# Basis vectors for Minkowski diagram

I have been studying the geometry of special relativity through the Minkowski diagrams. I wanted to derive the figure shown in an article by using simple change of basis methods using the Lorentz transformation $\mathbf{x}'=\Lambda \mathbf{x}$ for observer $S'$ moving at speed $v$ relative to $S$ (in one dimension),

$$\begin{pmatrix} ct' \\ x' \end{pmatrix} = \begin{pmatrix} \gamma_v & -\beta \gamma_v \\ -\beta \gamma_v & \gamma_v\end{pmatrix} \begin{pmatrix} ct \\ x \end{pmatrix}$$

I wanted to define the basis vectors in $S$ to be the Euclidean basis vectors $$\mathbf{s}_t = \mathbf{s}_0 = (0,1) \; \text{and} \; \mathbf{s}_x = \mathbf{s}_1 = (1,0)$$

But under the transformation, $$\mathbf{s}'_0=\Lambda \mathbf{s}_0 = (-\beta\gamma_v,\gamma_v)$$ $$\mathbf{s}'_1=\Lambda \mathbf{s}_1 = (\gamma_v,-\beta\gamma_v)$$

This basis has the same angle $\alpha$ between the two axes as given in the figure, namely $\tan \alpha = \beta$. However, this transformation would make the $S'$ basis have axes that are at an obtuse angle relative to the $S$ basis, whereas most Minkowski diagrams I've seen have $S'$ axes within the $S$ axes at an acute angle. Am I doing something wrong here? Thanks.

Compare the matrix of pure rotation:

\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{pmatrix}

which rotates both axes anti-clockwise by $\theta$.

Now

$$\Lambda=\sqrt{\sec 2\alpha} \begin{pmatrix} \cos \alpha & -\sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{pmatrix}$$

which rotates the $x$-axis anti-clockwise by $\alpha$ whereas the $y$-axis clockwise by $\alpha$.

Further points to be noticed:

• $0\le \alpha < 45^{\circ}$

• $(x',ct')=\sqrt{\cos 2\alpha}(1,0) \iff (x,ct)=(\cos \alpha, \sin \alpha)$

• $(x',ct')=\sqrt{\cos 2\alpha}(0,1) \iff (x,ct)=(\sin \alpha, \cos \alpha)$

• $\det \Lambda=1$

• $\Lambda^{-1}=\sqrt{\sec 2\alpha} \begin{pmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}$

• Great answer, thanks. However, the first matrix you have directly takes the Cartesian unit vectors and rotates them counterclockwise, so that using the matrix from a vector x in S to x' in S' we can construct the new basis vectors for the S' coordinate. Here, however, you're starting with the basis vectors in S'. What is the reasoning behind this? – ChrisL Aug 30 '18 at 3:52
• Basing on your choice of illustration, $S'$ frame is not orthogonal, so the coordinate grids are parallelograms instead of rectangles. It'll be difficult to convert $(x,y)=(1,0)$ to $(x',y')$, etc. Alternatively, we may use the eigenvectors of the Lorentz matrix, which are $x'+ct'$ and $x'−ct'$, as the bases. See another question here. – Ng Chung Tak Aug 30 '18 at 5:36