How does a 'field' become observer dependent?

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Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #500 on: 13/01/2014 13:03:59 »
That would make consciousness into something holographic, wouldn't it?
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Re: How does a 'field' become observer dependent?
« Reply #501 on: 13/01/2014 13:14:03 »
It should leave with two definitions of information. One that is meaningful, obeying 'c' representative (with 'forces') for the universe we can measure on. The other going in a unmeasurable direction, also defined by 'c' but now as 'gestalts' or instants of patterns. Then we find new 'holographic' structures, or projections created by our local definition of that local ground beat. But, how would it allow inertia to become gravity? I need that ground beat, don't I? If I want gravity to be a result of 'c', but I also need a way of making it fit different mass. I'll ignore 'energy for it, for now.
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Re: How does a 'field' become observer dependent?
« Reply #502 on: 13/01/2014 13:25:20 »
The ground beat should represent Inertia, don't you agree? Giving us a constant, sort of. So what would mass represent from such an idea? mass is acting 'side way'. All interactions that we would be able to measure should act such, in this weird universe.
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Re: How does a 'field' become observer dependent?
« Reply #503 on: 13/01/2014 14:23:21 »
Hmm, it's not a unmeasurable direction. It's the local arrow we all share. But treated as sheets it becomes a sheet of its own, a 'dimension' if you like, called time. What's new to me is the way I decided to define the sheet as a pattern of 'instants'. Then defining it such as on this other 'measurable sheet' we have only one direction to measure in, sideways, following the sheet. The other sheet is not measurable in the same way, although it gives us a locally definable direction in time. It's old ideas I've had, it's just that I look at it differently now, as I formulated the question differently.
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Re: How does a 'field' become observer dependent?
« Reply #504 on: 13/01/2014 14:30:57 »
What I then would use for gravity is the constant update by 'c', of the local 'sheet'. Without the direction representing a 'ground beat' we are left with a local property I call inertia. And thinking of each sheet as a static pattern, gravity becomes what defines that pattern.
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Re: How does a 'field' become observer dependent?
« Reply #505 on: 13/01/2014 14:34:10 »
So we come back to mass. That's what defines gravity's measurable difference's. How would it do it? Assuming the sheet to be defined by local constants proper mass becomes a new parameter.
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Re: How does a 'field' become observer dependent?
« Reply #506 on: 13/01/2014 15:31:05 »
You have the Higgs field, to me representing inertia too, as it discus it as a result of forces acting upon you in a acceleration.

"The Higgs field applies only to the electro-weak sector of the Standard Model. The mass of ordinary matter is overwhelmingly due to the protons and neutrons in the nuclei of atoms. Protons and neutrons are comprised of the two lightest quarks: the up and down quarks. The rest masses of their constituent quarks (approx. 0.005 and 0.010 GeV/c2 for the up and down quarks respectively) which could be attributed to the Higgs field comprise only about one percent of the masses of the protons and neutrons (0.938 and 0.940 GeV/c2 respectively). The remainder of the proton and neutron masses would have to be attributed to contributions from the gluon field strong interaction energies plus smaller electromagnetic and weak fields contributions which would not be affected by a Higgs field.

The origin of inertial mass of ordinary matter is thus a wide open question."

" "Some particles interact with the Higgs field more than others,  which is why the particles in the Standard Model all have different  masses. For light particles such as electrons and neutrinos, traveling through the Higgs field is like running down the street. Heavier particles, such as the electron’s larger cousins, muon and tau, experience more resistance, as though they were running in a  swimming pool full of water. For the top quark, which is by far the heaviest particle in the Standard Model, traveling through the  Higgs field might feel like wading through a vat of molasses.....

A very important detail is that the speed of light in a transparent material is slightly different for each wavelength (i.e., momentum of the photons). For instance, considering visible light in water: So “yellow” photons travel through water faster than blue, and red even faster. We could say that blue photons have more problems to move in water than yellow and red. In this way, the blue photons act like as if they had more “inertia”, i.e., more “mass”. Refractive index gives a measure of the interaction between photons and a material medium through which they travel, but, somehow, it could be also considered an “index of mass”, since the bigger the value the smaller the speed of the photons.

