Gravitational Potential in GR
Edward Schaefer
How are gravitational potential and gravitational potential energy
defined in GR? I have been struggling with that concept lately
but have not hit upon a satisfactoty answer.
EMS
Nathan Urban
> How are gravitational potential and gravitational potential energy
> defined in GR?
Easy: they aren't! (Certain classes of spacetimes may be described
by an effective potential, but in general you can't do that in GR.
I forget which ones can be written in terms of a effective potential.
The asympotically flat ones, maybe? Or the static ones or stationary
ones? I need to check my books, unless someone else knows; I'm guessing
the first.)
Some of these issues are discussed in the Relativity FAQ entry, "Is
Energy Conserved in General Relativity?", at:
http://math.ucr.edu/home/baez/physics/energy_gr.html
(That entry might answer the previous question; I haven't looked at it
recently.)
Edward Schaefer
Nathan Urban wrote:
>
> In article <34EC8A...@plansys.com>, Edward Schaefer <scha...@plansys.com> wrote:
>
> > How are gravitational potential and gravitational potential energy
> > defined in GR?
>
> Easy: they aren't! (Certain classes of spacetimes may be described
> by an effective potential, but in general you can't do that in GR.
Perhaps so in general, but I am dealing in the static spacetime
surrounding a massive object. The conservation of energy should
apply there. After all, the Newtonian version of it keeps the
planets in orbit, and tells you how much fuel you need to send
a spacecraft between them.
> I forget which ones can be written in terms of a effective potential.
> The asympotically flat ones, maybe? Or the static ones or stationary
> ones? I need to check my books, unless someone else knows; I'm guessing
> the first.)
>
> Some of these issues are discussed in the Relativity FAQ entry, "Is
> Energy Conserved in General Relativity?", at:
>
> http://math.ucr.edu/home/baez/physics/energy_gr.html
>
> (That entry might answer the previous question; I haven't looked at it
> recently.)
I've got it, and will take a closer look at it. Thanks.
EMS
Thomas H. Hoovler
On Thu, 19 Feb 1998, Edward Schaefer wrote:
> How are gravitational potential and gravitational potential energy
> defined in GR? I have been struggling with that concept lately
> but have not hit upon a satisfactoty answer.
>
> EMS
Gravitational potential energy is defined as a component of the metric
tensor, namely g_oo.
Quezron
"The most incomprehensible thing about the universe
is that it can be comprehended." -- A. Einstein
http://GeoCities.com/CapeCanaveral/Lab/5932
Edward Schaefer
Thomas H. Hoovler wrote:
>
> On Thu, 19 Feb 1998, Edward Schaefer wrote:
>
> > How are gravitational potential and gravitational potential energy
> > defined in GR? I have been struggling with that concept lately
> > but have not hit upon a satisfactoty answer.
> >
> > EMS
>
> Gravitational potential energy is defined as a component of the metric
> tensor, namely g_oo.
>
You may have just sprung me with that answer, even though it is not
totally correct.
Wrt a distant observer, it is Mc^2(1 - g_00), where M is the local
rest mass of the object. Therefore, where g_00 = 0, all of the
rest mass of the object has been transformed into potential energy,
which isconsistant with both standard GR (or rod-GR as I like to
call it), and my paralactic-GR.
However, I have been using g_00 in my calculations based on the same
knee-jerk misconception that you just used. So that you for showing
me my mistake.
EMS
Steve Carlip
Edward Schaefer (scha...@plansys.com) wrote:
: > > How are gravitational potential and gravitational potential energy
: > > defined in GR?
: Wrt a distant observer, it is Mc^2(1 - g_00), where M is the local
: rest mass of the object.
In the weak field approximation, it's Mc^2(1-g_00)/2. For
strong fields, gravitational interactions can't be described
by a single potential. There are various definitions of
"quasilocal energy" inside a region, which might do for
"gravitational potential energy," but they all involve
fairly complicated manipulations of the metric.
: Therefore, where g_00 = 0, all of the
: rest mass of the object has been transformed into potential energy,
When g_00=0, the weak field approximation is not valid, and
the question doesn't quite make sense.
Steve Carlip
car...@dirac.ucdavis.edu
Tom Roberts
Thomas H. Hoovler wrote:
> Gravitational potential energy is defined as a component of the metric
> tensor, namely g_oo.
That's BLATANTLY false. g_00 gives the contribution of dt to the proper time.
In SR, g_00 is 1, so it simply CANNOT be a gravitational potential.
In certain approximations in GR (I forget the details), IIRC only the g_00
component of the metric differs from the Lorentz metric, and is given by
g_00 = (1 + 2*Phi/c^2)
Here Phi is the Newtonian gravitational potential, and the acceleration due
to gravity is -Grad Phi.
Beware -- there are all sorts of conditions and limitations of this. This
is certainly NOT covariant....
