Is Energy Conserved in General Relativity?

[DW]

Steve Carlip is 100% right in this argument that the Newtonian potential concept you are arguing over breaks down and is not general relativistic at all. Also relativistic mass as I have already proven is not the same thing as inertial mass. In fact it has no place in modern relativity.

 

[pervect]

I also recall seeing it in MTW and in Wald but I can't locate it at the moment

If you can find it in MTW or Wald, I'll conceed the point. It's certainly not in the index of either one of them (of course MTW's index **** ###). I really don't see the point of giving the metric coefficients Yet Another Name, though. It's easy enough for me to remember that when you say "relativistic mass" I think "energy", and when you say "gravitational force" I think "Christofel symbol", but I'm starting to build up quite a dictionary here :-(.

You understand that by weak I do not mean Newtonian, right? Why do you think that the weak field approximation is not part of general relativity.

The issue we are talking about, energy, is trivial in PPN, but much less so in the general theory. Unforutnately PPN theory doesn't illuminate much of the general problem.

Its my turn to ask you something - A particle in a gravitational field has energy P₀ where P is the 4-momentum of the particle. Do you think that a particle in a gravitational field has rest energy?

My view is that when we talk about the energy-momentum 4 vector of a particle, we are talking about the energy relative to some specific observer and some specific coordinate system. We need to define the coordinate system and the observer to measure the energy in GR, just as we need to define the "frame" we are talking about when we measure the energy of a body in Newtonian physics.

I also believe that given an energy-momentum 4-vector, we have an invariant mass^2 that's equal to g_{uv} E^u E^v -- of course we do to know the metric coefficients as well as the energy-momentum 4-vector to compute the mass

I'll leave to you to do the translation into your own terminology, I hope my meaning is clear.

The other point I want to make is that issues do arise when attempting to parallel transport the energy-momentum 4-vectors in order to do some sort of intergal to get "total energy". So having a definition of the energy-momentum 4 vector at a point isn't all that's needed to be able to define the energy of a system. For many applications, if the region is small enough, the parallel transport issue can be ignored, but this isn't always true.

 

[pmb_phy]

If you can find it in MTW or Wald, I'll conceed the point.

Okay, but that seems to be a bit of a narrow view. The GR literature uses the same terminology. I'm just letting you know what's out there. But I found where I saw it in MTW. Its on page 434 where they are quoting Hilbert. I've e-mailed someone to find it in Wald. I'll let you know the result.

To see where Einstein used it in the most notable place, its in his Nobel lecture. See - http://www.nobel.se/physics/laureates/1921/einstein-lecture.pdf

It's certainly not in the index of either one of them (of course MTW's index **** ###).

Indexes don't always list all terms used in a text

I really don't see the point of giving the metric coefficients Yet Another Name, though.

John Stachel wrote a paper which touched on this point in How Einstein Discovered General Relativity: A Historical Tale with some Contemporary Morals. Stachel is a famous GRist and GR historian. He points out

The main difficulty at this stage was to grasp the dual nature of the metric tensor: it is both the mathematical object which represents the space-time structure (chronogeometry) and the set of 'potentials,' from which represents the gravitational field (Christoffel symbols) and the tidal 'forces' (Riemann tensor) may be derived.

I'm doing something bad right now, i.e. I'm sitting and typing against docs orders so I can't get into all the details of that paper and the context. If you want more then I'll get to it at a later date.

But there is the question of "Why would I want to know that?" wherein I'd respond "To understand what you're reading when you encounter it." For example. See - http://arcturus.mit.edu/8.962/notes/gr6.pdf

This was written by Edmund Bertschinger, MIT cosmologist (GR prof). What do you think he means when he says

Although the gravitational potentials represent physical metric perturbations,...

What are these "gravitational potentials" that he refers to?

It's easy enough for me to remember that when you say "relativistic mass" I think "energy", and when you say "gravitational force" I think "Christofel symbol", but I'm starting to build up quite a dictionary here :-(.

That's a good thing.

If you have MTW then see what they say about the stress-energy pseudo-tensor of the gravitational field. It's in MTW page 465 Eq. (20.18).

My view is that when we talk about the energy-momentum 4 vector of a particle, we are talking about the energy relative to some specific observer and some specific coordinate system.

That is inaccurate. The 4-vector itself has a geometric signifigance which has nothing to do with coordinates. When you wish to choose a coordinate system then you've chosen a basis and a particular value of the energy. Once you do that, my question remains - Is the rest energy of the particle part of that energy?

