[yor_on (OP)]

Hope you like this one.

" Astrophysicists Alexei Filippenko at the University of California, Berkeley and Jay Pasachoff at Williams College explain gravity's negative energy by way of example in their essay, "A Universe From Nothing": "If you drop a ball from rest (defined to be a state of zero energy), it gains energy of motion (kinetic energy) as it falls. But this gain is exactly balanced by a larger negative gravitational energy as it comes closer to Earth’s center, so the sum of the two energies remains zero." "

Aha, 'zero'

At the center of a perfect sphere of a perfectly distributed density too?
what do you think?

( There the ball will be weightless as far as I get, so how does this work? :)

[guest47899]

imagine the big bang occurring at the top of a sphere. As the expansion progresses it fall outward along the outside of the sphere. another analogy to clarify, the big bang occurring at the top of a sphere. it is a sprinkler head spewing water in a 360 pattern downward along the outside of the sphere. The water is weightless .as it falls along the spherical contour of the sphere. The negative Gravity of the sphere holds the water to by attractional force to the sphere's contour. falling momentum bound by a negative gravity force along a spherical contour. simple

[yor_on (OP)]

It's weird, can you talk about the center of Earth as being of the largest gravitational potential?
what do you think

[Bill S]

I’ve had a quick look through the paper, and my initial impression is that they make a good argument for the Universe having arisen from a quantum fluctuation.  I think this is weakened by the (IMO) unnecessary introduction of the “free lunch” idea.  Could it have been a ruse to give them an eye-catching title?

Their assertion that “ The meaning of “nothing” is somewhat ambiguous here” is significant.  How can “nothing” be ambiguous?

If you drop a ball from rest (defined to be a state of zero energy), it gains energy of motion (kinetic energy) as it falls.

Surely, if you can drop the ball from rest, it must be in a position in which it already has GPE?  If you define this as “a state of zero energy”, this may be valid in terms of whatever theoretical point you are making, but has little relevance to “reality”

[Bill S]

It is negative because it is finite in one direction (in no gravity well) but infinite in the other.  There is no limit to how far you can fall, so there is nowhere to put the origin in that direction.

Let’s make sure I understand this before trying to go further. 

I would have expected: “ It is negative because it is finite in one direction (in [a] gravity well)”; because the source of the gravity is at a finite distance away.

I would see the “nowhere to put the origin” as being in the direction away from the gravitating body.

Where have I gone astray?

[yor_on (OP)]

I write at the weirdest times nowadays due to my operation, and me still on drugs.  Don't be surprised if I commend TNS on their exquisite choice of psychedelic colors :) And buy me a used Wolksvagen Buss.

Anyway, all of it seems to go out from treating gravity as a 'force field' of some sort. Furthermore it doesn't really treat it as a field, because a field must have a field strength, and 'gravity' is in my eyes presumed to exist even in a 'flat space', as you write Halc. And if we to that add that it is 'negative taking out, if I may say so, 'positive energy' which then must be everything of matter, and other 'fields', it doesn't add up to me?

Is it presumed to be some sort of 'anti field, if so, how?

[Halc]

I would have expected: “ It is negative because it is finite in one direction (in [a] gravity well)”; because the source of the gravity is at a finite distance away.

You can get a lot lower than being say at the center of Earth.  xkcd put out the best picture of it I've ever seen:
https://xkcd.com/681_large
That 'depth' is actually plotted in kilometers, at Earth gravity.

There's no limit in fact in the 'down' direction.  There is a limit on how far 'up' you can get.
From the most remote place I can think of, a rock falling to Earth's surface might achieve a speed of one or two thousand km/sec which corresponds to a depth of perhaps 100000 km at Earth gravity.  No more.  That's finite energy, sort of.  I say sort of because the most remote place anywhere is still in a gravity well.

Furthermore it doesn't really treat it as a field, because a field must have a field strength, and 'gravity' is in my eyes presumed to exist even in a 'flat space', as you write Halc.

Still a field, but a depth field, not a force field.  The latter (force) dictates how much an object placed at any given point will accelerate.  If the field is flat, that net force is zero and the object stays inertial.
A depth field on the other hand determines gravitational time dilation, not acceleration.

Um... Is this true?  Equivalence principle says gravity is equivalent to (locally indistinguishable from) acceleration, and there is no acceleration in a deep but flat field.  That's ok.  It's not like time dilation can be detected with a local test, and pure acceleration similarly does not cause time dilation, so I see no violations here.

[yor_on (OP)]

Hmm, and when you turn course in a 'flat space', from where would you define the inertia aka 'gravity' created in it Halc? Would you call it a relation to the 'non existent gravitational field' defined as 'flat space' or?

I also think you need to define what you mean by a 'depth field' as contrasted to a 'force field' there.
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Ahh okay. But I'm still not sure?
A Depth field would be what?

the way you find yourself 'time dilated'`?
How would you do that if so?

