- #1
gentsagree
- 96
- 1
So, I'm trying to show that by duality [itex]B_{i}\rightarrow E_{i}[/itex], using tensor notation. I've done it in a different way, and it works (starting from [itex]\overline{F}_{ij}[/itex], the dual of F_ij), but I would like to get it from B_i directly. Where am I going wrong?
This is what I did:
[tex]B_{i}=\frac{1}{2}\epsilon_{ijk}F^{jk}\rightarrow B'_{i}=\frac{1}{2}\epsilon_{ijk}(i\overline{F}^{jk})=\frac{1}{2}i\epsilon_{ijk}(-\frac{1}{2}i\epsilon^{jk\rho\sigma}F_{\rho\sigma})[/tex]
[tex]=\frac{1}{4}\epsilon_{ijk}\epsilon^{jk\rho\sigma}F_{\rho\sigma}=\frac{1}{4}\epsilon_{0ijk}\epsilon^{jk0i}F_{0i}[/tex]
where in the last line I have inserted an extra 0-index in the Levi-Civita symbol (although I am not sure I know how to deal with zeros with epsilon), and made the substitution [itex](\rho,\sigma)\rightarrow(0,i)[/itex].
However I calculate this to be [itex]-\frac{3}{2}F_{i0}[/itex] when it should be just [itex]F_{i0}=E_{i}[/itex]
Any advice?
This is what I did:
[tex]B_{i}=\frac{1}{2}\epsilon_{ijk}F^{jk}\rightarrow B'_{i}=\frac{1}{2}\epsilon_{ijk}(i\overline{F}^{jk})=\frac{1}{2}i\epsilon_{ijk}(-\frac{1}{2}i\epsilon^{jk\rho\sigma}F_{\rho\sigma})[/tex]
[tex]=\frac{1}{4}\epsilon_{ijk}\epsilon^{jk\rho\sigma}F_{\rho\sigma}=\frac{1}{4}\epsilon_{0ijk}\epsilon^{jk0i}F_{0i}[/tex]
where in the last line I have inserted an extra 0-index in the Levi-Civita symbol (although I am not sure I know how to deal with zeros with epsilon), and made the substitution [itex](\rho,\sigma)\rightarrow(0,i)[/itex].
However I calculate this to be [itex]-\frac{3}{2}F_{i0}[/itex] when it should be just [itex]F_{i0}=E_{i}[/itex]
Any advice?