(t,s)&q(t)<s<t\\\
-s+V(t,s)&s>t\end{cases}$ (5)

where the functions $T_{\pm}$ are defined by

$T_{\pm}(q(x)\pm x)=x$ (6)

for all $x$, i.e. $T_{\pm}$ is the inverse function to $q(x)\pm x$ (which
exists and is $C^{1}$ so long as $|\dot{q}|<1$), and

$V(t,s)=\frac{1}{2}(V_{0}(s-t)+V_{0}(s+t))+\frac{1}{2}\int_{s-t}^{s+t}V_{1}(y)dy.$
(7)

Furthermore, $u(t,s)$ is at least $C^{1}$ away from the path $s=q(t)$.

###### Proof.

Define $\Psi(t,s)$ and $\Phi(t,s)$ as solving the following equations:

$\begin{cases}\partial_{t}^{2}\Phi-\partial_{s}^{2}\Phi=0\\\
\Phi(0,s)=-\frac{a}{2}\lvert s\rvert+V_{0}(s)\\\
\partial_{t}\Phi(0,s)=V_{1}(s)\\\
\end{cases}\qquad\begin{cases}\partial_{t}^{2}\Psi-\partial_{s}^{2}\Psi=a\delta(s-q(t))\\\
\Psi(0,s)=0\\\ \partial_{t}\Psi(0,s)=0.\\\ \end{cases}$ (8)

Furthermore, define $V(t,s)$ and $U(t,s)$ as solving the following equations:

$\begin{cases}\partial_{t}^{2}V-\partial_{s}^{2}V=0\\\ V(0,s)=V_{0}(s)\\\
\partial_{t}V(0,s)=V_{1}(s)\\\
\end{cases}\qquad\begin{cases}\partial_{t}^{2}U-\partial_{s}^{2}U=0\\\
U(0,s)=-\frac{a}{2}\lvert s\rvert\\\ \partial_{t}U(0,s)=0.\\\ \end{cases}$ (9)

We thus have that $\Phi=U+V$ and $u=\Psi+\Phi$. Note that because $V_{0}$ and
$V_{1}$ are smooth functions, $V(t,s)$ is smooth as well. Hence, $V(t,s)$
contains no singularities.

We can solve for $V$ and $U$ (and hence $\Phi$) by using d’Alembert’s formula.
We then have the following:

$V(t,s)=\frac{1}{2}(V_{0}(s-t)+V_{0}(s+t))+\frac{1}{2}\int_{s-t}^{s+t}V_{1}(y)dy,$
(10)

$U(t,s)=-\frac{a}{4}(\lvert s-t\rvert+\lvert
s+t\rvert)=\begin{cases}\frac{a}{2}s&s\leq-t\\\ -\frac{a}{2}t&-t<s<t\\\
-\frac{a}{2}s&s\geq t.\end{cases}$ (11)

We can solve for $\Psi$ with Duhamel’s Principle. Define $W(t,s,\tau)$ as
follows:

$\Psi(t,s)=\int_{0}^{t}W(t-\tau,s,\tau)d\tau.$ (12)

It follows that:

$\begin{cases}W_{tt}-W_{ss}=0\\\ W(0,s,\tau)=0\\\
W_{t}(0,s,\tau)=a\delta(s-q(\tau)).\\\ \end{cases}$ (13)

To solve for W, we apply d’Alembert’s formula. We have:

$W(t,s,\tau)=\frac{1}{2}\int_{s-t}^{s+t}a\delta(y-q(\tau))dy=\frac{a}{2}\chi_{[s-t,s+t]}(q(\tau))$
(14)

where $\chi$ is the characteristic function, i.e.:

$\chi_{[a,b]}(x)=\begin{cases}1&a\leq x\leq b\\\
0&\mbox{otherwise}.\end{cases}$ (15)

To integrate $W$ to get $\Psi(t,s)$, we consider the backward light cone of
the event $(t,s)$. The $\tau$ for which $(\tau,q(\tau))$ is in this light cone
will contribute $\frac{a}{2}d\tau$ to the integral. See Fig. 1.

Because $q(0)=0$ and $c=1$, $q(t)$ is inside the forward light cone drawn from
$(0,0)$. As a result of this, $\Psi(t,s)=0$ when $s>t$ and $s<-t$. Inside the
forward light cone of the origin, it is certainly true that the backward light
cone of the event $(t,s)$ will intersect with $s=q(t)$. Moreover, it
intersects exactly once (going from time 0 to $t$, once $q(t)$ leaves the
backward light cone of the event $(t,s)$, it cannot re-enter due to the fact
that $c=1$). We must determine the point at which it intersects, the so-called
retarded time. To the left of $q(t)$, this retarded time $\tau_{2}$ is the
solution to $q(\tau_{2})+\tau_{2}=s+t$, or $T_{+}(s+t)$ for short. The
solution is hence $\frac{a}{2}T_{+}(s+t)$. To the right of $q(t)$, this
retarded time $\tau_{1}$ is the solution to $q(\tau_{1})-\tau_{1}=s-t$, or
$T_{-}(s-t)$ for short. The solution is hence $\frac{a}{2}T_{-}(s-t)$.

Figure 1: Retarded time $T_{+}$

We then get the following expression for $\Psi$:

$\Psi(t,s)=\frac{a}{2}\begin{cases}0&s<-t\\\ T_{+}(s+t)&-t<s<q(t)\\\
T_{-}(s-t)&q(t)<s<t\\\ 0&s>t.\end{cases}$ (16)

We see from here that $\Psi$, like $U$, is $C^{0}$ but not $C^{1}$ because it
has two singularities at $s+t=0$ and $s-t=0$, i.e. along the light cone of the
origin. We will see in Proposition 2 that when we add $\Psi$ and $U$, the
singularities along the light cone cancel each other out.

