Tom Van Flandern wrote:
[This replies to Saul and Steve Carlip.]

[...]
Fixing all these poorly worded statements in the way I just described (a
risky thing to do, I admit, because this assumes I know what you were trying
to say), I think your main point is that the potential field as a whole is
unchanging as long as the source mass is unchanging, and that therefore
gradients of that potential field everywhere are likewise unchanging. That
much is trivially true, but ignores the case that matters. The gradient of
that same potential field as seen and experienced by a moving target body is
ever varying in direction,

Yes.

and therefore has instantaneous and retarded directions that differ.

No. At a given point in space and a given time, a potential has a unique
gradient.

[Carlip]: Note that when the orbiting object is at a position (x,y,z), the
force is determined by the gradient of the potential at (x,y,z), at the
time the object is at that location.

What you say is certainly true for a non-moving target body. But why
would the gradient of the potential in the source mass???s frame be more
important to a moving target body than the gradient of the same potential as
seen and experienced by the moving target body?

An ordinary "potential" is a scalar. It is independent of coordinates, i.e.,
independent of frame. The gradient of a scalar is a vector. Its direction and
magnitude are also independent of coordinates. (The components of a vector
may change depending on what basis I use, but the vector itself doesn't --
a vector in the plane doesn't magically change direction if I use polar
coordinates rather than Cartesian coordinates to describe it, and a vector
in spacetime doesn't magically change if I use a moving frame rather than
a stationary one to describe it.)

[...]
[Carlip]: Please tell me what the "retarded gradient" is.

The same mathematical function calculated in the frame of the moving
target body, assuming the gradient must always be toward the source mass.

The gradient of a function is frame-independent. The phrase "assuming the
gradient must always be toward the source mass" makes no sense -- given a
function, you can calculate its gradient, and that *tells* you the direction.
You don't calculate and then independently decide what direction the vector
points!

Because the source mass direction varies as seen from the target body, so
goes the field gradient direction vary from moment to moment. The spatial
partials change with time.

Yes, of course. That's not the issue -- the issue is what, at any given time,
the gradient is.

[Carlip]: For example, here's a function: F(x,y,z,t) = 1/sqrt{ (x-at)^2 +
(y-bt)^2 + (z-ct)^2} I know how to compute its gradient at any position
and time. Please write down its "retarded gradient."

You have mentioned only one coordinate system. The question itself
betrays that you don???t understand retarded gradients.

Tell me the "retarded gradient" in the coordinate system I've given. Or choose
whatever coordinate system you want, transform to that system, and tell me
the "retarded gradient" there.

Now if we imitated the
case of a target body on a circular orbit, then we would introduce the
target???s body???s frame wherein the source mass has coordinates (X,Y,Z,t), and
X = r sin nt, Y = r cos nt, Z = 0, r^2 = x^2 + y^2 + z^2, n = angular
velocity of source mass around target body. Then when we do a coordinate
transformation from (x,y,z,t) to (X,Y,Z,t), your function F becomes a
function of time, and so does any derivative of F taken in the (X,Y,Z)
frame.

No kidding. The function I gave depends on time -- see the little "t" in it?
What's its "retarded gradient"?

[TomVF]: *After* you determine the geodesics in that metric, you must
still compute a gradient (or take the equivalent spatial partials) to get
the 3-space force/acceleration.

[Carlip]: This statement demonstrates clearly that you don't understand
basic general relativity. The geodesics of the metric *are* the paths of
the target bodies. Once you know the path -- the position as a function of
time -- you can use any coordinate system you like, and any definition of
acceleration you like. The answer is uniquely determined.

Maybe all our issues are terminology. The (time-like) geodesic equations
are not the path, but merely allow the path to be determined by taking
partials to form a gradient and *assuming* that this gradient represents a
force.

You evidently don't know what the geodesic equation is.

Otherwise, the target body would remain in its initial state relative
to the source mass, as it must if no force acts on it. 3-space dynamics are
not present in the geodesic equations.

You evidently don't know what the geodesic equation is. It is a set of
coupled second order ordinary differential equations for the second derivative
of the position. Given an initial position and an initial velocity, the geodesic
equation determines a unique path (x(t),y(t),z(t)). No further assumptions
are needed.

[...]
[Carlip]: The geodesic *is* the path. It's (x(t),y(t),z(t)). Once you give
an initial position and velocity, this path is completely and uniquely
determined by the geodesic equation. No further assumptions are needed.

Are you talking about trivial cases such as space-like geodesics or null
geodesics here?

No.

Are you talking about a spacetime path instead of a 3-space path?

No.

