If a geodesic is a curvature in the geometry of space-time caused by the
presence of a massive body...

If that body is moving with constant velocity is the geodesic spherical?

Or does the geodesic reflect the distirbution of the mass of the object?

And in either case, if that body is under constant acceleration is the
geodesic displaced relative to the centre of the mass?

For example, if the body were spherical and under constant acceleration
would the geodesic become elliptical with the centre of mass at the foward
focal point (WRT the direction of motion)?

"Mark Hendy" wrote in message
...
If a geodesic is a curvature in the geometry of space-time caused by the
presence of a massive body...

No. A geodesic is a path of "extremal length." It is not curvature. These
are very different concepts. You can have an object in accelerated motion in
a gravitational field and have no spacetime curvature in the region in which
the particle is moving.

Pmb

Subject: Elliptical geodesics?
From: "Pmb"
Date: 9/27/03 5:22 AM US Mountain Standard Time
Message-id:


. You can have an object in accelerated motion in
a gravitational field...


A geodesic is a path of zero four-vector acceleration.

Mark Hendy wrote:
If a geodesic is a curvature in the geometry of space-time caused by the
presence of a massive body...

But a geodesic is not that. A geodesic is a PATH that extremizes its
path-length (however defined). In Euclidean geometry, geodesics are
straight lines. In GR, a timelike geodesic is the path a test particle
takes when there are no external forces impressed upon it (gravitation
is NOT a force).


If that body is moving with constant velocity is the geodesic spherical?

A geodesic is a 1-dimensional line, and "spherical" cannot apply.


Or does the geodesic reflect the distirbution of the mass of the object?

Hmmm. The geometric structure of spacetime determines the geodesics at
each point, and the distribution of mass-energy determines the geometric
structure (and is affected by that structure, so this is a nonlinear
problem).


And in either case, if that body is under constant acceleration is the
geodesic displaced relative to the centre of the mass? [...]

I have no idea what you are trying to ask.



Tom Roberts

(WaiteDavid137) wrote in message ...
Subject: Elliptical geodesics?
From: "Pmb"
Date: 9/27/03 5:22 AM US Mountain Standard Time
Message-id:


. You can have an object in accelerated motion in
a gravitational field...


A geodesic is a path of zero four-vector acceleration.

No **** **** sherlock. Tell me something I don't know. Why are you
even bothering to tell me something so obviously true? My post was
explaining why that persons notion of a geodesic is wrong - i.e. why
geodesics do not imply spacetime curvature. Sheesh! You sure love to
waste people's times with your nonsense.

If you want to impress someone the respond to the poster who asked the
question.


Mr. Pmb - waite's relativity teacher

Tom Roberts wrote in message ...
Mark Hendy wrote:
If a geodesic is a curvature in the geometry of space-time caused by the
presence of a massive body...

But a geodesic is not that. A geodesic is a PATH that extremizes its
path-length (however defined). In Euclidean geometry, geodesics are
straight lines.

The term "Euclidean geometry" seems not to have a unique meaning. Some
seem to say that it's a geometry in which Euclid's 5th postulate
holds. Others seem to say it is a geometry in which, when we choose
locally orthogonal coordinates (x1,x2,x3), the metric is

ds^2 = dx1^2 + dx2^2 + dx3^2

(See - "Relativity, The Special and General Theory -15th Edition,"
Albert Einstein, Three Rivers Press, (1959), page 102). When this is
the metric then its said "..the space of reference is said to be
Euclidean, and the co-ordinates Cartesian."
See - http://www.geocities.com/physics_wor...id_vs_flat.htm

I guess it's a difference between local and global?

Pmb

"Tom Roberts" wrote in message
...
Mark Hendy wrote:
If a geodesic is a curvature in the geometry of space-time caused by the
presence of a massive body...

But a geodesic is not that. A geodesic is a PATH that extremizes its
path-length (however defined). In Euclidean geometry, geodesics are
straight lines. In GR, a timelike geodesic is the path a test particle
takes when there are no external forces impressed upon it (gravitation
is NOT a force).


If that body is moving with constant velocity is the geodesic spherical?

A geodesic is a 1-dimensional line, and "spherical" cannot apply.


Or does the geodesic reflect the distirbution of the mass of the object?

