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Is Gravity energy? Or equivalent ??
« on: 13/03/2011 00:44:15 »
Invariant mass is equivalent to energy. Thinking of a black hole and its gravity it seems reasonable to assume energy to be equivalent to gravity (infinite energy becomes infinite gravity) ..

But, is gravity then equivalent to energy? In Einsteins universe gravity is no force, but, then we have 'gravitational waves', that also is thought to 'radiate' according to some sources, containing a 'energy'? Eh, not that I know exactly what 'energy' in itself should be defined as :) But that one can wait for this question to be answered.

So?

Is gravity equivalent to 'energy'?
What do you think, and how did you get there?

Give me your thoughts on it, because I'm stuck :) Yep, and my headache is a multitudal one now :)
To me it can't be a radiating energy. And furthermore, if a energy, then unlike any other definition I know of energy, at least if we want to stay inside Einsteins universe.

Well, as I see it?
Hope I was understandable?

« Last Edit: 13/03/2011 00:46:14 by yor_on »


 

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Is Gravity energy? Or equivalent ??
« Reply #1 on: 13/03/2011 03:33:48 »
Let's look at my understanding of the quadrupole (gravitational) 'radiation', and don't you laugh :)
And yes, they are, naturally, all in Greek letters :)

 (Hey, we're educated here, well, somebody should? hopefully:)

A monopole defines just one thing, like a planets total amount of its mass. That planets dipole could be, for example, vector(s) describing how its magnetic field is distributed off-center from the planets natural centers (North/south pole which also could be seen as one center only, drawn through the inside of the earth, possibly?), where the dipole will be null, and also which way the magnetic field is oriented. Yep, they're all vectors, needing both magnitude and direction. There is no way (as I see?) of using a dipole for describing just one planets invariant mass, so I had to correct that one, hopefully it makes slightly more sense now.

The quadrupole describes how stretched out, along some axis, the planets mass is, and if that planet is a perfect sphere then it has no quadrapole as its mass is perfectly centered (think Newtons spheres for that one, with all gravity belonging to the center of any sphere). But if our planet would have its mass unevenly distributed, as most planets I think, then it can have a quadrupole 'radiation' as I understands it.

This 'radiation' differs from EM in that you only have one 'sign' for gravitation, positive (attracting). So a rod have a quadrapole 'radiation' as you can see it as sphere stretched out in a direction. But no invariant mass can have a dipole radiation, again as I understands it, as that presumes that you by rotating two masses around each other will get a change in the gravitational field (earth and the moon). But as they both are 'positive' they cancel each other out, and so you won't get a gravitational dipole 'radiation' from it, contrasting it from EM in where you have two signs ( - & + ).

"Gravitational radiation is proportional to the second time derivative of the mass quadrupole, and possible higher order terms".

And "for a changing position { x } its time derivative { x . } is its velocity, and its second derivative with respect to time, { x .. } is its acceleration. A common occurrence in physics is the time derivative of a vector, such as velocity or displacement. In dealing with such a derivative, both magnitude and orientation may depend upon time."

Finally gravity waves is expected to have a energy and propagate at the speed of light.  And here my head start to hurt "The gravitational equivalent of the photon has zero rest mass. It has an 'effective' mass that is equal to its energy divided by c^2.  This 'effective' mass is very small in comparison to the masses involved in generating the gravitational radiation." A indirect proof for it, and the only one I know of is, the Hulse-Taylor binary. And, I don't know if I think that's enough?

What gravitational equivalent?
Did Einstein state that?

So he proposed gravitons then? 

I don't like this explanation, the derivation should be Einsteins but I haven't found a good explanation for it on his terms. There should be one somewhere?
==

Is there a better way to describe it?
« Last Edit: 13/03/2011 04:36:03 by yor_on »
 

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Is Gravity energy? Or equivalent ??
« Reply #2 on: 13/03/2011 13:12:06 »
Nobody except me that wonders about it?

I have two questions about it. What does it do to the idea of the gravitational metric of SpaceTime. In a theory of particles you can have all kinds of 'pressure' involved. And if you do not explain that you should at least consider 'why?', as all other 'particles' except virtual, and then only when 'not existing inside our arrow', do have it?

But maybe?

After all, we're discussing 'energy' here. One of the weirdest ideas physics have as I see it. Something annihilating in a photon, still 'conserved'. Something able to 'radiate' gravitationally without destroying GR?
==

In fact it makes 'anti gravity' viable, did you know that?



