In our blog post on February 18, 2021 we calculated (but made calculation errors) that we can explain the Larmor formula if we assume that the electron elastic electric far field uses its inertia to pump out the maximum possible power output from an oscillating charge.
The accurate calculation should be done numerically, but we believe that the rubber plate model is right in this. Joseph Larmor in 1897 derived his formula from classical electromagnetism. Electromagnetic radiation is not a quantum phenomenon, and the rubber plate model sits well in the setup as a general model of wave phenomena.
The problem is general in classical physics: what kind of power output does oscillation of an elastic medium produce? It is often hard to calculate, but we do get qualitative understanding of the process when we know we are dealing with an elastic medium.
We should check Larmor's original calculation. Does he assume the electric field to be elastic? Has anyone calculated the power output in a general elastic medium?
Power output for a tense long rope
The power output is easy to calculate in the case of a tense long rope. Let the end of the rope do circular motion of a radius r at the speed of v. Let the mass of the rope be m per unit of length. Let the wavelength be L.
The wave energy of one wavelength is
E = L m v^2.
The radial acceleration is
a = v^2 / r.
The person rotating the rope at the end must do the work E per one rotation of length 2 π r. He must use a force
F = L m v^2 / (2 π r).
If we write
F = M a,
we get
M = L / (2 π) * m.
In this case, the inertial mass which pumps energy from the movement is the mass of
1 / (2 π)
of the wavelength of the rope. That is, the mass of one radian of the wave in the rope.
The result sounds reasonable. If the wavelength is 6.28 meters, then it is the inertia of one meter of rope which is pumping the maximum energy from the person rotating the rope.
The mass of the rope is uniform per length, and infinite for an infinite rope.
The elastic electric field: the electromagnetic wave gets "detached" from the static electric field at the distance of sqrt(3) / 2 radians
The mass-energy of the electric field of a charge has a formula 1 / R^2 per unit of the radius R.
At larger distances, the electromagnetic wave lives on the combined electric and magnetic fields. The situation is quite different from a tense rope whose mass per meter quickly approaches zero.
Also, the tension in the rope model should be larger for small R.
Maybe we can approximate the generation of an electromagnetic wave with a rope whose mass per length for small R is 1 / R^2, but constant for large R? The tension is larger for small R and constant for large R.
The power drain is one radian of the wave in the large R zone pumping the maximum energy using its inertia.
The Larmor formula corresponds to the far field more than 3/4 radians away using its whole inertia to pump out the maximum energy.
Let us use one radian of the wave as the unit of length. Let initially the mass of the rope per unit of length be 1 / R^2. The total mass M of the rope farther than 3/4 radians away would be
M = 4/3.
We get the following model: at the distance of R >= sqrt(3) / 2 radians we let the mass of the rope per unit length be constant. For smaller R, the rope still has mass density 1 / R^2.
The rope model indicates that the electromagnetic wave is "detached" from the static electric field of the charge already at a distance of sqrt(3) / 2 radians of the electromagnetic wave.
In radio antenna technology, the field > 1 radian away is considered the far field (Tom Lecklider, 2005).
We get a more detailed qualitative rubber plate model. The far field starts already at the distance 1 radian, or even closer.
The electromagnetic wave compared to the transverse static electric field
Let us do more calculations. Let an electron do a circle whose radius is r, at a speed v. Then
a = v^2 / r.
The transverse component of the static electric field at a distance R is
E = e r / (4 π ε_0 R^3).
The Larmor formula says that the power output is
P = e^2 a^2 / (6 π ε_0 c^3).
The energy density of the power output is
e^2 a^2 / (24 π^2 ε_0 c^4 R^2).
That corresponds to a transverse dynamic electric field
E' = e a / (sqrt(12) π ε_0 c^2 R).
E' = E if
v^2 / (r sqrt(12) c^2 R) = r / (4 R^3)
<=>
v^2 / (sqrt(12) c^2) = r^2 / (4 R^2)
<=>
v / (sqrt^2(12) / 2 * c) = r / R
<=>
R = r / v * 0.9306 c
= 0.1481 * 2 π r / v * c.
The transverse static electric field is equal to the transverse dynamic electric field at the distance of 0.1481 wavelengths. That is 0.9306 radians. The value is close to the radio technology far field distance which is 1 radian.
Determination of the "lagging mass" in the Larmor formula
Let us do the February 18, 2021 calculation again, this time correctly. We want to find m such that
P = e^2 v^4 / (6 π ε_0 c^3 r^2)
= m v^3 / r = m a v = F v
<=>
m c^2 = e^2 v / (2 π r c * 3 ε_0)
= e^2 / (3 ε_0 λ).
The energy density of the electron static field at a distance R is
E_d(R) = e^2 / (32 π^2 ε_0 R^4).
The radial density is
e^2 / (8 π ε_0 R^2).
The energy of the field farther than R is
E(R) = e^2 / (8 π ε_0 R).
Let us solve R from m c^2 = E(R). We get
R = 3 / (8 π) λ,
or 0.1194 wavelengths, or 3/4 radians.
The energy of the static electric field outside the classical radius r_e is only HALF of the mass-energy m_e of the electron
On February 18, 2021 we were thinking that the mass-energy of the static field of the electron outside the classical radius r_e is m_e. But it is only 1/2 m_e.
The potential of an electron at the distance r_e from a proton is -m_e c^2.
Thus, the classical radius r_e, in a sense, corresponds to the distance of two unit charges, but is 2X "too large" for the mass-energy of the electric field of a unit charge.
Conclusions
The far field of an oscillating charge turns out to start surprisingly close, only at 3/4 ... 1 radians from the charge, where the wavelength of the electromagnetic wave is denoted as 2 π radians.
The electromagnetic wave lives on the oscillating magnetic and electric fields and does not need the static electric field of the charge. We can say that the wave "breaks off", or is detached from the static field, at the distance of 1 radian.
We can explain the Larmor formula by saying that the static electric field farther than 3/4 radians uses its entire inertia to pump the maximum energy out of the oscillation acceleration.
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