# Lorentz transformation matrix and its meaning in Minkowski diagram

I have sucessfully derived the Lorentz matrix for the boost in $x$ direction and its inverse. So I know how to get these two matrices:

$$\Lambda = \begin{bmatrix} \gamma & 0 & 0 & -\beta \gamma\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta \gamma & 0 & 0 & \gamma \end{bmatrix}$$

$$\Lambda^{-1} = \begin{bmatrix} \gamma & 0 & 0 & \beta \gamma\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \beta \gamma & 0 & 0 & \gamma \end{bmatrix}$$

What I lack is a geometrical meaning of those two. I have read, that it is some sort of a rotation, so I have already drawn the Minkowski diagram for steady observer 1 and one for the moving observer 2 as observer 1 sees it. Examples are drawn for $\beta = 0.5$. I have noticed that:

• base cell is a square (picture 1) for a steady observer and a parallelogram (picture 2) for a moving observer
• light travels only diagonals of base cells
• axis $x'$ has coefficient $\beta$
• axis $ct'$ has coefficient $1/\beta$

question 1: Is there a way I can help myself with what I have figured out so far to show, how the Lorentz matrix deforms a square into a parallelogram and vice versa?

question 2: I have heard of the invariant quantity $\Delta s$ in relativity for which I have stumbled upon an equation $\Delta s^2 = \Delta x^2 - c \Delta t^2$. Is it possible $\Delta s$ is an area of a base cell?

question 3: Why is it so necessary for a scalar product to be defined with a minus sign for Minkowski spacetime?

picture 1: picture 2: Thank you.

Your questions 1 and 2 can be answered if you consider one more geometrical aspect of the group of the one-dimensional Lorentz bosts and that is the orbit of a single point. In other words, the set obtained by transforming one fixed point, say, (0, 1), by Lorentz boosts of all possible $\beta$'s. If you would apply several transformations successively, this is a trajectory the point would follow and never leave, so we have an important invariant.
You should be getting one arm of a hyperbola with asymptotes given by the boundaries of a light cone. The interval $\Delta s$, measured from the reference point in the origin, indexes the possible hyperbolas: it is their semi-major axis. This helps to understand why $s$ is defined the way it is, its geometrical meaning and its conservation under boosts.
The semi-axis of a hyperbola going through a given point is also one half of its so-called hyperbolic coordinates. The other component is the hyperbolic angle, which, unlike polar angle, can reach any value in $\mathbb R$. Any Lorentz boost is just a shift of the hyperbolic angle—it adds a constant $\kappa$ to the h.a. of any transformed point. This constant can be found in the boost matrix if you rewrite it like $$\begin{pmatrix} \cosh{\kappa} & 0 & 0 & -\sinh{\kappa} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\sinh{\kappa} & 0 & 0 & \cosh{\kappa} \\ \end{pmatrix}.$$ (This is always possible as $\gamma^2 - (\gamma\beta)^2 \equiv 1$.) All these facts, together with the similarity of the above matrix with a rotational matrix, justify calling boosts "some sort of" rotations in the $(t, x)$ plane.
• One question. What is $\kappa$ in terms of hyperbola? – 71GA Oct 21 '12 at 11:01
• It works as a means of parametrizing points on the hyperbola, namely in your case, the farthest vertex of the parallelogram. Its meaning is depicted in this picture: phil4.com/images/hyperbolic_functions.jpg (called $\varphi$ in the link). The object is filled because $\varphi$ not only serves as an argument to the hyperbolic cosine and sine but also corresponds to the blue area. – The Vee Oct 21 '12 at 14:27
• Is $\kappa$ an area? Is it possible to draw it like a normal angle? – 71GA Nov 3 '12 at 19:03