Quick Calculation 3.4

A First Course in String Theory

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Show that

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

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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}  \\    }

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\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}       }

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\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}       }

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\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

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2022.05.06 Friday (c) All rights reserved by ACHK