Sample Calculations using Weber Force Lawand Free Charges

Kurt Nalty

July 21, 2011

Abstract

Weber and Gauss (1831 and later) extended Ampere’s magneticforce law to include a potential formulation for the force between twocharged particles. For macroscopic engineering work, their formulais indistinguishable from the Biot-Savart law or Ampere law. Thisnote examines the Weber law as a candidate for particle to particleinteraction.

Ampere’s and Weber’s Force Laws

Wilhelm Eduard Weber literally set the standard in fundamental electro-magnetic experimental work, from measuring terrestrial magnetism to set-ting standards for resistance, current, magnetic fields and voltage. Weber(and Gauss) were also the first, to my knowledge, to achieve a potentialformulation for electromagnetic interactions.

Prior to Weber, explanations for electromagnetic forces presented theappropriate force laws, consistent with measurement, and the matter wassettled. From an engineering point of view, this is a very reasonable approach.However, with the concept of the conservation of energy, and the originationof forces as gradients in energy density, Weber wanted to do more. Startingwith Ampere’s law, he examined different configuration of charge carriers,and using the principle of virtual work, found a potential formulation thatleads to Ampere’s law.

1

Using MKS units, Ampere’s differential force law between two infinitesi-mal current elements is

d2 ~F

ds ds′=

µII ′

4πr2

(2r

∂2r

∂s ∂s′− ∂r

∂s

∂r

∂s′

)~ar

=µII ′

4π

1

r2[3 (~ar · ~u) (~ar · ~u′)− 2 (~u · ~u′)]~ar

where ds and ds′ are differential lengths along the two conductors, r is theseparation between the current elements, ~u and ~u′ are unit tangents to thecurves, ~ar is the unit vector in the direction between the two current elements,and I and I’ are the current values.

Dropping from current elements to elementary charges and time deriva-tives, Weber’s force law is

~F =qiqj4πε0

~rijr3ij

(1−

r2ij

2c2+rij rijc2

)where ε = 8.854 · 10−12 F/m, r is the radial separation, r is the time rate ofchange of radial separation, r is the second time derivative of the separation,and c = 2.998 · 108 m/s is the speed of light.

We can re-write this using vector notation. Define relative positions,velocities and accelerations.

~R = ~r1 − ~r2

~ar =~R√~R · ~R

~V = ~v1 − ~v2

~A = ~a1 − ~a2

Calculate time derivatives of radial separation.

~R · ~R = r2

d

dt~R · ~R =

d

dtr2

2~R · d~R

dt= 2~R · ~V = 2r

dr

dt= 2rr

r = ~ar · ~V

2

Now for the second derivative.

d

dtr =

d

dt

(~R · ~Vr

)=

1

r

(~R · ~A+ ~V · ~V

)− 1

r2

(~R · ~V

) drdt

r =1

r

(~R · ~A+ ~V · ~V −

(~ar · ~V

)2)

As an aside, the last two terms in the parenthesis can be spotted as thesquared magnitude of ~ar × ~V .

Substituting these in the Weber force equation, we have (in vector com-ponent format)

~F =qiqj4πε0

~rijr3ij

(1−

r2ij

2c2+rij rijc2

)

=qiqj4πε0

~rijr3ij

1 +

(V

c

)2

− 3

2

(~ar · ~Vc

)2

+~R · ~Ac2

Weber’s potential, from which the above is derived by radial gradient, is

U =qiqj4πε

1

rij

(1− ˙rij

2

2c2

)or in vector notation

U =qiqj4πε

1

r

(1− (~R · ~V )2

2r2c2

)

Historical Loss of Interest in Weber’s Force Law

Weber’s force law fell into neglect for several reasons.Weber used Fechner’s hypothesis that current was carried in conductors

by positive and negative charges both carrying half the current. We know thishypothesis to be false, as the electrons are mobile but the nucleons fixed formetals. With the dismissal of the Fechner hypothesis, many people dismissed

3

the Weber formula as also necessarily invalid. This was unfortunate. AndreAssis, in his book Weber’s Electrodynamics (1994) demonstrates that theonly necessary requirement to achieve Ampere’s law from Weber’s law isthat of neutral conductors.

Weber’s force law takes more effort for engineering calculations than Biot-Savart. For engineering calculations, the three Weber interactions (e1-e2,p1-e2,e1-p2) simplify to Ampere’s law. Ampere’s law, in turn, requires a fewmore add/multiplies than Biot-Savart. In the engineering regime, when re-sults are identical, the calculators will use the simplest formula.

Weber’s formula gave strange answers at high velocity. Helmholtz inte-grated the trajectories of charged particles in a charged, insulator sphericalshell, and demonstrated run-away solutions for speed. This was deemed afatal flaw. However, Weber pointed out that Helmholtz’ initial conditionswere already super-luminal (tachyonic).

