View Single Post
Feb28-13, 07:20 AM
P: 10
I'm having some trouble calculating the stress tensor in the case of a static electric field without a magnetic field. Following the derivation on Wikipedia,

1. Start with Lorentz force:
[tex]\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})[/tex]

2. Get force density
[tex]\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}[/tex]

3. Substitute using Maxwell's laws
[tex]\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 \mathbf{E}}{\partial t} \times \mathbf{B}[/tex]

4. Replace some curls and combine
[tex]\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)[/tex]

5. Get the tensor
[tex]\sigma_{i j} = \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)[/tex]

6. Assuming B=0:
[tex]\sigma_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right)[/tex]

7. Assume flat surface with perpendicular field (z-direction)
[tex]\sigma_{z z} = \epsilon_0 \left(E^2 - \frac{1}{2} E^2\right)=\frac{\epsilon_0}{2} E^2[/tex]

This is the formula given in e.g. The Feynman Lectures in Physics Vol. 2 (Page 31-14), and some other text books.

However, this derivation seems to assume a magnetic field until the final steps. Since most terms in eq. 4 result from the initial v x B term (even those that depend only on E, [itex](\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E}[/itex] and [itex]\frac{1}{2} \boldsymbol{\nabla}\epsilon_0 E^2[/itex] ), these should not be present in my case, and in fact eq 4 should be as simple as
\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]

Tensor calculus is not my strong point. To me it is not clear how to get from eq 4 to eq 5, and how modifying eq 4 alters the resulting stress tensor. Will it really still be the same as eq 6? To me it seems strange that removing terms would not affect the result, yet this seems to be what many text books claim. Or is there some reason why the initial v x B term can not be removed, even when there is no magnetic field?
Phys.Org News Partner Physics news on
A new, tunable device for spintronics
Watching the structure of glass under pressure
New imaging technique shows how cocaine shuts down blood flow in mouse brains