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Author Topic: How does a 'field' become observer dependent?  (Read 200531 times)

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #150 on: 16/12/2013 15:19:08 »
We call it geometry. But what make it allowed to exist?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #151 on: 16/12/2013 19:52:14 »
What is probability?

How does a electron become a probability? Is a electron itself 'isolated' existing at all?
Does the moon exist when you're not looking?

Relations defines it. Well, as my assumption for this :)
Think of the universe we observe as a description of probabilities, relations defining those probabilities. The moon don't care if you look, it's you that do that :)

And that becomes a geometry, relations defining a reality. The real question, and the one I'm truly confused about, is in what way one frame can communicate with another? In my universe that is :) A lot of stuff is easy to explain thinking of a universe this way. A particle becoming a wave, becoming a particle, depending on relations, in this case meaning your experiment for measuring. And entanglements? Well, we have a situation in where we have a limit for communication, 'c'. But the entanglement in itself? the idea of a instant 'spooky action at a distance'. Depends on how you look at it, what was it I suggested at a microscopic level? That it was no use trying to define a position, if defining it such as a arrow disappear there? There are no 'positions' to be found. You might also want to consider, a superposition maybe?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #152 on: 16/12/2013 19:56:25 »
A superposition, without a geometry? That's pretty weird.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #153 on: 16/12/2013 19:58:22 »
We look at QM, the really, really, 'small', through our macroscopic definitions. Our local clock, decoherence, ruler, etc etc.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #154 on: 16/12/2013 20:03:41 »
And the 'really small' ignore our geometry. Your frame of reference 'force you' to define it from 'dimensions', doesn't it :) You entangle two particles, place them at different positions, measure one, then the other, finding them to have a 'opposite relation', falling out the same way every time you repeat it (ideally this is, entanglements are hard to set up practically as I understand).
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #155 on: 17/12/2013 19:16:30 »
Why not think of it in terms of relations? A entanglement craves a 'setup' before you can get to it. Fulfilling the setup correctly should give you a high probability of it falling out. That should then either mean that most particles aren't entangled, or that we fail to see how they are.
 

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Re: How does a 'field' become observer dependent?
« Reply #156 on: 18/12/2013 11:02:04 »
How far can one take a entanglement?

Let's assume that at a 'Big Bang' everything should be entangled. What happens after that? As particles bounce each other? Do they find new 'entanglements'? Whatever they do, do you expect them to keep the original entanglement? A Big Bang was a lot of energy, wasn't it? In a geometric 'point', that somehow became a lot of 'points', assuming a inflation faster than light. How can we assume a 'ftl', without defining it such as there is a origin geometrically?

We can't.
 

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Re: How does a 'field' become observer dependent?
« Reply #157 on: 18/12/2013 11:03:48 »
Can you see how our archetypes constantly come into play, creating riddles for us. If there was no 'origin', then there was no 'ftl' either.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #158 on: 18/12/2013 19:15:39 »
So what am I suggesting? Some perpendicular plane to our universe that act on it, giving us a illusion of propagation? I don't think so, using such a analogue ftl exist, even if not in the plane we exist and observe. And this one also goes back to how a point communicate with another, if I now got it right, that way creating our dimensions. I have good reasons to prefer a universe defined locally, but defining 'dimensions' from such an idea is trickier. Either one assume a geometry free from locality, but that's not true. If you trust relativity, Lorentz contractions must exist, and they are observer dependent. If they are you can't sponsor a geometry isolated from the observers. Or you have to find a way to describe a geometry from whatever change its description. That would then, as I see it, be all types of motion, mass, and that undefinable quality 'energy'. What's good with defining it from a observer is that you don't need to argue what is more 'real', what you measure at rest with earth relative what you would measure moving relativistically. But 'dimensions' won't be the same after such a change, neither will what makes them. The universe in such a description don't care of your mass, if we just use relativistic motion for now. One gram or one tonne, it doesn't matter for the frame of reference moving relativistically. If it measures it should see the same contraction, loosely defined. And as all measurements it does tells it is true, then, from 'localitys' point of view, it is true.
=