I want to emphasize that all of the equivalent Maxwell potentials such asGrassman, Biot-Savart, Whittaker and Weber, are non-relativistic.

Motivation for Re-examination

I like the ideas of purely electromagnetic mass, inertia based upon collectiveradiation reaction, and magnetic moments as an artifact of internal trajecties.I am re-examining historical force laws, looking for such solutions.

Low Speed Examination of Weber’s Potential

The goal in this section is to accept the Weber formulations, and discussthem in the context of 1840-1870 electrical technology.

Verification of Engineering Equivalency

http://www.kurtnalty.com/Weber.cpp demonstrates the equivalence of Am-pere, Biot-Savart, Whittaker and Weber force laws for forces involving closedcurrent loops. A relevant snippet follows. Weber force calculations (F8) aredone between source and sensor electrons, as well as between source elec-trons and sensor nucleons, and source nucleon and sensor electrons. Thenucleon-nucleon calculation is not necessary for stationary coils.

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//Ampere’s Law

F4 += ir2*(2.0*dot(v1,v2) - 3.0*dot(v1,ar)*dot(v2,ar))*ar;

// Biot-Savart

F5 += ir2*(dot(v1,v2)*ar - dot(v2,ar)*v1);

//Whittaker Riemann

F6 += ir2*(dot(v1,v2)*ar - dot(v2,ar)*v1 - dot(v1,ar)*v2);

//Weber e- e- proves inadequate

F7 += -ir2*(dot(V,V)-1.5*dot(ar,V)*dot(ar,V)+dot(R,A))*ar;

// Three interacting charge collections. Works

//Weber e- e-

F8 += -ir2*(dot(V,V)-1.5*dot(ar,V)*dot(ar,V)+dot(R,A))*ar;

//Weber p+ e-

F8-=-ir2*(dot(-v2,-v2)-1.5*dot(ar,-v2)*dot(ar,-v2)+dot(R,-a2))*ar;

//Weber e- p+

F8-=-ir2*(dot(v1,v1)-1.5*dot(ar,v1)*dot(ar,v1)+dot(R,a1))*ar;

As far as being consistent with existing observations, Weber’s law gavethe same engineering results as Ampere’s or Biot-Savart laws, with only threetimes the previous effort. As a calculator, you would prefer to use Biot-Savartform, as this was the simplest formula of these.

Description of Terms

From the point of view of particle interactions, the Weber law is a modifedGauss law for electric fields.

~F =qiqj4πε0

~rijr3ij

(1−

r2ij

2c2+rij rijc2

)The first term in parenthesis is the Coulomb force law. The second term

provides a magnetic field correction which always works against the Coulomb

5

term. The third term is the acceleration term, which can alternate fromassisting to opposing the Coulomb term.

Because the force is purely radial, this force law cannot change angularmomentum. This means that if we are not aimed directly at another charge,we will not collide, but rather orbit or scatter.

Closed Form Solutions with Elliptic Integrals

This section is based upon Assis, Chapter 7, with initial solution credited toSeegers in 1864.

Radial forces inherently preserve angular momentum.

~L = ~R× ~P

= m~R× ~V

d~L

dt= m~V × ~V +m~R× ~A = 0

Since acceleration ~A is parallel to radial separation ~R, both cross productszero out. Consequently, the angular momentum becomes a constant of mo-tion, and the particle’s orbit is in a plane.

For a particle in the plane with polar coordinates

~L = m~R× ~V~L = m~R× (Vr~ar + Vθ~aθ)

~L = m~R× (r~ar + rθ~aθ)

~L = m~R× rθ~aθ~L = mr~ar × rθ~aθLz = mr(rθ) = mr2θ

Place the origin at the center of mass. The two masses M1 and M2 arealways connected by a line through this origin. If the separation between thetwo masses is r, then the two lengths as measured from the center of massare

r1 = r ∗M2/(M1 +M2)

r2 = r ∗M1/(M1 +M2)

r = r1 + r2

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Some authors, for example Bradbury Theoretical Mechanics, will use a nega-tive sign on r2 to emphasize that the two points are on opposite sides of thecenter of mass.

The total angular momentum is

L = M1 ∗ r21 ∗ θ +M2 ∗ r2

2 ∗ θ

= θ

(M1 ∗ r ∗

M2

M1 +M2

∗ r1 +M2 ∗ r ∗M1

M1 +M2

∗ r2)

= θM1M2

M1 +M2

(r ∗ r1 + r ∗ r2)

= mr2θ

with the reduced mass defined as

m =M1M2

M1 +M2

The total energy for this system becomes

U =m

2

(r2 + r2θ2

)+q1q24πεr

(1− r2

2c2

)The first term above is the kinetic energy, and is always positive. The

second term, the modified Coulomb potential, has an interesting behavior.When the radial separation speed exceeds c

√2, we change from electrical

repulsion to attraction, or vice versus. Choosing initial conditions with pos-itive masses, positive charges, but high separation rates leads to a boundsystem (negative total energy), which I will address in the later section.