Tom Roberts tjro...@lucent.com
Edward Schaefer
Steve Carlip wrote:
>
> Edward Schaefer (scha...@plansys.com) wrote:
>
> : > > How are gravitational potential and gravitational
> : > > potential energy defined in GR?
>
> : Wrt a distant observer, it is Mc^2(1 - g_00), where M is the local
> : rest mass of the object.
>
> In the weak field approximation, it's Mc^2(1 - g_00)/2.
*Ouch*. And thank you.
> For strong fields, gravitational interactions can't be described
> by a single potential. There are various definitions of
> "quasilocal energy" inside a region, which might do for
> "gravitational potential energy," but they all involve
> fairly complicated manipulations of the metric.
Could you E-mail me some references for future use?
> : Therefore, where g_00 = 0, all of the
> : rest mass of the object has been transformed into potential energy,
>
> When g_00=0, the weak field approximation is not valid, and
> the question doesn't quite make sense.
Check out Ohainian (or maybe Wald) and you will see what I am
refering to. (I'll get you a page # for Ohanian tomorrow, and
for Wald if I can find what I'm looking for there.)
EMS
Edward Schaefer
Tom Roberts wrote:
>
> Thomas H. Hoovler wrote:
> > Gravitational potential energy is defined as a component of the metric
> > tensor, namely g_oo.
>
> That's BLATANTLY false. g_00 gives the contribution of dt to the proper time.
> In SR, g_00 is 1, so it simply CANNOT be a gravitational potential.
>
> In certain approximations in GR (I forget the details), IIRC only the g_00
> component of the metric differs from the Lorentz metric, and is given by
>
> g_00 = (1 + 2*Phi/c^2)
IIRC? I get the gist, but lack the words.
>
> Here Phi is the Newtonian gravitational potential, and the acceleration due
> to gravity is -Grad Phi.
>
> Beware -- there are all sorts of conditions and limitations of this. This
> is certainly NOT covariant....
Why do you say "NOT covariant"?
EMS
Edward Schaefer
Between Steve and Tom I have gotten what I need.
1) The total energy of an object descending into a gravitational
field is Mc^2, where M is its rest mass when it is distant from
any gravitational fields.
2) g_00 = 1 + 2 * Phi/c^2, where Phi is the Newtonaian
gravitational potential, meaning that an object free-falling
from infinity to Phi will have a speed of v = sqrt (-2 * Phi)
when it gets there.
3) The relativistic energy of an object moving at v is:
m = m_r /sqrt (1 - v^2/c^2)
so:
4) The rest mass of an object at Phi is
m_r = M * sqrt (1 + 2*Phi/c^2)
From 2) we find that 2 * Phi = c^2 (g_00 - 1)
Therefore m_r = M * sqrt (g_00)
5) Gravitational potential energy is therefore:
Mc^2 [1 - sqrt (g_00)]
Figures. It was too easy.
EMS
Tom Roberts
Edward Schaefer wrote:
> IIRC? I get the gist, but lack the words.
IIRC is defined, along with other common internet acronyms, in the new users
FAQs for netnews/USENET.
IIRQ = If I Reacll Correctly
> Why do you say "NOT covariant"?
Because the metric tensor is normally covariant (i.e. it IS a tensor).
The expression I gave is only valid for coordinates which are nearly
Lorentzian. That is, it is NOT valid for all coordinate systems =>
it is not covariant.
Tom Roberts tjro...@lucent.com
Edward Schaefer
In that case, what is the general rule for the Phi - g_00
relationship? I'm curious because I can make that rule work
all the way down to g_00 = 0, giving me a straight conservation
of energy in the process.
(Note: The gravitational field is found to not contain any
energy in that case, but paralactic-GR still gives me T_uv /= 0
even with T_00 = 0.)
EMS
Tom Roberts
Edward Schaefer wrote:
> > Because the metric tensor is normally covariant (i.e. it IS a tensor).
> > The expression I gave is only valid for coordinates which are nearly
> > Lorentzian. That is, it is NOT valid for all coordinate systems =>
> > it is not covariant.
> relationship? I'm curious because I can make that rule work
> all the way down to g_00 = 0, giving me a straight conservation
> of energy in the process.
I don't know what you mean by a "general rule" here. The approximation I
gave is appropriate for nearly-Lorentzian (i.e. small gravitational field)
situations. It is valid only for those cases, and only in coordinates which
are themselves near Lorentzian. There are other conditions, as I have said
before, and you had better look them up before using any of this....
If g_00 = 0, then the metric is singular. That CANNOT make sense for a
near-Lorentzian situation.
And, as I have said before, if you are NOT in a near-Lorentzian situation,
then the concept of gravitational potential (Phi) is completely
inadequate.
> (Note: The gravitational field is found to not contain any
> energy in that case, but paralactic-GR still gives me T_uv /= 0
> even with T_00 = 0.)
