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

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 partially resolved this problem by using the momentum balance between the field and the particles, extracting an equation for the force that is well-defined everywhere, 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 for ever. We will prove the stability of the static solution for particles with positive bare mass by showing that any incoming radiation 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.


Introduction and Main Result
Classical electromagnetism seems to have a fundamental problem: For a charged 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 depends on the field, force is also undefined where it's needed, i.e. for the particle's equations of motion to be well-defined. Kiessling [2,3] showed that if one postulated local conservation laws to hold for the total (field + particle) energy tensor, 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 1 . In one space dimension, for a single particle with trajectory s = q(t), this balance law takes the form where f (t, s) is the force exerted by the field at a generic space-time point (t, s) ∈ R 1,1 , π is the field momentum density and τ is the field stress. Here, π = T 01 and τ = T 11 are components of the energy(-density-momentum-densitystress) tensor. It is of interest to study the effect of the Kiessling force on the motion of the charge. To simplify matters 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 [4] 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: The field satisfies t u − ∂ 2 s u = aδ(s − q(t)) u(0, s) = − a 2 |s| + V 0 (s) ∂ t u(0, s) = V 1 (s) (2) while the equations of motion for the particle are (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, compactly supported incoming radiation. Thus V 0 , V 1 ∈ C ∞ c (R \ {0}). Real constants a and m represent the charge and the (bare) rest mass of the particle. Equations (3) are simply Newton's Laws of Motion, amended by Einstein's special-relativistic relation between momentum and velocity. The initial conditions in (3) can always be satisfied by going into the initial rest frame of the particle. We will use equation (1) 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 ≡ 0 and V 1 ≡ 0. In that case, u = − a 2 |s| is clearly a time-independent solution to (2). We shall see that in this case, F ≡ 0, so that we end up with q(t) = 0 for all t as well, i.e. the particle remains at rest forever. In this paper we will prove: (a) Suppose m > 0. For all smooth, compactly supported initial data (V 0 , V 1 ) for (2), there exists a solution (u(t, s), q(t)) to the joint field-particle evolution problem (2)(3), with the property that: (i) the field u is Lipschitz everywhere and the particle trajectory q is C 1 .
(ii) u is at least C 1 away from the particle path s = q(t), and (iii) for all > 0, ∃T > 0 such that t > T implies |q(t)| < .
, and a solution (u(t, s), q(t)) to the joint field-particle evolution problem (2)(3), with the property that: (i) the field u is Lipschitz everywhere and the particle trajectory q is C 1 .
(ii) u is at least C 1 away from the particle path s = q(t), and (iii) the particle reaches the speed of light in finite time, i.e. ∃T > 0 s.t. |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 instability claim in Section 6. We conclude in Section 6 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 |q| < 1, q(0) = 0, anḋ q(0) = 0, the following initial value problem has the unique solution: where the functions T ± are defined by for all x, i.e. T ± is the inverse function to q(x) ± x (which exists and is C 1 so long as |q| < 1), and Furthermore, u(t, s) is at least C 1 away from the path s = q(t).
Proof. Define Ψ(t, s) and φ(t, s) as solving the following equations: Furthermore, define V (t, s) and Γ(t, s) as solving the following equations: We thus have that φ = Γ + V and u = Ψ + φ. 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 Γ (and hence φ) by using D'Alembert's formula. We then have the following: We can solve for Ψ with Duhamel's Principle. Define W (t, s, τ ) as follows: It follows that: To solve for W, we apply d'Alambert's formula. We have: where χ is the characteristic function, i.e.: To integrate W to get Ψ(t, s), we do the following. Given (t, s), backward light cone. The τ for which (τ, q(τ )) is trapped in this light cone will contribute a 2 dτ to the integral. See Fig. 1 Because q(0) = 0 and c = 1, q(t) is trapped inside the forward light cone drawn from (0, 0). As a result of this, Ψ(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 will intersect with s = q(t). Moreover, it intersects exactly once (going from time 0 to t, once it goes out, it cannot go back in 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 τ 2 is the solution to q(τ 2 ) + τ 2 = s + t, or T + (s + t) for short. The solution is hence a 2 T + (s + t). To the right of q(t), this retarded time τ 1 is the solution to q( We then get the following expression for Ψ: We see from here that Ψ, like Γ, 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 that when we add Ψ and Γ, the singularities along the light cone cancel each other out. The full solution u(t, s) is thus as follows: where V is as defined in Equation (12).
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. 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: We will first show w(t, s) is C 1 across s = −t. We have: Recall that T + (s + t) = τ 2 where τ 2 solves the following: Using implicit differentiation by s and t yields: respectively. At the line s = −t, τ 2 = 0, soq(τ 2 ) =q(0) = 0. Hence: We thus have: and 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.

Finding the Kiessling force
We will now combine equation (5) for u(t, s) with equation (3) to find a system of ODEs for q(t). To do this, we will work out the Kiessling force.
Recall that π and τ depend on the derivatives of the field u: 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 (1) is where [·] s=q(t) denotes the jump across the path.
Proof. Note that the assumptions imply that π and τ are bounded and can at most have a jump discontinuity across the path, so that the right-hand side of (27) is well-defined. Let > 0 and T 2 > T 1 ≥ 0. We will integrate over the region Ω = {(t, s)|T 1 ≤ t ≤ T 2 , q(t) − ≤ s ≤ q(t) + } and then take the limit as goes to 0. After integrating and taking the limit, the right-hand side of equation (1) becomes:  (1) becomes: Because π is locally integrable, the first two terms go to 0. Taking the limit as goes to 0 gives us: for the left-hand side. Because T 1 and T 2 are arbitrary: Proposition 4. Assume u(t, s) is a solution to the joint evolution problem (2-3). The Kiessling force in (3) is given by: Proof. By Prop. 3, the force is: using u(t, s) as given by equation (18). Substituting in u = V + w gives us and [ 1 2 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: Inserting these values into equation (45) gives us Note that the first term represents of the force that the external field is exerting on the particle. That is, the term 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.

