Derive Minkowski metric from Lorentz transformation

Question

I am trying to learn special relativity. My goal is to prove that given the fact that a 4-vector $\mathbf{x}$ is transformed as $\mathbf{Lx}$, between two inertial reference frames where $\mathbf{L}$ is Lorentz transformation matrix, there exists a dual-vector $\mathbf{x'}$ (a.k.a. the covariant vector) related to $\mathbf{x}$ by a linear transform $\mathbf{\eta}$ s.t. the dot product of $\mathbf{x}$ and $\mathbf{x'}$ remains invariant between the two reference frames. i.e.,

$\mathbf{x^{'T}}\mathbf{x} = (\mathbf{Lx'})^T (\mathbf{Lx})$

The LHS is the dot product between $\mathbf{x}$ and $\mathbf{x'}$ in frame 1 and the RHS is the dot product measured in frame 2. I use the matrix notation in computer science and linear algebra books as that is the one I am comfortable with. But this gives me:

$(\mathbf{\eta x})^Tx = (\mathbf{L \eta x})^T(\mathbf{Lx})$

$\mathbf{x^T} \mathbf{\eta ^ T} x = \mathbf{x^T} \mathbf{\eta ^ T} \mathbf{L^T} \mathbf{L} \mathbf{x} $

requiring $\mathbf{L^T} \mathbf{L} = \mathbf{I}$ or equivalently stated $\mathbf{L^T} = \mathbf{L^{-1}}$ i.e., $\mathbf{L}$ is a rotation matrix. And there are no constraints on $\mathbf{\eta}$. The hope was to derive that $\mathbf{\eta} = \textrm{diag}(-1, 1, 1, 1)$. Also we know that $\mathbf{L}$ is not a rotation matrix so above is saying it is not possible to find a dual-vector s.t. the dot product will be invariant in the two reference frames. Where have I gone wrong?

Answer 1

The problem is over here in the RHS:

$\mathbf{x'^T x} = \mathbf{(Lx')^T (Lx)}$

This is not what is meant when we say that dot product of a vector remains invariant between reference frames. You don't take the dot product of the transformed vector and the transformed co-vector. i.e., don't do $\mathbf{dot(L(x'), L(x))}$. What you do is this:

1. Get the transformed vector $\mathbf{Lx}$
2. Apply $\mathbf{\eta}$ to above get the dual-vector $\mathbf{\eta L x}$
3. Dot 1 and 2

So the RHS is $\mathbf{(\eta L x)^T(Lx)} = \mathbf{x^T L^T \eta^T L x}$. Now let's equate it to LHS to see what we get:

$ \mathbf{x^T \eta^T x} = \mathbf{x^T L^T \eta^T L x} $

requiring that,

$ \mathbf{\eta ^ T} = \mathbf{ L^T \eta^T L } $

Now as exercise, plug in the Lorentz transformation matrix with $c=1$ and verify that

$\eta = \textrm{diag}(-1,1,1,1)$ (the Minowski metric or matrix depending on how you want to look at it :)

is a solution.