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Science Forum Index » Physics - Research Forum » Double counting gravitational potential energy
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| Jonathan Scott |
 Posted: Wed Feb 14, 2007 8:27 am |
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I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount. Each mass is effectively
moved to a different potential within the field of the other. That
means that the total change in the potential energy seems to be
exactly twice the work done to move the masses apart. This doesn't
seem to add up unless some other energy is transferred from somewhere
else at the same time, for example out of the field, but I'm not aware
of this normally being considered necessary in a Newtonian model.
Have I missed something obvious?
In a relativistic weak field model, the rest energy of each mass has
been multiplied by a factor (1-Gm/rc^2) where the m refers to the
other mass. This has exactly the same effect as the potential energy
decrease above, so the problem still occurs. |
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| Guest |
| Posted: Wed Feb 14, 2007 2:38 pm |
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In article <1171474036.443689.18910@p10g2000cwp.googlegroups.com>,
Jonathan Scott <jonathan_scott@vnet.ibm.com> wrote:
Quote: I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount.
Here's the cure for this malady. Teach yourself never to say "the potential
energy of a mass", but only to say "the potential energy of the system".
The expression you give is the potential energy for the two-body system,
not the potential energy of each of the two masses.
-Ted
--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |
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| NoEinstein |
| Posted: Sat Feb 17, 2007 12:40 am |
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On Feb 14, 1:27 pm, Jonathan Scott <jonathan_sc...@vnet.ibm.com>
wrote:
Quote: I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount. Each mass is effectively
moved to a different potential within the field of the other. That
means that the total change in the potential energy seems to be
exactly twice the work done to move the masses apart. This doesn't
seem to add up unless some other energy is transferred from somewhere
else at the same time, for example out of the field, but I'm not aware
of this normally being considered necessary in a Newtonian model.
Have I missed something obvious?
In a relativistic weak field model, the rest energy of each mass has
been multiplied by a factor (1-Gm/rc^2) where the m refers to the
other mass. This has exactly the same effect as the potential energy
decrease above, so the problem still occurs.
Dear Jonathan: Moving two bodies apart requires a greater force at
the beginning, and a lesser force (actually ¼ as much) at the radius
2r. Summing those forces along the move will give the total potential
energy gained.
You are right that there "appears" to be a double counting. However,
you will always have both a reaction, and an equal and opposite
reaction. You can't pull on a rope and develop a resistance without
there being a reaction on both ends. Similarly, you can't push two
masses apart without there being a force applied to both masses. The
lone exception to that would be if a force were to be applied to one
mass so quickly, that the inertia of the other mass was sufficient to
keep it from moving, via gravity, to "follow" the other mass.
Ted is both right and wrong. The two masses are "a system" because
the attraction can't occur without both masses contributing their
equal gravitational attractions. But he is wrong, if you wish to
consider either of your masses as your "point of reference". Then,
each mass DOES have a potential energy just its own. But if you
change your point of reference to the other mass, there isn't this
sudden DOUBLING of the potential energy. As with my rope analogy,
there is only ONE stress! Having two points of view doesn't "double
the potential energy" it only DOUBLES the number of points of view!
NoEinstein
__________ |
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| Greg Egan |
| Posted: Sun Feb 18, 2007 11:54 am |
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In article <1171474036.443689.18910@p10g2000cwp.googlegroups.com>,
Jonathan Scott <jonathan_scott@vnet.ibm.com> wrote:
Quote: I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount. Each mass is effectively
moved to a different potential within the field of the other. That
means that the total change in the potential energy seems to be
exactly twice the work done to move the masses apart. This doesn't
seem to add up unless some other energy is transferred from somewhere
else at the same time, for example out of the field, but I'm not aware
of this normally being considered necessary in a Newtonian model.
Have I missed something obvious?
Your formula for potential energy is perfectly correct, but then I think
you confuse matters by treating the two masses as if they were *test
masses* in the gravitational field of some much larger object, rather
than two parts of a single system which has to be treated as such.
If we're discussing, say, the potential energy of a stone in the Earth's
gravitational field, then we can pretend that the Earth is an immovable
object and the source of the entire field. We can also pretend that the
potential energy belongs to the stone itself, rather than the system as a
whole. And if we have two stones, we can compute their potential
energies (or changes of potential energy) independently, and then just
add them up.
That's all fine when one of the masses is dominant and the others
negligible, but when you have two bodies of comparable mass, you can't
talk about the potential energy of each body separately. You just have
the two-body system, and its potential energy as a whole. You have the
correct formula for that. What you can't do is then switch back to the
"test mass" model twice, taking turns to call each mass the test mass,
and then add up the results. |
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| Jonathan Scott |
| Posted: Sun Feb 18, 2007 11:55 am |
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On 15 Feb, 00:38, e...@lfa221051.richmond.edu wrote:
Quote: ...
Here's the cure for this malady. Teach yourself never to say "the potential
energy of a mass", but only to say "the potential energy of the system".
The expression you give is the potential energy for the two-body system,
not the potential energy of each of the two masses.
I agree that the energy which was added or removed from the system
is the standard potential energy for the system, and in basic
Newtonian
theory you can just claim vaguely that this energy cannot be
localized.
However, in basic relativistic gravity, the explanation is that the
time rate
of each body is red-shifted by the potential of the other, which has
the
effect of changing the total potential energy of the two bodies by
exactly
twice the expected value.
I already have an explanation for this effect, which is that whenever
the
overall potential energy between two objects changes by a certain
amount, the potential energy of each object changes by that same
amount
but half of this change is matched by an equal and opposite change to
the energy of the field, so overall energy is conserved. With this
model,
the energy density of the field is mathematically in exactly the same
form
as the Maxwell energy density of an electrostatic field (at least in a
weak
field approximation) except that it has a positive sign because the
field
energy is opposite to the potential energy in this case.
(In the electrostatic case, the field energy is simply another way of
looking
at the potential energy, and there is no change to the mass/energy of
the actual objects).
However, this explanation seems to suggest that a basic relativistic
model
requires energy to be located in the field, which contradicts GR, so
I'm
wondering whether anyone is about to spring to the defence of GR and
point out some flaw.
I'm quite happy about this personally, because I have my own Machian
gravity
theory which in its simplest form doesn't give the right PPN beta and
is
hence ruled out by experiment, but if the field in empty space has the
energy density given by the above explanation, then the PPN beta value
comes out exactly right (and as far as I know, all the other PPN
parameters
should too, although I don't yet have exact field equations for the
theory).
Jonathan Scott |
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| Thomas Smid |
| Posted: Sun Feb 18, 2007 11:56 am |
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On 14 Feb, 18:27, Jonathan Scott <jonathan_sc...@vnet.ibm.com> wrote:
Quote: I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount. Each mass is effectively
moved to a different potential within the field of the other. That
means that the total change in the potential energy seems to be
exactly twice the work done to move the masses apart. This doesn't
seem to add up unless some other energy is transferred from somewhere
else at the same time, for example out of the field, but I'm not aware
of this normally being considered necessary in a Newtonian model.
Have I missed something obvious?
Your expression -G m_1 m_2 (1/r_2 - 1/r_1) is the change in potential
energy for one particle, not the total energy change. Assuming the
kinetic energies are kept constant, the total energy change is thus
-2*G m_1 m_2 (1/r_2 - 1/r_1), so everything adds up as it should.
Thomas |
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| Jonathan Scott |
| Posted: Sun Feb 18, 2007 11:56 am |
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On 17 Feb, 10:40, NoEinstein <noeinst...@bellsouth.net> wrote:
Quote: Dear Jonathan: Moving two bodies apart requires a greater force at
the beginning, and a lesser force (actually =BC as much) at the radius
2r. Summing those forces along the move will give the total potential
energy gained. ...
Of course. That's how the Newtonian result for the potential energy
for
a system is defined.
In pure Newtonian theory, the doubling can be treated as an illusion.
However, in simple relativistic gravity theory, each of the bodies
causes
a "red shift" of the other by a factor of the form (1 - Gm/rc^2) where
m
is the other mass, causing the effective rest energy of each of the
two
bodies to change by exactly the same amount of potential energy
as the overall change, and hence requiring a transfer of the same
amount of energy to the field to compensate. I can't see any way for
this to be wrong without violating the well-known assumptions about
how
gravity affects clock rates. I've now investigated this further, and
it is
looking very promising as a basis for a new way for handling gravity
in a semi-Newtonian model.
Jonathan Scott |
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| Thomas Smid |
| Posted: Sun Feb 18, 2007 11:56 am |
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On 14 Feb, 18:27, Jonathan Scott <jonathan_sc...@vnet.ibm.com> wrote:
Quote: I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount. Each mass is effectively
moved to a different potential within the field of the other. That
means that the total change in the potential energy seems to be
exactly twice the work done to move the masses apart. This doesn't
seem to add up unless some other energy is transferred from somewhere
else at the same time, for example out of the field, but I'm not aware
of this normally being considered necessary in a Newtonian model.
Have I missed something obvious?
Hi Jonathan,
Your expression -G m_1 m_2 (1/r_2 - 1/r_1) is the potential energy
change of one particle, not the total energy change (as you are
claiming). So, assuming the kinetic energies remain constant, the
total energy change for both particles is twice this values i.e. -2*G
m_1 m_2 (1/r_2 - 1/r_1). So everything adds up as it should.
Thomas |
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| Jonathan Scott |
| Posted: Sun Feb 18, 2007 11:57 am |
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After some helpful discussions on the Physics Forums, I think I now
have a much simpler illustration of how the paradox arises, involving
only one body, so it can be matched with the Schwarzschild solution.
Consider assembling a thin shell of mass m and radius r by lowering
the mass bit by bit from infinity (in a spherically symmetrical way),
extracting the potential energy in the process. The work done is the
integral of Gm/r dm which is Gm^2/2r. When this is complete, the whole
shell is at a red-shift of the potential (1-Gm/rc^2) so the effective
energy differs from the original energy by Gm^2/r, which is twice as
much. This seems to mean that the shell has lost twice as much energy
as was extracted, hence the paradox.
My suggestion is that the missing energy must have gone "into the
field", which seemed a meaningful thing to suggest in a semi-Newtonian
model, and the amount exactly matches the amount for a similar Coulomb
model except for the sign, so it suggests that the mathematics of this
"field" should be similar to that for electrostatics.
However, the Schwarzschild solution seems to deny the existence of
this energy, in that Einstein's vacuum equations say there is nowhere
it could be outside the shell. So where did it go? |
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| Guest |
| Posted: Tue Feb 20, 2007 4:09 am |
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In article <1171529525.194321.164630@m58g2000cwm.googlegroups.com>,
Jonathan Scott <jonathan_scott@vnet.ibm.com> wrote:
Quote: On 15 Feb, 00:38, e...@lfa221051.richmond.edu wrote:
...
Here's the cure for this malady. Teach yourself never to say "the potential
energy of a mass", but only to say "the potential energy of the system".
The expression you give is the potential energy for the two-body system,
not the potential energy of each of the two masses.
I agree that the energy which was added or removed from the system
is the standard potential energy for the system, and in basic
Newtonian
theory you can just claim vaguely that this energy cannot be
localized.
Or if you prefer (still within Newtonian gravity) you can define an
energy density of the gravitational field in a manner analogous to the
energy density of the electric field in electrostatics. At least, I
think you can -- I don't see any reason the usual argument in
electrostatics wouldn't go through. The energy density will come out
negative everywhere, but in a classical Newtonian context I don't
particularly see why that needs to bother us.
Quote: However, in basic relativistic gravity, the explanation is that the
time rate
of each body is red-shifted by the potential of the other, which has
the
effect of changing the total potential energy of the two bodies by
exactly
twice the expected value.
I don't understand this description at all. Fundamentally, of course,
there's no such thing as gravitational potential energy in a
relativistic context. In certain limited contexts, we can define
something that behaves like a gravitational potential energy, and when
we do, we always define it in such a way that it behaves correctly in
the Newtonian limit. If you've adopted a definition of gravitational
potential energy that behaves incorrectly by a factor of 2 in the
Newtonian limit, then you've simply made a poor choice of definition.
-Ted
--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |
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| Guest |
| Posted: Tue Feb 20, 2007 4:09 am |
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In article <1171627815.714023.163290@q2g2000cwa.googlegroups.com>,
Thomas Smid <thomas.smid@gmail.com> wrote:
Quote: Your expression -G m_1 m_2 (1/r_2 - 1/r_1) is the change in potential
energy for one particle, not the total energy change. Assuming the
kinetic energies are kept constant, the total energy change is thus
-2*G m_1 m_2 (1/r_2 - 1/r_1), so everything adds up as it should.
