Re: Special Relativity fails a simple algebraic test

On 24 ene, 13:34, Randy Poe <poespam-t...@xxxxxxxxx> wrote:
On Jan 24, 8:04 am, Albertito <albertito1...@xxxxxxxxx> wrote:



The general expresion for composition of velocities is

a*b = sqrt((a^2 + b^2 + 2ab cos(alpha)) -
(ab sin(alpha))^2/c^2 ) /(1 + ab cos(alpha)/c^2),

where alpha is the angle between the velocities a and b.

If alpha=0, then the binary operator * reduces to

a*b = (a + b) /(1 + ab/c^2)

if alpha = pi/2, then

a*b = sqrt(a^2 + b^2 - (ab)^2/c^2)

if alpha = pi, then

a*b = (a - b) /(1 - ab/c^2)

Expressed more compactly:

(Galilean rule)
|a * b|^2 = |a + b|^2

(SR rule)

|a * b| = [ |a+b|^2 - |a x b|^2 ] / [1 - |a.b|/c^2]^2

Since the operator * is defined as a function of alpha, we see it is
not unique.

What do you mean it's not unique? alpha is fixed
for any vectors a and b, as is the sum a+b,
the cross product a x b and the dot product
a.b.

"Not unique" would imply that you could have different
values for a given a and b.

Wrong again, as usual.

- Randy

I mean by 'not unique' that the law of parallelogram for addition
of velocities in SR would be only valid to a first aproximation.
If a and b are two vectors, then the norm of its sum vector is
|a+b| = sqrt(|a|^2 + |b|^2 + 2|a| |b| cos(alpha)).

So, if the norm of its relativistic composition is
|a*b| = sqrt(( |a|^2 + |a|^2 + 2 |a||b| cos(alpha)) -
( |a||b| sin(alpha))^2/c^2 ) /(1 + |a||b| cos(alpha)/c^2),

We have

|a*b|^2 = (|a+b|^2 - B)/A,

where
A = (1 + |a| |b| cos(alpha)/c^2)^2 , and
B = ( |a| |b| sin(alpha)/c)^2.

SR deforms the law of paralleogram.
.