The other point I want to make is that issues do arise when attempting to parallel transport the energy-momentum 4-vectors in order to do some sort of intergal to get "total energy".

I have no idea what you're talking about. What is it that you're adding? Total energy of what? I was referring to the energy of a single particle in a G-field.

So having a definition of the energy-momentum 4 vector at a point isn't all that's needed to be able to define the energy of a system.

Who was speaking of the energy of a system? I was speaking of the energy of a particle in a g-field. Those are two different topics.

Pete

 

[pervect]

When you wish to choose a coordinate system then you've chosen a basis and a particular value of the energy. Once you do that, my question remains - Is the rest energy of the particle part of that energy?

As far as "rest energy" goes, do you mean something other than the invariant associated with the 4-vector? We both agree that the energy-momentum 4-vector exists, and that it has an associated invariant. It seems to me that that should be sufficient.

Who was speaking of the energy of a system? I was speaking of the energy of a particle in a g-field. Those are two different topics.

Most of my remarks in this thread have been addressed to the issue of energy conservation in GR. This requires that one consider the notion of the nergy of a system as well as the energy of a particle.

 

[pmb_phy]

As far as "rest energy" goes, do you mean something other than the invariant associated with the 4-vector?

I don't mean something other. I'm referring to rest energy, i.e. the product of a particle's proper mass with c².

We both agree that the energy-momentum 4-vector exists, and that it has an associated invariant. It seems to me that that should be sufficient.

For what? The topic here is energy in GR, correct? There are different subtopics on that topic. One subtopic is P₀ = energy of a particle in a g-field. Another is the mass-energy that creates the g-field, i.e, T⁰⁰, and then there is the self energy of the g-field which is represented be a pseudo-tensor (which I'm not all that familiar with).

Most of my remarks in this thread have been addressed to the issue of energy conservation in GR. This requires that one consider the notion of the nergy of a system as well as the energy of a particle.

So do you think that energy is not conserved?

By the way, I just looked at the index to Wald. It does not contain the term "invariant." However I don't think that Wald considered it to be useless in GR.

Pete

 

[pmb_phy]

Pete, I have an idea for you

Go to this thread here at PhysicsForums
https://www.physicsforums.com/showthread.php?t=26625

look at post #15

here is the direct link to it:
https://www.physicsforums.com/showthread.php?p=227804&posted=1#post227804

PF has "eaten" your old Alma Mater newsgroup sci.physics.research

The look and feel is nicer in the PF version

Now you can enjoy corresponding with Steve Carlip in the comfort of PF accomodations! Have fun!

Okay. I've changed my mind. Since its a moderated forum I'll take a look. But Carlip and I have discussed something similar before. It turned into a discussion of semantics. But he obviously knows GR better than most of us so I recommend that people pay close attention to what he says.

Pete

 

[selfAdjoint]

Good show, Pete. We'll be paying attention for sure.

 

[pmb_phy]

Good show, Pete. We'll be paying attention for sure.

Well, I didn't want to appear closed minded. This is a topic that I'm highly interested in. One problem in addressing it is that energy itself has never been a well-defined quantity so when one says that its not well-defined in GR I'm not really surprised. But I do want to learn more about these problems and the meaning of the stress-energy-momemtum pseudo-tensor of the gravitational field.

Pete

 

[kurious]

If energy is not conserved in GR but it is conserved in QM this would make it difficult to see how QM and GR can be reconciled in one theory.Energy is conserved in every other branch of physics so why should GR be the exception? Can't the energy of microwave background photons for example be absorbed by something else in the universe? This has got to be more likely than a lack of energy conservation.
And if one looks at the entropy of the universe as a whole,are we to believe that
heat energy is lost from the universe? This would imply that the second law of thermodynamics can be violated or that the energy passes into another universe.
There is no proof for either of these.

 

[pervect]

For what? The topic here is energy in GR, correct? There are different subtopics on that topic. One subtopic is P₀ = energy of a particle in a g-field. Another is the mass-energy that creates the g-field, i.e, T⁰⁰, and then there is the self energy of the g-field which is represented be a pseudo-tensor (which I'm not all that familiar with).
So do you think that energy is not conserved?

OK, we've discussed most of the other subtopics, but we haven't discussed how the stress-energy tensor arises from the energy-momentum 4 vector in this thread.

