Using F'μν = (∂ x^α)/(∂ x'^μ)(∂ x^β)/(∂ x'^ν)Fαβ with t' = γ(t - β x), x' = γ(x - β t), y' = y,

Question

Using F&#x27;μν = (∂ x^α)/(∂ x&#x27;^μ)(∂ x^β)/(∂ x&#x27;^ν)Fαβ with t&#x27; = γ(t - β x), x&#x27; = γ(x - β t), y&#x27; = y,

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Nicole Hoffman · Answer

So this question wants me to simplify the function, doing F of X plus H minus F of X all over H.

Using $F'_{\mu\nu} = \frac{\partial x^{\alpha}}{\partial x'^\mu}\frac{\partial x^{\beta}}{\partial x'^\nu}F_{\alpha\beta}$ with $t' = \gamma(t - \beta x)$, $x' = \gamma(x - \beta t)$, $y' = y$, $z' = z$, \newline find the non-zero components of the tensor $F'_{\mu\nu}$