# Helmholtz Decompositon Of EM Field

According to Helmholtz's theorem, any given vector field may be decomposed into a curl-free component and a divergence-free component:

In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as the fundamental theorem of vector calculus, states that any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field; this is known as the Helmholtz decomposition. It is named after Hermann von Helmholtz.
This implies that any such vector field F can be considered to be generated by a pair of potentials: a scalar potential φ and a vector potential A.

Since the electromagnetic field can be described as a vector field, we should be able to identify the two components of the Helmholtz decomposition and thus come to a description of the electromagnetic fields. Heretofore, the magnetic field is considered to be divergence free in all circumstances. We shall show that this is not the case.

In his book Theory of Wireless Power Eric Dollard describes induction in space, using a transmission line model of a coil, whereby he distinguishes a (divergence-free) transverse component as well as a (curl-free) longitudinal component, although he does not make a distinguishment between the transverse and longitudinal selfinductance and selfcapacitance components of the coil.

In order to make such a distinguishment, we shall derive both components from an idealized model of a coil, decomposed into curl-free and a divergence-free elements, which we subsequently superpose to derive a complete model of a solenoid.

For the transverse component, we consider the inductance associated with the current flowing trough the windings, along the direction of the coil wire, whereby the parasitic self capacitance is distributed between the wires and thus is considered to be perpendicular with respect to the direction of the current. This component can thus be modelled as dual-wire transmission line consisting of a series of parallel LC circuits as shown here:

This component is described by the telegrapher's equations whereby the distributed inductance and capacitance are considered as a series of elemental parallel LC circuits as shown here:

From this, we can note the following characteristics for the transverse component:

• There is a return path for the current
• There is a net rotating current trough the inductance and the capacitance
• A static, DC current is possible trough the inductances in series
• A static, DC voltage is possible across the capacitances in parallel
• Since voltage (E) and current (B) are perpendicular with respect to one another, the Poynting vector is unequal to zero and thus this component represents a rotational field in infinitesimal consideration.

For the longitudinal component, we consider the inductance associated with the current flowing around the windings, perpendicular to the direction of the coil wire, whereby the parasitic self capacitance is distributed between the wires and thus is considered to be in the same direction as the direction of the (net) current. This component can thus be modelled as single-wire transmission line consisting of a series of series LC circuits as shown here:

From this, we can note the following characteristics for the longitudinal component:

• There is no return path for the current
• There is no net rotating current trough the inductance and the capacitance
• A static, DC current is impossible trough the capacitances in series
• A static, DC voltage is impossible across the inductances in series
• Since voltage (E) and current (B) are parallel with respect to one another, the Poynting vector is equal to zero and thus we can only take a dot product and therefore this component represents a scalar, rotation-free field in infinitesimal consideration.

## Derivation of the distributed transverse and longitudinal inductance of a wire

In order to come to a model with which we can derive both the longitudinal and transverse versions of the telegrapher's equations, we consider the fields within and around an ideal loss-free wire with radius $r_w$.

## The transverse component

As we have seen, the transverse component is associated with a current in the length direction of the wire. In order to investigate this component, we must consider such a configuration that we can be sure our derivation only accounts for a divergence free resulting B-field. We can do that by considering an infinitely long wire with radius $r_w$ for which we can derive a distributed inductance $L_T$ in Henry/m. The first step is to derive $\bar B$ as a function of the current I trough the wire, at a distance $r_v$ to the wire:

Ref:Elektro-magnetisch veld 1 pagina's 3.15 - 3.16

$\bar B$ in vacuum is given by:

 $\bar B = \frac{\mu_0 I}{2 \: \pi \: r_v} \bar e_{\phi} \: \: [T]$ (1)

The second step is to derive the energy stored in the B-field, also per meter of wire length.

