# Relative Velocity (Or: The Velocity-Addition Formula)

With the knowledge that the speed of light is constant and the same for every reference frame, and that no object can travel at the speed of light or faster than the speed of light in any reference frame, we ask ourselves the question: what happens when two spaceships leave from a space station in opposite directions and both reach a constant speed of $0.95c$ (with $c$ the speed of light) relative to that space station? In the space station’s reference frame, the two ships travel at $2 \cdot 0.95c = 1.9c$ relative to each other. This doesn’t violate relativity, as neither spaceship is actually going at or faster than the speed of light. But, ignoring the effects of relativity, in the reference frame of either spaceship, the other ship would appear to travel at $1.9c$. This would violate relativity!

Applying the equations we found before, we will find out what actually happens.

# Deriving the Lorentz Factor

The Lorentz Factor is defined as $\gamma = 1 / \sqrt{1 - v^2 / c^2}$ with $v$ the relative velocity between two reference frames (or two objects, if you will) and $c$ the speed of light. This is the factor with which length contracts and time dilates between two reference frames. For example, if you’d sit in a rocket that was traveling at 80% the speed of light relative to your family on Earth, the Lorentz Factor would be $\gamma = 1 / \sqrt{1 - 0.8^2} = 1 \frac{2}{3}$.

What this means is that your rocket and everything in it — including you — would be shortened along the direction of motion for your family relative to the length you observe the rocket to be. In fact, its length would be contracted with factor $\gamma$; if the rocket’s length is $x$ as measured by you, your family would measure it to be $x' = x / \gamma$. In this case, they would measure the rocket to be 60% of the length your measure it to be!

Calculating the time dilation is a similar process. In your reference frame, a clock in the rocket would tick every second. If your family back on earth are able observe that clock, it would appear to tick slower. For one second to pass on your clock, they would have to wait $1 \cdot \gamma = 1 \frac{2}{3}$ seconds. In general, if a time of $t$ has passed in your frame of reference, a time of $t' = \gamma t$ has passed for your family.

In this post, I will follow Einstein’s steps in deriving this factor.