Question

S = 1/(?0) E X B

How was the poynting vector derived? Please give a detailed derivation of equations that leads to the Poynting Vector.

Answer 1

Poynting's theorem[edit]

Considering the statement in words above - there are three elements to the theorem, which involve writing energy transfer (per unit time) as volume integrals:[2]

Since u is the energy density, integrating over the volume of the region gives the total energy U stored in the region, then taking the (partial) time derivative gives the rate of change of energy:

U=\int_V u dV \ \rightarrow \ \frac{\partial U}{\partial t} = \frac{\partial}{\partial t} \int_V u dV = \int_V \frac{\partial u}{\partial t} dV .

The energy flux leaving the region is the surface integral of the Poynting vector, and using the divergence theorem this can be written as a volume integral:

\oiint\scriptstyle \partial V\mathbf{S}\cdot d\mathbf{A}=\int_V\nabla\cdot \mathbf{S} dV .

The Lorentz force density f on a charge distribution, integrated over the volume to get the total force F, is

\mathbf{f} = \rho\mathbf{E}+\mathbf{J}\times\mathbf{B} \ \rightarrow \ \int_V \mathbf{f} dV = \mathbf{F} = \int_V (\rho\mathbf{E}+\mathbf{J}\times\mathbf{B} )dV ,

where ? is the charge density of the distribution and v its velocity. Since \mathbf{J} = \rho \mathbf{v}, the rate of work done by the force is

\mathbf{F}\cdot\frac{d \mathbf{r}}{dt} = \mathbf{F}\cdot \mathbf{v} = \int_V (\rho\mathbf{E}\cdot\mathbf{v}+\rho\mathbf{v}\times\mathbf{B}\cdot \mathbf{v} )dV \ \rightarrow \ \mathbf{F}\cdot \mathbf{v} = \int_V \mathbf{E}\cdot \mathbf{J}dV .

So by conservation of energy, the balance equation for the energy flow per unit time is the integral form of the theorem:

-\int_V\frac{\partial u}{\partial t}dV = \int_V\nabla\cdot\mathbf{S}dV+\int_V\mathbf{J}\cdot\mathbf{E}dV,

and since the volume V is arbitrary, this is true for all volumes, implying

- \frac{\partial u}{\partial t} = \nabla\cdot\mathbf{S} + \mathbf{J}\cdot\mathbf{E},

which is Poynting's theorem in differential form.

Poynting vector[edit]

Main article: Poynting vector

From the theorem, the actual form of the Poynting vector S can be found. The time derivative of the energy density (using the product rule for vector dot products) is

\frac{\partial u}{\partial t} = \frac{1}{2}

\left(\mathbf{E}\cdot\frac{\partial \mathbf{D}}{\partial t}

+ \mathbf{D}\cdot\frac{\partial \mathbf{E}}{\partial t}

+ \mathbf{H}\cdot\frac{\partial \mathbf{B}}{\partial t}

+ \mathbf{B}\cdot\frac{\partial \mathbf{H}}{\partial t}\right)=

\mathbf{E}\cdot\frac{\partial \mathbf{D}}{\partial t}

+ \mathbf{H}\cdot\frac{\partial \mathbf{B}}{\partial t},

using the constitutive relations

\mathbf{D} = \epsilon_0 \mathbf{E},\quad \mathbf{B} = \mu_0 \mathbf{H}.

The partial time derivatives suggest using two of Maxwell's Equations. Taking the dot product of the Maxwell