Re: the basis of relativity

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)

> 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, 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, 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).

> 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."

>> Ken, you are making elementary mistakes about elementary Riemannian
>> geometry.

> I understand the classical approach

No, you clearly don't. Ken, please stop deluding yourself that you
understand things when you very clearly don't. Go take a real class
or two. At least, stop insulting people who point out, correctly,
your mistakes.

Steve Carlip
.