Therefore, in vacuum all the photons travel with identical speed, but if the Universe were filled with water photons corresponding to different wavelengths would travel with different speeds. As it has been said before, they would have “different masses”. So we would be passing from a symmetrical situation to a non-symmetrical one. This is what in Particle Physics is called symmetry breaking phenomenon.

ow we are ready to establish our comparison. Initially, all the particles would be travelling through an “empty” Universe with the maximum speed permitted. So they would all be massless, and from this point of view the Universe would be symmetric. That is what SM originally states. But obviously the Universe does not work in this way.

The current SM suggests that all the particles had no mass just after the Big Bang, but as the Universe cooled and the temperature fell below a critical value, an invisible field called the ‘Higgs field’ appeared filling all the space. We could also say that the Higgs field was born in the begining of the Universe, but it only showed its influence once the Universe cooled down enough.

Unlike magnetic or gravitational fields, which vary from place to place, the Higgs field is exactly the same everywhere. What varies is how the different fundamental particles interact with it. That interaction is what gives particles mass. Of course, other kinds of interaction, such as the electromagnetic, weak or strong interaction may contribute significantly to the resulting mass. Moreover, the degree of resistance of the Higgs field is different depending on the fundamental particle, and this generates, e.g., the difference in mass between an electron and a quark.

Now, suppose a quark or electron moving (making up composite particles such as proton, neutron, or various atoms) in this uniform Higgs field. If these atoms (or molecules) change their velocities, that is, if they accelerate, then the Higgs field is supposed to be exerting a certain amount of resistance or drag, and that is the origin of the inertial mass." "
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Re: How does a 'field' become observer dependent?
« Reply #507 on: 13/01/2014 15:34:53 »
Weird isn't it. Higgs presented as a solution to mass? Let's see, 'energy' is mass, that's Einstein. Higgs = 'acceleration is 'inertial mass' :) Ok, what about the rest of the mass then, and what about uniform motions and proper mass?
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Re: How does a 'field' become observer dependent?
« Reply #508 on: 13/01/2014 15:57:05 »
If I instead define inertia as a initial property then? Using a arrow to define it as gravity? Then I need to introduce the magnitude of a displacement as one variable, mass as another. Should we give gravity several definitions? Higgs does it, Einsteins equivalence principle does too actually, differently though, and if we want to be strict. Because the equivalence principle is about comparing proper mass in a uniform motion to what a accelerometer defines in a constant uniform acceleration, finding those experiences to be equivalent, ignoring tidal forces (spin).

But I don't think Einstein looked at it that way. Generally speaking I would expect him to have considered all forms of acceleration and deceleration to represent 'gravity', although, only constant uniform acceleration becoming the one giving us a equivalent phenomena to a Earthly gravity. It seems rather reasonable to assume all sorts of accelerations to give us a same equivalence. The difference between the one being constant and the one varying is how it act in time. You know, the time that doesn't exist :) Freezing a instant of a acceleration should still give you a 'gravity', preferably for my needs as defined from mass, displacement (magnitude of acceleration) and some inertial constant, preexisting.
« Last Edit: 13/01/2014 16:03:43 by yor_on »
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Re: How does a 'field' become observer dependent?
« Reply #509 on: 13/01/2014 16:00:05 »
Can't use magnitude of 'motion' for that one, can I? :) As Einstein defined it we only have one type of acceleration, but I will chance on all types.
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Re: How does a 'field' become observer dependent?
« Reply #510 on: 13/01/2014 16:11:01 »
How would you define the energy for one rocket, accelerating at one constant gravity? let it be three of different proper mass, all accelerating at one constant uniform G.
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Re: How does a 'field' become observer dependent?
« Reply #511 on: 13/01/2014 17:02:55 »
They are not 'identical' as they have different mass, and so they must expend different amounts of fuel, and that's expending energy. Do you expect there to be a proportionality then? You do, right? And so do I. We have three proper mass, three different amount of spent fuel, and a same 'gravity'. If you expect there to be a proportionality you also expect there to be a logic. So what is the logic to their 'gravity' being the same?

Does their mass have a relation to their local definition of one G? It should, shouldn't it?

Do you expect the vacuum to treat them differently, depending on mass.
It shouldn't? I know no experiments proving a vacuum to act differently on different proper mass.

But they have different relativistic masses. Where, or what, would you define that relativistic mass too?
Their acceleration? Does their relativistic mass disappear in a uniform motion then, no acceleration?

