I have recently been introduced to Lorentz transformations in the form of 4-Vectors, I have been told the following: $$\Lambda^T\eta\Lambda=\eta,$$ Where $\eta$ is the matrix: \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} The relation supposedly holds for any transformation, $\Lambda$, but I am attempting to show that it works for: $$\Lambda=\begin{bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix},$$ Which is just the Lorentz boost in x.
My Problem is when it comes to the matrix multiplication I see that because $\Lambda$ is symmetric: $$\Lambda^T=\Lambda,$$ by multiplication: $$\Lambda^T\eta= \begin{bmatrix} \gamma & 0 & 0 & 0 \\ 0 & -\gamma & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix},$$ Multiplying The above with $\Lambda$ gives: $$\Lambda^T\eta\Lambda= \begin{bmatrix} \gamma^2 & 0 & 0 & 0 \\ 0 & -\gamma^2 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix},$$ This is clearly not correct, as $\Lambda^T\eta\Lambda$ should be: $$\Lambda^T\eta\Lambda= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}.$$ I have repeated the question many times and can not get to the correct answer. I initially thought it was something to do with the ordering of my multiplication, but even when I retry the calculation and taking the product of $\Lambda$ and $\Lambda^T$ first, I still get the incorrect answer.
If anyone can let me know where I am going wrong, that'd be great. Thanks.
You should have: \begin{align*}\Lambda^T \eta&=\left[\begin{matrix}\gamma &\beta\gamma &0 &0\\-\beta\gamma &-\gamma &0 &0 \\ 0 &0 &-1 &0\\ 0 &0 &0 &-1\end{matrix}\right]\\ \Lambda^T\eta\Lambda&=\left[\begin{matrix}\gamma^2-\beta^2\gamma^2 &0 &0 &0\\ 0 &-\gamma^2+\beta^2\gamma^2 &0 &0\\0 &0 &-1 &0\\ 0 &0 &0 &-1\end{matrix}\right].\end{align*} If you simplify $\gamma^2-\beta^2\gamma^2$, you'll find that the LHS does indeed equal the RHS.