Mirror, mirror on the spaceship...

Listen, I've never aspired to be a poet.

Anyway.

You now have some idea of spacetime distance, and it's time to get more familiar with it. Luckily, you're chilling in your space station, in orbit around the planet when you see something strange. Two spaceships are playing ping-pong with a laser beam? This is the perfect situation for developing a deeper understanding of special relativity and spacetime distance! 

The spaceships have equal velocity and are moving to the left with respect to the space station, a fixed distance \(L'\) apart. They are equipped with mirrors that reflect laser beams.

We see here that we have two frames of reference, the spaceships and the space station. We will use a primed coordinate system \((x',y')\) for the spaceship frame (the leftmost spaceship at the origin) and an unprimed coordinate system \((x,y)\) for the space station (at the origin).

The leftmost spaceship emits a laser beam which results in the following events:

Again, we want to use these events to explore special relativity, beginning with time differences in your own frame of reference, and using this to gain a bit of intuition about how the events will look in a different frame of reference.

Let's begin by comparing the time it takes for the laser beam to move from left to right, \(\Delta t'_{AB}\), and the time it takes to go from right to left, \(\Delta t'_{BD}\). We know that the speed of light is constant, and as both ships are in the same frame of reference, they are not moving relative to one another - the beam always travels the same distance between them. It follows that \(\Delta t'_{AB} = \Delta t'_{BD}\). We also know that at \(t'_{B}\) a random explosion happens at the space station.

But as we've seen, time is relative. You may not see the same events at the same time from your space station. To you, it will look as if the spaceships are moving towards the left, and therefore the beam will have a longer way to travel towards spaceship A, than towards spaceship B. Spaceship B is, after all, meeting it on the way, while A is speeding away! This means that you will see that \(\Delta t_{AB}<\Delta t_{BD}\).

Let's see a visualization of how the time intervals change depending on the velocity of the spaceships:

Towards the end, the beam takes a much longer time from A to B than the other way!

The speed of light might be a bit difficult to visualize, however, so let's look at a more relatable case, before we get back to the actual space station frame.

You are playing ping pong with your friend while on a trolley going 50 km/h, with a ball moving back and forth at 80 km/h with respect to you (keep in mind that this velocity is with respect to the planet)! This is a regular activity for most people, so you should easily be able to visualize it. The situation is similar to the laser - we expect the ball to take longer traveling to the right, towards you as you are moving away from it, (provided you are also moving to the right).

Does this mean that an observer on the trolley and a passerby will observe different travel times for the ball, as we do for the light beam? If we define what is reffered to as spacetime as \(\Delta s^2_{AB} = \Delta t^2_{AB} - \Delta x^2_{AB} \), that is, spacetime is the time interval minus the space interval (squared), we can say that the spacetime interval is the same in both frames of reference, \(\Delta s_{AB}' = \Delta s_{AB}\). As we have \(\Delta t'_{AB} = \Delta t'_{BD}\), the relative distance between the laser in the two frames is \(\Delta x'_ {AB} = \Delta x_ {AB} \). However, we know that \(\Delta t_{AB} \neq \Delta t_{BD}\) which requires that there is a change in the speed of light, which just isn't okay! The speed of light is constant!

Well, there is one difference between light and a ball that may allow for a different outcome than our the spaceships have: A ball does not travel at a constant velocity no matter the frame of reference it's viewed in. Therefore, we can set the velocty of the ball in the trolley frame of reference not equal to the velocity in the earth frame of reference, \(v'\neq v\), and we see that \(\frac{\Delta t'_{AB}}{\Delta t'_{BD}} = \frac{\Delta t_{AB}}{\Delta t_{BD}} \).

Okay, back to space. It's time we figured out what order these events happen in the space station frame. We know that from the spaceship frame event B and C happen simultaneously. Imagine you are in the middle of the spaceships, in their frame of reference. In the space station frame the spaceships move towards the left and so the light has to travel a shorter distance to the right spaceship, meaning that in order for the light to reach you the second time, the reflection has to hit before the explosion does, i.e. event B is before event C.

Using this we can create a set of formulas for the behavious of the systems. The velocity of the spacecraft in the planet frame is 0.65c.

Again, as we did last time, we want to find the time intervals, \(\Delta t_{AB}\) and \(\Delta t_{BD}\). Perhaps it may be easier to begin with the spacetime intervals, \(\Delta s_{AB}\) and \(\Delta s_{BD}\)? After all spacetime, like light, is constant no matter which frame of reference you're in. This means that we can set \(\Delta s_{AB}^2=\Delta s_{AB}'^2\):

\(\sqrt{\Delta t^2_{AB} - \Delta x^2_{AB}} = \sqrt{\Delta t'^2_{AB} - \Delta x'^2_{AB} }\)

\(\sqrt{(t_B-t_A)^2 - (x_B-x_A)^2} = \sqrt{ (t_B'-t_A')^2 - (x_B'-x_A')^2}\)

Inserting values from the table, \(t'_A=0\), \(x'_A=0\), \(t_A=0\), \(x_A=0\), \(t_B'=\frac{L'}{c}\) and \(x'_B=c\Delta t = ct'_B=\frac{L'}{c}\). We rewrite:

\(\sqrt{t_B^2 - x_B^2} = \sqrt{L'^2 - L'^2}\)

where we have used that \(c=1\). As the expression \( \sqrt{L'^2 - L'^2} = 0\), we have to have that \(t_B^2 = x_B^2\), which makes sense as light travels at a constant speed and so the distance it travels is pretty much the time it takes to get there, and the time it takes to get somewhere, is the distance.

We repeat the process for \(\Delta s_{AC} = \Delta s'_{AC}\), \(\Delta s_{BC} = \Delta s'_{BC}\), and \(\Delta s_{AD} = \Delta s'_{AD}\) (we could have used other intervals, but this is the easiest as the variables for A are all zero). We get that \(t_C = 1.01ms\), \(t_B^2 = x_B^2\), and so \(t_B = 0.61ms\), and finally \(t_D =3.5ms\).

You see how we can use the spacetime invariance in order to find the time and place of individual events?

We make a table to properly showcase all of the information above:

We can easily find \(\Delta t_{AB}=t_B-t_A=0.61ms\) and \(\Delta t_{BD} = t_D - t_B=2.89ms\). We also note that the time it takes from event A to event C is longer than the time between event A and B, meaning that we were right in our reasoning that event C happens after event B in the space station frame!

So, by finding the time and place of all the events in both frames of reference, we've been able to find how much time there is between each event in the planet frame- without ever looking at this frame, we have a good idea on how things will happen.

Event A happens first, then the beam is reflecred from spaceship B (event B). Next, the random explosion happens (event C), and finally the laser reaches spaceship A and is reflected again (event D).

Finally, we can view the events from the space station, as you would see it.

More relativity to come!