I'm not sure if we need to talk about the stress energy tensor or not. I would describe it informally as the density per unit volume of the energy-momentum 4 vector. I could make this description more formal if there is a point to doing so, but I won't bother unless there is some interest and/or flames about my informal description. Well, I will mention that one can measure volume with both a three-form and with the dual (Hodges dual) of a three form, which is a formal way of saying that any volume element defines a unique perpendicular time-like vector. Hopefully that terse description may be useful in communicating my meaning.

The fact that the self energy of the field is represented by a pseudo-tensor and not a real tensor means that the notion of self-energy of a gravitational field isn't a geometric object. Different observers define it differently. So, when we require a pseudo tensor to measure system energy, "red flags" should be raised, because we are combining geometric objects, and something that's not geometric. Unfortuately, we can't simply ignore or drop the self-energy term and retain a conserved energy.

There's an amusing quote from MTW on this topic, I'll try and find it, after I resolve some technical difficulties with my computer - look for a short separate post, hopefully coming soon, unless I can't find it - did I mention that the index in MTW **** ###?.

In any event, defining energy with the pseudo-tensor approach is, at least in my opinion, a little bit like adding together all the physical energies, plus the well-known "Finnagle Factor" from engineering, where the "Finnagle Factor" takes on whatever value is required to make the answer come out right. It's not surprising that one can come up with a conserved quantity by adding the Finnagle number F to *any* general quantity, conserved or not, but it's not particularly physically significant to be able to do so.

Fortunately, there are a number of more physicaly motivated ways of defnining the total energy of a system in a manner which retains the desired conservation property. One way of doing this is to use the approach of Cartan moments, described in MTW. Another way of doing this, also mentioned in MTW, is to look at the behavior of an object orbiting the system "a long ways away" from the system, and to use agreement with Newtonian theory to define the energy of the system.

A more modern approach ito defining energy in GR s to focus on the behavior of the gravitational field at infinity, the behavior of the field at null infinity gives the Bondi energy, and the behavior of the field at spacelike infinity gives the ADM energy. Look at my previous posts in the thread for a reference to the pages in Wald where these are discussed. Wald discusses the relationship between the Bondi and ADM energy as well, according to Wald they eventually turn out to be closely related, though it takes a considerable amount of work to establish the relationship. I believe that the Bondi energy is also closely related to the first two definitions of energy I mentioned in the previous paragraph, but I'm not absolutely positive on this partciular point (I'm still working on undertanding Wald's discussion more fully, it's pretty technical).

The unifying thread to all of the methods for dealing with the energy problem that I"ve mentioned is the notion of the observer at infinity. Apparently this isn't quite the last word, though, based on the comments of another poster. It's a bit scary to think that there are probably more definitions of energy in GR than I've posted even in this very long post - but probably accurate.

 

[Garth]

In general energy is not conserved in GR - energy-momentum is. This is a result of the EEP and the carrying forward of the SR "non-preferred frame of reference" interpretation of physical observations into GR.
It is energy-momentum that is invariant over transformation between frames. Energy is frame independent, not just in the SR sense of the variation of kinetic energy between non-co-moving frames, but also because of curvature effects.

Is energy described by P^0 or P_0? Is it naturally a contravariant or covariant quantity?
Weinberg uses P^0 e.g. G&C, pg 182 eq. 8.2.16 as do MTW, Gravitation, e.g. pg 158 and yet it is P_0 that is the conserved quantity when the gravitational potentials are not time dependent. The covariant form P_0 is often defined as energy because of that. Yet this usage is confusing. It only works because in the asymptotic limit of flat space-time P^0 = P_0, and it is conserved with the condition specified.
In fact energy is neither covariant nor contravariant, its value is the scalar time component of the norm of the energy-momentum vector, E^2 = P^0.P_0, (see MTW pg 463 eq. 20.10) and this is not conserved except under special conditions i.e. flat space-time.
Might it be that only when the non-conservation of energy in GR, as against its conservation in quantum theory, is fully recognised that the problems at their interface in quantum gravity may be resolved?

Energy may be defined in GR but it is not general conserved; if it is demanded that energy should be conserved then it can not be localised, because the energy density has to be integrated over the entire field out to the flat space-time asymptotic limit.

 

[pmb_phy]

Is energy described by P^0 or P_0, is it naturally a covariant or contravariant quantity?

The energy of a particle which is moving in a gravitational field is proportional to P₀ while the inertial mass is proportional to P⁰.

