How Can We Derive the General Stress Tensor Without Assuming Linear Media?

[Alexandra]

Homework Statement

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

Homework 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

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.

 

[mark726]

Hello Alexandra,

Welcome to the forum! It's great to have you here. I understand your question and I will try my best to help you out.

Firstly, I would like to clarify that the stress tensor, also known as the Maxwell stress tensor, is defined as:

$\sigma_{ij} = \epsilon_{0}E_{i}E_{j} + \mu_{0}H_{i}H_{j} - \frac{1}{2}\delta_{ij}\left(\epsilon_{0}E^{2} + \mu_{0}H^{2}\right)$

This tensor is used to determine the forces and stresses on a material due to the presence of electric and magnetic fields. It is derived from the Lorentz force law, which is given by:

$\textbf{f} = q(\textbf{E} + \textbf{v} \times \textbf{B})$

In your attempt at a solution, you have used the macroscopic Maxwell equations and the Lorentz force law to derive an expression for the force per unit volume. However, this expression does not involve the stress tensor and therefore, it cannot be used to determine the general stress tensor.

To obtain the general stress tensor, we need to start with the Lorentz force law and use the constitutive equations for electric and magnetic fields:

$\textbf{f} = q\left(\epsilon\textbf{E} + \textbf{P} + \textbf{v} \times \mu(\textbf{H} + \textbf{M})\right)$

Now, we can use the product rule of derivatives and Faraday's law to rewrite this equation as:

$\textbf{f} = q\left[\epsilon\textbf{E} + \mu\left(\textbf{H} + \textbf{M} + \textbf{v} \times (\nabla \times \textbf{H})\right) + \textbf{v} \times \left(\nabla \times \textbf{M}\right)\right]$

Using the identity ##\nabla \times (\textbf{a} \times \textbf{b}) = \textbf{a}(\nabla \cdot \textbf{b}) - \textbf{b}(\nab