Equations of motion as a dynamical system
We can now look at the equations of motion for the particle, which are the following: We substitute the value ofq into the equation ofṗ, which results in the following equation:ṗ = aV s (t, q(t)) − a 2 2 In addition to this, let us rewrite V s (t, q(t)). Recall that: Hence, we have that: Let us further define F (x) and G(x) as follows: From our definitions of F and G, we can update our equations of motion, specifically the value ofṗ. It now becomes : We can simplify our equation using a change of variable so that we can get rid of the square roots. We will let p m = tan θ, so our new equations become: To get an autonomous system, we define new unknowns: and write the system of three equations as follows: To solve this, we need to look for a solution (b, d, θ) such that: These are the consequences of our initial conditions q(0) = 0 andq(0) = 0. Furthermore, we know the following limits for each variable: 0 < d(t) < 2t, −2t < b(t) < 0, −π/2 ≤ θ(t) ≤ π/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.

Proof of stability for positive bare mass
In the case of positive bare mass, we are concerned with (53) with m > 0. Recall that because F (d) and G(b) is defined as in (49), they must be compactly supported. Furthermore, note that it must always be the case thatḋ = sin θ + 1 ≥ 0 and thatḃ = sin θ − 1 ≤ 0. Hence, it suffices to look only at the region where b ≤ 0 and d ≥ 0. We will define Based on the fact thaṫ d ≥ 0 andḃ ≤ 0, Figure 3 shows the rough trajectory of the solution on 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 (53) reduces to: One can check that the unique solution to (55) with initial conditions in (54) is: Figure 3: Projection of trajectory of the solution to (53) in the (b,d) plane Hence, the particle will have the following conditions entering the second region.
Proof. θ(t) = 0 is a trivial solution which satisfies θ(t 2 ) = 0. Noting that − a 2 2m sin θ and its derivative with respect to θ is continuous everywhere, we see that such a solution is unique. Hence, (a) follows.
With this, we have proved the stability of the solution for positive bare mass. In the case where θ 2 = 0, we see that θ(t) = 0 for t > t 2 . Recalling thaṫ q = sin θ, we haveq(t) = 0 for t > t 2 . In the case where θ 2 = 0, we see that lim t→∞q (t) = lim t→∞ θ = 0. In either caes, the speed of the particle goes to zero.

Proof of instability for negative bare mass
In the case of negative bare mass, we are concerned with (53) with m < 0. We will show that a specific case of radiation leads to an unstable solution. Assume that the radiation is purely incoming: where β ∈ R. We can ignore the d equation and are left with a system of two equations.
We will rewrite (63) by replacing m with −|m|.
In addition, it will be useful to work with the reverse flow of the system: Note that β = 0 corresponds to the static solution. To see this, note that (65) becomes: The solution to this with initial conditions given in (66) is: Since θ = 0 corresponds to v = 0, (69) is the static solution. To get a better sense of the system of ODEs in (65) for non-zero β, see Figure 4. The interval [−3, −1] represents the particle becoming perturbed by the incoming radiation.
A particle with initial conditions given in (66) will end up with b = −1 and θ = 0. To see this, note that outside of [−3, −1], G β (b) = 0, and (65) reduces to (68). Hence, in the interval [−1, 0], the solution to (65) with initial conditions given in (66) is (69). At t 1 = 1, b(t 1 ) = −1 and θ(t 1 ) = 0. At this point, we can take the system of ODEs in (66) and modify the initial conditions as follows: 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 (θ = 0 at b = −3, represented by the grey line in Figure 4) or be left with some speed (θ = 0 at b = −3). In the former case, the particle will remain at rest. In the latter case, the particle will go toward θ = ± π 2 . Recalling thatq = sin θ, this meansq = ±1. That is, the particle will reach the speed of light in finite time. This is proved formally in the next proposition.
The solution to this is simply: We have: scalar field would be governed by the Klein-Gordon equation rather than the wave equation:      ∂ 2 t u − ∂ 2 s u + µ 2 u 2 = aδ(s − q(t)) u(0, s) = a 2µ e −µ|s| + v 0 (s) ∂ t u(0, s) = v 1 (s) 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 the particle still gets knocked 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 (even though of course the Cauchy problem still breaks Lorentz invariance.) 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 [1].
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 would necessitate the use of differential-delay equations rather than simple ODEs, which require much more intricate analysis. and thus in this case The energy tensor for the particle is defined such that: Sinceq(t) is a contravariant vector quantity, from (3) we see that f is a contravariant vector as well. Meanwhile, f ν is a covariant vector. Hence, we have f = η 1ν f ν = (0)(f 0 ) + (−1)(f 1 ) = −f 1 . Since f is concentrated on the trajectory s = q(t) of the particle, f = f (t, s)δ(s − q(t)). Going back to the energy tensor for the field, we have for ν = 1: Using equation (80), we have: Rearranging this gives us the momentum-balance law: