Double counting gravitational potential energy
Jonathan Scott
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.
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@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.
>
>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 na...@domain.edu, as opposed to na...@machine.domain.edu.]
NoEinstein
wrote:
Dear Jonathan: Moving two bodies apart requires a greater force at
the beginning, and a lesser force (actually ΕΊ 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
__________
Jonathan Scott
> ...
> 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
Thomas Smid
> 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
Jonathan Scott
> Dear Jonathan: Moving two bodies apart requires a greater force at
> 2r. Summing those forces along the move will give the total potential
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
Thomas Smid
> 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
Jonathan Scott
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?
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>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.
>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.
eb...@lfa221051.richmond.edu
Thomas Smid <thoma...@gmail.com> wrote:
>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.
John C. Polasek
<noein...@bellsouth.net> wrote:
>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.
>> 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
Jonathan Scott
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.
Jonathan Scott
wrote:
> 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
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>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?
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."
It's just false to think that the potential energy of a two-body
system can be written as the potential energy of body A plus the
potential energy of body B. To calculate the potential energy of the
Earth-Moon system, you can *either* consider the potential energy of
the Moon in the gravitational potential of the Earth, *or* the
potential energy of the Earth in the gravitational potential of the
Moon, but not both. That's true in a Newtonian context, and it's true
in a relativistic context (if you're in a situation in which it makes
sense to talk about gravitational potential energy at all).
Jonathan Scott
> ...
> 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.
Gerry Quinn
eb...@lfa221051.richmond.edu says...
> In article <1172054600.7...@s48g2000cws.googlegroups.com>,
> Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>
> >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?
>
> 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."
One way to look at it is to think about what happens if you try to
extract the potential energy by letting one mass fall onto the other.
First you drop A onto B to extract A's potential energy. Then you drop
B onto... oh, wait.
- Gerry Quinn
Jonathan Scott
I've been reminded that there are ways of representing the energy of
the gravitational field in GR as a pseudotensor relative to a chosen
set of background coordinates. One such method is the Landau-
Liftshitz pseudotensor, which I'm guessing might be analogous to the
g^2/8piG field energy in Newtonian theory. If so, can anyone tell me
what sign it has for the field energy density? I should be able to
work it out eventually from MTW section 20.3, but I've studied special
relativity a lot (and even GR a little) using different sign
conventions from MTW, so I often get terribly confused when working
with GR.
>From my semi-Newtonian explanation where the double potential energy
deficit is matched by a positive field energy density, I'd would
expect the effective total energy to be given by the local energy
tensor multiplied by the local clock rate term of the potential plus a
positive field energy density roughly matching the Newtonian g^2/8piG.
a student
wrote:
> 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 have a possibly relevant (possibly not!) comment that a number of GR
effects rely on both local time dilation effects (clock rate) and
global curvature effects. The bending of starlight by the sun is a
classic example, where each effect contributes equally. So maybe your
approach is not taking spatial curvature effects into account ? (eg,
when bringing mass in from infinity).
For example, suppose one has a universe containing a single straight
cosmic string, with mass per unit length mu. The metric is then
locally flat everywhere outside the string, but with an angle deficit
proportional to mu (i.e., in cylindrical polar coordinates, with the
string lying along the z-axis, the metric is the usual
ds^2 = dt^2 - dz^2 - dr^2 - r^2 (sin^2 theta) d theta^2,
but where theta only runs from 0 to 2pi - k mu). One can equivalently
throw away the z-dimension, and instead consider a particle of mass
proportional to mu in 2+1 GR, instead of the string in the usual 3+1
GR.
Now, if one brings in mass from infinity, thus increasing mu, no
clock rates change at all. The only change is in the global angle
deficit.
I'm not sure what a 'semi-Newtonian' approach looks like for 2+1 GR,
but the above suggests that some of the 'energy' in your 3+1 case may
be involved in global curvature effects. In particular, bringing in
mass from infinity doesn't just add to the mass already there, it also
bends space.
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
[...]
>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).
I agree with you that it's *possible* to do this; what I don't
understand is why you'd want to do this. I didn't understand all the
details of this post, unfortunately, so maybe you explained it and I
just don't get it. Oh, well.
Just to be clear, you are talking about defining a total energy to be
Kinetic energy + "Potential energy" + "Field energy",
where in the Newtonian limit your potential energy is precisely twice
the standard expression, and your field energy is precisely minus
the standard expression, right? To be specific, by "the standard
expression," I mean
U = (1/2) Integral of rho Phi dV
where rho is the mass density and Phi is the gravitational potential.
If I've got that right, then I just have two obvious observations:
1. You'll have to clearly demonstrate that there are advantages to
writing things this way if you want to win any converts.
2. I think that you should clearly signpost that factor of 2 when
describing this to people, lest others be confused in the same way
that I was.
Jonathan Scott
> [...]
> Just to be clear, you are talking about defining a total energy to be
>
> Kinetic energy + "Potential energy" + "Field energy",
>
> where in the Newtonian limit your potential energy is precisely twice
> the standard expression, and your field energy is precisely minus
> the standard expression, right? To be specific, by "the standard
> expression," I mean
>
> U = (1/2) Integral of rho Phi dV
>
> where rho is the mass density and Phi is the gravitational potential.
Yes, that's exactly the idea, thanks.