Hmmm. The geometric structure of spacetime determines the geodesics at
each point, and the distribution of mass-energy determines the geometric
structure (and is affected by that structure, so this is a nonlinear
problem).


And in either case, if that body is under constant acceleration is the
geodesic displaced relative to the centre of the mass? [...]

I have no idea what you are trying to ask.

Tom Roberts

Ok it looks like I blew the definition of geodesic! I hate that!! Then lets
drop that name from the discussion because my real question is the last one
so I'll put it another way.

If a uniformly massive spherical body (which would cause a spherical
curvature in space-time) were under constant acceleration, is the
curvature in space-time displaced relative to the centre of the mass such
that the curvature becomes elliptical with the centre of mass at the foward
focal point of the ellipse (WRT the direction of motion)?

Gauge wrote:
The term "Euclidean geometry" seems not to have a unique meaning. Some
seem to say that it's a geometry in which Euclid's 5th postulate
holds. Others seem to say it is a geometry in which, when we choose
locally orthogonal coordinates (x1,x2,x3), the metric is

ds^2 = dx1^2 + dx2^2 + dx3^2

The usual definition of an N-dimensional Euclidean manifold is that it
has a Euclidean metric[#] everywhere, and has the topology of R^N. These
are sufficient to prove Euclid's postulates as theorems, and his
postulates are sufficient to prove these conditions as theorems.

So the two meanings are equivalent.

[#] One can find coordinates for which
ds^2 = dx^2 + dx2^2 + ... + dxN^2
I believe one can combine the two requirements into one
by requiring that these coordinates cover the manifold.


I guess it's a difference between local and global?

No. There's no difference when one uses the full definition. You forgot
the topological requirement, which is a global property of the manifold.
For instance, the cylinder SxR is not a 2-d Euclidean manifold, even
though one can apply a Euclidean metric to it. Note that one cannot
apply a Euclidean metric to any manifold with topology S^N for N1 --
that's why there are all those different projections for maps of the
earth's surface, which is S^2 (the paper onto which we project is
Euclidean).


Tom Roberts

Mark Hendy wrote:
If a uniformly massive spherical body (which would cause a spherical
curvature in space-time) were under constant acceleration, is the
curvature in space-time displaced relative to the centre of the mass such
that the curvature becomes elliptical with the centre of mass at the foward
focal point of the ellipse (WRT the direction of motion)?

This is a complex problem, and you have not given sufficient information
to solve it, even in principle. Any agency capable of imparting a
constant acceleration to a mass M must necessarily not be negligible
compared to M, and will therefore affect the geometry of spacetime. So
you must specify in detail precisely how the aceleration is imparted to
the mass.

In addition, you must specify what boundary conditions you want to
apply. IOW: how does the manifold behave "at infinity"?

In practice, such complicated situations have no analytical solution.
And numerical integration of the field equation has thorns....


Tom Roberts tjroberts2lucent.com

"Tom Roberts" wrote in message
...
Gauge wrote:
The term "Euclidean geometry" seems not to have a unique meaning. Some
seem to say that it's a geometry in which Euclid's 5th postulate
holds. Others seem to say it is a geometry in which, when we choose
locally orthogonal coordinates (x1,x2,x3), the metric is

ds^2 = dx1^2 + dx2^2 + dx3^2

The usual definition of an N-dimensional Euclidean manifold is that it
has a Euclidean metric[#] everywhere, and has the topology of R^N.

Who was refering to a manifiold? Do you have a source of these definitions?

[#] One can find coordinates for which
ds^2 = dx^2 + dx2^2 + ... + dxN^2
I believe one can combine the two requirements into one
by requiring that these coordinates cover the manifold.


I guess it's a difference between local and global?

No. There's no difference when one uses the full definition. You forgot
the topological requirement, ..

I didn't forget anything. I've never seen that as a requriement. Please
provide a source for this definition. Either from this list
http://www.geocities.com/physics_world/books.htm
or from the internet. Perhaps in Thorne and Blanchards new text?
http://www.pma.caltech.edu/Courses/p...002/index.html


which is a global property of the manifold.
For instance, the cylinder SxR is not a 2-d Euclidean manifold,

What is "the cylinder SxR"??? SxR????

Again - Who was talking about a manifold? I was speaking in general.
Submanifolds are obviously Euclidean. And on that note see Wald page 386

Pmb