This article shows how to use a Green function to calculate the reaction force based in future external forces instead of using the time derivative of the acceleration, which leads to wrong results.

1. A Programma c Way to Calculate the Electromagne c Reac on
Force (In a Simple Case)
Sergio Prats López, June 2023
You can never plan the future by the past. - Edmund Burke, quoted
when reaching future technologies on Civiliza on VI.
Abstract
The Electromagne c Reac on Force (ERF) depends on the me derivate of the accelera on, 𝑎̇.
When simula ng the mo on of a point par cle on a classic electromagne c field, the addi on
of ERF may cause the algorithm to reach catastrophic results if the accelera on changes fast
enough provided that 𝑎̇ is calculated from the previous values of accelera on. The problem lies
in the fact that increments in the accelera on and its me derivate feedback one another leading
to divergent results.
The reason of this is that, as Paul Dirac showed when he solved the mo on of an electron
disturbed by a pulse, the ERF 𝑎̇ term is influenced by the future external accelera ons and not
by the past ones, this is the reason I have added Edmund Burke quote a er the tle: ERF is not
determined by the past external influences by for the future ones, therefore it cannot be
"planned" by the past.
The simula ons I did was very simple, a light par cle was approaching to another infinitely heavy
par cle, both with the same charge sign, the par cle was approaching either frontally in X
direc on or with some offset in the Y direc on. The heavy par cle will be all the me at rest
while the light par cle will accelerate, that means that the EM field will only depend on the
posi on but not on me.
The method exposed here to calculate the reac on force requires moving backwards the par cle
in the me so that the future accelera on can be computed to calculate the ERF. Since the heavy
par cle does not accelerate at all, calcula ng its EM field is straigh orward, if that par cle did
accelerate, that would increase a lot the complexity of the algorithm since the retarded posi on
is known la er when processing backwards in me.
As a measure of success for this algorithm, the difference between the final and ini al kine c
energies was compared against the Larmor energy radiated by the light par cle.

2. A func on to calculate the reac on force from the external accelera on
Dirac showed that for a point-par cle ini ally at rest, a pulse of external accelera on that would
cause an accelera on 𝒂𝒆𝒙𝒕 = 𝛿(𝑡)𝑥 would transform into an accelera on:
𝒂 = 𝑏 ∗ 𝑒 𝑥 if 𝑡 < 0
𝒂 = 𝟎 if 𝑡 ≥ 0
Where 𝑏 = ∗ = 𝑚 𝑘. The most remarkable fact is that the accelera on happens
before the external force affects the par cle, however, the external force is clearly the cause of
the reac on force although it is not "cause" in a me-fashion since it happens la er.
Since the external force it is the cause of the reac on for, I tried to find a "Green func on" for
the reac on force based on the external forces happening on the par cle. Let the non-covariant
reac on force be something like this:
𝑭(𝑡) = 𝑭𝒆𝒙𝒕(𝑡) + ∫ 𝑔(𝜏)𝑑 𝑭𝒆𝒙𝒕(𝑡 − 𝜏)𝑑𝜏 [1]
Where the first term is the external EM field Lorentz force and the second term is the reac on
force, this means that we want the me derivate of the accelera on to be:
𝒂̇ = ∫ 𝑔(𝜏)𝑑 𝑭𝒆𝒙𝒕(𝑡 − 𝜏)𝑑𝜏 [2]
On the other hand, the me derivate of the accelera on must be consistent with [1], therefore:
𝑚 ∗ 𝒂̇ = 𝑑 𝑭 = 𝑑 𝑭𝒆𝒙𝒕 + ∫ 𝑑 𝑔(𝜏) ∗ 𝑑 𝑭𝒆𝒙𝒕(𝑡 − 𝜏)𝑑𝜏 [3]
We need to find a solu on for 𝑔(𝜏) sa sfying [2] and [3]. We should not include past forces, we
can try some term such as 𝑒 𝑢(−𝜏) where 𝑢(−𝜏) is the heaviside func on and whose me
deriva ve is: 𝑑 𝑒 𝑢(−𝜏) = −𝛿(𝜏) + 𝜏𝑢(−𝜏)𝑒
This leads to the following solu on for the Green func on:
𝑔(𝜏) = 𝑒 𝑢(−𝜏) [4]
With this solu on the me derivate of the accelera on, defined as (𝑭𝒆𝒙𝒕 + 𝒂̇ ) is the same
that what we have defined to be 𝒂̇ , hence, the accelera on must be correct except for a
constant that we can fix to zero.
The total force will be:
𝑭 = 𝑭𝒆𝒙𝒕 + ∫ 𝑢(−𝜏)𝑒 𝑑 𝑭𝒆𝒙𝒕(𝑡 − 𝜏)𝑑𝜏 [5]
And its me deriva ve as:
𝑚 ∗ 𝒂̇ ≡ 𝑭̇ = ∫ 𝑘 ∗ 𝑢(−𝜏)𝑒 − 𝛿(𝜏) 𝑑 𝑭𝒆𝒙𝒕(𝑡 − 𝜏)𝑑𝜏 = 𝑑 𝑭𝒆𝒙𝒕 + 𝑑 𝑭𝒓 [6]
By expressing the reac on force as an integral on future external forces, the infinites will
banish, however it is challenging to figure out the future accelera ons when you do not know
yet the future posi on for the par cle, while this is a big obstacle, the validity of this formula

