Gravity affecting gravity?

[Runner 1]

I was thinking about how a compressed spring weighs more than the same spring uncompressed, and it got me wondering about the earth and moon system.

Basically, if you separate the two and weigh them individually, the sum is not going to weigh as much as if you weighed them as a system because of the potential energy bound in the gravitational field. Well, this potential energy has mass too, right? Which means if we want to calculate completely how the earth and moon interact, we have to take a third mass into consideration.

However, this third mass (I'll just call it the field) interacts itself with the earth and moon, and this interaction is itself another potential (or simply modifies the field from what we assumed it was)!

It seems to me this recursive process goes unto infinity, so you can't really calculate the interacting system directly -- you have to iterate and converge onto the correct solution.

Is this right? Or am I way off base?

[1mmorta1]

The energy in the system doesn't "have" mass, it "is" mass. So you don't add in the mass of the energy, and then recalculate what the total energy of the new system in addition to the sum of the energy and mass of the old system would be.
Too wordy? Or does that make sense?

[Runner 1]

The energy in the system doesn't "have" mass, it "is" mass.

You know what I meant. I didn't think I had to state in every single post "mass is energy".

I'm approaching it from a calculating point of view, where you start with an initial inaccuracy and converge. How else would you calculate it?

[1mmorta1]

Well, just from what you said there, you automatically know that you won't have a recursive process. I don't know what the potential energy of the gravitational relationship between the moon and the earth is, so as far as calculating goes....
But what I AM saying is that you take the total mass of the earth-moon system, find the potential energy of the relationship, and find the sum for the total mass(as you know), which probably will be several place values off from significant. That is it.
My only point is that, if you find the potential energy of THAT system, you are creating energy.

[1mmorta1]

When it WOULD be significant, and maybe even large enough in some cases to have an effect, is if you introduce a third gravitational body.
If two black holes were orbitting each other, and you introduce a third black hole, the total mass of the two orbitting black holes may be enough to alter gravitational pull. Good luck with the three body problem though ;)

[SteamKing]

Compressed spring weighs more than uncompressed spring?!?

What strange new physics is this?

[1mmorta1]

Steamking, a compressed spring has more energy than an uncompressed spring. E = mc^2, therefore the spring has more mass as well. Unless you were being sarcastic...in which case stop being sarcastic haha.

[1mmorta1]

Keep in mind that this is not a discussion of proper mass, but relativistic mass, which I think would apply since we are talking about bodies in motion. I believe the same goes for the spring, it only weighs more when in motion...though I am not a physics doctorate and would gladly defer to one on that matter :)

[SteamKing]

A compressed spring is not in motion. It is compressed.

[1mmorta1]

Hence my confusion: Does the compressed spring actually weigh more or does it have to be being compressed, or decompressing?
I know that a body with greater energy(i.e. motion or heat) will have greater relativistic mass. However, I can't remember if potential energy actually increases relativistic mass...especially considering the common use of the term "rest mass," which would seem to indicate that an object at rest has minimal mass. BLAH! There is so much to remember, and so so much to learn....

[Runner 1]

But what I AM saying is that you take the total mass of the earth-moon system, find the potential energy of the relationship, and find the sum for the total mass(as you know), which probably will be several place values off from significant. That is it.
My only point is that, if you find the potential energy of THAT system, you are creating energy.

I think you're missing my point a little bit. I'm saying the potential energy stored in the gravitational field itself is not a simple function of the masses of the earth, the moon and their distance apart. Because the potential energy of the field is itself mass, there is, for lack of a more semantically astute way of expressing it, a gravitational force to the gravitational field itself. You see where the recursive effect comes in? I'm not saying there's a series of events that happens or that energy is magically generated -- I'm just saying that to calculate the potential of the interaction requires some sort of recursive function.

The effects of course diminish very rapidly. But this isn't a practical question -- it's purely theoretical. When someone is trying to do relativistic calculations, you wouldn't tell them to just use Newton's law because the difference is insignificant at slow speeds. How would you ever learn the theory then?

[Runner 1]

Hence my confusion: Does the compressed spring actually weigh more or does it have to be being compressed, or decompressing?
I know that a body with greater energy(i.e. motion or heat) will have greater relativistic mass. However, I can't remember if potential energy actually increases relativistic mass...especially considering the common use of the term "rest mass," which would seem to indicate that an object at rest has minimal mass. BLAH! There is so much to remember, and so so much to learn....