To see where I'm going with this comparison just translate mass into 'energy'. Doesn't matter what 'energy' you have, or spend. To get to a relativistic motion, (well, we know it does, mass, relativistic and restmass, do have a role but we're ignoring that fact for now) one gram 'energy' or one tonne, the contraction of your universe should be the same (loosely defined). So a good question here is to ask yourself how this ever can be true?
« Last Edit: 18/12/2013 19:45:43 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #159 on: 22/12/2013 03:20:18 »
For me it comes back to a observer dependent geometry, defined by 'c'. Assume :) that we're living in a projection, it's not a new idea. Do I then need 'dimensions' from it to start with? Holography needs it, but that we can prove to be a illusion. What does a universe need? You need some sort of 'space' for those points to interact in, creating a universe, but if the points themselves create the connections, giving us the dimensions we define? Do those 'points' need to be separated?
=

It's also a question of 'energy'. Assume that we have a equilibrium, assume that the only thing that happens is transformations, of a constant unchanging magnitude of 'energy' existing as a universe. Isn't that a 'free lunch'? Nothing gets lost, it just transforms.
==

Then look at a inflation, and a 'accelerating' expansion of a vacuum. Is that a transformation too? Where from? Either you define a 'outside' of some type for this, or you define it from a 'inside'. If you do the last you need to take 'something', to deliver something new. As it all should be about a equilibrium of 'energy', transforming from one state to another.
==

Or it's a projection, in which case our definitions probably hold true anyway. The 'equilibrium of energy' transforming. But I don't think it to define a 'inside' anymore, if that would be the case, although it still will be/seem so to us, measuring.
« Last Edit: 22/12/2013 03:36:13 by yor_on »
 

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Re: How does a 'field' become observer dependent?
« Reply #160 on: 22/12/2013 03:40:46 »
The way we measure a 'loss', is using a arrow. We define useful energy relative non useful. The non useful energy is defined as 'heat'. Heat is radiation, interacting with matter.
 

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Re: How does a 'field' become observer dependent?
« Reply #161 on: 22/12/2013 03:50:08 »
For heat to be non useful it has to stop being heat, wouldn't you agree? If it interacts, how can I define it as not being 'useful'? Assume a universe of radiation, will that too be in a equilibrium? If it will, where do we get to a density? Can I assume 'hot spots' in such a universe? giving a restmass? If it does, do it transform? A Big Bang then? Would you define a constricted geometry for that? A pinpoint? Won't work with a definition of a 'instant inflation', in where there is no defined origin. If you on the other hand define a origin, a pin point, then? Well, then astronomy should be all wrong.
=

I'm usually referring to a point, becoming a lot of points during a inflation. And that can't be true, as it both assumes a arrow, as well as it assumes a 'speed'. I better stop defining it that way.
« Last Edit: 22/12/2013 03:59:58 by yor_on »
 

Offline CPT ArkAngel

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Re: How does a 'field' become observer dependent?
« Reply #162 on: 22/12/2013 07:13:10 »
A Big Bang then? Would you define a constricted geometry for that? A pinpoint? Won't work with a definition of a 'instant inflation', in where there is no defined origin. If you on the other hand define a origin, a pin point, then? Well, then astronomy should be all wrong.

Inflation is not based on any real physical process. It has absolutely no basis, no root. It is a floating model which you can modify almost as you wish.

Einstein himself said that GR is not a finished theory. Einstein searched for a better theory until his death.

Before Hubble's observations, the universe was thought to be extended to our own galaxy alone. Now, we think it is much larger but we really don't know how large it is. Most physicists think there is only one big bang, and this big bang is the creation of the universe...

If the big bang occurred in a larger universe, then, its origin must have coordinates. In no way it disproves astronomy, just the cosmological model. The distances measured are based on the standard candle. As long as the standard candle used is reliable, you have a reliable distance, unless there is a major flaw.

You must take care more about the data than the interpretations because the interpretations are based on the actual cosmological model. They are filtered by it...

Our universe is not well known beyond a few billion light years. When we look at a galaxy at 10 billion ly, you don't see the entire galaxy so it is quite difficult to know what type it is. Interpretation will be biased.

Dark energy doesn't fit in the actual cosmological model. This model has been created more than 80 years ago, based on Hubble's observations. Recently, Planck's data showed some anisotropy, which is very unexpected. Dark energy has not been measured in the southern hemisphere (there is a current mission on this). The true cosmological model may mimic the actual model to great distances. If inertia is mediated by photons, after some time, the matter of our big bang will be uniformly distributed, especially if it is older than 13 bly.

Our visible universe is not in equilibrium because entropy grows. Matter is searching equilibrium...

