Poynting's theorem

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Poynting's theorem is a statement due to John Henry Poynting about the conservation of energy for the electromagnetic field. It relates the time derivative of the energy density, u to the energy flow and the rate at which the fields do work. It is summarised by the following formula:

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

where S is the Poynting vector representing the flow of energy, J is the current density and E is the electric field. Energy density u is (symbol ε₀ is the electric constant and μ₀ is the magnetic constant):

$u = \frac{1}{2}\left(\epsilon_0 \mathbf{E}^2 + \frac{\mathbf{B}^2}{\mu_0}\right).$

Since the magnetic field does no work, the right hand side gives the negative of the total work done by the electromagnetic field per second·meter³.

Poynting's theorem in integral form:

$\frac{\partial}{\partial t} \int_V u \ dV + \oint_{\partial V}\mathbf{S} \ d\mathbf{A} = -\int_V\mathbf{J}\cdot\mathbf{E} \ dV$

Where $\partial V \!$ is the surface which bounds (encloses) volume $V \!$ .

In electrical engineering context the theorem is usually written with the energy density term u expanded in the following way, which resembles the continuity equation:

$\nabla\cdot\mathbf{S} + \epsilon_0 \mathbf{E}\cdot\frac{\partial \mathbf{E}}{\partial t} + \frac{\mathbf{B}}{\mu_0}\cdot\frac{\partial\mathbf{B}}{\partial t} + \mathbf{J}\cdot\mathbf{E} = 0$

Where $\mathbf{S}$ is the energy flow of the electromagnetic wave, $\epsilon_0 \mathbf{E}\cdot\frac{\partial \mathbf{E}}{\partial t}$ is the power consumed for the build-up of electric field, $\frac{\mathbf{B}}{\mu_0}\cdot\frac{\partial\mathbf{B}}{\partial t}$ is the power consumed for the build-up of magnetic field and $\mathbf{J}\cdot\mathbf{E}$ is the power consumed by the Lorentz force acting on charge carriers.

[edit] Derivation

The theorem can be derived from two of Maxwell's Equations. First one to consider is the Faraday's Law:

$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$

Taking the dot product of this equation with $\mathbf{B}$ yields:

$\mathbf{B} \cdot (\nabla \times \mathbf{E}) = - \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t}$

Another one to consider is the Ampère-Maxwell law equation:

$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \epsilon_0 \mu_0 \frac{\partial \mathbf{E}}{\partial t}$

Taking dot product of this equation with $\mathbf{E}$ yields:

$\mathbf{E} \cdot (\nabla \times \mathbf{B}) = \mathbf{E} \cdot \mu_0 \mathbf{J} + \mathbf{E} \cdot \epsilon_0 \mu_0 \frac{\partial \mathbf{E}}{\partial t}$

Subtracting the first dot product from the second yields:

$\mathbf{E} \cdot (\nabla \times \mathbf{B}) - \mathbf{B} \cdot (\nabla \times \mathbf{E}) = \mu_0 \mathbf{E} \cdot \mathbf{J} + \epsilon_0 \mu_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} + \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t}$

Finally, by the product rule, as applied to the divergence operator over the cross product (described here):

$- \nabla\cdot\ ( \mathbf{E} \times \mathbf{B} ) = \mu_0 \mathbf{E} \cdot \mathbf{J} + \epsilon_0 \mu_0 \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} + \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t}$

Since the Poynting vector $\mathbf{S}$ is defined as:

$\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$

This is equivalent to:

$\nabla\cdot\mathbf{S} + \epsilon_0 \mathbf{E}\cdot\frac{\partial \mathbf{E}}{\partial t} + \frac{\mathbf{B}}{\mu_0}\cdot\frac{\partial\mathbf{B}}{\partial t} + \mathbf{J}\cdot\mathbf{E} = 0$