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How does a 'field' become observer dependent?
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How does a 'field' become observer dependent?
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yor_on
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Re: How does a 'field' become observer dependent?
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Reply #60 on:
29/10/2013 20:35:47 »
Scales becomes something its own right there. In one way able to use as we normally use it, 'magnifying and shrinking', but also becoming an idea in where it is our limits of observation that defines what a scale means. As seen from the 'inside' of a 'universe', scales must exist. From the 'eye of a God', or 'outside' any such definition, scales should become an approximation of what a dimensional system contain, but a meaningless concept when described over 'it all'.
If it didn't presume dimensions, a canvas would be a good concept for describing it all.
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Re: How does a 'field' become observer dependent?
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Reply #61 on:
31/10/2013 21:41:02 »
Let's consider it from temperature.
A temperature is 'vibrations' by matter. Motion of some sort. A 'perfect vacuum' is the absolute absence of matter at some geometrically defined 'space'. Does that 'space' exist? What about Heisenberg's uncertainty principle (HUP) defining it such as you can't know both a particle's position and its momentum, exactly. Using that on a perfect vacuum, will that create particles? No, it won't.
"Virtual Particles
In many decays and annihilations, a particle decays into a very high-energy force-carrier particle, which almost immediately decays into low-energy particle. These high-energy, short-lived particles are virtual particles.
The conservation of energy seems to be violated by the apparent existence of these very energetic particles for a very short time. However, according to the above principle, if the time of a process is exceedingly short, then the uncertainty in energy can be very large. Thus, due to the Heisenberg Uncertainty principle, these high-energy force-carrier particles may exist if they are short lived. In a sense, they escape reality's notice.
The bottom line is that energy is conserved. The energy of the initial decaying particle and the final decay products is equal. The virtual particles exist for such a short time that they can never be observed."
You need interactions by 'real particles' to define 'virtual particles'. But let us consider a vacuum as a 'fluid' then? Not good to me, it presumes that a vacuum has properties relating it to matter. Ever seen the statement that a vacuum can do ftl? If it can it must be a fluid, and of a extremely strange sort, as it then goes against relativity's statement that nothing surpass the speed of light, in a 'perfect vacuum'. In fact invalidating relativity, as we now have defined a 'absolute nothing' as possible to define a position in/to it, excepting any and all matter. Because that is what it state. That you can define a point in a universe consisting of a perfect vacuum, then follow its 'propagation' through that same vacuum, to find it 'move', and also move 'ftl'?
A temperature is matters motion, or interactions. So does a vacuum have a temperature? Not a perfect vacuum, unless you introduce mass, which is what you must do, to measure any possible temperature. You could imagine that a vacuum contain 'properties of matter' without creating it, unless you present it with real matter, such matter measurable over time as existing.
A 'motion' of matter is measurable displacements over time. A virtual particle is a displacement, not measurable over time, as I see it.
A perfect vacuum can not, in and by itself, be defined and constricted to displacements over time. All geometrical definitions of points need real matter, as your anchors, from where you define a motion of 'something', be it a 'hole' or a 'particle'.
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Last Edit: 31/10/2013 21:43:41 by yor_on
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Re: How does a 'field' become observer dependent?
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Reply #62 on:
31/10/2013 21:56:53 »
Maybe you can shorten this to the question: Does a perfect vacuum have properties, except geometrically? Can you scale a vacuum down? If you would, do you expect to reach some ultimate scale from where it can't be scaled down further? If you want 'bits', and if you want some common smallest nominator scale wise for both particles and 'vacuum', where do you expect it to exist? Planck scale?
Does a vacuum need to 'exist'? Matter exist, measurably so. Bosons and fermions exist, measurably so, a vacuum though? From a point of a universe, as a four dimensional container of it all, measuring from an 'inside' you might want to consider particles of all kinds as dancing on a 'energy' of the vacuum, defined through a arrow.
From a point of particles 'properties' creating the four dimensional universe we see, and measure on though? What would that vacuum become then?
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Re: How does a 'field' become observer dependent?
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Reply #63 on:
31/10/2013 22:15:59 »
So, what is 'energy'
From a point of 'properties' defining a universe, creating dimensions, energy becomes understandable to me. It's fermions and bosons properties, defining 'my universe', observer dependently, although taking observer dependencies further than what we normally means. From a point of a 'cosmic container' it has no simple answer to me. You're free to define it as only consisting of 'one container' or 'several containers' co-existing, alternatively 'splitting' (many worlds) etc etc. And what the two definitions also differs in is the way they treat a arrow. The first one assumes a arrow as a property, equivalent to 'c', existing in all 'points'. The second one?
The first one defines a universe by measuring, locally.