The spring simply needs to be compressed -- it does not have to be in motion. I was under the impression you knew more about this area of physics than me, but as this seems to not be the case, see this Wikipedia article for an introduction to the subject: http://en.wikipedia.org/wiki/Mass-energy_equivalence (http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence). It has the spring example, as well as a few others.

[1mmorta1]

Runner1, its hard to remember certain things sometimes, and I apologize. I'm not a doctor, and so don't claim to be an expert(I admittedly have struggled with some aspects of relativity in school), but I do have valuable knowledge to share with you. I do not need an introduction, but I appreciate your sincere gesture.
However, regarding your question on the recursive properties of gravity calculations....do you feel they have been answered or are you still confused?

[Runner 1]

Oh no offense was meant at all! I hope you didn't take it that way. I forget a lot myself ;)

Regarding my question, I'm really just wondering if I'm right, and if so, how one goes about doing these calculations. I'm kind of hoping an expert in this area will respond...

[1mmorta1]

Yes I see what your saying, and I suppose you could call it recursive. The problem is that relativity is background independent. SO, when you calculate the masses of the Earth and Moon, your fine and dandy. Then, when you calculate the potential energy of the relationship between the two, your still fine and dandy, because it is the potential energy of one object relative to another, regardless of background. Agreed?
Now, next step: You find the mass of the energy in the system described above. Once you do this, you have the mass of the entire system, which could be floating off in oblivion and be the whole extent of the universe, and your calculations are, once again, fine and dandy.
In order to calculate the Gravity of the gravity, you must introduce another body. This extends your universe, and you can once again calculate the mass of gravity, HOWEVER this time you will calculate the mass of the gravity of the entire 3 body system, which I have never done and is no doubt entirely too complicated.
THIS process can go on and on and on until you've calculated the mass of the entire universe....What I'm trying to say is that, in your two body system, you run out of objects which can be used for relativistic calculations.

[1mmorta1]

Sorry, got caught up for a bit. Long story short:

The POTENTIAL gravitational energy between two objects could be found by taking the negative gravitational constant(google it), multiplying it by (the masses of your two objects divided by the distance between them) and adding the constant of integration.

You have a constant of integration because the formula above is a simplified way of integrating the gravitational force. (In case your unfamiliar with calculus, which it seems like you are, this should maybe provide you some closure as to your feelings of a recursive problem, although I hope that you see the gravity doesn't actually compound.)

[1mmorta1]

Then, naturally, you can convert the potential energy into mass using Einstein's equation :) Don't keep doing that though, haha.

[Runner 1]

In case your unfamiliar with calculus, which it seems like you are

How does it seem that I am unfamiliar with calculus? I got a 5 on Calc BC in high school, and A's in Calc II, Calc III/linear algebra, Physics I, Physics II, and Diff eq my past 4 years in college. My questions on PhysicsForums are more for conceptual understanding. The math follows pretty easily.

Does anyone else want to chime in on this thread or is it just us two that have an interest in this? For all I know, we could be exchanging all sorts of inaccuracies.

[1mmorta1]

How does it seem that I am unfamiliar with calculus? I got a 5 on Calc BC in high school, and A's in Calc II, Calc III/linear algebra, Physics I, Physics II, and Diff eq my past 4 years in college. My questions on PhysicsForums are more for conceptual understanding. The math follows pretty easily.

Does anyone else want to chime in on this thread or is it just us two that have an interest in this? For all I know, we could be exchanging all sorts of inaccuracies.

Oh no! That's not at all what I meant to say! I apologize, I meant to say something more along the lines of "just IN CASE you aren't familiar with calculus, though it seems that you are...."

I just threw it in there on the odd chance that you were an enthusiast and not familiar with the math :( Sorry.

[khemist]

I am pretty sure potential energy does not have a "mass." Mass is mass, energy is energy, though they can be converted. The gravity that comes from the earth or the moon is a result of the mass. When two charges (say, electrons) are near each other, does their electric fields cause them to actually have a greater charge? There may be a force, which will cause them to move, but their charge certainly does not increase.

However I could be mistaking...

[Delta Kilo]

Is this right?General Relativity is in fact non-linear, unlike say, Maxwell equations.

