Poynting's theorem

Volume element notation

Given the prominent role played by voltage in this discussion, perhaps the use $dV$ to denote a volume element should be avoided. A series of commonly used replacements are shown below. Here we adopt the latter in this list: The volume element is $d{\mathcal {V_{ol}}}$.

$dV\rightarrow dxdydz\leftrightarrow d\tau \leftrightarrow d^{3}x\leftrightarrow d{\mathcal {V}}_{\text{ol}}$

(1)

Ohmic power and energy density

Table 1.

$I={\vec {J}}\cdot d{\vec {A}}$ (current)	$V=E_{\parallel }d\ell$ (voltage)	- - -(2)
$IV=\left({\vec {J}}\cdot {\vec {E}}\right)d{\mathcal {V}}_{\text{ol}}$ (electromagnetic power)		- - -(3)

Power $P$ is delivered when a current $I$ passes across a voltage drop $V$. To keep the argument simple we assumed a small cylinder of area $dA$ and length $d{\vec {\ell }}$, with the current and electric field parallel to the axis. The current and voltage, and power generated for this small volume, $d{\mathcal {V}}_{\text{ol}}=d\ell \,dA$, are shown in the table.

Power is also the derivative of energy $U$ with respect to time: $P=dU/dt$. When the ideas of (2-4) are incorporated into integrals, the assumption is made that the variables $({\vec {E}},{\vec {J}},u)$ are nearly constant inside the small differential volume $d{\mathcal {V}}_{\text{ol}}$. It is proper to use the partial derivative $(\partial /\partial t)$ when operating on energy density $u$ because it is also a function of position.

Vector calculus

Our "proof" of Poynting's theorem relies on a simple vector identity, two of Maxwell's equations, and the divergence theorem. This vector identity is valid for any pair of vector fields for which the curl and divergence are well-behaved.

${\vec {\nabla }}\cdot ({\vec {E}}\times {\vec {B}})={\vec {B}}\cdot ({\vec {\nabla }}\times {\vec {E}})-{\vec {E}}\cdot ({\vec {\nabla }}\times {\vec {B}})$

(4)

Table 2.

$u={\frac {\varepsilon _{0}}{2}}E^{2}+{\frac {1}{2\mu _{0}}}B^{2}$ field energy density	- - -(7)
${\vec {J}}\cdot {\vec {E}}$ work on moving charge	- - -(8)
${\begin{aligned}{\vec {S}}&={\frac {1}{\mu _{0}}}\left({\vec {E}}\times {\vec {B}}\right)\\&=\varepsilon _{0}c^{2}\left({\vec {E}}\times {\vec {B}}\right)\end{aligned}}$ Poynting vector	- - -(9)

Only two of Maxwell's equations are required for this proof,

${\vec {\nabla }}\times {\vec {E}}=-{\frac {\partial {\vec {B}}}{\partial t}}\quad$ (Faraday's law)

(5)

$\mu _{0}{\vec {J}}={\vec {\nabla }}\times {\vec {B}}-\mu _{0}\varepsilon _{0}{\frac {\partial {\vec {E}}}{\partial t}}\quad$ (Ampère's law with displacement current)

(6)

Table 2 introduces the key terms of what will become an electromagnetic $({\mathcal {EM}})$ energy conservation law. Plausibility arguments for electric and magnetic energy densities at (7) were developed earlier when we calculated the power $(IV)$ required to generate uniform fields in a capacitor or inductor. At (8) the power density involves only the work done by the electric field ${\vec {E}}$ because a magnetic field does no work on a charged particle. At (9) the Poynting vector ${\vec {S}}$ first appeared in the vector identity (4), and will

We create a four-term expression using (4) and (5). Then we move the terms involving energy density and the Poynting vector to the LHS to obtain:

$\overbrace {\mu _{0}\underbrace {{\vec {J}}\cdot {\vec {E}}} _{nq{\vec {E}}\cdot {\vec {v}}}+\nabla \cdot \underbrace {\left({\vec {E}}\times {\vec {B}}\right)} _{\mu _{0}{\vec {S}}}} ^{\text{IV power + energy flux out}}=$$\;\overbrace {{\vec {B}}\cdot \underbrace {\left({\vec {\nabla }}\times {\vec {E}}\right)} _{-\partial {\vec {B}}/\partial t}-\mu _{0}\varepsilon _{0}\underbrace {\left({\frac {\partial {\vec {E}}}{\partial t}}\cdot {\vec {E}}\right)} _{{\frac {1}{2}}(\partial /\partial t){\vec {E}}\cdot {\vec {E}}}} ^{\text{reduction in field energy}}$

