Re: SR question

First, thanks for the reply.

Baugh wrote:
> lzwnews@xxxxxxxxx wrote:
> > Dear group,
> >
> > I'm a mathematics student who's studying SR on my own. Here's a
> > question which has plagued me for quite some time.
> >
> > Suppose O is an observer in a fixed inertia frame. Taking Minkowski
> > coordinates with respect to him, suppose we have:
> >
> > (1,0,0,0), (0,1,0,0), (0,0,1,0) (0,0,0,1)
> > v1 v2 v3 v4
> >
> > The first 3 spacelike vectors form an orthonormal basis of his
space of
> > "simultaneous events" (correct me if there's a better term). The
last
> > vector is his time vector.
> >
> > Now suppose P is travelling (wrt O) at a rate of (x,y,z). Hence,
> >
> > P's position : (xt, yt, zt, t).
> > P's timeline : (x,y,z,1) (of norm^2 = 1-x^2-y^2-z^2).
> >
> > My question is: is there a natural "projection" of v1,v2,v3 onto
P's
> > "space of simultaneous events"? I could take, for instance,
> >
> > (1,0,0,x), (0,1,0,y), (0,0,1,z).
> >
> > but these are not orthonormal!
> >
> > Thanks. This is so that I can add two velocity vectors given in the
> > form (x,y,z).
> >
>
> The answer is yes. You use the same projection formula as for
standard
> Euclidean geometry but keeping the indefinite metric in mind.
> To get V's projection orthogonal to u=(x,y,z,1) you subtract out
> V's projection onto the direction u:
>
>
> Vperp = V - u{<u|V>/<u|u>}
> < | > being the "dot product" in the indefinite metric.
>
> Note that <u|Vperp> = <u|V> - <u|u><u|V>/<u|u> = <u|V>-<u|V> = 0.
>
> So for v1 above you have:
> v1perp =
> (1,0,0,0) - (x,y,z,1)*<(x,y,z,1)|(1,0,0,0)>/<(x,y,z,1)|(x,y,z,1)>
> = (1,0,0,0) - (x,y,z,1)*x/{1-x^2-y^2-z^2}
> = (1-2x^2-y^2-z^2 , -xy , -xz , -x )/ {1-x^2-y^2-z^2}.
>
> However when you are done the three orthogonal vectors will not
> project onto three orthogonal vectors. This doesn't happen in
> Euclidean geometry either. Note that if you take a piece of wire
> bent at a right angle and turn its corner away from your eyes
> the flat picture you see will be an obtuse angle.
>

Hmm, but as you pointed out, the projections of the vectors v1,v2,v3
are no longer orthonormal (as you've pointed out, in Euclidean geometry
one can project a right angle onto a straight line!).

My motivation for looking at this:

If we add one-dimensional velocities v1 and v2, we get v1+v2/(1+v1.v2).
In this case, since space is 1-dimensional, there's no need for us to
choose coordinates. However, in 3+1 space-time, does it make sense to
say addition of velocity vectors (x,y,z) & (x',y',z')?


> This projection procedure also works for the Hermitian metrics
> of Hilbert spaces and the anti-symmetric metrics of Symplectic
> spaces (such as phase space (p,q)). You must be careful in these
> cases to take the 'dot product' in the correct order.

.

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