The standard Newtonian approximation for GR (or any similar
relativistic gravity theory) already says that the effective rest
energy of an object within a potential Phi (whether caused by itself
or another object) is reduced by a potential energy given by the
integral of rho Phi dV (without any factor of a 1/2). This leads
directly to the factor of 2 paradox, and hence requires some
compensating positive energy to balance the books.
It is obvious from standard electrostatic potential theory that the
classical field energy would exactly match the classical potential
energy for any combination of sources, and hence that in the Newtonian
approximation, the field energy with a switched sign would exactly
compensate for the double loss in the potential energy of the actual
objects, regardless of configuration. I am therefore investigating
the hypothesis that this is actually a viable physical model of what
happens, which among other things would explain exactly where
gravitational energy was located at least to this level of
approximation. However, I'm also interested in whether there is some
other alternative explanation of this paradox.
In GR, there is a conventional expression for the effective mass of a
stationary system containing a central object, which is known as the
"Komar mass". In this expression, the compensating positive energy
expression is already present, and is in the form of 3 * integral of
pressure dV, which is equivalent to the sum of the other diagonal
terms of the stress-energy tensor. This gives the right answer
mathematically in the single-object case (and does so as well in a
Newtonian approximation), but I don't so far see how this could
actually correspond to physical energy at that location, especially in
the Newtonian approximation.
In the case where there are multiple bodies, things are much more
complicated in GR, and I'm not sure whether it's possible to get any
meaningful answers from GR at all, especially given my limited skills
and shortage of time to look at this. However, I am fairly sure that
the 3 * pressure integral as in the Komar mass cannot fix the paradox
in this case, because the potential energy deficit is proportional to
potential times mass, but in the two-object case, the pressure does
not build up in the same way as in the self-energy case. Even if the
object is held fixed in space against the field the resulting pressure
within the object will only be proportional to the field, not to the
potential of the other object, and if it is in free fall, any pressure
would only arise from tidal effects. I therefore suspect that the
conventional "Komar mass" calculation in GR cannot consistently
describe the location or total amount of energy even in a trivial
system containing two static masses (held apart by a massless rod if
necessary), even though my simple Newtonian idea using positive field
energy would give a sensible answer in such a case.
There are however other approaches to energy conservation in GR,
involving non-tensors and typically expressed in terms of a
"background" coordinate system. It might well be that one of these
already matches my idea above. However, these seem to be quite
advanced topics which are the subject of current research.
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>The standard Newtonian approximation for GR (or any similar
>relativistic gravity theory) already says that the effective rest
>energy of an object within a potential Phi (whether caused by itself
>or another object) is reduced by a potential energy given by the
>integral of rho Phi dV (without any factor of a 1/2). This leads
>directly to the factor of 2 paradox, and hence requires some
>compensating positive energy to balance the books.
Can you give me a reference to this "standard Newtonian
approximation"? It seems, frankly, like a silly approximation for
exactly this factor-of-two reason. To whom is it "standard"?
When I try to go from GR to a Newtonian approximation (which I only do
in an asymptotically flat, weak-field regime), I define energy in such
a way that kinetic plus potential energy is conserved, not in such a
way that it's not conserved and I have to postulate another form of
energy!
Jonathan Scott
> In article <1172678974.702079.300...@s48g2000cws.googlegroups.com>,
>
> Jonathan Scott <jonathan_sc...@vnet.ibm.com> wrote:
> >The standard Newtonian approximation for GR (or any similar
> >relativistic gravity theory) already says that the effective rest
> >energy of an object within a potential Phi (whether caused by itself
> >or another object) is reduced by a potential energy given by the
> >integral of rho Phi dV (without any factor of a 1/2). This leads
> >directly to the factor of 2 paradox, and hence requires some
> >compensating positive energy to balance the books.
>
> Can you give me a reference to this "standard Newtonian
> approximation"? It seems, frankly, like a silly approximation for
> exactly this factor-of-two reason. To whom is it "standard"?
That's just a matter of applying the square root of the 00 component
of the metric, that is the factor 1/sqrt(1-2Gm/rc^2), to the rest
energy of each part of the body as observed in the local frame in
order to evaluate the energy of the same part as seen in the
coordinate system. This is for example exactly how red-shift and the
Pound-Rebka experiment are explained for objects near the surface of
the Earth.
MTW box 23.1 claims to derive the total mass-energy by a different
method which gives the expected potential energy. However, I find
this derivation quite puzzling. In the simplified case where the
internal energy is assumed to be negligible, it appears that this
integral claims to be integrating the average baryon mass in the rest
frame times the baryon number density n in the rest frame times the
proper volume. This is surely equal to the average baryon mass times
the total number of baryons in the object, which should be exactly
equal to the original rest mass without any potential factor,
shouldn't it? Have I missed something here too?
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>On 1 Mar, 21:30, e...@lfa221051.richmond.edu wrote:
>> In article <1172678974.702079.300...@s48g2000cws.googlegroups.com>,
>> Can you give me a reference to this "standard Newtonian
>> approximation"? It seems, frankly, like a silly approximation for
>> exactly this factor-of-two reason. To whom is it "standard"?
>
>That's just a matter of applying the square root of the 00 component
>of the metric, that is the factor 1/sqrt(1-2Gm/rc^2), to the rest
>energy of each part of the body as observed in the local frame in
>order to evaluate the energy of the same part as seen in the
>coordinate system. This is for example exactly how red-shift and the
>Pound-Rebka experiment are explained for objects near the surface of
>the Earth.
