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I have to show that $$q \equiv \int d^{3}\vec{x}\, J^0 = -\int d^{3}\vec{x}\ \partial_i F^{0i} = -\int \frac{1}{2} d^{3}\vec{x}\ \varepsilon^{ijk} \partial_{i}G_{jk} \tag{1}$$ (this is exactly what is given in the book) where $J^{\nu} = -\partial_{\mu} F^{\mu \nu}$ coming and $G^{\mu \nu} = -i\tilde{F}^{\mu \nu}$. The latter, $\tilde{F}^{\mu \nu}$ is the dual of $F^{\mu \nu}$ and it is given as $$\tilde{F}^{\mu \nu} = -\frac{i}{2} \varepsilon^{\mu \nu \rho \sigma}F_{\rho \sigma} \tag{2}$$

My attempt:

We have $G^{\mu \nu} = -i\tilde{F}^{\mu \nu} = -i (-\frac{i}{2} \varepsilon^{\mu \nu \rho \sigma}F_{\rho \sigma} ) = -\frac{1}{2}\varepsilon^{\mu \nu \rho \sigma}F_{\rho \sigma}$. Now if we fix $\sigma=0$ we have $-\frac{1}{2}\varepsilon^{jki}F_{i0}$ and inverting the indices in the field strength we obtain $$ G^{jk} = \frac{1}{2}\varepsilon^{jki}F_{0i}. $$ Therefore we see that $$2G^{jk}\varepsilon_{jki} = F_{0i}$$ but this is the opposite of what we want, i.e. $F$ must be proportional to half of $G$ and not vice versa. So, where is the mistake? Any help?

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  • $\begingroup$ Are you intentionally neglecting the 0 superscripts in $\epsilon$? $\endgroup$ – Kyle Kanos Jan 5 '15 at 18:04
  • $\begingroup$ it's not clear why you're fixing $\sigma$ which is a dummy repeated index, hence involved in a sum from 0 to 3. Moreover, what is that thing that you claim equal to $-\frac12\epsilon^{jki}F_{i0}$? $\endgroup$ – Phoenix87 Jan 5 '15 at 18:09
  • $\begingroup$ Yes I am fixing $\sigma$ since I know that one of the indices has to be zero. Then, in order the $\varepsilon$ not to be zero we know that all other indices must be spacial hence it can be replaced by the 3d tensor. $\endgroup$ – Marion Jan 5 '15 at 18:48
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The electromagnetic tensor can be written in the form $$F = \begin{bmatrix}0 & -\mathbf E\\\mathbf E & \star\mathbf B\end{bmatrix},$$ where $\star\mathbf B$ is the Hodge dual of $\mathbf B$, which in vector notation can be realised as the operator $\mathbf v\mapsto \mathbf B\times\mathbf v$ (I'm totally unsure about the signs, but everything should work if $F$ is skewsymmetric, only problem is that the physics might get a wee messed up). In this notation, its dual is then given by (again up to signs) $$\tilde F = \begin{bmatrix}0 & \mathbf B\\-\mathbf B & -\star\mathbf E\end{bmatrix},$$

Now $-\partial_i F^{0i} =\nabla\cdot\mathbf E$, which is $\rho$ by the first Maxwell's equation. $\frac12\epsilon^{ijk}G_{jk}$ is the Hodge dual of $-\star E$ which is $-\star\star\mathbf E = -\mathbf E$, hence again $-\frac12\partial_i\epsilon^{ijk}G_{jk} = \nabla\cdot\mathbf E=\rho$. Finally $$q = \int\rho\ \text d^3\mathbf x,$$ whence the result by substitution.

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  • $\begingroup$ Hi and thanks for your answer. I struggle a little bit trying to translate everything in index notation. Any help would be appreciated, even more if you d let me know where I am wrong. $\endgroup$ – Marion Jan 5 '15 at 18:50
  • $\begingroup$ I believe it is clear from above that $F^{0i} = E^i$. The problem with your steps is that you fix an index, $\sigma$, which is actually running from 0 to 3. As for the expression involving the constitutive tensor $G$, the operation $-\frac12\epsilon^{ijk}G_{jk}$ is reversing the definition of $G$ in terms of $F$ and spitting out $E^i$ again. You can see this by going through the computation by doing all the sums explicitly. $\endgroup$ – Phoenix87 Jan 5 '15 at 20:26
  • $\begingroup$ I am confused. Should I not fix one index to be zero in order to get the conserved charge? Your answer is ok (btw thanks for it) but I cannot make it work with what I have. $\endgroup$ – Marion Jan 6 '15 at 1:42
  • $\begingroup$ You can fix $\mu$ or $\nu$ because they're not repeated $\endgroup$ – Phoenix87 Jan 6 '15 at 9:51
  • $\begingroup$ I got it. It is really simple and I got it thanks to your answer. It is one line and it is even simpler than what you wrote. Thanks a lot $\endgroup$ – Marion Jan 6 '15 at 10:11

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