Weinberg uses P^0 e.g. G&C pg 182 eq. 8.2.16 and MTW Gravitation e.g. pg 158 and yet it is P_0 that is the conserved quantity when the gravitational potentials are not time dependent.

Weinberg uses P⁰ there to represent the mass/energy of the body which generates the gravitational field. That M is the proportional to the time component of the 4-momentum of the gravitating body. So you can think of that as a particle. He shouldn't have called it that but such is life. I don't see what you're referring to on that page in MTW since there is no mention of what you said there.

But, as I recall, MTW do explain this elsewhere. They explain that it is P₀ which is the conserved quantity when the particle is in a time independant g-field (i.e. g_uv do not exlicitly depend on time).

The contravariant form P_0 is often defined as energy because of that. Yet this usage is confusing. It only works because in the asymptotic limit of flat space-time P^0 = P_0, and it is conserved with the condition specified.

That is incorrect. It works under all conditions where one would expect the energy of a particle to be conserved, i.e. in a time independant g-field.
For proof see -
http://www.geocities.com/physics_world/gr/conserved_quantities.htm

In fact energy is neither covariant nor contravariant, its value is the scalar time component of the norm of the enegy-momentum vector, E^2 = P^0.P_0, (see MTW pg 463 eq. 20.10) and this is not conserved except under special conditions i.e. flat space-time.

That is incorrect. What you've just described is not energy. Energy is not a scalar since a scalar is a tensor of rank zero and therefore is independant of the system of coordinates. You've just described proper energy, E₀. Energy is P₀.

Also, what you've quoted is not from MTW. That equation in MTW you just quoted is really

M = (-P^{\mu}P_{\mu})^{1/2}

where they use M to mean the proper mass of the object. You've made the mistake of confusing proper mass with energy. The correct relationship between the two is given by E₀ = Mc².

You should never expect the energy of a particle to be constant when the field is time dependant. That doesn't even hold in classical mechanics or electrodynamics.

Pete

 

[DW]

The energy of a particle which is moving in a gravitational field is proportional to P₀ while the inertial mass is proportional to P⁰.

Weinberg uses P⁰ there to represent the mass/energy of the body which generates the gravitational field. That M is the proportional to the time component of the 4-momentum of the gravitating body. So you can think of that as a particle. He shouldn't have called it that but such is life. I don't see what you're referring to on that page in MTW since there is no mention of what you said there.

But, as I recall, MTW do explain this elsewhere. They explain that it is P₀ which is the conserved quantity when the particle is in a time independant g-field (i.e. g_uv do not exlicitly depend on time).

That is incorrect. It works under all conditions where one would expect the energy of a particle to be conserved, i.e. in a time independant g-field.
For proof see -
http://www.geocities.com/physics_world/gr/conserved_quantities.htm
That is incorrect. What you've just described is not energy. Energy is not a scalar since a scalar is a tensor of rank zero and therefore is independant of the system of coordinates. You've just described proper energy, E₀. Energy is P₀.

Also, what you've quoted is not from MTW. That equation in MTW you just quoted is really

M = (-P^{\mu}P_{\mu})^{1/2}

where they use M to mean the proper mass of the object. You've made the mistake of confusing proper mass with energy. The correct relationship between the two is given by E₀ = Mc².

You should never expect the energy of a particle to be constant when the field is time dependant. That doesn't even hold in classical mechanics or electrodynamics.

Pete

Actually no, energy is most commonly defined as the time element of the contravariant momentum four-vector, but neither the time element of the contravariant or covariant expression of the vector is what is conserved in general. In general it is energy and momentum parameters that are conserved instead and only in special cases is the energy parameter equal to the energy or the time element of the covariant momentum four-vector. It is also wrong to call the mass m proper mass because it is the length of p^{\mu } according to any frame, not just the proper frame. It is also not equal to the relativistic mass M which shouldn't even be used anymore.

 

[Garth]

Pete:

The energy of a particle which is moving in a gravitational field is proportional to P₀ while the inertial mass is proportional to P⁰.

E₀ = Mc².

So P₀ = P⁰c² then? I think not; actually when the metric is symmetric
P₀ =g₀₀P⁰c² and therefore E₀ = Mc² is only true in Minkowski space-time in the absence of curvature.

I don't see what you're referring to on that page in MTW since there is no mention of what you said there.

Perhaps page 462 equations 20.6, 20.7, & 20.9 would have been a better example of MTW using P⁰ as energy.