[1mmorta1]

I am pretty sure potential energy does not have a "mass." Mass is mass, energy is energy, though they can be converted....
However I could be mistaking...

Look up "rest mass" or "proper mass" and "relativistic mass."

I think that runner1 is aware that relativity is non linear. That would sure make things easy :)

[Naty1]

Basically, if you separate the two and weigh them individually, the sum is not going to weigh as much as if you weighed them as a system because of the potential energy bound in the gravitational field.

This is backwards; gravitational potential energy is negative. It will be zero when the two objects are at infinite distance, and a finite negative value when they are close. In other words, positive work would have to be done to separate two gravitationally attractive objects.

Gravity IS non linear...it IS self interacting as suggested in the original post. That is what in part caused Einstein to take ten years developing GR....he did not initially know the proper
mathematics...anyway, that characteristic is taken into account
by the Einstein Field equations (tensors).

[1mmorta1]

This is backwards; gravitational potential energy is negative. It will be zero when the two objects are at infinite distance, and a finite negative value when they are close. In other words, positive work would have to be done to separate two gravitationally attractive objects.

Gravity IS non linear...it IS self interacting as suggested in the original post. That is what
in part caused Einstein to take ten years developing GR....he did not initially know the proper
mathematics...

But not self compiling....is the method I gave for calculating the potential energy not accurate? Integrating the gravitational force?

[1mmorta1]

Hey runner, lets collaborate a little here. I've been suggesting the use of E = mc^2 to calculate the mass of energy. This is Einsteins equation for particles at REST. His equation for particles in MOTION is different. Since, in GR, when discussing gravity we're not discussing particles, do you have any thoughts on which we should be using to convert potential energy to mass?

[Runner 1]

I'm pretty sure that equation applies to anything, whether it's in motion or not. Say you're given an electron with an energy E. Divide by c^2 to get m. This m is the sum of the particle's fundamental rest mass as well as it's kinetic energy.

[1mmorta1]

Are you familiar with the following: E = (m(naught)c^2)^2 + (pc)^2 where p is momentum? This is the equation for finding the energy of an object in motion.

[1mmorta1]

False. I forgot to make that E = (that whole mess)^(1/2)

[Runner 1]

Yes, I was just about to rearrange it for you. "That whole mess" equals rest energy + kinetic energy.

[1mmorta1]

Precisely, and are we certain that we treat gravitational potentional energy as an object at rest? I don't see how we could regard gravitational potential energy as being in motion, but I thought I'd at least throw the thought out there. (Since no experts are chiming in, and neither of us are capable of being individual Einsteins, maybe if we put our heads together we can cover all of our bases)

[Runner 1]

(Since no experts are chiming in...)

Weird isn't it? They normally jump on a chance to correct people.

Anyway, I'm fairly certain you don't need motion to have energy. Motion is just a form of it. Energy is also contained in fundamental (rest) masses, as well as fields (strong, weak, electromagnetic, gravitational). That's all I can think of -- maybe there's some more.

[1mmorta1]

That sounds correct to me, in fact I know your correct. I just wanted to at least bring up the fact that there exists another relativistic formula for calculating the energy of an object(Which is why photons don't have infinite mass).
Did you see how many views this page has? Maybe we've opened a can of worms that the experts don't want to be a part of.

[1mmorta1]

Speaking of lack of participation, did you see my post on the next step after the higgs(its on this board)? No one has anything to say on that either(maybe no one likes us) but I'd be interested on any knowledge/opinions you have on the matter.

[Runner 1]

It's really the same formula (and will give you the same result), but in a different form. For example, a photon has mass. Just not rest mass. I don't know why they've made the terms so confusing, but I remember being confused by it in high school.

[Runner 1]

I was researching my original post a little more, and found that energy is indeed carried away from systems in the form of gravity waves. Most physicists believe these to exist, however, they are having a very hard time detecting them because the measurements must be so precise.

[1mmorta1]

Your a better man than I am, I didn't study general relativity in highschool ;) Keep in mind that you can't use the first formula for objects in motion though. Duh...you know that(its just me being picky about them being the same formula)
That would be a nice solution to our dilemma, it would mean there's no need to calculate "extra" mass because its being carried away from the system. I'll have to look into these.
I have an uncle who worked on a project for NASA, he was a cryogenics engineer, detecting gravity waves by building the most precise gyroscopes ever created(and the most perfectly spherical objects in the universe)

[Runner 1]

Oh really? That sounds pretty cool! Ask him about it.

