The bilinear form is:
\(u^TMv\)
The transformations which preserve this are:
\(P^TMP=M\)
If the metric is:
\(M=\begin{bmatrix}-1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\end{bmatrix}\)
Then we have the indefinite orthogonal group \(O(3,1)\)
Where \(n=m\) we have the split orthogonal group.
\(O(n,n,F)\)
The Lorentz group is the \(O(1,3)\) group.
We can do the usual \(3\) rotations, however there are additional \(3\) symmetries, making the Lorzentz group \(6\)-dimensional.
These are the Lorentz boosts.
A symmetry has:
\(t'^2 - x'2 - y'^2 - z'^2 = t^2 - x^2 - y^2 - z^2\)
We consider the case where we just boost on \(x\), so \(y = y'\) and \(z = z'\).
\(t'^2 - x'2 = t^2 - x^2\)
Or with \(c\):
\(c^2t'^2 - x'2 = ct^2 - x^2\)
\(s^2 = t^2 - x^2 - y^z - z^2\)
\(s'^2 = t'^2 - x'^2 - y'^2 - z'^2\)
\(ds^2 = s'^2 - s^2\)
\(ds^2 = (t'^2 - x'^2 - y'^2 - z'^2) - (t^2 - x^2 - y^z - z^2)\)
\(ds^2 = (t'^2 - t^2) - (x'^2 - x^2) - (y'^2 - y^2) - (z'^2 - z^2)\)
\(ds^2 = dt^2 - dx^2 - dy^2 - dz^2\)
boost: \(s^2 = c^2t^2 - x^2 - y^z - z^2\)
we want new t and x where distance is same \(c^2t'^2 - x'^2 - y^z - z^2 = c^2t^2 - x^2 - y^z -z^2\) \(c^2t'^2 - x'^2 = c^2t^2 - x^2\)
We know that both transformations are linear [WHY??], therefore \(x' = Ax + Bt\) \(t' = Cx+Dt\)
we transform to \(x' = 0\). so \(Ax + Bt = 0\)
We define \(v = \dfrac{x}{t}\)
So: \(x = vt\)
We can plug these in: \(Avt + Bt = 0\) \(Av + B = 0\) \(\dfrac{A}{B} = -v\)