Those cases are irrelevant to orbital motion, our topic here. So
please look at the geodesic equations at
http://metaresearch.org/cosmology/gravity/spacetime.asp and tell me how to
compute a simple circular orbit from those equations without making some
assumption about force.

Are you sure referred to the right place? The site you cite doesn't have the
geodesic equation on it at all. If you *really* don't know what the geodesic equation
is, try http://en.wikipedia.org/wiki/Geodesi..._relativity%29.

Geodesic equations per se contain no 3-space dynamics. They merely tell
us how proper time differs from coordinate time along any path.

OK, now I'm convinced -- you don't know what the geodesic equation is.

[...]
I???m guessing you must have never understood what the
Einstein-Infeld-Hoffmann or Robertson-Noonan or Damour-Deruelle papers,
deriving 3-space equations of motion, are all about. Why do those papers
exist if the geodesic equations contain all the information we need about
orbits?

Those papers show that one can *derive* the geodesic equation from the field
equations of general relativity, rather than postulating it separately.

[Carlip]: You claim that there is a mathematical operation called
"retarded gradient." Define it! Given a field, as a function of position
and time in a given coordinate system, tell me the mathematical procedure
for computing its "retarded gradient."

For a moving target body, the instantaneous gradient points toward the
instantaneous direction of the source mass, and the retarded gradient points
toward the retarded direction of the source mass. Surely you don???t need me
to write the corresponding ASCII equations for you to grasp such a simple
physics concept.

So you're saying that to find either of the "gradients" at the location of, say,
the Earth, I can't just look at the potential -- I have to break it up into
"the piece due to the Sun," "the piece due to the Moon," "the piece due to
Jupiter," etc., and calculate all their directions separately? That may be how
you do gravity, but it has nothing to do with general relativity.

Tom, before we fo further, I suggest that you take a few days and learn what the
geodesic equation is.

Steve Carlip

Ken S. Tucker wrote on Mon, 09 Jun 2008 12:46:45 -0700:

On Jun 9, 3:57 am, "Juan R." González-�lvarez

I think you're best to postpone a book until you have things
clarified.

What book Ken? I did not speak about any book. I think you dream that
sometimes as you dream the nonsense of 1908 :-)

So you never studied Minkowski's 1908 SPACETIME article. Where did you
study science?

Not in Tucker University :-)

Are you sure you are replying me? or who you are? or do you just
dream that and typed? :-)

Oh, I stand corrected, I thought you were publishing a book.

Ha ha ha :-)
I recommend you to read other's posts *before* replying.

LOL, when I see evidence that you understand physics, I'd be happy to
review your article whatever it is.
There are a lot of kooks who can never under- stand relativity and
therefore regard it as wrong. OTOH Juan, you seem rather bright, so I'm
willing to review your work, as it compares to Minkowski's 1908
analysis, etc.

Oh! But i did *not* said the relativity is wrong. I already explained
several times that SR and GR are recovered as special case from a more
general post-relativity (PR) theory.

Ken, sincerely, read posts *before* reply.

--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org

On Jun 18, 1:22 pm, "Tom Van Flandern" wrote:

[...] Because that is so plainly true, I am mystified how you can say "no" to
my conclusion that the gradient has instantaneous and retarded directions
that differ. Anything changing continuously will look one way now and
another way one light-time ago...

Who now gives a damn?

So, this is what peer review is all about. All parties each having
each own agenda just shout at the opposite direction without trying to
understand each other. It is so sad like a marriage leading to
eventual failed path.

As a moderator or a referee, I say (also pointed out by Dr. Van
Flandern) Professor Carlip’s argument of the aberration of gravity is
utterly wrong because it only involves the speed of the gravitating
mass relative to the center of mass. As Dr Van Flandern has pointed
out, the speed of the gravitated mass must also be included into
consideration. In doing so, it is going to make the matter even
worse.

Without the aberration of gravity, the speed of gravity must be
several billion times the speed of light in order to explain the
stability of the solar system. This is all in the very simple
mathematics involved. With aberration of the relative speed of the
gravitated and the gravitating masses, the speed of gravity must be
several thousand times on top of that. There appears to be no miracle
for Professor Carlip to pull this one out.

However, this episode does not show Dr. Van Flandern to be ever so
wiser than other physicists. Dr. Van Flandern should have shut his
mouth after claiming the speed of gravity is several billions time the
speed of light. Instead, he went on to promote LT as a general case
to the Galilean transform to justify his claim. As easily shown
mathematically, LT does not even degenerate into the Galilean
transform at low speed. Thus, Dr. Van Flandern should not be
construed as someone having superior intellect or knowledge than
Professor Carlip.