[1mmorta1]

Ha. Its hard to get a hold of the guy these days. He's always working...I don't even know where he lives anymore, it was NASA's gravity probe b. Yuck...engineering. Don't think I could do it.

[DaleSpam]

Wow, this thread really ran amok.

OK, in the OP there are a couple of valid points:
1) The mass of a bound system is, in general, different from the mass of the individual components. There is a good Wikipedia article:
http://en.wikipedia.org/wiki/Binding_energy#Mass_deficit

2) Gravity gravitates. This is because the Einstein field equations are non-linear:
http://archive.ncsa.illinois.edu/Cyberia/NumRel/EinsteinEquations.html

[?????]

how one goes about doing these calculations.

Potential energy cannot gravitationally attract itself.

If you want to calculate the gravitational attraction force of the earth for the moon, then select your gravitational model and do the calculation. Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit. This potential energy has no place in this particular gravitational force calculation.

If you want to know the earth-moon pair gravitational force to the sun, then you take can include the earth-moon gravitational attraction into this calculation. In this case, the sun-moon force and the sun-earth force calculated separately will not give you the same answer as the earth-moon-potential energy composite attraction force to the sun. The two calculations will only differ very slightly. The gravitational potential energy of the earth-moon pair is very small compared to the mass of the material of those two bodies.

[1mmorta1]

Potential energy cannot gravitationally attract itself.

If you want to calculate the gravitational attraction force of the earth for the moon, then select your gravitational model and do the calculation. Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit. This potential energy has no place in this particular gravitational force calculation.

If you want to know the earth-moon pair gravitational force to the sun, then you take can include the earth-moon gravitational attraction into this calculation. In this case, the sun-moon force and the sun-earth force calculated separately will not give you the same answer as the earth-moon-potential energy composite attraction force to the sun. The two calculations will only differ very slightly. The gravitational potential energy of the earth-moon pair is very small compared to the mass of the material of those two bodies.

This is your response on how? Thank you, but did you read the rest of the thread....we are trying to figure out HOW, not looking for the conceptual process. Can you back the information above with some equations and directions?

[Runner 1]

Potential energy cannot gravitationally attract itself.

I don't believe this statement is true. I could be wrong though. Have you got a source (like a textbook) perhaps?

[zonde]

I don't believe this statement is true. I could be wrong though. Have you got a source (like a textbook) perhaps?
Did you read about mass deficit in the link given by DaleSpam?
I think it explains the problem very clearly.
1) The mass of a bound system is, in general, different from the mass of the individual components. There is a good Wikipedia article:
http://en.wikipedia.org/wiki/Binding_energy#Mass_deficit

Potential energy does not add or subtract energy from the system. If just specifies how much of the energy that the system has already can be radiated away.

[?????]

This is your response on how? Thank you, but did you read the rest of the thread....we are trying to figure out HOW, not looking for the conceptual process. Can you back the information above with some equations and directions?

Perhaps I am wrong and you are right. I did read the thread and I thought I was answering the original question.

By the way, my original statement "Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit." is in error. The correct statement would be "Now move the moon down to the earth and this will give you the potential energy gain of the moon in its original orbit."

As long as the OP wants to consider energy in gravitational calculations, I would suggest adding the kinetic energy of the moon in its orbit around the earth when calculating the moon-earth-potential energy combination gravitational attraction for the sun. And you might as well throw in the kinetic energy of the earth rotating on it's axis too.

[Runner 1]

Did you read about mass deficit in the link given by DaleSpam?
I think it explains the problem very clearly.

Yes, I did read it. That's why I followed it up with a question.


Potential energy does not add or subtract energy from the system. If just specifies how much of the energy that the system has already can be radiated away.

I never said anywhere that potential energy added or subtracted additional energy to/from the system. What I am saying is that, assume for a second that gravitational fields have no relationship to energy at all. They just lock objects into place. If you calculate the system dynamics of the earth and the moon now, it will be DIFFERENT than how it is in reality.

Take the spring example again. Let's assume potential energy does NOT obey Einstein's formula -- meaning it does not contribute to the gravitational attraction of the spring. This means a compressed spring will weigh the same as an uncompressed spring. This is obviously not true -- there is a difference. By analogy, I'm trying to quantify that sort of difference with the earth and moon system.

