# Minkowski Diagrams and Important Basics You Should Know

## Table of contents

- Equilocal and simultaneous events in spacetime Here you will learn what spacetime events are and when two events occur at the same position or time.
- World lines - as stories of the particles Here you will learn what world lines are in a Minkowski diagram, what the world line of a photon looks like.
- Understanding Relativistic Velocity Addition with the Minkowski Diagram Here you learn why there can be no relative velocity greater than the speed of light.
- World planes: World lines of extended objects Here you will learn how to represent extended bodies in a Minkowski diagram.
- Coordinate system of a moving observer Here we construct a coordinate system for a moving observer, learn about a light rectangle and unit hyperbola.
- Example: Time dilation Here we look at time dilation in a spacetime diagram.
- Example: Length contraction Here we look at length contraction in a spacetime diagram.
- Light cones in two-dimensional Minkowski diagrams In two-dimensional Minkowski diagrams, photon world lines represent so-called light cones.
- Space-like, time-like and light-like events Here you will learn when two events are space-like, time-like, or light-like to each other using a 2d spacetime diagram.

Minkowski diagrams can be used to illustrate the phenomena of special relativity (SRT). The Einstein's theory is based on the **principle of relativity** and the constancy of the **speed of light**. You will therefore also encounter these principles in the Minkowski diagrams.

A Minkowski diagram is first of all a space-time diagram. A spacetime diagram has a **time axis** \(c\, t\) and a **spatial axis** \(x\), if we consider a one-dimensional motion. To describe a motion in a plane, two spatial axes (\(x\), \(y\)) are used. We first look at the one-dimensional motion.

In Minkowski diagrams, the convention is that time is plotted on the *vertical* axis and position is plotted on the *horizontal* axis (see Illustration 1). We also write on the time axis not \(t\), but \(c\,t\). Why? You will understand soon.

The spacetime diagrams (time-position coordinate systems) will represent any *observers* in the following (we can also call them *systems*). For example, an observer can be a stationary spacecraft. Such a stationary observer, we call a **system at rest** and assign a *orthogonal* coordinate system to it (see illustration 1). Of course, you could also declare any oblique coordinate system to be the system at rest. But why should we make the problem unnecessarily complicated? We will not do that here.

## Equilocal and simultaneous events in spacetime

A point in the Minkowski diagram is determined by a position and time value and is called an **event**. An event thus indicates *when* and *where* something takes place.

So that the respective event is unique, you must refer the given time and position to a *determined coordinate system*. So if you specify an event, you must also say in which coordinate system you measured the time and position. Only then we are able to look at an event from *different* "perspectives" (coordinate systems) and to compare position and time values with each other. You should keep that in mind!

### Equilocal events: Parallels to the time axis

Consider a system at rest represented by a two-dimensional orthogonal coordinate system. Let's choose any location \( \class{blue}{x_1} \) to the right of the system at rest. At this location the rest system sees three explosions \( \class{blue}{E_1} \), \( \class{blue}{E_2} \) and \( \class{blue}{E_3} \) passing in succession.

All three events happen from the point of view of the rest system at the *same* location, but at different times. Such events, which happen for the rest observer at the same location, are called **equilocal events**.

If you connect equilocal events, you get a *vertical* straight line. The time axis of the rest observer, for example, is also such a vertical straight line. It is the sum of all events which have taken place, are taking place and will take place at the location \( x = 0 \). The time axis is the history, i.e. past, present and the future of the position \( x = 0 \), where the rest observer is located.

### Simultaneous events: Parallels to the spatial axis

A rest observer sees two events \(\class{red}{E_0}\) and \(\class{red}{E_1}\) happen at time \( c\, \class{red}{t_1}\). A little later, at time \(c\, \class{blue}{t_2}\), he sees three other events \(\class{blue}{E_2}\), \(\class{blue}{E_3}\) and \(\class{blue}{E_4}\) happen.

If events do not take place parallel to the time axis, but parallel to the spatial axis, then they take place *simultaneously* relative to the rest observer. Such events are called **simultaneous events**. For example, the events \(\class{blue}{E_2}\), \(\class{blue}{E_3}\) and \(\class{blue}{E_4}\) are simultaneous events.

If you connect simultaneous events with each other, then you get a *horizontal* straight line. The spatial axis of the system of rest is for example also such a horizontal straight line. All events on the spatial axis happen simultaneously at the time \( c\, t = 0 \).

