Tom Roberts wrote:

Sure they do. The 4-vector itself is a geometic quantity with
physical
relevance and meaning -- specifically it can be intrinsic to an
object,

The norm of the 4-vector is 1, it's just a definition,
relating space & time.

and can appear in physical equations that manifestly satisfy the PoR.

The components of a 4-vector depend explicitly on the coordinate
system
on which the 4-vector is projected -- this makes it impossible for
them
to be intrinsic to any object, or to appear in physical equations
except
in certain combinations with the components of other 4-vectors.

But there is an isomorphism present that permits sloppy
authors to confuse a vector with its components and
get away with it, as long as one does not look too
carefully. But your original question was "too careful"
for the sloppiness to hold up.

Tom you and your side-kick Hooba are confused,
the rest of the worlds reputable GRist's aren't.
You've been cranking your kooky version of
differential calculus religiously, simply because
your ilk simply does not understand tensor analysis
and so you decry it, just as the standard cranks
decry relativity.
You've posted Weinberg and Einstein both Nobel
prize winners use sloppy notation. What you utterly
fail to realize is why components U^u and U_u are
different. I've even tried to help you, you're DOA,
because you have pre-determined Weinberg is sub-
standard, because you don't understand it.

I've been fortunate because some of the world's
finest mathematician's have taken the time with
me, face to face, and one on one, to dicuss the
ignorance a student has about the ragged edges
of mathematics. In particular, take very good
care of your assumptions, they may become your
prison, by prejudice.
Regards
Ken S. Tucker













































Tom Roberts

"Ken S. Tucker" wrote in message ups.com...

[snip]


I've been fortunate because some of the world's
finest mathematician's have taken the time with
me, face to face, and one on one, to dicuss the
ignorance a student has about the ragged edges
of mathematics.

Would that be this particular "student":
http://groups.google.co.uk/groups?th....dstc.edu.a u
http://groups.google.co.uk/groups?th...23i7@ 4ax.com
?

Dirk Vdm

VD, the institute allowed you access to
a computer, isn't it wonderful that that the
first person you choose was me.
Seriouslly have you figure out U_i yet?
Ken

"Ken S. Tucker" wrote in message
ups.com...
Tom Roberts wrote:

Sure they do. The 4-vector itself is a geometic quantity with
physical
relevance and meaning -- specifically it can be intrinsic to an
object,

The norm of the 4-vector is 1, it's just a definition,
relating space & time.

and can appear in physical equations that manifestly satisfy the PoR.

The components of a 4-vector depend explicitly on the coordinate
system
on which the 4-vector is projected -- this makes it impossible for
them
to be intrinsic to any object, or to appear in physical equations
except
in certain combinations with the components of other 4-vectors.

But there is an isomorphism present that permits sloppy
authors to confuse a vector with its components and
get away with it, as long as one does not look too
carefully. But your original question was "too careful"
for the sloppiness to hold up.

Tom you and your side-kick Hooba are confused,
the rest of the worlds reputable GRist's aren't.
You've been cranking your kooky version of
differential calculus religiously, simply because
your ilk simply does not understand tensor analysis
and so you decry it, just as the standard cranks
decry relativity.
You've posted Weinberg and Einstein both Nobel
prize winners use sloppy notation. What you utterly
fail to realize is why components U^u and U_u are
different. I've even tried to help you, you're DOA,
because you have pre-determined Weinberg is sub-
standard, because you don't understand it.

I've been fortunate because some of the world's
finest mathematician's have taken the time with
me, face to face, and one on one, to dicuss the
ignorance a student has about the ragged edges
of mathematics. In particular, take very good
care of your assumptions, they may become your
prison, by prejudice.
Regards
Ken S. Tucker


LOL! Four coordinates do not a vector make.
Androcles.

Tom Roberts wrote:
Philo wrote:

Tom Roberts wrote:

Work with the real 4-vector and your confusions disappear.


Sorry, you divert from the topic.


No, I didn't. The distinction between a 4-vector and its components is
_essential_ here. But you don't seem to know the difference, and quote
books that make the same mistake and promote the same confusion.

I know very well the distinction between a vector as an algebraic object
and his components. The distinction is not relevant for the problems,
which I have pointed out here.

If we define a basis in a vector-space, then a vector is uniquely
defined by his components. If in R^2 no other basis is given explicitly,
it is clear for every mathematician, that all components are related to
the standard-basis (1,0), (0,1). Rindler is not confused and I'm not
confused.

"Philo" wrote in message
...
Tom Roberts wrote:
Philo wrote:

Tom Roberts wrote:

Work with the real 4-vector and your confusions disappear.


Sorry, you divert from the topic.


No, I didn't. The distinction between a 4-vector and its components is
_essential_ here. But you don't seem to know the difference, and quote
books that make the same mistake and promote the same confusion.

I know very well the distinction between a vector as an algebraic object
and his components. The distinction is not relevant for the problems,
which I have pointed out here.

If we define a basis in a vector-space, then a vector is uniquely
defined by his components.

That is not the same as saying the vector is its components in a particular
basis. It is not. Because something is isomorphic to something does not
make it the same as the thing it is isomorphic to. Indeed that is the whole
point - a vector is isomorphic to many representations in different basis so
obviously it can not be the same as every one of those reorientations - that
would physically be a silly way to look at the objects.