I don't know what you mean by this at all. You seem to be implying
that there can be momentum transfer and/or stress in a region with
ZERO energy density. Offhand, I cannot see how that can occur (but I
am not familiar enough with T and its physical meanings to know for
sure (:-)).
Tom Roberts tjro...@lucent.com
Edward Schaefer
Tom Roberts wrote:
>
> Edward Schaefer wrote:
> > > Because the metric tensor is normally covariant (i.e. it IS a tensor).
> > > The expression I gave is only valid for coordinates which are nearly
> > > Lorentzian. That is, it is NOT valid for all coordinate systems =>
> > > it is not covariant.
> > In that case, what is the general rule for the Phi - g_00
> > relationship? I'm curious because I can make that rule work
> > all the way down to g_00 = 0, giving me a straight conservation
> > of energy in the process.
>
> I don't know what you mean by a "general rule" here.
If Phi = -M * (1 - sqrt(g00)) does not apply *throughtout* the
gravitational field, then what does?
> The approximation I
> gave is appropriate for nearly-Lorentzian (i.e. small gravitational field)
> situations. It is valid only for those cases, and only in coordinates which
> are themselves near Lorentzian. There are other conditions, as I have said
> before, and you had better look them up before using any of this....
>
> If g_00 = 0, then the metric is singular. That CANNOT make sense for a
> near-Lorentzian situation.
>
> And, as I have said before, if you are NOT in a near-Lorentzian situation,
> then the concept of gravitational potential (Phi) is completely
> inadequate.
This I disagree with, but that is a matter of philosophy.
> > (Note: The gravitational field is found to not contain any
> > energy in that case, but paralactic-GR still gives me T_uv /= 0
> > even with T_00 = 0.)
>
> I don't know what you mean by this at all. You seem to be implying
> that there can be momentum transfer and/or stress in a region with
> ZERO energy density. Offhand, I cannot see how that can occur (but I
> am not familiar enough with T and its physical meanings to know for
> sure (:-)).
The simple (but probably inadequate) answer is that my ideas
call for the Lenz-Schiff Arguemnt[1] to hold (in normal or
rod-GR it does not[2]). Becauase of this, it is demanded in
a "good" system of isometric coordinates that A(r) = 1/B(r)
given the metric form ds^2 = A(r)dt^2 - B(r)(dr^2 + r^2dO^2),
where O is spherical surface coordinates surrounding a
spherically symmetric non-rotating massive object.
When you work through the math, you get a variation on the
isometric form of the Schwarzschild solution(*), and a stress/
energy tensor with non-zero terms in T_11, T_22, and T_33. So
I find that the field *does* contain stress even without energy.
As for what that means, I'm not quite sure yet. I'm about to
post the details, and see what the group has to say.
(*) You cannot have the isometric Schwarzschild solution becuase
A(r) /= 1/B(r) in it.
[1] L. I. Lenz, American Journal of Physics, 28,340 (1960)
[2] W. Rindler, American Journal of Physics, 36,540 (1968)
EMS
Steve Carlip
Edward Schaefer (scha...@plansys.com) wrote:
: If Phi = -M * (1 - sqrt(g00)) does not apply *throughtout* the
: gravitational field, then what does?
Nothing does. The concept of gravitational potential energy is
a Newtonian concept; it has no equivalent in general relativity.
The best you can do is to define an approximate gravitational
potential energy in regions in which Newtonian gravity is a
good approximation.
The whole concept of gravitational energy in general relativity is
quite tricky. The principle of equivalence implies that there can
be no covariant gravitational stress-energy tensor---one can always
choose coordinates in which the geodesics are arbitrarily close to
straight lines in a small region, which implies that the gravitational
energy in that region is arbitrarily close to zero; but a tensor that
vanishes in any coordinate system vanishes in every coordinate system.
(For a fleshed-out version of this argument, see section 20.4 of
Misner, Thorne, and Wheeler.)
The best one can hope for is a "quasilocal gravitational energy,"
energy defined in a finite region. A nice proposal for such an
energy is given by Brown and York, Phys. Rev. D47 (1993) 1407. It
can be described in terms of the metric, but not in any particularly
somple way. It *certainly* can't be given in terms of a particular
component of the metric---that's a highly non-covariant procedure.
Steve Carlip
car...@dirac.ucdavis.edu
Tom Roberts
Edward Schaefer wrote:
> Tom Roberts wrote:
> > And, as I have said before, if you are NOT in a near-Lorentzian situation,
> > then the concept of gravitational potential (Phi) is completely
> > inadequate.
>
> This I disagree with, but that is a matter of philosophy.
I meant it in the sense that a single function Phi is inadequate to express
the complexity of gravitation in GR, which requires the ten functions of
the metric tensor. This is not philosophy, but mathematics -- one function
cannot possibly represent or contain the complexity of ten!
Tom Roberts tjro...@lucent.com
Edward Schaefer
First of all, I will check out those reference ASAP.