Under mundane conditions, positive energy means unbound systems. Forexample, repulsive charges, or the combination of attractive charges buthigher kinetic energy. For inverse square force laws, negative energy is asso-ciated with a maximum radial separation (this being a turning point), andleading to the rule of thumb of bound systems.

Our roadmap for solving this problem begins with eliminating θ fromthe energy equation using the angular momentum relationship above. Next,we change variables, leaving the time domain and instead finding an orbitequation θ(r). We determine maximum and minimum values for r, then findthe angle corresponding to each radial value. Finally, knowing θ and r, wecan integrate the angular momentum relationship for θ to recover θ(t) andr(t)

7

Eliminate θ

θ =L

mr2

U =m

2

(r2 + r2θ2

)+q1q24πεr

(1− r2

2c2

)=

m

2

(r2 +

L2

m2r2

)+q1q24πεr

(1− r2

2c2

)Find Turning Points

U = r2

(m

2− q1q2

4πεr

1

2c2

)+

L2

2mr2+q1q24πεr

r2 =U − L2

2mr2− q1q2

4πεrm2− q1q2

8πεrc2

Turning points occur where the numerator equals zero.

0 = U − L2

2mr2− q1q2

4πεr

= r2 − r q1q24πεU

− L2

2mU

rturn =q1q2

8πεU±√( q1q2

8πεU

)2

+L2

2mU

For L = 0, we get a collision as expected in the attractive case (chargesigns differ), and a point of closest approach in the repulsive case (chargesigns match) using the positive root.

For the next few cases, we need to be aware that the system energy canbe positive, negative or zero.

When U = 0, we cannot determine a turning point, but we can sayUrturn = 0, meaning rturn is finite. Probably, this system does not evolve,but is static.

For L 6= 0, for the repulsive case (charge signs match and U is positive),we can use the positive root to find the radius of closest approach.

For L 6= 0, for the attractive case (charge signs differ), we have two cases.For bound systems, where U < 0, I believe we use the negative root. Forunbound systems, where U > 0, we use the positive root.

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Solving the Orbit Equation

Start with the energy equation. (We are still following the approach in Assis.)

U =m

2(r2 + r2θ2) +

α

r

(1− r2

2c2

)where

α =q1q24πε

Change variables.

x2 = 1− α

mc2r

2xdx

dt=

α

mc2r2

dr

dtdx

dt=

α

2mc2r2x

dr

dtdx

dt

dt

dθ=

α

2mc2r2x

dr

dt

dt

dθdx

dθ=

α

2mc2r2x

dr

dt

mr2

Ldr

dt=

2c2xL

α

dx

dθ=c2L

α

d(x2)

dθ

r =α

mc21

1− x2

1

r=

mc2(1− x2)

αα

r= mc2(1− x2)

9

Start substituting into the energy equation

U = r2

(m

2− q1q2

4πεr

1

2c2

)+

L2

2mr2+q1q24πεr

= r2

(m

2− α

r

1

2c2

)+

L2

2mr2+α

r

U =

(2c2xL

α

dx

dθ

)2(m

2− α

r

1

2c2

)+

L2

2mr2+α

r

U =

(2c2xL

α

dx

dθ

)2(m

2− m(1− x2)

2

)+

L2

2mr2+mc2(1− x2)

=

(2c2xL

α

dx

dθ

)2(mx2

2

)+

L2

2mr2+mc2(1− x2)

=

(2c2xL

α

dx

dθ

)2(mx2

2

)+L2

2m

(mc2(1− x2)

α

)2

+mc2(1− x2)

2α2U = 4c4x4L2m

(dx

dθ

)2

+ L2(mc4(1− x2)2

)+ 2α2mc2(1− x2)

Continuing after a short typesetting break,

4c4x4L2m

(dx

dθ

)2

= 2α2U − L2mc4(1− 2x2 + x4)− 2α2mc2(1− x2)

Divide both sides by 4c4x4L2m(dx

dθ

)2

=2α2(U/m)− L2c4(1− 2x2 + x4)− 2α2c2(1− x2)

4c4x4L2

=x4(−L2c4) + x2(2L2c4 + 2α2c2) + (2α2(U/m)− L2c4 − 2α2c2)

4c4x4L2

=x4(−c4) + x2(2c4 + 2α2c2L−2) + (2α2L−2(U/m)− c4 − 2α2L−2c2)

4c4x4

=x4(−1) + x2(2 + 2α2c−2L−2) + (2α2L−2c−4(U/m)− 1− 2α2L−2c−2)

4x4

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We can de-clutter a bit by introducing

β2 =α2

L2c2(dx

dθ

)2

=x4(−1) + x2(2 + 2β2) + (2β2c−2(U/m)− 1− 2β2)

4x4

=x4(−1) + x2(2 + 2β2) + (−1− 2β2 (1− (U/ (mc2))))

4x4

Let’s factor the numerator.