To keep the current theories, we need more free parameters. It is a symptom that something is wrong.

http://physicsworld.com/cws/article/news/2013/dec/04/mystery-of-neutron-lifetime-discrepancy-deepens
http://pirsa.org/displayFlash.php?id=13080001
« Last Edit: 22/12/2013 08:46:57 by CPT ArkAngel »
 

Offline CPT ArkAngel

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Re: How does a 'field' become observer dependent?
« Reply #163 on: 22/12/2013 07:21:41 »
If there is a center to our big bang, transverse velocity will affect the formation of galaxies. The further in time you look and the further you look from the center, the younger galaxies will look. It will take more time for galaxies to form where initial transverse velocities are greater.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #164 on: 22/12/2013 22:58:19 »
Hmm. I use it, and find it explaining some really weird stuff CPT, looking out on a universe. You can check out https://en.wikipedia.org/wiki/Cosmic_inflation#Observational_status for some of the things it explains. But you're perfectly correct in that it is a theory, or hypothesis. Based on my own thoughts it fits a place of sorts, where what's 'stable' comes from locality, 'points' interacting, creating and defining dimensions to us. I have no real problems with it so far.
=

A standard candle is defined as "celestial objects with well-defined absolute magnitudes which are assumed to not vary with age or distance. Type I and II Cepheids and RR Lyraes are all examples". Stars who we assume to be of the same energy output, aka radiation, no matter where we find them.

"To help understand what is meant by standard candle, we first need to have a basic understanding of how distances are measured in astronomy. For small distances such as from the earth to the moon, lasers are used. Moving further out to Mercury, Venus or Mars, we use radar. Leaving our solar system and measuring to nearby stars, we use semi-annual parallax. And out to 500 parsecs (pc), spacecraft (e.g. Hipparcos) are used with measurements computed trigonometrically.

We refer to these as direct methods of measurement.

At distances greater than 500 pc, the error in the parallax measurement is too great and not usable. Indirect methods are used based on stellar properties such as luminosity,radii, the effective temperature and others. Distances are determined from relationships connecting these properties, including the period-luminosity relation for Cepheid variables. [Illingworth & Clark 2000]

While it is difficult to find a ‘pure’ definition for STANDARD CANDLE, reliable sources provide enough information to define it as saying there is no single object used for a Standard Candle; there are collections of stellar objects with known luminosities that allow them to be used to determine distances. The Standard Candle object used depends on the distance being measured. The brightest Cepheids, for example, can be seen out to about 60 megaparsecs (Mpc). For distances of 150 and 250 Mpc, red and blue supergiants can be used, respectively. For distances even greater, a galaxy’s HII region or brightness of its globular clusters are used. Beyond 900 Mpc, astronomers rely on supernovae. In all measurements, as the distance increases, the accuracy decreases. [Kaufmann & Freedman 1999, Illingworth & Clark 2000]"

From STANDARD CANDLES by RONALD E. MICKLE Denver, Colorado 80005
=

And one more, for those getting stuck on where parsecs come from. http://csep10.phys.utk.edu/astr162/lect/distances/units.html

Light years are a easier definition to us laymen :)
« Last Edit: 22/12/2013 23:22:28 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #165 on: 22/12/2013 23:32:28 »
One thing I do get a little thoughtful about though is the cosmological redshift, due to a accelerating expansion. Not because it redshifts, as that to me is a question of the detector relative what it detects, relations defining it, but when comparing it to a gravitational redshift? Very similar, aren't they? And that one is then also described in 'relative motion'(s). So three types (causes), same effect.

We don't know what gravity is, do we?
And motion is weird too :)
=

you can actually get to four causes, if you split motion into two parts, uniform motion also called 'relative motion', versus acceleration/deceleration.  Although if relative motion indeed is purely relative, you then have to define that from a 'black box' (local) perspective. In a 'commonly same container universe' relative motion can't be relative, 'globally' defined we can prove different uniform motions, as soon as we involve more than two objects, measuring their 'speeds', using a local clock and ruler.
=

Or five? A constant uniform acceleration, equivalent to a gravity, and so different from all other types of accelerations/decelerations :) Depends on how strict you want to be, right?
Heh..
==

