This time, a spaceship is moving with a speed close to the speed of light and emitting two laser beams.
We have two frames of reference:
- The planet frame (x,y)
- The spaceship frame (x',y')
We can define three events:
| Event A | The first laser beam is emitted |
|---|---|
| Event B | The second laser beam is emitted |
| Event B' |
The position of the first laser beam when the second laser beam is emitted |
We know that the speed of light is constant for every observer, and that the laser beams are moving at the speed of light. In the spaceship frame of reference the spaceship will stand still, and the planet will be moving.
If someone in the spaceship, let's say Luke, is trying to find the velocity of light by looking at the beams, he will get \(2.9757 \cdot10^{8}\) m/s. This is a pretty good approximation considering a couple of estimations were made. We could attempt to use classical physics in order to find this, from the planets frame of reference, but in that case we would see that the velocity is faster than the speed of light. In a Galilean transformation we add the velocity of the frame of reference to the objects speed in that frame of reference, meaning that a spaceship moving at a velocity \(0.99c\) emitting a beam with velocity \(c\) in the spaceship's frame of reference will have a velocity of \(1.99c\) in the observers frame of reference, which is of course impossible!
To get a clearer picture of how the laser beams move through spacetime, we make spacetime diagrams of how the laser beams move in the planet's frame of reference and the spaceship's frame of reference.
Before we go further into the relativistic effects of high velocities, we find the velocities of the spaceship and laser beams in each frame of reference.
| Laser beam velocity | Spaceship velocity | |
|---|---|---|
| Planet frame of reference | \(2.9979 \cdot 10^{8} \, m/s\) | \(2.74909 \cdot 10^{8} \, m/s\) |
| Spaceship frame of reference | \(2.9757 \cdot 10^{8} \, m/s\) | \(-2.74909 \cdot 10^{8} \, m/s\) |
The negative velocity of the "spaceship" isn't really the spaceship's velocity, as it's at rest within its frame of reference, but the velocity of the planet which is disappearing behind the spaceship, hence the negative sign.
For someone on the planet, it will look like the laser beam is just slightly quicker than the spaceship, as the spaceship is nearly moving at the speed of light. For someone in the spaceship, the laser beam will move very fast compared to them.
If you're sitting on the planet and want to find out how fast the laser beam moves in your frame of reference, you can use the Lorentz transformation for velocities. You know how fast the spaceship moves and how fast a spaceship observer measures the light speed.
The Lorentz transformation is given by
\(u = \frac{u'+v}{1 + \frac{u'v}{c^2}}\)
where \(u'\) is the speed of the laser beam in the spaceship's frame of reference, and \(v\) is the speed of the spaceship in the planet's frame of reference. Inserting values, we get \(u=2.99895 \cdot 10^8 \, m/s\). This is the speed of light, which should come as no surprise!
As we can see from the videos, the distance between the laser beams looks different in the
different frames of reference, but how different are they really? We find the distance in each frame of reference and compare:
\(\Delta x_{planet} =90.739 \, km\)
\(\Delta x_{spaceship} =505.497 \, km\)
To compare the two, let's look at the ratio between the two distances:
\(\frac{\Delta x_{planet}}{\Delta x_{spaceship} } = \frac{90.739}{505.497} = 0.1795\)
The distance between the laser beams in the planet's frame of reference will only seem like 17% of the distance experienced by someone on the spaceship. This makes sense, as someone on the spaceship will see themselves as at rest, while someone on the planet will se the spaceship following the laser beams at nearly the same velocity as the beams.
What if we put a stick between the two laser beams and measured the length of it? Let's say that the person on the spaceship measures the length to be \(L_0\). What would someone on the planet measure it to be? Let's see if we can use the formula for length contraction!
The formula is given by \(L = \frac{L_0}{\gamma}\) , where \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\). As the laser beams are moving at \(v=c\), the denominator is equal to 0, and the formula breaks down. In other words, the formula simply does not apply to this situation.
And at last, please chase you dreams. You will never catch the lasers anyway.