# LOG#025. Minkovski diagrams.

“(…)The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.(…)”
— Hermann Minkowski, in ‘Space And Time’, a translation of an address delivered at the 80th Assembly of German Natural Scientists and Physicians, at Cologne, 21 Sep 1908. In H.A. Lorentz, H. Weyl, H. Minkowski, et al., The Principle of Relativity: A Collection of Original Memoirs on the Special and General Theory of Relativity (1952), 74.

In Special Relativity, via Lorentz transformations, space and time are intrinsically united. Therefore, it makes sense to call the abstract set of space and time together spacetime, as we have seen before in our relativistic thread. Hermann Minkovski, a former professor of Albert Einstein guessed a quite simple and pictorical representation of “events” in “space-time”. That representation shares very similarities with usual cartesian geometry in the plane, but with some differences. Let me explain it better. Generally, the position of a point P in the real plane is expressed by Cartesian ( also called parallel or rectangular) coordinates $P(x_0,y_0)$. But one could just as well use two oblique-angled axes, $x'$ and $y'$ instead, i.e., we can choose “a rotated frame”. In order to determine the new coordinates $x'_0,y'_0$ of any point P in such coordinate system, we draw parallels to the oblique axes through P, and then, we can calculate $x'_0,y'_0$ at the intersection points with the new frame. This result is completely general, and the units of length on the new axes don’t need to be the same as those we used in the old coordinates. See the following diagram:

Lorentz transformations are what mathematicians call “affine transformations”, i.e., a one-to-one (a.k.a. bijective) mapping of a plane on itself that preserves parallelism and rectilinearity of straight lines! For a 2D Minkovski spacetime transformation, being general, we have

$ct'=x'_0=L_{00}ct+L_{01}x+b_0$

$x'=L_{10}ct+L_{11}x+b_1$

Setting that $b_0=b_1=0$ fixes a common origin of coordinates (i.e. an unchanged point of origin). It is very simple to calculate the direction and units of length of the new primed axes in the coordinate system of the unprimed (old) coordinates. We have to set

$ct'=0, x'=1$

into the inverse 2d Lorent transformations above, and we obtain the expected result $ct=\gamma$, $x=\beta \gamma$. It is a common practice to represent the time in the vertical axis, and the position in the horizontal axis. Curiously, it is the opposed practice in the more conventional plane motion with which we are more familiar,  and where we use to representa horizontally the time and vertically the position of some objetct. Of course, that is only a conventional election, we could draw the axis as we wanted to. We only have to establish the “rules”. And we will follow the common practices here. Furthermore, a Lorentz transformation does not cause the coordinate axes to rotate unidirectionally, since with the rotation of the coordinate frame around its origin, they will rotate counterdirectionally! Elementary trigonometry helps us now ( we reviewed it in a previous log), and the angle between the primed (new) and the unprimed (old) axes is showed to be:

$\boxed{\tan \delta=\beta=\dfrac{v}{c}}$

The units of length, L, on the new reference frame in spacetime can be calculated with a simple application of the Pythagoras theorem:

$L=\sqrt{\gamma^2+\beta^2\gamma^2}$

and thus

$\boxed{L=\sqrt{\dfrac{1+\beta^2}{1-\beta^2}}}$

This stuff can be reviewed in the following cool diagram:

Example: the relative velocity between S and S’ is $c/2$. It provides $\tan \delta=\beta=1/2$ and $L=\sqrt{5/3}\approx 1.29$. Therefore, coming back to the new reference frame, the spacing of the “tick” marks (the scale) is stretched by a factor of 1.29. Suppose in addition that the units of length are given in light-years (ly). Then, time coordinates will be given too in light-years since we have $ct, ct'$ in the temporal axes. Finally, an important remark to relativity beginners: the obliquity of the x’-axis in a Minkovsky spacetime diagram DOES NOT IMPLY that there is an angle between the x and the x’ axes IN SPACE. The angles between frames are between space-time axis in general, it is not necessary that the x-x’ axes form an angle a priori.

Some practical exercises will aid us to experiment the power of the Minkovski diagramas to solve commonly given examples in Special Relativity.

Exercise 1. Events in space-time.