The full solution $u(t,s)$ is thus as follows:

$u(t,s)=\frac{a}{2}\begin{cases}s+V(t,s)&s<-t\\\
T_{+}(s+t)-t+V(t,s)&-t<s<q(t)\\\ T_{-}(s-t)-t+V(t,s)&q(t)<s<t\\\
-s+V(t,s)&s>t\end{cases}$ (17)

where $V$ is as defined in Equation (10). ∎

We expect $u(t,s)$ to have singularities only at $s=q(t)$. We pause for a
moment to show that $u$ has no singularities along the lines $s=-t$ and $s=t$.

######  Proposition 2.

For $u(t,s)$ given by Equation (5), $u(t,s)$ is $C^{1}$ across $s=-t$ and
$s=t$.

###### Proof.

Because $V_{0}$ and $V_{1}$ are smooth, $V(t,s)$ is smooth. Hence, it suffices
to look at the singular part of $u(t,s)$. Let:

$w(t,s)=U(t,s)+\Psi(t,s)=\frac{a}{2}\begin{cases}s&s<-t\\\
T_{+}(s+t)-t&-t<s<q(t)\\\ T_{-}(s-t)-t&q(t)<s<t\\\ -s&s>t.\end{cases}$ (18)

We will first show $w(t,s)$ is $C^{1}$ across $s=-t$. We have:

$\partial_{s}w|_{(s=-t)^{-}}=\partial_{s}(\frac{a}{2}s)=\frac{a}{2},\quad\partial_{t}w|_{(s=-t)^{-}}=\partial_{t}(\frac{a}{2}s)=0.$
(19)

Recall that $T_{+}(s+t)=\tau_{2}$ where $\tau_{2}$ solves the following:

$q(\tau_{2})+\tau_{2}=s+t.$ (20)

Using implicit differentiation by $s$ and $t$ yields:

$\partial_{s}\tau_{2}\
\dot{q}(\tau_{2})+\partial_{s}\tau_{2}=1,\quad\partial_{t}\tau_{2}\
\dot{q}(\tau_{2})+\partial_{t}\tau_{2}=1$ (21)

respectively. At the line $s=-t$, $\tau_{2}=0$, so
$\dot{q}(\tau_{2})=\dot{q}(0)=0$. Hence:

$\partial_{s}\tau_{2}=\partial_{t}\tau_{2}=1.$ (22)

We thus have:

$\partial_{s}w|_{(s=-t)^{+}}=\frac{a}{2}\partial_{s}T_{+}(s+t)=\frac{a}{2}\partial_{s}\tau_{2}=\frac{a}{2}$
(23)

and

$\partial_{t}w|_{(s=-t)^{+}}=\frac{a}{2}(\partial_{t}T_{+}(s+t)-1)=\frac{a}{2}(\partial_{s}\tau_{2}-1)=0.$
(24)

Comparing with (19) shows that $w$ is $C^{1}$ across $s=-t$.

The proof that $w(t,s)$ is $C^{1}$ across $s=t$ is completely analogous.

∎

##  3 Computing the Kiessling force

We would like to combine the solution (5) for $u(t,s)$ with equation (3) to
find a system of ODEs for $q(t)$. To do this, we need to work out the
Kiessling force $f$ in (3).

We begin by recalling that Kiessling postulates the local conservation of
total energy-momentum for the field and particle system:

$\partial^{\mu}T^{\mbox{\tiny total}}_{\mu\nu}=0.$ (25)

Here, $T^{\mbox{\tiny total}}_{\mu\nu}$ is the energy density-momentum
density-stress tensor (or energy-momentum tensor, for short) for the field and
particle system. To find the energy-momentum tensor for the field, we start
with the Lagrangian. The Lagrangian for the massless scalar field is:

$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}\partial_{\mu}u\partial_{\nu}u.$ (26)

Here, $\eta=\mbox{diag}(1,-1)$ is the Minkowski metric. The energy-momentum
tensor for the field is defined as:

$T_{\mu\nu}^{\mbox{\tiny
field}}=2\frac{\partial{\mathcal{L}}}{\partial{\eta^{\mu\nu}}}-\eta_{\mu\nu}\mathcal{L},$
(27)

and thus in this case

$T_{\mu\nu}^{\mbox{\tiny
field}}=\partial_{\mu}u\partial_{\nu}u-\frac{1}{2}\eta_{\mu\nu}\partial_{\alpha}u\partial^{\alpha}u.$
(28)

Since $u$ is expected to be singular on the worldline of the particle, the
above is only well-defined away from the particle path, but our assumptions on
the field are such that $T^{\mbox{\tiny field}}$ can be continued into the
particle path as a spacetime distribution.

The energy-momentum tensor for the particle on the other hand is defined as a
distribution on spacetime that is concentrated on the worldline
$x^{\mu}=z^{\mu}(\tau)$ of the particle ($\tau$ is the arclength parameter):

$T^{\mbox{\tiny
particle}}_{\mu\nu}:=m\int\mathbf{u}_{\mu}\mathbf{u}_{\nu}\delta^{(2)}(x-z(\tau))d\tau=\frac{m}{\mathbf{u}^{0}}\mathbf{u}_{\mu}\mathbf{u}_{\nu}\delta(s-q(t)),$
(29)

where $\mathbf{u}$ is the unit tangent to the worldline of the particle

$\mathbf{u}^{\mu}:=dz^{\mu}/d\tau,\qquad\mathbf{u}_{\mu}\mathbf{u}^{\mu}=1.$
(30)

The definition of $T^{\mbox{\tiny particle}}$ is such that:

$\partial^{\mu}T_{\mu\nu}^{\mbox{\tiny
particle}}=\mathbf{f}_{\nu}(t)\delta(s-q(t))$ (31)

holds, where the spacetime covector $\mathbf{f}_{\nu}$ is the 2-force acting
on the particle (cf. [7], eq. 72).