## World lines - as stories of the particles

The motion, but also the mere existence of a particle or another body considered as point-like (for example spaceship) generates a so-called **worldline** in the Minkowski diagram. It represents the trajectory of that body. The world line is, so to speak, a sequence of all events that happen to this body as time goes by.

### Examples of different world lines

In the illustration 4 seven different world lines are shown, which belong to different bodies (particles, observers, ...). Only one \(x\)-axis is drawn, which is why only one-dimensional movements are possible: i.e. movements either to the left or to the right. The black coordinate system in illustration 4 belongs to the rest observer. This rest observer sees different bodies moving differently:

**World line 1**- this body moves with constant speed away from the body to which the world line 2 belongs. It is the only body that moves to the left.**World line 2**- this body is like the rest observer at rest, at a certain location. For him the time passes, however, his location remains the same. This body has a fixed distance to the rest observer.**World line 3**- this body moves with constant speed away from the body to which the world line 2 belongs. According to its world line, it moves to the right like bodies 4, 5 and 6.**World line 4**- this body has the same speed as the body to which world line 3 belongs. Bodies 4 and 3 move to the right at a fixed distance from each other.**World line 5**- this body is quite fast and accelerates almost to the speed of light. At the event S it meets with the body 6 at a certain time at a certain location.**World line 6**- this body flies almost with speed of light and decelerates until it stops at some point, because the world line becomes a vertical straight line with the time. At the event S it meets the body 5.**World line 7**- this body oscillates back and forth from right to left. It accelerates to high speed and decelerates at the reversal points (at the tops of the hills and valleys) to change the direction of motion.

### World lines of light particles

The **speed of light** is perceived the same in every reference frame. No matter whether the reference frame moves or not. The velocity of the light particles (photons) is always *constant*. This is one of the two postulates on which special theory of relativity is based.

For this reason, we choose the scales of the two axes in the Minkowski diagram in such a way that the world line of a photon is a **angle bisector of the space and time axes**. For the photon world line to always be such a bisector, three things must be satisfied:

Do not plot \( t \) but \( c \, t \) on the time axis. Thus you scale the time with the speed of light \( c \). The \( c\,t\) axis has now the dimension of a length, just like the \(x\) axis!

Both axes must be in the same unit of length. For example, light years must be entered on both axes, but not light years on one axis and light seconds on the other axis.

Both axes must be scaled equally. For example, if one light year on the \(ct\) axis corresponds to one centimeter on paper, then one light year on the \(x\) axis must also correspond to one centimeter on paper.

Now, if a photon is emitted at time \(c\,t = 0 \) from location \( x = 0 \) in the positive \(x\) direction (to the right), then the world line of this photon will always enclose a 45 degree angle with the time axis and the spatial axis.

The world line of a photon is never bent in any way because of the constant speed of light, but always straight. But it can be very well in zigzag shape, if for example the photon is reflected back and forth at any mirrors. However, the zigzag shape cannot be arbitrary. The "reflected world line" must have exactly an angle of inclination of 45 degrees to the space and time axis. This angle is guaranteed in any case if at the refection points the world line is reflected by 90 degrees.

### Examples of world lines of particles with different velocities

The speed of light is the maximum speed in the universe and can be reached only by massless photons. Therefore the world lines of objects with a mass can never be flatter than the world line of a photon! Their speed can approach the speed of light, but never reach it or even exceed it.

In illustration 6 you can see different world lines, which are inclined differently. Let's assume that these world lines belong to different spaceships. From the point of view of the stationary spaceship 1 (black coordinate system) all spaceships start at the time \( c\,t = 0 \) and all at the position \(x = 0\). From there they all move to the right, but with different velocities:

**Spaceship 2**- has a very steep world line, it moves very slowly to the right.**Spaceship 3**- moves faster than spaceship 2 to the right.**Spaceship 4**- moves faster than spaceship 3 and at about half the speed of light. (Note: Half speed of light does not mean that you should halve the angle of the world line of light (22.5 degrees). This is wrong, if you look at the relation2

between the angle and the speed).**Spaceship 5**- moves almost at the speed of light.**Spaceship 6**- moves at faster than the speed of light and is therefore not allowed.

