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Suppose we have a vector $A^\mu$. This vector has 4 components (0 through 3). If we consider the Lorentz's transformation of each component, we could simplify them into matrix notation, ${A^\prime}^\mu=L^\mu{}_{\nu} A^\nu$, where:

$$L^\mu{}_{\nu}= \left( \begin {matrix}\gamma & -v\gamma & 0 & 0\\ -v\gamma & \gamma & 0 & 0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix}\right)$$

where $\gamma=(1-\frac{v^2}{c^2})^{-\frac{1}{2}}$

Knowing that $A_{\mu}= \eta_{\mu\nu} A^\nu$ and

$$\eta_{\mu\nu}= \left( \begin{matrix} -1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix}\right)$$

we could say that $A_\mu^\prime= M_{\mu}^\nu A_{\nu}$. According to my calculations:

$$M_{\mu}^\nu= \left( \begin{matrix} -\gamma&v\gamma&0&0\\ v\gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix}\right)$$

But according to Leonard Susskind's lectures: $M=\eta L\eta$

which doesn't agree with my calculations.

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    $\begingroup$ I think the problem might be in wanting to interpret expressions as $M_\mu^\nu A_\nu$ as matrix-vector products, which they are not. $$\tilde{A}_{\mu}=\eta_{\mu\nu} \tilde{A}^{\nu}=\eta_{\mu\nu}L^{\nu}_{\alpha}A^{\alpha}=\eta_{\mu\nu}L^{\nu}_{\alpha}\eta^{\alpha\beta}A_{\beta}=\eta_{\mu\beta}L^{\beta}_{\alpha}\eta^{\alpha\nu}A_{\nu}$$ and thus $$M_{\mu}^{\nu}=\eta_{\mu\beta}L^{\beta}_{\alpha}\eta^{\alpha\nu}.$$ Now try to figure out what matrix these components would make. If you ask me, I would try to work with components rather than matrices and only use those to display components. $\endgroup$
    – Adomas Baliuka
    Commented Sep 25, 2016 at 23:05
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    $\begingroup$ @AdomasBaliuka, why not make that an answer? $\endgroup$
    – Alfred Centauri
    Commented Sep 26, 2016 at 0:48
  • $\begingroup$ Thanks! I now understand the derivation but i still don´t get why my matrix assumption is wrong. I calculated $L$ using a matrix-vector multiplication consideration and it worked. Why doesn´t it work for $M$? $\endgroup$
    – Duarte Costa
    Commented Sep 26, 2016 at 6:28
  • $\begingroup$ Never mind... I realized now what I'm doing wrong. Thank you @AdomasBaliuka $\endgroup$
    – Duarte Costa
    Commented Sep 26, 2016 at 9:04

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