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-   -   Maxwell Stress Tensor in the absence of a magnetic field (http://www.physicsforums.com/showthread.php?t=675141)

 pafcu Feb28-13 07:20 AM

Maxwell Stress Tensor in the absence of a magnetic field

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,

$$\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$$

2. Get force density
$$\mathbf{f} = \rho\mathbf{E} + \mathbf{J}\times\mathbf{B}$$

3. Substitute using Maxwell's laws
$$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$

4. Replace some curls and combine
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\nabla}\left(\epsilon_0 E^2 + \frac{1}{\mu_0} B^2 \right) - \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)$$

5. Get the tensor
$$\sigma_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)$$

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

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

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, $(\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E}$ and $\frac{1}{2} \boldsymbol{\nabla}\epsilon_0 E^2$ ), 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?

 Meir Achuz Feb28-13 09:30 AM

Re: Maxwell Stress Tensor in the absence of a magnetic field

The first two terms of Eq. (4) should be
$$\mathbf{f} = \epsilon_0\left[ \boldsymbol{\nabla}\cdot( \mathbf{E}\mathbf{E}) + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right]$$.
Then use $$(\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} =(1/2)\nabla({\bf E\cdot E })$$.
Writing these two terms with indices gives Eq. (5).
The other terms don't enter with no B field. dE/dt can't enter because this would imply a B field.

 pafcu Feb28-13 09:51 AM

Re: Maxwell Stress Tensor in the absence of a magnetic field

I don't understand how you get
$$\mathbf{f} = \epsilon_0\left[ \boldsymbol{\nabla}\cdot( \mathbf{E}\mathbf{E}) + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right]$$

Looking at eq. 3
$$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$

and assuming B = 0 gives me
$$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]$$

Can you clarify?

 Meir Achuz Feb28-13 12:59 PM

Re: Maxwell Stress Tensor in the absence of a magnetic field

Quote:
 Quote by pafcu (Post 4289462) I don't understand how you get $$\mathbf{f} = \epsilon_0\left[ \boldsymbol{\nabla}\cdot( \mathbf{E}\mathbf{E}) + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right]$$ Looking at eq. 3 $$\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}$$ and assuming B = 0 gives me $$\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E}\right]$$ Can you clarify?
$$(\boldsymbol{\nabla}\cdot \mathbf{E} ){\bf E}= \nabla\cdot({\bf EE})-{(\bf E\cdot\nabla)E}$$.