1. In this work are investigated details of an elementary electric and magnetic dipole radiation at distances much greater than size of the emitting element. Debatable conclusions are drawn.

Formulas are given in a cylindrical coordinate system (ρ,φ,z)
r2 = ρ2 + z2
Given this, it is possible to write expressions differently for ρ and z
For example, 2 - 3 · ρ2 / r2 = 3 · z2 / r2 - 1
For all values ∂/∂φ = 0 (cylindrical symmetry)
Time derivatives are denoted by a quote '

Electric elementary dipole

Charge oscillates along z-axis near zero point with frequency ω, amplitude of dipole moment is P0.

Dipole moment:
Pz = P0 · cos(ω·t)

Auxiliary functions:
COS = cos(ω·(t - r/c)), SIN = sin(ω·(t - r/c))

Scalar potential:
a = P0 / (4·π·ε0) · z / r2 · (1 / r · COS - ω/c · SIN)
a' = - P0 / (4·π·ε0) · ω · z / r2 · (ω/c · COS + 1 / r · SIN)

Vector potential:
Az = - P0 · μ0/(4·π) · ω / r · SIN
Az' = - P0 · μ0/(4·π) · ω2 / r · COS
div A = ∂Az/∂z = P0 · μ0/(4·π) · ω · z / r2 · (ω/c · COS + 1 / r · SIN)
a' = - c2 · div A

∂a/∂ρ = P0 / (4·π·ε0) · ρ · z / r3 · {(ω2/c2 - 3 / r2) · COS + ω/c · 3 / r · SIN}
∂a/∂z = P0 / (4·π·ε0) / r2 · {1 / r · (ω2/c2 · z2 + 1 - 3 · z2 / r2) · COS + ω/c · (3 · z2 / r2 - 1) · SIN}

Magnetic induction:
Bφ = - ∂Az/∂ρ = - P0 · μ0/(4·π) · ω · ρ / r2 · (ω/c · COS + 1 / r · SIN)
Bφ' = - P0 · μ0/(4·π) · ω2 · ρ / r2 · (1 / r · COS - ω/c · SIN)

Electric field:
Eρ = - ∂a/∂ρ = - P0 / (4·π·ε0) · ρ · z / r3 · {(ω2/c2 - 3 / r2) · COS + ω/c · 3 / r · SIN}
Ez = - Az' - ∂a/∂z = P0 / (4·π·ε0) / r · {(ω2/c2 · ρ2 / r2 - 1 / r2 + 3 · z2 / r4) · COS + ω/c / r · (1 - 3 · z2 / r2) · SIN}

Electric field annular curl:
∂Eρ/∂z - ∂Ez/∂ρ = P0 / (4·π·ε0) · ω2/c2 · ρ / r2 · (1 / r · COS -ω/c · SIN)

Bφ' = - (∂Eρ/∂z - ∂Ez/∂ρ)
as it should be in equations of electromagnetic field.

div E = ∂Eρ/∂ρ + Eρ / ρ + ∂Ez/∂z = 0 (checked)

Magnetic field curl:
Jρ = - 1/μ0 · ∂Bφ/∂z = - P0 / (4·π) · ω · ρ · z / r3 · {ω/c · 3 / r · COS - (ω2/c2 - 3 / r2) · SIN}
Jz = 1 / μ0 · (∂Bφ/∂ρ + Bφ / ρ) = P0 / (4·π) · ω / r · {ω/c / r · (1 - 3 · z2 / r2) · COS - (ω2/c2 · ρ2 / r2 - 1 / r2 + 3 · z2 / r4) · SIN}

Eρ' = - P0 / (4·π·ε0) · ω · ρ · z / r3 · {ω/c · 3 / r · COS - (ω2/c2 - 3 / r2) · SIN} = Jρ0
Ez' = P0 / (4·π·ε0) · ω / r · {ω/c / r · (1 - 3 · z2 / r2) · COS - (ω2/c2 · ρ2 / r2 - 1 / r2 + 3 · z2 / r4) · SIN} = Jz0
as it should be in equations of electromagnetic field.

Magnetic dipole

An annular current with small radius R changes direction according to periodic law.

Magnetic moment is directed along z-axis:
Mz = M0 · cos(ω·t), где M0 = π · R2 · I0, I0 is current amplitude.