That is incorrect. It works under all conditions where one would expect the energy of a particle to be conserved, i.e. in a time independant g-field.
For proof see -
http://www.geocities.com/physics_world/gr/conserved_quantities.htm

Your notes are useful but (more simply) they have shown that a free falling particle's covariant four velocity U_a is invariant if the metric components are not dependent on x^a, where
U_a = g_abU^b. Multiplying by mass gives you P_awhich is also invariant and which therefore you define as energy. However this is a very special case and, as I have said, GR is an improper energy theorem. To introduce the concept of energy conservation is very attractive but it is not consistent with the general theory, it is 'bolting on' a classical (i.e. pre-GR) principle.

I was using "scalar" in its general and not specific meaning, i.e. as a mathematical 'object' having only magnitude and not as a "scalar invariant". I agree it is important to be specific about language and I should have been more accurate in terminology.

 

[pmb_phy]

Pete:
So P₀ = P⁰c² then?

Nothing I've ever posted ever hinted at that being true. You were using the term M to mean proper mass and you then claimed that the magnitude of 4-momentum was energy. That is not the case. I said that correct relationship between energy and M is given by E₀ = Mc².

I think not; actually when the metric is symmetric
P₀ =g₀₀P⁰c² and therefore E₀ = Mc² is only true in Minkowski space-time in the absence of curvature.

That is incorrect. In the first place the metric is always symmetric. P₀ is related to a particle's 4-momentum through the relation

P_0 = g_{0\mu}P^{\mu}

Perhaps page 462 equations 20.6, 20.7, & 20.9 would have been a better example of MTW using P⁰ as energy.

Yeah, I don't know why they do that. Seems strange and doesn't make sense.

Your notes are useful but (more simply) they have shown that a free falling particle's covariant four velocity U_a is invariant if the metric components are not dependent on x^a, where
U_a = g_abU^b. Multiplying by mass gives you P_awhich is also invariant and which therefore you define as energy. However this is a very special case and, as I have said, GR is an improper energy theorem.

I know that's what you think. You're repeating yourself since we've discussed this point several times in the past. Do you want to rehash that whole discussion we had earlier on this one point?

Note: It appears that you're using the term invariant to mean "does not change with time" when, in relativity, it is always used to mean "remains unchanged by a change in coordinates".

Pete

 

[pervect]

Here's a definition from Wald for the total mass in a stationary, asymptotically flat space with a vacuum at infinity. It expresses the mass in terms of the time translation symmetry, represented by the killing vector \xi, and a volume intergal.

<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 \Sigma, and \xi^b is the Killing vector representing the time translation symmetry of the static system.

This is about as simple of a definition of the concept of mass as I've seen in GR, which is why I picked it.

One can also write the intergal in terms of the Riemann tensor

<br /> M = \frac{1}{4 \pi}\int_{\Sigma} R_{ab} n^a \xi^b dV<br />

 

[Garth]

In the first place the metric is always symmetric.
Pete

Of course, you are quite correct, I meant "diagonal",

Garth

 

[hellfire]

According to Noether's theorem, there is a procedure to determine the stress-energy tensor (as the Noether current associated to spacetime translations) of a given action. What happens if one applies this procedure to the Einstein-Hilbert action? Does the resulting energy tensor fulfill the requirements imposed by general relativity? and how far does it have a physical meaning as the energy associated to the gravitational field? I have not seen this aspect discussed in this thread, isn't it relevant? Could anyone elaborate on this?

Thanks.

 

[pervect]

According to Noether's theorem, there is a procedure to determine the stress-energy tensor (as the Noether current associated to spacetime translations) of a given action. What happens if one applies this procedure to the Einstein-Hilbert action? Does the resulting energy tensor fulfill the requirements imposed by general relativity? and how far does it have a physical meaning as the energy associated to the gravitational field? I have not seen this aspect discussed in this thread, isn't it relevant? Could anyone elaborate on this?

Thanks.

I may be way off base here, but if I'm interpreting my textbook properly, this tensor is known as the canonical energy-momentum tensor (Wald, pg 457). The term stress-energy tensor is reserved for T_ab, your tensor gets a different name.

Apparently, when you add terms to the Einstein-Hilbert action to generate the gravity associated with electromagnetic fields, you find that the cannonical energy momentum tensor depends on the Maxwell gauge, and is not symmetric. There are apparently some other problems with it as well.