0 = x4 + x2(−2− 2β2) +(1 + 2β2

(1−

(U/(mc2

))))x2 = (1 + β2)±

√(1 + β2)2 − (1 + 2β2 (1− (U/ (mc2))))

= (1 + β2)±√β4 + (2β2U/(mc2))

Define

x2A = (1 + β2) +

√β4 + (2β2U/(mc2))

x2B = (1 + β2)−

√β4 + (2β2U/(mc2))

Then, our differential equation is

dx

dθ= ± 1

2x2

√(x2

A − x2)(x2 − x2B)

dθ

dx= ± 2x2√

(x2A − x2)(x2 − x2

B)

θ(x) = ±∫ xA

x

2x2dx√(x2

A − x2)(x2 − x2B)

= ±2xAE(φ, k)

by Gradshteyn, p247 formula 3.153.8, where

φ = sin−1

(√x2A − x2

x2A − x2

B

)

k =

√x2A − x2

B

x2A

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Calculating and Plotting an Orbit

So, now that we have a formula, how do we get an orbit, or better still, atime history?

The program WeberForces.c, at http://www.kurtnalty.com/WeberForces.c, which uses http://mymathlib.webtrellis.net/c_source/functions/

elliptic_integrals/legendre_elliptic_integral_second_kind.c solvesfor the orbit of an electron in a hydrogen atom environment at a potentialslightly above ground level.

Given initial conditions of charges, masses, positions and velocities, wecalculate a reduced mass and center of gravity. We calculate some constantsof convenience, α = (q1q2)/(4πε) and β2 = (α2)/(L2c2). We next calculatetwo limiting parameters, x2

A and x2B. The two parameters x2

A and x2B above

provide our information about the minimum and maximum radius. Theseradii match the previous rmin and rmax from the energy equation.

rmin = ((alpha)/(m*c*c))*((1.0)/(1.0-xA2));

rmax = ((alpha)/(m*c*c))*((1.0)/(1.0-xB2));

We also get the angular extent for this semi-orbit.

phi = asin(sqrt((xA2-xB2)/(xA2-xB2))); // asin(1) = 90 degrees

theta_high = 2.0*sqrt(xA2)

*Legendre_Elliptic_Integral_Second_Kind(phi,’k’,k);

phi = asin(sqrt((xA2-xA2)/(xA2-xB2))); // asin(0) = 0 degrees

theta_low = 2.0*sqrt(xA2)

*Legendre_Elliptic_Integral_Second_Kind(phi,’k’,k);

For the example studied, Theta high = 180.005 degrees, and Theta low= 0 degrees. We see the precession amount at this energy and eccentricity isquite small.

For the simulation, I start with the lower value of xB2, calculate R fromx2, then calculate phi and theta from x2. I then step 1% toward xA2, andrepeat until we reach xA2.

output = fopen("R_theta_x_y.txt","w");

dx2 = 0.01*(xA2 - xB2); // increment 1%

for (i=0;i<101;i++) {

12

x2 = xB2 + i*dx2;

rFromX = ((alpha)/(m*c*c))*((1.0)/(1.0-x2));

phi = asin(sqrt((xA2-x2)/(xA2-xB2)));

theta = 2.0*sqrt(xA2)*

Legendre_Elliptic_Integral_Second_Kind(phi,’k’,k);

fprintf(output,"%g %g %g %g \n",rFromX,

theta, rFromX*cos(theta), rFromX*sin(theta) );

}

fclose(output);

The output file has four columns, being radius, radian angle, x coordinateand y coordinate. I used the first two columns to plot a radial impulsediagram for this simulation using GnuPlot

Plot with GnuPlot - commands:

set term postscript

set output "R_theta.ps"

set polar

set size square

set xrange[-1e-10:1e-10]

set yrange[-1e-10:1e-10]

plot "R_theta_x_y.txt" using 2:1 title "r(theta)" with impulses

The postscript image is turned into a pdf using ps2pdf, then included asFigure 1.

Bypassing x2 and using r

Given the simple relations between x2 and r, and the especially simple answerformat, I now eliminate the use of x2, and simply present the same resultsparameterized in terms of r.