It doesn't. Or? Would you like to define it as a acceleration being a local definition, experimentally measurable, with all uniform motions becoming locally undifferentiated?

So in a acceleration we have something locally measurable, in a uniform motion it disappear?

"As an object's speed approaches the speed of light from an observer's point of view, its relativistic mass increases thereby making it more and more difficult to accelerate it from within the observer's frame of reference.

The energy content of an object at rest with mass m equals mc2. Conservation of energy implies that, in any reaction, a decrease of the sum of the masses of particles must be accompanied by an increase in kinetic energies of the particles after the reaction. Similarly, the mass of an object can be increased by taking in kinetic energies.."
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Re: How does a 'field' become observer dependent?
« Reply #512 on: 13/01/2014 17:08:36 »
That one shot a big hole in that balloon, didn't it? What that says is that a relativistic mass should be defined relative a velocity, or speed. As I read it, it also states that it doesn't matter for this if we define that motion as accelerating, or as a uniform motion. Or do you read it differently? Do you expect there to be a difference if you stop accelerating at some point, uniformly coasting for a while, enjoying the sights, to then start accelerating again?
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Re: How does a 'field' become observer dependent?
« Reply #513 on: 13/01/2014 17:10:11 »
So where do you store that relativistic mass in a uniform motion?
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Re: How does a 'field' become observer dependent?
« Reply #514 on: 13/01/2014 17:18:44 »
"Mass–energy equivalence is a consequence of special relativity. The energy and momentum, which are separate in Newtonian mechanics, form a four-vector in relativity, and this relates the time component (the energy) to the space components (the momentum) in a nontrivial way. For an object at rest, the energy–momentum four-vector is (E, 0, 0, 0): it has a time component which is the energy, and three space components which are zero. By changing frames with a Lorentz transformation in the x direction with a small value of the velocity v, the energy momentum four-vector becomes (E, Ev/c2, 0, 0). The momentum is equal to the energy multiplied by the velocity divided by c2. As such, the Newtonian mass of an object, which is the ratio of the momentum to the velocity for slow velocities, is equal to E/c2."
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Re: How does a 'field' become observer dependent?
« Reply #515 on: 13/01/2014 17:24:19 »
Assuming the vacuum to be 'neutral', not acting on the rocket in any measurable manner as a 'resistance'. Also as all uniform motions imply in a flat space where your geodesic never ends, no matter what speed you define to it.
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Re: How does a 'field' become observer dependent?
« Reply #516 on: 13/01/2014 17:28:28 »
Let us look at the rocket, if you have a light bulb in it, will that be brighter the faster you go? Will the rocket hull start to glow in a perfect vacuum as you accelerate? Well, it will meet light from stars won't it? And depending on your velocity those 'beams' will blue shift into the gamma sector. But if we turn of those suns then?

Will there be a locally stored energy measurable in your rocket?
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Re: How does a 'field' become observer dependent?
« Reply #517 on: 13/01/2014 17:37:14 »
We agree on the vacuum behaving the same, measurably, don't we? To do otherwise would be to give a vacuum a resistance. And I hope we agree on that there will be no measurable changes locally, proving a 'stored energy' locally measured.

Using two inertial observers then? At rest with your rockets origin (Earth). One placed at the fore of your motion/acceleration, the other inertial observer placed at the stern, of your rockets motion, will we find the 'apparent' blue and red shift they measure to cancel out when compared?
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Re: How does a 'field' become observer dependent?
« Reply #518 on: 13/01/2014 17:40:30 »
Where is the relativistic book keeping stored?
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Re: How does a 'field' become observer dependent?
« Reply #519 on: 13/01/2014 17:43:41 »
In all accelerations you have ever-growing displacements. So using a light bulb situated in the middle of the rocket, two observers, fore and stern inside the rocket, they will be able to measure a blue respective red shift. They will also find a gravity, situated outside the stern. (constant uniform acceleration now)
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Re: How does a 'field' become observer dependent?
« Reply #520 on: 13/01/2014 17:52:58 »
Not in space, measurably so. Not in the rocket, at least not as defined from a uniform motion. And even though we find a blue/red shift inside that accelerating rocket you can move that light bulb anywhere you want inside it, to find a same effect. And remember that the relativistic mass is a definition relative a velocity/speed. Doesn't matter if you stop the acceleration for an hour, no resistance to a geodesic, no loss of velocity/speed in a perfect vacuum. The relativistic mass as you start to displace the rocket doesn't become less just because you stopped accelerating for a little while.

