(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Hi everyone! My name is Alexandra, and I'm new in this forum. I am trying to determine the mentionated tensor without the assumption of linear media or vacuum ( ## \textbf{D} = \epsilon \textbf{E} ## and ## \textbf{B} = \mu \textbf{H} ##). What I want to obtain is the "general" stress tensor with constitutive equations ## \textbf{D} = \epsilon \textbf{E} + \textbf{P} ## and ## \textbf{B} = \mu ( \textbf{H} + \textbf{M} ) ##

2. Relevant equations

##\textbf{f} = \rho \textbf{E} + \textbf{j} \times \textbf{B} ## Force per unit volume

## \textbf{D} = \epsilon \textbf{E} + \textbf{P} ##

## \textbf{B} = \mu ( \textbf{H} + \textbf{M} ) ##

Macroscopic Maxwell Equations

3. The attempt at a solution

##\textbf{f} = \rho \textbf{E} + \textbf{j} \times \textbf{B} ##

I reeplace the charge/current densities by the fields E and B, usindg Gauss and Ampere's law:

##\textbf{f} = \nabla . \textbf{D} \textbf{E} + \nabla \times \textbf{H} \times \textbf{B} - \dfrac{\partial \textbf{D}}{\partial t} \times \textbf{B}## (1)

Now, using product rule of derivaties, and Faraday Law(I won't use textbf):

##\dfrac{\partial}{\partial t}(D \times B) = \dfrac{\partial D}{\partial t} \times B + D \times \dfrac{\partial B}{\partial t} = \dfrac{\partial D}{\partial t} \times B - D \times \dfrac{\partial E}{\partial t}##

Using this result on (1):

##f = (\nabla . D )E + (\nabla \times H) \times B - \dfrac{\partial}{\partial t} (D \times B) - D \times \dfrac{\partial E}{\partial t}##

I know that isn't much, but if I add a term ##(\nabla . B) B ## it doesn't solve anything, because I can't eliminate the curves, like in the most books/papers that use the mentionated before. If someone can help me I will be grateful.

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# Macroscopic Stress Tensor

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