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Jonathan Scott

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If i separate two rocks,the total weight of them will increase? If so,the extra mass stored in the gravitational field between them?

If you separate two rocks, the total energy (and effective rest mass) of the system increases. However, the location of the additional energy is not clear in General Relativity.

In a semi-Newtonian equivalent, one apparently mathematically self-consistent solution is that each of the rocks increases in energy by the potential energy gained relative to the other rock, and the field loses the same amount of energy as either rock, so the total energy gained is the change in Newtonian potential energy of the system as a whole. In this model, for a field [itex]g[/itex] the energy density is [itex]g^2/(8 \pi G)[/itex] which is analogous to the Maxwell energy density [itex]\epsilon_0 E^2/2[/itex] in electromagnetism, where [itex]\epsilon_0[/itex] is matched with [itex]1/(4 \pi G)[/itex]. The field energy is maximized when mass is concentrated together, and decreases to zero as the masses are separated to infinity.

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When you're dealing with gravitational waves, you can get around this by averaging over multiple wavelengths.

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Jonathan Scott

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If i separate two rocks,the total weight of them will increase

...because you have done work....that is, put energy into the system, done work to overcome the attractive gravitational force. But it's clearer to say the energy rather than the mass has increased....and that seems to be a modern preference/convention as discussed here in other threads.

A similar situation exists when a relaxed spring is compressed and held...the jack in the box example.....there is more energy in the coiled than the uncoiled spring....

and a hot blob of metal contains more energy than the same blob when cold....heat added is a form of energy....

You'll also observe more energy in a faster moving mass than when it is slowed....

Thanks to Einstein!!!

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That field energy would decrease if you break each of them into smaller rocks and separate those, and so on.

Which field energy...seems like both would increase???...according to your earlier post:

If you separate two rocks, the total energy (and effective rest mass) of the system increases.

The second quote is, I believe, the correct one....

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( I'm confined at home because of a bad foot tonight - I'm glad the forum isn't deserted )

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Jonathan Scott

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As I seem to remember mentioning recently in some other thread, there are two ways of extending Newtonian gravity to include energy in the field.

If you just assume in a non-relativistic way that the masses are not affected by gravity, then the potential energy has to be assumed to be in the field, which then has a negative field density, [itex]-g^2/8 \pi G[/itex], because the total energy of the system decreases as masses are brought together, giving a more intense outward field overall (although the field between the masses cancels, of course).

If however you include the known relativistic effect of gravity on time and hence on energy, then each mass is effectively modified by the potential due other masses. Since this occurs both ways, this means that the effective total mass is modified by TWICE the potential energy, but mathematically this cancels out nicely if you assume that there is also a positive energy density in the field of [itex]g^2/8 \pi G[/itex], the same as in the original Newtonian model except that it has the opposite sign.

Note also that this is only a model in a particular frame of reference. General Relativity tells us that this model is probably incomplete because it only works for certain frames of reference.

Regardless of these details, the total effective mass of a system of objects is decreased slightly when they are brought together (in a static configuration) and increased again if they are moved apart. The increase tends to a limit as the separation goes to infinity. If all of the components are separated as far apart as possible, the external field of the system as a whole is minimized.

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If i separate two rocks,the total weight of them will increase? If so,the extra mass stored in the gravitational field between them?

Yes the gravitational field has energy just as a fast moving ball has kinetic energy. We do not say the ball stores energy though I suppose that is correct. Likewise for fields.

If the masses are increasing then energy is being drawn out of somewhere else. Maybe the person or object that is doing the work to move the masses apart is supplying the energy?

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Yes the gravitational field has energy just as a fast moving ball has kinetic energy. We do not say the ball stores energy though I suppose that is correct. Likewise for fields.

If the masses are increasing then energy is being drawn out of somewhere else. Maybe the person or object that is doing the work to move the masses apart is supplying the energy?

If you ever want a practical and dangerous demonstration of the "storage" of energy, buy a flywheel. Spin that sucker up, then break it FAST, and as the shards pass through your body, you'll realize that the explosive 'breakup' of the wheel is a result of the 'stored' potential kinetic energy becoming not-so-potential anymore.

Given that nothing else in the room explodes, I have to assume the potential energy is a property of the dynamic system that is the flywheel, and if there is "storage" it's in the medium which is in motion relative to a rest frame.

