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The questions:

Our Prof wrote the following expression that confused me, \begin{align} \Lambda^0_{\,\,\,i}&=\eta^{00}\eta_{ij}\Lambda_{0}^{\,\,\,j}\\ &=-\eta_{ij}\Lambda_{0}^{\,\,\,j}\\ &=v_i\gamma \end{align} Now, as far as I understand, this could only be true, if $$\Lambda_0^{\,\,\,j}=-\Lambda^j_{\,\,\,0}$$ because we know (see below in the Background section, where I have written down the derivation starting from as early as possible), $$\Lambda^j_{\,\,\,0}=\gamma v^j$$ But the relation $\Lambda_0^{\,\,\,j}=-\Lambda^j_{\,\,\,0}$ can be true? Isn't the Lorentz transformation matrix symmetric? Moreover, if the Lorentz matrix really is antisymmetric, why then the last line in the following expression, \begin{align} \eta_{\mu\nu}&=\eta_{\alpha\beta}\Lambda^\alpha_{\,\,\,\mu}\Lambda^\beta_{\,\,\,\nu}\\ &=\left(\Lambda^T\right)^{\,\,\,\alpha}_{\mu}\eta_{\alpha\beta}\Lambda^\beta_{\,\,\,\nu} \end{align} does not pick up a minus sign?

To summarize, I have actually two questions:

  1. What is the symmetric/antisymmetric properties of the Lorentz transformation matrix in the various situations, like, when both of its indices are down; or both are up; or one up and one down; or the upper index come before the lower index or the lower index come before the upper one etc.?

  2. As evident from question 1, I am extremely confused regarding this upper index coming before the lower index or the lower index coming before the upper one business. Can someone explain in details or refer to some good student friendly source on this?

The background:

\begin{align} \eta_{00}=-{c^2},\quad\eta_{ij}=\delta_{ij} \end{align} \begin{align} \eta_{\mu\nu}=\eta_{\alpha\beta}\Lambda^\alpha_{\,\,\,\mu}\Lambda^\beta_{\,\,\,\nu} \end{align} For $00$ components, \begin{align} \eta_{00}&=\eta_{\alpha\beta}\Lambda^\alpha_{\,\,\,0}\Lambda^\beta_{\,\,\,0}\nonumber\\ &=\eta_{00}\Lambda^0_{\,\,\,0}\Lambda^0_{\,\,\,0}+\eta_{ij}\Lambda^i_{\,\,\,0}\Lambda^j_{\,\,\,0}\nonumber\\ -{c^2}&=-{c^2}\left(\Lambda^0_{\,\,\,0}\right)^2+\sum_{i=1,2,3}\left(\Lambda^i_{\,\,\,0}\right)^2\nonumber\\ {c^2}\left(\Lambda^0_{\,\,\,0}\right)^2&={c^2}+\sum_{i=1,2,3}\left(\Lambda^i_{\,\,\,0}\right)^2\nonumber\\ \Lambda^0_{\,\,\,0}&=\sqrt{1+\frac{1}{{c^2}}\sum_{i=1,2,3}\left(\Lambda^i_{\,\,\,0}\right)^2}\tag{1}\label{eq:Lorentderivationone} \end{align} In the last line we have chosen the positive solution only. Such a choice is known as the proper Lorentz transformation.\par Consider two frames, \begin{equation} \begin{aligned} &\text{Frame } S && \text{Frame } \bar{S}\\ &\text{Event A: } ({c} t, x,y,z)\qquad &&\text{Event A: } ({c} \bar{t}, \bar{x},\bar{y},\bar{z})\\ &\text{Event B: } ({c} (t+dt), x,y,z) &&\text{Event B: }({c} (\bar{t}+d\bar{t}), \bar{x}+d\bar{x},\bar{y}+d\bar{y},\bar{z}+d\bar{z}) \end{aligned} \end{equation} The vector transformation rule, \begin{align} d\bar{x}^\mu=\Lambda^\mu_{\,\,\,\nu}dx^\nu \end{align} For $\mu=0$, \begin{align} d\bar{x}^0&=\Lambda^0_{\,\,\,\nu}dx^\nu\nonumber\\ &=\Lambda^0_{\,\,\,0}dx^0+\Lambda^0_{\,\,\,i}dx^i\nonumber\\ &=\Lambda^0_{\,\,\,0}{c} dt \end{align} For $\mu=i$, \begin{align} d\bar{x}^i&=\Lambda^i_{\,\,\,\nu}dx^\nu\nonumber\\ &=\Lambda^i_{\,\,\,0}dx^0+\Lambda^i_{\,\,\,j}dx^j\nonumber\\ &=\Lambda^i_{\,\,\,0}{c} dt\quad(i=1,2,3) \end{align} Now, the relative velocity, \begin{align} \frac{d\bar{x}^i}{d\bar{t}}=\frac{\Lambda^i_{\,\,\,0}}{\Lambda^0_{\,\,\,0}}\equiv v^i,\quad(i=1,2,3) \end{align} Using (\ref{eq:Lorentderivationone}) we get, \begin{equation} \begin{gathered} \Lambda^0_{\,\,\,0}=\sqrt{1+\frac{1}{{c^2}}\left(\Lambda^0_{\,\,\,0}\right)^2\sum_{i=1,2,3}\left(v^i\right)^2}\nonumber\\ \Lambda^0_{\,\,\,0}=\sqrt{1+\frac{1}{{c^2}}\left(\Lambda^0_{\,\,\,0}\right)^2\left(\mathbf{v\cdot}\mathbf{v}\right)}\nonumber\\ \left(\Lambda^0_{\,\,\,0}\right)^2=1+\frac{1}{{c^2}}\left(\Lambda^0_{\,\,\,0}\right)^2\left(\mathbf{v\cdot}\mathbf{v}\right)\nonumber\\ \left(\Lambda^0_{\,\,\,0}\right)^2\left(1-\frac{\mathbf{v\cdot}\mathbf{v}}{{c^2}}\right)=1\nonumber\\ \Lambda^0_{\,\,\,0}=\frac{1}{\sqrt{1-\left(\frac{\mathbf{v\cdot}\mathbf{v}}{{c^2}}\right)}}\equiv\gamma \end{gathered} \end{equation} And $$\Lambda^i_{\,\,\,0}=\gamma v^i$$