If in R^2 no other basis is given explicitly,
it is clear for every mathematician, that all components are related to
the standard-basis (1,0), (0,1).

Which basis would that be? Draw a Cartesian coordinate system. Rotate it
and place it anywhere you like and you have a valid 'standard' coordinate
system from which to get 'standard' basis vectors.

Rindler is not confused and I'm not confused.

In R2 X = a*X1 + b* X2 where X1 and X2 are basis vectors ie a vector is the
sum of vectors that form the basis. Saying that the vector is determined by
a and b in that basis is not the same as saying the vector is [a,b]. The
vector is an object not dependant on a particular basis. In R2 for example
a vector is defined as a line with magnitude and direction - such exists
independent on any coordinate system. The same in SR - events exist
independent of any particular coordinate system but it is assumed they form
a vector space - and it is pretty obvious physically they do. Indeed look
at the definition of a vector space from linear algebra - the objects of
that definition contain nothing about basis - they are things introduced
later for mathematical convenience.

Bill

Bill Hobba wrote:

Philo wrote:

If in R^2 no other basis is given explicitly,
it is clear for every mathematician, that all components are related to
the standard-basis (1,0), (0,1).

Which basis would that be?

I can't believe it. Do you know the definition of the vector-space R^2?
The elements of R^2 are pairs of real numbers. The vector (1,0) is a
pair of numbers: ONE and ZERO. These numbers are NOT coordinates. But
the coordinates of the vector (1,0) in the standard basis are also 1 and 0.

Draw a Cartesian coordinate system. Rotate it
and place it anywhere you like and you have a valid 'standard' coordinate
system from which to get 'standard' basis vectors.

Not true. The elements of the new basis are pairs of new numbers, not
(1,0) and (0,1). But her coordinates are 1 and 0, respectively 0 and 1.

Mr. Philo
(got a real name?)

Anyway, it doesn't take long to figure Hooba
to be our local idiot.

The 4-velocity norm is a unit=1,

U_u U^u =1 , summed {u=0,1,2,3}

I think in spacetime this means everything has that characteristic.

I know you and all are likely to reject my argument, but I think we all
have one universe composed of different relative experiences .

I sense the commonality is U_u U^u=1
standing for that one universe we share.

Now, I have found a need to eliminate
absolute spatial motion, and get yelled
at alot in this NG...so to do that I
thought the guy's would agree with,

U_i U^i =0 , {i=1,2,3}

For complicated reasons

U_i=0

U^i = relative displacement.

Regards
Ken S. Tucker

"Philo" wrote in message
...
Bill Hobba wrote:

Philo wrote:

If in R^2 no other basis is given explicitly,
it is clear for every mathematician, that all components are related to
the standard-basis (1,0), (0,1).

Which basis would that be?

I can't believe it. Do you know the definition of the vector-space R^2?

Sure do. Look up the definition of vector space -
http://mathworld.wolfram.com/VectorSpace.html. R^2 is the Euclidian 2
dimensional sapce. Such a space is isomorphic to [a, b] but is not
necessarily [a, b] - eg the set of all line segments in the Euclidian plane
forms a vector space of dimension 2 and is not [a, b].

The elements of R^2 are pairs of real numbers.

Not necessarily - as any book on vector analysis or even linear algebra will
tell you.

The vector (1,0) is a
pair of numbers: ONE and ZERO. These numbers are NOT coordinates. But
the coordinates of the vector (1,0) in the standard basis are also 1 and
0.

Philo wrote:
Tom Roberts wrote:
No, I didn't. The distinction between a 4-vector and its components is
_essential_ here. But you don't seem to know the difference, and quote
books that make the same mistake and promote the same confusion.

I know very well the distinction between a vector as an algebraic object
and his components.

But the crucial distinction is between a 4-vector as a _PHYSICAL_ object and its
components. Physics is not math, and "algebraic objects" are not necessarily
physical objects, or physical properties of physical objects.


If we define a basis in a vector-space, then a vector is uniquely
defined by his components.

Sure. So what? A set of four real numbers (presumably the components of a
4-vector) is not a 4-vector. As I keep saying, it is the 4-vector itself that
has physical relevance, not any of its many sets of components. shrug

Careful authors do not ignore an isomorphism, such as the fact
that "If we define a basis in a vector-space, then a vector is
uniquely defined by his components" -- 'defined by' is not at
all 'the same as'.


If in R^2 [...]

This is supposed to be physics, not math.

In theoretical physics we work in a manifold M that is a model of the real
world, and in SR/GR it is a semi-Riemannian manifold of 3+1 dimensions.
Attempting to work in R^2 is hopeless. To do what you did in M requires first
that one establish a diffeomorphism from M to R^4, and use it to pull back the
standard basis of R^4 to TM* (the cotangent spaces of M). It clearly makes no
sense to claim that (1,0,0,0) is a basis 4-covector in TM* [the corresponding
basis 4-covector is the pullback of (1,0,0,0)]....

This last is inherently contravariant; if you want a covariant basis
you need a metric to establish the usual isomorphism between vectors
and covectors in TM and TM*.... I am ignoring some details here....

Note this is all just basic differential geometry....


Bottom line: if you want to discuss the details of the foundations of any
physical theory, you need to be precise in your language and mathematics. This
is the case if you want to discuss the physical relevance of different
mathematical structures.


Tom Roberts