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| Jamie Vicary |
 Posted: Thu Feb 03, 2005 12:09 pm |
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Hi,
Is the field x^mu a valid vector field in a general curved
spacetime, where x^mu are the coordinates? Along the same lines, is
a^mu x_mu a true covariant scalar in a general spacetime, where a^mu is
a constant vector?
Thanks,
Jamie. |
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| tj Frazir |
| Posted: Sat Feb 05, 2005 5:33 pm |
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UP is still a gain in mass.
gravity is a push to less mass.
Gravity is the energy gain pushing the atom. ,,tell us how
gravity moves mass. Im doing it and proving it with the math . Your too
stupid to understand atoms push them selves to less mass. So then how
could you do the math ? You did not know the mass gain was doing
the pushing or how. All you had is a bunch of idias.
F is identical to the mass gain pushing the wieght of the atom.
Identical as in THE LAW.
Re: gravitons are for dumbasses.
The gain in mass is F pushing te wieght of the atom.
Newton had is right, F = dp/dt is right on!
"
F is the gain in mass and up is a gain in mass SAM . M is the
wieght
of the atom befor the gain in mass .
Gravity is the energy slope across the atom. all the mass of the
atom falls twards its center. Atoms change mass at C. So the tom
has more mass falling twards its center from one side than the other and
the atom pushes its self down the energy slope . V will be the
same for evry atom as the mass gain pushes the wieght and the gain is
allways proportinal to the mass. SAM wrote
F
= ma
Here F is the applied force, m is the mass of the
particle, and a = dv/dt is the particle's acceleration, with v being
the particle's velocity. This equation, together with the principle
that bodies act symmetrically on one another--so that the force
particle A feels from particle B is equal to the force B feels from
A--is the basis for understanding particle dynamics".
"Newton's law completely describes all the phenomena
of classical mechanics...." |
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| Edward Green |
| Posted: Sun Feb 06, 2005 12:04 am |
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Androcles wrote:
Quote: "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?
<...>
Quote: 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)
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>
Quote: 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.
I have no idea what you are talking about, but a tolerable confidence
that it is wrong, if that.
Quote: 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.
No. Spacetime is, within the context of the theory of special
relativity, simply a name given to the set of all points R,t, and in
the context of general relativity, a similar object with some
additional properties, and a dropping of the requirement that R, even
in a particular coordinate representation, be isomorphic to E^3.
<snip more silliness> |
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| Androcles |
| Posted: Sun Feb 06, 2005 12:59 am |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:1107666277.799138.315570@c13g2000cwb.googlegroups.com...
[snip silliness]
Quote: 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
[snip silliness]
Quote: 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.
[snip silliness]
Quote: 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
All silliness has been snipped. Nothing left to discuss.
Done.
Androcles. |
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| Bjoern Feuerbacher |
| Posted: Sun Feb 06, 2005 9:39 am |
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Ken S. Tucker wrote:
Quote: 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.
Then you'd only demonstrate again that you don't understand
GR.
As Steve Carlip said: the x^mu does *not* transform like
a vector field.
Quote: Slewing a bit, consider the following Invariant,
x_u dx^u = r*dr = s*ds
What makes you think that this is an invariant?
What do you mean with r, and with s?
Quote: Let r be in a Euclidian space so that the usual,
r^2 = x^2 +y^2.... hold
Err, we were talking about a *general* coordinate
system. Not only about Euclidian spaces.
Quote: 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?).
[snip remaining rant]
Bye,
Bjoern |
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| Ken S. Tucker |
| Posted: Sun Feb 06, 2005 5:08 pm |
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Bjoern Feuerbacher wrote:
Quote: 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. That group is important in unified field
theory, that's exterior to your interest, so there
is no need to concern yourself with it.
Ken |
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| Y.Porat |
| Posted: Mon Feb 07, 2005 6:02 am |
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BA in matemathics does sometimes
zero physicsts and pompous with big pretentions - like you
Y.Porat
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| Dirk Van de moortel |
| Posted: Mon Feb 07, 2005 6:36 am |
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"Y.Porat" <maporat@012.net.il> wrote in message news:1107774148.691843.102310@z14g2000cwz.googlegroups.com...
Quote: BA in matemathics does sometimes
zero physicsts and pompous with big pretentions - like you
Y.Porat
It would help if you would quote the message you are replying to.
I have no idea in which context you are saying this.
Of course I could try to find out, but since I have never seen
you write anything coherent in the past, I will not try to find
out. If you have something interesting to say, try to say it in
context, and I might reply.
Dirk Vdm |
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| Androcles |
| Posted: Mon Feb 07, 2005 12:03 pm |
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"Y.Porat" <maporat@012.net.il> wrote in message
news:1107774148.691843.102310@z14g2000cwz.googlegroups.com...
Quote: BA in matemathics does sometimes
zero physicsts and pompous with big pretentions - like you
Y.Porat
----------------------------------
Was that an attempt at communication?
Androcles. |
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| Edward Green |
| Posted: Thu Feb 10, 2005 6:48 pm |
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carlip-nospam@physics.ucdavis.edu wrote:
Quote: 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) <...
Yes. In general there is no valid claim of the coordinates to represent
vectors, but one might still mention for completeness a case where a
special set of coordinates in fact has that property; linear
coordinates in a flat spacetime. Linear coordinates look like -- in
fact are -- representations of displacement vectors from a given
origin: true vectors.* **
The coordinates still do not form a vector "field", implying a
collection of vectors each living in its own appropriate tangent space,
but can at least be argued to all be vectors in the single tangent
space associated with the origin of coordinates.
*By "linear" here I mean a slight generalization of "Lorentzian",
wherein the metric is not required to be diag(-1,1,1,1) -- loosely,
where the axes are not required to be orthogonal.
**[One might add some qualifying language like "the global coordinate
system induced by extension of the local representation of displacement
vectors" to satisfy the requirement that "vectors" have transformation
properties under arbitrary changes of spacetime coordinates; but mainly
as an i-doting academic flourish.] |
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| Bjoern Feuerbacher |
| Posted: Fri Feb 11, 2005 4:42 am |
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Ken S. Tucker wrote:
Quote: 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.
Quote: 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.
x_u dx^u can be calculated for any metric. So what group
are you talking about?
[snip]
Bye,
Bjoern |
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| Ken S. Tucker |
| Posted: Fri Feb 11, 2005 3:45 pm |
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Guest
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Bjoern Feuerbacher wrote:
Quote: 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 |
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| Bjoern Feuerbacher |
| Posted: Sat Feb 12, 2005 12:19 pm |
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Ken S. Tucker wrote:
Quote: 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?
Which part of what I said? The first or the second?
Bye,
Bjoern |
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| Ken S. Tucker |
| Posted: Sat Feb 12, 2005 9:19 pm |
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Guest
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Bjoern Feuerbacher wrote:
Quote: 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.
Ken |
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| Bjoern Feuerbacher |
| Posted: Mon Feb 14, 2005 5:05 am |
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Ken S. Tucker wrote:
Quote: 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.
Sorry. My original statement was not right.
Please rephrase it as:
"x_u dx^u makes perfects sense for any form of the
curvature tensor. It's just not invariant under general
coordinate transformations."
Quote: BTW, I trust you read Mr. Green's and Mr. Carlips
responses.
Yes. Apparently, you haven't, since you apparently
still insist that x_u dx^u *is* invariant.
Bye,
Bjoern |
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