Observe the following Minkovski diagram

The event E is located 5 years in the future and 3 ly on the right of the origin (0,0), as it is given in the S frame. The S’ frame moves with velocity $\beta=0.5$ to the right. The question is: when and where does event E happen in the S’ frame? A simple calculation shows it:

$\beta=0.5\rightarrow \gamma = \dfrac{2}{\sqrt{3}}$

Therefore, the coordinates of the event E in the primed frame will be:

$ct'=\gamma (ct-\beta x)=\dfrac{2}{\sqrt{3}}(5ly-0.5\cdot 3ly)\approx 4.04ly$

$x'=\gamma(x-\beta c t)=\dfrac{2}{\sqrt{3}}(3ly-0.5\cdot 5ly)\approx 0.59ly$

In the primed reference frame S’, the event E happens in such a way that E’ is about 4.04 years in the future and about 0.58 ly on the right of the origin! Fascinating and elementary, from our limited knowledge at least. Consider a variation of this exercise, where the event E is located 2 years in the past and 2 ly on the left of the origin, as given in the S-frame. The S’-frame moves with velocity $\beta=-0.6$ to the left. Again, the question is: when and where does the event E happen in the S’-frame? Looking at the new Minkovski diagram:

We can proceed to make a simple calculation as before we did:

$\beta=-0.6\rightarrow \gamma =\dfrac{5}{4}$

And thus, the new coordinates will be boosted by Lorentz transformations:

$ct'=\gamma(ct-\beta x)=\dfrac{5}{4}(-3ly-(-0.6)(-2ly))=-4ly$

$x'=\gamma (x-\beta ct)=\dfrac{5}{4}(-2ly-(-0.6)(-2ly))=-4ly$

In the primed reference frame S’, the event is located 4 years in the past and 4 light-years (ly) on the left of the common origin of both frames.

Exercise 2. Simultaneity in action.

We will suppose now that $\beta=0.5$. All the events that are located on a straight line parallel to the x-axis (dashed in the next Minkovski diagram) are simultaneous in S.

Therefore, every event located on a straight line parallel to the x’-axis (dotted in our Minkovski diagrama) are simultaneous in S’. It is evident and obvious that two arbitrary events, $E_1, E_2$, that are simultaneous in one inertial frame can NOT be simultaneous in any other inertial frame. This fact drives many of the “paradoxes” of Special Relativity. They are not really paradoxes in general, since usually we can solve the apparent contradictions with our language using a proper language and a right physical insight.

The relativity of simultaneity can also be understood with other Minkovski diagram for different events:

Exercise 3. The world lines in space-time.

Choose any inertial reference frame, e.g., the S-frame and place some object in space with spacetime coordinates $E(x,ct)=(x_0,ct_0)$. This point of the spacetime will be denoted as $E_0$. In a posterior moment, the object moves towards any other arbitrary location, e.g., $x=x_1$ and $ct=ct_1$. We define the event $E_1(x,ct)=(x_1,ct_1)$.

If we track the object through the spacetime, we obtain a sequence of events! This sequence of events is called “worldline”.

The worldline of any particle/object in nothing but a path-time diagram, i.e., the diagram length-time of the particle, with the path plotted to the right an the time plotted towards the top. Moreover, the worldline of any object AT REST in the S-frame is parallel to the ct-axis. The faster the object is in the S-frame, the flatter its worldline is. If the object moves at the speed of light, the worldline is a straight line with certain slope that can be only either +1 or -1, since it moves one unit in the x-direction (1 ly) in one unit of the ct-direction ( 1year times the speed of light). No material object can move faster than the speed of light in Special Relativity (i.e., under the hypothesis of Special Relativity, any body is restricted to move wih velocities lower than “c”. In addition, worldlines intersecting the x-axis at angles less than 45º are thus excluded ( unless we allow the notion of tachyons, to be discussed in the near future in this blog).

These notions can be understood with the following diagram, sketching the worldline:

Imagine a light flash propagating from the event $E(ct,x)=(2ly,2ly)$ in a spherical shell expanding in all directions of space. In the one dimensional representation of space, given by the following Minkovski spacetime diagram, the spherical shell is, at any instant of time, reduced to two points ( intersection of a line with the spherical shell). As time flows, the two points turn into two diverging world lines, we drawed them dashed. We can read from the Minkovski diagram that the light reaches the observer at $t=4yr$ in the S-frame, and $t'\approx 2.3 yrs$ in the S’-frame ($\beta=0.5$). The calculation for S is trivial, but for S’ some harder calculation is to be done.

Exercise 4. Time dilation via diagrams.

Again we take $\beta=0.5$. The event $P'(ct',x')=(3ly,0ly)$ has oblique coordinate axis. In the cartesian frame at rest, S, we have $ct\approx 3.5 ly$. Explanation: a process extending from the origin to E’ and it takes 3 years in the S’-frame, while it will table about 3.5 years in the S-frame. The observer at rest in S concludes that the “clocks” of the S’-observer are slower than his own clocks.

