Is Energy Conserved in General Relativity?

[juju]

Hi all,

I know very little of the math specific concepts of GR, but is it possible that the missing energies are bound up in the space/time curvature itself.

juju

 

[Garth]

Hi all,

I know very little of the math specific concepts of GR, but is it possible that the missing energies are bound up in the space/time curvature itself.

juju

Yes that is the normal way it is explained, the energy is absorbed by the gravitational field, but it then ceases to appear in the "accounts", it has disappeared, and will re-appear again when the process is reversed. The question is whether this is an appropriate way to deal with energy.
Garth

 

[juju]

The question is whether this is an appropriate way to deal with energy.
Garth

I would think this is an OK way to deal with energy.

It would be just like a battery or capacitor. You store the energy in a potential field and it is not manifest until drawn upon.

Thanx Garth.

juju

 

[Garth]

It would be just like a battery or capacitor. You store the energy in a potential field and it is not manifest until drawn upon.
juju

That is a good example of what I mean. As you charge a capacitor the charge on the plates increases and the electric potential between the plates. The energy does show in the electric 'potential' "accounts".
However with the gravitational field in GR this is not so clear.
The Schwarzschild solution describes a spherically symmetric and static gravitational field around a central mass at any radius from that centre which depends on that central mass alone. So long as there is spherical symmetry it does not matter how the mass is distributed. However you can redistribute the mass, keeping spherical symmetry, using up/releasing energy in the process and the exterior field will not change at all. There will not even be gravitational waves generated by such a process. The Sun could turn into a black hole and the Earth's orbit would not be affected.
So, using a 'skyhook', lift a shell around a spherical mass and prop it up high off the ground. According to GR the total energy of the system, rest mass and gravitational energy is equal to P^{0}, which is equal to M before and after the process. The energy you used has simply "disappeared"! It has been absorbed into the field somehow. The standard explanation is to say that energy cannot be localised, it is a mistake to try to locate it. However this is just an example of the fact

GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which is different (generally).

Garth

 

[juju]

So long as there is spherical symmetry it does not matter how the mass is distributed. However you can redistribute the mass, keeping spherical symmetry, using up/releasing energy in the process and the exterior field will not change at all.

But don't the internal gravitational potentials change as a result of this redistribution. This would change the potenial energy structure internal to the mass,

I think I have read of experiments that have been done that say they have proven that this internal structure affects the effective gravitational mass of the object. I believe I read this in Scientific American but I am not sure.

juju

 

[Garth]

You find in the Schwarzschild solution that internally the mass M that appears in the gravitational potential is redefined and normalised so that its value is as determined by Kepler at 'infinity', i.e. determined to agree with Kepler's period of an orbit in a region approaching flatness.
Garth

 

[juju]

Hi Garth,

But then doesn't this changing and renormalizing of the mass M, actually prevent apriori any possible energy accounting. And isn't this a bad way of going about this.

juju

 

[Garth]

The key issue is that energy is not conserved in GR - energy-momentum is, which is different. There is no other way of defining the M you put in the gravitational potentials, is there?
Garth

 

[juju]

Hi Garth,

I think I understand what you are saying. Since the mass and the symmetry don't change, the external fields do not change. I get that. However, the internal gravitational binding energy of the particles making up the mass does change.

juju

 

[pervect]

However you can redistribute the mass, keeping spherical symmetry, using up/releasing energy in the process and the exterior field will not change at all. There will not even be gravitational waves generated by such a process. The Sun could turn into a black hole and the Earth's orbit would not be affected.
So, using a 'skyhook', lift a shell around a spherical mass and prop it up high off the ground. According to GR the total energy of the system, rest mass and gravitational energy is equal to P^{0}, which is equal to M before and after the process. The energy you used has simply "disappeared"! It has been absorbed into the field somehow. The standard explanation is to say that energy cannot be localised, it is a mistake to try to locate it. However this is just an example of the fact
Garth

The specific example here is flawed, I think. Neither P^{0} or P_{0} represents the total energy of a system as I have been trying to explain for some time. Neither one of them is even coordinate independent! It's true that P₀ is an invariant of motion for a particle following a geodesic in a static space-time. But this doesn't mean that summing up the P₀ makes a good measure of system energy.

