# Minkowski-spaces/Special relativity theory/Section
## Definition

A real vector space of dimension $n$, endowed with a bilinear form $\left\langle -,-\right\rangle$ of type $(n-1,1)$, is called a
Minkowski-spaces provide a simple model for special relativity theory, we talk also about an Einstein-Minkowski-space, and the bilinear form on it is also called Minkowski-form or Lorentz-form. The classical space-time-world is of the form $\mathbb {R} ^{3}\times \mathbb {R}$, where the three-dimensional component represents the space, and the one-dimensional component represents time. In this world, every motion of a point to another point is possible, as long as the second point is later than the first point. Accordingly, the points in a four-dimensional Minkowski-space represent the relativistic world points (the events); however, the separation into space and time depends on the observer; the theory yields also a definition of what an observer is; see below. Of special importance is the set of vectors
    ${\left\{v\in V\mid \left\langle v,v\right\rangle =0\right\}};$
this set is called the light cone. This means the set of all light rays which start (or end) in a world point. This light cone is, according to special relativity theory, independent of any observer (an absolut notion), and exactly this is modelled the Minkowski-spaces. We allow every dimension, because many important phenomena are already visible for $n=2,3$. The Minkowski-Form, given on $\mathbb {R} ^{n}$ with respect to the standard basis by the Gram matrix
    ${\begin{pmatrix}1&0&\cdots &0&0\\0&1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&0\\0&0&\cdots &0&-1\end{pmatrix}},$
is called Minkowski-standard-form. Due to Sylvester' law of inertia, every Minkowski-form can be brought to this form with respect to a suitably scaled orthogonal basis (called a Minkowski-basis).

## Definition

Let $V$ be a Minkowski-space, endowed with the Minkowski-form $\left\langle -,-\right\rangle$. A vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle =0\,$
is called lightlike, a vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle <0\,$
is called timelike, and a vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle >0\,$
Caution! These properties do not define linear subspaces, the sum of two spacelike vectors is in general not spacelike again.
In this space-time-light-world, not all vectors (or linear motions) are realizable by a (material) observer; to the contrary, the following restriction is a substantial part of this world model.

## Definition

Let $V$ be a Minkowski-space, endowed with a Minkowski-form $\left\langle -,-\right\rangle$. A vector $v\in V$ fulfilling
    $\left\langle v,v\right\rangle =-1\,$
is called Observer vector or
We can imagine a person being the observer, but this does not mean any kind of subjectivity. The observer has a clock, a meterstick, and a protractor, and every observer doing the same motion obtain the same measurements. Instead of using the condition $=-1$, an observer vector is often defined using the condition $=-c$, where $c$ denotes speed of light. This is just a rescaling.
The observer vectors are in particular timelike, since every observer grows older; time is also running for an observer "not moving in space“. The line $\mathbb {R} v$ is a linear subspace of dimension $1$; the bilinear form restricted to is becomes negative definite. Let $U\subseteq V$ be the linear subspace orthogonal (with respect to the Minkowski-form)to this line. This is a three-dimensional space, and the bilinear form restricted to it becomes positive definite. This linear subspace is the space $V_{v}$ for this observer (or $V_{B}$, if $B$ denotes the observer), and $\mathbb {R} v$ is the time axis of the observer. For an observer, we have a decomposition of the total world of the form $\mathbb {R} \times \mathbb {R} ^{3}$; but this decomposition depends on the observer. This decomposition is called the reference frame of the observer. The positive definite restriction of the Minkowski-form to his space component is an inner product; with this, the observer can measure lengths and angles, and he can also fix an orthonormal basis. In particular, for an observer with an allowed four-velocity $v$, there exists an orthogonal basis $e_{1},e_{2},e_{3},v$ with
    $\left\langle e_{j},e_{j}\right\rangle =1\,$
and
    $\left\langle v,v\right\rangle =-1\,.$
With respect to such a Minkowski-basis, the Minkowski-form is described simply by the Gram matrix
    ${\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{pmatrix}}.$
In this model, a substantial amount of relativistic phenomena is exhibited if we consider a base change between two such bases (a change of reference frame), the essential point being the change in the decomposition into space component and time component.
If $v$ is an observer vector, then, by definition, also $-v$ is an observer vector. This observer moves into the opposite time direction. The set of all observer vectors is divided into two sheets; we define one to by the future sheet. In the same way, the light cone is divided into two cones, the future light cone and the past light cone (we choose one to be the future light cone).

