

The Privileged Character of 3+1 Spacetime (2011) - Bootvis
http://tetrahedral.blogspot.com/2011/12/privileged-character-of-31-spacetime.html

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drostie
Another fun fact is that the +−−− metric allows relativity to be described in
a spinor calculus (which are in turn isomorphic to quaternions).

The basic idea looks like this: annex the identity matrix as σ_0 = onto the 3
Pauli spin matrices σ_1, σ_2, σ_3 and then form the product with any 4-vector
v^μ to find that every 4-vector is a 2x2 Hermitian matrix,

    
    
        V = v^μ σ_μ.
    

Of course there's nothing special about the +−−− metric here; but when you
calculate the determinant of V you find:

    
    
        det V = v^μ v_μ
    

so that the determinant is the Lorentz metric. This means that Lorentz
_transforms_ preserve determinants in this system. There is the 4-reversal C =
\v → -v as well as the parity transform P = \v → (det v) v^(-1), with time
reflection T = CP as usual. Lorentz transforms which do not have those
reflections in them instead have some complex matrix det L = 1 such that they
take the form

    
    
        \v → L v L†.
    

Now think about the vectors in this system. A good way to poke inside the
vectors is to look at matrices which are _projections_ , det V = 0 so that

    
    
        V = [α] [α* β*] = u u†
            [β]
    

If you actually work it out, things like α/β work out to be stereographic
projections of that null vector on the celestial sphere. The Lorentz transform
takes the vector u above to the vector L u. So when you look at, say, incoming
light rays to a point (say, yourself), those points are shifted by this L
matrix, performing a bilinear/Mobius transform of the incoming light rays.

By ignoring rotations (which are unitary matrices!) and going for boosts, you
can quickly convince yourself that as you boost in a direction, the stars all
"tilt forwards", and you can dream that maybe if you were a photon, in
addition to living a timeless existence, maybe only the event which emitted
you is behind you, while the entire universe lies in front of you.

~~~
HCIdivision17
This is a classic example of something I feel I should be able to follow and
understand, but in fact cannot interpret as anything other than technobabble.
I'm going to point to this as an example of how there are different levels of
familiarity and expertise.

And to be sure: this seems to be a fantastic comment. I'm going to spend some
time slowly working through it until I grok the terms.

~~~
drostie
Okay, so, four-vectors have four components: v0, v1, v2, v3. The last three
components are "spatial" and the first is "temporal". An example of a four-
vector is a 4-position of a particle moving from position (0, 0, 0) with speed
u in the x-direction:

    
    
       r0 = c * t
       r1 = u * t
       r2 = 0
       r3 = 0
    

This is actually a family of points, one for each number t.

The 4-vector is a thing which everyone can agree on. (Mind you, you might
disagree on the _components_ of the 4-vector, but everyone agrees on the
"arrow" in 4d space which the vector represents. That's a little abstract and
philosophical but it means that there is one objective reality which people
are seeing in different ways. In particular if one person says "these two
particles collided at this place and time," everybody else has _some_ place
and time where they also saw those two particles colliding.)

In particular, everyone agrees on a number based on the 4-vector, which is the
number v0² − v1² − v2² − v3². This is called the "magnitude" of V under the
"Lorentz metric". In our case above, the magnitude is (c² − u²) t².

In the particle's own "rest frame" (v1 = v2 = v3 = 0) it must agree with these
coordinates so we must find that its coordinates in that frame are: [√(c² −
u²) t, 0, 0, 0]. But of course this must be equal to [c τ, 0, 0, 0] for the
particle's own time coordinate τ (Greek letter tau), because everybody agrees
on the speed of light c and in this formalism it's a conversion factor between
space-dimensions and time dimensions. The conclusion is that t = τ / √(1 −
u²/c²). This conversion factor occurs so much that it's usually just written
with a γ (Greek letter gamma) as t = γ τ. The general transform which
preserves the Lorentz metric in this direction is the "x-boost":

    
    
        v0' = γ * (v0 − (u/c) * v1)
        v1' = γ * (v1 − (u/c) * v0)
        v2' = v2
        v3' = v3
    

We can perform this "boost" twice, once with speed u = +w, then again with
speed u = -w, to find that v0'' = v0, v1'' = v1, which proves that the
mathematics is totally consistent. We can also compute (v0')² − (v1')² to
confirm that it is the same as v0² − v1². There are other possibilities too
for preserving the Lorentz metric: there are 3d space rotations with no change
to v0, and boosts in other directions, and combinations of the two (what are
called 4-screws). There are also reflections (flipping vi to -vi). But that's
about it. When we want to ignore the reflections we talk about "proper"
Lorentz transforms, with the idea that reflections are somehow improper
(mostly because they're not a continuous family of transforms).