Here's one of the definitions from Wald for the total mass in a stationary, asymptotically flat space with a vacuum at infinity, which I've located from earlier in the thread and cut & pasted. It's expressed as a volume intergal of the stress-energy tensor.

<br /> M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV<br />

Here T_ab is the stress-energy tensor, g_ab is the metric tensor, n^a is the unit future normal to the volume element, and \xi^b is the Killing vector representing the time translation symmetry of the static system, sutiably normalized so that |\xi^a \xi_a| equals unity at infinity.

Note that

T_{ab} n^a dV

is going to give the energy-momentum 4-vector P, because the stress-energy tensor is the density of 4-momentum per a vector-valued unit volume, and n^a dV is just that vector-valued unit of volume.

In a static space time, \xi^b is going to be a unit vector. This means that

T_{ab} n^a \xi^b dv

is going to pick out P_{0} in a coordinate system that's flat at asymptotic infinity.

However, note that
<br /> M = \int_{\Sigma} T_{ab} n^a \xi^b dV<br />

is not the system energy! Because of the second term in the intergal, the total system energy does change when you "lift up" mass from a region with a different metric curvature g₀₀.

To put this all in words, you have to integrate the following quantity:

twice the energy momentum P₀ minus the trace of T_{ab} multiplied by g₀₀.

to get the system energy.

You can, I think, find a similar formula in MTW, if you have MTW and not Wald I can _probably_ find it for you (but I'm not going to bother to look through that book's terrible index system unless you both have the book and are interested enough that finding it would be worthwhile).

 

[pervect]

Please define your term covariant energy.

That's the covariant energy component of the energy-momentum 4-vector, i.e. P₀

Since when??
That is not quite correct. One must demand that their choice of "scaling" leads one to obtain the Newtonian values in the Newtonian limit.

The Newtonian limit exists when you have an asymptotically flat space-time- then you can look at the "far field". But you're trying to get energy without asymptotic flatness, so it's not clear that the Newtonian limit exists.

But what you say is the same in non-relativistic mechanice,. Or do you know a reason why I can't call W = 13*mv² + 26*U the "energy" of a particle in non-relativistic mechanics? Its still a constant isn't it?
One does not need asymptotic flatness since one can always go to the weak field limit by choosing coordinates such that h_ab << 1 at any event in spacetime.

Unfortunately, the value of energy you get is going to depend on which point in space-time you choose to make this happen at, i.e. at what point in space_time is g₀₀ going to be nearly equal to 1.

To put it another way that may (or may not) be simpler, energy is conjugate to time, but the clocks are all running at different rates. Usually we pick the clock at infinity - without asymptotic flatness, we don't have this ability.

 

[gptejms]

The problem in GR is that GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which, when space-time curvature is introduced, is different. GR is an example of one of Noether's improper energy theorems.

The desire to conserve energy in GR is where confusion may arise.

Garth

Thanks all of you for your inputs.I searched on the internet and found an article "Energy not conserved in GR"(by Baez,I guess) which explains that the stress-energy tensor doesen't satisfy Gauss's law if there is spacetime curvature.Is this not in contradiction with your statement above "GR conserves energy-momentum"?

Another question:- One of you(I guess pmb_phy) has said that energy is conserved in static fields. Why so?----a static field would also have accompanying spacetime curvature and Gauss's law wouldn't be satisfied.

 

[pmb_phy]

What's potential energy in a non-inertial frame?Potential energy in a gravitational field is fine,but we are talking of non-inertial frames here.

Recall the equivalence principle: A uniformly accelerating frame of reference is equivalent to a uniform gravitational field.