## Lemma

Let $V$ be a Minkowski-space, endowed with a Minkowski-form
$\left\langle -,-\right\rangle$
* For every observer vector $v\in V$, we have a direct sum decomposition where the restriction of the Minkowski-form to $\mathbb {R} v$ is negative definite, and where the restriction of the Minkowski-form to $V_{v}=(\mathbb {R} v)^{\perp }$ is positive definite. Here, $V_{v}$ consists of spacelike vectors.
    $V=\mathbb {R} v\oplus (\mathbb {R} v)^{\perp }\,,$
* For two observer vectors $v,w\in V$ from the same half cone, we have
    $\left\langle v,w\right\rangle <0\,.$
* For a timelike vectors $v,w\in V$, we have
    $\left\langle v,w\right\rangle ^{2}\geq \left\langle v,v\right\rangle \cdot \left\langle w,w\right\rangle \,.$

### Proof

exercise, exercise, and exercise.
$\Box$
The condition, that all velocities of an observer fulfill $\left\langle v,v\right\rangle =-1$, is a strong restriction on the set of the possible motions. If a Minkowski-basis is fixed, then ${\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}$ is an observer vector if and only if
    $y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-t^{2}=-1\,$
(and $t\geq 0$, this is enforced by the future direction) holds.

## Example

In a four-dimensional standard-Minkowski-space, we want to move unformly from the point $P={\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\\r\end{pmatrix}}$ to the point $Q={\begin{pmatrix}q_{1}\\q_{2}\\q_{3}\\s\end{pmatrix}}$. In the classical framework, we would just take the vector
    ${\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}:={\begin{pmatrix}q_{1}\\q_{2}\\q_{3}\\s\end{pmatrix}}-{\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\\r\end{pmatrix}}\,.$
However, this is in general not an observer vector, and then the looked-for motion is not realizable. If $y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-t^{2}$ is negative, which means that we have a timelike vector, then we can rescale this vector to obtain an observer vector
    ${\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\\u\end{pmatrix}}={\frac {1}{\sqrt {-\left\langle {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}},{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}\right\rangle }}}{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}\,.$
Then, the mapping
    $x\longmapsto {\begin{pmatrix}p_{1}\\p_{2}\\p_{3}\\r\end{pmatrix}}+x{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\\u\end{pmatrix}}$
describes a motion, which, for $x=0$, starts at the point $P$, and ends, for $x={\sqrt {-\left\langle {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}},{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\t\end{pmatrix}}\right\rangle }}$, at the point $Q$, and which is realizable in the physics world.

## Remark

For a four-velocity $v$ of an observers $B$ with the decomposition
    $V=V_{v}\oplus \mathbb {R} v\,,$
the set of points (events) of the form $sv+V_{v}$ with a fixed $s\in \mathbb {R}$ is called the space at the time $s$. The points from this set are called simultaneous for the observer $B$. For another observer $C$ with the four-velocity $w$, these points are not simultaneous. The concept of simultaneity of the observer $A$ depends on his decomposition of the Welt $V$ into space component and time component. If, for example, the four-velocity of the second observer is given, with respect to a Minkowski-basis of the first observer, by ${\begin{pmatrix}{\frac {3}{4}}\\0\\0\\{\frac {5}{4}}\end{pmatrix}}$, then
    ${\frac {15}{4}}{\begin{pmatrix}{\frac {1}{3}}\\0\\0\\{\frac {1}{5}}\end{pmatrix}},{\begin{pmatrix}0\\1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\\0\end{pmatrix}}$
is an orthonormal basis of the space component of the second observer. For the first observer, the events ${\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}$ and ${\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}$ are simultaneous, but they are not simultaneous for the second observer, because the first vector has the same description, and the second vector equals
    ${\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}={\frac {75}{16}}{\begin{pmatrix}{\frac {1}{3}}\\0\\0\\{\frac {1}{5}}\end{pmatrix}}-{\frac {3}{4}}{\begin{pmatrix}{\frac {3}{4}}\\0\\0\\{\frac {5}{4}}\end{pmatrix}}\,.$
The time component of the second event with respect to the second observer vector is $-{\frac {3}{4}}$.
We compare now velocities of different observers.

