Timeline for Faraday tensor, antisymmetric rank two
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Apr 26, 2013 at 19:32 | comment | added | FraSchelle | As an interesting far aside, you then have the proof that any quantity $H^{\mu \nu \xi ...}H_{\mu \nu \xi ...}$ is also Lorentz invariant, as long as the two quantities (called tensor) $H$ have the same indices. This is called a contraction in tensor calculation. The value of $F_{\mu \nu}$ in terms of the fields $E_{\mu}$ or $B_{\mu}$ are found using the definition $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ with $A\equiv \left(\phi , \mathbf{A} \right)$ the potentials. | |
| Apr 26, 2013 at 18:03 | comment | added | Ana S. H. | @JishnuRay I just added some details. | |
| Apr 26, 2013 at 18:02 | history | edited | Ana S. H. | CC BY-SA 3.0 |
Adding mathematical details of the answer.
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| Apr 26, 2013 at 16:36 | comment | added | Ana S. H. | It's actually the same. | |
| Apr 26, 2013 at 6:01 | comment | added | Jishnu Ray | Actually from physicspages.com/2013/03/15/… I got $F_{\mu \nu}F^{\mu \nu}$. But I want $F^{\mu \nu}F_{\mu \nu}$. | |
| Apr 26, 2013 at 5:50 | comment | added | Jishnu Ray | How is $F^{\mu \nu}F_{\mu \nu}$ related to $F_{\mu \nu}F^{\mu \nu}$ | |
| Apr 26, 2013 at 5:11 | comment | added | Jishnu Ray | Please Please write the explicit expressions . | |
| Apr 26, 2013 at 0:12 | comment | added | Ana S. H. | If you need more explicit expressions please tell me, I'll write them down. | |
| Apr 25, 2013 at 12:21 | history | answered | Ana S. H. | CC BY-SA 3.0 |