Right. And it's a perfectly sensible thing to do in those contexts --
specifically, it's a perfectly sensible thing to do when considering
test masses in an external geometry. I claim that it's not a sensible
thing to do when considering situations in which the gravity of the
"test mass" itself (which is then no longer a test mass) is important.
Proof: Doing so leads to a factor-of-two error in the total energy,
which requires strange contortions to fix. ["Doctor, it hurts when I
do this." "Then don't do that."]
So I still have the same question: Is there anyone to whom it is
"standard" to define energy in the way you do, in the case where we're
not just considering test masses (i.e., masses whose effect on the
spacetime geometry can be neglected) in a given external potential?
If not, why on earth would you want to define energy in this way,
given that it then forces you to "fix" a factor of two error that
need never have arisen in the first place?
>MTW box 23.1 claims to derive the total mass-energy by a different
>method which gives the expected potential energy. However, I find
>this derivation quite puzzling. In the simplified case where the
>internal energy is assumed to be negligible, it appears that this
>integral claims to be integrating the average baryon mass in the rest
>frame times the baryon number density n in the rest frame times the
>proper volume. This is surely equal to the average baryon mass times
>the total number of baryons in the object, which should be exactly
>equal to the original rest mass without any potential factor,
>shouldn't it? Have I missed something here too?
I don't have MTW in front of me right now; I'll look at it later.
Jonathan Scott
hours of the previous article (March 4th) but although Google said it
had been posted (to the moderators), it has never turned up but neither
did I receive any rejection. This happened once before in the same
thread. I'm therefore posting a shorter reply, and saving a copy in
case it gets lost too.]
I'm still wondering about the basic principle here. Why should it not
be correct to assume that the effective rest mass as seen from a
distance of a massive object in a potential due to another object is to
the first order reduced by exactly the usual Newtonian potential for the
other object, just as for a test object as in the Pound-Rebka
experiment? The g_00 potential is normally assumed to add linearly at
this sort of level of weak field approximation, and I can't see any
reason why it should not for the effect of two sources on on another.
If this isn't correct, what is the Newtonian approximation to the effect
of two massive bodies on one another's clock rate?
I've replaced my paper on this subject on my web site with a modified
one that takes into account various discussions. It can be found at the
following URL:
http://pws.prserv.net/jonathan_scott/physics/doublepe.pdf
Apart from that, I've now understood that the mass expression in MTW box
23.1 is one which is derived from the requirement for consistency with
total energy conservation, and is physically a bit odd, in that it is
the integral of the proper mass density with respect to coordinate
space, which doesn't make much sense but gives the right factor to make
the potential energy come out right.
Thomas Smid
> In article <1171627815.714023.163...@q2g2000cwa.googlegroups.com>,
>
> Thomas Smid <thomas.s...@gmail.com> wrote:
> >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.
artificial concept in order to make the energy a conserved quantity
(and it is probably worth noting here that even the notion of 'energy'
in the first place is a redundant one in classical mechanics; one can
solve any mechanical problem without even mentioning the word 'energy'
simply by integrating the equations of motion for the given
interaction forces; of course, for certain questions (and for checking
the correctness of the integration) the 'energy' point of view can be
very convenient, so I am not saying that it doesn't make sense).
So basically, as long as the equations add up, everything should be
all right. But I admit that my factor 2 would be inconsistent with
general conventions, so I should have divided everything by two in the
first place.
However, I don't agree with your statement that 'potential energy' has
to be understood as the potential energy of a system only, but not as
that of a particle:
consider for instance a gravitationally bound system of N masses (with
the same mass m) with a radius R. The work needed to move one of the
masses from the edge of the volume to infinity is G*(N-1)*m^2/R, and
this is exactly the potential energy of one of the masses (at the edge
of the volume). The point here is that if you remove this mass, you
are reducing the mass of the remaining system, so the potential energy
of each of the remaining masses is then only G*(N-2)*m^2/R (assume
that you do it quickly enough so that the radius doesn't change in the
meanwhile). Now if you continue this procedure until only one mass is
left, you have spent an overall work of 0.5*G*N*(N-1)*m^2/R (because
the sum of n from 1 to N-1 is N*(N-1)/2). This is were the factor 1/2
comes from. It is simply a consequence of the fact that in a self-
interacting system (i.e. if the force is not an external force), the
test mass (i.e. the mass acted upon) is also part of the acting mass
that produces the field, which means that the overall potential energy
of the system is only half that of the sum of the N individial
potential energies.
So it all depends on your question in the first place. If you are
asking for the potential energy of the overall system, then this is
different to the question what the potential energy of an individual
particle is.
Thomas
Jonathan Scott
>...
> So it all depends on your question in the first place. If you are
> asking for the potential energy of the overall system, then this is
> different to the question what the potential energy of an individual
> particle is.
However you define it, I have no doubt about the amount of
gravitational potential energy in a system, and neither should
anyone else who understands the basic Newtonian theory.
That side of the "paradox" can be assumed to be a given.
It is the other side which creates the puzzle, in that the total
rest energy as normally defined (by multiplying the proper
rest energy by the time dilation) is decreased relative to the
original proper rest energy by exactly twice the potential
energy (in all cases, regardless of configuration). This is
unaffected by whether the original potential energy has
been converted to some other form of internal energy
within the system, such as kinetic energy, or has been
extracted entirely from the system.