Let us take a second to consider how this relates to the $f(t,q,\dot{q})$, the
force on the particle, that appears in (3). There, $f(t,q,\dot{q})$ is clearly
the spatial component of a spacetime (contravariant) vector. We therefore must
have

$f(t,q(t),\dot{q}(t))=\mathbf{f}^{1}(t)=-\mathbf{f}_{1}(t)$ (32)

since we have chosen the signature $(+,-)$ for the Minkowski metric.

Hence, setting $\nu=1$:

$\partial^{\mu}T^{\mbox{\tiny particle}}_{\mu
1}(t,s)=-f(t,q(t),\dot{q}(t))\delta(s-q(t)).$ (33)

Going back to the energy-momentum tensor for the field, we have for $\nu=1$:

$\partial^{\mu}T^{\mbox{\tiny field}}_{\mu 1}=\partial^{0}T^{\mbox{\tiny
field}}_{01}+\partial^{1}T^{\mbox{\tiny
field}}_{11}=\partial_{t}\pi-\partial_{s}\tau$ (34)

Using equation (25), we have:

$0=\partial^{\mu}T^{\mbox{\tiny total}}_{\mu 1}=\partial^{\mu}T_{\mu
1}^{\mbox{\tiny field}}+\partial^{\mu}T_{\mu 1}^{\mbox{\tiny
particle}}=\partial_{t}\pi-\partial_{s}\tau-f(t,q,\dot{q})\delta(s-q(t)).$
(35)

Rearranging this gives us the momentum-balance law:

$\partial_{t}\pi-\partial_{s}\tau=f(t,q,\dot{q})\delta(s-q(t)).$ (36)

From (28) we have

$\pi(t,s)=T^{\mbox{\tiny field}}_{01}=u_{s}u_{t},$ (37)
$\tau(t,s)=T^{\mbox{\tiny field}}_{11}=\frac{1}{2}(u_{s}^{2}+u_{t}^{2}).$ (38)

######  Proposition 3.

Assume that the field $u$ is Lipschitz continuous in a tubular neighborhood of
the $C^{1}$ path $(t,q(t))$ of the particle, and that $u$ is $C^{1}$ on either
side of the path. Then the force appearing in (36) is

$f(t,q(t),\dot{q}(t))=-\dot{q}[\pi(t,s)]_{s=q(t)}-[\tau(t,s)]_{s=q(t)}$ (39)

where $[\cdot]_{s=q(t)}$ denotes the jump across the path.

###### Proof.

Figure 2: Region of integration for the momentum-balance equation

Note that the assumptions imply that $\pi$ and $\tau$ are bounded and can at
most have a jump discontinuity across the path, so that the right-hand side of
(39) is well-defined. Let $\epsilon>0$ and $T_{2}>T_{1}\geq 0$. We will
integrate (36) over the region $\Omega=\\{(t,s)|T_{1}\leq t\leq
T_{2},q(t)-\epsilon\leq s\leq q(t)+\epsilon\\}$ and then take the limit as
$\epsilon$ goes to $0$. After integrating and taking the limit, the right-hand
side of equation (36) becomes:

$\int_{T_{1}}^{T_{2}}f(t,q(t),\dot{q}(t))dt.$ (40)

After integrating and using Green’s Theorem, the left-hand side of (36)
becomes:

$\displaystyle\int_{q(T_{2})-\epsilon}^{q(T_{2})+\epsilon}\pi(T_{2},s)ds$
$\displaystyle-$
$\displaystyle\int_{q(T_{1})-\epsilon}^{q(T_{1})+\epsilon}\pi(T_{1},s)ds$
$\displaystyle-$
$\displaystyle\left.\int_{T_{1}}^{T_{2}}\dot{q}\pi(t,s)+\tau(t,s)\right|_{s=q(t)+\epsilon}dt$
$\displaystyle+$
$\displaystyle\left.\int_{T_{1}}^{T_{2}}\dot{q}\pi(t,s)+\tau(t,s)]\right|_{s=q(t)-\epsilon}dt.$

Because $\pi$ is locally integrable, the first two terms go to $0$ as
$\epsilon\to 0$. Taking the limit as $\epsilon$ goes to $0$ in the other two
terms gives us:

$-\int_{T_{1}}^{T_{2}}[\dot{q}\pi(t,s)+\tau(t,s)]_{s=q(t)}dt$ (41)

for the left-hand side. Because $T_{1}$ and $T_{2}$ were arbitrary, we
therefore have:

$f(t,q(t),\dot{q}(t))=-\dot{q}[\pi(t,s)]_{s=q(t)}-[\tau(t,s)]_{s=q(t)}.$ (42)

∎

######  Proposition 4.

Assume $u(t,s)$ is a solution to the joint evolution problem (2-3). The
Kiessling force in (3) is given by:

$f(t,q,\dot{q})=aV_{s}(t,q)-\frac{a^{2}}{2}\frac{\dot{q}}{1-\dot{q}^{2}}$ (43)

###### Proof.

By Prop. 3, the force is:

$f(t,q(t),\dot{q}(t))=-\dot{q}[\pi]_{s=q(t)}-[\tau]_{s=q(t)}=-\dot{q}[u_{s}u_{t}]_{s=q(t)}-\frac{1}{2}[u_{s}^{2}+u_{t}^{2}]_{s=q(t)}$
(44)

using $u(t,s)$ as given by equation (5). Substituting in $u=V+w$ gives us

$[u_{t}u_{s}]_{s=q(t)}=V_{t}[w_{s}]_{s=q(t)}+V_{s}[w_{t}]_{s=q(t)}+[w_{s}w_{t}]_{s=q(t)}$
(45)

and

$[\frac{1}{2}u^{2}_{t}+\frac{1}{2}u^{2}_{s}]_{s=q(t)}=\frac{1}{2}[w^{2}_{t}]_{s=q(t)}+\frac{1}{2}[w^{2}_{s}]_{s=q(t)}+V_{s}[w_{s}]_{s=q(t)}+V_{t}[w_{t}]_{s=q(t)}$
(46)

where we’ve used the fact that $V$ is smooth. To determine the necessary
values, we will first compute $w_{s}|_{s=q(t)^{-}}$, $w_{s}|_{s=q(t)^{+}}$,
$w_{t}|_{s=q(t)^{-}}$, and $w_{t}|_{s=q(t)^{+}}$. We have:

$w_{s}|_{s=q(t)^{-}}=\partial_{s}(\frac{a}{2}(T_{+}(s+t)-t))=\frac{a}{2}\frac{1}{\dot{q}(t)+1},$
(47)

$w_{s}|_{s=q(t)^{+}}=\partial_{s}(\frac{a}{2}(T_{-}(s-t)-t))=\frac{a}{2}\frac{1}{\dot{q}(t)-1},$
(48)

$w_{t}|_{s=q(t)^{-}}=\partial_{t}(\frac{a}{2}(T_{+}(s+t)-t))=-\frac{a}{2}\frac{\dot{q}(t)}{\dot{q}(t)+1},$
(49)

$w_{t}|_{s=q(t)^{+}}=\partial_{t}(\frac{a}{2}(T_{-}(s-t)-t))=-\frac{a}{2}\frac{\dot{q}(t)}{\dot{q}(t)-1}.$
(50)

Thus, our final results for $[w_{s}]_{s=q(t)}$, $[w_{t}]_{s=q(t)}$,
$[w_{s}w_{t}]_{s=q(t)}$, $[w_{s}^{2}]_{s=q(t)}$, and $[w_{t}^{2}]_{s=q(t)}$
are:

$[w_{s}]_{s=q(t)}=\frac{a}{2}(\frac{1}{\dot{q}(t)-1}-\frac{1}{\dot{q}(t)+1})=\frac{a}{\dot{q}(t)^{2}-1},$
(51)

$[w_{t}]_{s=q(t)}=-\frac{a\dot{q}(t)}{2}(\frac{1}{\dot{q}(t)-1}-\frac{1}{\dot{q}(t)+1})=-\frac{a\dot{q}(t)}{\dot{q}(t)^{2}-1},$
(52)

$[w_{s}w_{t}]_{s=q(t)}=-\frac{a^{2}\dot{q}(t)}{4}(\frac{1}{(\dot{q}(t)-1)^{2}}-\frac{1}{(\dot{q}(t)+1)^{2}})=-\frac{a^{2}\dot{q}(t)^{2}}{(\dot{q}(t)^{2}-1)^{2}},$
(53)

$[w^{2}_{t}]_{s=q(t)}=\frac{a^{2}}{4}((\frac{\dot{q}(t)}{\dot{q}(t)-1})^{2}-(\frac{\dot{q}(t)}{\dot{q}(t)+1})^{2})=\frac{a^{2}\dot{q}(t)^{3}}{(\dot{q}(t)^{2}-1)^{2}},$
(54)

$[w^{2}_{s}]_{s=q(t)}=\frac{a^{2}}{4}((\frac{1}{\dot{q}(t)-1})^{2}-(\frac{1}{\dot{q}(t)+1})^{2})=\frac{a^{2}\dot{q}(t)}{(\dot{q}(t)^{2}-1)^{2}}.$
(55)

Inserting these values into equation (44) gives us

$f(t,q(t),\dot{q}(t))=aV_{s}(t,q(t))-\frac{a^{2}}{2}\frac{\dot{q}(t)}{1-\dot{q}(t)^{2}}.$
(56)

∎

Note that the first term represents the force that the external field is
exerting on the particle. That is, the first term is usually taken to be the
force acting on a scalar particle. The second term represents the self-force
(the force the particle exerts on itself), which here is in the opposite
direction of the motion.

##  4 Equations of motion as a dynamical system

We can now look at the equations of motion for the particle, which are the
following:

$\begin{cases}\dot{q}=\frac{p}{m\sqrt{1+\frac{p^{2}}{m^{2}}}}\\\
\dot{p}=aV_{s}(t,q(t))-\frac{a^{2}}{2}\frac{\dot{q}(t)}{1-\dot{q}(t)^{2}}.\\\
\end{cases}$ (57)

We substitute the expression for $\dot{q}$ into the equation of $\dot{p}$,
which results in the following equation:

$\dot{p}=aV_{s}(t,q(t))-\frac{a^{2}}{2}\frac{p}{m}\sqrt{1+\frac{p^{2}}{m^{2}}}.$
(58)

In addition to this, let us rewrite $V_{s}(t,q(t))$. Recall that:

$V(t,s)=\frac{1}{2}(V_{0}(s+t)+V_{0}(s-t))+\frac{1}{2}\int_{s-t}^{s+t}V_{1}(x)dx.$
(59)

Hence, we have that:

$V_{s}(t,s)=\frac{1}{2}(\dot{V_{0}}(s+t)+\dot{V_{0}}(s-t)+V_{1}(s+t)-V_{1}(s-t)).$
(60)

Let us further define $F(s)$ and $G(s)$ as follows:

$\begin{cases}F(s)=\dot{V_{0}}(s)+V_{1}(s)\\\ G(s)=\dot{V_{0}}(s)-V_{1}(s).\\\
\end{cases}$ (61)

From our definitions of $F$ and $G$, we can rewrite our equations of motion,
specifically the expression for $\dot{p}$. It now becomes:

$\begin{cases}\dot{q}=\frac{p}{m\sqrt{1+\frac{p^{2}}{m^{2}}}}\\\
\dot{p}=\frac{a}{2}(F(q+t)+G(q-t))-\frac{a^{2}}{2}\frac{p}{m}\sqrt{1+\frac{p^{2}}{m^{2}}}.\\\
\end{cases}$ (62)

We can further simplify our equations using a change of variables so that we
can get rid of the square roots. We will let $\frac{p}{m}=\tan\theta$, so our
new equations become:

$\begin{cases}\dot{q}=\sin\theta\\\
\dot{\theta}=\frac{a}{2m}(F(q+t)+G(q-t))\cos^{2}\theta-\frac{a^{2}}{2m}\sin\theta.\\\
\end{cases}$ (63)

To get an autonomous system, we define new unknowns:

$\begin{cases}d(t)=q(t)+t\\\ b(t)=q(t)-t\\\ \end{cases}$ (64)

and write the system of three equations as follows:

$\begin{cases}\dot{d}=\sin\theta+1\\\ \dot{b}=\sin\theta-1\\\
\dot{\theta}=\frac{a}{2m}(F(d)+G(b))\cos^{2}{\theta}-\frac{a^{2}}{2m}\sin\theta.\\\
\end{cases}$ (65)

To solve this, we need to look for a solution $(b,d,\theta)$ such that:

$\begin{cases}b(0)=0\\\ d(0)=0\\\ \theta(0)=0.\\\ \end{cases}$ (66)

These are the consequences of our initial conditions $q(0)=0$ and
$\dot{q}(0)=0$. Furthermore, we know the following limits for each variable:
$0<d(t)<2t$, $-2t<b(t)<0$, $-\pi/2\leq\theta(t)\leq\pi/2$.

We will split this into two cases: one where the bare mass, $m$, is positive
and one where it is negative. For the first, we will prove that the solution
will always be stable. For the second, we will show a case where the solution
is unstable.

##  5 Proof of stability for positive bare mass

In the case of positive bare mass, we are concerned with (65) with $m>0$.
Recall that because $F(d)$ and $G(b)$ are defined as in (61), they must be
compactly supported. Furthermore, note that it must always be the case that
$\dot{d}=\sin\theta+1\geq 0$ and that $\dot{b}=\sin\theta-1\leq 0$. Hence, it
suffices to look only at the region where $b\leq 0$ and $d\geq 0$. We will
define $[b_{L},b_{R}]$ such that $-\infty<b_{L}<b_{R}\leq 0$ and $G(b)=0$ for
$b$ outside $[b_{L},b_{R}]$. Similarly, we will define $[d_{L},d_{R}]$ such
that $0\leq d_{L}<d_{R}<\infty$ and $F(d)=0$ for $d$ outside $[d_{L},d_{R}]$.
Based on the fact that $\dot{d}\geq 0$ and $\dot{b}\leq 0$, Figure 3 shows a
rough sketch of the trajectory of the solution projected onto the $(b,d)$
plane.

Figure 3: Projection of trajectory of the solution to (65) in the (b,d) plane

Based on this, we will divide this analysis into three regions: before the
radiation, during the radiation, and after the radiation. In the first region,
$G(b)=F(d)=0$, so (65) reduces to:

$\begin{cases}\dot{d}=\sin\theta+1\\\ \dot{b}=\sin\theta-1\\\
\dot{\theta}=-\frac{a^{2}}{2m}\sin\theta.\\\ \end{cases}$ (67)

One can check that the unique solution to (67) with initial conditions in (66)
is:

$\begin{cases}d(t)=t\\\ b(t)=-t\\\ \theta(t)=0.\\\ \end{cases}$ (68)

Hence, the particle will have the following conditions entering the second
region:

$\begin{cases}b(t_{1})=b_{1}\\\ d(t_{1})=d_{1}\\\ \theta(t_{1})=0\\\
\end{cases}$ (69)

where $t_{1}>0$. In the second region, note that when $\theta=\frac{\pi}{2}$,
$\dot{\theta}=-\frac{a^{2}}{2m}<0$. Similarly, when $\theta=-\frac{\pi}{2}$,
$\dot{\theta}=\frac{a^{2}}{2m}>0$. Hence, the trajectory can never cross
$\theta=\frac{\pi}{2}$ and $\theta=-\frac{\pi}{2}$. The particle will thus end
up with the following conditions entering the third region:

$\begin{cases}b(t_{2})=b_{2}\\\ d(t_{2})=d_{2}\\\ \theta(t_{2})=\theta_{2}\\\
\end{cases}$ (70)

where $t_{2}>0$ and $-\frac{\pi}{2}<\theta_{2}<\frac{\pi}{2}$. Hence, the
third interval will amount to solving the following set of ODEs with
conditions specified in (70):

$\begin{cases}\dot{b}=\sin\theta-1\\\ \dot{d}=\sin\theta+1\\\
\dot{\theta}=-\frac{a^{2}}{2m}\sin\theta.\\\ \end{cases}$ (71)

We solve the third ODE explicitly in Proposition 5.

######  Proposition 5.

Suppose we have (71) with initial conditions in (70).

  1. (a)

If $\theta_{2}=0$, $\theta(t)=0$ for $t>t_{2}$.

  2. (b)

If $\frac{\pi}{2}>\theta_{2}>-\frac{\pi}{2}$ and $\theta_{2}\neq 0$, then
$\underset{t\rightarrow\infty}{lim}\theta(t)=0$.