As for the Principle of Equivalence, there is a fly in that ointment:
An observer in a box DOES have a way of determining whether the
gravitational field that he is in has curvature of not, based on the
tidal effects of a free-falling ball coated with a liquid. (See H.
Ohanian and R. Ruffini, "Gravitation and Spacetime", 2nd ed. (1994).)
The percentage of distortion caused by those effects do not vanish
as the size of the ball goes to zero. Therefore, there is
*something* about the field that causes those differences. My
current candidate is stress.
EMS
Edward Meisner
Edward Schaefer wrote:
If there is no concept of potential energy in GR, then it would be
incapable of describing many phenomena. There would be no equations to
describe the work done in raising a mass to a greater height, for instance.
If GR can not produce equations that describe such situations, then what
good is it? We might as well go back to Newton because GR would be useless
and worthless. And a much more serious problem arises if there is no
potential energy in GR. When a charged particle is accelerated in a
gravitational field, there is an increase in magnetic energy, because, as
Dr. Sarfatti has pointed out, the magnetic field is the inertia of a charged
particle. Who knows how many charged particles are being accelerated in
gravitational fields at every instance through out the universe. If there is
no potential energy, that would mean that the energy in the universe is
increasing by billions upon billions of joules every second. So if GR can
not describe such phenomena, it would not only be incomplete but outright
wrong.
Chris Hillman
On Mon, 23 Feb 1998, Edward Meisner wrote:
> If there is no concept of potential energy in GR, then it would be
> incapable of describing many phenomena. There would be no equations to
> describe the work done in raising a mass to a greater height, for instance.
> If GR can not produce equations that describe such situations, then what
> good is it? We might as well go back to Newton because GR would be useless
> and worthless.
Don't worry, there's no need for such an extreme reaction :-) gtr is
certainly very useful (over a broader range of physical phenomena than
Newtonian theory), despite the nonwelldefinedness :-/ of "potential
energy". I think, however, you may have missed the point of what Steve
Carlip was saying...
Chris
Edward Schaefer
That I do agree with.
Think of me as being a blind man who has just come across an elephant.
I have to start SOMEWHERE. [At least I know that it is not just a
rope :-).]
EMS
Edward Schaefer
> incapable of describing many phenomena. There would be no equations to
> describe the work done in raising a mass to a greater height, for instance.
> If GR can not produce equations that describe such situations, then what
> good is it? We might as well go back to Newton because GR would be useless
> potential energy in GR. When a charged particle is accelerated in a
> gravitational field, there is an increase in magnetic energy, because, as
> Dr. Sarfatti has pointed out, the magnetic field is the inertia of a charged
> particle. Who knows how many charged particles are being accelerated in
> gravitational fields at every instance through out the universe. If there is
> no potential energy, that would mean that the energy in the universe is
> increasing by billions upon billions of joules every second. So if GR can
> not describe such phenomena, it would not only be incomplete but outright
> wrong.
Thank you. You have done a good job of expressing why I operate on
the basis that potential exists, and does so in a simple, coherent
form.
EMS
Todd Desiato
Edward Meisner <ode...@injersey.com> wrote in article
<34F23008...@injersey.com>...
Point well taken! IMO GR is not wrong, but is useful only to a handful of
cosmologists, professors and a few satellite designers. That is why I have
been trying to promote the idea of an electrodynamic model. Not as
something new and radical, but as an interpretation of how GR and EM mix on
an engineering level.
Todd Desiato
Tom Roberts
Edward Meisner wrote:
> If there is no concept of potential energy in GR, then it would be
> incapable of describing many phenomena.
You are assuming the "potential energy" is the ONLY way to describe such
phenomena.
> There would be no equations to
> describe the work done in raising a mass to a greater height, for instance.
That presupposes that the concept "greater height" has meaning. In a
arbitrarily-curved spacetime there is NOT necessarily such a concept....
And what do you mean by "work"? In curved spacetime it is a bit wobbly (to
say the least). That's the whole point. "Work" is a SINGLE number, and
in GR, gravitation is MUCH MORE COMPLEX than any single value can ever
express.
In Newtonian gravitation, the gravitational force is conservative. That is,
no matter what path you take to a give point, the amount of energy required
to get there is independent of the path taken. In GR that no longer holds;
the energy required (however defined) is in general path dependent. All
sorts of things are path dependent in GR, which in Newtonian physics are
path-independent. The curved geometry of spacetime has great ramifications
throughout one's theoretical outlook on the world.
> If GR can not produce equations that describe such situations, then what
> good is it?
_IF_ that were true, then the theory would not be much good....
But GR can produce equations to describe such situations. They just do
not use "potential energy", etc. They use geodesic equations. GR is
quite "good" (but admittedly mathematically cumbersome (:-)).