The linear approximation to GR which is used to calculate
the time dilation is valid even for arbitrary numbers of
sources, provided that it only involves weak fields and
non-relativistic velocities. At this level of accuracy, it
also does not seem that there should be any problem
with expecting conservation of energy relative to a flat
semi-Newtonian coordinate system, as this is of course
exactly what is done to calculate kinetic energy and
potential energy for orbits and trajectories within the
solar system.
Obviously, it would make a lot of sense for the overall
rest energy to be decreased by the potential energy.
That would provide an obvious explanation of the
location of potential energy. However, in this case the
overall rest energy is decreased by exactly twice
that amount, so for conservation of energy, that extra
lost energy must have gone somewhere else within
the system.
There are various forms this missing energy could take.
In the GR "Komar mass" expression for a single central
fluid mass, this missing energy density is mathematically
equal to three times the pressure density diagonal terms
in the stress-energy tensory. However, this does not make
sense physically and does not work for dynamic situations
involving multiple objects.
My suggestion is that the missing energy could be given
by half of the square of the field, as this clearly works
mathematically in all cases, and is physically similar to
the Maxwell electrostatic energy density (apart from
a change of sign). If this is the case, then gravitational
energy would have a location like other forms of energy
(at least relative to a background coordinate system)
and I think this field idea would allow it to flow
continuously under changes of configuration.
For more details, especially some mathematical details
(which don't work well in text) see my previously
referenced paper on my web site. Here's another link
to it:
http://pws.prserv.net/jonathan_scott/physics/doublepe.pdf
I should stress that I'm not trying to say that the potential
energy is any different from the standard expression.
I'm saying that the rest energy of a set of gravitationally
interacting masses is decreased by an amount of energy
which is exactly twice the potential energy, which means
that for energy to be conserved, the missing energy must
still be within the system in some other form, for which the
obvious candidate is field energy. That is:
Original rest energy - potential energy
=
Original rest energy * time dilation factor + missing energy
where mathematically the missing energy is exactly equal
to the potential energy (and is therefore mathematically
also equal to the field energy as in electrostatic theory)
and any other internal energy such as kinetic, thermal
or mechanical energy would appear on both sides of the
equation.
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>It is the other side which creates the puzzle, in that the total
>rest energy as normally defined (by multiplying the proper
>rest energy by the time dilation) is decreased relative to the
>original proper rest energy by exactly twice the potential
>energy (in all cases, regardless of configuration).
I'm still not convinced that this is something "normally defined".
Who besides you defines "total rest energy" in this way?
It's a sensible definition when you're talking about test masses in an
external potential. (Here by "test mass" I mean a body whose
contribution to the gravitational field / spacetime curvature can be
neglected). But in the situation you're considering, of a bunch of
bodies under mutual gravitational interactions, it's not a useful
definition.
Jonathan Scott
> In article <1176055920.698992.226...@d57g2000hsg.googlegroups.com>,
> Jonathan Scott <jonathan_sc...@vnet.ibm.com> wrote:
>
> >It is the other side which creates the puzzle, in that the total
> >rest energy as normally defined (by multiplying the proper
> >rest energy by the time dilation) is decreased relative to the
> >original proper rest energy by exactly twice the potential
> >energy (in all cases, regardless of configuration).
>
> I'm still not convinced that this is something "normally defined".
> Who besides you defines "total rest energy" in this way?
>
> It's a sensible definition when you're talking about test masses in an
> external potential. (Here by "test mass" I mean a body whose
> contribution to the gravitational field / spacetime curvature can be
> neglected). But in the situation you're considering, of a bunch of
> bodies under mutual gravitational interactions, it's not a useful
> definition.
OK, this is the key point. I agree that it is conventional only to
assume that the effective rest energy of an object in a gravitational
potential is decreased by the time dilation factor when discussing
test masses within a potential due to a central mass. However,
the maths of linearized GR says that provided we are dealing
with weak fields and non-relativistic speeds, the time dilation at any
location is given to first order by the linear sum of the potentials
due to each source, so the effective rest energy of any source
or part of a source should still be still given by the product of the
proper rest energy and the time dilation factor unless there are
any significant corrections caused by an increase in some form
of internal gravitational energy.
It is well known that in GR (for example as part of the Komar mass
expression) the time-dilated rest energy on its own doesn't tie up
with the potential energy, and there are various mathematical
modifications which fix up the result. For example, in MTW box
23.1, the correction is achieved by integrating the proper mass
density in coordinate space with respect to proper volume, which
is physically weird but introduces exactly the right correction.
Similarly, in the Komar Mass expression, an extra term is added
which is effectively three times the pressure, which gives the
right answer in the central case, but does not work for dynamical
cases involving multiple objects.
In cases where there are two separate objects involved which
are well separated, any internal energy terms due to pressure
or tidal effects caused by the other object would be negligible
compared with the first-order time-dilation effect, so neither of
the above corrections seems to be viable.
In contrast, my controversial suggestion that there could be a
positive energy density in the field, analogous to the Maxwell
electrostatic energy density, gives exactly the right correction
mathematically within the Newtonian model, regardless of the
number and configuration of the sources. So far, I do not
necessarily think that this has to contradict GR, in that this
energy may only be visible relative to a semi-Newtonian
background frame, as a non-tensor. However, it certainly
seems to contradict the way in which GR is conventionally
described, in that Einstein's vacuum equations say that there
is no energy in empty space (even though they allow the
propagation of gravitational radiation).