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Jonathan Scott

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I don't think any of those are needed. There are energy-momentum pseudotensors which say where the energy is from any one point of view, but in GR it gets mixed up with the shape of space-time, in that what appears to be a free fall geodesic from one point of view involves a change of momentum (and possibly energy) from another more global point of view. The non-linearity of GR seems to make this worse, in that if gravitational energy is also a source of gravity then it's tricky to understand how it can appear to be in different places for different observers.

Personally, I think GR isn't working very well except in the case of a single dominant central object with a fairly weak field, where it is obviously extremely successful. Despite the neatness of the idea, I think that it will probably turn out eventually that the concept is only accurate to first (Newtonian) order and the second-order fit in the solar system (with the PPN beta parameter) is a result of two or more equal and opposite second-order errors (for example relating to Machian variation with location and non-zero energy density in the field). I think it's likely that a future improved theory of gravity will eliminate the concepts of dark matter, dark energy and black holes, will fix the incompatibility with QM, and will clearly include some form of Mach's principle. I just wish someone would come up with it quickly; my own attempts have so far been unsuccessful.

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Hi,magnetar

If i separate two rocks,the total weight of them will increase?

No, "gravitational energy" does not hold energy that appear in Einstein's gravitation equation.

Total mass is considered to be reserved in spite of the separation, however the summation of rest mass in different location is physically meaningless.

Regards.

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I am not following please explain more. In the case of gravity the field goes as 1/r so it is not local? What exactly do you mean by local? Why non-conservation of energy? I like conservation of energy.

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I am not following please explain more. In the case of gravity the field goes as 1/r so it is not local? What exactly do you mean by local? Why non-conservation of energy? I like conservationof energy.

By non-local he is talking about "big N/L" Local/Non-Locality". As in, Entanglement, EPR, etc. He doesn't mean local in the Astronomical sense. As for conservation of energy, it's already only conserved in certain instances, and questions like "where does all energy come from" is analogous to, "what IS space-time?" or "Why is the universe the way it is?"

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Regardless of these details, the total effective mass of a system of objects is decreased slightly when they are brought together (in a static configuration) and increased again if they are moved apart.

says it all......

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It's not clear to me if this assumption (the one I coloured in blue) is arbitrary or if there are strong physical reasons to make it.[...]

If however you include the known relativistic effect of gravity on time and hence on energy, then each mass is effectively modified by the potential due other masses. Since this occurs both ways, this means that the effective total mass is modified by TWICE the potential energy, but mathematically this cancels out nicely if you assume that there is also a positive energy density in the field of [itex]g^2/8 \pi G[/itex], the same as in the original Newtonian model except that it has the opposite sign.

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Jonathan Scott

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It's not clear to me if this assumption (the one I coloured in blue) is arbitrary or if there are strong physical reasons to make it.

It's basically a simple solution to make the energy add up giving a continuous conservation law in this type of simplified model while still matching the time dilation of GR as a first-order approximation.

The maths is similar to that for the electrostatic energy density, but with a few things switched around.

In the electrostatic (Coulomb potential) case an integral by parts is combined with Gauss' law (to convert a volume integral to a surface integral) to show that the electric potential energy can either be attributed to charges residing in potentials (with a factor of 1/2) or to energy in the field, and that the two forms are mathematically equal. See for example Wikipedia article on "http://en.wikipedia.org/wiki/Electric_potential_energy" [Broken]", especially the section on "Energy stored in an electrostatic field distribution".

A similar scheme can be used for the non-relativistic Newtonian gravitational case, with a minus sign in the field energy.

In a more relativistic gravitational model, taking account of the way in which the potential operates primarily through time dilation, a similar integral says that that the effective energy of each part of the system is effectively decreased in potential energy by the potential due to all sources (including itself, without any factor of 1/2) but energy is still conserved locally and globally if a field with positive energy density adds back in the same amount of energy.

Someone suggested to me previously that the Landau-Lifgarbagez pseudotensor reduces to this model in the static weak case, but so far I've not had the time or brainpower to check this out. Can anyone confirm or refute this, please?

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My unease about this starts with the Newtonian gravitational field where we have to make the potential energy negative to get an attractive force. We can add a constant potential K, assumed to be very large without affecting the EOMs ( which is all we can measure). With this 'gauge' freedom, it is meaningless to ask 'how much energy is in the field'.

