# Derive Minkowski metric from Lorentz transformation

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?

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.