Thomas Smid
> My suggestion is that the missing energy could be given
> by half of the square of the field, as this clearly works
> mathematically in all cases, and is physically similar to
> the Maxwell electrostatic energy density (apart from
> a change of sign). If this is the case, then gravitational
> energy would have a location like other forms of energy
> (at least relative to a background coordinate system)
> and I think this field idea would allow it to flow
> continuously under changes of configuration.
>
> For more details, especially some mathematical details
> (which don't work well in text) see my previously
> referenced paper on my web site. Here's another link
> to it:
>
> http://pws.prserv.net/jonathan_scott/physics/doublepe.pdf
I had a look at your paper and it seems that your problem is caused by
two things: a) you are mixing up the notions of 'potential' and
'potential energy', and b) you are confusing the individual mass with
the total mass:
the gravitational *potential* of a mass m1 at distance R is
(1) V=-G*m1/R ,
and the *potential energy* of mass m2 at this potential is
(2) U=-G*m1*m2/R .
The latter is also the overall potential energy of the system as it is
the work required to move either of the masses out of the field of the
other. So if one has two equal masses m1=m2=m, the overall potential
energy of this two-particle system is
(3) U(2)=-G*m^2/R .
Note that m is the individual mass here, not the total mass.
However, for an arbitrary number N of particles with mass m the
overall
potential energy is (see my previous post)
(4) U(N)= -G*N*(N-1)*m^2/2/R ,
and for large N, this becomes therefore
(5) U(N)~ -G*N^2*m^2/2/R ,
and in terms of the total mass M=N*m
(6) U(N)~ -G*M^2/2/R .
The latter is equivalent to the result of the integration in your
paper,
but it holds strictly only if you let N go to infinity, because for
instance for a two-particle system you have M=2m, and from (6) you
would
get U(2)=-G*2*m^2/R, which is too large by a factor two (see Eq.(3)).
The exact Eq.(4) does indeed give the correct result here.
In any case, one should bear in mind that the change in potential
energy is *defined* as the negative of the change of the kinetic
energy. The sum of both has to be zero at any time, so any additional
'field energy' would violate energy conservation.
Thomas
eb...@lfa221051.richmond.edu
Jonathan Scott <jonatha...@vnet.ibm.com> wrote:
>> I'm still not convinced that this is something "normally defined".
>> Who besides you defines "total rest energy" in this way?
>>
>OK, this is the key point. I agree that it is conventional only to
>assume that the effective rest energy of an object in a gravitational
>potential is decreased by the time dilation factor when discussing
>test masses within a potential due to a central mass. However,
>the maths of linearized GR says that provided we are dealing
>with weak fields and non-relativistic speeds, the time dilation at any
>location is given to first order by the linear sum of the potentials
>due to each source, so the effective rest energy of any source
>or part of a source should still be still given by the product of the
>proper rest energy and the time dilation factor unless there are
>any significant corrections caused by an increase in some form
>of internal gravitational energy.
Let me focus in on the phrase "effective rest energy" in the last
sentence. As far as I can tell, that phrase means
mc^2 + (1/2) mv^2 + Potential energy due to all of the other masses.
Is that what you intend by this phrase?
If so, then I agree with this sentence, but I claim that you have no
right to add up the "effective rest energies" for a bunch of
gravitationally interacting particles and expect the result to behave
like a "total energy". The reason is that adding up these "effective
rest energies" involves adding up the potential energies of each mass
due to each other mass, which is double-counting. That's why defining
an "effective energy" in this way might make sense for test masses
but does not make sense for systems of non-test-masses.
In short, it still looks to me like the problem you're trying to solve
is entirely self-inflicted. Don't try to define energy in a way that
intrinsically involves double-counting, and you won't have a
double-counting problem to solve.
Jonathan Scott
> I had a look at your paper and it seems that your problem is caused by
> two things: a) you are mixing up the notions of 'potential' and
> 'potential energy', and b) you are confusing the individual mass with
> the total mass:
All of your points are still on the side of calculating or clarifying
the potential energy, and appear to me to be correct.
As I said before, I have no problem with the potential energy.
It's the other side (the time-dilation reduced rest energy) which
is the problem, in that for any system of masses, this quantity
is less than the original proper rest energy by exactly twice the
potential energy. This is obviously equivalent to omitting the
factor of two when evaluating the potential energy of a set of
masses in each other's potentials, but that's not what I'm
doing; I am instead applying the time-dilation factor to the
original (proper) rest energy of the objects to obtain the
effective rest energy. I can't see any reason why this should
not be valid, and since starting this thread I've checked that
in linearized GR there is no problem with adding up the
separate time-dilations due to the separate sources.
Wolfgang G. Gasser
:: = Ted (ebunn)
::: = Jonathan Scott
::: The standard Newtonian approximation for GR (or any similar
::: relativistic gravity theory) already says that the effective rest
::: energy of an object within a potential Phi (whether caused by itself
::: or another object) is reduced by a potential energy given by the
::: integral of rho Phi dV (without any factor of a 1/2). This leads
::: directly to the factor of 2 paradox, and hence requires some
::: compensating positive energy to balance the books.
::
:: Can you give me a reference to this "standard Newtonian
:: approximation"? It seems, frankly, like a silly approximation for
:: exactly this factor-of-two reason. To whom is it "standard"?
:
: That's just a matter of applying the square root of the 00 component
: of the metric, that is the factor 1/sqrt(1-2Gm/rc^2), to the rest
: energy of each part of the body as observed in the local frame in
: order to evaluate the energy of the same part as seen in the
: coordinate system. This is for example exactly how red-shift and the
: Pound-Rebka experiment are explained for objects near the surface of
: the Earth.
