Re: rapidity

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Regarding:

> Is there a conceptual/physical interpretation of this quantity (rapidity)?
>
> Bob

You can think of the rapidity of a Lorentz boost as the imaginary
'rotation angle' of the transformation divided by i=sqrt(-1). This comes
from the following mapping.  If you multiply the time by i this
effectively turns Minkowski spacetime into a 4-d Euclidean space, and
turns Lorentz transformations into rotations in this space.  Rotations
in the x-it 'plane' (where x signifies any direction in space) are
boosts.  Rotations in any plane involving 2 spatial directions are
ordinary rotations since they do not involve the modified time.  The
imaginary angle of the boost 'rotation' divided by i is the rapidity of
the transformation.

The slope m of a rotation angle [theta] is given by m = tan([theta]).
When writing a Lorentz boost in terms of the rapidity [psi] the velocity
v of the boost is given by v/c = tanh([psi]).  You can think of the
velocity of the boost as the slope of the hyperbolic 'rotation angle'
which is the rapidity.  The reason that the composition of the rapidities
of two parallel boosts 'add' in the usual way is the same reason as why
the rotation angles 'add' in the usual way when two coplanar rotations
are composed together to make a total rotation.  The slopes of the
rotation angles do not 'add' in a simple way, however.  Similarly, the
velocities of two parallel Lorentz boosts do not add simply either.  The
value of the boost velocity is bounded above by c, but the rapidity is
unbounded in magnitude.

In terms of the rapidity [psi] a Lorentz boost looks like:

   dx = cosh([psi])*dx' + sinh([psi])*d(ct')
d(ct) = sinh([psi])*dx' + cosh([psi])*d(ct')

This is a simple 'hyperbolic' analog of the usual rotation:

x = cos([theta])*x' - sin([theta])*y'
y = sin([theta])*x' + cos([theta])*y'

In fact the Lorentz boost above can be written in the Euclidean-like
rotation version as:

    dx = cos(i*[psi])*dx' - sin(i*[psi])*d(ict')
d(ict) = sin(i*[psi])*dx' + cos(i*[psi])*d(ict')

Here the imaginary 'rotation angle' [theta] is just i*[psi] where [phi]
is the rapidity.

I hope this helps with the conceptual/physical interpretation.

David Bowman
dbowman@georgetowncollege.edu