If it was so that you could cheat the relativistic book keeping that way, you should be able to reach (close to) 'c' expending a lot less 'energy' than when constantly uniformly accelerating.
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Re: How does a 'field' become observer dependent?
« Reply #521 on: 13/01/2014 17:56:25 »
So, where is it stored?
In time?
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Re: How does a 'field' become observer dependent?
« Reply #522 on: 13/01/2014 17:59:08 »
Relativistic mass must exist, all collisions are defined from velocities (speeds are without defined directions) and mass.
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Re: How does a 'field' become observer dependent?
« Reply #523 on: 13/01/2014 18:05:03 »
So what does relative motion mean, in what manner are they all equal?
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Re: How does a 'field' become observer dependent?
« Reply #524 on: 13/01/2014 18:10:43 »
They are equal locally measured. As for example relative a local scale that you stand on. You do not gain weight by doubling Earths uniform motion. And that weight scale and you are then loosely defined as being 'at rest' with Earth. Using distant suns the incoming light, meeting Earths motion, will become measurably blue shifted though.
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Re: How does a 'field' become observer dependent?
« Reply #525 on: 13/01/2014 18:24:39 »
In a comparison between two frames of reference. You on Earth measuring some comets uniform motion for example, you are free to state the comet to be standing still, or Earth, standing still, it's a equivalence of sorts in uniform motions. If one of you was accelerating through the vacuum though, there would be a locally measurable change, for the one accelerating.
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Re: How does a 'field' become observer dependent?
« Reply #526 on: 13/01/2014 18:27:05 »
So Earth is constantly uniformly accelerating at one Gravity. But you can double its uniform motion, without adding to a gravity.
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Re: How does a 'field' become observer dependent?
« Reply #527 on: 13/01/2014 18:30:27 »
Gravity is not relativistic mass, that we can see. If it was, then doubling Earths uniform motion should change your weight on that scale. But uniform motion must contain a relativistic mass, otherwise you can cheat the relativistic book keeping.
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Re: How does a 'field' become observer dependent?
« Reply #528 on: 13/01/2014 20:00:24 »
So how does the Higgs work?

It uses an idea of densities I would say. It defines different attachments or properties to different 'stable particles', then defines something we can't measure on, except by inductive logic, giving a probability of it existing, when measuring interactions at very high energies. It's ghosts passing you through acting on some of your particles, but only as you accelerate, giving you a inertial reaction equivalent to a gravity, if we then use the equivalence principle to define it.

in a way not so different from my own idea of inertia becoming gravity, but to me it presumes a container universe. You could argue that if observer dependencies are real, and if we exist, then this problem is no problem as we do exist and consist of densities, particles, fermions and bosons. I think this is what the Higgs theory will argue, finally backed into a corner. That is not sufficient, and it is not physics. New physics should answer old questions, and give new questions to ask. It seems for example that the Higgs boson is though to change the chirality, "Having a handedness or helicity, not having mirror symmetry" of particles, lefthandedness to righthandednes, and back, each time they interact. But this seems not to be true with neutrinos, that has been found to have a mass. So where does this mass come from?

So what is observer dependencies from a Higgs field'?
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Re: How does a 'field' become observer dependent?
« Reply #529 on: 13/01/2014 20:09:55 »
I can transform away the observer dependencies :) By stipulating that this universe always is locally defined. And to do so I just erase the container model. I do not stipulate any defined dimensions, more than using a sheet for describing the fractals I think a universe should be made of. Observer dependencies are very hard to understand from the idea of a defined four dimensional container, that also is observer dependent. If you instead assume connections, relations, creating a universe then all relations you ever will measure on will be defined by you, locally. And instead of dimensions we have the degrees of freedom we find something to have, which are four macroscopically.
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Re: How does a 'field' become observer dependent?
« Reply #530 on: 16/01/2014 22:58:37 »
Was discussing the idea that when you measure a entanglement you also inject it with a energy that must be duplicated at the receiving end. I said that either you then have to assume the energy to get split in two,(1 becoming .5) or you need to lend the far end of the entanglements 'energy', preferably then from a vacuum. Assuming entanglements to exist spontaneously, or as utilized by us, this should lead to a vacuum getting 'depleted of energy'.