There are other versions but this one gives you a clearer idea of what I was referikng to. In such a field (1) There here is somethikng which you can clearly call a "potential" and the total energy is a constant. In the weak field limit you can express the total energy as the sum of kinetic, potential and rest energy. I've never calculated the exact value so I'm not sure what it is in the strong field case. I'll get back to that on Friday when I'm near a computer again.

In the weak field limit of a uniformly accelerating frame of reference the gravitational potential is V(z) = mgz where g = gravitational acceleration as measured at z = 0.

But in all of relativity "energy" is a frame dependent concept and as such the value changes when you change frames.

Inspite of the equivalence principle,I don't see the concept of potential energy carrying over.

Why?

The problem in GR is that GR does not conserve energy. It doesn't intend to. It conserves energy-momentum, which, when space-time curvature is introduced, is different.

Wrong. GR does not say this. This is simply Garth's opinion and it can be shown to be false be mere calculation. Conservation of energy-momentum means T^ab_;b = 0. This equation contains both the conservation of energy and the conservation of momentum. In fact in order to prove that T^ab_;b = 0 one starts with the conservation of energy and the conservation of momentum and one derives[/sub] this expression. It simply puts two conservation theorems in one tiny package. But each, i.e. energy and momentum, are conserved independantly of each other.

That's the covariant energy component of the energy-momentum 4-vector, i.e. P₀

Yipes! That is an odd way to phrase that. Energy is energy and there is little reason to qualify it with "covariant". Energy is the time component of the energy-momentum 1-form.

The Newtonian limit exists when you have an asymptotically flat space-time- then you can look at the "far field".

Sufficient but not neccesary. Consider an infinitely long straight string of uniform mass density. The potential does not go to zero at infinity, its logarithmic. One defines the zero reference at a finite distanc, r₀, from the string. If you stay close to ₀ then you can choose r close to ₀ such that h_ab is as small as you like and therefore the weak field limit will apply - or even the Newtonian limit.

But you're trying to get energy without asymptotic flatness, so it's not clear that the Newtonian limit exists.

Energy is always there in a static g-field. There is no requirement for the g-field to go to zero for one to be able to use the weak field limit (where, similar to the Newtonian limit, one can define a potential energy term). I can always change my reference point to any location that I like so that if I stay near it the potential is small. The weak field limit only demands that h_ab is small. It doesn't demand asymptotic flatness. The string is a great example of this.

Unfortunately, the value of energy you get is going to depend on which point in space-time you choose to make this happen at, i.e. at what point in space_time is g₀₀ going to be nearly equal to 1.

Why is that unfortunate? Energy has always been a relative quantity and a quantity whose magnitude has never been of great importance. Only differences in magnitude are of any importance. And energy is always frame dependent in relativity.

To put it another way that may (or may not) be simpler, energy is conjugate to time, but the clocks are all running at different rates. Usually we pick the clock at infinity - without asymptotic flatness, we don't have this ability.

I'm sorry but I don't know what your point is here.

Thanks all of you for your inputs.I searched on the internet and found an article "Energy not conserved in GR"(by Baez,I guess) which explains that the stress-energy tensor doesen't satisfy Gauss's law if there is spacetime curvature.Is this not in contradiction with your statement above "GR conserves energy-momentum"?

You're speaking about something else. That is the total energy of the closed system consisting of the source of the g-field. We're talking anout the energy of a particle moving in a g-field. These are different topics.

Another question:- One of you(I guess pmb_phy) has said that energy is conserved in static fields. Why so?----a static field would also have accompanying spacetime curvature and Gauss's law wouldn't be satisfied.

First off - The presence of a gravitational field does not imply the existace of spacetime curvature. In the second place you're referring to the energy of the source of gravity and not the energy of a particle moving in a static gravitational field. In all cases where there is a static gravitational field P₀ is a constant. This constant is called the "energy" of the particle.

As to why this is a constant see the derivation at
http://www.geocities.com/physics_world/gr/conserved_quantities.htm

Pete

 

[Garth]

The nature of the signature of the metric in SR/GR means that the energy time-component (squared) and the momentum space-component (squared) are subtracted from each other to obtain the norm (squared) of the energy-momentum.