## Understanding Relativistic Velocity Addition with the Minkowski Diagram

What if two spaceships are moving away from each other at 60% of the speed of light? Is their relative velocity then greater than the speed of light? In relativistic mechanics, it is not possible to get a relative velocity that is greater than the speed of light. Let's illustrate this with a Minkowski diagram.

We assign an orthogonal coordinate system to the earth (or to a rest observer on the earth). Two spaceships start from the earth and move away from with 60% of the speed of light in earth's reference frame. One spaceship moves to the left and the other one to the right. A naive velocity addition would show that the spaceships move away from each other with 120% of the speed of light from the point of view of one of the spaceships:

**Incorrect speed addition**$$ \begin{align} 0.6 \, c + 0.6 \, c = 1.2 \, c \end{align} $$

This is a naive classical velocity addition, which is only approximately valid at small relative velocities. Here, however, we have a very large relative velocity, therefore you must calculate relativistically here.

Let one of the spaceships send a light signal to the other spaceship at any time. The world line of this light signal is that of a photon, i.e. a shifted 45° straight line (see illustration 7). The light signal will arrive at the other spaceship *in any case*. At which time and at which location? Draw the photon straight line simply up to the intersection with the world line of the other spaceship. The intersection is the location and time where the light pulse arrives.

Since the signal travels at the speed of light and it eventually catches up with the other spacecraft moving away, it means that the second spacecraft is moving away from it slower than the speed of light. So it can't move away from the other spaceship at 120% of the speed of light at all. Otherwise the light signal would never arrive at the second spaceship!

You can calculate the correct **relative velocity** \(u\) between the two spaceships using special relativity as follows:

**Formula: Example for relativistic velocity addition**

&~=~ \frac{0.6\,c ~+~ 0.6\,c}{1 ~+~ \frac{ 0.6\,c ~\cdot~ 0.6\,c}{c^2}} \\\\

&~=~ \frac{1.2}{1 ~+~ 0.6 \cdot 0.6} \, c \\\\

&~=~ 0.88 \, c \end{align} $$

Here \(v_1\) is the velocity of the first spaceship and \(v_2\) the velocity of the second spaceship from the view of the terrestrial rest observer. The actual relative velocity of both spaceships is 88% and not 120% of the speed of light!

## World planes: World lines of extended objects

Up to now, you have only known world lines of bodies idealized as point-like or very very small by nature. However, every body is in reality not point-like, but *spatially extended*. That is, a body is made up of many particles occupying many locations at the same time, and we combine all of these particles into a single extended body.

Each of these particles generates its own world line in the Minkowski diagram. If you combine them all, you get a **world plane** - as a sequence of many world lines. This body could be for example a long thin rod.

A

*resting*rod of length \( L_1 \) generates a world surface which is parallel to the time axis. The world surface is a rectangle of width \(L_1\) which becomes longer with time along the time axis.A

*moving*rod, on the other hand, also produces a rectangular world surface, but it is tilted either to the left or to the right, depending on whether the rod is moving to the left or to the right.

## Coordinate system of a moving observer

The coordinate system of an observer at rest is orthogonal. How does the coordinate system of a *moving* observer look in comparison? Let's assume: You are the rest observer who owns a black orthogonal coordinate system (see illustration 9). Next to you (in the coordinate origin) is a spaceship XxX at the time \( t = 0 \). There you are at the same time at the same location. Now the spaceship starts to fly unaccelerated to the right. Thereby it creates a world line in your coordinate system, which runs tilted to the right.

### Construct time axis of the moving observer

According to the principle of relativity, the spaceship XxX can consider itself to be at rest. The spaceship has its own coordinate system \( (c \, t_{\text R}, x_{\text R}) \) and it is located in the coordinate origin \( x_{\text R} = 0 \). The world line of a rest observer XxX located at the origin runs along its \(c \, t_{\text R}\) time axis (see illustration 9). The spacecraft's rest \( (c \, t_{\text R}, x_{\text R}) \)-system has an tilted time axis from the point of view of the \( (c \, t, x) \)-system.

### Construct spatial axis of the moving observer

The time axis of the spacecraft XxX is tilted by a certain angle from the point of view of the black coordinate system.

*What does the spatial axis of the spaceship XxX look like?*

You have learned that *simultaneous* events in the Minkowski diagram must be *parallel to the spatial axis*. We take advantage of this. In addition to spaceship XxX, we take two more identical spaceships: left spaceship LR and right spaceship RR. All three fly in the same direction with the same speed. The two new spaceships keep an equal distance to XxX.

