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

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

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Martin Cohen

Posts: 350
Registered: 12/4/04
Re: Is there a compact form for n-tuple relativistic additions of
velocities?

Posted: Jun 3, 2010 4:25 PM
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On Jun 2, 8:50 pm, mjc <mjco...@acm.org> wrote:
> On Jun 1, 10:30 am, stargene <starg...@sbcglobal.net> wrote:
>
>
>
>
>

> > > 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.

>
> > Thanks for your response.
>
> > Indeed, I looked it up and you are right. However, as I am
> > often preternaturally dense, I don't understand how knowing
> > that makes my task of adding, say, n = 10^6 identical little
> > velocities, where each one is v = Co / 10^6, (identically one
> > millionth the velocity of light) any easier without steam
> > shooting out of my ears.

>
> > Eg: How would taking the tanh of both sides of your relation
> > help me, and how does that take into account the totality of
> > all my n= 10^6 additions? And when n = 10^9 or more? :-(

>
> > Can you please enlighten me?
> > Gene

>
> Let T(a) = tanh(a) and AT(a) = atanh(a), for brevity.
>
> If we write vsum(x, y) = (x+y)/(1+xy),
> then, if x = T(u) and y = T(v),
> vsum(x, y) = (T(u)+T(v))/(a+T(u)T(v))
> so AT(vsum(x, y)) = AT((T(u)+T(v))/(a+T(u)T(v)) = AT(T(u)) + AT(T(v))
>                   = u + v = AT(x) + AT(y).
> To get the vsum of 3 variables,
> AT(vsum(vsum(x, y), z)) = AT(vsum(x, y)) + AT(z) = AT(x)+AT(y)+AT(z).
> Note that this shows that we can write vsum(x, y, z) since the result
> holds for and permutation of x, y, and z.
> Continuing, by induction, AT(vsum(x, y, z, a, b, ...)) = AT(x)+AT(y)
> + ....
> Setting all of the variables equal, if there are n of then,
> AT(vsum(n copies of x)) = n*AT(x)
> or vsum(n copies of x) = T(n*AT(x)).
>
> We now need to look at T(x) for large values of x (= n*AT(y) for large
> n).
> Writing E(x) = exp(x),
> T(x) = (E(x)-E(-x))/(E(x)+E(-x)) = (E(2x)-1)/(E(2x)+1)
>      = (E(2x)+1-2)/(E(2x)+1) = 1 - 2/(E(2x)+1)
>      ~ 1 - 2/E(2x) for large x. ("~" means "is close to")
>
> Putting this in the above,
> vsum(n copies of x) = T(n*AT(x)) ~ 1 - 2/E(2*n*AT(x)).
>
> You can rewrite this in various ways. One is
> vsum(n copies of x) = 1-2/(2*AT(x))^n, showing that
> (1-vsum(n copies of x))/(1-vsum(n-1 copies of x)) ~ 1/(2*AT(x)).
> This shows how fast the SR sum of n copies of x approaches 1.- Hide quoted text -
>
> - Show quoted text -


A correction:

vsum(n copies of x) = 1-2/E(2*AT(x))^n (I lost the E)

A small addition:

If x is small, both T(x) and AT(x) are about x. so
vsum(n copies of x) ~ 1-2/E(2*x)^n
~ ~ 1-2/(1+2*x)^n
Since E(x) + 1+x for small x.




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