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Re: How does a 'field' become observer dependent?
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Reply #64 on:
31/10/2013 22:20:57 »
And constants then?
That should be the properties, shouldn't it?
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Re: How does a 'field' become observer dependent?
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Reply #65 on:
31/10/2013 22:35:08 »
What I mean saying "And what the two definitions also differs in is the way they treat a arrow. The first one assumes a arrow as a property, equivalent to 'c', existing in all 'points'. The second one?" Is that from the first definition a arrow must be what creates a measurable universe. We live inside a arrow of time, defining us and our measurements.
What you can't measure on won't exist for you. That's your 'container' of sorts, but not as seen from the normal point of view of a universal container, as that 'commonly shared universe' we think we observe. This universe is defined by what you can measure, possibly infinitely 'co-existing' with other definitions and properties we don't know how to, and possibly never can, measure. It takes on another aspect to me, 'co-existing', if you think of it as defined by measurements. And scales, as well as all other properties we might find a universe to consist of, then becomes a result of those properties. We becomes one emergence, of many.
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Re: How does a 'field' become observer dependent?
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Reply #66 on:
31/10/2013 22:37:24 »
What it does is to simplify things for me, and I like it as simple as I can get it.
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Re: How does a 'field' become observer dependent?
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Reply #67 on:
31/10/2013 22:45:25 »
Then again, if 'energy' is bosons and fermions only, what would those other possibilities be? The ones 'co-existing' with us, although not measurable? I guess I would call that 'energy' too? That we can't measure energy by itself, only measure its transformations, supports that view, I think? 'Energy' is still a slippery thing, although as defined from inside a dimensional system perfectly simple, following the principles, constants, properties, defining transformations, which is what we measure.
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Re: How does a 'field' become observer dependent?
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Reply #68 on:
31/10/2013 22:57:54 »
'c', and that equivalent, locally measured arrow? At what ultimate scale could I expect to pinpoint them? Light is without a 'dimension', can you see how that could be, from a view in where those 'particles' create those dimensions? Maybe 'dimensions' is just our construct.
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Re: How does a 'field' become observer dependent?
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Reply #69 on:
31/10/2013 23:11:25 »
Because a scale needs dimensions, just as a fish needs water, ahem. One gets defined by the other so to speak
If you are able to define a system, using whatever properties it have to decide a dimensionality, creating interactions following a arrow of time, defining degrees of freedom, then you have made a universe, and you should be able to scale it.
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Re: How does a 'field' become observer dependent?
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Reply #70 on:
31/10/2013 23:22:40 »
And such a universe explains Lorentz contractions better to me, because those are observer dependent, and there is no 'commonly same universe' as some fishbowl containing us all. What 'contains us' are shared principles, constants, and properties. And the same must then go for that complementary time dilation, as what we really share, isn't what we measure on, as that rocket at that black hole, 'standing still' to us, forever. What we share is 'locality'.
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Re: How does a 'field' become observer dependent?
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Reply #71 on:
31/10/2013 23:26:22 »
Not 'locality' as in 'action and reaction'. That's a classically valid definition, but not the one I'm referring to. I'm referring to the way we measure, always locally. I'm referring to shared principles, constants and properties. Those are in a sense not 'observer dependent', although they will define 'observer dependencies', as your locally defined arrow, and 'c'.
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Re: How does a 'field' become observer dependent?
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Reply #72 on:
31/10/2013 23:56:28 »
So what about Hawking radiation, defined as a virtual particle pair created at the event horizon of a black hole, one part wandering inside annihilating some other particle inside it, the other then materializing inside a arrow, outside the event horizon, as needed by conservation of energy, creating that Hawking radiation?
Where was the matter involved there? If my definition of the procedure now is correct? It came about spontaneously, from a virtual pair production, didn't it? Doesn't that state that a vacuum must exist, as a real entity in itself?
Don't know, maybe? Or maybe you could define it as where particles 'energy' gets excited, due to properties as relative mass/ relative motion accelerations, rest mass gets created? All of it assuming that Hawking radiation exist. You can find a similar result in a cloud chamber, creating more particles than what went into the collision.
It's not that a vacuum isn't there, it's what it should consist of.
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Re: How does a 'field' become observer dependent?
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Reply #73 on:
01/11/2013 00:47:54 »
The mass of a black hole has a proportion to its geometry, or better expressed, it's when the proportion breaks down you get to a black hole.
"The Chandrasekhar limit (named after Subrahmanyan Chandrasekhar) is the maximum nonrotating mass which can be supported against gravitational collapse by electron degeneracy pressure. It is commonly given as being about 1.4 or 1.44 solar masses. Electron degeneracy pressure is a quantum-mechanical effect arising from the Pauli exclusion principle. Since electrons are fermions, no two electrons can be in the same state, so not all electrons can be in the minimum-energy level. Rather, electrons must occupy a band of energy levels.
