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I have the followwing Lagrangian for the free electromagnetic field,

$$\mathcal{L} = -\frac{1}{4} F^{\mu \nu}F_{\mu \nu},$$

and the canonical stress tensor is,

$$T^{\alpha \beta}=\frac{\partial \mathcal{L}}{\partial \left(\partial_{\alpha} A^{\lambda}\right)}\partial^{\beta} A^{\lambda}-g^{\alpha \beta}\mathcal{L}.$$

My $T^{0i}$ is, $T^{0i} = (\vec{E}\times \vec{B})_{i}$

but it's the symmetric stress tensor, and I need to find it:

$$T^{0i} = (\vec{E}\times \vec{B})_{i} + \nabla \cdot (A_{i}\vec{E})$$

I don't understand where this divergence term came from. How can I do this asymmetric $T^{0i}$ tensor?

(Just for reference, see Jackson - Classical electrodynamics, Third ed.- page 606)

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1 Answer 1

Answering my own question

We have:

$T^{ab} = -\frac{1}{4\pi} g^{ac}F_{cd}\partial^b A^d - g^{ab}L_{elec}$

$g^{0i} = 0$. In the first term we have, (using $F_{00} = 0$):

$T^{0i} = -\frac{1}{4\pi} g^{0c}F_{cd}\partial^i A^d - g^{0i}L_{elec}$

we need $c=0$, because $g^{00}=1$:

$T^{0i} = -\frac{1}{4\pi} g^{00}F_{0d}\partial^i A^d$

with $d=i,j,k$

$F_{0d}=(E_i + E_j + E_k)$

then, whe have:

$T^{0i} = -(E_i\partial^i A^i + E_j\partial^i A^j + E_k\partial^i A^k)$

$T^{0i}= - \frac{1}{4 \pi} \left( E_x \partial^x A^x + E_y \partial^x A^y + E_z \partial^x A^z \right )$

using $- B^j = \partial_k A^i - \partial_i A^k $:

$T^{0i}= \frac{1}{4 \pi} \left(E_i \partial_i A^i + E_j \partial_j A^i + E_j B_k - E_k B_j +E_k \partial_k A^i \right ) $

with $ \nabla \cdot \vec{E} = 0 $:

$ T^{0i} = \frac{1}{4\pi} \left( ( \vec{E} \times \vec{B} )_i + \nabla \cdot (A_i \vec{E}) \right)$

QED.

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