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## How to add velocities in special relativity

As I mentioned in this blogpost, in special relativity any observer will measure the speed of light in a vacuum to be $c$, irrespective of whether the observer is moving towards or away from the source of light. We can think of the speed of light as a cosmic speed limit, nothing can travel faster than it.

But, let us suppose that we have two reference frames $S$ and $S^{\prime}$ moving relative to each other with a speed of $v=0.9c$, 90% of the speed of light. Surely, if someone in frame $S^{\prime}$ fires a high-speed bullet at a speed of $u^{\prime}= 0.6c$, an observer in frame $S$ will think that the bullet is moving away from him at a speed of $u = v + u^{\prime} = 0.9c + 0.6c = 1.5c$, which seemingly violates the comic speed limit.

What have we done wrong?

We cannot simply add velocities, as we would do in Newtonian mechanics. In special relativity we have to use the Lorentz transformations to add velocities. How do we do this? Let us remind ourselves that the Lorentz transformations can be written as The Lorentz transformations to go either from reference frame $S \text{ to } S^{\prime}$, or to go from $S^{\prime} \text{ to } S$.

## Calculating a velocity in two different reference frames

To calculate the velocity $u$ of some object moving with a velocity $u^{\prime}$ in reference frame $S^{\prime}$ we need to use these Lorentz transformations.

We start off by writing $x = \gamma \left( x^{\prime} + vt^{\prime} \right) \text{ (1) }$

and $t = \gamma \left( t^{\prime} + \frac{x^{\prime}v}{c^{2} } \right) \text{ (2) }$

We will now take the derivative of each term, so we have $dx = \gamma \left( dx^{\prime} + vdt^{\prime} \right)$

and $dt = \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right)$

We can now write $dx/dt = u$ (the velocity of the object as seen in frame $S$) as $\frac{dx}{dt} = \frac{ \gamma \left( dx^{\prime} + vdt^{\prime} \right) }{ \gamma \left( dt^{\prime} + \frac{dx^{\prime}v}{c^{2} } \right) }$

The $\gamma$ terms cancel, and dividing each term on the right hand side by $dt^{\prime}$ gives $\frac{dx}{dt} = \frac{ \left( dx^{\prime}/dt^{\prime} + vdt^{\prime}/dt^{\prime} \right) }{ \left( dt^{\prime}/dt^{\prime} + \frac{dx^{\prime}v}{c^{2} dt^{\prime} } \right) } = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) }$ $\boxed{ u = \frac{ \left( u^{\prime} + v \right) }{ \left( 1 + \frac{ u^{\prime} v}{c^{2} } \right) } }$

where $u^{\prime}$ was the velocity of the object in reference frame $S^{\prime}$.

Going back to our example of $v = 0.9c$ and $u^{\prime} = 0.6c$, we can see that the velocity $u$ as measured by an observer in reference frame $S$ will be $u = \frac{ 0.6c + 0.9c }{ \left( 1 + \frac{ (0.6c \times 0.9c) }{ c^{2} } \right) } = \frac{ 1.5c }{ 1 + 0.54 } = \frac{ 1.5c }{ 1.54} = \boxed {0.974c}$, not $1.5c$ as we naively calculated.

## The constancy of the speed of light

What happens if a person in reference frame $S^{\prime}$ shines a light in the same direction as $S^{\prime}$ is moving away from $S$? In this case, $u^{\prime}=1.0c$. Putting this into our equation for $u$ we get $u = \frac{ 0.6c + 1.0c }{ \left( 1 + \frac{ (0.6c \times 1.0c) }{ c^{2} } \right) } = \frac{ 1.6c }{ 1 + 0.6 } = \frac{ 1.6c }{ 1.6} = 1.0c$

So they both agree that the light is moving away from them with the same speed $c$!