13

Figure 1 - Radius versus Angle for Electron Near Proton

1e-10

5e-11

0

5e-11

1e-10

1e-10 5e-11 0 5e-11 1e-10

r(theta)

14

α =q1q24πε

β2 = (α2)/(L2c2)

U =m

2(r2 +

L2

m2r2) +

α

r

(1− r2

2c2

)rmin =

α

2U−√( α

2U

)2

+L2

2mU

rmax =α

2U+

√( α

2U

)2

+L2

2mU

Define

ρ = − α

mc2

This number just happens to be the classical electron radius. Using ρ and r,we can write the expressions for k, φ and θ.

k =

√ρ(rmax − rmin)

rmax(rmin + ρ)

φ(r) = sin−1

√1− (rmin/r)

1− (rmin/rmax)

θ(r) = ±2√

1 + (ρ/rmin)E(φ, k)

The agreement between these two formula sets is shown in Figure 2.The above formulas have a nice geometrical interpretation. E(φ, k) is the

distance travelled along an ellipse of unit major radius, travelling to angle φstarting at the minor radius

√1− k2. The scale factor in from of E is can

be a scaling factor for the ellipse, or the rolling circle, or split among both.A unit rolling circle covers an angle θ during this event. Of course, if we aredriving θ, then the ellipse rolls correspondingly.

Now, the next step in making this simulation user friendly, is to invert θ(r)to provide r(θ). Getting φ from E(φ, k) seems to be a task that just hasn’tbeen widely done. This inversion can be done graphically, by rotating theimage of E(φ, k) around the 45 degree diagonal. This inversion can probably

15

Figure 2 - X,Y Overlay for R versus x2 Calculations

-1e-10

-5e-11

0

5e-11

1e-10

-1e-10 -5e-11 0 5e-11 1e-10

r(theta) using x^2r(theta) using r

16

be done by inverting the series expansion. Whenever I figure this out, I’llrevise and finish this section.

At any rate, now that we’ve verified reasonable behavior at low speeds,let go to outrageous speeds next!

Science Fiction Aspects of Weber’s Potential

The goal in this section is to continue to literally accept the Weber formu-lations, and look at microscopic and transluminal extrapolations as well asBoscovitch-like behavior.

Roger Joseph Boscovich, also known as Ruggero Giuseppe Boscovich(1711-1787) proposed a force law for solids which is attractive at large dis-tances, yet turns repulsive or even oscillatory at close range. He also discussedvariations, where the repulsive forces can arise due to velocity or accelerationterms between the particles.

Coulomb’s law for electrostatics, U = q1q2/(4πε), is alway repulsive or at-tractive depending upon the charges involved. Biot-Savart’s non-relativisticextensions for magnetics, U = q1q2(1−~v1 ·~v2/c

2)/(4πε), reduces the Coulombforce between parallel currents at low speeds, nulls out the Coulomb for atlight speed, and overwhelms repulsion at superluminal speeds.

Weber’s potential, U = q1q2(1 − r2/2c2)/(4πε), by contrast, reduces theCoulomb force between approaching/retreating charges, rather than parallelcharges. This magnetic term does nulls out, then overwhelms, the Coulombforce as the radial speed exceeds 1.414 c.

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Weber’s Potential with Angular Momentum Substitution

In reduced mass, center of gravity coordinates, we have the Weber energy as

U =m

2(r2 + r2θ2) +

α

r

(1− r2

2c2

)where

α =q1q24πε

L = mr2θ so

rθ =L

mr

U =m

2(r2 +

L2

m2r2) +

α

r

(1− r2

2c2

)U =

m

2r2 +

L2

2mr2+α

r− 1

r

α

2c2r2

Superluminal Bonding for Like Polarity Charges

Look at a case where we want two negative charges to be bound. This canbe done with the Weber force law if we make the system energy negative.Looking at the first three terms on the right hand side, we have strictlypositive quantities. The only chance for a negative contribution comes fromthe fourth term, which incidently does not care about approach or retreat.We see that the minimum radial speed for attraction is r >

√2c.

Let’s rearrange this equation looking at r2.

U =m

2r2 +

L2

2mr2+α

r− 1

r

α

2c2r2

r2 =U − L2

2mr2− α

rm2− 1

rα

2c2

The left hand side of the equation, as a square, is always positive. Thenumerator has three always negative terms, and so is always negative. Con-sequently, we require the denominator to be always negative. This in turn,

18

sets a maximum radius for this system.

m

2− 1

r

α

2c2< 0

m

2<

1

r

α

2c2

r <α

mc2

Special Case Where L = 0

For the special case where L = 0, as r → 0, our expression for r2 limits to

r2 =U − α

rm2− 1

rα

2c2

=2rc2U − 2αc2

mrc2 − α= 2c2 when r = 0

We have the interesting situation, where we hit our minimum speed possibleat r = 0. In effect, we slow down to just above 1.414c as we hit zero. We nowask, what happens next? A simple minded approach indicates that r nowgoes negative, and this, in turn, has profound consequences for our potentialsand forces.

In the expressions above, we have α/r = q1q2/(4πεr). The consequenceof r going negative is indistinguishable from one of the charges changingpolarity. We now have a new simulation, of opposite charges at very closeseparation, and a separation speed at 1.414c.

Let’s return to the energy equation, for L = 0.