In the particular case when energy is conserved and momentum is conserved then energy-momentum is conserved.

However in the general case both energy and momentum individually may not be conserved and yet their difference of squares, the norm squared of energy-momentum, may still be conserved.

Therefore the conservation of energy-momentum does not require the conservation of both energy and momentum. In fact in most cases of freely falling frames of reference through space-time with curvature there are no Killing vectors and individually they are not conserved.

The desire to conserve energy is a natural one but it does not sit lightly with the principles of GR.

Garth

 

[pmb_phy]

The nature of the signature of the metric in SR/GR means that the energy time-component (squared) and the momentum space-component (squared) are subtracted from each other to obtain the norm (squared) of the energy-momentum.

That is only true in SR. It is quite false in GR.

In the particular case when energy is conserved and momentum is conserved then energy-momentum is conserved.

Yep.

Your comments are confusing since you switch topics in a heartbeat with no mention that you're doing so. On the one hand there is the energy-momentum of the source of gravity. That is described by the energy-momentum tensor. Then there is the energy-momentum of a particle which is moving in the G-field. That is something quite different and is described by the 4-momentum.

However in the general case both energy and momentum individually may not be conserved and yet their difference of squares, the norm squared of energy-momentum, may still be conserved.

So? That only means that the proper energy of a particle is constant, e.g. not a function of proper time. That is something quite different than referring to conservation of energy. The term conservation of energy refers only to the total energy of a particle and not part part of the total energy of which proper energy is. You're referring to proper energy which is part of total energy. And even then you're speaking of SR only (inertial frame) and then seeming to claim that it somehow applies to GR. The magnitude of 4-momentum is proportional to proper energy. The square of proper energy = E^2 - (pc)^2 only when you're in an inertial frame of reference, i.e. when you're using Lorentz coordinates.

Therefore the conservation of energy-momentum does not require the conservation of both energy and momentum.

That is totally wrong. The very phrase "conservation of energy-momentum" quite literally means "conservation of inertial energy and conservation of 3-momentum".

In fact in most cases of freely falling frames of reference through space-time with curvature there are no Killing vectors and individually they are not conserved.

So what? What does that have to do with the meaning and definition of the terms we're discussing?

Pete

 

[gptejms]

You're speaking about something else. That is the total energy of the closed system consisting of the source of the g-field. We're talking anout the energy of a particle moving in a g-field. These are different topics.

Ok.

First off - The presence of a gravitational field does not imply the existace of spacetime curvature.

Why?

In the second place you're referring to the energy of the source of gravity and not the energy of a particle moving in a static gravitational field. In all cases where there is a static gravitational field P₀ is a constant. This constant is called the "energy" of the particle.

Pete

Ok.

 

[pervect]

Thanks all of you for your inputs.I searched on the internet and found an article "Energy not conserved in GR"(by Baez,I guess) which explains that the stress-energy tensor doesen't satisfy Gauss's law if there is spacetime curvature.Is this not in contradiction with your statement above "GR conserves energy-momentum"?

Another question:- One of you(I guess pmb_phy) has said that energy is conserved in static fields. Why so?----a static field would also have accompanying spacetime curvature and Gauss's law wouldn't be satisfied.

I would assume that you found the sci.physics.faq about energy in GR

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

a good source of information.

What you'll need to appreciate the argument here is in the FAQ, look for the line:

In certain special cases, energy conservation works out with fewer caveats. The two main examples are static spacetimes and asymptotically flat spacetimes.

Pete's been talking abut static space-times. There is indeed a conserved quantity that can be defined for a static space-time, even one that is highly curved, one that is not asymptotically flat. The only slightly odd thing you run into is that it's hard to set the right "scale factor" if you have a static space-time without asymptotic flatness. (AFAIK it's impossible to find any generally preferred scale factor for this conserved quantity without asymptotic flatness).