###### Proof.

$\theta(t)=0$ is a trivial solution which satisfies $\theta(t_{2})=0$. Noting
that $-\frac{a^{2}}{2m}\sin\theta$ and its derivative with respect to $\theta$
is continuous everywhere, we see that such a solution is unique. Hence, (a)
follows.

For (b), assume $\theta_{2}>0$. The proof is similar for $\theta_{2}<0$. If
$\theta=0$ for some time $t_{3}>t_{2}$, we are left with (a), and
$\theta(t)=0<\epsilon$ for $t>t_{3}$. Hence, assume $\theta>0$ at all times.
Then $\sin\theta\neq 0$, and we can separate the third equation of (71):

$\frac{1}{\sin\theta}d\theta=-\frac{a^{2}}{2m}dt.\\\ $ (72)

Integrating, we have:

$-\ln|\csc(\theta)+\cot(\theta)|=-\frac{a^{2}}{2m}t+C_{0}\\\ $ (73)

or

$\csc(\theta)+\cot(\theta)=C_{1}e^{\frac{a^{2}}{2m}t}.$ (74)

We get rid of the absolute value by choosing the sign for $C_{1}$. Here,
$C_{1}>0$ since $\theta(t_{2})>0$ implies $\csc(\theta)+\cot(\theta)>0$. For
any $\epsilon>0$, we can choose $t_{3}$ such that
$\csc(\epsilon)+\cot(\epsilon)=C_{1}e^{\frac{a^{2}}{2m}t_{3}}$. Then for
$t>t_{3}$,
$\csc(\theta(t))+\cot(\theta(t))=C_{1}e^{\frac{a^{2}}{2m}t}>C_{1}e^{\frac{a^{2}}{2m}t_{3}}=\csc(\epsilon)+\cot(\epsilon)$.
Thus, $\theta(t)<\epsilon$. ∎

With this, we have proved the stability of the solution for positive bare
mass. In the case where $\theta_{2}=0$, we see that $\theta(t)=0$ for
$t>t_{2}$. Recalling that $\dot{q}=\sin\theta$, we have $\dot{q}(t)=0$ for
$t>t_{2}$. In the case where $\theta_{2}\neq 0$, we see that since
$\underset{t\rightarrow\infty}{\text{lim}}\theta(t)=0$ and
$\dot{q}=\sin\theta$, $\underset{t\rightarrow\infty}{\text{lim}}\dot{q}(t)=0$.
In either case, the speed of the particle goes to zero.

##  6 Proof of instability for negative bare mass

In the case of negative bare mass, we are concerned with (65) with $m<0$. We
will show that a specific case of radiation leads to an unstable solution.
Assume that the radiation is purely incoming: $F\equiv 0$. Set
$G_{\beta}(x)=\beta\sin(\pi x)\chi_{[-3,-1]}$ where $\beta\in\mathbb{R}$. We
can ignore the $d$ equation and are left with a system of two equations:

$\begin{cases}\dot{b}=\sin\theta-1\\\
\dot{\theta}=\frac{a\cos^{2}\theta}{2m}(G_{\beta}(b))-\frac{a^{2}}{2m}\sin\theta,\\\
\end{cases}$ (75)

$\begin{cases}b(0)=0\\\ \theta(0)=0.\\\ \end{cases}$ (76)

We will rewrite (75) by replacing $m$ with $-|m|$:

$\begin{cases}\dot{b}=\sin\theta-1\\\
\dot{\theta}=-\frac{a}{2|m|}G_{\beta}(b)\cos^{2}\theta+\frac{a^{2}}{2|m|}\sin\theta,\\\
\end{cases}$ (77)

$\begin{cases}b(0)=0\\\ \theta(0)=0.\\\ \end{cases}$ (78)

In addition, it will be useful to work with the reverse flow of the system:

$\begin{cases}\dot{b}=-\sin\theta+1\\\
\dot{\theta}=\frac{a\cos^{2}\theta}{2|m|}(G_{\beta}(b))-\frac{a^{2}}{2|m|}\sin\theta.\\\
\end{cases}$ (79)

Note that $\beta=0$ corresponds to the static solution. To see this, note that
(77) becomes:

$\begin{cases}\dot{b}=\sin\theta-1\\\
\dot{\theta}=\frac{a^{2}}{2|m|}\sin\theta.\\\ \end{cases}$ (80)

The solution to this with initial conditions given in (78) is:

$\begin{cases}b(t)=-t\\\ \theta(t)=0.\\\ \end{cases}$ (81)

Since $\theta=0$ corresponds to $\dot{q}=0$, (81) is the static solution.

To get a better sense of the system of ODEs in (77) for non-zero $\beta$, see
Figure 4. The interval $[-3,-1]$ represents the particle becoming perturbed by
the incoming radiation. A particle with initial conditions given in (78) will
end up with $b=-1$ and $\theta=0$. To see this, note that outside of
$[-3,-1]$, $G_{\beta}(b)=0$, and (77) reduces to (80). Hence, in the interval
$[-1,0]$, the solution to (77) with initial conditions given in (78) is (81).
At $t_{1}=1$, $b(t_{1})=-1$ and $\theta(t_{1})=0$. At this point, we can take
the system of ODEs in (78) and modify the initial conditions as follows:

$\begin{cases}b(t_{1})=-1\\\ \theta(t_{1})=0.\\\ \end{cases}$ (82)

After entering the interval $[-3,-1]$, Figure 4 suggests that the particle
will oscillate. The question is whether or not the particle will go back to
rest ($\theta=0$ at $b=-3$, represented by the grey line in Figure 4) or be
left with some speed ($\theta\neq 0$ at $b=-3$). In the former case, the
particle will remain at rest. In the latter case, the particle will go toward
$\theta=\pm\frac{\pi}{2}$. Recalling that $\dot{q}=\sin\theta$, this means
$\dot{q}=\pm 1$. That is, the particle will reach the speed of light in finite
time. This is proved formally in the next proposition.

