Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: yor_on on 10/01/2019 16:42:22

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? :)

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

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

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 eyecatching 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”

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”
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.
That said, it is an interesting exercise to compute exactly how deep into a gravity well an object is when sitting on the surface of Earth. An object in a flat gravitational field (such as exists between galaxies??) is hardly a location with zero negative potential energy.

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?

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?

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.

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?

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

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?
Sorry, I cannot parse this.
I also think you need to define what you mean by a 'depth field' as contrasted to a 'force field' there.
The acceleration at any one point is the vector sum of all gravitational accelerations due to all masses acting on that point. I suppose it should then be called an acceleration field, not a force field. The force is the acceleration field multiplied by the mass of the object at that point, which has not really been defined.
The depth field is the same thing, but the scalar sum of all masses acting on that point.
the way you find yourself 'time dilated'`?
How would you do that if so?
No local test for that. I can tell by comparing my clock to a different one that is more or less dilated.

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.
Proper time is what your wristwatch says, hardly a standard for anybody else.
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.
GR does not posit 'universal time'. An absolutist does, but I notice that they all fail to compute it. You'd think they'd publish a value, but they don't because they're embarassed by it.

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 uptake here, but, physically, what sets the limit?
What would set a theoretical limit?

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 uptake 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.

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?

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.

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?

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 protonsize 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.

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. :)

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?

" 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'.

The definition of a geodesic is somewhere where no 'forces' are acting upon you.
One can define proper time as a equivalence of 'c',
Maybe I could stick with the physics definitions of these things. It hampers communication if you redefine all the words, especially to very different meanings.

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/140physics/thetheoryofrelativity/generalrelativity/1059ifgravityisntaforcehowdoesitaccelerateobjectsadvanced

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

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.

This is the sentence getting me stuck " But this gain is exactly balanced by a larger negative gravitational energy as it comes closer to Earth’s center, "
Well, with a greater 'negative energy' at the surface it doesn't ring right to me. Unless you ignore Relativity and just use Newton where 'gravity' is a force, but even then you need to explain how that 'gravitational 'negative' energy'' increase with the ball 'at rest' in the middle of Earth. And using Newton for it can't be right.

http://curious.astro.cornell.edu/physics/140physics/thetheoryofrelativity/generalrelativity/1059ifgravityisntaforcehowdoesitaccelerateobjectsadvanced
Undoubtedly there is some good stuff in this article, I intend looking when I have time. However, I find the opening assertion offputting:
Einstein said there is no such thing as a gravitational force.
Lots of other people have said it, but not, I understand, Einstein.

You have to put a distinction between the theory and what Einstein wanted Bill. EM is a 'force' and Einstein wanted relativity to become one too. But he failed to incorporate it, and so have everybody else.
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Maybe it's better to say he wanted it to be a field, thinking of it.

That you see it used as a 'force' doesn't imply that relativity agree. We want it to be fields, so we interpret it that way. But a geodesic is something without 'forces' acting on it according to the way I see it, and as I think, relativity. We see the apple fall and we call it a force, it 'accelerates' with time. But not in relativity, not from a 'black box' perspective, in that one you're unable to find any acceleration, and you're 'weightless'.
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Let's see what more we do, ok?
We give it a 'potential energy' that doesn't exist inside that black box.
We call it (gravity) a 'negative energy' as it works against the 'fields' we've already found to exist.
I'm sure I can think up more examples of how we treat it.
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And the whole point of this thought exercise is why I started to differ between what I call 'global' versus 'local' interpretations. It seems I'm in a minority doing so, but I still think it's as important as your 'frame of reference'.

That you see it used as a 'force' doesn't imply that relativity agree.
There are certainly experts who are adamant that gravity is a force. I’ve quoted Dr Baird before: “Yes, gravity is a force. But it is a force that is more completely described by spacetime curvature and not Newton's law.”
Are you saying this is wrong?
Just a thought from one whose maths/geometry is abysmal:
A geodesic is defined as the most direct route in curved spacetime. Thus, a geodesic seems to have a lot in common with a straight line. Could it be that a straight line in Euclidian geometry is equivalent to a curved line in the nonEuclidian geometry of curved spaces?
Would it follow from that that, given an appropriate change in geometry, no force, and therefore no expenditure of energy, is required to alter the course of an object?

Yes, I think that might be a third interpretation, don't know how many that use it as it contain a contradiction in terms. It joins forces with gravity without defining how. https://en.wikipedia.org/wiki/Force
" In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newtons and represented by the symbol F. "
'gravity' neither push on you or pull in a geodesic, and there are a untold amount of geodesics in one single patch of space depending on uniform motion (relative speeds) and mass. It's called 'curved SpaceTime' but there are no 'forces' curving it, in contrast to f.ex a stream that you force to change direction. A acceleration is a force, uniform motion is not.

Well, with a greater 'negative energy' at the surface it doesn't ring right to me. Unless you ignore Relativity and just use Newton where 'gravity' is a force, but even then you need to explain how that 'gravitational 'negative' energy'' increase with the ball 'at rest' in the middle of Earth. And using Newton for it can't be right.
At rest just means it has no kinetic energy. At the center of earth, the maximum energy is needed to get away from Earth. Escape velocity is about 11.2 km/s at the surface of Earth, but greater from its center since you need more energy just to get the object to the surface.