I don't like it specifically. But as it seems to fit in, and as it fits the idea of 'quanta' I will have to make allowances, won't I :) But I'm not happy, no Siree, not happy at all in fact. I find it very hard to see how it can allow gravity not to be a 'force' if it 'radiates'. So those of you knowing something about this idea are hereby invited to clear up my confusion and comment. And those of you that just see it as a proof of everything being radiation, EM or ?? Please wait a sec, and let us see if we have some knowledgeable people here that have wondered about it before.
« Last Edit: 13/03/2011 15:12:10 by yor_on »
 

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Is Gravity energy? Or equivalent ??
« Reply #3 on: 13/03/2011 14:47:14 »

So now I've two different solutions/suggestions, both finding anti-gravity viable, or even necessary?  from his equations? And I still don't now if he ever thought this way, or if this all are later assumptions. It seems a lot of assumptions today builds on others interpretations than Einstein himself.

Like you saying "It's a beautiful night." in a que, only to hear the last person repeating it as something entirely different. But it's very human of course, take a look at this example.

" Einstein’s equivalence principle is very different from the version formulated by Pauli [10, p.145],

“For every infinitely small world region (i.e. a world region which is so small that the space- and time-variation of gravity can be neglected in it) there always exists a coordinate system K0 (X1, X2, X3, X4) in which gravitation has no influence either in the motion of particles or any physical process.”

Note that in Pauli’s misinterpretation, gravitational acceleration as a physical cause is not mentioned, and thus Pauli’s version, which is now commonly but mistakenly regarded as Einstein’s version of the principle, actually is not a physical principle. Based on Pauli’s version, it was believed that in general relativity space-time coordinates have no physical meaning. In turn, diffeomorphic coordinate systems are considered as equivalent in physics not just in certain mathematical calculations. However, according to Einstein’s calculations this is simply not true (see section 3).

The initial form of the equivalence principle is a relation between acceleration and gravity. However, in the above clarification, the role of acceleration is not explicitly shown. One may ask if acceleration does not exist for a static object, would the equivalence principle be satisfied? One must be careful because a geodesic may not represent a physical free fall.

There are three physical aspects in Einstein’s equivalence principle as follows [3]:

1) In a physical space, the motion of a free falling observer is a geodesic.
2) The co-moving local space-time of an observer is Minkowski, when 1) is true.
3) A physical transformation transforms the metric to the co-moving local Minkowski space.

Point 3) must indicate that this physical local coordinate transformation is due to the free fall alone. In other words, the physical validity of the geodesic 1) is a prerequisite for the satisfaction of the equivalence principle, and validity of 3) is an indication of such a satisfaction. Thus, a satisfaction of the equivalence principle is beyond the mathematical tangent space."

So, what conclusions can one draw?
Better get your facts as straight as you can before posting that theory. And trusting that last person in the que to know what the first guy said, that is most probably wrong. Check your basics.
« Last Edit: 13/03/2011 15:15:22 by yor_on »
 

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Is Gravity energy? Or equivalent ??
« Reply #4 on: 13/03/2011 17:12:58 »
"GR takes pride in treating all coordinate systems equally.  Mathematicians invented tensors precisely to meet this sort of demand — if a tensor equation holds in one coordinate system, it holds in all.  Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles.  In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects.  (See the FAQ entry on black holes for some examples.)

These pseudo-tensors have some rather strange properties.  If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime.  By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation.  For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy.

One other complaint about the pseudo-tensors deserves mention.  Einstein argued that all energy has mass, and all mass acts gravitationally.  Does "gravitational energy" itself act as a source of gravity?  Now, the Einstein field equations are

            Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below.  Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy". 

So one can argue that "gravitational energy" does NOT act as a source of gravity. 

On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory). 

So one can argue that "gravitational energy" IS a source of gravity."

Well, in my 'world' gravity is no force. So then I have to define 'gravity waves' as SpaceTime deformations I think. And if that is right my first question still stands. Is gravity 'energy'. If it is I need to reconsider how I look at energy I suspect, although I don't think anyone knows what it is if I'm honest. But a Black Hole assumes a 'infinite energy' which makes sense to me.

The truth is that we live in a 'finite' universe from our macroscopic viewpoint. We expect most of everything in it to be countable, and we don't like when it becomes 'infinite'. But if the universe is approximation of something existing in a equilibrium, then it's also a island. That this island to us have uncomfortable qualities like, possibly, being 'infinite' is just a definition from where we stand and look at it. Maybe it's all 'densities' in a wider meaning? And maybe the 'densities' comes to bear through 'time'? Like a gas inside a gas inside a .. But in a four dimensional continuum in where the arrow may behave differently and even 'disappear' .. Each representation seeing itself, unable to observe the other representations? Maybe the way we explore QM is our way to connect what we can't see to what is observable? I don't know, but personally I prefer this idea before the one in where our universe is some mechanical clock, its spring set by ??




« Last Edit: 13/03/2011 17:15:31 by yor_on »
 

Offline JP

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Is Gravity energy? Or equivalent ??
« Reply #5 on: 13/03/2011 18:09:38 »
Now, the Einstein field equations are

            Gmu,nu = 8pi Tmu,nu

To answer your original question in the threat topic, no.  The equals sign in that equation tells you that the deformation of space-time is given by momentum and energy and the flow of momentum and energy through space and time.  But gravity is the curvature of space-time combined with rules of geodesic motion.  So gravity is not equivalent to energy.

Gravitational fields, however, do contain energy, whether you want to think of them in terms of general relativity or Newtonian gravity, but containing energy is not the same as being equivalent to energy.
 

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« Reply #6 on: 13/03/2011 20:45:16 »
What do you mean by the "rules of geodesic motion" JP?
The way Space warps and so change them, and they change Space (with motion and 'gravitational relations' changing in a 'system')?

Or is it what formalism one use, like coordinate-system?
Or something entirely different perhaps :)

« Last Edit: 13/03/2011 22:49:18 by yor_on »
 

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« Reply #7 on: 14/03/2011 00:04:48 »
Lets use two bodies rotating about each other. Will they have a quadrupole moment if I assume that they both are perfect spheres? And if they're not, will they then start to 'radiate'?

And when they do so they will lose 'energy', how? oscillating gravitation creating some sort of inertial effects or? What happens to them, is it their intrinsic energy, but then their mass should decrease, shouldn't it? If it is the relation 'gravity' has between those two planets then the 'energy lost' still will have to come from somewhere?

If I assume that it is the 'stress energy tensor', and further assume that in our 'two planet system' this should be defined as all of the space in where they move relative each other. The planets should then still lose some energy it seems? But as the surrounding space has the larger area, most will be milked out of the 'space' joining them, depending on the vectors the planets have relative each other, moving in time and space? Would that be reasonable. Or is this just plain wrong?

Its such a weird concept :) and it can't really be the planets invariant mass that change, can it? It has to be their angular momentum/speed around a joined common barycenter, and so 'energy'. But not measurable? Or maybe you could consider them more 'compressed' the further down in a gravity well they are? And so losing some mass too?

Frankly I'm having trouble seeing how the gravity waves can be produced? And then there is my idea that as the energy weakens they shouldn't fall into each other but instead wander away from each other, slowing down. Why do they attract each other by losing 'energy' and how do that create gravity waves?

(And another thing, why do we say we 'gain energy' by firing our rocket, gaining height? We lose energy as I see it, exchanging that energy for motion, I know, just a thought, but it bugs me:)


 

Offline JP

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Is Gravity energy? Or equivalent ??
« Reply #8 on: 14/03/2011 00:44:45 »
Yor_on, I think you're making this question way too complicated for an easy answer by jumping immediately to gravitational radiation.  :)

Let's go back to the very basic, first question you asked, which is whether gravity is equivalent to energy.  Like I said, it isn't equivalent.  There are two equations that tell you what gravity does:

1) Einstein's field equations, which you posted above, and which tell you how space-time bends due to a stress-energy tensor.  The stress-energy tensor contains information about the amount and flow of momentum and energy of masses and electromagnetic fields in space and time.

2) Geodesic motion, which says that when you put a mass in space-time, it will follow geodesics unless it's accelerated somehow. 

When you combine these, you have a description of how space bends and how objects move within that bending.  This describes gravity.

The only equation which could be said to be due to energy is the first one--the bending of space-time.  If you really wanted to confuse people, you could say that the bending of space time is "equivalent" to the stress-energy tensor, since Einstein's field equations have the bending terms on one side and the stress-energy tensor on the other.  I think it's much less confusing if you think about the curvature as being due to the presence of mass.  But even in that case, all you've done is to say that the presence of a mass causes a gravitational field due to the mass's inherent energy (E=mc2), or more properly, the mass's stress-energy tensor.