[1mmorta1]

Perhaps I am wrong and you are right. I did read the thread and I thought I was answering the original question.

By the way, my original statement "Now move the moon out to infinity and this will give you the potential energy gain of the moon in its original orbit." is in error. The correct statement would be "Now move the moon down to the earth and this will give you the potential energy gain of the moon in its original orbit."

As long as the OP wants to consider energy in gravitational calculations, I would suggest adding the kinetic energy of the moon in its orbit around the earth when calculating the moon-earth-potential energy combination gravitational attraction for the sun. And you might as well throw in the kinetic energy of the earth rotating on it's axis too.

It is true that if we were looking at the total mass of the system we would have to observe that. We were looking into the mass of the system with regards to the original masses of the earth and moon plus the mass of gravitational potential energy, and runner1 is curious as to whether or not the new mass considered in the system would result in more gravity potential, and thus more potential energy...in a recursive process.

Runner1, i think in taking the equations for finding gravitational energy potential(double check mine to make sure they're right), and using our original conversion rate of e = mc2(I should never have brought up relativistic mass), and converting that potential energy to mass which is added to the mass of the Earth and the Moon, we have answered our questions.

(In this instance though, we are ignoring the rest of the universe and the fact that the system "gravitates" that energy.)

[?????]

I don't believe this statement is true. I could be wrong though. Have you got a source (like a textbook) perhaps?

If two compressed springs are sitting in open space, the potential energy's stored in each spring will gravitationally attract each other (and the mass of the material in each spring). But a single compressed spring in open space does not attract itself, in a kind of endless increasing spiral.

[zonde]

Yes, I did read it. That's why I followed it up with a question.

I never said anywhere that potential energy added or subtracted additional energy to/from the system. What I am saying is that, assume for a second that gravitational fields have no relationship to energy at all. They just lock objects into place. If you calculate the system dynamics of the earth and the moon now, it will be DIFFERENT than how it is in reality.
I don't understand your question.

Take the spring example again. Let's assume potential energy does NOT obey Einstein's formula -- meaning it does not contribute to the gravitational attraction of the spring. This means a compressed spring will weigh the same as an uncompressed spring. This is obviously not true -- there is a difference. By analogy, I'm trying to quantify that sort of difference with the earth and moon system.
No, when you compress the spring you put "real" energy into the spring. This "real" energy of course contribute to the mass. So it's not potential energy that contributes to the mass.
Now if you release the spring you can take away some "real" energy from the spring. But if you release the spring and let it oscillate eventually converting energy into the heat then the energy is still in the spring and it's mass wouldn't change.

[zonde]

It is true that if we were looking at the total mass of the system we would have to observe that. We were looking into the mass of the system with regards to the original masses of the earth and moon plus the mass of gravitational potential energy, and runner1 is curious as to whether or not the new mass considered in the system would result in more gravity potential, and thus more potential energy...in a recursive process.
This is wrong. Gravitational potential energy is negative. You have less mass in the system after system has released energy that is equivalent to gravitational potential energy.

[bahamagreen]

If gravity effected gravity, this interactive gravitational energy for a particular mass would have to accumulate by some function to the energy of that mass. This would mean that this interaction of gravity with itself would change that thing's effective gravitational mass; but would not change that thing's inertial mass.

The strong equivalence principle holds that moving things within a gravitational field should only be influenced by their position and not be influenced by what they might be made of or how much matter comprises them.

This suggests that gravity affecting gravity violates the strong equivalence principle?

[jfy4]

After reading, I'm actually still not sure what it is this thread is about :rolleyes: But, it sounds like perhaps you are interested in the Weyl Tensor. It describes the curvature due intrinsically to the gravitational field. Here is the Wiki article for your convenience.

Weyl Tensor (http://en.wikipedia.org/wiki/Weyl_tensor)

[Sam Gralla]

Guys, the reason why you are arguing is because gravitational energy can't be localized. You can only talk about the total energy of the system. You can't split it up into "matter energy" and "field energy" (or, put another way, there are too many inequivalent ways to try to do this--hence all your arguing). This is a standard topic covered in GR books. The most challenging part about learning GR is learning the right questions to ask.