U =m

2r2 +

α

r− 1

r

α

2c2r2

U − m

2r2 =

α

r

(1− r2

2c2

)

The left hand side is always negative. For the r > 0 case, to get the rightside negative required r2 > 2c2. Now, for the r < 0 case, we require just theopposite, namely r2 < 2c2, and slowing down. How far out do we go before

19

hitting a turning point? Setting r = 0,

U =α

r

r =α

Umaximum negative radius

After hitting this turning point, we reverse, and accelerate heading backtoward zero. We hit r = 0 at 1.414c, as before, and now are outbound,positive radius, and accelerating.

We know from previously, that there will be a maximum radial separation.This system is going to have a hard bounce at rmax = α/(mc2). How fastwill we be at the hard bounce point? We are going infinitely fast due tothe divide by zero. These results are similar in spirit to tachyon calculations,where instead of a critical speed c, we have a critical speed c

√2. For tachyons,

minimum kinetic energy occurs at infinite speed, and lower, finite speeds area higher kinetic energy.

Knowing the overall behavior, let’s now solve the radial velocity differen-tial equation. As an added bonus, since L = 0, this is also the speed andvelocity equations as well.

r2 =2rc2U − 2αc2

mrc2 − α

=2

m

[rc2U − αc2

rc2 − α/m

]=

2

m

[rU − α

r − α/(mc2)

]

Define

ρ =α

mc2

Then

r2 =2

m

[rU − αr − ρ

]=

2

m

[rU − ρU + ρU − α

r − ρ

]=

2

m

[U +

ρU − αr − ρ

]20

The constant ρ is the maximum separation in the positive radial case wesaw earlier. I’ll call this distance “the wall”, as it is the place of reflectionin the high speed case. We see when r = ρ, we have our infinite speed case.We also see when r = α/U , (a negative radius, by the way), that r = 0, asbefore.

We now make a change of coordinates. Let x = ρ − r be the distancefrom the wall. This x will be a positive number between xmin = 0 (the wall)and xmax = ρ− U/α. During most of this time, r is negative. Keep in mindthat ρ is a constant.

r2 =2

m

[U +

ρU − αr − ρ

]r2 =

2

m

[U +

α− ρUρ− r

](d(ρ− r)

dt

)2

=2

m

[U +

α− ρUρ− r

]x2 =

2

m

[U +

α− ρUx

]=

2U

m+

2(α− ρU)

m

1

xDefine

a = −2U

mU always negative so a is positive

b =2(α− ρU)

malways positive

Our differential equation is then

x2 = −a+ (b/x)

x2 = (b/x)− adx

dt=

√(b/x)− a

dt =dx√

(b/x)− a

=

√xdx√b− ax

t =

∫ x

0

√y

√b− ay

dy

21

Substitute

y =b

asin2 θ

t =

∫ x

0

dy√(b/y)− a

becomes

t =b

a√a

∫ φ

0

(1− cos (2θ)) dθ

φ = sin−1

(√xa

b

)t =

b

a√a

[φ− 1

2sin (2φ)

]

If you want a longer formula without trig substitutions,

t =b

a√a

[sin−1

(√xa

b

)−√xa

b

√1− xa

b

]So, now that we have our formulas, let’s plot x(t) and v(t). The file

“SimulateTwoWeberBoundElectronsAt-13.6V.c”, available at http://www.

kurtnalty.com/SimulateTwoWeberBoundElectronsAt-13.6V.c models twosuperluminally bound electrons.

We have two regimes. From x = 0 to x = ρ we are inbound from infinitespeed having hit the wall at x = 0. We have two electrons, both withpositive radial separation, at very superluminal speeds. This regime onlylasts 8.68721e-24 seconds, and only covers 5.55642e-15 meters. We dropfrom infinite speed to 4.26402e+08 (c

√2) m/s during this time. Figure 3

provides distance versus time, while Figure 4 provides speed versus time forthis regime. I suspect outside observers would see two negative electronsduring this very short interval.

The next regime goes from x = ρ to x = ρ − rmax, which is the nega-tive radial separation regime. This regime lasts a much longer 4.69246e-17seconds, and covers a huge 1.04692e-10 meters. We drop from 4.26402e+08(c√

2) m/s to zero during this time. Figure 4 provides distance versus time,while Figure 6 provides speed versus time for this regime. I suspect out-side observers could not distinguish what they see from an electron and apositron during this interval, excepting the difficulty of assigning charge to

22

Figure 3 - Inbound Position Versus Time

0

1e-15

2e-15

3e-15

4e-15

5e-15

6e-15

0 1e-24 2e-24 3e-24 4e-24 5e-24 6e-24 7e-24 8e-24 9e-24

Pos

ition

Time

Position Versus Time Positive Radius Inbound

23

Figure 4 - Inbound Velocity Versus Time

0

5e+08

1e+09

1.5e+09

2e+09

2.5e+09

3e+09

3.5e+09

4e+09

4.5e+09

0 1e-24 2e-24 3e-24 4e-24 5e-24 6e-24 7e-24 8e-24 9e-24

Vel

ocity

Time

Velocity Versus Time Positive Radius Inbound

24

Figure 5 - Outbound Position Versus Time

0

2e-11

4e-11

6e-11

8e-11

1e-10

1.2e-10

0 5e-18 1e-17 1.5e-17 2e-17 2.5e-17 3e-17 3.5e-17 4e-17 4.5e-17 5e-17

Pos

ition

Time

Position Versus Time Negative Radius Rebound

the individual elements. I suspect the outside world would see a neutral sys-tem during this time. (The third particle observor simulations are anotherpriotity task.)