It doesn't look like it. If all the derivatives in the second term are zero, there's no contribution from gJonathan Scott said:Someone suggested to me previously that the Landau-Lifgarbagez pseudotensor reduces to this model in the static weak case...

The Lagrangian of the gravitational field in GR is ( see e.g. Weinstein in 'Strange Couplings')

[tex]

(16\pi G)^{-1}c^4\sqrt{-g}R

[/tex]

I wonder if the integral of this energy over all space diverges for most solutions of the EFE ?

[Edit : I changed this while Jonathan was posting - I hope nothing he has said below is affected. ]

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Jonathan Scott

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Jonathan, I agree mostly with what's in your post #12. I would add that GR is pretty successful with cosmological solutions also.

edpell, by non-locality I mean action-at-a-distance ( asFrame Draggersaid ).

My unease about this starts with the Newtonian gravitational field where we have to make the potential energy negative to get an attractive force. We can add a constant potential K, assumed to be very large without affecting the EOMs ( which is all we can measure). With this 'gauge' freedom, it is meaningless to ask 'how much energy is in the field'.

Even in Newtonian gravity, this isn't a problem because the potential energy reaches a limit as the separation goes to infinity.

The ambiguity of the additive potential disappears in relativistic gravity.

For weak fields and non-relativistic speeds, the Newtonian potential is simply replaced by the time dilation factor, which is approximately of the form (1 - Gm/rc

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Have you come across the 'Newtonian metric' ?

[tex]ds^2= (1-2m/r)dt^2-dr^2-\sqrt{(1-2m/r)}dtdr - r^2d\theta^2- r^2\sin(\theta)d\phi^2

[/tex]

I have some trouble interpreting this. It's a vacuum solution because the Ricci tensor is all zero ( according to Maxima). It doesn't look spherically symmetric either.

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I think it is incorrect to compare 2 far objects VS 2 objects closer to each other. Yes, objects sink in the gravitational field and they lose mass, BUT: if you start from 2 distant objects, they would accelerate, falling towards each other. So, you must STOP them. Ultimately, they can crash into each other and energy will be converted into radiation.

So, there a 2 cases:

1. G energy is at first converted into movement, then into some form of radiation and it is radiated away.

2. 2 objects form a black hole and nothing is radiated away (except the G waves?)

Lets talk about #1.

I believe the very good calculations about the semi-newtonian approach about how much energy is stored ignore the amount of mass/energy which MUST be radiated away.

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Jonathan Scott

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I think it is incorrect to compare 2 far objects VS 2 objects closer to each other. Yes, objects sink in the gravitational field and they lose mass, BUT: if you start from 2 distant objects, they would accelerate, falling towards each other. So, you must STOP them. Ultimately, they can crash into each other and energy will be converted into radiation.

So, there a 2 cases:

1. G energy is at first converted into movement, then into some form of radiation and it is radiated away.

2. 2 objects form a black hole and nothing is radiated away (except the G waves?)

Lets talk about #1.

I believe the very good calculations about the semi-newtonian approach about how much energy is stored ignore the amount of mass/energy which MUST be radiated away.

No, it all adds up as in Newtonian theory. If you lower two masses towards each other in such a way that you extract energy, the total effective mass changes by that amount. If they are however freely falling towards one another, there is no overall energy change. If they collide, the energy gets converted to some combination of heat and mechanical or chemical energy, but is still there initially. It may of course be radiated away as heat. If for example they collide in such a way as to compress a spring reversibly, then the total energy of the system still remains constant.

If the field isn't weak and the speeds non-relativistic, then in the free-fall case you can also lose a little energy through gravitational radiation, but that's outside the scope of this approximation.

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Jonathan Scott

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Have you come across the 'Newtonian metric' ?

[tex]ds^2= (1-2m/r)dt^2-dr^2-\sqrt{(1-2m/r)}dtdr - r^2d\theta^2- r^2\sin(\theta)d\phi^2

[/tex]

I have some trouble interpreting this. It's a vacuum solution because the Ricci tensor is all zero ( according to Maxima). It doesn't look spherically symmetric either.

From first impressions, it looks literally like a metric which describes the Newtonian potential, in that it gives the Newtonian acceleration to all objects but static space isn't curved. However, I'm not an expert on such things. It certainly looks spherically symmetrical to me, too. I might look it up some time, but as it clearly doesn't match nature, it's not very interesting.

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