The simplest cases, BTW, like the Schwarzschild black hole, are both static AND asymptotically flat. That's the simple case I've been presenting the formulas for.

Of the two concepts, asymptotic flatness is, in my opinion, by far the most useful, and you'll find that that's how Wald, for instance, approaches the problem of energy in GR. MTW takes a different route to the same path - they discuss energy in terms of pseudo-tensors (also mentioned in the FAQ) - an approach that winds up working if and only if you have asymptotically flat space-times.

The two concepts do not conflict - when they both apply, they yield the same answer.

So Garth is right when he says that energy is not always conserved in GR - you do need some additional conditions, you can't define energy for an arbitrary space-time. But the specific example he gave to illustrate this was flawed - rather than going through the details, saying that we know that energy is conserved in asymptotically flat space-times should be enough to illustrate why the example is flawed.

Pete is also right when he says that energy is conserved if you have a static space-time (though I'm unclear if he's ever acknowledged that energy can also be conserved if you have asymptotic flatness in a non-static space-time.) Regardless of whether Pete acknowledges it or not, energy is conserved in this case.

 

[Garth]

pervect - Agreed. But about your post #100 in this thread my understanding of the total system energy was taken from Weinberg, eq. 8.2.16 pg. 182.

Pete - From your post #103

Originally Posted by pervect
That's the covariant energy component of the energy-momentum 4-vector, i.e. P_{0}



Yipes! That is an odd way to phrase that. Energy is energy and there is little reason to qualify it with "covariant". Energy is the time component of the energy-momentum 1-form.

Elsewhere you define energy as P_{0} as it is this time element of four-momentum that is conserved if the metric time component is static. Yet a 1-form is the contravariant form of a vector.

Garth

 

[gptejms]

Pete is also right when he says that energy is conserved if you have a static space-time (though I'm unclear if he's ever acknowledged that energy can also be conserved if you have asymptotic flatness in a non-static space-time.) Regardless of whether Pete acknowledges it or not, energy is conserved in this case.

Thanks pervect for your answer.So,energy is conserved in the case of a star collapsing to a black hole(non-static spacetime with asymptotic flatness).(right?).
What about RW spacetimes?(I've picked up these terms reading the discussions here!)

 

[pmb_phy]

Why?

Why would it? Miknd you - I'm going by Einstein's definition of "gravitational field" and nobody elses. Spacetime curvature, aka tidal force (i.e. Non-vanishing Riemman tensor) only means that there is a gravitational field present which can't be transformed away in a finite region of spacetime. The presence of a gravitational field, aka gravitational acceleration, is dictated by the non-vanishing of the Christoffel symbols (when the spatial coordinates are expressed in Cartesian coordinates). This does not require spacetime curvature. Thus a gravitational field can be "produced" by a change in spacetime coordinatges. But this is not true when the spacetime is curved, i.e. I can't produce spacetime curvature by changing coordinates.

..though I'm unclear if he's ever acknowledged that energy can also be conserved if you have asymptotic flatness in a non-static space-time.

In the first place I think that you're referring to the conservation of the mass of source of the gravitational field and not the conservation of the energy of a particle moving in the gravitational field. I was speaking of the later. In the second place I have not acknowledged what you speak of since I have not proved it to myself nor have I seen a proof ... and if its in Wald then I doubt I'd be able to follow it since that is an extremely hard text to follow. Someday.

Pete - From your post #103 ... Elsewhere you define energy as P_{0} as it is this time element of four-momentum that is conserved if the metric time component is static.

I understand that and yes, that how its defined. By the way - why do you say that its how "I" defined it? Don't assume that it was I who presented this definition to the world of GR.

I simply don't think its a good idea to call it "covariant" energy or something like that - its simply "energy" and calling it something else like "covariant energy" gives the impression that its something different than "energy".

Yet a 1-form is the contravariant form of a vector.

That is incorrect. A 1-form is a very different animal than a vector so its not a good idea to think of it as being a component of a different "form" of a vector. What you said here is like saying that a ket is a different form of a bra and that is certainly a bad way to look at bras and kets.