(10)

In the first term on the LHS we used the fact that current density, ${\vec {J}}$, equals $nq{\vec {v}},$ where $n$ is number density, $q$ is charge, and ${\vec {v}}$ is drift velocity. Power is force times velocity: $P=N{\vec {F}}\cdot {\vec {v}}$, where $N=n{\mathcal {V}}_{ol}$, is the number of particles inside a given volume. This leads to the interpretation of ${\vec {J}}\cdot {\vec {E}}$ as the power density, and:

${\vec {\nabla }}\cdot \underbrace {\overbrace {{\tfrac {1}{\mu _{0}}}\left({\vec {E}}\times {\vec {B}}\right)} ^{\vec {S}}} _{\text{Poynting vector}}+\underbrace {\overbrace {{\vec {J}}\cdot {\vec {E}}} ^{nq{\vec {E}}\cdot {\vec {v}}}} _{\begin{matrix}{\text{power}}\\{\text{density}}\end{matrix}}+{\frac {\partial }{\partial t}}\underbrace {\overbrace {\left({\tfrac {\varepsilon _{0}}{2}}E^{2}+{\tfrac {1}{2\mu _{0}}}B^{2}\right)} ^{u_{E}\;+\;u_{B}}} _{{\mathcal {EM}}{\text{ energy density}}}=0$

(11)

Just as Gauss's law informs us that positive (negative) charge are the sources (sinks) of electric field lines, the Poynting vector ${\vec {S}}$ terminate on sources and sinks associated with electromagnetic energy. To see this we apply the divergence theorem to (11) for any arbitrary volume (and its closed surface) yields,

$\oint _{\partial {\mathcal {V}}}{\vec {S}}\cdot d{\vec {A}}+\int _{\mathcal {V}}\left({\vec {J}}\cdot {\vec {E}}\right)d{\mathcal {V}}_{ol}+{\frac {\partial }{\partial t}}\int _{\mathcal {V}}\left({\tfrac {\varepsilon _{0}}{2}}E^{2}+{\tfrac {1}{2\mu _{0}}}B^{2}\right)d{\mathcal {V}}_{ol}$

(12)

Does the energy flow through the wires...or through space?

A DC circuit consisting of a battery (V) and w:resistor (R), showing the direction of the Poynting vector (S, blue arrows) in the space surrounding it, along with the fields it is derived from; the electric field (E, red arrows) and the magnetic field (H, green arrows). In the region around the battery the Poynting vector is directed outward, indicating power flowing out of the battery into the fields; in the region around the resistor the vector is directed inward, indicating field power flowing into the resistor. Across any plane P between the battery and resistor, the Poynting flux is in the direction of the resistor. The magnitudes (lengths) of the vectors are not shown accurately; only the directions are significant.

This image from wikipedia:Poynting vector seems to suggest that the energy from a battery flows through space into the resistor. This might be true, and is caused by the fact that much of the magnetic field required to produce Poynting's vector lie outside the wire, battery, or resistor. In this figure, H is used to denote magnetic field. The electric field required to produce, S= ExH, is associated with the voltage drop between the elements. The top portion of the image represents positive voltage and carries a very small positive charge that they don't tell you about in the electronics books because the charge on a wire is insignificant compared with the amount charge that flows through the wire every second.

It might be true that ${\vec {S}}$ describes the physical flow of energy. But nowhere in the derivation of (12) was it necessary to assume that ${\vec {S}}$ actually represents the location of energy. As Feynman put it:

“

Before we take up some applications of the Poynting formulas, ... we would like to say that we have not really “proved” them. All we did was to find a possible “u” and a possible “S.” ... There are, in fact, an infinite number of different possibilities for u and S, and so far no one has thought of an experimental way to tell which one is right!

”

Examples

An excellent set of examples can be found at wikipedia:Poynting vector#Examples_and_applications. Permalinks to those presently on this article are:

1. Resistive dissipation
2. Coaxial cable
3. Plane waves
4. Radiation pressure
5. Static fields

Footnotes

• An essay to explore the difficulties in ascertaining the energy density of a transverse wave on a string is under construction at String vibration.

---