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    $\begingroup$ A generic Lorentz transformation matrix is not symmetric. $\endgroup$ – Prahar Mitra Oct 24 '20 at 14:29
  • $\begingroup$ @Prahar Sorry, I should have been more clearer. I am mainly concerned about boosts. $\endgroup$ – Faber Bosch Oct 24 '20 at 14:42
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    $\begingroup$ It's important to note that $\Lambda_\nu{}^\mu = (\Lambda^{-1})^\mu{}_\nu$ so what you are proving is $(\Lambda^{-1})^i{}_0 = - \Lambda^i{}_0$. $\endgroup$ – Prahar Mitra Oct 24 '20 at 14:47
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There's no sign discrepancy at all. Start from $\Lambda^0_{\:i}=\eta^{0\mu}\Lambda_{\mu i}=\eta^{0\mu}\eta_{i\nu}\Lambda_\mu^{\:\nu}$. Since $\eta$ doesn't mix space with time in Cartesian coordinates (a fact I'll use hereafter without comment), this simplifies to $\eta^{00}\eta_{ij}\Lambda_0^{\:j}$. The next calculation shows your professor is using $-+++$, for which $\eta^{00}=-1,\,\eta_{ij}=\delta_{ij}$. To unite $\Lambda_{0i}=-\gamma v_i$ with $\Lambda_0^{\:j}=\gamma v^j$, note$$\Lambda_0^{\:j}=\eta_{0\mu}\Lambda^{\mu j}=\eta_{00}\Lambda^{0j}=-\Lambda^{0j}=-\Lambda^0_{\:\mu}\eta^{\mu j}=-\Lambda^0_{\:i}\eta^{ij}=\gamma v_i\eta^{ij}=\gamma v_i\delta^{ij}=\gamma v^j.$$As you say, $\Lambda$ is symmetric, so similarly $\Lambda^j_{\:0}=\gamma v^j$.

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I don't assume this will be a complete answer, just several suggestions that I hope could be helpful.

So you are using the $(-,+,+,+)$ metric, I'm using the opposite $(+,-,-,-)$, sorry. Consider that a generic Lorentz transformation is subjected to the following \begin{gather*} \Lambda^T \mathbb{G} \Lambda = \mathbb{G} \end{gather*} where $\Lambda=({\Lambda^\alpha}_\beta)$ while $\mathbb{G}=(\mathbb{G}_{\alpha\beta})$ so that everything's fine with that equation \begin{equation*} {\Lambda^\beta}_\alpha \mathbb{G}_{\beta\gamma} {\Lambda^\gamma}_\delta = \mathbb{G}_{\alpha\delta} \end{equation*} There is not a general simmetric or antisimmetric property for Lorentz matrices, in fact consider that for a boost transformation on the $x$-axis \begin{equation*} \Lambda = \begin{pmatrix} \cosh{\beta}&\sinh{\beta}&0&0 \\ \sinh{\beta}&\cosh{\beta}&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{pmatrix} \end{equation*} while for a $x/y$ rotation \begin{equation*} \Lambda = \begin{pmatrix} 1&0&0&0 \\ 0&\cos{\alpha}&-\sin{\alpha}&0 \\ 0&\sin{\alpha}&\cos{\alpha}&0 \\ 0&0&0&1 \end{pmatrix} \end{equation*} What you can say from the first formula is that $\det\Lambda=\pm 1$ and when the sign is positive and $\Lambda_{00}\geq 1$ (maybe $\Lambda_{00}\leq -1$ in your case) you speak about proper Lorentz transformation.

In general indices are uppered or lowered by the metric so \begin{equation*} {\Lambda_\alpha}^\beta = \mathbb{G}_{\alpha\gamma} {\Lambda^\gamma}_\delta \mathbb{G}^{\delta\beta} \end{equation*} so that, defining $({\Lambda_\alpha}^\beta)\doteq\tilde{\Lambda},(\mathbb{G}^{\delta\beta})\doteq\tilde{\mathbb{G}}$ you will have (be careful with indices representing rows and that representing colums, to write the right matrices products, with the right order: consider also that metric tensor is symmetric) \begin{equation*} \tilde{\Lambda} = \mathbb{G} \Lambda \tilde{\mathbb{G}} \end{equation*} This is what it came to my mind, hope you find inspiration in something and that are not big mistakes. Good luck

P.S. If you are looking for a brief discussion on special relativity but very student-friendly I suggest you "D'Auria, Trigiante - From Special Relativity to Feynman Diagrams"

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