Pete

 

[pervect]

Thanks pervect for your answer.So,energy is conserved in the case of a star collapsing to a black hole(non-static spacetime with asymptotic flatness).(right?).
What about RW spacetimes?(I've picked up these terms reading the discussions here!)

Basically, yes. This works in the real world because our spacetime appears to be reasonably flat (on a global scale, it's obviously curved near massive bodies). So if you have a black hole that's reasonably isolated, energy is conserved as it collapses, when you add together the energy of the BH itself, any ejecta (looking at the Crab nebula as an example, there will be a fair amount of this in an actual collapse), and the energy of the radiation it emits (both electromagnetic and gravitational). All of this is measured from the viewpoint of a far-away observer, one where space-time is reasonably flat. A non-rotating black-hole wouldn't emit gravitational radiation as it collapses, but it rotating black hole is expected to emit gravitational radiation as it collapses.

BY RW space-time, you mean "Robertson-Walker"?
[edit]
I'm deleting my previous response here, I actually have to *think* about this one more before I answer!
[end edit]

 

[pmb_phy]

I'm *think* that the spatially flat Robertson-Wlaker universe is also asymptotically flat, but I'm not 100% positive.

The spacetime curvature for the RW spacetime has the same value everywhere. Since the spacetime is not flat anywhere it can't be asymptotically flat.

Pete

ps - I say "has the same value everywhere" but I'm not sure what that means per se. If I change frames from one inertial frame to another I will change the value of the spacetime curvature. But due to the homogeneity of the spacetime I think the curvature will still be the same everywhere. Yes? No? Maybe? To much to drink last night?

 

[pervect]

I wasn't quite fast enough on the "edit" button, I see...

I am currently tending to think that energy is not conserved in Robertson-Walker space-times. The reason I think it's not conserved is cosmological redshift. If you emit some light, wait a bit, and measure it's energy in isotropic coordinates, you find that the light has lost energy.

If you don't use isotropic coordinates, you find that masses are accelerating away from each other which is also no good for energy conservation.

However, I can't quite pin down the reason why a spatially flat Robertson-Walker space-time wouldn't be asymptotically flat in the sense needed for energy conservation. It's very clear that there is a conformal isometry to flat space-time, because the metric is just -dt^2 + a^2(t)(dx^2 + dy^2 + dz^2).

However, there are 5 additional conditions that the conformal isometry must satisfy for a metric to be considered asymptotically flat. I suspect one of them is violated, but I can't pin down which one.

 

[juju]

Hi,

It may just be that energy/momentum conservation is the basic conservation rule everywhere, rather than two separate rules one for energy and one for momentum.

Since in most cases energy and momentum are both conserved separately in most cases, this might be an appropriate idea.

In this way the lack of energy conservation on its own is no big deal.

This seems to suggest to me that the concept of forces arises from one continuous energy/momentum field, whose mode of energy/momentum transfer is a function of the basic structure of space/time in the neighborhood where such forces arise.

juju

juju

 

[Haelfix]

"However, I can't quite pin down the reason why a spatially flat Robertson-Walker space-time wouldn't be asymptotically flat in the sense needed for energy conservation."

Bingo, you came around to my original posts back on page 3 of this thread. Look up what k asymptotic freedom means, things will become clearer.

But yes in general there is no ADM solution to FRW solutions. Let me say that again..

Global energy (defined canonically from the hamiltonian for the pedants) does not make sense in *general* in GR! Only in very specific formulations of asymptotic flatness where we can far field it, or in stationary metrics.

Worse.. gravitational energy due to matter is not relative in the strict sense we are used too classically. Why? B/c matter particles fix their definition of energy at infinity relative to the ground state vacuum (otherwise you get infinity for say the harmonic oscillator for a scalar particle). But in curved spacetime, there is no good way to even define a vacuum at some fixed point in space. Quantum mechanics bites us in the ass again.