Let's see Einstein's field equations again:
Quote
            Gmu,nu = 8pi Tmu,nu
The left-hand side describes the bending of space-time and the right-hand side is the stress-energy tensor, which includes contributions from things in space-time, but not space-time itself: the information about space-time and it's bending is entirely contained in the left-hand side of the equation.

Now, you can chop up the left-hand side into terms so that one of them looks like the "energy" of the gravitational field, and move it over to the right hand side, so that you now have a "stress-energy" term for gravity, then it appears that curvature depends on the energy of the gravitational field.  But the problem is that the right-hand side is no longer reference frame invariant, meaning that this physical law varies depending on where you are in space.  So you can't really point to a spot in space-time and say "I know the energy there."

-----------------------------------------------------------------

As I understand it, though, there are some cases where you can talk about the energy due to the bending of space-time.  One of those is when you're far away from a gravitating object, so space-time near you is nearly flat.  All observers far away will agree with you on your definition of energy.  I believe this is how they get to gravitational radiation carrying away energy.  Gravitational waves escape from the region around the gravitating objects and propagate far away from them.  In this far-away region, you can define an energy being carried by these waves, and then clearly, the energy of the total field near the objects will decrease as a result of gravitational radiation.

What happens as a result of this energy loss?  Newton's laws should still approximately hold, and in that case, energy loss leads to decaying orbits.  Therefore in the gravitational wave case, the orbits should decay, or in other words, they begin to orbit closer to each other, which leads to a configuration of space-time that has less total energy as you measure it from far away...


And all that comes with my usual disclaimer about not being an expert, of course.  :)
 

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« Reply #9 on: 14/03/2011 01:34:13 »
Nice JP, got to admit that I wondered how Einstein reasoned his way to such a concept. And now I will need to reread you :)
 

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« Reply #10 on: 14/03/2011 01:53:47 »
Okay we have two different 'types' acting on 'binary systems'. When it comes to the moon Earth they separate due to the 'tidal forces' created between them, creating Earths bulges/tides. So here, as the 'system' loses energy it will move apart due to the 'conservation of angular momentum'. In the case you're discussing we have those stars orbiting each other very close, they have a shared center of 'mass' and their speeds relative each other make their orbits very elliptical, meaning that they will 'radiate' away 'energy' as they change their speeds constantly depending on where they are, and the oscillations (gravitational radiation) also steal 'energy' forcing them closer and closer.

==

What I mean there is that they can't have stable orbits, they are too close, going faster the whole time, that forces them to accelerate when close in their orbits to each other, and to slow down when further away. And that may be how they start to oscillate away that gravitational radiation too? But I..absolutely insist on it being a deformation of SpaceTime, highly localized and propagating at 'c' :) Now if someone would prove to me that gravity waves transfered a measurable 'jiggling' of atoms etc or a higher invariant mass? Well, then I would go away to mutter :)
==

Would that be a closer approximation?

Anyway, I still don't know how to think of this 'radiation'. I want it to be a 'deformation' propagating in SpaceTime, not 'energy'. But to make it so it needs not to be able to 'compress' it seems? Or maybe it's just me looking at 'energy' the wrong way? Ah well, in time so to speak, all in good time :)
« Last Edit: 14/03/2011 02:21:59 by yor_on »
 

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« Reply #11 on: 14/03/2011 05:33:15 »

and their speeds relative each other make their orbits very elliptical, meaning that they will 'radiate' away 'energy' as they change their speeds


I don't think so Yoron. The graviatational effect will cause a loss of energy due to friction in the orbiting bodies. 
 

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« Reply #12 on: 14/03/2011 12:41:20 »
Sorry, I was thinking of the gravitational energy there Geezer. But I'm not sure how two orbiting body's in space lose their 'energy'. I know they will speed up as they get closer to each other, but that they do so (getting closer) must be the result of them somehow feeling the result of inertia, or 'jerks' as I think Jartza called it. And that has to come in their unstable orbits as they constantly change speed as I think of it now. That they are constantly more unstable must be a result of them getting more and more elliptical? I think?
==

And the 'speeding up' we see as they close up will not result in any more energy as I see it as they only are following geodesics, constantly broken by inertia that then 'steal' some of their angular momentum. At least I think that this is the way it works? That is if we define those geodesics as being true geodesics, but as we actually see them happening they should be as 'real' as it is possible to get? There ways to test a if a geodesic is real looking at the paths a object take that you throw from the surface of what you want to check out, if I remember right. But, that's another subject :)