> We might as well go back to Newton because GR would be useless
> and worthless. And a much more serious problem arises if there is no
> potential energy in GR. When a charged particle is accelerated in a
> gravitational field, there is an increase in magnetic energy, because, as
> Dr. Sarfatti has pointed out, the magnetic field is the inertia of a charged
> particle.
Hmmm.
> Who knows how many charged particles are being accelerated in
> gravitational fields at every instance through out the universe. If there is
> no potential energy, that would mean that the energy in the universe is
> increasing by billions upon billions of joules every second.
That doesn't follow at all. If there is no concept of "potential energy",
then you cannot use it in your last phrase....
> So if GR can
> not describe such phenomena, it would not only be incomplete but outright
> wrong.
IMHO, GR would only be "outright wrong" if you can come up with experimental
observations which disagree with its predictions. Gedankens like you present
here are NOT refutations of any theory.
Tom Roberts tjro...@lucent.com
Edward Schaefer
Tom Roberts wrote:
>
> Edward Meisner wrote:
> > If there is no concept of potential energy in GR, then it would be
> > incapable of describing many phenomena.
>
> ever express.
Yes it is, and that is beside the point. A scalar value which can
be referred to as the gravitational potential in a static field
surrounding a spherically massive object wrt a distant observer can
be expressed.
If I asked you to recite the First Amendment, would you answer me
by saying "No. I can't do that. The Constitution is *much* more
complex than just one amendment."?
> In Newtonian gravitation, the gravitational force is conservative.
> That is, no matter what path you take to a give point, the amount
> of energy required to get there is independent of the path taken.
That is news to me. The last I heard, the delta-v used to make
your rocket go along a given path in a given time very much
determines the energy used. The above statement holds in
Newtonian physics only for the exchange of kinetic & potential
energy in an object following a Keplerian orbit (in the 2-body
case with one object having much less mass than the other). The
equivalent case is an object orbitting a massive object along a
geodesic of spacetime. I find the idea of its not not exhibiting
a speed behavior (wt a distant observer) that can be
characterized as a kinetic/potential energy exchange to be very
odd.
> In GR that no longer holds; the energy required (however defined)
> is in general path dependent. All sorts of things are path
> dependent in GR, which in Newtonian physics are path-independent.
> The curved geometry of spacetime has great ramifications
> throughout one's theoretical outlook on the world.
The potential energy of an object at a given point in spacetime
wrt to given observer is NOT path dependent. That is not to say
that you can describe GR as a series of scalar functions. You
only get yourself bitten in the arse when you try that (and if
you have following my attempts to produce a mathematically
coherent theory based on the premises of paralactic-GR you will
know that I say so from experience), but all the same, it does
not keep those scalars from existing.
EMS
TelNet Communication
Todd Desiato <tde...@san.rr.com> wrote:
: Edward Meisner <ode...@injersey.com> wrote in article
: > If there is no concept of potential energy in GR, then it would be
: > incapable of describing many phenomena. There would be no equations to
: > describe the work done in raising a mass to a greater height, for
: > instance.
: > increasing by billions upon billions of joules every second. So if GR can
: > not describe such phenomena, it would not only be incomplete but outright
: > wrong.
: Point well taken! IMO GR is not wrong, but is useful only to a handful of
: cosmologists, professors and a few satellite designers. That is why I have
: been trying to promote the idea of an electrodynamic model. Not as
: something new and radical, but as an interpretation of how GR and EM mix on
: an engineering level.
: Todd Desiato
Todd, please address the Principle of Equivalence,
the whole point of General Relativity is to remove the
concept of gravity acting as a "force" on free moving
(freefalling and orbiting bodies).
Obviously there is a force acting on objects
on the surface of the Earth, but not in freefall or
orbit (if more than one body in freefall or orbit are
close together then geodesic deviation causes the
appearance of forces acting (tidal forces) because
each object has it's own geodesic to follow.
Edward's objection is answered best by the
fact that for a given massive body such as the Earth,
the acceleration at a given altitude (potential) is
a constant K = GM, and not much math is needed to
go from there for any r^2.
It seems everybody is trying to force the
existence of a "force" theory, even the aether nuts,
and even the GR proponents who are really followers
of Newton, and not true relativists.
Synge seems to me to be different from others,
and his GR text gives examples of the worldline of
an apple before and after the stem breaks, and the
geodesic of the apple after the stem breaks relative
to the non-inertial worldline of the tree branch and
the surface of the Earth.
I agree that em is the cause of gravity, but
not as a propagation of particles, or waves, and not
as an interaction of freemoving objects with a medium
bay any name (aether, space-time, or what-have-you).