######  Proposition 6.

Suppose for some $t_{0}>0$, $b(t_{0})=-3$.

  1. (a)

If $\theta_{0}=\theta(t_{0})=0$, $\theta(t)=0$ for $t>t_{0}$.

  2. (b)

If $\frac{\pi}{2}>\theta(t_{0})>0$, $\theta(t_{1})=\frac{\pi}{2}$ for some
$t_{1}>t_{0}$.

  3. (c)

If $-\frac{\pi}{2}<\theta(t_{0})<0$, $\theta(t_{1})=-\frac{\pi}{2}$ for some
$t_{1}>t_{0}$.

###### Proof.

Because $\dot{b}=$ $\sin\theta-1$, it is always true that $\dot{b}\leq 0$.
Hence, for $t>t_{0}$, $b<-3$, and $G(b)=0$. We can then rewrite the ODEs in
(77) as (80). $\theta(0)=0$ is a trivial solution and satisfies
$\theta(t_{0})=0$. Since such a solution is unique, (a) follows.

We will now show (b). Because $\theta(t_{0})>0$ and $\dot{\theta}>0$ if
$\theta>0$, it follows that $\theta(t_{1})>0$ for $t_{1}>t_{0}$. Hence,
$\sin\theta\neq 0$, and we can separate the second equation of (80):

$\frac{1}{\sin\theta}d\theta=\frac{a^{2}}{2|m|}dt.\\\ $ (83)

Integrating, we have:

$-\ln|\csc(\theta)+\cot(\theta)|=\frac{a^{2}}{2|m|}t+C_{0}\\\ $ (84)

or

$\csc(\theta)+\cot(\theta)=C_{1}e^{-\frac{a^{2}}{2|m|}t}.$ (85)

We get rid of the absolute value by choosing the sign for $C_{1}$. Here,
$C_{1}>0$ since $\theta(t_{0})>0$ implies $\csc(\theta)+\cot(\theta)>0$. Now,
for $0\leq\theta\leq\pi$, $0\leq\csc(\theta)+\cot(\theta)<\infty$, is
monotonously decreasing, and is invertible. Let
$\Theta:[0,\pi]\longrightarrow[0,\infty)$ be the inverse of
$\csc(\theta)+\cot(\theta)$. We can now write the solution explicitly as:

$\theta(t)=\Theta(C_{1}e^{-\frac{a^{2}}{2|m|}t}).$ (86)

The initial conditions tell us:

$C_{1}=\frac{\csc(\theta(t_{0}))+\cot(\theta(t_{0}))}{e^{-\frac{a^{2}}{2|m|}t_{0}}}=\frac{C_{2}}{e^{-\frac{a^{2}}{2|m|}t_{0}}}.$
(87)

where $C_{2}=\csc(\theta(t_{0}))+\cot(\theta(t_{0}))$. Because
$0\leq\theta(t_{0})<\frac{\pi}{2}$, $C_{2}>1$. Substituting $C_{1}$ gives us:

$\theta(t)=\Theta(C_{2}e^{-\frac{a^{2}}{2|m|}(t-t_{0})}).$ (88)

Consider $t_{1}=-\frac{2|m|}{a^{2}}\ln(\frac{1}{C_{2}})+t_{0}>t_{0}$. Using
the fact that $\Theta(1)=\frac{\pi}{2}$ shows us that
$\theta(t_{1})=\frac{\pi}{2}$.

For part (c), repeat the proof for part (b), except $C_{1}<0$, and we define
$\Theta:[-\pi,0]\longrightarrow(-\infty,0]$ instead. ∎

Figure 4: Hypothetical solutions to the system of ODEs

We show instability by proving the existence of a solution to (77) which
satisfies the conditions of (c). To do this, we first work with the backward-
flow defined in (79) and make a change of variables.

######  Proposition 7.

Assume $G(x)=\beta\sin(\pi x)\chi_{[-3,-1]}$. There exists an $\epsilon>0$
such that $0<\beta<\epsilon$ implies that the system of ODEs in (77) with
initial conditions at $t_{1}$:

$\begin{cases}b(t_{1})=-1\\\ \theta(t_{1})=0\\\ \end{cases}$ (89)

has a unique solution $(b(t),\theta(t))$ such that at some $t_{2}>t_{1}$:

$\begin{cases}b(t_{2})=-3\\\ \theta(t_{2})<0.\\\ \end{cases}$ (90)

###### Proof.

To start, consider the backward flow (79) with initial conditions

$\begin{cases}b(t_{1})=-3\\\ \theta(t_{1})=0\\\ \end{cases}$ (91)

and make the following change of variables:

$\begin{cases}y=\theta\\\ x=b+2.\end{cases}$ (92)

Using the fact that $\frac{dy}{dx}=\frac{\dot{\theta}}{\dot{b}}$ gives us the
following ODE:

$\begin{cases}\frac{dy_{\beta}}{dx}=\frac{a}{2|m|}(1+\sin(y_{\beta}))(\beta\sin(\pi
x)-a\sec(y_{\beta})\tan(y_{\beta}))\\\ y_{\beta}(-1)=0.\end{cases}$ (93)

By Lemma 1 below, there exists an $\epsilon>0$ such that $0<\beta<\epsilon$
implies $y_{\beta}(1)>0$. Because of the uniqueness of solutions for first-
order ODEs, a $\beta$ satisfying the previous statement implies that the
solution for the forward-flow with initial conditions specified in (89) ends
up below $0$. This can be intuitively seen in Figure 5. ∎

Figure 5: Relation between forward and backward solution

######  Lemma 1.

Assume that $a>0$. Suppose $y_{\beta}:\mathbb{R}\longrightarrow\mathbb{R}$ is
a function satisfying:

$\begin{cases}\frac{dy_{\beta}}{dx}=\frac{a}{2|m|}(1+\sin(y_{\beta}))(\beta\sin(\pi
x)-a\sec(y_{\beta})\tan(y_{\beta}))\\\ y_{\beta}(-1)=0.\end{cases}$ (94)

There exists an $\epsilon>0$ such that $0<\beta<\epsilon$ implies
$y_{\beta}(1)>0$.