In fact, anyone who promotes a general transform that does not
degenerate into the Galilean transform at low speeds is utterly
ignorant of basic physics if you ask for my humble opinion. Any
conjecture as a general case to the Newtonian world of physics must
degenerate into the Newtonian world of physics. This should be a no-
brainer. shrug

"Koobee Wublee" writes:

[Wublee]: ... claiming the speed of gravity is several billions time the
speed of light. ... promote LT as a general case to the Galilean transform
to justify his claim.

If special relativity is correct, then nothing can propagate faster than
light in forward time. So it is necessary to show that there exists a viable
alternative to special relativity that passes all experimental tests and yet
has no universal speed limit. Lorentzian relativity
(http://metaresearch.org/cosmology/gravity/LR.asp) is just such a model.

[Wublee]: As easily shown mathematically, LT does not even degenerate into
the Galilean transform at low speed.

True, but irrelevant, because there is no connection. Lorentzian
relativity changes the physics behind the Lorentz transformation. “T” is no
longer time, but is instead the readings on an atomic clock. As anyone
familiar with GPS and other modern experiments knows, clocks do slow down
with speed and with stronger gravitational potentials. But nothing whatever
happens to space or time, so the Galilean transformations still apply the
same as always in classical physics.

In Lorentzian relativity, the Lorentz transformations simply describe
how much speed affects the rate of ticking of certain kinds of clocks. This
is exactly analogous to increasing temperature causing a pendulum clock to
slow down. Changing the rate of ticking of a clock does not imply that
anything has happened to the rate of passage of time.

[Wublee]: In fact, anyone who promotes a general transform that does not
degenerate into the Galilean transform at low speeds is utterly ignorant
of basic physics if you ask for my humble opinion.

One should not be so focused on a principle as to misunderstand the
message. -|Tom|-


Tom Van Flandern – Sequim, WA - see our web site on frontier astronomy
research at http://metaresearch.org

"Tom Van Flandern" wrote in message
...
| "Koobee Wublee" writes:
|
| [Wublee]: ... claiming the speed of gravity is several billions time the
| speed of light. ... promote LT as a general case to the Galilean transform
| to justify his claim.
|
| If special relativity is correct, then nothing can propagate faster
than
| light in forward time. So it is necessary to show that there exists a
viable
| alternative to special relativity that passes all experimental tests and
yet
| has no universal speed limit. Lorentzian relativity
| (http://metaresearch.org/cosmology/gravity/LR.asp) is just such a model.
|
| [Wublee]: As easily shown mathematically, LT does not even degenerate
into
| the Galilean transform at low speed.
|
| True, but irrelevant, because there is no connection. Lorentzian
| relativity changes the physics behind the Lorentz transformation. “T” is
no
| longer time, but is instead the readings on an atomic clock. As anyone
| familiar with GPS and other modern experiments knows, clocks do slow down
| with speed and with stronger gravitational potentials. But nothing
whatever
| happens to space or time, so the Galilean transformations still apply the
| same as always in classical physics.
|
| In Lorentzian relativity, the Lorentz transformations simply describe
| how much speed affects the rate of ticking of certain kinds of clocks.
This
| is exactly analogous to increasing temperature causing a pendulum clock to
| slow down. Changing the rate of ticking of a clock does not imply that
| anything has happened to the rate of passage of time.
|
| [Wublee]: In fact, anyone who promotes a general transform that does not
| degenerate into the Galilean transform at low speeds is utterly ignorant
| of basic physics if you ask for my humble opinion.
|
| One should not be so focused on a principle as to misunderstand the
| message. -|Tom|-
|
|
| Tom Van Flandern – Sequim, WA - see our web site on frontier astronomy
| research at http://metaresearch.org

One should not be so focused on Lorentzian crap as to misunderstand the
data. -|Androcles|-

Androcles – England - see our web site on frontier astronomy
research at http://www.androcles01.pwp.blueyonder.co.uk/

Tom Van Flandern wrote:
Steve Carlip writes:

[...]
[Carlip]: The geodesics of the metric *are* the paths of the target
bodies. Once you know the path -- the position as a function of time --
you can use any coordinate system you like, and any definition of
acceleration you like. The answer is uniquely determined.

[TomVF]: Maybe all our issues are terminology.

[Carlip]: You evidently don't know what the geodesic equation is. It is a
set of coupled second order ordinary differential equations for the second
derivative of the position. Given an initial position and an initial
velocity, the geodesic equation determines a unique path (x(t),y(t),z(t)).
No further assumptions are needed.

My terminology conjecture seems confirmed. We use different terminology
in celestial mechanics from what relativists are using these days. In fact,
"geodesic equation" has been abandoned in my field for most purposes because
the orbit is obviously not the shortest distance (the original meaning of
"geodesic") between two points or events in 3-space.

