Re: GR1916 question about g=1?

On Nov 25, 9:16 pm, Koobee Wublee <koobee.wub...@xxxxxxxxx> wrote:
On Nov 25, 4:58 pm, JanPB <film...@xxxxxxxxx> wrote:



On Nov 24, 11:33 am, Tom Roberts wrote:
A nit, but an important one in this discussion: that is NOT the metric,
that is the line element. The line element expresses the length of a
differential element (ds) of a line in terms of the coordinate
differentials of the element and the metric COMPONENTS projected onto
that coordinate basis.

Please convince me otherwise but I am unaware of any interpretation of
ds^2 in terms of what you've just said that does _not_ involve ill-
defined metaphors like "infinitesimal".

The only well-defined meaning of the symbol "ds^2" that I'm aware of
is that it's a rank-2 covariant tensor (the metric itself), with each
of the dq^i being a coordinate 1-form and multiplication of the dq's
(the multiplication sign being usually omitted) being the symmetric
tensor product.

Clearly, ds^2 is a scalar mathematically with rank-0. It is
impossible to fudge it into a ran-2 tensor or even a matrix.
Professor Roberts is very correct on this one. <shrug>

If ds^2 were a scalar then it would have a value at each point. So
tell us what is the value of ds^2 at (x,y,z,t) = (0,0,0,0) in the
Minkowski space? Remember, you've just said that it was a number -
what is it?

Thanks for using my symbols to describe coordinate (not to be confused
with displacement).

Dream on :-)

(This is simply a precise way of saying that the metric is a function
which to every manifold point assigns a symmetric bilinear form of
tangent vectors at that point - namely, their dot product.)

This is done because infinitesimals are illegal mathematical objects
(ignoring the super-special-idiosyncratic cases like non-standard
analysis) and the only legal interpretation of infinitesimal is as
either "tangent vector" or "1-form", depending on the requirement at
hand.

Have you taken calculus? After all, the mathematics involved is at
least 200 years old.

Infinitesimals are not well-defined mathematical objects (except in
certain esoteric mathematical contexts that so far have not been seen
in physics) and have status similar to that of the delta function
before Laurent Schwartz. Is this news to you? In mathematics/physics
the actual tool used whenever one talks about infinitesimals employs
various types of limiting processes (which are well-defined). It is
the notion of limit which is encapsulated in the concept of "tangent
vector" and "1-form" (aka. "tangent covector").

I see no way of avoiding this without letting poetry in. (Nothing
wrong with poetry but one must know which is which.)

There cannot be any poetry in mathematics. <shrug>



The symbol ds^2 is used (although it's not a square of anything unless
pulled back to a curve) because it has the right connotations: given a
curve gamma on the manifold, the metric ds^2 pulled back to the curve
is:

ds^2 = g_ij dgamma^i/dt dgamma^j/dt dt^2

(a 1-by-1 matrix in front of dt^2 - curves are 1-dimensional)

...whereas the 1-volume element on gamma (i.e. the element of length)
is:

dvol = sqrt(det(< d/dt, d/dt >)) dt

...where < , > is the metric g. On the curve the matrix <d/dt,d/dt> is
a 1-by-1 matrix so "det" is a no-op, while:

<d/dt,d/dt> = g_ij dgamma^i/dt dgamma^j/dt

Give it up and study. Coordinate is as simple as it can be that it
cannot be described any simpler. It is observer's choice of ruler.

You have no idea what I said above, so why do you comment on it?
Weird.

It is so by definition. It has nothing to do with the curvature in
space or spacetime --- not the least with gamma whatever that is.

It is not so. Coordinates are not choice of ruler.

...so we see that the element of 1-volume (length) dvol is precisely
what's usually denoted by "ds" - i.e., a "square root" of ds^2.

None of this is meant to deny the intuition the "line element"
interpretation of ds^2 provides.

If you are passing coded messages to Professor Roberts, I'd appreciate
if you do so through your private mail. Thanks in advance.

Just refrain from posting if you don't know what to say.

--
Jan Bielawski
.

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