If you will make the unification of em and
gravitation the priority rather than seeking the
concurrence of others on a proffered model then you
will have a pleasurable task that can keep you busy
in idle time for quite a while (but it won't make
you rich, and maybe not famous, you have to work
with black hole theory for that). :-)
Ken Fischer
---
TelNet Communication
Edward Schaefer <scha...@plansys.com> wrote:
: Edward Meisner wrote:
: > gravitational fields at every instance through out the universe. If
: > no potential energy, that would mean that the energy in the universe is
: > increasing by billions upon billions of joules every second. So if GR can
: > not describe such phenomena, it would not only be incomplete but outright
: > wrong.
:
: the basis that potential exists, and does so in a simple, coherent
: form.
: EMS
cow stands on the railroad track, the potential is
in the speeding train!
Ken Fischer
---
Thomas H. Hoovler
On Tue, 24 Feb 1998, Edward Schaefer wrote:
> Tom Roberts wrote:
> >
> > Edward Meisner wrote:
> > > If there is no concept of potential energy in GR, then it would be
> > > incapable of describing many phenomena.
> >
> > in GR, gravitation is MUCH MORE COMPLEX than any single value can
> > ever express.
>
> Yes it is, and that is beside the point. A scalar value which can
> be referred to as the gravitational potential in a static field
> surrounding a spherically massive object wrt a distant observer can
> be expressed.
The problem is that this "scalar" value you refer to sits in a tidy little
corner of the metric tensor in even the best of systems where it can be
interpreted as such. Hence, it will not be a scalar in GR, but part of
the second rank metric tensor. There have been some relativistic attempts
at scalar gravity theories in the past (The Pais Einstein biography is a
good source of some of these), but nothing apparently came of them.
>
> If I asked you to recite the First Amendment, would you answer me
> by saying "No. I can't do that. The Constitution is *much* more
> complex than just one amendment."?
>
> > In Newtonian gravitation, the gravitational force is conservative.
> > That is, no matter what path you take to a give point, the amount
> > of energy required to get there is independent of the path taken.
>
> That is news to me. The last I heard, the delta-v used to make
> your rocket go along a given path in a given time very much
> determines the energy used. The above statement holds in
> Newtonian physics only for the exchange of kinetic & potential
> energy in an object following a Keplerian orbit (in the 2-body
> case with one object having much less mass than the other). The
> equivalent case is an object orbitting a massive object along a
> geodesic of spacetime. I find the idea of its not not exhibiting
> a speed behavior (wt a distant observer) that can be
> characterized as a kinetic/potential energy exchange to be very
> odd.
>
I think you might be confusing the notion of impulse with energy, here.
That's probably because impulse is J = Int(F dt), while conservative
forces are defined by work W = Int(F.dx) being path independent, as indeed
it must be for Newtonian gravity.
> > In GR that no longer holds; the energy required (however defined)
> > is in general path dependent. All sorts of things are path
> > dependent in GR, which in Newtonian physics are path-independent.
> > The curved geometry of spacetime has great ramifications
> > throughout one's theoretical outlook on the world.
>
> The potential energy of an object at a given point in spacetime
> wrt to given observer is NOT path dependent. That is not to say
> that you can describe GR as a series of scalar functions. You
> only get yourself bitten in the arse when you try that (and if
> you have following my attempts to produce a mathematically
> coherent theory based on the premises of paralactic-GR you will
> know that I say so from experience), but all the same, it does
> not keep those scalars from existing.
>
> EMS
The potential energy of an object in Newtonian physics indeed is not path
dependent, but in GR, with a background curvature tensor, the notion gets
a little skewed. I forget the exact formula, but integrations over the
background curvature are never path independent in a curved space. Such
scalars as you mention might exist, but they would have to true scalars in
four dimensional curved space-time.
Quezron
"The most incomprehensible thing about the universe
is that it can be comprehended." -- A. Einstein
http://www.geocities.com/CapeCanaveral/Lab/5932
http://freenet.buffalo.edu/~bx238
http://members.tripod.com/~Vibrations
Edward Schaefer
Thomas H. Hoovler wrote:
>
> On Tue, 24 Feb 1998, Edward Schaefer wrote:
>
> > Tom Roberts wrote:
> > >
> > > Edward Meisner wrote:
> > > > If there is no concept of potential energy in GR,
> > > > then it would be
> > > > incapable of describing many phenomena.
> > >
> > > in GR, gravitation is MUCH MORE COMPLEX than any single value can
> > > ever express.
> >
> > Yes it is, and that is beside the point. A scalar value which can
> > be referred to as the gravitational potential in a static field
> > surrounding a spherically massive object wrt a distant observer can
> > be expressed.
>
> The problem is that this "scalar" value you refer to sits in a
> tidy little corner of the metric tensor in even the best of
> systems where it can be interpreted as such.
That is the key to this thread. I am focussing on a small piece
of a much bigger problem, and acknowledge it as such.
> Hence, it will not be a scalar in GR, but part of the second
> rank metric tensor.
As noted elsewhere, the simplistic answer is Phi = 1 - sqrt (g_00),
which prety well illustrates your point.