So let me get this straight. We're talking about general relativity, I refer
to the geodesic equation, and you don't know what that means. Well, OK...

Now that we're agreed on what "geodesic" means in general relativity --
i.e., that it means what you'll find in any textbook on GR -- do you also
agree that according to general relativity, small bodies move along
geodesics? (If you aren't sure of this, there's a very nice recent proof by
Gralla and Wald, http://arxiv.org/abs/0806.3293.)

Then we can get back to the example I raised earlier: a gravitating
object -- call it A -- moving at a constant velocity suddenly stops. What
happens to the motion of a test body B a distance R away from the point
that A stops?

In general relativity, you solve this problem as follows:

1. Write down a stress-energy tensor for the gravitating source. (Of
course, you have to include all sources -- if A stops because it hits a
wall, you'd better include the field of the wall as well.)

2. Solve the Einstein field equations to determine the metric, given
this stress-energy tensor. (There are nice existence and uniqueness
theorems, going back to Yvonne Choquet-Bruhat's work in the '50s,
that guarantee that this can be done, although in practice you often
need an approximation procedure.)

3. Given the metric from step 2, write down the geodesic equations.
(Once you have the metric, these are unique.)

4. Solve the geodesic equations to determine the motion of body B.
(Here, the existence and uniqueness theorems are centuries old; given
an initial position and velocity for B, the equations uniquely determine
its future motion.)

So now, calculate. Here are some hints:

1. Step 1 can be easy or hard depending on how you want to get your
source to stop moving. General relativity has internal consistency
requirements -- you can't just declare that it should stop by magic,
but have to put something in to stop it. Fortunately, there's an easy
way to do this, using Kinnersley's "photon rocket" ("Field of an
Arbitrarily Accelerating Point Mass," Phys. Rev. 186 (1969) 1335),
in which body A accelerates arbitrarily by emitting electromagnetic
radiation.

2. Finding exact solutions to the field equations is very hard in general,
but if you use the Kinnersley stress-energy tensor, he's already done
the work for you. Feel free to use a different stress-energy tensor, of
course, but it will be harder.

3. Step 3 is easy -- you'll find the general connection in my paper
gr-qc/9909087, eqn. (2.2).

4. Step 4 may be easy or hard, depending on the initial velocity of B
and on the way you make A stop. But you don't need the full solution
-- it's enough to look at the acceleration, and see how it changes.

If you do this, you will find that object B will initially continue to
accelerate toward the "extrapolated" position of object A, even though
A has stopped and never reaches that extrapolated position. After a
finite time t=R/c, the acceleration of B will abruptly change, to point
toward the position at which A has come to rest.

There is no room for ambiguity here. Given the mass, composition,
and motion of A, the stress-energy tensor is unique. Given the stress-
energy tensor, the metric is unique (up to extraneous gravitational
waves coming in from infinity). Given the metric, the geodesic equation
is unique. Given the geodesic equation and the initial position and
velocity of B, the motion of B is unique.

You may not like this answer. That's fine. But then *stop claiming that
you agree with the mathematics of general relativity*, when you very
clearly don't.

(I understand, of course, why you don't want to do this. GR is a very
successful theory, and if you reject it, you'll have to come up with a
different explanation of why, for example, the last term of line 2 of
eqn. (39.64) in MTW -- the equation you like to cite -- has a coefficient
of -3/2.)

[...]
[TomVF]: I'm guessing you must have never understood what the
Einstein-Infeld-Hoffmann or Robertson-Noonan or Damour-Deruelle papers,
deriving 3-space equations of motion, are all about. Why do those papers
exist if the geodesic equations contain all the information we need about
orbits?

[Carlip]: Those papers show that one can *derive* the geodesic equation
from the field equations of general relativity, rather than postulating it
separately.

But that "derivation" is not simply "turning the crank". One of the
assumptions that must be made along the way is the direction of the
gradient - instantaneous or retarded.

Unlike you, I have actually done this derivation, starting from the Einstein
field equations. No such assumption is needed. You are just making this up.

Steve Carlip

On Jul 4, 2:39*pm, wrote:
Tom Van Flandern wrote:

Steve Carlip writes:

[...]

[Carlip]: The geodesics of the metric *are* the paths of the target
bodies. Once you know the path -- the position as a function of time -- *
you can use any coordinate system you like, and any definition of
acceleration you like. The answer is uniquely determined.
[TomVF]: Maybe all our issues are terminology.
[Carlip]: You evidently don't know what the geodesic equation is. It is a
set of coupled second order ordinary differential equations for the second
derivative of the position. Given an initial position and an initial
velocity, the geodesic equation determines a unique path (x(t),y(t),z(t)).
No further assumptions are needed.
*My terminology conjecture seems confirmed. We use different terminology
in celestial mechanics from what relativists are using these days. In fact,
"geodesic equation" has been abandoned in my field for most purposes because
the orbit is obviously not the shortest distance (the original meaning of
"geodesic") between two points or events in 3-space.