3. can be assessed by compu ng the mo on of a par cle backwards in me, provided that the
external fields are well known, for example for being sta c.
Extrapola ng the Green func on to the covariant case
The reac on force is an effect that only becomes relevant at rela vis c veloci es, therefore,
using a non-covariant expression for it is not very useful, fortunately, the previous future- me
integral can be adapted to the Reac on Force covariant expression:
𝐹 =
2
3𝑘
𝑎̇𝜇 +
1
𝑐2
𝑎2
𝑣𝜇 [7]
The covariant formula on is needed, let v be the velocity in natural units (c=1) and 𝛾 =
√
,
we will use here the following 4-vectors: 𝑣 = 𝛾(1, 𝒗) would be the four-velocity.
𝑎 = 𝑑 𝑣 ≡ 𝑣̇ is the 4-accelera on, it turns out 𝑚 𝑎 = 𝛾[𝑭 · 𝒗, 𝑭] where F is the force
measured in the lab, the last four-vector to consider is the proper me deriva ve of the
accelera on 𝑎̇ = 𝑑 𝑎 . For the sake of simplicity from now on I will work with normalized
mass 𝑚 = 1 to make the following expressions simpler.
Equa on [1] can be rewri en as:
𝑎 = 𝑎 + 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 + 𝑎𝜇𝑣 [8]
Where 𝑎 = 𝛾[𝑭𝒆𝒙𝒕 · 𝒗, 𝑭𝒆𝒙𝒕] plays the role of the external force, the 𝑎2
𝑣𝜇 term is part of
the reac on force but should be pulled out of the current integral since it depends on the actual
accelera on, no ce that 𝑎2
is a nega ve vector since 𝑎2
= 𝛾2((𝑭 · 𝒗)2
− 𝐹2
). Also no ce that
it is enough with ge ng the space components for 𝑎 (the force), since the power is completely
determined by the force and the velocity: 𝑃 = 𝑭 · 𝒗.
The integral is something that must be equal to 𝑎̇𝜇. We expect to have a Green func on 𝑔(𝜏)
similar to the non-rela vis c case based on the external force but this me the overall force
depends also on 𝑎2
𝑣𝜇 so we need to modify the former 𝑑 𝑭𝒆𝒙𝒕 term to match this equa on:
𝑑 𝑎 + 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 + 𝛾𝑎𝜇𝒗 = 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 [9]
The solu on for this is to add a new 𝑎 𝑣 together with the external force into the integral:
𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 ≡ 𝑎̇ = ∫ 𝑔(𝜏)𝑑 𝑎 (𝑠 − 𝜏) + 𝑎 𝑣 𝑑𝜏 [10]
The me deriva ve for the new term is: 𝑑 𝑎 𝑣 = 𝑎 𝑎 + 2 ∗ 𝑎 · 𝑎̇ 𝑣
Using this integral for 𝒂̇ , the reac on force can be calculated programma cally (at least when
moving backwards). The fact that 𝑎̇ is inside the integral defining 𝑎̇ does not cause a great
impact since the Integral so ens the effect of it.
References
1. Classical theory of radia ng electrons - P. A. M Dirac, 1938.
h ps://royalsocietypublishing.org/doi/10.1098/rspa.1938.0124