## Re: Have space-time and the metric tensor finally run their course?

<jpolasek@cfl.rr.com> wrote:

[snip]
> The ei on the input vector have a 1:1 neat correspondence with 1 1 1
> 1, the first component being representative of time. The output dual
> vector likewise must have a 1:1 correspondence, it's not just 4
> numbers and the correspondence now is $e0 = -1$ as in $-1 1 1 1$.
> is 1111. What are we to make of this misfit $-1111$? The first element
> must be time and it's reversed.
> Admit that you would rather have it other, but "we're getting by OK".

No, I wouldn't "rather have it other", because I know that timelike
vectors are different from spacelike vectors, and as I've said the
"misfit" you keep complaining about reflects precisely that physical
reality.

GR and SR do not claim that time is "just another dimension". The
difference is built into the structure of the theories from the start.
What they *do* insist on is that different observers are entitled to
different definitions of time, just as different observers are entitled to
different definitions of "left-right, forwards-backwards, up-down". What
they do *not* allow is anyone's "left-right" axis to be someone else's
"future-past" axis.

> Greg Egan wrote:
>
> >What follows from this is that if you feed the vector v as an argument to
> >its own dual vector, the number you get is:
> >
> > $-(v^0)^2 + (v^1)^2 + (v^2)^2 + (v^3)^2$
> >
> >i.e. the quadratic form that gives the squared magnitude of v. The minus
> >sign on the first coordinate allows there to be three distinct classes of
> >vectors: those for whom this quadratic is $-ve, +ve$ or zero. This
> >corresponds very usefully to three classes of vectors observed in
> >physical reality: timelike, spacelike, and lightlike or null.
> >
> >There are no mathematical or logical difficulties following from this,
> >and rather than GR trying to pretend that there's nothing special about
> >time, the mathematical aspect you complain about is actually what
> >supports the distinction between timelike and spacelike vectors.

>
> Excuse me for not following up your original thread because I thought
> it would be easier to prove the defect in g. And it is simple: g00
> should be $e0*e0$. Don't tell me it is because a defective matrix was
> used to make the covector. $3D +$ time is far superior to STR and GTR,
> getting all the results without the clutter.

I have no idea what you mean by "$e0*e0$" -- is e0 meant to be a vector
here? A component of a vector? What kind of product is "*" here?

You'll get a more productive conversation if you agree to use some
standard notation which makes things clearer: distinguish upper and lower
indices, which can be done in ASCII by prefixing them with $^$ and $_$.

What *should* be true, and is true, is that:

$$e_0 . e_0 = g_{00}$$

i.e. the dot product of the timelike basis vector with itself will be
equal to $g_{00}$. This follows from the definition of the dot product in
terms of a general metric; for vectors v and w it is:

v . $w = v^i w^j g_{ij}$

which in the case $v=w=e_0$ becomes:

$$e_0 . e_0 = \delta^i_0 \delta^j_0 g_{ij}= g_{00}$$

And what is $e_0 . e_0 ?$ Because $e_0$ is a timelike vector, it must be -1.
This has got nothing whatsoever to do with time "running backwards" or
having "reversed polarity". It is about timelike vectors being different
from spacelike ones.

You are free to prefer some other formalism, if you can actually make it
mathematically consistent and produce predictions that match up with
physical reality. But every time you assert that there's a "defect" in
the use of a metric tensor just because you have some purely aesthetic
difficulty with the fact that its signature is ${-1,1,1,1}$ rather than
${1,1,1,1},$ you're just demonstrating that you don't understand general
tensors. There's a world of difference between you finding something
annoying, and it being logically inconsistent or self-contradictory. The
things that you keep asserting that "must be true" about tensors are
simply wrong.

Possibly you read some definitions in a textbook that was dealing only
with tensors in the context of purely spatial dimensions. If, for
example, you're dealing with a manifold that is embedded in ordinary
Euclidean space, then you can write a general point of that manifold as a
vector in the embedding space as a function of the coordinates on the
manifold: $p(x_1,....,x_n)$ in $R^m,$ where $R^m$ is the Euclidean space in
which the manifold is embedded. Then there are coordinate basis vectors
$*for$ the manifold*, in $R^m:e_i = dp/dx_i$

and the metric has coordinates in this basis of:

$$g_{ij} = e_i . e_j$$

where "." is the conventional Euclidean dot product in $R^m$.

What's happening here is that the metric of the embedded manifold is
induced by the flat Euclidean metric on $R^m$. With this method, you can
find, for example, the appropriate metric to use on the surface of the
Earth.

What this won't ever give you is a manifold with timelike dimensions. But
GR does not assert that spacetime is embedded in a larger Euclidean space,
so don't expect formulas and assumptions from a limited version of tensor
calculus that deals with that context to apply to the more general
situation.

--
Greg Egan