This is false. There is no factor of two. Suppose I raise a 1-kg
rock up one meter at constant speed in the Earth's gravitational
field. That costs me 9.8 joules of energy, not 19.6 joules. The same
is true if I try to move the Earth one meter in the gravitational
field of the rock, or if I move them both 50 cm away from each other.
In Newtonian gravity, potential energy is potential energy of the system,
not of each particle. I promise.
-Ted
--
[E-mail me at name@domain.edu, as opposed to name@machine.domain.edu.] |
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| John C. Polasek |
| Posted: Tue Feb 20, 2007 12:08 pm |
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On Sat, 17 Feb 2007 10:40:28 +0000 (UTC), NoEinstein
<noeinstein@bellsouth.net> wrote:
Quote: On Feb 14, 1:27 pm, Jonathan Scott <jonathan_sc...@vnet.ibm.com
wrote:
I thought I knew quite a lot about gravity, but I've just totally
confused myself when thinking about potential energy in good old
Newtonian gravity.
It's a standard result in Newtonian gravity that if two masses m_1 and
m_2 are moved apart from distance r_1 to distance r_2, then the total
energy transferred to the masses is -G m_1 m_2 (1/r_2 - 1/r_1). An
obvious way to see this is to integrate the force between them over
the distance moved.
Think of it this way. The masses do not gain energy nor does a
potential energy *exist*. The *potential energy* for m1 is the
integral of the force m1MG/r^2*dr being simply a measure of the
effort expended against gravity, which in this case has force negative
to the direction. Neither mass is affected one whit by the experience.
The potential energy of M can be computed without m1 and it is as
energy per unit mass.
In other words the measurables are the force function and the
displacement. Energy cannot be measured non-destructively. It is a
value only derived by multiplying measureable quantities and the
product is not a measurement, it's a derived value.
I don't know if that helps.
Quote: However, as far as I can see, the potential energy of each mass
changes by the exactly the same amount. Each mass is effectively
moved to a different potential within the field of the other. That
snip
John Polasek |
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| Jonathan Scott |
| Posted: Tue Feb 20, 2007 12:08 pm |
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Cracked it!
After being very helpfully reminded on the Physics Forums to reread
MTW box 23.1 more carefully, I now understand that the source of the
paradox is entirely in the internal energy of the pressure term need
to maintain equilibrium. In the case where the central object is
stable, this pressure term gives rise to exactly the same energy as
the external potential energy, balancing out the double decrease in
the rest mass.
What really confused me in this case is the fact that I had been
taught long ago that the nominal first order "rest mass" of the
central object is m(1-Gm/rc^2), without mentioning that this EXCLUDES
the internal energy. The overall effective rest mass INCLUDING the
internal energy is however m(1-Gm/2rc^2), matching the potential
energy loss in forming the body. |
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| Jonathan Scott |
| Posted: Thu Feb 22, 2007 4:56 am |
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On 20 Feb, 22:08, "Jonathan Scott" <jonathan_sc...@vnet.ibm.com>
wrote:
Quote: Cracked it!
After being very helpfully reminded on the Physics Forums to reread
MTW box 23.1 more carefully, I now understand that the source of the
paradox is entirely in the internal energy of the pressure term need
to maintain equilibrium. In the case where the central object is
stable, this pressure term gives rise to exactly the same energy as
the external potential energy, balancing out the double decrease in
the rest mass.
Grrrrr. No I haven't. I was just feeling so intimidated by the
"factor of two error" responses here and elsewhere that as soon as I
realized I'd made an error (in ignoring the internal energy) I thought
it must be the explanation of this paradox, and caved in.
Unfortunately, I don't think that the internal energy cannot be enough
to make the difference. If for example the body is formed from a
nearly-incompressible medium, then I don't think there is any internal
energy worth mentioning. The pressure will be the same anyway, but
the stored energy related to pressure is given by the pressure times
the change in volume, and for a nearly-incompressible medium that
change would be negligible.
MTW box 23.1 and exercise 23.7 are entirely on the "actual potential
energy" side of the way of looking at it, and are therefore obviously
entirely self-consistent, as one would obviously expect. I'm also
completely happy that the effective total energy of the system is the
same as the original energy of the mass minus the gravitational
potential energy removed from the system, despite the fact that most
of the replies are still trying to convince me of this obvious fact.