All of the stuff about the Pauli matrices is basically about forming the 2x2
complex matrix V =

    
    
        [v0 + v3     v1 − i v2]
        [v1 + i v2   v0 − v3  ]
    

The determinant of a matrix [a b; c d] is the product of its eigenvalues and
can be calculated as a * d − b * c. This has a nice property: the determinant
of a product of matrices is the product of the determinants; det(A * B) =
det(A) * det(B). This matrix is also nice because it is _Hermitian_ , which
means that it is its own conjugate transpose. Hermitian matrices are complex
matrices which have real eigenvalues.

Every proper Lorentz transform can be thought of as a 2x2 complex (non-
Hermitian in general) matrix L. We take its conjugate-transpose L† and convert
the matrix V above to be V' = L V L†. This arrangement keeps V' Hermitian, and
we can work out v0, v1, v2, and v3 from its above components. The only problem
is that the Lorentz metric may have changed, unless det(V') = det(L) det(L†)
det(V) = det(V). This amounts to demanding that det(L) = ±1. However, we can
throw away the det(L) = -1 solutions as redundant, because if det(L) = -1 then
det(i * L) = +1, but when we complex-conjugate that we get:

    
    
        L V L† = 1 * L V L† = (-i * i) L V L† = (i L) V (i L)†
    

so they have the exact same effect on these four-vector matrices. The Lorentz
boost in the v3 direction has a very nice Lorentz matrix which looks like:

    
    
        [e^(k/2)     0   ]
        [  0     e^(-k/2)]
    

Running through the motions gives v0' \+ v3' = e^k (v0 + v3), v0' − v3' = e^-k
(v0 − v3), which works out to v0' = cosh(k) v0 + sinh(k) v3 while v3' =
sinh(k) v0 + cosh(k) v3, where sinh and cosh are the hyperbolic sines/cosines.
Conclusion: the v3 boost by a speed u is defined by that matrix with k =
atanh(-u/c).

Again, lightlike vectors (det V = 0) are projections of some 1x2 complex
vector called a spinor, V = s s†. Expanding V out with s = [α; β] we find that
β/α = (v1 + i v2) / (v0 + v3).

This is essentially a stereographic projection. To fill in the gaps, let us
assume that we've got future-pointing vectors (v0 > 0) which are "coming in
towards us" and need to be projected to a point on a sphere at radius 1 from
which they came. After scaling the null vector by dividing by R it has some
components (1, -x/R, -y/R, -z/R) where (x, y, z) is the point on the sphere of
radius R which the null vector seems to be "coming out of". Then -β/α is an
actual stereographic projection of (x, y, z) onto the complex plane; if we
take the last formula with R = 1 we get back just (x + i y) / (1 - z).

Lorentz matrices then transform these components directly. In the case of the
v3 boost, we're transforming -β/α to (-β/α) * e^k, so we're just scaling the
complex plane by a constant, k = -atanh(u/c).

If u > 0 then we are boosting towards the North pole and k is negative, so all
of the stars that we see (lightlike vectors) get compressed in the
stereoscopic projection towards z=0 on the complex plane. In other words, all
of the stars tilt in the direction we're going.

Final freebie: if something has a round outline when it is stationary, then it
has a round outline when it's moving past you, because bilinear transforms map
circles to circles. The image that you see within that circle is distorted
(you can imagine a bullseye target, the interior circles "shift" towards one
of the walls of the outer circle uniformly) but the outline is still circular.
In this respect, Lorentz length contraction is "invisible".

------
transfire
The problem with this analysis is that it is basing its anthropic conclusion
on anthropic physical theory. In other words, it compares what would happen if
the number of dimensions changed using theories devised for a 3+1 universe.
Obviously if the dimensions changed, the theories have to change. And in an
alternate universe with a different number of dimensions the theories would
not be the same.

~~~
nhaehnle
I agree. I like this topic, but I have to say that there are better, more
fundamental arguments in favour of 3 space dimensions. For example, two space
dimensions only allow planar graphs, which seems pretty limiting for the
development of complex structures. On the other hand, as you increase the
number of spatial dimensions, the curse of dimensionality becomes more
important and space in general becomes "emptier". Also, knots are only
interesting in 3 dimensions.

Arguing about time dimensions is harder, because time is tied more strongly to
our subjective experience of consciousness, which is something that we really
don't understand yet.

------
abecedarius
Greg Egan has a whole lot to say about 4+0 spacetime:
[http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGON...](http://gregegan.customer.netspace.net.au/ORTHOGONAL/ORTHOGONAL.html)

