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Author Topic: How does a 'field' become observer dependent?  (Read 201450 times)

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?
 

Offline yor_on

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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.
 

Offline yor_on

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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.
 

Offline yor_on

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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.
 

Offline yor_on

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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.
 

Offline yor_on

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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.
 

Offline yor_on

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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." "
 

Offline yor_on

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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?
 

Offline yor_on

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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 »
 

Offline 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.
 

Offline yor_on

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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.."
 

Offline yor_on

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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?
 

Offline yor_on

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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."
 

Offline yor_on

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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.
 

Offline yor_on

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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?
 

Offline yor_on

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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?
 

Offline yor_on

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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)
 

Offline yor_on

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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 #524 on: 13/01/2014 18:10:43 »

 

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