Compression of the electron gas increases the number of electrons in a given volume and raises the maximum energy level in the occupied band. Therefore, the energy of the electrons will increase upon compression, so pressure must be exerted on the electron gas to compress it. This is the origin of electron degeneracy pressure."
Passing this you get a black hole, defined by all paths leading to the same place passing its event horizon, and it is at that event horizon we have this virtual particle formation. So there is an abundance of mass, creating those virtual particles, or/and depending on definition, energy. And mass defines what vacuum, or 'space' you will find existing measuring.
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Re: How does a 'field' become observer dependent?
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Reply #74 on:
01/11/2013 00:58:13 »
We get our forms from the 'relation' between mass and vacuum, don't we? Without a space no form. So? Energy?
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Re: How does a 'field' become observer dependent?
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Reply #75 on:
01/11/2013 09:23:37 »
It's quite nice, isn't it? Just consider how things connect to things, consider 'forces'. The forces we talk about, at a small scale, create a form macroscopically. It's not about some universal container inside where we find mass and radiation, it's about connections, communication, and how it can create a measurable universe. We 'talk' with that universe constantly, through those forces, defining ourselves relative it, with the universe defining itself relative us. And we connect in so many ways simultaneously, communication existing as an abstract reality, as with speech.
And yes, I think simultaneously exist, locally defined
The problem being defining what locally should mean there. Is it a question of some smallest scale? Or is it locally shared properties, principles and constants? Would you say a constant have a scale?
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Re: How does a 'field' become observer dependent?
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Reply #76 on:
01/11/2013 10:22:23 »
There is a problem with it, there always is. The question if what we see as forces must be what defines everything? And by everything I mean all possible connections, those that are measurable as well as those that might not be measurable. Assuming that quantum computer can exist, assuming that it reach a outcome 'instantly', would you say this involve a arrow? Logically some questions would take too long to answer normally, or be just impossible, but we do expect a quantum computer to be able to crack them. As asymmetric encryption.
So, did it 'think' about it? Did it 'test' for all possibilities?
What is 'energy'? Is it only those forces we find that defines it? Simplest solution is to define it as what we measure is what exist.
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Re: How does a 'field' become observer dependent?
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Reply #77 on:
01/11/2013 10:31:49 »
But if I define a arrow to 'c', then you might be able to argue that a quantum computer set this principle aside, as it reaches its outcome 'instantly'. If that is so, then I'm either wrong in finding the arrow equivalent to 'c' (locally defined) or it should be a indication of there existing possibilities outside our definition of a linearly timed universe, following different principles.
What would 'energy' be without a arrow?
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Re: How does a 'field' become observer dependent?
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Reply #78 on:
08/11/2013 17:34:58 »
And then we have 'c' itself. The speed of light in a vacuum, as locally measured, is the same everywhere. Doesn't matter what velocity you may define a object too, a light beam leaving that object should be at 'c', as measured from your frame, or any other non accelerating frame. And that light has a momentum is not synonymous with it having a rest mass, there exist no 'rest frame' for light in where you can measure a mass.
And 'c builds on two assumptions, one is a arrow of time existing, locally equivalent everywhere. The other is our definition of a distance, as locally measured. The really interesting thing with both local time and a distance are that, no matter time dilations and Lorentz contractions, they implicitly are assumed to have equivalent values locally measured. And, assuming they aren't, we can't prove any repeatable experiments anymore, unless you have a way to superimpose their 'frames of reference' upon each other, them becoming exact replicas, as 'one and the same'. And that really mean they have to be the exact same, inseparable from each other.
So we build science, and repeatable experiment, on a arrow of time, locally equivalent for all of us. You want to retract that arrow, you also retract our definition of a repeatable experiment
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Re: How does a 'field' become observer dependent?
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Reply #79 on:
08/11/2013 17:55:13 »
A 'virtual particle' is a 'possible particle'. Its definition rest on HUP (Heisenberg's Uncertainty Principle). The idea is that as its actual 'existence' is so short, it finds itself able to express a wide variance of energy levels (time versus energy, or, position versus momentum). What the duration should be for such a particle I don't know, but presuming its existence it should be of interest to define it. One Planck time is the time it takes for light to propagate one Plank length, so maybe you could find it around there?
The interesting idea here is the assumption that we can reach a 'end' of our arrow, scale wise. We scale it down and as we do another vista opens itself, free from our constraints. You can also ignore the idea of virtual particles for it I think, instead using indeterminacy for pointing out a same phenomena.
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