Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Is there a compact form for n-tuple relativistic additions of
velocities?

Replies: 7   Last Post: Jun 4, 2010 6:08 PM

 Messages: [ Previous | Next ]
 Martin Cohen Posts: 350 Registered: 12/4/04
Re: Is there a compact form for n-tuple relativistic additions of
velocities?

Posted: May 31, 2010 10:00 AM

On May 29, 12:00 pm, stargene <starg...@sbcglobal.net> wrote:
> I seek advice on whether a certain very long function can be made
> significantly more compact and therefore easier to compute.
>
> I am interested in calculating the value of a function which resembles
> both a continued fraction and an iterated function, where the basic
> unit of the iteration is the right hand side of the formula for
> relativistic
> addition of two (collinear) velocities:
>
> (1)  W_1 = (u + v) / (1+uv) .
>
> u and v are the summed velocities, given as decimal fractions of the
> velocity of light C.  Ie: u,v range from 0.0 to 1.0 .  Their resultant
> velocity is W_n.
>
> The essential idea is to divide C into n <equal> velocities (v) and
> then
> add them sequentially.  My procedure is to take the first sum, W, and
> then add to it a third identical velocity v, which gives a second
> resultant
> velocity
>
> (2)  W_2 = (W_1 + v) / (1 + vW_1) .
>
> This is an iteration of the basic form in (1).
>
> Similarly, a fourth identical velocity v is added to W_2, giving
>
> (3)  W_3 = (W_2 + v) / (1+ vW_2) ,
>
> and so on and on...
>
> This is to be repeated n times, where n becomes very large and
> v commensurately smaller.  I wish to know the values of this function
> as n becomes extremely large.  I especially want to know what the
> function converges to as n ---> infinity and C/n ---> zero velocity.
>
> A partial glimpse of this procedure as a single large relation
> resembles
> an unusual variety of continued fraction.  I omit rendering it in ASC
> II
> form here since it rapidly becomes very confusing and opaque.
>
> Any feedback will be very appreciated.
>
> Thanks,
> Gene

iirc, the addition formula for the inverse hyperbolic tangent is the
same as the addition formula for velocities in SR.
i.e., atanh(x)+atanh(y) = atanh((x+y)/(1+xy)).
Take tanh of both sides and have fun.

Date Subject Author
5/29/10 stargene@sbcglobal.net
5/31/10 Martin Cohen
6/1/10 stargene@sbcglobal.net
6/2/10 Martin Cohen
6/3/10 Martin Cohen
6/4/10 stargene@sbcglobal.net
6/4/10 Ilya Zakharevich
6/4/10 Martin Cohen