So let me get this straight. *We're talking about general relativity, I refer
to the geodesic equation, and you don't know what that means. *Well, OK....

Now that we're agreed on what "geodesic" means in general relativity --
i.e., that it means what you'll find in any textbook on GR -- do you also
agree that according to general relativity, small bodies move along
geodesics? *(If you aren't sure of this, there's a very nice recent proof by
Gralla and Wald,http://arxiv.org/abs/0806.3293.)

Then we can get back to the example I raised earlier: a gravitating
object -- call it A -- moving at a constant velocity suddenly stops. *What
happens to the motion of a test body B a distance R away from the point
that A stops?

In general relativity, you solve this problem as follows:

1. *Write down a stress-energy tensor for the gravitating source. *(Of
course, you have to include all sources -- if A stops because it hits a
wall, you'd better include the field of the wall as well.)

2. *Solve the Einstein field equations to determine the metric, given
this stress-energy tensor. *(There are nice existence and uniqueness
theorems, going back to Yvonne Choquet-Bruhat's work in the '50s,
that guarantee that this can be done, although in practice you often
need an approximation procedure.)

3. *Given the metric from step 2, write down the geodesic equations.
(Once you have the metric, these are unique.)

4. *Solve the geodesic equations to determine the motion of body B.
(Here, the existence and uniqueness theorems are centuries old; given
an initial position and velocity for B, the equations uniquely determine
its future motion.)

So now, calculate. *Here are some hints:

1. *Step 1 can be easy or hard depending on how you want to get your
source to stop moving. * General relativity has internal consistency
requirements -- you can't just declare that it should stop by magic,
but have to put something in to stop it. *Fortunately, there's an easy
way to do this, using Kinnersley's "photon rocket" ("Field of an
Arbitrarily Accelerating Point Mass," Phys. Rev. 186 (1969) 1335),
in which body A accelerates arbitrarily by emitting electromagnetic
radiation.

2. *Finding exact solutions to the field equations is very hard in general,
but if you use the Kinnersley stress-energy tensor, he's already done
the work for you. *Feel free to use a different stress-energy tensor, of
course, but it will be harder.

3. *Step 3 is easy -- you'll find the general connection in my paper
gr-qc/9909087, eqn. (2.2).

4. *Step 4 may be easy or hard, depending on the initial velocity of B
and on the way you make A stop. *But you don't need the full solution
-- it's enough to look at the acceleration, and see how it changes.

If you do this, you will find that object B will initially continue to
accelerate toward the "extrapolated" position of object A, even though
A has stopped and never reaches that extrapolated position. *After a
finite time t=R/c, the acceleration of B will abruptly change, to point
toward the position at which A has come to rest.

There is no room for ambiguity here. *Given the mass, composition,
and motion of A, the stress-energy tensor is unique. *Given the stress-
energy tensor, the metric is unique (up to extraneous gravitational
waves coming in from infinity). *Given the metric, the geodesic equation
is unique. *Given the geodesic equation and the initial position and
velocity of B, the motion of B is unique.

You may not like this answer. *That's fine. *But then *stop claiming that
you agree with the mathematics of general relativity*, when you very
clearly don't.

(I understand, of course, why you don't want to do this. *GR is a very
successful theory, and if you reject it, you'll have to come up with a
different explanation of why, for example, the last term of line 2 of
eqn. (39.64) in MTW -- the equation you like to cite -- has a coefficient
of -3/2.)

[...]

[TomVF]: I'm guessing you must have never understood what the
Einstein-Infeld-Hoffmann or Robertson-Noonan or Damour-Deruelle papers,
deriving 3-space equations of motion, are all about. Why do those papers
exist if the geodesic equations contain all the information we need about
orbits?
[Carlip]: Those papers show that one can *derive* the geodesic equation
from the field equations of general relativity, rather than postulating it
separately.
* * But that "derivation" is not simply "turning the crank". One of the
assumptions that must be made along the way is the direction of the
gradient - instantaneous or retarded.

Unlike you, I have actually done this derivation, starting from the Einstein
field equations. *No such assumption is needed. *You are just making this up.

Steve Carlip

The curvature of motion taken by matter is steeper than the actual
space curvature taken by light.

Mitch Raemsch

On Jul 4, 3:39 pm, wrote:
Tom Van Flandern wrote:

My terminology conjecture seems confirmed. We use different terminology
in celestial mechanics from what relativists are using these days. In fact,
"geodesic equation" has been abandoned in my field for most purposes because
the orbit is obviously not the shortest distance (the original meaning of
"geodesic") between two points or events in 3-space.