Ref: Elektro-magnetisch veld 1 pagina's 5.15 - 5.17

The magnetic energy stored in a magnetic field is given by:

 $W_m = \frac{1}{2} \int \int \int \frac{B^2}{\mu_0 \: \mu_r} dV \: \: [J]$ (2)

We can work this out by substituting (1) into (2):

 $W_m$ $= \frac{1}{2} \int \int \int \frac{(\mu_0 I)^2}{(2 \: \pi \: r_v)^2 \: \mu_0 \: \mu_r} dV \: [J]$ (3) $= \frac{1}{2} \int \int \int \frac{\mu_0 I^2}{4 \: \pi^2 \: r_v^2 \: \mu_r} dV$ $= \frac{1}{2} \frac{\mu_0 I^2}{4 \: \pi^2 \: \mu_r} \int \int \int\: \frac{1}{r_v^2 } dV$ $= \frac{\mu_0 I^2}{8 \: \pi^2 \: \mu_r} \int \int \int\: \frac{1}{r_v^2 } dV$

Since we are interested in deriving a formula for the distributed inductance $L_T$ we use cylindrical coordinates and integrate over $r_v$ and $\phi$:

 $W_m$ $= \frac{\mu_0 I^2}{8 \: \pi^2 \: \mu_r} \int_{r_w}^{\infty} \int_0^{2\pi} \: \frac{1}{r_v^2 } d\phi d{r_v} \: \: [J/m]$ (4) $= 2\pi \frac{ \mu_0 I^2}{8 \: \pi^2 \: \mu_r} \int_{r_w}^{\infty} \: \frac{1}{r_v^2 } d{r_v}$ $= \frac{ \mu_0 I^2}{4 \: \pi \: \mu_r} \frac{1}{r_w }$ $= \frac{ \mu_0 I^2}{4 \: \pi \: \mu_r \: r_w}$

The energy per meter for a inductance $L_T$ [H] is given by:

 $W_m$ $= \frac{1}{2} L_T I^2 \: \: [J]$ (5)

Equalling (4) and (5) gives us the value for $L_T$:

 $L_T = \frac{ \mu_0 }{2 \: \pi \: \mu_r \: r_w} \: \: [H/m]$ (6)

## The longitudinal component

As we have seen, the longitudinal component is associated with a current rotating around the wire. In order to investigate this component, we must consider such a configuration that we can be sure our derivation only accounts for a rotation free resulting B-field, which thus has to be a homogeneous B-field. We can do that by considering an infinitely long cylinder with radius $r_w$ for which we can derive a distributed inductance $L_T$ in Henry/m. The first step is to derive $\bar B$ as a function of the current density around the wire, written as a surface current density $\lambda_s$ in [A/m] along the surface of the wire/cylinder.

Ref: Elektro-magnetisch veld 1 pagina 3.29

For this configuration $\bar B$ within a wire is given by:

 $\bar B = \mu_0 \mu_r \lambda_s \bar e_{z} \: \: [T]$ (7)

The field outside of the wire/cylinder is 0.

We can work this out by substituting (7) into (2):

 $W_m$ $= \frac{1}{2} \int \int \int \frac{(\mu_0 \mu_r \lambda_s)^2}{\mu_0 \: \mu_r} dV \: [J]$ (8) $= \frac{1}{2} \int \int \int \mu_0 \mu_r \lambda_s^2 dV$ $= \frac{1}{2} \mu_0 \mu_r \lambda_s^2 \int \int \int dV$

Before we continue, we must consider the surface current density $\lambda$ more closely. Since in our consideration the longitudinal component of the current density $\lambda_l$ is considered be net rotation free and flowing trough the centre of the wire/cylinder, we can calculate the current surface density $\lambda_s$ as follows:

 $\lambda_s$ $= \frac {2 \: r}{\pi \: r} \lambda_l \: [A/m]$ (9) $= \frac {2}{\pi} \lambda_l$

Substituting (9) into (8) gives:

 $W_m$ $= \frac{1}{2} \frac {4}{\pi^2} \mu_0 \mu_r \lambda_l^2 \int \int \int dV \: [J]$ (10)

The energy for an inductance $L_L$ [H] is given by:

 $W_m$ $= \frac{1}{2} L_L \lambda_l^2 \: \: [J]$ (11)

Equalling (10) and (11) gives us the value for $L_L$:

 $L_L = \frac {4}{\pi^2} \mu_0 \mu_r \int \int \int dV \: \: [H]$ (12)

Integrating this over the surface area of the wire gives us the value for the distributed selfinductance $L_L$ in [H/m]:

 $L_L = \frac{4}{\pi} \: \mu_0 \: \mu_r \: {r_w}^2 \: \: [H/m]$ (13)