Maxwell stress tensor

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The tensor

T_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0}  \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,.

Lorentz force

In order to discuss the energy and momentum content of the electromagnetic field, we consider the force on a charge distribution is given by

\mathbf{F} = \int_V\, \rho \mathbf{E} + \mathbf{J} \times \mathbf{B}\,.

Using Maxwell's equations, this can be written in terms of the fields, as follows

\mathbf{F} = \int_V\, \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}\,.

Note that

\frac{\partial}{\partial t} (\mathbf{E}\times\mathbf{B}) = \frac{\partial\mathbf{E}}{\partial t}\times \mathbf{B} + \mathbf{E} \times \frac{\partial\mathbf{B}}{\partial t} = \frac{\partial\mathbf{E}}{\partial t}\times \mathbf{B} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E})\,

using Faraday's law of induction. Therefore the force per unit volume is

\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right) - \epsilon_0 \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E})\,,

or

\mathbf{f} = \epsilon_0\left[  (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} -  \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right)  \right]
+ \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,,

where we have used \boldsymbol{\nabla}\cdot \mathbf{B} = 0\,. Using the curl identity

\tfrac{1}{2} \boldsymbol{\nabla} A^2 = \mathbf{A} \times (\boldsymbol{\nabla} \times \mathbf{A}) + (\mathbf{A} \cdot \boldsymbol{\nabla}) \mathbf{A} ,

we find that

\mathbf{f} = \epsilon_0\left[  (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right)
+ \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,

Maxwell stress tensor and Poynting vector

Introduce the Maxwell stress tensor

T_{ij} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0}  \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,,

and the Poynting vector

\mathbf{S} \equiv \frac{1}{\mu_0} \mathbf{E}\times\mathbf{B}\,.

Then \mathbf{f}\, has components

f_j = \sum_i \nabla_i T_{ij} + \epsilon_0 \mu_0 \frac{\partial S_j}{\partial t}\,.

back to Poynting's theorem
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