Then again, assuming my own ideas of scaling a entanglement should be special relation, allowing it to presented at two 'points' in a positional SpaceTime, both having 1, without any 'lending' being involved in it, if it is correct.
=

Spelling and words :)

btw. That would then mean that you can get 'free energy', if the injection is found to work? So maybe I'm wrong with that one? Although you might also see it as them being the same photon, doubly represented? The arrow is after all a local definition (defined over frames of reference). Ouch, don't know how to think of that one :)
« Last Edit: 16/01/2014 23:14:30 by yor_on »
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Re: How does a 'field' become observer dependent?
« Reply #531 on: 16/01/2014 23:18:58 »
The important point (Well, I think?:) there is if it is correct assuming that you can 'inject' a annihilating photon with energy, just because it's entangled? Can you?
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Re: How does a 'field' become observer dependent?
« Reply #532 on: 16/01/2014 23:23:45 »
Whatever you inject it with should 'propagate' at 'c', as it seems to me? For example, photons should be able to annihilate in a sun, and assuming them able to get entangled in there this should lead to a excess of energy, either 'lent' from a vacuum, or 'split the injection' in two.  I think we can rule out a 'injection' as lending from a vacuum at the 'far end', don't you?
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Re: How does a 'field' become observer dependent?
« Reply #533 on: 16/01/2014 23:30:44 »
Maybe the first question should be.

Can the sun act as a beam splitter?
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Re: How does a 'field' become observer dependent?
« Reply #534 on: 17/01/2014 08:23:00 »
I don't see how a sun can avoid entanglements, for both photons/waves and electrons? And if you now assume that a annihilation inject a energy at both sides we get to a proposition in where the sun lends energy from a vacuum.
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Re: How does a 'field' become observer dependent?
« Reply #535 on: 17/01/2014 23:04:59 »
I know, this has very little to do with relativity, but if you imagine relativity scaled down to meet QM, which it properly done should be able to then I guess everything has a relevance relative relativity, I'm very pleased with the last sentence there, by the way.. Quietly imposing, sort of :)

