1
$\begingroup$

Recently I have came across a new term called rapidity in relativistic kinematics. We are using it in relativistic kinematics due to its additivity whereas velocity is not a additive quantity in relativistic case. But I am wondering that, is it introduced only to make our calculation easier or it does has any physical significance?

$\endgroup$

2 Answers 2

0
$\begingroup$

Rapidity is especially meaningful in the context of Galileo's relativity principle. That is, if you ride an inertial frame, and use the same physics to impart the same impulse to yourself to accelerate to a new, colinear moving frame, then this "standard physical manoeuvre" will always lead to the same rapidity change relative to your initial frame.

As a mind picture, imagine a set of travellators, like those you find in airports to convey passengers to flight gates. Each travellator moves at some constant speed $\Delta v$ relative to its immediate neighbor. Thus travellator 1 moves at speed $\Delta v$ relative to the airport, travellator 2 moves at the same speed increment $\Delta v$ faster than travellator 1 and in the same direction, travellator 3 moves at the same speed increment $\Delta v$ faster than travellator 2 and in the same direction and so on.

Then the rapidity of the motion on any given travellator relative to the airport frame is proportional to the aisle number $N$ of the travellator you're riding on. It will have rapidity $\eta = N\,\Delta v$, where we use natural units wherein $c=1$.

Put it another way: stepping from one travellator to the next always imparts the same boost matrix $B$ to one's co-ordinates. So the boost matrix linking the $N^{th}$ travellator's co-ordinates to the airport's is $\exp(N\,\log B)$. You can make the rapidity more and more like a continuous variable by imagining the speed $\Delta v$ to be smaller and smaller and imagining more and more travellators.

$\endgroup$
0
$\begingroup$

Rapidity has a number of related advantages, as you and the accepted answer have pointed out. It also has a geometric interpretation. See this SE answer (but note that that answer answers a somewhat different question).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.