Why are the coordinate axes of a moving frame $K'$ tilted in spacetime diagrams?

Question

I am currently trying to self-teach special relativity (if relevant, I am still in school). I think that I already have a good intuitive understanding of, for example, time dilation and length contraction as a result of constant $c$.

I now want to understand (by which I mean, actually understand, not just learn!) how the mathematics of special relativity work - or in other words, I want to understand how the formulas for time dilation, length contraction, etc. are derived.

In this paper, special relativity is explained by using spacetime (Minkowski) Diagrams.

I am somewhat confused by this image (from the linked paper):

where $x'$ and $ct'$ are the coordinate axes of a moving (relative to a "stationary" frame $K$) inertial frame of reference $K'$ (the speed is $v$; $tan(\theta)= \frac{v}{c}$).

The dotted red line represents anything travelling with the speed of light (as the time axis is chosen to be $ct$ instead of $t$ only, light will always be at a 45° angle, as I understood).

How is it possible that $x'$ is "below" the red line, i.e. faster than $c$ (which obviously cannot be the case, so I must have misunderstood something)?

It kind of makes sense that the coordinate axes of $K'$ are at an angle to $K$ which depends on the relative speed, but I don't know why the axes are the way they are (I also read some introductions to spacetime diagrams, but none of them explained this).

In a later image, some IFORs are shown like this:

which makes sense to me as they are receding slower than $c$ from $K$. However, the coordinate axes are not shown - so how does one get to the first image, which seemingly contradicts $c$ being the "speed limit"?

It would be great if you could provide a mathematical (i.e. explain why the axes are tilted at $\theta=arctan \frac{v}{c}$) and an intuitive (it does not actually have to be intuitive, but rather using logic instead of math) explanation or some combination of both. I apologize if my question is trivial.

Answer 1

Let's use units where c = 1 (ex: x = light-years and t = years) to get rid of the c's.

I suppose that the blue line ct' is clear. At $t = t'= 0$ both frames are in the same location $(x = x' = 0)$. For later times, the location of the origin of the moving frame in the units of the stationary frame is $x = vt$. Being the origin of the moving frame, this line corresponds to the points where $x' = 0$.

The blue line x' is the result of synchronization of clocks. The Einstein method is to send 2 light rays (or radio signals) simultaneously from 2 points A and B to a middle point C. For the stationary frame it is easy to understand that if C receives the signals at the same moment, the clocks are synchronized.

For the moving frame, the points are A', B' and C'. A' send a signal to the right at t = t' = 0. The signal will reach C' at the point indicated as O. We must find when the signal to the left has to be send by B' to reach O (that is to reach C' at the same time). It is a problem of analytic geometry to get the equations of the straight lines, and find that the interception point P has the property of $t = vx$.

By changing B' (and C'), we get the line of simultaneity $t = vx$. That is the axis $t' = 0$ for the moving frame. That are the meaning of the moving frame axis.

Answer 2

Here's a variation of the answer provided by @Claudio.

Imagine constructing a "longitudinal light clock" (whose mirror-separation is parallel to the direction of relative motion).

Suppose one tick begins at event P and ends at event R along the traveler's worldline. Light-signals from event P (along the future light-cone of P) reflect off distant mirrors [not shown] whose worldlines are parallel to the traveler's worldline and are received at event R (along the past light-cone of R).

The intersection events F and G are equidistant from the traveler's worldline and are simultaneous according to that traveler. FG defines the X' axis for the event at the midpoint of PR. Note that this X'-axis is parallel to the X'-axis through event O.

The intersection of the future of P and the past of R (when P and R are timelike related) is called the "causal diamond of PR". The spacelike diagonal FG of that diamond is Minkowski-perpendicular to PR.
("[An observer's sense of] space is Minkowski-perpendicular to [an observer's sense of] time".)
In fact, FG and PR have the same magnitude. More importantly, the area of the causal diamond is a lorentz-invariant and is proportional to the square-interval of PR.

I used these ideas to develop a method of calculation on rotated graph paper.
(ref: my "Relativity on Rotated Graph Paper", American Journal of Physics 84, 344 (2016); https://doi.org/10.1119/1.4943251
See also: https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ )
This was inspired by the papers by Mermin that @ZeroTheHero referenced in the comment to your post.

Answer 3

After some further research, thinking and with the help of the two existing answers, here is my own answer which might or might not be more intuitive that the existing ones (which are great!).

Let's start with the $ct'$ axis of $K'$. For now, consider you are in $K$, thus "stationary" (as seen from your perspective). Since you do not move relative to $K$ (which always has to be the case, since $K$ is defined to be stationary relative to you), your worldline coincides with $ct$.

Now imagine some other person (or object, but it does not matter) travelling away from you along your $x$ axis with constant velocity $v$. As seen from $K$, the moving observer traces a worldline which at some point (here at $t=0=t'$) coincides with your worldline (or the $ct$ axis).

Now let's say that the moving observer is in some inertial frame of reference $K'$ which is similar to $K$ in that it is stationary compared to some observer. So we may say that $K'$ is receding from $K$ with velocity $v$. Since the observer in $K'$ does not move relative to $K'$, we may, similarly to $K$, define the observer's worldline to be the $ct'$ axis.

As we scaled the time axis to be $ct$, the speed of light $c$ will always be represented by a line at 45° angle (one could also use scalings such as 1 light-second for distance and 1 second for time to get the same result and maintain a "normal" $t$ axis for time instead of $ct$).

Since $c$ is the same for all inertial frames of reference, it must be at at 45° angle in all inertial frames of reference. Since a line of angle 45° is equivalent to a line which has the same angle to both perpendicular axes, we can conclude that the $x'$ axis must be rotated by the same amount to $x$ as $ct'$ is rotated to $ct$.

You might now say that the line which resembles $c$ is not at 45° angle compared to $x'$ and $ct'$. This is because as viewed from $K$, $K'$ moves, thus the coordinate axes appear to be non-perpendicular (of course, in reality, they are).

The angle $\theta=\arctan \frac{v}{c}$ describes how much the axes of $K'$ appear to be tilted when "viewed" from $K$. I cannot really explain how to get there, but it does make sense. Consider $K'$ to be moving with the speed of light. It is then $$\theta=\arctan \frac{v}{c}=\arctan \frac{c}{c}=\arctan(1)=45°$$

which is the line representing the speed of light. Practically, $K'$ cannot move with the speed of light, but this example shows that the worldline of an observer in $K'$ gets closer to that of light as $K'$ approaches $c$.