And it has bugged for quite some while, the idea of a injection in a entanglement, and still does btw, it's all assumptions I make, can't go into the sun and present a experiment proving it one way or another.
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Re: How does a 'field' become observer dependent?
« Reply #536 on: 17/01/2014 23:14:00 »
On the other hand, I look at a entanglement as something 'whole' in both ends, and then make it fit my proposition from scales, that you will lose a arrow, locally as you get down there. The first problem with my proposition is that you, to measure on it, use a local clock and ruler. As well as we have this fuzzyness and HUP scaling down. Then again, if relativity is an idea of frames of reference interacting, defined by a local clock and ruler. Then QM is an idea of a local definition also being possible, best expressed through the expectation of quanta. If quanta (qbits etc) exist, then they should be a very local definition, even though you are involved too, with that local clock and ruler.
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Re: How does a 'field' become observer dependent?
« Reply #537 on: 17/01/2014 23:23:26 »
Then we come to the original definition of a locality, that one is you throwing a stone in a pond watching the rings spread and interact. I don't think of it in that way, although I do :) The stone you throw in, is you scaling something down, the pond that stone scaling down meet should lose its arrow. And if there is no arrow (always strictly locally defined, remember) then our ideas of motion distance etc, disappear. And there is one more difference introduced through this idea, it allows for a entanglement. So you have our macroscopic reality, and 'under it' a microscopic ending in something unmeasurable. Someone wrote that Black Holes was 'censured' by the cosmos, I would say the same for scaling, you should meet infinites there too.
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Re: How does a 'field' become observer dependent?
« Reply #538 on: 17/01/2014 23:25:50 »
And that makes constants the most interesting idea you can have. What is a constant, and what is not?
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Re: How does a 'field' become observer dependent?
« Reply #539 on: 17/01/2014 23:34:20 »
Because that is what you have left. Constants, properties, and principles, creating a macroscopic universe. And it is made from a local definition of a ideal clock and ruler, everything else defined relative it. And that ideally defined clock and ruler must become a constant in 'my universe' :) as well as in Einsteins relativity. If I give 'c' a equivalence to the clock, then I will do the same for the ruler. And I need to lock it down somehow, that's Plank scale to me.
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Re: How does a 'field' become observer dependent?
« Reply #540 on: 17/01/2014 23:43:37 »
That's also why I hold out on defining what a vacuum really, really, is. Energy exist, we see it transformations, although on its own it becomes non measurable, as with objects in uniform motion. There is no extra measurable energy locally in those objects, it's all about relations between frames of reference creating the energy you find in incoming light for example. And a vacuum 'on its own' does not make for a definition of dimensions, and neither do I think it will give degrees of freedom. You need mass.
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Re: How does a 'field' become observer dependent?
« Reply #541 on: 17/01/2014 23:50:05 »
A vacuum is a form-able thing in my thoughts, proper mass giving us the dimensions we define. And it goes back to the idea of relations, locally defined by you, defining a universe. Just as a inflation and expansion has no defined 'origin' so it should be with 'locality', it's as valid everywhere, and now you can define that 'sidereal universe' you act on, that acts on you.
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Re: How does a 'field' become observer dependent?
« Reply #542 on: 17/01/2014 23:53:18 »
So dimensions are questionable things to me. Degrees of freedom is not, they are exactly what they are defined as, the degrees of freedom you find something to have, measured locally. So, to me, they becomes a better definition of how a universe acts than dimensions, because you can apply that same point of view at any scale.
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Re: How does a 'field' become observer dependent?
« Reply #543 on: 17/01/2014 23:58:54 »
And then a idea of inertia fits in so well, because a inertia is a unwillingness of motion, a 'resistance' to motion. If you are able to accept the idea of Earth gravitationally accelerating, as locally measured by a accelerator (scale). then you can split it in inertia and a arrow, giving you 'gravity'.
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Re: How does a 'field' become observer dependent?
« Reply #544 on: 18/01/2014 00:01:17 »
That makes Inertia into a constant too, I think? You could define it as a property of mass, but the way I think of it, I think it should behave as a constant, like 'c'.
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Re: How does a 'field' become observer dependent?
« Reply #545 on: 18/01/2014 00:06:16 »
And it seems correct if you assume that we are continuously connected to that scaled down 'time less' reality. That's also what I mean by assuming that we always are as close, (loosely, and locally, defined now) to Planck scale. Doesn't matter where you are.
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Re: How does a 'field' become observer dependent?
« Reply #546 on: 18/01/2014 00:07:44 »
And down there the rules change.
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Re: How does a 'field' become observer dependent?
« Reply #547 on: 18/01/2014 00:13:14 »
So gravity and the arrow disappear, but 'time' and 'inertia' should be able to exist, as constants.
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Re: How does a 'field' become observer dependent?
« Reply #548 on: 18/01/2014 00:19:08 »
The arrow and the ruler, would then be a phenomena similar to the idea of decoherence. Always locally defined, and locally equivalent everywhere. The arrow and the ruler being equivalent to 'c', which we then define to 'propagate' at a set speed in a vacuum, equivalent for all frames of reference. At least as I see it.
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Re: How does a 'field' become observer dependent?
« Reply #549 on: 18/01/2014 13:01:50 »
Let us do it this way. You have two things. 'Yin and yang' :)

The soft and the hard principle in life, as one way I think the Chinese thought of it. You can translate it to a cosmos too, although it becomes rather confusing to define which is which. You can also translate it into a description of frames of reference, acknowledging that relativity is described through two frames, yours, relative what you observe, comparing.

Not that hard to get to, is it? :)  You don't really need to use that idea though, as long as you acknowledge that we are not the first humans on this planet, wondering about it. The ancient Greeks did too, probably we can find that most all have thought and wondered about it, at some time.

So you have a vacuum, and proper mass. Fermions and bosons. Two objects, a sun and a planet interacting. Depending on their relative motion giving a planetary observer different definitions of that suns rays 'energy'.

The energy is correct, but where is it contained? It differs, you can measure on that, but is it contained in the vacuum? That would be a rather weird proposition, as I can add how many suns I like, in different relative motion versus Earth. Depending on what sun you measure on you know will find that this vacuum then should have different 'energy values'. Do you think that is true?

So where is that energy stored?
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