If we now consider the spaceship XxX to be at rest according to the principle of relativity, then also from its point of view the two spaceships are at rest.

The spacecraft XxX sends two light signals in opposite directions to the spacecraft at time \( t_{\text R} = 0 \).

Both light signals will arrive at both spaceships *simultaneously*. Why simultaneously? Because the LR and RR spaceships are at the same distance from XxX. So we have two simultaneous events:

\( E_{\text{LR}} \):"

*Light signal has arrived at the left spacecraft*"\( E_{\text{RR}} \):"

*Light signal has arrived at the right spaceship*"

Connect the two events \( E_{\text{LR}} \) and \( E_{\text{RR}} \). This connection line represents a straight line on which only simultaneous events take place from the point of view of XxX. Simultaneous events are always parallel to the spatial axis of the system for which the events take place simultaneously. Thus, the connecting line is parallel to the spatial axis of the spaceship XxX.

Just move the connecting line into the origin to get the spatial axis of the coordinate system of XxX. So the spaceship XxX has its own spatial axis (let's call it \( x_{\text{R}} \)-axis) and it is not allowed to use the \( x \) axis!

### Simultaneity and Equilocality

After you have constructed the coordinate system of the moving system, you can observe a phenomenon of the relativity theory, namely:

*Simultaneous events for an observer at rest are not simultaneous for an observer in motion!*

Let's do a little example of simultaneity:

Take two events that happen simultaneously in the black coordinate system, i.e. lie on a parallel to the spatial axis.

Now look whether the two events also happen simultaneously in the moving system (spaceship XxX). You will find out that this is not the case. Why? Because their connecting line is

*not parallel*to the spatial axis of the moving system.

The same is true for equilocality. The lines of simultaneity and equilocality of the two systems do not coincide.

### Light rectangle and angle between spatial and time axes

The angle between the light world line and the \(c\,t\) time axis is 45 degrees. The angle between the light-world line and the \(x\) axis is also 45 degrees.

*By what angle is the location and time axis of the moving coordinate system inclined with respect to the stationary coordinate system?*

Let's reflect the two light pulses emitted from the coordinate origin to the two spaceships RR and LR back to the spaceship XxX. The two reflections represent the events \( E_{\text{LR}} \) and \( E_{\text{RR}} \). The reflected light pulses arrive again simultaneously at XxX.

By emitting and reflecting both light pulses, a **light rectangle** has been created in the space-time diagram. It is literally a rectangle, because the light world lines enclose a 90 degree angle "at the corners".

The *steeper diagonal* in the light corner, i.e. the \( c\,t_{\text{R}} \) axis, encloses with the \(c\,t\) axis the same angle as the \(x_{\text R}\)-axis and the \(x\)-axis. The light represents not only an angle bisector of the axes of the resting system, but also of the axes of *moving* observers!

The enclosed angle \(\alpha\) between the time axes of the stationary and moving observer is opposite cathetus \( x = v\,t\) divided by adjacent cathetus \(c\,t\):

**Enclosed angle is velocity divided by speed of light**

&~=~ \frac{v}{c} \end{align} $$

From the formula 2

and the fact that the two angles are equal, you can conclude: The faster the spaceship moves from the point of view of a resting observer, the closer the space and time axis of the spaceship approach the light world line; until they *collapse* in the case of the light-fast spaceship. Do space and time merge into one dimension? Does everything in the universe happen simultaneously for a photon? What do you think?

From the equation 2

you can also see that the larger the velocity \( v \) of an observer, the larger the angle \( \alpha \) becomes; until the velocity reaches the speed of light \( c \) - then the quotient is:

**Ratio of velocities is equal to one**

The angle \(\alpha\) is then: \( \alpha ~=~ 45^\circ \) resp. \( \alpha ~=~ \frac{\pi}{4} \) (because of: \( \arctan(1) ~=~ \frac{\pi}{4} \)).

### Scaling of the time axis of the moving observer

So far, we have only constructed the *alignment* of the spatial and time axes of the moving system. Now we still have to care about the *scaling* of the axes to get a complete coordinate system of a moving observer. After all, you cannot be sure whether one second in the moving system corresponds to one second in the resting system. So we want to clarify the following question:

*How many time units in the moving system correspond to one time unit in the rest system?*

According to special relativity, the following **distance squared** \( s^2 \) remains the same in any system:

**Distance squared from SR**

Distance squared \( s^2 \) is the squared time coordinate minus the squared space coordinate.

