"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:1107623649.284479.43630@g14g2000cwa.googlegroups.com...
Androcles wrote:

"Jamie Vicary" <jamievicary@gmail.com> wrote in message
news:cttlrs$f42$1@gemini.csx.cam.ac.uk...

Is the field x^mu a valid vector field in a general curved
spacetime, where x^mu are the coordinates?

2.0 "Coordinates", i.e. n-tuples, neither form nor fail to
form
vectors nor vector fields, though in some cases they may form
_representations_ of these objects.

n-tuples may form a vector such as (x,y,z)

2.1 The notation "x^mu" in general is unjustified. Coordinate
n-tuples are neither covariant nor contravariant, which names label
tangent vectors concepts: the variation of point coordinates under
change of coordinates lies outside the scope of this nomenclature.
The
coordinate label of a point on the manifold under a given
coordinate
system does not in general represent a tangent vector.

Even the smartest person in the universe may fail to recognize that
Einstein's assertion (tau0 + tau2) = tau1
doesn't mean tau() is a function of (x,0,0,0)
or even (x',0,0, x'/(c-v)) as he claims.
But, there are plenty of fruitcakes.

2.2 We must however immediately point on the exception to this
rule, as other in effect have but nebulously, that at least one
vector
space associated with each point in spacetime has the appearance of
a
space of "coordinate vectors" as requested, and that is the space
of
displacement vectors.


There is no such animal as spacetime. You are assigning properties to

the empty set.
"Spacetime" is a hype word used to sell books to unsuspecting
teenagers.

I was being pedantic. In this context, a vector is a geometric
object,
and the best an n-tuple may do is form a representation of it.

snip silliness

Even the smartest person in the universe may fail to recognize that
Einstein's assertion (tau0 + tau2) = tau1
doesn't mean tau() is a function of (x,0,0,0)
or even (x',0,0, x'/(c-v)) as he claims.
But, there are plenty of fruitcakes.

I have no idea what you are talking about, but a tolerable confidence
that it is wrong, if that.

There is no such animal as spacetime. You are assigning properties to

the empty set.
"Spacetime" is a hype word used to sell books to unsuspecting
teenagers.

[snip silliness]
snip more silliness

To Jamie and Steve ... et al...


Is the field x^mu a valid vector field in a general curved
spacetime, where x^mu are the coordinates?


If the condition, "in a general curved spacetime",
could be modified to one "in a spacetime consistent
with General Relativty" I'd say yes, but with hesitation.

Slewing a bit, consider the following Invariant,

x_u dx^u = r*dr = s*ds

Let r be in a Euclidian space so that the usual,

r^2 = x^2 +y^2.... hold

It follows that r*dr is easily calculated, and is
a specialized CS satisfying x_u dx^u.

Ken S. Tucker wrote:
It follows that r*dr is easily calculated, and is
a specialized CS satisfying x_u dx^u.

r*dr is not a coordinate system (that's what you meant
with "CS", right?).
Bjoern

BA in matemathics does sometimes
zero physicsts and pompous with big pretentions - like you
Y.Porat

BA in matemathics does sometimes
zero physicsts and pompous with big pretentions - like you
Y.Porat
----------------------------------

Jamie Vicary <jamievicary@gmail.com> wrote:

Is the field x^mu a valid vector field in a general curved
spacetime, where x^mu are the coordinates?

No. There are various technical ways of saying this (vector fields
live on the tangent bundle to the spacetime, while coordinates are
functions on the spacetime itself) <...

Bjoern Feuerbacher wrote:


Ken S. Tucker wrote:

It follows that r*dr is easily calculated, and is
a specialized CS satisfying x_u dx^u.


r*dr is not a coordinate system (that's what you meant
with "CS", right?).
Bjoern


Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,

that of course
has an infinite number of CS's.
But that is not the limitation of the *group*
where x_u dx^u can be calculated it's merely an
example.

Ken S. Tucker wrote:
Bjoern Feuerbacher wrote:


Ken S. Tucker wrote:

It follows that r*dr is easily calculated, and is
a specialized CS satisfying x_u dx^u.


r*dr is not a coordinate system (that's what you meant
with "CS", right?).
Bjoern


Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,

It even makes perfect sense when the curvature tensure
has any other form. It's just not an invariant then.

Bjoern Feuerbacher wrote:

Ken S. Tucker wrote:

Bjoern Feuerbacher wrote:



Ken S. Tucker wrote:


It follows that r*dr is easily calculated, and is
a specialized CS satisfying x_u dx^u.


r*dr is not a coordinate system (that's what you meant
with "CS", right?).
Bjoern


Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,

It even makes perfect sense when the curvature tensure
has any other form. It's just not an invariant then.


Can you prove that?

Ken S. Tucker wrote:

Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,

It even makes perfect sense when the curvature tensure
has any other form. It's just not an invariant then.

Can you prove that?

Which part of what I said? The first or the second?

Bjoern Feuerbacher wrote:

Ken S. Tucker wrote:


Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,

It even makes perfect sense when the curvature tensure
has any other form. It's just not an invariant then.

Can you prove that?

Which part of what I said? The first or the second?


The latter, it need not be rigorous a hint
on how I could do it would helpful, then I'll
try to work it out myself.

BTW, I trust you read Mr. Green's and Mr. Carlips
responses.

It's been pointed out to me that flat spacetime maps more appropriately
to an affine space.