Is Energy Conserved in General Relativity?

[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.

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

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

$$M = \frac{1}{4 \pi}\int_{\Sigma} R_{ab} n^a \xi^b dV$$

 

[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

 

[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.

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

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
$$M = \int_{\Sigma} T_{ab} n^a \xi^b dV$$

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 dependant 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 dependant 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