> There have been some relativistic attempts at scalar gravity
> theories in the past (The Pais Einstein biography is a good
> source of some of these), but nothing apparently came of them.
That hardly surprises me. Even as I question part of GR, I still
cannot help but respect Einstein for all that he accomplished with
it 80 years ago; and that respect has only grown as I delve more
and more into GR.
Besides, I was not looking for a scalar theory, but instead a
scalar desciption of a corner of the tensor theory.
EMS
Edward Schaefer
If thge train is accelerating, then that is indeed the case. To
an observer on the train, the cow is being accelerated towards it
by a "gravitational field" created by the work of the engine. I
admit that this is a very odd field: One in which the Andromeda
galaxy experienced the acceleration from thr field 2 million years
ago for example. None-the-less, it does serve to describe the
connection between the theories of Einstein and Newton, and how
the concept of potential can be translated over. (I admit that
it is a loose translation, but of some use none-the-less.)
EMS
Edward Meisner
Tom Roberts wrote:
> Edward Meisner wrote:
> > If there is no concept of potential energy in GR, then it would be
> > incapable of describing many phenomena.
>
> You are assuming the "potential energy" is the ONLY way to describe such
> phenomena.
>
> > There would be no equations to
> > describe the work done in raising a mass to a greater height, for instance.
>
> That presupposes that the concept "greater height" has meaning. In a
> arbitrarily-curved spacetime there is NOT necessarily such a concept....
> And what do you mean by "work"? In curved spacetime it is a bit wobbly (to
> say the least). That's the whole point. "Work" is a SINGLE number, and
> in GR, gravitation is MUCH MORE COMPLEX than any single value can ever
> express.
>
> In Newtonian gravitation, the gravitational force is conservative. That is,
> no matter what path you take to a give point, the amount of energy required
> to get there is independent of the path taken. In GR that no longer holds;
> the energy required (however defined) is in general path dependent. All
> sorts of things are path dependent in GR, which in Newtonian physics are
> path-independent. The curved geometry of spacetime has great ramifications
> throughout one's theoretical outlook on the world.
>
> > If GR can not produce equations that describe such situations, then what
> > good is it?
>
> _IF_ that were true, then the theory would not be much good....
> But GR can produce equations to describe such situations. They just do
> not use "potential energy", etc. They use geodesic equations. GR is
> quite "good" (but admittedly mathematically cumbersome (:-)).
>
Obviously I am no expert in these matters, so I will defer to your greater
knowledge. Nevertheless, this is reassuring. I am glad to hear that GR can
describe these things. When people say there is no potential energy in GR, I get
alarmed because that would exclude practically the whole of physics that came
before it from its domain. For example, how would GR describe what happens when
you throw a projectile upward? Newton handles this with no problem and makes very
accurate predictions. I thought that GR could not handle this because it had no
potential energy and therefore could not predict the trajectories of projectiles,
for instance. Nothing could be more clear cut than the fact that work must be done
to increase the distance between two masses. Why does it take an expenditure of
energy to increase the distance? What is happening? Every other natural process
that involves a transformation of energy can be described in a mathematically
rigorous fashion. Why couldn't GR do this with gravity? I realize now that I was
too hasty in my judgment and that my understanding was too simplistic. I have only
one question. Can GR correctly model energy transformations, or is this still too
simplistic? To make this more clear, what is happening to, say, the energy of the
combustion process in a rocket. My hypothesis would be that the energy is going
into curving space and that the field of the rocket is interacting with the field
of the earth to produce its trajectory. Before you say that GR is nonlinear and
that therefore the fields cannot interact, remember that gravitomagnetic fields
can interact act with gravitational fields to produce chiral fields.
Edward Meisner
ROB wrote:
> Edward Schaefer wrote:
>
> > How are gravitational potential and gravitational potential energy
> > defined in GR? I have been struggling with that concept lately
> > but have not hit upon a satisfactoty answer.
> >
> > EMS
>
> Ed,
> It's hardly surpising that you've been struggling with such a concept
> because the concepts of potential and energy are negated within the
> gravitation model.
What if a one ton meteorite flattened me into a pancake? Would GR
have nothing to say about the eneregy transfer that caused this? What if
you pumped a trillion gallons of water to a mountain top 2 miles high?
What was expended to do the work? You don't mean to say that the heat
generated, and the wear on the parts are illusions do you?
Steve Carlip
Edward Schaefer (scha...@plansys.com) wrote:
: As for the Principle of Equivalence, there is a fly in that ointment:
: An observer in a box DOES have a way of determining whether the
: gravitational field that he is in has curvature of not, based on the
: tidal effects of a free-falling ball coated with a liquid. (See H.
: Ohanian and R. Ruffini, "Gravitation and Spacetime", 2nd ed. (1994).)
: The percentage of distortion caused by those effects do not vanish
: as the size of the ball goes to zero.
The effect you're describing certainly exists. The question
is whether it can be used to define a gravitational potential.
