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Re: Is there a compact form for n-tuple relativistic additions of
velocities?
Posted:
Jun 2, 2010 11:50 PM
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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.
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