Re: the basis of relativity

carlip-nospam@xxxxxxxxxxxxxxxxxxx wrote:
> Ken S. Tucker <dynamics@xxxxxxxxxxxx> wrote:
> > Hi Steve
> > I'll include a bit more detail (word salad).
>
> > carlip-nospam@xxxxxxxxxxxxxxxxxxx wrote:
> >> Ken S. Tucker <dynamics@xxxxxxxxxxxx> wrote:
>
> >> > Ok, then define the curvature scalar.
> >> > Does one exist, and don't trim this question!!!!
> >>
> >> R = R^a_{bab}
>
> >> Equivalently, let e^i be an orthonormal basis. Then
> >> R = sum_i K(e^i,e^i)
>
> My typo: should be R = sum_{i,j} K(e^i,e^j)

Ok, is that last Eq. to hold at a point or over a
finite region in a g-field. The former is a Minkowski
limit, the latter is interesting.

> > The above doesn't apply to a g-field, wherein the
> > basis is nonorthogonal because g_00 =/=g^00 therein,
>
> This isn't just "word salad," it's a fundamental misunderstanding
> of math. There is no such thing as "the basis" -- there are an
> infinite number of possible bases. Furthermore, the claim that
> g_00 =/=g^00 has no objective meaning,

below...

>because it holds only in
> a particular choice of coordinates -- in *any* spacetime, one can
> *always* find coordinates in which g_00=g^00 (Gaussian normal
> coordinates), and in any spacetime,

below...

> including flat spacetime, one
> can *always* find coordinates in which g_00 =/= g^00. *Furthermore*,
> whether g_00=g^00 or not, in some particular coordinate system,
> has nothing to do with orthogonality, even orthogonality of the
> coordinate basis (though it is relevant to orthonormality).

Once you choose a CS, the general laws still hold.
To simplify I'll set the 3D spatial motion to zero,
and only examine the clock rates given by dx^0 and
dx_0 when the spatial 3D dx_i and dx^i are zero.

The general law requires

ds^2 = dx_0 dx^0

and by association it follows (when both dx i =0)

dx_0 dx^0 = g_00 dx^0 dx^0 = g^00 dx_0 dx_0 .

If one mistakenly sets g_00 = g^00 in a g-field,
then one also sets dx_0 = dx^0, and then ds=dx^0,
IOW's it transforms away the effect of a g-field
on a clock's rate.
Steve if you need more help with that problem,
I think I could explain that more deeply.

> > so I don't get the point of going to an orthonormal
> > basis as you used it.
>
> Yes, it's clear that you don't get the point.
>
> >> Notice that this is a *sum* of Gaussian curvatures. It is -- I hope
> >> -- obvious that a sum can be zero even if the individual items being
> >> added aren't.
>
> > No prob, but the relevance in nonorthogonal spacetime
> > escapes me.
>
> There is no such thing as a "nonorthogonal spacetime."

Sure there is, have a look at Dover's "PoR" pg.184 Eq.(8),
"as always with static problems...g_14 = g_24 = g34 =0",
those 3 metric look like velocity dependant nonorthogonal
spacetime metrics to me.

> >> Ken, you are making elementary mistakes about elementary Riemannian
> >> geometry.
>
> > I understand the classical approach
>
> No, you clearly don't.

Steve, I joined Mensa with a score of 99%+, I wrote
an employment exam and have IQ 15 points higher than
university profs so I'm disqualified from even being
a prof.

> Ken, please stop deluding yourself that you
> understand things when you very clearly don't.
> Go take a real class or two.

Steve, if I was ever a student in your class, I'd
end up doing the teaching, been there, done that.
If you want to aspire to become a theoretician
you focus on your ignorance, and ask yourself the
hard questions.
But I'd love to sit in on one of your lectures,
incognito.

> At least, stop insulting people who point out, correctly,
> your mistakes.

Aside from trivial ascii, there was no mistakes,
I followed Weinberg.
As I said Roberts threw the 1st spit-ball, he
does that when given a ref he doesn't understand,
that's his way.

> Steve Carlip

Regards
Ken S. Tucker

.

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