That said, the physical intuition in the OP is sound. If you view the theory perturbatively off of flat spacetime then there is a sense in which this "iterative" procedure described in the OP does take place. In fact there is a famous derivation (really more of a plausibility argument) due to Feynman whereby you begin with a linear equation and add "gravity gravitates" to boostrap your way up to the full nonlinear Einstein equations.

[Runner 1]

If you view the theory perturbatively off of flat spacetime then there is a sense in which this "iterative" procedure described in the OP does take place. In fact there is a famous derivation (really more of a plausibility argument) due to Feynman whereby you begin with a linear equation and add "gravity gravitates" to boostrap your way up to the full nonlinear Einstein equations.

There we go! That's exactly what I wanted to know. What are these equations called? (I don't know anything about general relativity -- is it the same thing?)

[jfy4]

Guys, the reason why you are arguing is because gravitational energy can't be localized. You can only talk about the total energy of the system. You can't split it up into "matter energy" and "field energy" (or, put another way, there are too many inequivalent ways to try to do this--hence all your arguing). This is a standard topic covered in GR books. The most challenging part about learning GR is learning the right questions to ask.


I don't think this is true... The Ricci tensor, or equivalently the stress-energy-momentum tensor, is precisely the object which describes the energy not due to the gravitational field, However, a separate object exists which does include that energy.

See here Stress-energy Pseudo-tensor (http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor)

[jfy4]

There we go! That's exactly what I wanted to know. What are these equations called? (I don't know anything about general relativity -- is it the same thing?)

if you want to ask gravity, and relativity questions, you should study GR, since this is exactly what it covers.

[pervect]

I don't think this is true... The Ricci tensor, or equivalently the stress-energy-momentum tensor, is precisely the object which describes the energy not due to the gravitational field, However, a separate object exists which does include that energy.

See here Stress-energy Pseudo-tensor (http://en.wikipedia.org/wiki/Stress-energy-momentum_pseudotensor)

Because it's not a true tensor, the stress-energy tensor isn't coordinate independent (and IIRC it even depends on the gauge). In any event, it won't give you a unique answer for the distribution of energy, because it's not a true tensor and hence not coordinate independent.

So Sam's answer is spot-on, and very well written.

[Runner 1]

if you want to ask gravity, and relativity questions, you should study GR, since this is exactly what it covers.

Well, I didn't realize it was a GR question. My post just got moved to this forum.

[jfy4]

Because it's not a true tensor, the stress-energy tensor isn't coordinate independent (and IIRC it even depends on the gauge). In any event, it won't give you a unique answer for the distribution of energy, because it's not a true tensor and hence not coordinate independent.

So Sam's answer is spot-on, and very well written.

I'll have to do some looking into that, I'm sorry to doubt, but wiki gives an explanation as to why what you said isn't true, and I need more time because of my naivete to be sure.

Thanks all the same.

EDIT: also, I guess I'm a little confused by your post since you segued with "because". Are you saying that the stress-energy-psudeo-tensor does reflect an object that represents the energy of the gravitational field. It seems like you wrote "the SEM-pseudotensor is that object 'because'..." but your post seems to object to what I said. Could you clarify what you meant up above.

[ZealScience]

Perhaps you are talking about gravitational waves. Gravitational waves carry energy just like electromagnetic waves, and it is gravitating.

[zonde]

Guys, the reason why you are arguing is because gravitational energy can't be localized.
And why do you think that there is any "gravitational energy" at all?

You can only talk about the total energy of the system. You can't split it up into "matter energy" and "field energy" (or, put another way, there are too many inequivalent ways to try to do this--hence all your arguing).
If you can attribute all of the total energy of the system to "matter energy" plus radiation energy without anything left then there is no reason to think that there anything like "field energy".

[Sam Gralla]

And why do you think that there is any "gravitational energy" at all?

If you can attribute all of the total energy of the system to "matter energy" plus radiation energy without anything left then there is no reason to think that there anything like "field energy".

There is such a thing as "gravitational energy" because gravitational waves carry it. You can precisely characterize the rate at which an isolated system is losing energy to gravitational-wave emission. What you can't do is say that so much of the energy was at one place in spacetime and so much of the energy was elsewhere. or whether the energy "came from" matter or field energy. (Of course, you could try to make such a statement using a "psuedotensor" as somebody brought up, but somebody else could come along with another inequivalent pseudotensor and claim that *that* one was the true gravitational energy. What everyone will agree about is the total energy in the system as well as its rate of change.) People do use "stress-energy pseudotensors" in computations, but that's mainly for convenience in performing some specific computation. All claims about a "true" local gravitational energy density have disappeared from the literature at this point. (There is a community still looking for "quasi-local" gravitational energy, but that doesn't seem to work very well either.)