Let us deal with the gravitational red shift on earth surface
(assuming the earth far away from other masses). The weak field
approximation is then the gamma-factor of the escape velocity v_e:
v_e = sqrt(2GM/r) = 11,2 km/s^2
gamma(v_e) = 1/sqrt(1-v_e^2/c^2) = 1/sqrt(1-2Gm/rc^2)
Now let us assume that the earth consists of alternating thin equal
layers of matter and antimatter isolated by some kind of mechanism,
and that neighbouring layers are fully transformed into radiation
if isolation in between is removed.
If the layers start radiating away from the surface, the radiation
generated by the uppermost layers will certainly be redshifted
corresponding to gamma(v_e) wrt infinity. The last, innermost
layers however will radiate away with almost no redshift.
If we calculate the potential energy of the earth by integrating
the loss of energy because of redshift of all layers, we get
E_pot = 0.5 integral of rho Phi dV
(where E_pot is potential energy, rho the mass density and Phi the
gravitational potential).
So I don't see why "red-shift and the Pound-Rebka experiment"
should somehow support the formula
E_pot = integral of rho Phi dV
An intesting point however is: if all layers are transformed into
radiation at the same time, then all photons are redshifted
according to half their gravitational potential, i.e. according
to 0.5 Phi instead of Phi.
I think, the concept "total potential energy = integral of mass
density and potential" suffers from the same defect as the concept
"market capitalisation = number of shares times current price" in
economics. If all the shares have to be sold, only the first get
the "current price" whereas the last could turn out to be almost
worthless.
In the starting posting of this thread
http://groups.google.com/group/sci.physics.research/msg/41767593fca80fbd
Jonathan Scott 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.
I confused myself in a similar way around twenty years ago without
really having resolved the issue for myself until now.
I remember vaguely to have once read that Einstein had lost some
time by doubting of his equations of general relativity (or of
some reasonings leading to them) because of a factor 2 problem.
Am I mistaken? Could this have been the issue of this thread? Does
somebody know more about it?
Cheers, Wolfgang
Jonathan Scott
>...
> If we calculate the potential energy of the earth by integrating
> the loss of energy because of redshift of all layers, we get
>
> E_pot = 0.5 integral of rho Phi dV
>
> (where E_pot is potential energy, rho the mass density and Phi the
> gravitational potential).
Agreed.
> So I don't see why "red-shift and the Pound-Rebka experiment"
> should somehow support the formula
>
> E_pot = integral of rho Phi dV
That's not what I'm saying. I'm not saying that's the potential
energy.
The right hand side is the decrease in the effective rest energy
of the masses in the system as calculated by applying the time
dilation to the amount of rest energy present as seen in the local
frame, and the paradox is of course that it's NOT equal to the
potential energy, but is rather exactly twice the potential energy,
indicating that some other energy is required to balance the
books.
Relative to a conventional background coordinate system,
the scalar potential which determines the time dilation is
simply the linear sum of the potentials due to each source.
This means that energy at that location is red shifted by
the total time dilation, even if part of that time dilation is
due to the same object itself.
John C. Polasek
<noein...@bellsouth.net> wrote:
>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.
>>
>> 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 ΒΌ 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
>
Let F be the force between M1 and M2. Always the energy is
2E = M1V1^2 + M2V2^2 = E1 + E2
>From impulse and momentum
Fdt = M1dV1 = M2dV2 therefore
V1/V2 = M2/M1 always
Then E1/E2 = M1V1^2/M2V2^2 x (M2^2/M1^2) = M2/M1
Thus the potential is distributed, inversely as the mass
E1/E2 = M2/M1
They share. Each gets a part of the potential.
John Polasek
__________
Wolfgang G. Gasser
>> = Wolfgang G. Gasser in news:evgsvo$jrs$1...@atlas.ip-plus.net
There is an 'early-stage general-relativistic' derivation of
E = mc^2 by means of a diffential equation starting with
mass-energy-equivalence (relativistic mass)
E = k * m
and using the energy definition
dE = m * dPhi
A test body 'o' with a negligible mass is accelerated by the
gravitational field of e.g. the earth between two given potentials
Phi and Phi_down with the INFINITESIMAL difference dPhi.
Phi ----- o | v
v
dPhi
Phi_down ----- | v_down = v + dv
v
If the test body is at rest at Phi, then aften having fallen
to Phi_down, its speed is
v = sqrt(2 * dPhi)
Until now I probably would have assumed:
- Potential energy of the test body has been changed into the
same quantity of kinetic energy of the test body.
But now I assume that the following version is correct:
- Potential energy of the test body plus the same amount of
potential energy of the field-generating mass have been
changed into kinetic energy of the test body alone.
The E=mc^2-derivation proceeds this way: We ask what happens if
the test body already has a speed v (in direction to the field
generating mass) at Phi before falling to Phi_down. In this
case the energy transformed from potential energy to kinetic
energy is by definition
dE(v) = m(v) * dPhi
where m(v) is the relativistic mass at v. The decisive question
is therefore: what is the speed v_down of the test body when
reaching Phi_down. According to classical mechanics (because of
v * dv = dPhi) this speed would simply be :
v_down_classic = v + dPhi/v
But according to classical mechanics, photons moving at c would
also accelerate to
c_down_classic = c + dPhi/c
If we take into account that c is the constant benchmark then
we get:
v_down = v_down_classic * (c / c_down_classic)
= v + dv where dv = dPhi/v * (1 - v^2/c^2)
Thus, if the test mass at Phi already moves at speed v, then at
Phi_down its speed is v_down = v + dv.
