[Migration] Space-time Map: Taking a one-way journey to Sirius as an example

Recently, I watched "From One to Infinity" and was very interested in the Minkowski spacetime diagram. There happened to be an example of traveling to Sirius in it, so I wanted to draw it using a spacetime diagram. To simplify the problem, let's assume that the distance from Sirius to Earth is 9 light-years, which is actually about 8.6 light-years. At the same time, in order to facilitate drawing, the vertical axis of the time axis is measured in years times the speed of light, and the horizontal axis of the space axis is measured in light-years. Use geogebra for plotting. (This article is an old blog post migrated to xlog)

Spacetime diagram before departure#

As shown in the figure, line segment g (CAD) represents the spaceship, and point A is the center of mass. Line f represents the world line of Sirius B. Ray h (AE) represents A's light cone. Ray i (AF) represents the future world line of the spaceship, with a velocity of tanα=0.9 times the speed of light.

Spacetime diagram after departure#

As shown in the figure, in order to facilitate drawing, α has been reduced. Line AF and line AL form a new spacetime coordinate system with the spaceship as the reference frame.

Line AF with x' = 0 is the new vertical axis, and line AL with t' = 0 is the new horizontal axis, both symmetric about the AE axis.
The geometric position of AB in the new coordinate system is (AL, AK), and the physical position is (ALβ, AKβ).

Some interesting findings#

The spacetime distance of AB was originally 9, and now the spacetime distance of AB' is still 9, that is, $9^2 + (0i)^2 = 20.65^2 + (-18.58i)^2$
The geometric lengths corresponding to the units on AF and AL have become 1/β of the original lengths, that is, $\lvert\overrightarrow{AF}\rvert\beta = t'_{ \overrightarrow{AF'}} \quad \lvert\overrightarrow{AL}\rvert\beta = x'_{ \overrightarrow{AB'}}$
For the people on the spaceship, the journey only took 4.36 years, less than 10 years, that is, $t'_{ \overrightarrow{AF'}} = 4.36 \lt 10 = t_{ \overrightarrow{AF}}$
For the people on the spaceship, the distance to Sirius is 3.92 light-years, less than 9 light-years. The Sirius at point B' is a phantom from 18.58 years ago, and the current Sirius is the Sirius at point M', that is, $x'_{ \overrightarrow{AM'}} =(\frac{ \lvert\overrightarrow{AB}\rvert}{cos\alpha})\beta = 3.92 \lt 9 = x_{ \overrightarrow{AB}}$
For the people on the spaceship, the speed at which Sirius approaches is v = 3.92/4.36 = 0.9 times the speed of light, consistent with the speed at which the spaceship is observed to fly by the outside world.
The perceived spatial length $x_{\overrightarrow{AB}} = 9$ of the spaceship has become the spatial length $x'_{\overrightarrow{AB'}} = \lvert\overrightarrow{AL}\rvert\beta = 20.65$ , which has increased by a factor of γ. The ruler in the outside world has become longer.
The perceived time length $t_{\overrightarrow{AF}} = 10$ of the spaceship has become the time length $t'_{ \overrightarrow{AF'}} = \lvert\overrightarrow{AF}\rvert\beta = 4.36$ , shortened to 1/γ, and time in the outside world has become faster.
The spaceship thinks that the outside world's "ruler is longer and clock is faster", so the outside world thinks that the "ruler on the spaceship is shorter and clock is slower".
According to the invariance of four-dimensional spacetime distance, the travel time perceived by the spaceship can be easily calculated. The spacetime length of AF' in the spaceship's reference frame is the same as the spacetime length of AF in the outside world's reference frame, that is, $(i\lvert\overrightarrow{AF'}\rvert)^2 = (i\lvert\overrightarrow{AF}\rvert\beta)^2=AB^2+(iBF)^2$