A geodesic is defined as the most direct route in curved spacetime. Thus, a geodesic seems to have a lot in common with a straight line. Could it be that a straight line in Euclidian geometry is equivalent to a curved line in the nonEuclidian geometry of curved spaces?
I cannot explain this coherently either. I fire a bullet at a target, or lob a softball. Both get from me to the target by paths in curved spacetime, but they take different paths. How can both follow a deodesic? I think the answer is that both do not get to the same target event since if the meshooting/throwing is the same event, the arrival event at the target is not.
Geodesic BTW describes the trajectories that airliners take to get from A to B on the nonEuclidean surface of Earth, which appear to be curved lines on a flat map, but in fact can be determined by pulling a string from A to B on a globe. Similarly, a geodesic path in spacetime is such a straight line.
Would it follow from that that, given an appropriate change in geometry, no force, and therefore no expenditure of energy, is required to alter the course of an object?
How do you expect geometry to change without force being involved? As it is, geometry in spacetime is fixed. Geometry in space might not be, but in spacetime, time is part of the geometry, so there is no 'changing' it.
If you're talking about space, then paths are not straight. Force is involved, and can be changed by rearranging the objects that exert gravity.

We see the apple fall and we call it a force, it 'accelerates' with time. But not in relativity, not from a 'black box' perspective, in that one you're unable to find any acceleration, and you're 'weightless'.
This is wrong. Gravity (without acceleration) is equivalent to acceleration (without gravity), so the apple falls in both black boxes, and in a straight line to boot. Nothing is weightless in those two cases. That occurs in freefall, with or without gravity, which are indistinguishable from the box.

No Halc, you need to read up on relativity. A black box inside a gravitational potential /'field' has a observer dependent 'gravity', just like that apple 'accelerating'. That's why we differ between a accelerations and a 'gravitational acceleration', in the first case it will be unequivocal acceleration to all observers, the second is observer dependent. I really wish you started checking your sources.

A black box inside a gravitational potential /'field' has a observer dependent 'gravity', just like that apple 'accelerating'. That's why we differ between a accelerations and a 'gravitational acceleration', in the first case it will be unequivocal acceleration to all observers, the second is observer dependent.
You need to read up on relativity. Equivalence principle is not about outside (nonlocal) observers of the box. It is about the one local observer inside the box. The principle says the later cannot tell the difference. I really wish you started checking your sources.

It's called a transformation, you 'transform' the 'gravitational field' by choosing frame of reference. Inside the box (ignoring spin) you're 'weightless' following a geodesic. From a observer on f.ex Earth you're in a gravitational field and also gravitationally accelerating even if ever so slowly. So from inside that box there are no 'forces' acting upon you, although from the observer on Earth it will seem like it.
" In Einstein's theory of general relativity, gravitation is an attribute of curved spacetime instead of being due to a force propagated between bodies. In Einstein's theory, masses distort spacetime in their vicinity, and other particles move in trajectories determined by the geometry of spacetime. The gravitational force is a fictitious force. There is no gravitational acceleration, in that the proper acceleration and hence fouracceleration of objects in free fall are zero. Rather than undergoing an acceleration, objects in free fall travel along straight lines (geodesics) on the curved spacetime. " https://en.wikipedia.org/wiki/Gravitational_acceleration

Ok, the question still stands?
How should I think of this 'negative energy' at the middle of that sphere. From the balls point of view it's in a geodesic, no forces acting upon it. And from the idea on a 'gravitational field'' then?
If I use weight as a measure then the surface has a stronger 'field' and so more 'negative energy'.
So?
Anyone that has an idea here?
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You know, this 'negative energy' of gravity seems to me pretty weird. It's not the same as f.ex Dirac sea.
" The origins of the Dirac sea lie in the energy spectrum of the Dirac equation, an extension of the Schrödinger equation that is consistent with special relativity, that Dirac had formulated in 1928. Although the equation was extremely successful in describing electron dynamics, it possesses a rather peculiar feature: for each quantum state possessing a positive energy E, there is a corresponding state with energy E. This is not a big difficulty when an isolated electron is considered, because its energy is conserved and negativeenergy electrons may be left out. However, difficulties arise when effects of the electromagnetic field are considered, because a positiveenergy electron would be able to shed energy by continuously emitting photons, a process that could continue without limit as the electron descends into lower and lower energy states. Real electrons clearly do not behave in this way.
Dirac's solution to this was to turn to the Pauli exclusion principle. Electrons are fermions, and obey the exclusion principle, which means that no two electrons can share a single energy state within an atom. Dirac hypothesized that what we think of as the "vacuum" is actually the state in which all the negativeenergy states are filled, and none of the positiveenergy states. Therefore, if we want to introduce a single electron we would have to put it in a positiveenergy state, as all the negativeenergy states are occupied. Furthermore, even if the electron loses energy by emitting photons it would be forbidden from dropping below zero energy. " https://en.wikipedia.org/wiki/Dirac_sea
Then again, you have this idea of being able to fall past the event horizon, if we assume no tidal forces acting upon you. But that is a geodesic? So, yeah it's weird.