So let me get this straight. We're talking about general relativity, I refer
to the geodesic equation, and you don't know what that means. Well, OK...

I believe you that Dr. Van Flandern does not understand the geodesic
equations.

Then we can get back to the example I raised earlier: a gravitating
object -- call it A -- moving at a constant velocity suddenly stops. What
happens to the motion of a test body B a distance R away from the point
that A stops?

GR aka the set of field equations does not address this issue despite
the hand-waving of Rob Low[e]. shrug

In general relativity, you solve this problem as follows:

1. Write down a stress-energy tensor for the gravitating source. (Of
course, you have to include all sources -- if A stops because it hits a
wall, you'd better include the field of the wall as well.)

Yeah, most of the time, this mystical quantity is null as in vacuum.
shrug

Actually, the field equations do not address this issue. It is only
so as an added interpretation. Dr. Van Flandern merely does not
support that interpretation of yours. shrug

2. Solve the Einstein field equations to determine the metric, given
this stress-energy tensor. (There are nice existence and uniqueness
theorems, going back to Yvonne Choquet-Bruhat's work in the '50s,
that guarantee that this can be done, although in practice you often
need an approximation procedure.)

Yes, that yields an infinite set of solutions where each solution or
metric confined already to a particular (and no others) set of
coordinate system must describe a different and unique universe.
shrug

3. Given the metric from step 2, write down the geodesic equations.
(Once you have the metric, these are unique.)

Yes, I do not dispute this point.

4. Solve the geodesic equations to determine the motion of body B.
(Here, the existence and uniqueness theorems are centuries old; given
an initial position and velocity for B, the equations uniquely determine
its future motion.)

This is the most difficult part.

So now, calculate. Here are some hints:

1. Step 1 can be easy or hard depending on how you want to get your
source to stop moving. General relativity has internal consistency
requirements -- you can't just declare that it should stop by magic,
but have to put something in to stop it. Fortunately, there's an easy
way to do this, using Kinnersley's "photon rocket" ("Field of an
Arbitrarily Accelerating Point Mass," Phys. Rev. 186 (1969) 1335),
in which body A accelerates arbitrarily by emitting electromagnetic
radiation.

Conceptually, it is irrelevant if a hard stop or a soft stop. shrug

All the special treatment that you are applying to the field equations
can easily be done so with the Newtonian law of gravity. Have you
thought about that?

2. Finding exact solutions to the field equations is very hard in general,
but if you use the Kinnersley stress-energy tensor, he's already done
the work for you. Feel free to use a different stress-energy tensor, of
course, but it will be harder.

Yes, in general, it is impossibly hard to find a solution after a set
of field equations is identified in which the choice of coordinate
system is already cast in concrete. However, conveniently, the
Newtonian world represents the simplest form of the field equations.
Is that coincidental or philosophical just like a discussion about the
sex of God?

3. Step 3 is easy -- you'll find the general connection in my paper
gr-qc/9909087, eqn. (2.2).

I thought Dr. Van Flandern has already pointed out your error in only
including the aberration of the gravitating body. The principle of
relativity demands the relative velocity between the gravitating and
the gravitated bodies.

4. Step 4 may be easy or hard, depending on the initial velocity of B
and on the way you make A stop. But you don't need the full solution
-- it's enough to look at the acceleration, and see how it changes.

Solving the geodesic equations is in general very hard. However, the
scenario we can easily correspond to is exponentially easier. shrug

If you do this, you will find that object B will initially continue to
accelerate toward the "extrapolated" position of object A, even though
A has stopped and never reaches that extrapolated position. After a
finite time t=R/c, the acceleration of B will abruptly change, to point
toward the position at which A has come to rest.

This means you also have to include the aberrational effect of the
gravitated body. shrug

There is no room for ambiguity here. Given the mass, composition,
and motion of A, the stress-energy tensor is unique.

The stress-energy tensor is zero in most of discussions. So, you call
that unique in which I would not because I believe there must be a
separation of shamanistic interpretation and logical understanding.
shrug

Given the stress-
energy tensor, the metric is unique (up to extraneous gravitational
waves coming in from infinity).

Yes, each metric is unique. However, each metric using the same set
of coordinate system is a valid solution to the field equations.
shrug

Given the metric, the geodesic equation
is unique. Given the geodesic equation and the initial position and
velocity of B, the motion of B is unique.

Yes.

You may not like this answer. That's fine. But then *stop claiming that
you agree with the mathematics of general relativity*, when you very
clearly don't.

I actually agree with half of what you are saying except the mysticism
part.

