Steve Carlip wrote in message ...
In sci.physics Starblade Darksquall wrote:

[...]
Meaning if you're an observer at a fixxed distance from their center
of mass by accelerating likewise you will have to accelerate with a
greater power (I'm taking accelerations to be in the spaceship's frame
of reference, BTW) in order to remain at the same position.

No, you won't. (There may be slight differences depending on
your exact location relative to the line between the bodies, but
on the average there will be no change. If instead of talking about
two bodies you consider a collapsing spherical shell, there will be
no change.)


Well I was under the impression that two bodies was simpler than a
spherical shell... but we could do it your way.

As the two
large bodies approach, there will be more gravitational attraction
because they will gain net total energy but remain with the same net
total momentum, or at least from your reference frame they will.

No. Energy is conserved. In the Newtonian picture you seem
to like, the change in their kinetic energy will be canceled by
the change in gravitational potential energy. Although the
separation between different types of energy is more difficult
in general relativity, the same general idea holds.


In GR, mass is invariant, but in my frame of reference there will be
an overall gain in energy but no overall gain in momentum, meaning
that the energy momentum squared norm increases, thus increasing
gravity. Furthermore, in a frame of reference where you are falling
with them, while you can cancel out the gained energy of one particle
you cannot do this for another. Therefore, in all reference frames
gravity increases, however slightly.

This is not newtonian, but GR. In classical physics the potentail
energy would have to cancel out the kinetic energy, but in GR there is
no potential energy, nor anything analagous to it, therefore there is
no cancelation of energies.

Also, how does GR deal with the time lag involved with gravity?
[...]

How does GR figure this? Does it measure how gravitational
fields change with time? I know that it does
specify that it takes time for a change, differeing based on the
metric, but what I'm not sure of is if anybody's actually done the
mathematics for this so that it can actually be calculated.

For the detailed mathematics in one pretty general setting, see
http://arxiv.org/abs/gr-qc/9909087

Steve Carlip

Alright, I will go there.

(...Starblade Riven Darksquall...)

Starblade Darksquall wrote:

Steve Carlip wrote in message ...
In sci.physics Starblade Darksquall wrote:

One thing I want to know is... if two bodies are falling toward
eachother, doesn't the system increase in gravity due to the added
energy? At least with respect to an outside observer?

I don't understand what ``increase in gravity'' means. ``Gravity''
is not a scalar. What, exactly, do you expect to ``increase''? If two
objects fall toward each other, the form of the gravitational field
around them will change, of course. But how, exactly, do you
want to decide whether such a change is an ``increase''?


Meaning if you're an observer at a fixxed distance from their center
of mass by accelerating likewise you will have to accelerate with a
greater power (I'm taking accelerations to be in the spaceship's frame
of reference, BTW) in order to remain at the same position. As the two
large bodies approach, there will be more gravitational attraction
because they will gain net total energy but remain with the same net
total momentum, or at least from your reference frame they will.


A pertinent related scenario is enlightening. A space ship starts off with
zero velocity. It is located a radial distance away from, say, a spherical
non-spinning central mass which emits a gravitational Schwarzschild field. It
would immediately fall towards the central mass, but the ship fires its
engines in the exact way to just sit there. Then, I think you are saying that
the "going away" acceleration by the engines has to exactly compensate the
"going in" acceleration due to the ship being in the field, and this amount
can be computed from Newtonian mechanics and GR (Schwarzschild) mecahnics and
the GR amount is greater?

Let's look at some equations for radial acceleration in a Schwarzschild field.
Please see

http://www.mindspring.com/~sb635/cip.htm

On page 2, eq.(4) is given the Newtonian radial acceleration. In this "falling
straight in" scenario, no angular motion is ever performed, so dphi/dt = 0,
and the Newtonian gravitational acceleration takes the familiar form,
d^2r/dt^2 = a = -GM/r^2. Eq. (9) on p. 3 shows the "proper time" relativistic
radial acceleration (I know some folks object to calling these
"accelerations," but I think it's pretty obvious why I call them
accelerations.) It is a derivative w.r.t. proper time, or the time (tau) shown
on the face of a clock attached to the ship. The "coordinate time" t is the
time on a clock infinity far away, or the time shown on the face of a clock at
a "close" distance, but it's just sitting there, and is exactly compensating
for its own gravitational time dilation, so that it does show a time as if
infinity far away. (If you wish, place yourself at this location.) If the ship
is just sitting there firing its engines, it has no radial velocity, so that
eq. (9) falls out as a function of the time dilation dt/dtau. Eq. (13) shows
the time dilation, and with dphi/dt = dr/dt = 0, it takes the special case:

dt/dtau = ((r - r_s)/r)^-1/2

This shows how, even if just sitting motionless in a GR field, time is still
dilated. How much is a function of how far out in the field. r_s is the
Schwarschild radius for the amount of central mass and equals 2GM/c^2 where M
= the amount of central mass. Inserting the above into (9), we see

d^2r/dtau^2 = -GM/r^2

This is exactly equal to the non-relativistic Newtonian acceleration. One of
the most important differences between Newtonian mechanics and GR mechanics is
that Newtonian acceleration is only a function of position, but the GR
acceleration is not only a function of position but also of velocity. In a
spherically symmetric Schwarzschild field, if you attach zero velocity vectors
to each position vector, the radially directed acceleration equals simply the
Newtonian acceleration. But, if a test particle flies through some point with
a non-zero velocity, its GR acceleration is then different (stronger) than its
Newtonian acceleration. Hence, if the ship is just sitting there, (we do not
care how it got there yet) its engines compensate for no more than a Newtonian
acceleration.

Let us shut off the rocket engines. The ship begins to fall straight towards
the central mass. One second later, it flies through some closer in radial
point. It has a velocity vector pointing straight towards the central mass at
that point. It then instantaneously fires its engines, and stops dead,
hoovering once again. This time, because it was *moving* through this second
closer in point, it needed more fire power to stop dead as compared to what
Newtonian mechanics alone would state, even considering the velocity involved.

----------
Steve Bell

Astroimaging/Physics homepage: http://www.mindspring.com/~sb635

Starblade Darksquall wrote:

Steve Carlip wrote in message ...
In sci.physics Starblade Darksquall wrote:
Steve Carlip wrote in message
...

[...]
No. Energy is conserved. In the Newtonian picture you seem
to like, the change in their kinetic energy will be canceled by
the change in gravitational potential energy. Although the
separation between different types of energy is more difficult
in general relativity, the same general idea holds.

In GR, mass is invariant, but in my frame of reference there will be
an overall gain in energy but no overall gain in momentum, meaning
that the energy momentum squared norm increases, thus increasing
gravity. Furthermore, in a frame of reference where you are falling
with them, while you can cancel out the gained energy of one particle
you cannot do this for another. Therefore, in all reference frames
gravity increases, however slightly.

This is not newtonian, but GR.

No, it is not.

Yes it is GR. It's GR because in GR there is no potential energy, and
as I said there is no potential energy.


Please look at eqs. (20) & (21) on p. 17 at

http://www.mindspring.com/~sb635/pap4.htm

It is the total relativistic Kerr energy of a test particle in orbit about a
spherical and rotating central body. To me, this total energy is made up of
the test particle's "internal energy" (the E = mc^2 part), its kinetic energy
which is based on its velocity, and what's left is the potential energy of the
Kerr field at that point. This itself involves the test particle's velocity,
so there is a "feedback" between the state (pos,vel) of the test particle and
the actual Kerr-based acceleration it is subjected to. But the "feedback" is
not infinitly recurrsive in the "who came first, the chicken or the egg"
scenario. Given the test partcile's position and velocity, the Kerr
acceleration to which it is instantaneoulsy subjected (the relativistic
version of the simple Newtonian acceleration) is fixed and exact at that point
and velocity. For the simpler Schwarzschild case where the central body is not
rotating, the same is true.

----------
Steve Bell

Astroimaging/Physics homepage: http://www.mindspring.com/~sb635

Steve Bell wrote:


Please look at eqs. (20) & (21) on p. 17 at

http://www.mindspring.com/~sb635/pap4.htm


Sorry, that eqs. (21) and (22).


----------
Steve Bell

Astroimaging/Physics homepage: http://www.mindspring.com/~sb635

In sci.physics Starblade Darksquall wrote:

[...]
So, you're saying that because gravity gravitates, the field stays the
same. So the gravitational field loses energy in the exactly right
amount to counterbalance the gain in KE?

Yes, roughly. GR is a nonlinear theory, and there's no unambiguous
way to separate out what piece of a field comes from what source.
Besides the kinetic energy, you need to take into account the energy
of the gravitational field; the energy due to the gravitational interaction
between the gravitational field energy and the mass and kinetic energy
of the shell; the energy due to the gravitational interaction between the
energy due to the gravitational interaction between the gravitational
field energy and the mass and kinetic energy of the shell and the mass
and kinetic energy of the shell; the energy due to the gravitational
interaction between the energy due to the gravitational interaction
between the gravitational field energy and the mass and itself; etc.,
ad infinitum. When you add everything up, you find that the net
field outside a collapsing shell does not change.

