Why is rapidity additive?

Question

With the rapidity $\phi$ defined so that $\frac{v}{c}=\tanh{\phi}$, say you have 3 parallel moving reference frames $S$, $S'$ and $S''$ with a constant but different velocity/rapidity.

If the rapidity of $S'$ in relation to $S$ is $\phi_1$,
and the rapidity of $S''$ in relation to $S'$ is $\phi_2$,
then the rapidity of $S''$ in relation to $S$ is $\phi_1+\phi_2$

I don't really see why this is the case, it's probably something really simple I'm missing because there's no further explanation in my syllabus.

Answer 1

Relativistic velocity addition takes the form $$ V = \frac{v+u}{1+uv/c^2} $$ The rapidity is defined so that $$ v = c \tanh \eta $$ Plugging this into the the velocity addition formula and noting that $V = c \tanh N$, we find $$ c \tanh N = c \frac{ \tanh \eta + \tanh \eta' }{1 + \tanh \eta \tanh \eta'} = c\tanh(\eta+\eta') $$ Thus, the new rapidity is $$ N = \eta + \eta' $$

Answer 2

Alright here's a very crude heuristic(definitely not a proof) on why(without using velocity composition) rapidity is additive.

Impose these two constraints:

1) the constancy of speed of light in all frames. This means that the rapidity of light $\tanh\phi_c=1$ is the same in all frames of reference.

Consider this case:

You observe a frame of reference $S'$ moving with $\tanh \phi_1=v$ with respect to $S$ followed by a light beam with $\tanh\phi_c=1$. we know the rapidity of light in the $S'$ frame is the same as in $S$. Given this fact we'd like to find out what is the relation $ f(\phi_1,\phi_c)$ between the old and new rapidities in the new frame $S'$ such that: $$\tanh( f(\phi_1,\phi_c))=1$$

2) The second constraint is that the rapidity of an object at rest is zero. So if there's an object moving with velocity $v$ in $S$, it's at rest in $S'$. That is $$\tanh( f(\phi_1,\phi_1))=0$$

Then it's reasonable to conjecture that $f(\phi_1,\phi_2)=\phi_2 -\phi_1$

As you can check yourself using the hyperbolic identity $$\tanh (\phi_1 +\phi_2)=\dfrac{\tanh \phi_1+ \tanh \phi_2}{1+\tanh \phi_1 \tanh \phi_2} $$

Update: I was asked in the comments by user Prahar that I jumped into the conclusion too quickly. why for instance the functions $$(\phi_2 -\phi_1)^2$$ Or say $$(\phi_2 -\phi_1)^n$$ Or say $$\ln(\dfrac{\phi_2}{\phi_1})$$ are invalid?Although they satisfy the first two constraints?

The reason has to do with another constraint:

3)The principle of relativity dictates the following: If you're $S$ and $S'$ is moving with $\tanh \phi_1 =v$ according to $S$, then according to $S'$, $S$ is moving with $-v$ so that $$\tanh (f(\phi_1,0))=-v$$ This is only satisfied for $$\phi_2 -\phi_1$$

Because for $$(\phi_2 -\phi_1)^2$$ or any other $n$, We have $$\tanh ( (f(\phi_1,0))= \tanh (0 -\phi_1)^2= \tanh \phi_1^2 $$ which is evidently not equal to $-v$ because $\tanh \phi_1^2 $ is positive(whereas it should have been negative) and also because it's never equal to $v$, since $\tanh$ is a one to one function.

Answer 3

While looking into it again I've also found this simple solution to the problem.

Rewriting the Lorentz-Transformations with the rapidities gets us this: $$x'=x\cosh{\phi}-ct\sinh{\phi}$$ $$y'=y$$ $$z'=z$$ $$ct'=-x\sinh{\phi}+ct\cosh{\phi}$$

From this, it's easy to find:

$$ct'\pm x'=e^{\mp\phi}(ct\pm x)$$

Simply writing this out for $S\rightarrow S'$ and $S'\rightarrow S''$ and then substituting will get you:

$$ct''\pm x''=e^{\mp(\phi_2+\phi_1)}(ct\pm x)$$

Answer 4

In a hindsight, here's another proof of rapidity being additive(without using velocity composition again).

Consider this spacetime diagram in which the black axis represents $S$, the blue is the worldline of $S'$ and we also have another object in relative motion to both of them(the red one).

Rapidity is defined as $\tanh \phi=v$ (where $c=1$), which is $\dfrac{x}{t}$, that is the hyperbolic angle between the time axis and the trajectory of an object. $S'$(blue) moves with $\tanh \phi_1$ and the red with $\tanh \phi_2$. We wanna know what is the rapidity of the red object in $S'$. it's simply the hyperbolic angle between the time axis of $S'$(blue) and the red worldline which is $$\phi_2 -\phi_1$$

Answer 5

First, consider rotations in 2D. The analogous quantity to rapidity is $\theta = \tan^{-1}(y/x)$, and this quantity adds in successive rotations because it's just the rotation angle.

One heuristic way to consider boosts in Minkowski space is to imagine the time coordinate $t$ is an imaginary number; i.e. write $t = i t'$. Then the quantities $t'$, $x$, $y$, and $z$ act just like 4D Euclidean space, because the metric is $x^2 + y^2 + z^2 + t'^2 = x^2 + y^2 + z^2 - t^2$.

Then the "rotation angle" for a boost is $\tan^{-1}(x/t') \approx \tan^{-1}(iv)$. Since hyperbolic tangent is just tangent with a factor of $i$ in the argument, this means $\tanh^{-1}(v)$ is additive.

(Actually thinking about special relativity as having an imaginary time coordinate isn't very helpful, but it's an easy way to pop out formulas.)