(I understand, of course, why you don't want to do this. GR is a very
successful theory, and if you reject it, you'll have to come up with a
different explanation of why, for example, the last term of line 2 of
eqn. (39.64) in MTW -- the equation you like to cite -- has a coefficient
of -3/2.)

Dr. Van Flandern is very shallow in which you and I can see right
through, but what you are doing to promote the mystical nature of the
speed of gravity is rather priestly. It takes faith to accept that.
shrug

But that "derivation" is not simply "turning the crank". One of the
assumptions that must be made along the way is the direction of the
gradient - instantaneous or retarded.

Unlike you, I have actually done this derivation, starting from the Einstein
field equations. No such assumption is needed. You are just making this up.

The field equations do not support a unique and finite speed of
gravity. Thus, you are merely making this up to justify your faith in
the validity of the field equations. shrug

Steve Carlip writes:

[Carlip]: a gravitating object -- call it A -- moving at a constant
velocity suddenly stops. What happens to the motion of a test body B a
distance R away from the point that A stops?

This has nothing to do with the issue on the table, the propagation
speed of gravitational force. It concerns only the propagation speed of
changes in the gravitational potential field, about which there is no
dispute -- it is speed c. So let's stay on topic, please.

We don't need A to be moving, then stop, as in your example. The issue
of relevance here is present even when A is permanently at rest and its
field is completely static. The direction of the source mass as sensed by an
orbiting target body is toward its true instantaneous position when the
target body or field point is at rest. And it is toward the source mass's
retarded position (retarded by the speed of gravitational force) when the
target body is orbiting. That's elementary physics.

The exact same statement is equally true if the source mass is moving,
then stops (your example). That "move, then stop" distraction just makes a
simple problem more complicated. Address your attention back to the static
field problem, a much simpler one, and we will start to make progress.

[Carlip]: In general relativity, you solve this problem as follows ...

Most of your message was about this irrelevancy. But we have no issues
between us about the math. It is only the physics behind the math that has
been corrupted in modern GR texts to avoid uncomfortable conclusions such as
gravitational force having to propagate c. When we straighten out that
physics for those who hide behind the math, we get a simpler, revealing
picture of the origin and nature of gravity -- neatly, one lacking in
paradoxes.

Indeed, many of the most significant paradoxes in modern physics
disappear in a single stroke once the light-speed limit is lifted. No more
quantum "great smoky dragons", no more questioning of deep reality, no more
miracles required. It's really a very pleasing picture once anyone gets past
the mental blocks put there by teachers who knew no better.

[Carlip]: There is no room for ambiguity here. Given the mass,
composition, and motion of A, the stress-energy tensor is unique. Given
the stress-energy tensor, the metric is unique (up to extraneous
gravitational
waves coming in from infinity). Given the metric, the geodesic equation
is unique. Given the geodesic equation and the initial position and velocity
of B, the motion of B is unique.

This again ignores the relevant point here, namely, the aberration of A
relative to B when B has a transverse motion. Show me where you see the
aberration of the source mass in the equations of motion for the target
body. You can't, because it isn't there. Instead of straining all
plausibility and your personal credibility by trying to invent some logical
reason for the missing aberration, why not accept the obvious, classical
physics reason: If the speed of (3-space) gravitational force propagation is
c, aberration is driven to zero.

[Carlip]: GR is a very successful theory, and if you reject it, you'll
have to come up with a different explanation of why, for example, the last
term of line 2 of eqn. (39.64) in MTW -- the equation you like to cite --
has a coefficient of -3/2.)

How very ironic. I take it you have no idea what each of these terms
means physically. You picked the only term that has no observable
consequences for any known astrophysical system, and is only non-zero
(although still negligible) when more than two massive bodies are involved.

That said, what cryptic point were you trying to make?

[TomVF]: I'm guessing you must have never understood what the
Einstein-Infeld-Hoffmann or Robertson-Noonan or Damour-Deruelle papers,
deriving 3-space equations of motion, are all about. Why do those
papers exist if the geodesic equations contain all the information we
need about orbits?

[Carlip]: Those papers show that one can *derive* the geodesic equation
from the field equations of general relativity, rather than postulating
it separately.

[TomVF]: But that "derivation" is not simply "turning the crank". One of
the assumptions that must be made along the way is the direction of the
gradient - instantaneous or retarded.

[Carlip]: Unlike you, I have actually done this derivation, starting from
the Einstein field equations. No such assumption is needed. You are just
making this up.