In order to link the changes in energy dE(v) = m(v) * dPhi to
the speed changes dv (in function of v), we must replace dPhi:
dE(v) = m(v) * dPhi = m(v) * v/(1 - v^2/c^2) * dv
This amount of potential energy dE(v) is used to accelerate the
test body between Phi and Phi_down from v to v+dv.
In order to establish a differential equation at potential Phi
we must bring the test body moving at Phi_down with speed v_down
back to Phi leaving its new speed v_down unchanged. For this,
the same amount of energy dE(v) which has been transformed from
potential energy to kinetic energy during the infinitesimal
fall must be added.
This means that by adding the energy dE(v) to the test body
moving with v at Phi, its speed can be increased from v to
v + dv at the same potential Phi.
So we get the differential equation valid at potential Phi
E'(v) = dE(v)/dv = m(v) * v/(1 - v^2/c^2)
which together with
E(0) = k * m(0)
results in
k = c^2
E(v) = c^2 * m(0) * sqrt(1 - v^2/c^2) = m(v) * c^2
In this derivation of E=mc^2 it does not matter whether we
attribute the potential energy which is converted to kinetic,
fully to the test mass or to both the test mass and the field
generating mass.
>> So I don't see why "red-shift and the Pound-Rebka experiment"
>> should somehow support the formula
>>
>> E_pot = integral of rho Phi dV
>
> That's not what I'm saying. I'm not saying that's the potential
> energy.
> The right hand side is the decrease in the effective rest energy
> of the masses in the system as calculated by applying the time
> dilation to the amount of rest energy present as seen in the local
> frame, and the paradox is of course that it's NOT equal to the
> potential energy, but is rather exactly twice the potential energy,
> indicating that some other energy is required to balance the
> books.
As far as I understand, by "applying the time dilation" to "the
effective rest energy" you mean this: time dilation of an atom
must depend in the same way on Phi as the "effective energy" loss
of the atom. In fact however the relative change in "effective
energy" is only half the relative change in time caused by
time dilation.
If such a proportional change of time and "effective rest energy"
actually were a legitimate claim of GR, thus leading to an until
now unnoticed energy conservation violation (where "the change in
the total energy" is "equal to twice the change in potential
energy"), then I would rather assume that GR is wrong in this
respect than introduce a new form of "positive energy density".
> Relative to a conventional background coordinate system,
> the scalar potential which determines the time dilation is
> simply the linear sum of the potentials due to each source.
Yes, and the "effective rest energy" loss is simply the mass
times HALF the "linear sum of the potentials", or not?
Cheers, Wolfgang
Jonathan Scott
> ...
> There is an 'early-stage general-relativistic' derivation of
> E = mc^2 by means of a diffential equation starting with
> mass-energy-equivalence (relativistic mass)
This all looks a bit hand-waving to me, and I'm not necessarily
convinced, but I don't think it's relevant to the paradox.
> Until now I probably would have assumed:
>
> - Potential energy of the test body has been changed into the
> same quantity of kinetic energy of the test body.
>
> But now I assume that the following version is correct:
>
> - Potential energy of the test body plus the same amount of
> potential energy of the field-generating mass have been
> changed into kinetic energy of the test body alone.
That's not what I would assume. In any sort of Newtonian
approximation, the total potential energy depends on the
configuration of the system, but it can be apportioned to
individual components according to how they move, for
example using the simple concept that energy is force
times distance.
Considerations about two objects affecting each other
by the same amount only apply to the other side of the
paradox, relating to the time dilation effect on energy.
> In this derivation of E=mc^2 it does not matter whether we
> attribute the potential energy which is converted to kinetic,
> fully to the test mass or to both the test mass and the field
> generating mass.
Agreed.
> As far as I understand, by "applying the time dilation" to "the
> effective rest energy" you mean this: time dilation of an atom
> must depend in the same way on Phi as the "effective energy" loss
> of the atom. In fact however the relative change in "effective
> energy" is only half the relative change in time caused by
> time dilation.
You are using "effective energy" for two different things here.
The time dilation dependency on Phi is determined by GR,
not by Newtonian considerations or conservation laws.
This leads to one way of calculating the effective energy
of a set of masses.
The potential energy is calculated in other ways which
involve conserving the total energy, which we assume
to be always valid, and this gives a different result.
This means that when we calculate the effective
energy using the time dilation, we must be missing
some energy somewhere.
The time dilation is a property of the GR metric. For any
theory which is even approximately similar to GR, the time
dilation can be calculated from the linear sum of the
Newtonian potentials due to each source. It says that
clocks, frequencies and of course differences in frequencies
as observed in the observer background coordinate system
are multiplied by a certain scalar factor relative to their values
seen locally. For purposes of the Pound-Rebka experiment,
the time dilation factor is assumed to act as a scale factor on
any energy values, which can then be compared between any
two different potentials. In this experiment, the
gravitational source is assumed to be static, so the potential
energy difference between two different levels is the same
as the energy difference due to the time dilation factor.
The effect of the potential of the "test mass" on the "source
mass" is normally ignored, but everything seems to indicate
that this would result in the effective energy of the source
mass being reduced by exactly the same amount, hence
the paradox.