A good starting point is the Newtonian theory. The same kind
of distortion occurs there---it's proportional to the matrix
of second derivatives of the potential. That is, if I denote
the gradient by d_i and the potential by F, the distortion of
a droplet is described by the 3x3 matrix (d_i d_j F). The
three eigenvectors of this matrix determine three axes, and
the corresponding eigenvalues give the rate of stretching
along these axes.
In general relativity, a similar quantity exists, though it
depends on a choice of reference frame. If an observer has a
four-velocity U^a, the distortion matrix is (U^a R_{abcd} U^d).
(This looks like a 4x4 matrix, but the components in the U^a
direction, the "time" components, are identically zero.) To
use this to define a gravitational potential, you'd have to
show that it was a matrix of second derivatives of a single
function F, which would then be your relativistic potential.
Next, a mathematical diversion. Suppose you have a vector v_i,
and you want to know whether it can be written as the gradient
of a function f, v_i = d_i f. Suppose this is the case. Then
since partial derivatives commute, it must be true that
d_j v_i = d_j d_i f = d_i d_j f = d_i v_j,
i.e., d_i v_j - d_j v_i = 0 (or curl v = 0). This is known as
an integrability condition; if it is not satisfied, then v_i is
*not* the gradient of any scalar.
Similarly, for (U^a R_{abcd} U^d) to be a matrix of second
derivatives of a single potential function, it must obey a set
of integrability conditions. In general, it does not, and the
distortion you've described cannot be attributed to a potential.
In the special case of a nearly flat spacetime, in the rest frame
of the droplet (that is, U^i = 0 for i=1,2,3), it happens that
(U^a R_{abcd} U^d) *is* proportional to the matrix of second
derivatives of g_00, and a description in terms of potentials
is possible. But this is *only* true in the Newtonian limit,
and then only in the rest frame of the droplet. In almost all
other situations, the distortion you're describing is simply too
complicated to be described by a single potential.
Steve Carlip
car...@dirac.ucdavis.edu
ROB
Edward Schaefer
Steve Carlip wrote:
>
> Edward Schaefer (scha...@plansys.com) wrote:
>
> : As for the Principle of Equivalence, there is a fly in that ointment:
> : An observer in a box DOES have a way of determining whether the
> : gravitational field that he is in has curvature of not, based on the
> : tidal effects of a free-falling ball coated with a liquid. (See H.
> : Ohanian and R. Ruffini, "Gravitation and Spacetime", 2nd ed. (1994).)
> : The percentage of distortion caused by those effects do not vanish
> : as the size of the ball goes to zero.
>
> The effect you're describing certainly exists. The question
> is whether it can be used to define a gravitational potential.
>
>
> In the special case of a nearly flat spacetime, in the rest frame
> of the droplet (that is, U^i = 0 for i=1,2,3), it happens that
> (U^a R_{abcd} U^d) *is* proportional to the matrix of second
> derivatives of g_00, and a description in terms of potentials
> is possible. But this is *only* true in the Newtonian limit,
> and then only in the rest frame of the droplet. In almost all
> other situations, the distortion you're describing is simply too
> complicated to be described by a single potential.
>
That much I can accept (including what I snipped).
A related question: I have now painted myself into the following
corner: I have a metric which is the Scwarzschild Solution in
isometric coordinates, except that g_00 = 1/g_11. For the
covariant intergral, T_u^v;v to be 0, the stress-energy tensor
gives the gravitational field 0 energy but also non-zero stress.
What are the physical ramifications of that?
EMS
Todd Desiato
Edward Schaefer <scha...@plansys.com> wrote in article
<34F46E...@plansys.com>...
IMO, that would be the correct answer. The electromagnetic energy tensor is
conserved. See E-print archive "Einstein's Energy-Free Gravitational Field"
by Kenneth Dalton, gr-qc/9512008, available on the LANL E-print server. A
very short but interesting and note-worthy paper, showing a possible error
in Einstein's interpretation of conservation of energy. There are also 2
other papers available by Dalton that I found interesting regarding GR, and
EM in general.
The reason I agree is that in my paper I describe gravitation and the
creation of gravitons as a result of reflection of the EM field within 1
virtual particle wavelength distance from the emitting particle. Therefore
the gravitational field has no energy of its own. All of the energy is
conjugate reflected energy of the EM field. The EM field "force" is
propagated by virtual photon wave functions that are real and in-phase
(repulsive) or out-of-phase (attractive). Gravitation is the interaction of
some of these virtual photons with the ones that are reflected back. The
coefficient of reflection would naturally be proportional to the density of
energy in the region. That is how the theory of quantum gravity unifies QED
and GR. This reaction is explained better in Part II, B of my paper, but in
effect it leads to increased permittivity, permeability and a correlated
spin 2 particle.
http://tjd-online.simplenet.com/Model/Model.htm
Todd Desiato