It was I who first pointed you to the original papers deriving the
equations of motion, not vice versa. If you think I'm just "making this up",
then show me where in any of these papers the aberration of the source mass
from the moving body is taken into account. Remember, setting that
aberration to zero, as is done in Newtonian gravity, means physically that
you are setting the speed of gravity to infinity because aberration is
simply the ratio of the transverse orbiting body speed to the force
propagation speed, and the only logical way it can fail to appear in the
foresaid equations for a non-zero orbit speed is by adopting an infinite
speed of gravity, thereby driving gravitational aberration to zero.

Obviously, there is no legitimate way out of this dilemma. That seems to
cause you to keep reverting to mathematical issues and ignoring the physics
behind the math. You are wrong to do so. The field equations ultimately tell
us only about the gravitational potential field. They can tell us nothing
about gravitational force without making some assumption about the
propagation speed of that force. Adopting zero aberration (as in GR) is
making an assumption that the force speed of infinite. -|Tom|-


Tom Van Flandern - Sequim, WA - see our web site on frontier astronomy
research at http://metaresearch.org

Tom Van Flandern wrote:
Steve Carlip writes:
[Carlip]: a gravitating object -- call it A -- moving at a constant
velocity suddenly stops. What happens to the motion of a test body B a
distance R away from the point that A stops?

This has nothing to do with the issue on the table, the propagation
speed of gravitational force. It concerns only the propagation speed of
changes in the gravitational potential field, about which there is no
dispute -- it is speed c.

We don't need A to be moving, then stop, as in your example. The
issue of relevance here is present even when A is permanently at rest
and its field is completely static. The direction of the source mass as
sensed by an orbiting target body is toward its true instantaneous
position when the target body or field point is at rest. And it is
toward the source mass's retarded position (retarded by the speed of
gravitational force) when the target body is orbiting. That's elementary
physics.

So you claim that the "gravitational force" in Steve's example is not
the "gravitational force" in yours (in general, abstracting away the
differences in physical situations). That is ridiculous.

You claim Steve's example is "propagation speed of
changes in the gravitational potential field". But the
gradient of the potential gives the force (in your
model), so "gravitational force" also "is speed c"
-- either that is true or you disbelieve mathematics.

Steve's example is a COUNTEREXAMPLE to your claim "The direction of the
source mass as sensed by [the] target body is toward [the source's] true
instantaneous position".

You seem to think that "orbiting" is somehow special, and your claims
apply only to that specific physical situation, and not others (such as
Steve's). That is ludicrous for what purports to be a general theory.

As has been repeatedly pointed out, for the situation you
discuss an approximation to GR is valid, and in that
approximation the "gravitational force" points directly
to the EXTRAPOLATED position of the source. For the
situations you consider, that EXTRAPOLATED position is
indistinguishable from its present position [#]. But for
Steve's situation they are different, and clearly show
the error in your claims, WHEN USING THIS APPROXIMATION
TO GR.

[#] This is why the experiments you cite do not refute GR.


The exact same statement is equally true if the source mass is
moving, then stops (your example). That "move, then stop" distraction
just makes a simple problem more complicated.

It is not too complicated for sensible people to think about. And it
clearly and succinctly refutes your claims about "speed of gravitational
force" and the direction of "gravitational force pointing to the
source's true instantaneous position", IN THIS APPROXIMATION TO GR.

I repeat: your basic problem is confusing NG with GR.
Indeed, you even confuse NG with this APPROXIMATION to GR.


Address your attention
back to the static field problem, a much simpler one, and we will start
to make progress.

It is not possible to ascribe a "speed" to a static field. That is, in a
static situation it simply is not possible to distinguish among theories
in which "gravitational force" propagates with different speeds, because
for any propagation speed whatsoever one obtains the same "gravitational
force" and its direction.


Your claims about the orbiting body are basic math: in the frame of the
source the "gravitational force" is central. Transform to the
instantaneous rest frame of the orbiting object and of course the
"gravitational force" will still point directly at the source.

The problem is: this is NOT the math of GR. It is the math of Newtonian
gravitation. It is also not the math of the approximation to GR that
Steve is discussing.


[Carlip]: In general relativity, you solve this problem as follows ...

Most of your message was about this irrelevancy.

If applying General Relativity is an "irrelevancy", then you clearly are
not doing GR. THAT WAS STEVE'S POINT. And mine.


[said to Steve]
That said, what cryptic point were you trying to make?

His point is: you do not understand GR. Which you repeatedly
demonstrate, but refuse to admit.


Obviously, there is no legitimate way out of this dilemma.

The only "dilemma" is YOURS -- why do you keep claiming you are using GR
when you QUITE CLEARLY are not?

You repeatedly claim Steve (and I) are ignoring the "physics behind the
math". The problem is YOURS, not Steve's or mine -- you are confusing
Newtonian gravitation with General Relativity. The physics is DIFFERENT.
Until you actually learn about GR, you will remain confused.


Tom Roberts