###### Proof.

Consider $y(x,\beta)=y_{\beta}(x)$ and let $Z=\frac{\partial
y}{\partial\beta}$. We can rewrite (94) as:

$\begin{cases}\frac{\partial y}{\partial
x}=\frac{a}{2|m|}(1+\sin(y))(\beta\sin(\pi x)-a\sec(y)\tan(y))\\\
y(-1,\beta)=0\\\ Z(-1,\beta)=0.\end{cases}$ (95)

Because $y(1,0)=0$, it suffices to show that $Z(1,0)=\frac{\partial
y}{\partial\beta}(1,0)>0$. Using the fact that $\frac{\partial Z}{\partial
x}=\frac{\partial^{2}y}{\partial
x\partial\beta}=\frac{\partial^{2}y}{\partial\beta\partial x}$, and
substituting $\beta=0$ (meaning $y=0$), we arrive at the following linear
differential equation for $Z$:

$\begin{cases}\frac{\partial Z}{\partial x}(x,0)=\frac{a}{2|m|}\sin(\pi
x)-\frac{a^{2}}{2|m|}Z\\\ Z(-1,0)=0.\end{cases}$ (96)

The solution to this is simply:

$Z(x,0)=\frac{a}{2|m|}e^{-\frac{a^{2}x}{2|m|}}\int_{-1}^{x}\sin(\pi
t)e^{\frac{a^{2}t}{2|m|}}dt.$

We have:

$Z(1,0)=\frac{a}{2|m|}e^{-\frac{a^{2}}{2|m|}}\int_{-1}^{1}\sin(\pi
t)e^{\frac{a^{2}t}{2|m|}}dt=\frac{2\pi
a|m|(1-e^{-\frac{a^{2}}{|m|}})}{4\pi^{2}m^{2}+a^{4}}.$ (97)

Because $a$ is assumed to be positive, we have $Z(1,0)>0$ as needed.

∎

##  7 Summary and Outlook

We have shown that the static solution to this problem, where the particle
remains at rest forever, is stable for particles with positive bare mass.
However, for particles with negative bare mass, the static solution is highly
unstable. That is, a small amount of radiation can cause the particle to
accelerate to the speed of light in finite time. Though this result is not
intuitive, it is also not very surprising when considering the model we used.
In the initial conditions for the wave equation, we took:

$u(0,s)=-\frac{a}{2}|s|+V_{0}(s).$ (98)

Recall that the field energy density is $\epsilon(t,s)=T^{\mbox{\tiny
field}}_{00}=\frac{1}{2}(u_{t}^{2}+u_{s}^{2})$. Therefore this initial
condition has an infinite amount of energy:

$\int_{-\infty}^{\infty}\epsilon(s,0)ds=\int_{-\infty}^{\infty}\frac{a^{2}}{4}+\dot{V}_{0}^{2}\
ds=\infty.$ (99)

Since the total energy of the system is conserved, there is an infinite amount
of energy available that can be transferred to the particle, allowing it to
accelerate to the speed of light. But another reason this can happen in finite
time is that in this model the Kiessling force $f(t,q,\dot{q})$ itself
diverges as $|\dot{q}|\to 1$.

Hence, looking forward, we would like to examine different models for the
joint evolution, in which such problems are not present. In one such model,
the scalar field would be governed by the Klein-Gordon equation rather than
the wave equation:

$\begin{cases}\partial_{t}^{2}u-\partial_{s}^{2}u+\mu^{2}u^{2}=a\delta(s-q(t))\\\
u(0,s)=\frac{a}{2\mu}e^{-\mu\lvert s\rvert}+v_{0}(s)\\\
\partial_{t}u(0,s)=v_{1}(s).\\\ \end{cases}$ (100)

This would add a mass term to the field equations and change the part of the
initial conditions that represents the static solution. In this model, the
particle would start with a finite amount of energy. It would be interesting
to see if a particle with negative bare mass still accelerates to the speed of
light. We are currently investigating this.

Another model to consider is one in which the field equations are fully
relativistic. The wave operator appearing in (2) is of course relativistic,
but the delta source on the right-hand side of the equation is not manifestly
so. It turns out that it is possible to modify this right-hand side so that
the equation itself becomes fully relativistic. It is possible to show that
for this modified equation for a massless scalar field, the Kiessling force
will not diverge if the particle velocity reaches the speed of light, and
stability of the static solution is restored. This result will appear
elsewhere [5].

Additionally, we would like to explore what would happen with two particles
instead of one. Mathematically, this would involve the sum of two Dirac delta
functions as the source of the wave equation. This may necessitate the use of
differential-delay equations rather than simple ODEs, which would require much
more intricate analysis.

##  8 Acknowledgements

We thank Vu Hoang and Maria Radosz for helping us correct the formula for the
particle self-force, and Lawrence Frolov for reading through the paper and
providing helpful comments. We are grateful to the anonymous referee for many
helpful suggestions and comments.

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