The point $Q(ct,x)=(3ly,0ly)$ in the cartesian S-frame has the time coordinate $ct'\approx 3.5 ly$ in the S’-frame. Meaning: a process, extending from the origin to Q, it takes 3 years in the S-frame and it will take about 3.5 years in the S’-frame. The S’-observer claims that the clocks of the S-observer are slower that his own clocks.

The diagram is:

Remember: there is no contradiction in the fact that both observers “measure” that their own clocks are both slower than the other, since, at last, the events P’ and Q are not happening in the same point in spacetime!

Exercise 5. Length contraction via diagrams.

Once again, $\beta=0.5$, and the Minkovski diagram is now this one

We put a rod extended from the origin to the point $P'(ct',x')=(0ly,3.5ly)$ and it will be initially at rest in the S’-frame. Its length in that frame equals $3.5 ly$. The position of its left end is a function of time, i.e., its worldline IS the ct’-axis while the worldline of its right end is parallel to the ct’-axis through P’ (marked as a dotted line). At any instant t’ in teh S’-frame, the reod is placed on a paralllel line to the x’-axis.

By the other hand, in the S-frame the ros is extended from the worldline of the left end (ct’-axis) to the worldline of the right end (dotted line in the diagram). However, at any moment t in the S-frame, the rod is placed on a parallel line to the x-axis. At time t=0, the rod is extending from the origin to the point $P(ct,x)=(0,x\approx 3)$. Thus, its length is approximately 3ly instead of 3.5 ly, and the rod is contracted.

In a similar way, a rod at rest in the S-frame, extending from teh origin to the point $Q(ct,x)=(0ly,3.5ly)$ will be contracted in the S’-frame to the distance from the origin to the point $Q'(ct',x')=(0ly,x\approx 3ly)$.

Remark: the phenomenon of length contraction has to do with simultaneity as well, length measurements of moving objects are reasonable only if th epositions of both ends can be measured simultaneously! Since S and S’ don’t agree on simultaneity, they cannot agree on the results of their length measurements.

Exercise 6. The past, the future and causality.
The next Minkovski diagram owns the coordinate axes of three different reference frames S, S’, and S”. The beta parameters are respectively $\beta=0, \beta '=-0.6, \beta ''=0.6$. There are two wordlines of light rays (flashes of light) passing through the origin ( with the dashed bisectors). The projections of the events P, Q and R on the different time axis are dotted and clearly distinguished from one to another:

We observe and distinguish:

1st. P is located in the future, corresponding to the S’-frame ct’>0, in the present according to the S-frame since ct=0, and in the past in the S”-frame because of the relationship ct”<0.

2nd. Q is located in the future in every frame, R is located in the past in every frame.

We can simplify the above diagram with a more simple Minkovski diagram

The 2 sectors on the left and right, in medium-gray colour, comprise zones where, depending on the frame, can be comprehended as past, present or future. Every worldline are at least as steep as the worldlines of light (dashed here), and everyworldline created from any event therein have to intersect the ct-axis above the origin. Some worldlines, created at P, have been added to the figure. No event in the medium-gray zones can affect the origin. This is a notion of causality. Furthermore, no event in the medium-gray zones can be reached or be influenced by a worldline from the origin. These points outside the lightcone are said to be out of the causal influence of the past and future.

Events contained in the darker zone are regarded as future by every frame. A world line from the origin can reach any event therein. In this way, we can call this zone “absolute future”. In the same fashion, events in the lower zone (the lighter gray) are past for every frame. They are absolute past and any event happening inside that zone can be connected to the origin by a worldline as well.

Remark: it is impossible, a priori, for a worldline to run from the upper to the lower zone, i.e., for events like Q to influence R is impossible (according to SR). By no means can be Q the cause of R. However, R CAN be the cause of Q. The sequence of cause and effect, causality, cannot be inverted by the special theory of relativity. It it happens, it would be an argument against the theory.

Exercise 7. Faster than the speed of light?

Consider the next gedanken experiment (thought or imaginary experiment). Let us suppose that information can be transmitted at a speed faster than light. Observe then the following diagram and once again set $\beta=0.5$

And next, imagine the next experiment:

1st. A transmission tower is at rest in the origin of the S-frame and a relay station is at rest in the S’-frame at $x'=x'_0$.

2nd. At the time $t=t'=0$, a signal will be transmitted from the common origin of the two frames to the relay station at speed 10c, as measured from the S-frame (square dots in the diagram).

3rd. It is being received by the relay station in the event A, and being re-emitted after a short period of time to the source at speed of -10c, according to the S’-frame (round dots in the diagram).

4th. The signal is being received in the event B at a time t<0, i.e., at the source, but BEFORE the signal was emitted there.

Conclusion: If the signal carries a destructive energy, it can destroy the transmitter. Cause and effect would occur in an inverted sequence!