Note that this isn't something that you can decide by intuition or
guesswork. In the end, you have to sit down and calculate, or at
least look up someone else's calculations.

Steve Carlip

Steve Bell wrote in message ...
Starblade Darksquall wrote:

Steve Carlip wrote in message ...
In sci.physics Starblade Darksquall wrote:
Steve Carlip wrote in message
...

[...]
No. Energy is conserved. In the Newtonian picture you seem
to like, the change in their kinetic energy will be canceled by
the change in gravitational potential energy. Although the
separation between different types of energy is more difficult
in general relativity, the same general idea holds.

In GR, mass is invariant, but in my frame of reference there will be
an overall gain in energy but no overall gain in momentum, meaning
that the energy momentum squared norm increases, thus increasing
gravity. Furthermore, in a frame of reference where you are falling
with them, while you can cancel out the gained energy of one particle
you cannot do this for another. Therefore, in all reference frames
gravity increases, however slightly.

This is not newtonian, but GR.

No, it is not.

Yes it is GR. It's GR because in GR there is no potential energy, and
as I said there is no potential energy.


Please look at eqs. (20) & (21) on p. 17 at

http://www.mindspring.com/~sb635/pap4.htm

It is the total relativistic Kerr energy of a test particle in orbit about a
spherical and rotating central body. To me, this total energy is made up of
the test particle's "internal energy" (the E = mc^2 part), its kinetic energy
which is based on its velocity, and what's left is the potential energy of the
Kerr field at that point. This itself involves the test particle's velocity,
so there is a "feedback" between the state (pos,vel) of the test particle and
the actual Kerr-based acceleration it is subjected to. But the "feedback" is
not infinitly recurrsive in the "who came first, the chicken or the egg"
scenario. Given the test partcile's position and velocity, the Kerr
acceleration to which it is instantaneoulsy subjected (the relativistic
version of the simple Newtonian acceleration) is fixed and exact at that point
and velocity. For the simpler Schwarzschild case where the central body is not
rotating, the same is true.

----------
Steve Bell

Astroimaging/Physics homepage: http://www.mindspring.com/~sb635

So, how exactly am I to look at 'gravitational potential energy'? This
doesn't seem to explain that in the slightest.

However, I think I understand it a bit better. All aspects of motion
are involved in the generation of the gravitational field. If, for
example, we have a collapsing spherical shell, then the laws of
physics simply knows better than to alter the gravitational field. The
fact that the spherical shell is falling works to undo the fact that a
net kinetic energy component is gained when it comes to gravitation.
Saying that the amount of 'internal energy' determines gravitation is
a gross generalization, and the truth is far more complicated.

That's also the reason why light, although it has no rest mass, can
influence gravitation, though it produces the field in a slightly
convoluted manner, in that parallel beams of plane waves going in the
same direction do not influence eachother through gravitation, but
parralel beams of plane waves going in the opposite direction do
influence eachother through gravitation.

GR is really weird. The only concept that makes total sense is the
idea of all reference frames of any type of motion, whether free or
induced, are equal, at least in the sense that the mathematics can be
worked out independant of ones reference frame.

So... what quantities are conserved in GR?

(...Starblade Riven Darksquall...)

(Starblade Darksquall) wrote in message


So, how exactly am I to look at 'gravitational potential energy'?

It is rather simple: if you raise something its internal energy
(mc^2) goes up since you move the object into space where
time runs a little bit faster and so does teh speed of light. But
since there is also a little less space there the speed of light
doesn't go up the same fast as the rate of time does (in which
case the internal energy of the raised object would raise with
square of distance) but a little bit slower than that. It results
in a linear increase of internal energy of the raised object.

When you drop the object, you recover your energy because
everything works in the opposite direction. If you just let the
object go its internal energy will change into kinetic energy and
increase a little bit the mass of the object until it hits
something and changes its kinetic enrgy into something else.

The inability of nature to create energy from nothing (and
inability to destroy energy) a.k.a. "the principle of
conservation of energy" asures that those processes
have to run flawlessly.

However, I think I understand it a bit better. All aspects of motion
are involved in the generation of the gravitational field. If, for
example, we have a collapsing spherical shell, then the laws of
physics simply knows better than to alter the gravitational field. The
fact that the spherical shell is falling works to undo the fact that a
net kinetic energy component is gained when it comes to gravitation.
Saying that the amount of 'internal energy' determines gravitation is
a gross generalization, and the truth is far more complicated.

