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

  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?

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3 Answers 3

up vote 1 down vote accepted

The stress tensor is defined in such a way that taking its divergence brings you back to the force density (apart from the cross product term which corresponds to the time derivative of the Poynting vector). In the case of $\mathbf{B}=\mathbf{0}$, we take the divergence in index notation, i.e.

$$f_j=\partial_i\sigma_{ij}=\epsilon_0\left(\partial_i(E_iE_j)-\frac12\delta_{ij}\partial_i(E_kE_k)\right)=\epsilon_0\left(\partial_iE_iE_j+E_i\partial_iE_j-\frac12\partial_j(E_kE_k)\right).$$

In order to see that this expression corresponds to the first term in your expression 3., we recast this in index-free notation:

$$\mathbf{f}=\epsilon_0\left((\boldsymbol{\nabla}\cdot\mathbf{E})\mathbf{E}+(\mathbf{E}\cdot\boldsymbol{\nabla})\mathbf{E}-\frac12\boldsymbol{\nabla}(\mathbf{E}\cdot\mathbf{E})\right).$$

Making use of the identity

$$\frac{1}{2} \boldsymbol{\nabla} \left( \mathbf{A}\cdot\mathbf{A} \right) = \mathbf{A} \times (\boldsymbol{\nabla} \times \mathbf{A}) + (\mathbf{A} \cdot \boldsymbol{\nabla}) \mathbf{A}$$

and the Maxwell equation (for $\mathbf{B}=\mathbf{0}$)

$$\boldsymbol{\nabla}\times\mathbf{E}=\mathbf{0},$$

we find that

$$\mathbf{f}=\epsilon_0(\boldsymbol{\nabla}\cdot\mathbf{E})\mathbf{E}.$$

From this we can see that it does not matter at which stage of the derivation the magnetic field is omitted.

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A much, much easier method would be to find the stress-energy tensor of the electromagnetic field from its definition:

$$T^\mu_{~~~\nu}=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi )}\partial_\nu \phi-\mathcal{L}\delta^\mu_{~~\nu}$$

where $\phi$ is the field whose stress-energy you're looking for, and $\mathcal{L}$ is the Lagrangian density for the field, i.e. in this case the electromagnetic Lagrangian. Then the Maxwell Stress-tensor will just be the spacial components of $T_{\mu \nu}$.

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1  
Ehhhh... This is slightly trickier than it looks because naive application of this formula yields a stress energy tensor that isn't symmetric, and you need to correct for this by adding some divergence-free part if I am not mistaken. –  alexarvanitakis Feb 28 '13 at 20:41

You can go full-on into special relativity, using the Faraday bivector $F = e_t \wedge E/c$ (since we said specifically there's no magnetic field).

The stress-energy tensor is always

$$\underline T(a) = -\frac{1}{2\mu_0} FaF = -\frac{1}{2\mu_0 c^2} e_t E a e_t E$$

Let $e_t e_t = -1$ and take $a = e_t$.

$$\begin{align*} \underline T(e_t) &= -\frac{\epsilon_0}{2} e_t E e_t e_t E \\ &= -\frac{\epsilon_0}{2} e_t E(-1)E \\ &= \frac{\epsilon_0}{2} e_t E^2 \end{align*}$$

Of course, if you don't know what the EM stress-energy tensor is in the first place, that may not be of much help. Regarding your derivation, knowing that your field is static and without any magnetic contribution, you can throw away all the magnetic terms (and time derivatives) as soon as they appear. There is no reason to keep them around except to save work later (if you expect you'll need a more general expression for some other problem).

See that $(\nabla \cdot E) E + (E \cdot \nabla) E = \frac{1}{2} \nabla E^2$. This follows from the product rule. From here, you can just imagine that $\underline \sigma(\nabla) = f$, where $\nabla$ acts on the terms inside the stress tensor.

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