> If such a proportional change of time and "effective rest energy"
> actually were a legitimate claim of GR, thus leading to an until
> now unnoticed energy conservation violation (where "the change in
> the total energy" is "equal to twice the change in potential
> energy"), then I would rather assume that GR is wrong in this
> respect than introduce a new form of "positive energy density".
I'm trying to find a semi-Newtonian picture which is accurate
enough to match GR in weak field, non-relativistic situations
but strictly conserves energy, preferably giving it a definite
location. Within this picture, it seems to me that the positive
energy density in the field is the only solution I can find to the
paradox.
I'm not in a position to say what is right or wrong about GR
itself. In GR, "energy" is usually used to mean the quantity
whose density is given by a stress-energy tensor in a local
inertial frame. A true local inertial frame is however a
free-fall frame in which gravity cannot even be observed,
so it is not surprising that GR cannot describe gravitational
energy in this way. There may be a non-tensor way of
mapping my picture into GR proper, and it might or might
not be compatible with GR.
Wolfgang G. Gasser
>> = Wolfgang G. in news:evocd9$5rh$1...@atlas.ip-plus.net
>> But now I assume that the following version is correct:
>>
>> - Potential energy of the test body plus the same amount of
>> potential energy of the field-generating mass have been
>> changed into kinetic energy of the test body alone.
>
> That's not what I would assume. In any sort of Newtonian
> approximation, the total potential energy depends on the
> configuration of the system, but it can be apportioned to
> individual components according to how they move, for
> example using the simple concept that energy is force
> times distance.
I don't think it makes sense to apportion the potential
energy "to individual components according to how they move".
Why should we in the case of two interacting bodies with the
same mass attribute different potential energies to the
bodies? They may have different (frame-dependent) kinetic
energies. Is there a sound (quantifiable) alternative to
the claim that both have lost the same amount of potential
energy? I don't think so.
Now let us assume for simplicity that all matter consists
of particles of the same mass (and ignore the effect of
the particles on themselves). The classical gravitational
potential of a given particle is then the superposition of
the potentials caused by all other particles. But in each
pairwise interaction of the given particle with an other
particle, both particles have equal rights and it makes sense
to attribute the lost potential energy equally to both. And
if we do this, then for each particle the potential energy
lost because of the other particles is its own mass times
half its own potential loss (caused by the other particles).
So we get dE = 0.5 integral of rho Phi dV independently
of the configuration of the system.
> The time dilation dependency on Phi is determined by GR,
> not by Newtonian considerations or conservation laws.
But GR also depends on Newtonian considerations and on
conservation laws. It is very easy to calculate the weak
field approximation of the frequency shift of the Pound-Rebka
experiment over 22.5 m by simply applying Einstein's concept
"relativistic mass" to the photon concept:
E = h * f ==> m = h * f / c^2
The change in energy of the photon over a hight of 22.5 m is
dE = m * dPhi = m * 9.8 m/s^2 * 22.5 m
The red-shifted frequency is therefore:
f_red = (E - dE) / h = (h*f - h*f/c^2*dPhi) / h
= f * (1 - dPhi/c^2) = f * (1 - 2.5*10^-15)
But also in this early-stage-GR approximation, only half of
dE comes from the photon whereas the other half comes from
the earth (whose potential changes when the photon comes
closer).
> This leads to one way of calculating the effective energy
> of a set of masses.
And it is exactly this way of calculating the effective
energy "using the time dilation dependency on Phi" which
is questionable. The other way, involving conservation of
total energy, hasn't put into question by any poster (as
far as I can see).
> For purposes of the Pound-Rebka experiment, the time
> dilation factor is assumed to act as a scale factor on
> any energy values, which can then be compared between any
> two different potentials.
I suppose you mean by this that in the Pound-Rebka experiment,
the total energy of a body at rest near the roof is reduced
by dE = m*c^2 - m*c^2 * (1-dPhi/c^2) = m*dPhi with respect
to an identical body in the basement.
This could only be derived in a correct way if we neglected
the change in POTENTIAL ENERGY impacted by test bodies or
photons on source masses. Because the change in POTENTIAL of
a source is inversely proportional to the small mass of the
test body, this change can be neglected.
POTENTIAL ENERGY of the source however is proportional to
its own mass, so the smallness of the change in potential of
the source due the test mass is compensated by the hugeness
of the mass of the source. It follows that the change in
potential energy of the source results in exactly the same
amount as the change in potential energy of the test body.
Cheers, Wolfgang
Jonathan Scott
>...
> I don't think it makes sense to apportion the potential
> energy "to individual components according to how they move".
> Why should we in the case of two interacting bodies with the
> same mass attribute different potential energies to the
> bodies? They may have different (frame-dependent) kinetic
> energies. Is there a sound (quantifiable) alternative to
> the claim that both have lost the same amount of potential
> energy? I don't think so.
In the Newtonian view, potential energy can usually be
assigned unambiguously in a frame-dependent way in
the same way as kinetic energy (although in general it
may be subject to an arbitrary additive constant which
may not behave in a way which is consistent with the
physical distribution of energy).
For example, the simple Newtonian two-body case can be
treated as if the two bodies m_1 and m_2 are independently
orbiting at distances r_1 and r_2 around a fixed body at the
center of mass, so m_1 r_1 = m_2 r_2. In this case, both the
kinetic and (negative) potential energy of each mass can be
considered to be proportional to the distance r, or
equivalently proportional to the other mass.
This situation only becomes symmetrical in the time-dilation
model, where the product of either potential with the other
mass gives the same energy, equal to the potential energy,
not half of it.