General Discussion of L = 0 Case

As the binding energy increases, the rmax = α/U distance on the negativeradial rebound decreases. Our system shrinks with increasing binding energy.From the formulas ρ = α/(mc2) and rmax = α/U , we see that the magnitudeof the wall bounce equals the negative radial excursion when the magnitudeof the bound energy is mc2. This may be significant.

The run-away solutions of Helmholtz, in my opinion, are a lost oppor-tunity of their time. I am perfectly happy with tachyonic behavior, wherekinetic energy decreases with velocity, and have no problem with reflectionat the maximal radius, since the equations give r2, which is conserved onreflection through infinite speed (the zero kinetic energy speed for tachyons).

So, summarizing, given a negative system energy, two equal polarity

25

Figure 6 - Outbound Velocity Versus Time

0

5e+07

1e+08

1.5e+08

2e+08

2.5e+08

3e+08

3.5e+08

4e+08

4.5e+08

0 5e-18 1e-17 1.5e-17 2e-17 2.5e-17 3e-17 3.5e-17 4e-17 4.5e-17 5e-17

Vel

ocity

Time

Velocity Versus Time Negative Radius Rebound

26

charges oscillate along a line through their centers. Regardless of initialenergy, the speed at impact is c

√2. The motion has two regimes, corre-

sponding to positive and negative radial separations. In the positive radialseparation mode, the charges separate at c

√2 at collision, rapidly accelerate

to infinite speed at a distance ρ = α/(mc2). At this distance, their tachyonickinetic energy is zero, and they reflect back toward the origin. During thefall, they decelerate from infinite speed down to c

√2 at collision. They pass

through each other at this speed, then have a change in their force law due tothe negative radial separation ala CPT. They continue to decelerate to zerospeed at rmax = α/U , then reverse direction at this point of zero ordinarykinetic energy, and the cycle repeats.

General Case where L 6= 0

When L 6= 0, we won’t have a collision, and there will be a minimum radiusof approach, as well as a maximum. The biggest difference between this caseand the previous system should be an absense of negative radial motion.

We begin by looking at the energy equation including angular momentum,and solve for the turning points.

U =m

2r2 +

L2

2mr2+α

r− 1

r

α

2c2r2

r2 =Ur2 − L2

2m− αr

m2r2 − r α

2c2

= 0 implies

0 = Ur2 − L2

2m− αr

0 = r2 − αr

U− L2

2Um

rturn =α

2U±√( α

2U

)2

+L2

2mU

We see we have two turning points (no surprise). Given that the systemis bound, meaning U < 0, the square root term will be less in magnitudethan the magnitude of α/(2U). For the case of an attractive force betweenopposite charges, where α < 0 and α/(2U) > 0, we get a reasonable set ofpositive radii similar to the low speed example dealt with in the first part ofthis paper.

27

So, let’s look the turning points for the case of similar charges with L 6= 0.Because α > 0 due to both charges being the same polarity, given U < 0,we have two negative radii. In essense, if I demand a bound system withsimilar charges, the mathematics cheerfully reply back ‘Quite Right Sir! Wehope you are happy with negative radii which are indistinguishable from areversed charge.’

A numerical illustration is useful. The left column is a positron/electroncombo, while the right column is an electron/electron combo at the sameenergy and momentum.

U = −2.17610−18 U = −2.17610−18

m = 4.55510−31 m = 4.55510−31

q1 = −1.610−19 q1 = −1.610−19

q2 = 1.610−19 q2 = −1.610−19

α = −2.3008610−28 α = 2.3008610−28

LMax = 7.4437310−35 LMax = 7.4437310−35

L = 5.9549810−35 L = 5.9549810−35

a = 5.2869110−11 a = −5.2869110−11

r1 = 8.4590510−11 r1 = −2.1147610−11

r2 = 2.1147610−11 r2 = −8.4590510−11

We can’t distinguish between bound opposite charges, or reverse boundsimilar charges. As a consequence, I will only look at the opposite chargecase.

Angular Momentum and Velocity

We see another interesting feature from the turning point equation. To keepthe terms under the square root positive, we will have a maximum possi-ble angular momentum for each energy level. This is not too surprising.Thinking about closed orbits. If we increase L too much, we do escape thebound system and scatter. If we are at a known energy level, there will be amaximum angular momentum for orbiting, rather than scattering.