Equation 4

for the distance squared corresponds to an **equilateral hyperbola**. The equation for a general hyperbola is:

**Equation of a general hyperbola**

In our case, \( y ~=~ c\,t \) and \(a\) and \(b\) are equal, resulting in *equilaterality*. In the following we consider only the case \( s ~=~ 1 \), which corresponds to a **unit hyperbola**:

**Equation for a unit hyperbola**

Now draw spaceships with different velocities and see where their world lines intersect the unit hyperbola. The distance from the origin to the point of intersection corresponds to a time unit for the respective clock inside the spaceship. And the faster the spaceship is, the more stretched a time unit becomes in this spaceship.

If you send two light pulses from the origin to the intersection point in such a way that they are reflected at two places and arrive at the intersection point at the same time, then you get *area-equal light rectangles*. If you apply the binomial formula to the distance square 6

, you get the product of two expressions that can be interpreted as the side lengths of a rectangle:

**Factored equation of the unit hyperbola**

The product itself then results in the area of this drawn light rectangle (see illustration 14). And this area must be the same in every system, because it corresponds to the rewritten \(s^2\).

### Scaling of the spatial axis of the moving observer

You can not only draw a unit hyperbola for equal times in different systems, but also for equal locations! The hyperbola equation is exactly the same, with the only difference that the distance square is not \(s^2 = 1 \), but \( s^2 = -1 \):

**Unit hyperbola for the scaling of the spatial axis**

Here you can also draw spatial axes of different systems and look at the intersection points with the hyperbola. Space axes of systems moving at different speeds are scaled differently: The faster the system moves, the more stretched a unit of length in this system becomes.

Now, in addition to the *orientation* of the axes of the moving system, you have also figured out the *scaling* of the axes. As you can see, the scaling is different for a moving system than a resting one.

## Example: Time dilation

With the acquired knowledge you can now illustrate the most important phenomena of special relativity! Let us now compare how much time has passed on the clock of a rest observer (system A) after one second has passed on the clock of a moving observer (system B). For the graphical representation of time dilation in a Minkowski diagram, we exploit the *lines of simultaneity*. These are, as you know, parallels to the spatial axis.

**From the point of view of System A** (see Illustration 16):

In the system A the **event** E takes place at the time \(c\,t_{\text A} = 1 \) second. Draw the line of simultaneity (parallel to the \(x_{\text A} \)-axis) through the event E. Consider the intersection of this line with the \(c\,t_{\text B}\) time axis of the moving system B. Let us denote the intersection as event E'. Thus, system A observes that event E' and thus E in system B has not yet occurred. Time in system B seems to pass more slowly for system A.

**From the point of view of system B** (see illustration 17):

Let us now consider an **event** E occurring at time \(c\,t_{\text B} = 1 \) second from the point of view of system B. Draw the line of simultaneity (parallel to the \(x_{\text B} \)-axis) through the event E. Consider the intersection of this line with the \(c\,t_{\text A}\) time axis of system A. Let us denote the intersection as event E'. Thus, system B observes that event E', and hence E, has not yet occurred in system A. Time in system A seems to pass more slowly for system B.

As you can see, the time dilation is a symmetrical effect. The moved clock goes slower. But, which clock is moved, depends on the chosen system (observer).

## Example: Length contraction

Besides time dilation, special relativity predicts **length contraction**. For the graphical representation of length contraction in a Minkowski diagram, we exploit the *lines of equilocality*. These lines are, as you know, parallels to the time axis.

**From the point of view of System A** (see Illustration 18):

A unit hyperbola was used to determine the scaling of the spatial axes for system A and B. A rod in system A has the length 1 meter. (By the way: For a measurement of the length to be correct, the two locations of the rod ends must be measured at the same time. Everything that happens for a system at the same time is on parallels to its spatial axis. In our case, at the time \(c\,t_{\text A} = 0\) we measure the two rod ends and get 1 meter). The event E marks the end of the rod. Draw a vertical line through the event E. Thus you find the intersection point E' with the \(x_{\text B} \)-axis. From the point of view of A the rod seems to be contracted in the moving system B.

