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:
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.?
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 $$\Lambda^i_{\,\,\,0}=\gamma v^i$$