Then there is one more thing that I'm not sure about. All of those effects are, to me, connected to how we define a geometry relative 'anchors', the detectors, matter. Can't see any other way to detect?
« Last Edit: 23/12/2013 00:04:26 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #166 on: 23/12/2013 00:13:43 »
I can walk around a globe finding stars all around me. If I was in a plane (two dimensional) then? I could still walk in a circle though, but what about 'stars'? I need at least three dimensions to find stars, or can I define it otherwise? Doesn't matter if I use anchors for this, does it? To see stars around me, everywhere, I should need three dimensions. It's not holography, the universe, or if it is? Then all dimensions should be able to be questioned. I like a projection better.
=

I'm wrong :)
There is no way to define anything if you don't have anywhere to go out from. So where ever I am, whatever dimensionality I define, I still need a 'detector', and that is matter. And imagining myself fitting into a one or two dimensional reality I should still be able to define a 'above' and a 'under', shouldn't I :) As a dot, on a line.
« Last Edit: 23/12/2013 00:21:27 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #167 on: 23/12/2013 00:29:23 »
The question is what defines a two dimensional, or one, reality? Either we have to assume that looking in some direction nothing exist (assuming myself existing 'inside' it) or you will find a seamless reality of 'stars', all around you, in all measurable directions. When referring to two dimensional objects, existing inside a three dimensional reality, the logic breaks down. You can prove if the object is two dimensional geometrically quite easily, as it then should 'disappear' from some angle, but from a 'inside' I don't find it that easy.
==

"You can prove if the object is two dimensional geometrically quite easily, as it then should 'disappear' from some angle, but from a 'inside' I don't find it that easy." can be broken down two ways.

Either we define three room dimensions, in which case you can prove logically how a two dimensional object should behave in it, 'disappearing'. That would then prove the idea of singular dimensions, 'knitted together' with our arrow of time.

Or we don't, we just define a 'inside'.
« Last Edit: 23/12/2013 00:36:19 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #168 on: 23/12/2013 00:40:08 »
A 'inside' is what can be proven, the first example does not exist, as far as I know? We can refer to 'forces' behaving two dimensionally, as in a lattice, but we can't prove it geometrically, by matter disappearing.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #169 on: 23/12/2013 01:07:44 »
That, in its own turn, breaks down to two assumptions for me. Either 'dimensions' aren't singular, or there is no way for a lower dimensional object to exist in a higher dimensionality. I prefer the first.
 

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Re: How does a 'field' become observer dependent?
« Reply #170 on: 23/12/2013 01:21:19 »
Then again, what would this 'inside' be, if it isn't defined by singular dimensions, 'knitted together' with a arrow? Tough one isn't it? A projection is the way I think of it, as a hologram but not the same. A projections dimensions are defined, and will exist in measurements, to me. It's not about one or two 'dimensions' being a illusion. A 'illusion' shouldn't be measurable in my thoughts, although I admit it's possible to argue about this one, a lot :) But let us define it as the difference between a mirage and shooting a bullet on that same mirage, finding it not to be stopped. And using this definition a Lorentz contraction is real, and so is all time dilations. But they can only exist with you comparing locally, over frames of reference.
 

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Re: How does a 'field' become observer dependent?
« Reply #171 on: 23/12/2013 01:35:14 »
Let's go back to the idea of one gram or one tonne making no difference for the contraction you observe of your universe, you moving relativistically relative incoming light. The energy spent reaching that velocity will differ between those two, but it still won't make any better sense, if you compare it to the energy needed for moving a whole universe, closer to you.

But if it is relations that defines it, parameters consisting of your velocity (speed), mass and 'energy', as defined 'inside' a universe then? You can scale down any position, imaginatively.

When you do, do you expect those positions, at some scale, to become equivalent?
I do.
=

I mean that, exchanging positions for 'points', there must be a equivalence locally defined, for/in all points. Also that what we define as positions stops making sense at that scale.
« Last Edit: 23/12/2013 01:38:12 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #172 on: 23/12/2013 01:41:15 »
How would we get to a Lorentz transformation if this wasn't true?
 

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Re: How does a 'field' become observer dependent?
« Reply #173 on: 23/12/2013 01:43:04 »
You can see it as a opposite definition of the idea that the laws of physics are the same, locally measured, anywhere, in a universe. Same thing really.
 

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Re: How does a 'field' become observer dependent?
« Reply #174 on: 23/12/2013 04:24:22 »
Btw.

As it is xmas, well almost xmas anyway :)
And as we're on a physics site.
And as I liked it.

http://www.strangehorizons.com/2003/20031222/december.shtml
 

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Re: How does a 'field' become observer dependent?
« Reply #174 on: 23/12/2013 04:24:22 »

 

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