**From the point of view of system B** (see illustration 19):

A rod in system A has the length 1 meter. The event E marks the end of the rod. Draw the line of equality through the event E. With this you find the intersection point E' with the \(x_{\text B} \)-axis. From the point of view of A the rod seems to be contracted in the moving system B.

As you see: Also the length contraction is a symmetrical effect like the time dilation and does not mark a "special" observer.

## Light cones in two-dimensional Minkowski diagrams

To be able to describe movements in a plane, you need another spatial coordinate \(y\) in addition to the \(x\) spatial coordinate. In a one-dimensional spacetime diagram, light could propagate only to the right or to the left. In a two-dimensional spacetime diagram, however, the light can propagate in any direction in a plane. The angle bisectors can now be rotated rotationally symmetrically around the time axis. This creates a **light cone** that is open to the top. A light cone opened downwards results analogously with the angle bisectors below the \(x\)-axis. The world lines of light, which enclose a 45 degree angle with the axes, now form the edge of the light cone.

We can draw such a light cone for arbitrary events (points in the space-time diagram). The shown light cone (illustration 21) refers to an event in the origin.

The light cone above the origin represents the

*future*light cone.The light cone below the origin represents the

*past*light cone.

We will see in a moment, which role this light cone plays, if we add further events beside the event in the origin.

## Space-like, time-like and light-like events

After drawing the light cones for the event \( \class{red}{E_0} \) at the origin, we consider three other events \( \class{blue}{E_1} \), \( \class{green}{E_2} \) and \( \class{brown}{E_3} \) in spacetime, which are located in different regions with respect to \( \class{red}{E_0} \): Inside the light cone, outside the light cone and exactly on the shell of the light cone.

For example, imagine the following events:

\( \class{blue}{E_1} \) is: "

*A star explodes*".\( \class{green}{E_2} \) is: "

*Black hole devours a star*".\( \class{brown}{E_3} \) is: "

*A moon crashes into a planet*".\( \class{red}{E_0} \) is: "

*Three people in three spaceships start their flights to the three mentioned events at the same speed to experience them nearby*".

Next, we ask ourselves:

*Which of the spaceships will arrive in time for the respective event?*

Spaceship 1, which started its flight to \( \class{red}{E_0} \) at event \( \class{blue}{E_1} \), will never reach event \( \class{blue}{E_1} \) in time, because it is

*outside*its light cone. The connection of the two events (\( \class{red}{E_0} \) and \( \class{blue}{E_1} \)) would result in a world line which would be flatter than the world line of a light particle. The spaceship would have to move with superluminal speed.Events \( \class{red}{E_0} \) and \( \class{blue}{E_1} \) are

**SPACE-LIKE**to each other. Events that are*space-like*to another event have a larger space distance (to the time axis) than time distance (to the x-y plane).All events that are

*simultaneous*for an observer (i.e., lie on parallels to the \(x\) axis) have a space-like distance from each other, because you could not draw a light cone to an event \( E_{\text{x1}} \) that contains an event \( E_{\text{x1}} \) simultaneous to \( E_{\text{x2}} \).Spaceship 2, which started its flight to \( \class{green}{E_2} \), will reach it in time because its destination \( \class{green}{E_2} \) is

*inside*its light cone. A connection of \( \class{red}{E_0} \) and \( \class{green}{E_2} \) is steeper than the world line of a light particle - the spaceship 2 does not exceed the speed of light. With sufficient speed it reaches \( \class{green}{E_2} \) in time.Events \( \class{red}{E_0} \) and \( \class{green}{E_2} \) are

**TIME-LIKE**to each other.The spaceship 3 which started its flight to \( \class{red}{E_0} \) at the event \( \class{brown}{E_3} \) will reach \( \class{brown}{E_3} \) only if it moves exactly with the speed of light. But since no mass-bearing object (and this includes spaceship 3) can ever be accelerated to the speed of light (it would need infinite energy to do so), the spaceship itself, can never reach the event \( \class{brown}{E_3} \) in time. But it has the possibility to send at least a light signal to \( \class{brown}{E_3} \) at its start event \( \class{red}{E_0} \), which will reach \( \class{brown}{E_3} \) in time, because light signals just propagate with light speed.

Events \( \class{red}{E_0} \) and \( \class{brown}{E_3} \) are

**LIGHT-LIKE**related. Events, which stand*light-like*to another event, have same time distance as space distance.