==

The friction you speak of might be this too? You could of course assume that they feel each others 'breath' and so also are traveling in a particle cloud, slowing them down. Maybe that was what you meant?
« Last Edit: 14/03/2011 12:51:18 by yor_on »
 

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« Reply #13 on: 14/03/2011 13:01:55 »
JP that's what I wonder about. I've been assuming for the longest time :) that gravity is a result of mass bending SpaceTime. But I'm also wondering if you could describe it mathematically as something regulated by mass, not caused but regulated. That and 'relative size' if I can use that expression. Size seems to come with the objects introduced?
==

If it was it would still be coupled to mass, as gravity follow mass. But if there was this possibility then gravity, in a way, would get a 'life' of its own. It's a notion I have, and I can't get it out of my head it seems :)
==

It's more subtle than mass just 'regulating' if so. It has to be a very close coupling and what we call gravity will need mass to express itself if so. Well, that's what I'm wondering.
« Last Edit: 14/03/2011 13:08:44 by yor_on »
 

Offline JP

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« Reply #14 on: 14/03/2011 14:40:34 »
I'd be careful about the word "regulating."  That implies that space-time is apt to bend on it's own, and that mass is somehow stepping in to control that process.  In fact, when left to it's own devices, space-time is flat.
 

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« Reply #15 on: 14/03/2011 15:31:53 »
That's okay I think :)

I'm not arguing that gravity is a 'force', more of a property coupled to invariant mass and in some more direct way, also to energy. Regulating means just what it says, making the property appear. Ah I see JP, no it's a result of invariant mass and energy, it's not something existing as 'gravity' on its own. but I'm sure it have a existence, just as 'energy' does. and I think it may have very much to do with 'energy' but that one I can't prove.

(If just someone could lend me a ounce of energy to weight it I would be more than pleased?:)
==

It's like we see it as 'gravity' in here, but that is a transformation of it, created by the couplings it express itself through. Also time must have to do with it, refining and defining it to us. And what I call 'relative size' the way everything seems 'fractal' and 'magnifying/contracting' instead of just being a simple measure in flat space between two objects.
==

You could also define it as existing on its own, I know I do at times, for that to work you will have to assume that without matter, or energy (concentrated into a 'point' as observed by us), you won't notice it. So then gravity will be a hidden property of space called forward by invariant mass and energy, and time. but as I don't expect the 'original' to exist as we can observe it I would also expect Space without gravity to not be there. There is a subtle difference in saying flat SpaceTime meaning that there is no gravity there or saying flat SpaceTime meaning that you can't measure a gravity.

I'm not entirely sure but my reasoning goes like this. If you have a 'gravity' in some area of space, you can't measure it, not unless you break your geodesic. That's what a planet does, it breaks your geodesic, and that's also why we say that we all 'accelerate' with 9.8 meters I think per second. Another way is to look at how you can measure gravity, by tidal forces and by matter (energy too) as I see it now? If you found a place in space without gravity, how would you prove the difference between it having no 'gravity' and having a gravity that equaled out, acting on you in all directions? In such a way that it became impossible for you to measure any? If we are talking particles pressuring on you, or if a aether, we should notice it in a motion, as a increased 'pressure/friction', but we don't. So whatever it is acting on you it's more of a property than a force, or a pressure. So I think space need 'gravity' to exist :) assuming that it is a indivisible part of SpaceTime. The other way is to treat it as a 'force', propagating in space, much the same way we expect light to do so.

There are other arguments I can use too, building on that any uniform speed following a geodesic should be impossible to differ from any other geodesic, no matter your 'relative speed'. The conclusion I come to is that all speeds, in a black box experiment is equal and impossible to differ from being at rest. Using this I can now question any idea of 'speed' not involving a acceleration, and also state that all uniform speeds are being at rest in a place without 'gravity'. So how exactly can we measure 'gravity'?

=
Or maybe you could combine both :)
So far its more of a assumption than anything else.
Or even pre-assumption maybe. I can argue for it, but to prove it?

(It won't change what we see. The only thing it does is to give gravity a gravity of its own, as elusive as 'energy' instead of being 'particles':)
« Last Edit: 15/03/2011 00:45:29 by yor_on »
 

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