 

[Chronos]

There is that little covariance issue to deal with, as well.

 

[pmb_phy]

I may be way off base here, but if I'm interpreting my textbook properly, this tensor is known as the canonical energy-momentum tensor (Wald, pg 457). The term stress-energy tensor is reserved for T_ab, your tensor gets a different name.

There is a procedure to make them identical. Its not a good idea to think of them as separate things. It is important to know the relationship though. This tensor is obtained from the Lagrangian of the field. It is a good way to obtain the stress-energy-momentum tensor when one only has the Lagrangian. This is how the stress-tensor if obtained for a cosmic string and a vacuum domain wall.

Pete

 

[pervect]

I was going to ask about the procedure for finding T_ab from the cannonical energy momentum tensor, but I ran across

http://bolvan.ph.utexas.edu/~vadim/Classes/2004f.homeworks/hw01.pdf

which seems to cover most of it. It's not clear how one finds the correct divergence to add to the Lagrangian to get a symmetric and gauge invaraint stress energy tensor, though, unless you happen to know the right answer already.

 

[gptejms]

I know too little of GR to make serious comments in such a discussion,but here's what I wish to say(tell me if I'm wrong):-In a non-inertial frame,there are pseudo forces ; these pseudo forces would do work on particles/objects present in such a frame and energy is bound not to be conserved in such a frame.But if I watch these obects from outside i.e. from an inertial frame,energy would come out to be nicely conserved.Now in Einstein's model,gravitation is equivalent to a set of non-inertial frames(EEP)--so it should be of no surprise that energy is not conserved.I read in one of the posts above that an observer in asymptotically flat regions sees energy to be conserved---well this is your observer in the inertial frame.Comments please.

Jagmeet Singh

 

[pmb_phy]

I know too little of GR to make serious comments in such a discussion,but here's what I wish to say(tell me if I'm wrong):-In a non-inertial frame,there are pseudo forces ; these pseudo forces would do work on particles/objects present in such a frame and energy is bound not to be conserved in such a frame.

That is incorrect. Work can be done on a particle and still leave the energy conserved. What changes is the potential energy and the potential energy. Its the sum that is constant. Whenever the components of the metric are not functions of time the energy of a particle in free-fall will remain constant.

I read in one of the posts above that an observer in asymptotically flat regions sees energy to be conserved---well this is your observer in the inertial frame.Comments please.

Jagmeet Singh

Any observer who is in a frame of reference such that the metric is not an explicit function of time will reckon the energy of a particle in free-fall to be constant, regardless of whether the is an asymptotically flat region. If we're speaking of a g-field like the Earth then no observer will be an an inertial frame unless that inertial frame is a local one. The spacetime curvature of such a field prevents one from finding an extended inertial frame of arbitrary size.

Pete

 

[pervect]

Asymptotic flatness is still important for defining energy in the special case Pete is talking about, the case of a particle with a time-like killing vector, i.e. a small particle "in free fall" following a geodesic in a space-time where the metric coefficients are all constant.

The covariant energy of this particle will indeed be constant, so one has a conserved quantity independent of asymptotic flatness. This is not surprising, because one has a timelike symmetry, therefore one expects a conserved energy.

The problem is that the magnitude of this conserved energy isn't well defined. Multiplying a conserved quantity by a constant gives another conserved quantity. So one winds up with a conserved quantity, but no way of scaling it.

The usual way of scaling it is to say that the metric coefficients go to +/-1 at infinity, i.e. the metric becomes Minkowskian at infinity. But without asymptotic flatness, we can't scale the metric in this way. The value of the component E₀ will depend directly on the scale factor of the metric. As one sets the overall "scale factor" for the metric, (one can think of this process as picking a particular point where g_00 = 1, a condition that one can set arbitrarily), the numerical value of the conserved "energy" will change, depending on how this choice is made.

Thus, without asymptotic flatness, there is no convenient way to set the scale factor of the metric, and hence the scale factor of the conserved conjugate quantity, the energy.

So yes, one can have a conserved energy in the special case of a timelike killing vector, even without asymptotic flatness. But in general one won't have any good way of setting the scale factor for this conserved quantity unless one also has asymptotic flatness.

If one does have asymptotic flatness, it is sufficient on its own to define energy, though the details are rather technical.

 

[pmb_phy]

The covariant energy ..

Please define your term covariant energy.

The problem is that the magnitude of this conserved energy isn't well defined.