[yor_on (OP)]

Yes, gravity in Relativity (GR) is defined as equivalent a 'constant uniform acceleration'. And a time dilation is a relation, it needs you and your wristwatch to define other frames of references 'time dilation'. So even thinking about considering ones own 'frame of reference' 'time dilated' will invalidate any measurement made by you, unless you make 'Proper Time' a standard from where it can be done.

A Lorentz transformation is a proof of there being a logic when it comes to those, it can be balanced out. But what it doesn't is to give you a proof of what the 'real (universal) time' is. Unless 'Proper Time' is something more than just a local standard.
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Spelling

[Bill S]

You can get a lot lower than being say at the center of Earth.  xkcd put out the best picture of it I've ever seen:  https://xkcd.com/681_large 

Interesting link, thanks.

I can see that there can be, and are, gravity wells much deeper than Earth’s, but aren’t all gravity wells caused by the presence of massive bodies? 
Wouldn’t such a body place a physical limit on potential depth? 
Would a well of “infinite” depth require an infinite mass to cause it?

There is a limit on how far 'up' you can get.

I’m probably being a bit slow on the up-take here, but, physically, what sets the limit?
What would set a theoretical limit?

[Halc]

You can get a lot lower than being say at the center of Earth.  xkcd put out the best picture of it I've ever seen:  https://xkcd.com/681_large 

Interesting link, thanks.

I can see that there can be, and are, gravity wells much deeper than Earth’s, but aren’t all gravity wells caused by the presence of massive bodies? 
Wouldn’t such a body place a physical limit on potential depth?

No, they don't.  Earth for instance prevents further depth because the ground is in the way, but if you compress it to allow a smaller radius, the gravity well gets deeper than the graph shows.  There is no limit to that.  If Earth is compressed to a black hole, I will lose infinite potential energy as I fall to it, and gain infinite kinetic energy in the process.  Net zero.

Would a well of “infinite” depth require an infinite mass to cause it?

I did it with Earth just then, so no.

There is a limit on how far 'up' you can get.

I’m probably being a bit slow on the up-take here, but, physically, what sets the limit?

If you removed all mass from the universe (or moved it infinitely far away, such as what almost happens at heat death), you'd be at zero gravitational potential.  That's the limit.  You can't get further 'up' than that.  Your watch would run at 'actual time' if you acknowledge the meaning of such a thing, which I don't, but some do, like David Cooper.

The practical limit is much less than that:  Be at the most remote spot possible, which is way out between galactic superclusters.  That is as far 'up' as you can possibly be without doing impossible things like getting away from it all.

[yor_on (OP)]

My point was a perfect sphere of a perfect density distributed. Then drill a hole through its center. let a free falling observer 'bounce', he will end up in the middle of that sphere, finding himself 'weightless' as far as I know? How will that fit the idea of 'negative energy' taking out the positive energy?

gravity is always 'highest' or if you like, of a greater magnitude, when you're a bit away from that 'center' as another point of interest :)

So how does it work?

[Bill S]

No, they don't.  Earth for instance prevents further depth because the ground is in the way, but if you compress it to allow a smaller radius, the gravity well gets deeper than the graph shows.  There is no limit to that.

Makes sense, but isn’t stating that “there is no limit to that” just theoretical speculation?  Is there any physical demonstration of a mass, of any magnitude, being compressed infinitely? 
OK, that’s probably a silly question, but I wouldn’t have asked it without “provocation”. 

  If Earth is compressed to a black hole,

Wouldn’t you have to change the laws of physics to compress the Earth to form a black hole?  Undoubtedly, in can be done in theory, but physically (?)  I suspect not.

I will lose infinite potential energy as I fall to it, and gain infinite kinetic energy in the process.  Net zero.

Have you encountered the dreaded singularity here?  When your calculations include answers involving infinity, isn’t that a sign that there is probably something amiss, somewhere?

[Halc]

Makes sense, but isn’t stating that “there is no limit to that” just theoretical speculation?  Is there any physical demonstration of a mass, of any magnitude, being compressed infinitely? 

Don't need infinite compression.  Just enough to get the thing down to its Schwarzschild radius, which for Earth is about 9mm.

Wouldn’t you have to change the laws of physics to compress the Earth to form a black hole?  Undoubtedly, in can be done in theory, but physically (?)  I suspect not.

Pile on enough elephants until it occurs, then take away the elephants. You're going to have an awful time pulling the elephants out at the last moment.
Point is, you have to apply more force than the collective outward forces (mostly EM) resisting you.  That doesn't violate physics, it just makes it difficult and not likely to occur naturally.  Another way is to put a proton-size black hole somewhere and let it suck the rest of Earth in.  Some people worry that the LHC will produce such an object, dooming us all.

Have you encountered the dreaded singularity here?  When your calculations include answers involving infinity, isn’t that a sign that there is probably something amiss, somewhere?