No. The truth is far more simple: when the shell collapses
the total internal energy can't change since whateger gets
into shell's kinetic energy is taken from its internal energy
(mc^2) and while looking form outside one sees the same
amount of energy in the shell all the time, collapsing or not.

The details of whatever happens are irrelevant since
conservation of energy asures that from outside the energy
looks always the same and so does the "gravitational field"
unless it is not symmetric system and it changes as in
stars rotating around one another, or when some energy
is radiated out.

Gravity is the simplest part of physics. Nothing really to
hang your thought on and that's probably why people who
are curious about the nature don't work in it and so it is
so tough to get good explanation from someone when
one does not understand something. And that's also why
there is stuff like black holes and expanding space. As
Feynman said "It is not that the subject is hard; it is that
the good men are occupied elsewhere. Remind me not
to come to any more gravity conferences!" Source:
http://www.geocities.com/wlodekj/sci/feynman.htm

-- Jim

Starblade Darksquall wrote:
So, how exactly am I to look at 'gravitational potential energy'?

If you are using Newtonian gravitation, use gravitational potential
energy in the usual way. If you are trying to use or understand General
Relativity, don't attempt to use it in any way at all -- in GR it is not
well defined; but that's OK because there is no NEED for it, anyway.


GR is really weird.

Actually, GR is quite natural, once one understands geometry; Newtonian
gravitation, however, was quite a kludge....

OK, I'll admit that understanding geometry in the detail
required is non-trivial (:-)).


So... what quantities are conserved in GR?

Anything corresponding to a symmetry of the Lagrangian.

There is one general symmetry (due to the Bianchi identities) that leads
directly to the local conservation of energy (i.e. the covariant
divergence of the energy-momentum tensor vansihes). But in general those
local equations are not integrable, and one only has something
resembling "global conservation of energy" for certain special cases.


NOTE: attempting to learn GR via random posting on the newsgroups is
utterly hopeless. You need to find a few good textbooks and study. The
FAQ has a booklist. I'd start with Geroch's
_General_Relativity_from_A_to_B_.


Tom Roberts

Jim Jastrzebski wrote:
(Starblade Darksquall) wrote in message

So, how exactly am I to look at 'gravitational potential energy'?

It is rather simple: if you raise something its internal energy
(mc^2) goes up since you move the object into space where
time runs a little bit faster and so does teh speed of light.

Einstein said "Simplify everything as much as possible, but no simpler."
Your simplistic description is over simplified, to the point of being
either meaningless or wrong. Specifically you will have great difficulty
defining "into space" and "time runs a little bit faster" and "the speed
of light is a little bit faster".


The inability of nature to create energy from nothing (and
inability to destroy energy) a.k.a. "the principle of
conservation of energy" asures that those processes
have to run flawlessly.

Hmmmm. GR has conservation of energy, but not as you suppose. In GR it
is a LOCAL property only, and the global sorts of "conservation" you
imagine do not apply.


[a collapsing spherical shell]
No. The truth is far more simple: when the shell collapses
the total internal energy can't change since whateger gets
into shell's kinetic energy is taken from its internal energy
(mc^2) and while looking form outside one sees the same
amount of energy in the shell all the time, collapsing or not.

Again you oversimplify. In particular, you have given what you think is
a general description, but it only gets the right answer for the case of
a SPHERICAL shell collapsing in a SPHERICALLY-SYMMETRIC manner. In
particular, for a "lumpy" collapse the gravitational field[#] at a given
point will vary in magnitude in a complicated manner....

[#] Yes, it's rather difficult to define this precisely in GR.
But any sensible definition will vary over time in this
situation.


The details of whatever happens are irrelevant since
conservation of energy asures [...]

No. "Conservation of energy", as you suppose it, does not apply. GR is
more complicated than you think, and the details DO matter....


You too could profit from my advice to Starblade13 in my recent post to
this thread:
Attempting to learn GR via random posting on the newsgroups is utterly hopeless.
You need to find a few good textbooks and study. The FAQ has a
booklist. I'd
start with Geroch's _General_Relativity_from_A_to_B_.


Tom Roberts

Jim Jastrzebski wrote:
: It is rather simple: if you raise something its internal energy
: (mc^2) goes up since you move the object into space where
: time runs a little bit faster and so does teh speed of light. But
: since there is also a little less space there the speed of light
: doesn't go up the same fast as the rate of time does (in which
: case the internal energy of the raised object would raise with
: square of distance) but a little bit slower than that. It results
: in a linear increase of internal energy of the raised object.

Please stop posting such nonsense, internal
energy does not change, and time does not cause anything.

The potential energy exists in General Relativity,
it is just not calculated the same as in Newtonian
gravitation.

Joe Fischer

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