Auxiliary functions:
COS = cos(ω·(t - r/c)), SIN = sin(ω·(t - r/c))

Vector potential:
Aφ = M0 · μ0/(4·π) · ρ / r2 · (1 / r · COS - ω/c · SIN)

Electric field:
Eφ = - Aφ' = M0 · μ0/(4·π) · ω · ρ / r2 · (ω/c · COS + 1 / r · SIN)
Eφ' = M0 · μ0/(4·π) · ω2 · ρ / r2 · (1 / r · COS - ω/c · SIN)

Magnetic induction:
Bρ = - ∂Aφ/∂z = - M0 · μ0/(4·π) · ρ · z / r3 · {(ω2/c2 - 3 / r2) · COS + ω/c · 3 / r · SIN}
Bz = ∂Aφ/∂ρ + Aφ / ρ = M0 · μ0/(4·π) / r2 · {(ω2/c2 · ρ2 / r + 2 / r - 3 · ρ2 / r3) · COS - ω/c · (2 - 3 ·ρ2 / r2) · SIN}

Bρ' = - M0 · μ0/(4·π) · ω · ρ · z / r3 · {ω/c · 3 / r · COS - (ω2/c2 - 3 / r2) · SIN} = - (- ∂Eφ/∂z)
Bz' = - M0 · μ0/(4·π) · ω / r2 · {ω/c · (2 - 3 · ρ2 / r2) · COS + (ω2/c2 · ρ2 / r + 2 / r - 3 · ρ2 / r3) · SIN} = - (∂Eφ/∂ρ + Eφ / ρ)
as it should be in equations of electromagnetic field.

Magnetic field curl:
Jφ = 1/μ0 · (∂Bρ/∂z - ∂Bz/∂p) = M0 · μ0/(4·π) · ω2/c2 · ρ / r2 · {1 / r · COS - ω/c · SIN}

Eφ' = Jφ0 (checked)
as it should be in equations of electromagnetic field.

Conclusions

Although divergence of electric field div(E) is zero everywhere (charge density is zero), scalar potential is urgently needed to describe radiation of electric dipole. To express time derivative a' is required vector potential A. At long distances, there is no question of lagging potentials of forcibly oscillating system, waves must propagate "by themselves" in wave zone. It begs the conclusion that potentials are objective physical reality, fundamental fields in vacuum, and are not mathematical abstractions. To describe dipole radiation, three fundamental fields are sufficient:

a' = - c2 · div A
A' = - E - grad a
E' = c2 · rot rot A

At the same time, Laplacian div grad (a) is fundamentally different from local charge density ε0 · div E, these are different quantities. Laplacian of scalar potential can be locally not zero in electric dipole radiation, unlike divergence of electric field. Formally, both of these quantities are "conserved", since it is possible to express derivatives in time as minus divergence of some known "flow" or current. But with respect to electric dipole, laplasian of scalar potential is preserved only globally, when positive density is emitted in one direction along z-axis, in opposite direction the same modulo negative goes. It cannot be said that scalar potential has significant value only in near zone of forced generation and lagging potentials. In far wave zone, its intensity, like time derivative, decreases proportionally to 1 / r along z-axis, the same applies to its gradient in some directions (ρ · z / r3).

Electric and magnetic field decrease on average with distance as 1 / r, respectively, energy density decreases as 1 / r2. That is, integral of energy density throughout space is infinite, and elementary dipoles cannot be used as basis for representing field objects with finite energy. The more time emitter works, more energy it loses with waves, without restrictions on final value.

Considerations about compact field objects moving at the speed of light are in topic:
Field objects moving at the speed of light

2.

3. Originally Posted by compuAI
That is, integral of energy density throughout space is infinite, and elementary dipoles cannot be used as basis for representing field objects with finite energy. The more time emitter works, more energy it loses with waves, without restrictions on final value.

Why go to all this trouble? By postulate your dipole radiates constant energy per unit time forever. The integrated energy therefore must be infinite.

Your conclusion that therefore one cannot use dipoles is absurd. Just don’t integrate forever!

4. Originally Posted by tk421

Why go to all this trouble? By postulate your dipole radiates constant energy per unit time forever. The integrated energy therefore must be infinite.

Your conclusion that therefore one cannot use dipoles is absurd. Just don’t integrate forever!
I think not everyone understands, for example, that electric field around an electron must have finite internal energy, and may try to represent it like "standing wave", consisting of some dipoles combination.

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