Quick Calculation 3.4

A First Course in String Theory

Show that

$\displaystyle{ \begin{aligned} F^{\mu \nu} &= - F^{\nu \mu} \\ F^{0 i} &= - F_{0 i} \\ F^{ij} &= F_{ij} \\ \end{aligned} }$

~~~

Eq. (3.15):

$\displaystyle{ F_{\mu \nu} \equiv \partial_\mu A_\nu - \partial_\nu A_\mu \\ }$

Eq. (3.29):

$\displaystyle{ F^{\mu \nu} = \eta^{\mu \alpha} \eta^{\nu \beta} F _{\alpha \beta} \\ }$

Eq. (3.16):

$\displaystyle{ F_{\mu \nu} = - F_{\nu \mu} \\ }$

$\displaystyle{ \begin{aligned} F_{\mu \nu} &= - F_{\nu \mu} \\ \eta^{\mu \alpha} \eta^{\nu \beta} F_{\mu \nu} &= - \eta^{\mu \alpha} \eta^{\nu \beta} F_{\nu \mu} \\ \eta^{\mu \alpha} \eta^{\nu \beta} F_{\mu \nu} &= - \eta^{\nu \beta} \eta^{\mu \alpha} F_{\nu \mu} \\ F^{\mu \nu} &= - F^{\nu \mu} \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} F^{\mu \nu} &= \eta^{\mu \alpha} \eta^{\nu \beta} F_{\alpha \beta} \\ F^{0 i} &= \eta^{0 \alpha} \eta^{i \beta} F _{\alpha \beta} \\ &= \eta^{0 0} \eta^{i \beta} F _{0 \beta} + \eta^{0 1} \eta^{i \beta} F _{1 \beta} + \eta^{0 2} \eta^{i \beta} F _{2 \beta} + \eta^{0 3} \eta^{i \beta} F _{3 \beta} \\ &= (-1) \eta^{i \beta} F _{0 \beta} + (0) \eta^{i \beta} F _{1 \beta} + (0) \eta^{i \beta} F _{2 \beta} + (0) \eta^{i \beta} F _{3 \beta} \\ &= (-1) \eta^{i \beta} F _{0 \beta} \\ &= - F _{0 i} \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} F^{\mu \nu} &= \eta^{\mu \alpha} \eta^{\nu \beta} F_{\alpha \beta} \\ F^{i j} &= \eta^{i \alpha} \eta^{j \beta} F_{i j} \\ &= \sum_\alpha \sum_\beta \eta^{i \alpha} \eta^{j \beta} F_{\alpha \beta} \\ &= \sum_\beta \eta^{i i} \eta^{j \beta} F_{i \beta} \\ &= \eta^{i i} \eta^{j j} F_{i j} \\ &= (1) (1) F_{i j} \\ &= F_{i j} \\ \end{aligned} }$

— Me@2022.05.05 07:49 PM

Physics Town

continues

Quick Calculation 3.4

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