## Definition

Let $V$ denote a Minkowski-space, and let $B$ and $C$ denote observers with the four-velocities $v$ and $w$. The vector
    $v_{BC}=-v-{\frac {1}{\left\langle v,w\right\rangle }}w\,$
is called the velocity vector of $C$ relative to $B$. The number
    $\rho _{BC}={\sqrt {1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}}}\,$
is called the relative velocity
The relative velocity vector is a vector. Note that, due to fact (3), we have $\left\langle v,w\right\rangle \geq 1$; therefore, the relative velocity is a well-defined nonnegative real number, bounded from above by $1$. The relative velocity is symmetric in $v$ and $w$; however,
    $v_{CB}=-w-{\frac {1}{\left\langle v,w\right\rangle }}v\,$
is, in general, different from $v_{BC}$.. Since the speed of light was normalized to be $1$, one should imagine this relative velocity to be quite small. In case $v=w$, the relative velocity equals $0$.

## Lemma

Let $V$ be a Minkowski-space, and let $B$ and $C$ denote observers from the same time sheet with den four-velocities
$v$
$w$
* The relative velocity vector $v_{BC}$ is orthogonal to $v$.
* The relative velocity vector $v_{BC}$ is spacelike, and we have
    $\Vert {v_{BC}}\Vert =\Vert {v_{CB}}\Vert =\rho _{BC}=\rho \,.$
* We have
    $\left\langle v,w\right\rangle =-{\frac {1}{\sqrt {1-\rho ^{2}}}}\,.$
* We have and this is the decomposition of $w$ into the space component and the time component of $B$.
    $w={\frac {v}{\sqrt {1-\rho ^{2}}}}+{\frac {v_{BC}}{\sqrt {1-\rho ^{2}}}}\,,$
* The time coefficient of $w$ with respect to $B$ is ${\frac {1}{\sqrt {1-\rho ^{2}}}}$.

### Proof

* we have so these vectors are orthogonal to each other. Therefore, $v_{BC}$ belongs to the space component of $B$.
    $\left\langle v,v_{BC}\right\rangle =\left\langle v,-v-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle =\left\langle v,-v\right\rangle +\left\langle v,-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle =1-1=0\,,$
* We have By part (1) (or because of fact (3)), this expression is non-negative. The square root of this number is the relative velocity $\rho$.
    ${\begin{aligned}\left\langle v_{BC},v_{BC}\right\rangle &=\left\langle -v-{\frac {1}{\left\langle v,w\right\rangle }}w,-v-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle \\&=\left\langle v+{\frac {1}{\left\langle v,w\right\rangle }}w,v+{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle \\&=\left\langle v,v\right\rangle +2{\frac {\left\langle v,w\right\rangle }{\left\langle v,w\right\rangle }}+{\frac {1}{\left\langle v,w\right\rangle ^{2}}}\left\langle w,w\right\rangle \\&=1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}.\end{aligned}}$
* This follows directly from the definition using
    $\rho ={\sqrt {1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}}}\,,$
    $\left\langle v,w\right\rangle <0\,.$
* From and (3), we can infer Because of part (1), the vector $v_{BC}$ belongs to the space component of $B$.
    $v_{BC}=-v-{\frac {1}{\left\langle v,w\right\rangle }}w\,$
    $w=-\left\langle v,w\right\rangle v-\left\langle v,w\right\rangle v_{BC}={\frac {v}{\sqrt {1-\rho ^{2}}}}+{\frac {v_{BC}}{\sqrt {1-\rho ^{2}}}}\,.$
* From (4) we can see directly that the time cefficient of $w$ with respect to $B$ equals ${\frac {1}{\sqrt {1-\rho ^{2}}}}$.
$\Box$
The principle formulated in the fifth statement of the preceding lemma is called time dilation. An observer measures for another observer a longer time than the observed person does in his reference frame.