Since when??

Multiplying a conserved quantity by a constant gives another conserved quantity. So one winds up with a conserved quantity, but no way of scaling it.

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

The usual way of scaling it is to say that the metric coefficients go to +/-1 at infinity, i.e. the metric becomes Minkowskian at infinity. But without asymptotic flatness, we can't scale the metric in this way.

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.

The value of the component E₀ will depend directly on the scale factor of the metric. As one sets the overall "scale factor" for the metric, (one can think of this process as picking a particular point where g_00 = 1, a condition that one can set arbitrarily), the numerical value of the conserved "energy" will change, depending on how this choice is made.

Why are you so hot and bothered about this scale factor? Do you think its a problem in Newtonian mechanics? The magnitude of energy is of no real use. All that one needs is the fact that it is constant for it to be useful. One can even add a constant since only changes are meaningful.

So yes, one can have a conserved energy in the special case of a timelike killing vector, even without asymptotic flatness. But in general one won't have any good way of setting the scale factor for this conserved quantity unless one also has asymptotic flatness.

This is clearly not the case. One does not need asymptotic flatness to use the Newtonian limit.

Pete

 

[gptejms]

That is incorrect. Work can be done on a particle and still leave the energy conserved. What changes is the potential energy and the potential energy. Its the sum that is constant. Whenever the components of the metric are not functions of time the energy of a particle in free-fall will remain constant.

Pete

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

 

[Garth]

The problem is we find the classical treatment of energy so useful and persuasive that it is hard to let it go. We do work and then expect the energy we have expended to show somewhere in the 'energy accounts', either as kinetic energy, heat (same thing on a microscopic scale) or potential energy. (Yes I know this isn't an exhaustive list) The classical treatment (and SR) balances these 'energy accounts' by conserving energy. Furthermore, SR gives mass an energy equivalent, enabling, in some cases, energy conversion to/from mass.

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.

It is true in a static field, where the metric components do not depend on time, that the covariant time element of a free-falling particle's four-velocity is conserved. Therefore it is tempting to multiply it by the mass (rest-mass to some) of a particle and obtain a conserved covariant time element of four-momentum. Because it is conserved it may then be used as a definition of energy.

However, as I posted above, you find several authorities, such as Weinberg and MTW, define energy, at times, as the contra-variant time element of four-momentum. As Pete said above

Yeah, I don't know why they do that. Seems strange and doesn't make sense.

Four-momentum is naturally described by a contra-variant vector (one form), hence if energy is its (frame dependent) time element then that too is more naturally described as a contra-variant element. However as such it is not conserved.
Quite.
“GR does not conserve energy. It doesn't intend to. It conserves energy-momentum”

The problem resolves itself in asymptotic flatness, because there the covariant and contra-variant elements of a four-vector converge. Hence system energy is said to be defined only at the null infinity of asymptotic flatness.
[But note this is a little artificial as the Schwarzschild solution is embedded in Minkowski, flat, space-time, whereas real gravitational systems are embedded in a cosmological background in which asymptotic flatness does not exist]

If energy is the time element of four-momentum and if the scalar value of four-momentum is given by its norm then an alternative definition of energy (frame-dependent) may be more consistently given by:

E = (-P^{0}P_{0})^{1/2},

because 3-momentum may be defined by

p = (-P^{i}P_{i})^{1/2} (i = 1,2,3),

and

|P|^{2} = p^{2} + E^{2} = -g_{\nu}_{\mu}P^{\nu}P^{\mu}.

However this E will not be conserved in GR, but then it shouldn't be; “GR does not conserve energy. It doesn't intend to. It conserves energy-momentum”
But note it is in SCC.
Garth

 

[meteor]

My knowledge of GR is quite meagre, but if it can helps, i was reading today the TASI lectures of cosmology, and equation 25 is referred as the "Energy conservation equation"
http://arxiv.org/abs/astro-ph/0401547

 

[Garth]

My knowledge of GR is quite meagre, but if it can helps, i was reading today the TASI lectures of cosmology, and equation 25 is referred as the "Energy conservation equation"
http://arxiv.org/abs/astro-ph/0401547

Yes meteor, but that equation is the cosmological case energy conservation equation in the special co-moving R-W cosmological frame, in which it is really a conservation of energy-momentum. (In the cosmological case they are equal - there are no local motions, just Hubble expansion) The above discussion is in the more general case of local gravitational fields and local definitions of energy.
Garth