I still think that there seems to be a valid alternative relativistic
way of looking at this situation which is that the clock rate change
(in the central case caused by its own potential, or in the two-object
case caused by the potential due to the other mass) should essentially
multiply the rest energy of the relevant object by a factor (1-Gm/
rc^2). I cannot see how to explain the factor of 2 difference between
this view and the total potential energy of the system (except of
course by my "energy in the field" suggestion mentioned previously,
which however doesn't seem to have an equivalent in the GR
interpretation).
I'm quite prepared to believe that the "clock rate" view may be
invalid for some reason, but especially in the two-object case I
believe it is normally considered to be a perfectly standard way of
considering the potential energy of a test mass in the potential of a
larger mass, so why shouldn't it also apply to two masses which affect
each other?
(I'll probably have changed my mind about this twice more by the time
this appears; I've been using this moderated newsgroup because I've
seen some very good thinking here and I assume that anyone with half a
brain probably isn't going to spend a lot of time wading through the
rubbish on the unmoderated newsgroups, but the technical side of the
moderation process seems to be a bit of a lottery, in that some posts
seem to be approved almost immediately yet posts sent earlier
sometimes turn up a day or two later if at all).
Jonathan Scott |
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| Jonathan Scott |
| Posted: Sat Feb 24, 2007 4:58 am |
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On 22 Feb, 22:54, e...@lfa221051.richmond.edu wrote:
Quote: ...
But that's just the point! Even in the Newtonian context, you always
make a factor-of-two error if you consider the potential energy of
each of two bodies due to the other. As the old joke goes, "Doctor, it
hurts when I do this." "So don't do that."
That's not what I'm doing! I've now made quite a bit of progress on
this and I think I can now explain what I mean somewhat more
meaningfully than at the start of the thread.
In the basic Newtonian context, it is possible by analogy with
electrostatics to define a field energy density of -g^2/(8 pi G) which
is conventionally conserved. The potential energy can then be
described entirely by a change in the field energy.
It is also possible to use a more relativistic semi-Newtonian model
where the loss of potential energy of an individual object is
calculated by multiplying its rest mass by its potential (or more
accurately integrating the rest mass density times the local
potential) but a separate positive energy density of g^2/(8 pi G) is
assumed to be present in the field (including any part of the field
which lies within the object). This means that the energy lost from
the mass of each relevant object is twice the potential energy change
of the system, but the gain in the field energy exactly compensates
for this so that the overall potential energy is correct.
This can be shown by Gauss's Law and a trivial integral to give
exactly the same mathematical result as the first model, at least when
integrated "over all space":
original mass - field energy = mass as affected by total potential +
field energy
In General Relativity, the Komar mass formula for a stationary system
similarly decreases the original rest mass by the total potential
(giving twice the potential energy loss) but then adds back in 3 times
the integral of the pressure over the body (that is, T_11 + T_22 +
T_33), which is yet another mathematical expression with the same
value as the potential energy. This expression works in the Newtonian
form as well (see MTW exercise 23.7).
In GR, this pressure term forms part of the stress-energy tensor and
is apparently interpreted as meaning that there is "really" that
amount of energy density due to pressure at that location. However,
in the Newtonian model, this looks like just another mathematical
expression that comes to the same value when integrated. Athough
internal pressure can store some internal energy in a Newtonian
system, that energy is the integral of the pressure times the change
in volume, so for a relatively incompressible material there is little
energy stored, yet this pressure term is 3 times the pressure
integrated over the whole volume. I'm still trying to understand this
term.
The semi-Newtonian model which includes positive field energy can
certainly be extended to handle an arbitrary number of objects (where
the mathematics of the field energy is identical to that for an
electrostatic field), and conserves energy globally under any sort of
gravitational interactions. I still need to check whether it
conserves energy locally as well; that is certainly true for the
trivial electrostatic analogy using negative field energy.
It is however obvious that a semi-Newtonian model based on the Komar
mass interpretation, with no energy located in the field, cannot
possibly conserve gravitational energy locally.
I'm therefore currently investigating whether the semi-Newtonian model
with positive field energy is a physically plausible and self-
consistent model at that level of approximation, and if so whether
there is some systematic way of mapping it to or from the GR way of
looking at things. |
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