You may be trying to call this sort of energy "radiation energy" so you don't need "field energy". But there are reasons for saying that even a non-radiating system has "gravitational binding energy" (that can't be localized). For example, in the Newtonian limit of GR the energy/mass of an isolated system can be written as an integral over the mass density of the matter. But as soon as you go back to GR (or just a post-Newtonian correction), this property is lost. What is the other contribution? It makes sense to think of it as energy in the field. But there's no way to write it as an integral over matter plus an integral over field, so again it makes sense to say "gravitational energy exists but can't be localized".

[pervect]

Do you happen to have MTW's textbook, "Graitation"? I know they have a good section on the topic of why you can't localize the energy of a gravitational field.

I'm not sure how much more clear I can be, but I'll try saying it again.

Covariance is an important physical principle. It boils down to saying that measurements made by different observers represnt the same underlying reality.

The sort of covariance we need for relativity is Lorentz covariance. Any four-vector, regardless of whether it is (time, distance) or (energy, momentum) must transform via the lorentz transforms to have a physical meaning that's independent of the coordinate system.

If you don't have covariance, your quantity cannot be defined in an observer independent way.

Pseudotensors, in the sense used in General Relatiavity (i.e. the energy pseudotensor you refer to) do not define energy in a way that's independent of the observer.

Some people have remarked, with some merit, that the "pseudotensors" in GR are really just non-tensors.

The thing that makes energy pseudotensors useful at all is that while they don't offer an observer-compatible defintion of energy, the total energy computed via them will transform properly given the proper conditions (usually asymptotic flatness).

So the pseudotensors themselves do not offer any physically meaningful way to localize energy because different observers will, as other posters have remarked, not have compatible views of how the energy is distributed.

They do allow you to come up with a total energy that everyone agrees on, however. (And they aren't the only method of doing it, there's a proof I think that the pseudotensor definition of energy matches the Bondi defiition).

[zonde]

There is such a thing as "gravitational energy" because gravitational waves carry it. You can precisely characterize the rate at which an isolated system is losing energy to gravitational-wave emission.
Isolated system can't loose energy.
Actually we don't have truly isolated systems so in order to model such a system we can imagine system in environment that mirrors all the outward interactions of the system with equivalent inward interactions.
This means that if the system is emitting gravitational waves then equivalent waves are directed toward the system from environment. And I would say that if gravitating system can emit these waves then it can absorb the same waves unless you have some strong arguments why this shouldn't be so.

What you can't do is say that so much of the energy was at one place in spacetime and so much of the energy was elsewhere. or whether the energy "came from" matter or field energy. (Of course, you could try to make such a statement using a "psuedotensor" as somebody brought up, but somebody else could come along with another inequivalent pseudotensor and claim that *that* one was the true gravitational energy. What everyone will agree about is the total energy in the system as well as its rate of change.) People do use "stress-energy pseudotensors" in computations, but that's mainly for convenience in performing some specific computation. All claims about a "true" local gravitational energy density have disappeared from the literature at this point. (There is a community still looking for "quasi-local" gravitational energy, but that doesn't seem to work very well either.)
What I say is that mass is moving down in gravitational potential when the system emits gravitational waves. So it's mass that is loosing this energy. The same way if you would try to reverse this process it's mass where you have to put this energy to move it upwards in gravitational potential.

You may be trying to call this sort of energy "radiation energy" so you don't need "field energy". But there are reasons for saying that even a non-radiating system has "gravitational binding energy" (that can't be localized). For example, in the Newtonian limit of GR the energy/mass of an isolated system can be written as an integral over the mass density of the matter. But as soon as you go back to GR (or just a post-Newtonian correction), this property is lost. What is the other contribution? It makes sense to think of it as energy in the field. But there's no way to write it as an integral over matter plus an integral over field, so again it makes sense to say "gravitational energy exists but can't be localized".
I have problems with that. "Gravitational binding energy" is negative energy. So we have missing energy not excess energy.

[jfy4]

Thanks pervect,

your helpful as always. I do have that book and I take a look at that section.