28

( α

2U

)2

+L2

2mU> 0

L2

2mU> −

( α

2U

)2

L2 < −2mU( α

2U

)2

= −mα2

2U

Lmax <

√−mα

2

2U

When we are at the turning points, r = 0, and our velocity at that instantis v = rθ = L/(mr). Since Lmax and rmin = r2 are known, we can calculatethe maximum speed expected.

vmax = rθ

=L

mr2where

r2 =α

2U−√( α

2U

)2

+L2

2mU

We now want to look at the special case where L = Lmax. At this energy,we have a circular orbit of radius r = α/(2U)).

L =

√−mα

2

2U

r =α

2U

v =L

mr=

√−2U

m

We can achieve any speed we wish simply by cranking down the bindingenergy. Nothing prevents superluminal behavior in this slowest of cases.

The more general case has L < Lmax. In this more general case, we willfind peak speeds faster the circular case for the same energy. Let a = L/Lmax.Then

29

r2 =α

2U−√( α

2U

)2

+L2

2mU

=α

2U

(1−√

1− a2)

v =L

mr=

aLmaxmα2U

(1−√

1− a2)

=a2U

√−mα2

2U

mα(1−√

1− a2)

=a(

1−√

1− a2)√−2U

m

=1 +√

1− a2

a

√−2U

m

The forefactor starts at one when a = 1, and shoots to infinity as a→ 0.Consequently, we can achieve any high speed we wish by choosing an energylevel, then choosing a low enough angular momentum. Notice the trick withthe forefactor on the last two lines above. We will use this below in a bit.

Examine the kinetic and potential energies of this orbiting system atminimum separation. We know the separation from above. We know r = 0at minimaxes, and we know v at this minimum.

KE =1

2mv2

=m

2

(1 +√

1− a2

a

)2 −2U

m

=m

2

(1 +√

1− a2

a

)(a

1−√

1− a2

)−2U

mnotice the trick?

= −U(

1 +√

1− a2

1−√

1− a2

)= −U

(2 + 2

√1− a2 − a2

a2

)or the initial approach

30

PE = α1

r

= α1

α2U

(1−√

1− a2)

=2U(

1−√

1− a2)

U = PE +KE

Nicely enough, none of the high speed cases requires any modification ofthe previous closed form solutions for closed orbits. Definitely, there will begreater precession at higher speeds. However, no special treatment is neededfor superluminal orbital solutions.

Speculation About the Capture Process

I am fairly sensitive toward the details of electron capture requiring maxi-mized angular momentum for the bound states. In my mind’s eye, I perceivea ‘bumper cars’ event, where an electron, which happens to be marginallyabove capture angular momentum scatters initially from the target. How-ever, radiation reaction brings the two together again at a closer match, untilcoupling occurs just at the maximum possible angular momentum.

Speculation About Energy Limits

All of this work has been a non-relativistic treatment which does not limitspeed.

However, we are fairly sure that speeds have a pole at c. If we ignoreincreasing effective mass effects, we can see that limiting v = c will providea limit on angular momentums available, and provide a basement for boundenergies.

As an example, from the previous Weber equation work,

Lmax =

√−mα

2

U= mrv

= m( α

2U

)c

31

This implies √−mα

2

U= m

( α

2U

)c

√−mUα2 =

mα

2c

√−mU =

m

2c

U = (mc2)/4

This corresponds to a base energy level of 256 keV for positronium, whichis much higher than measure values. However, the principle that relativisticspeed limits can limit angular momentum does have merit, in my opinion.

Commentary About this Exercise

The value of this exercise has been extrapolation of a massive, low speedforce law candidate to a high speed regime. While the Weber law doesnot satisfy our relativistic experience, it does demonstrate Boscovitch stylemodifications of forces from attractive to repulsive based upon some criterionother than separation. To provide a relativistic force, at a minimum, we needto include the Leonard Wiechert corrections for apparent charge, and providefield calculations bases upon retarded images of the source particles.

References

[1] J. C. Maxwell, Electricity and Magnetism, Vol II Cambridge UniversityPress, New York. 1873 reprinted 2011.

[2] Sir Edmund Whittaker, History of the Theories of Aether and Electric-ity, Vol II, Thomas Nelson and Sons, New York 1951.

[3] Andre Kock Torres Assis, Weber’s Electrodynamics, Kluwer AcademicPublishers, Boston, 1994.

[4] Ted Clay Bradbury, Theoretical Mechanics, Krieger Publishing Co., 1981reprint.

[5] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series and Prod-ucts, Academic Press, New York, 1980. ISBN 0-12-294760-6

32

[6] Ruggero Giuseppe Boscovich, Translated by J. M. Child A Theory ofNatural Philosophy, Nabu Press, 2010. ISBN 1-17-775930-6

33