 

[gptejms]

All of this is measured from the viewpoint of a far-away observer, one where space-time is reasonably flat.

So I was right after all when I said that energy was conserved from the viewpoint of an observer in an inertial frame(corresponding to the asymptotically flat region)looking at the non-inertial frames from outside--the reason being that there are no pseudo forces in the inertial frame.

I also said that energy is not conserved for observers in the non-inertial frames(regions where spacetime is not flat) because there are pseudo forces in such frames which do work.However I'm not sure of this latter statement since pmb_phy and pervect didn't seem to agree.

 

[gptejms]

The presence of a gravitational field, aka gravitational acceleration, is dictated by the non-vanishing of the Christoffel symbols (when the spatial coordinates are expressed in Cartesian coordinates). This does not require spacetime curvature.

Pete

You mean non-vanishing Christoffel symbols associated with a gravitational field do not necessarily imply spacetime curvature-----this is news to me.Do others agree?

 

[pervect]

You mean non-vanishing Christoffel symbols associated with a gravitational field do not necessarily imply spacetime curvature-----this is news to me.Do others agree?

If you take "space-time curvature" to mean a non-vanishing Riemann tensor, the statement is definitely true - an example is an accelerating rocket. The metric for an accelerating rocket is the Rindler metric

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

where g is a constant, the acceleration of the rocket. (This is in geometric units where c=1).

if you calculate the Riemann tensor and the Christoffel symbols for this metric, you'll find the Riemann is zero and the Christoffel symbols are not. (FYI some texts vary on exactly what they call the Rindler metric, but it doesn't matter to the point I"m making which uses the above metric as an example regardless of what name you choose to give it).

Cheat and use your favorite tensor program to calculate this sort of calculation, it's a pain to do by hand.

However, sometimes the term "space-time curvature" is loosely used to mean that the metric coefficients vary as a function of the coordinates. The metric above is curved in that lose sense, but not in the strict sense of having a non-zero Riemann curvature tensor.

 

[pervect]

Bingo, you came around to my original posts back on page 3 of this thread. Look up what k asymptotic freedom means, things will become clearer.

What text would I look this up in? I don't think it's in Wald :-(, his treatment of ADM mass is fairly superficial (his treatment of the Bondi mass is better).

I hope k asymptotic freedom is simpler than conformal infinity - the basic idea of making infinity a "place" is pretty clear, but the mathematical fine print is rather intricate.

 

[Garth]

So I was right after all when I said that energy was conserved from the viewpoint of an observer in an inertial frame(corresponding to the asymptotically flat region)looking at the non-inertial frames from outside--the reason being that there are no pseudo forces in the inertial frame.

I also said that energy is not conserved for observers in the non-inertial frames(regions where spacetime is not flat) because there are pseudo forces in such frames which do work.However I'm not sure of this latter statement since pmb_phy and pervect didn't seem to agree.

An inertial frame is a freely falling coordinate system. In such a frame of reference particles do not suffer accelerations unless there are specific non-gravitational forces acting on them. Such a frame can only be defined for a sufficiently small region around its origin, otherwise tidal forces will be experienced.

However, measured in such a frame, the energy of other particles outside this sufficiently small region will not be conserved in general. There is a confusion even in the definition of the energy of a freely falling particle in GR; as to whether it is P_{0} or P^{0}, because although P^{0} is the natural definition it is not conserved, even though no work is being done on the particle. Although P_{0} is conserved in the frame of reference of the centre of mass of a static system, in general it is not so and this identification with energy is restricted and rather artificial.

The overall insight is that GR does not in general conserve energy, it is an improper energy theorem, it conserves energy-momentum instead.
The principle of the conservation of energy-momentum is not a concatenation of the principle of the conservation of energy and principle of the conservation of momentum; energy-momentum is a geometric concept in its own right, invariant under Lorentz transformations.

Energy and momentum are frame dependent concepts; therefore it is necessary to define a frame of reference, a preferred frame. in order to restore the principle of the conservation of energy.

Garth