It takes time to fall into a bottomless pit.  Some say that objects never have time to get to the bottom.  I have no singularity in my posts.  Just that for a given depth of well, an object can indeed fall that far.  There is no bottom at which a zero can be assigned.

My point was a perfect sphere of a perfect density distributed. Then drill a hole through its center. let a free falling observer 'bounce', he will end up in the middle of that sphere, finding himself 'weightless' as far as I know? How will that fit the idea of 'negative energy' taking out the positive energy?

gravity is always 'highest' or if you like, of a greater magnitude, when you're a bit away from that 'center' as another point of interest

Your 'weight' on Earth is the result of the vector sum of gravity from all the parts of Earth, which is zero at its center.  The depth of the gravity well is the scalar sum of the gravity from those same parts, which is at a maximum at the center of earth.

[Bill S]

Don't need infinite compression.  Just enough to get the thing down to its Schwarzschild radius, which for Earth is about 9mm.

  I will lose infinite potential energy as I fall to it, and gain infinite kinetic energy in the process.

I have a problem with losing and gaining infinite energy, and potentially falling eternally, in the finite space enclosed by a 9mm Schwarzschild radius.

Pile on enough elephants until it occurs, then take away the elephants. You're going to have an awful time pulling the elephants out at the last moment.
Point is, you have to apply more force than the collective outward forces (mostly EM) resisting you.  That doesn't violate physics, it just makes it difficult and not likely to occur naturally.

Perhaps, as a pragmatist, I’ll always have a problem with theoretical physics. 

[yor_on (OP)]

Maybe?  I'm weightless, that should mean that I'm in a geodesic. The definition of a geodesic is somewhere where no 'forces' are acting upon you. But, what you're saying there is that there are 'forces' acting upon me?

[yor_on (OP)]

" No local test for that.  I can tell by comparing my clock to a different one that is more or less dilated." I don't agree there, no way to define it as long as there is no universal standard of what a 'correct time rate' should be. The only thing you can define is the ratio of difference between your clock and another. You just need two other 'clocks' in your vicinity to see that, all in a relative motion.

"Proper time is what your wristwatch says, hardly a standard for anybody else." depends, different uniform motions do not change a repeatable experiment, but will present different time rates. One can define proper time as a equivalence of 'c', and as 'c' is defined as a constant so can you do with 'proper time'.

[yor_on (OP)]

Yep, the one where I equalize it to 'c' is my own, the one with a geodesic as something having no forces acting upon it isn't though.
http://curious.astro.cornell.edu/physics/140-physics/the-theory-of-relativity/general-relativity/1059-if-gravity-isn-t-a-force-how-does-it-accelerate-objects-advanced

[yor_on (OP)]

And furthermore, nobody have given me a good answer yet.
I'm still waiting for a answer, or I'm wasting my time.

[yor_on (OP)]

Maybe I need to clarify it.

There are two ways to define gravity, one is as a 'force' (Newton) the other is by geodesics where no 'forces' are involved. Just read the link above if someone wonder how that works. When talking about gravcity as a 'force' people have invented the concept of it also being a 'negative energy'. The wording of that makes for all kind of interpretations, from imagining negative meeting positive annihilating each other, to 'anti particles meeting particles'. In the later case the energy released is always 'positive', in the first case?

Be that as it will, but when talking about net sums of 'gravity' it to me sounds as treating it like a force, and when talking about it acting on that ball it definitely becomes a 'force'. But in GR it's not a 'force', and there is no negative energy associated with gravity that I know of. That's where the example comes from, and if one want to reconcile those two interpretations of 'gravity' there should be a way to describe that make sense from both. So, the gravity acting on that ball is strongest at the surface, in the middle it disappear. From one side this should mean that the 'negative energy' is null, from the other that the ball is in a geodesic. Or we have to invent a new definition for this, not a geodesic at all. I'm wondering how one reconcile those two, if one can't then it reminds me of the inability of reconciling quantum gravity with relativity. And that's all I'm gonna write about it.
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Maybe it is the wording of it that makes me irritated?

If we define a geodesic as being 'at rest', then something resting on a table isn't at rest with a geodesic, but with the table. Calling those two equivalent doesn't sit right with me? Seems like a unholy mix of ideas there. That as from the point of view of GR the ball on the table actually is 'accelerating' together with that table and Earth. So the only way it is at rest is when defining it as a 'system' where alll parts are at rest with each other. As soon as it drops though it becomes a geodesic, and it continue to be so the whole way down and when 'stopping' too, as far as I can see. But as we said the 'gravity' acting upon it, and it upon Earth, is highest as the surface, disappear in the middle, that's why it stops there.

So is that a geodesic too? I'm still wondering there, it should be but it's now also at rest with earth, following a general geodesic of Earth instead of the one it had before. It's tricky. Thinking some more I have to say that it definitely is a geodesic the ball has followed bouncing back and forth, and finally being at rest with earths geneal geodesic 'motion'.

But it doesn't answer how the 'negative energy' works here. As pointed out strongest at the surface, negligible in the middle.