special relativity - How do I derive the expression for velocity in $S_n$ frame that has a velocity $v$ with respect to $S_{n-1}$ frame?

If there are $n$ frames and the $i$th frame has velocity $v$ with respect to $i-1$th frame. How do I derive the relation between velocity in $S_0$ and $S_n$ frame?

I found velocity in nth frame to be $u_n=\gamma^nu_0-v(\sum_{i=1}^n\gamma)$

What happens when n tends to infinity?

Here $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Answer

Suppose you have your base frame $S_0$ and a frame $S_1$ moving at a relative speed $v_1$. Then you have a second frame $S_2$ moving at $v_2$ relative to $S_1$. To calculate the speed of $S_2$ relative to your base frame you use the equation for the relativistic addition of velocities:

$$v_{02} = \frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}} \tag{1}$$

You could then use this to calculate $v_{02}$, then use it again to calculate $v_{03}$, and so on though that is rapidly going to get tedious. This is where the concept of rapidity mentioned by Ken G comes in.

Firstly let's write all our velocities as a fraction of the speed of light, $v/c$, in which case equation (1) simplifies to:

$$v_{02} = \frac{v_1 + v_2}{1 + v_1v_2} \tag{2}$$

Now suppose we take the inverse hyperbolic tangent of this. This seems a strange thing to do, but you'll see why this simplifies things. The atanh function is:

$$\text{atanh}(x) = \tfrac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$

If we take the atanh of equation (2) we get:

$$\text{atanh}(v_{02}) = \tfrac{1}{2}\ln\left(\frac{1+\frac{v_1 + v_2}{1 + v_1v_2}}{1-\frac{v_1 + v_2}{1 + v_1v_2}}\right)$$

This apparently horrendous equation simplifies very easily. We just multiply everything inside the $\ln$ by $1 + v_1v_2$ and gather terms and we get:

$$\begin{align} \text{atanh}(v_{02}) &= \tfrac{1}{2}\ln\left(\frac{(1+v_1)(1 + v_2)}{(1 - v_1)(1 - v_2)}\right) \\ &= \tfrac{1}{2}\ln\left(\frac{1+v_1}{1 - v_1}\right) + \tfrac{1}{2}\ln\left(\frac{1 + v_2}{1 - v_2}\right) \\ &= \text{atanh}(v_1) + \text{atanh}(v_2) \end{align}$$

So the $\text{atanh}$ of $v_{02}$ is calculated just by adding the $\text{atanh}$s on $v_1$ and $v_2$. Calculating the relative velocity for the third frame just means adding $\text{atanh}(v_{03})$:

$$\text{atanh}(v_{03}) = \text{atanh}(v_1) + \text{atanh}(v_2) + \text{atanh}(v_n)$$

And it should be obvious the general case is:

$$\text{atanh}(v_{0n}) = \sum_{i = 1}^n \text{atanh}(v_i)$$

The quantity $\text{atanh}(v)$ is called the rapidity, and this is what Ken means when he says the rapidities just add together.

The reason why we get this surprising behaviour is that a frame $S_1$ moving at $v_1$ is related to our rest frame by a hyperbolic rotation of a hyperbolic angle $\theta_1 = \text{atanh}(v_1)$. A second frame $S_2$ moving at $v_2$ relative to $S_1$ is rotated relative to $S_1$ by $\theta_2 = \text{atanh}(v_2)$, and the angles of rotation just add. So relative to use it is rotated by:

$$\theta_{02} = \theta_1 + \theta_2 = \text{atanh}(v_1) + \text{atanh}(v_2)$$

That's why the rapidities add.