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 #150 on: 16/12/2013 15:19:08 »
We call it geometry. But what make it allowed to exist?
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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?
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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.
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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.
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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).
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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.
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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 »
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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.
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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 »
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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 »

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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.

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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 »
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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 »
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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 »
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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 »
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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.
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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 »
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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 #175 on: 23/12/2013 19:35:09 »
Let's talk about multiverses :)

I have one, inside, here. You can expand that into several. I can't see anything stopping you from that. I don't have anything limiting the amount of 'points'. What limit our observations are 'c'. I can define a arrow, for each point, but I also have to accept 'time dilations', comparing between frames of reference.
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Re: How does a 'field' become observer dependent?
« Reply #176 on: 23/12/2013 19:40:30 »
There is one thing though. You have to accept a 'inflation', and, it has nothing to do with time. At least that's my presumption.
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Re: How does a 'field' become observer dependent?
« Reply #177 on: 23/12/2013 19:43:03 »
You see, either I can argue that everything is a result of a arrow, or I define a arrow to 'relations', creating dimensions, and inflation. I'm using scales.
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Re: How does a 'field' become observer dependent?
« Reply #178 on: 23/12/2013 19:47:41 »
I define it as 'relations' creating a arrow now. But it might also be so that it is possible to define 'relations' through a arrow. If you really want 'multi dimensionality' then this should be worth thinking about.
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Re: How does a 'field' become observer dependent?
« Reply #179 on: 23/12/2013 19:52:36 »
Then I have this vague idea about time. Time is not a arrow, it's a origin. I think of it as a property for now. Arguing that local arrows define universes, separated by 'c':s  limits though? Do I need a property of 'time', isn't the arrow everything there is?
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Re: How does a 'field' become observer dependent?
« Reply #180 on: 23/12/2013 19:55:31 »
Although, defining a arrow as equivalent for all points, can you get to different constants, for different areas? Doesn't matter if 'c' sets a limit for communication here. Neither does it matter if you measure something to have a different arrow from your local definitions, or contractions.
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Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #181 on: 23/12/2013 19:58:39 »
If you want to change constants, I would expect you to have to prove different arrows too. Because in Relativity you can't separate the room from the time.
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Re: How does a 'field' become observer dependent?
« Reply #182 on: 25/12/2013 15:35:11 »
Been thinking about what happens as you scale something down. Does gravity disappear? I define it for now as a 'down welling' in each point, putting points together, interacting, creating effects measurable 'sideways' in a universe, sloppy writing but I think you can get my drift. Because gravity is directed inwards, toward some 'center'.

So, does 'gravity' disappear? Or is it just unmeasurable? If it is unmeasurable, then it becomes a property, to me. If it really disappear then? It's about a vacuum too. And how to think of it. As a something? Consisting of points too, containing all properties I would define to something of matter? 'Energy'? And the the arrow, did inflation take a time?
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Re: How does a 'field' become observer dependent?
« Reply #183 on: 25/12/2013 15:48:34 »
There are some consequences worth mentioning, taking a approach in where dimensions gets created by (local) interactions. It doesn't really matter, as far as I see it now that is, if the universe is deemed 'flat', saddle formed, or as a ball, or any other thought up possibilities. It's an illusion. Real to us, but there are no definable limits to any shape, more than what properties you give each point. Although? What properties would you need to give us a octagonal universe? :) whatever that would be?
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Re: How does a 'field' become observer dependent?
« Reply #184 on: 27/12/2013 20:07:15 »
So what is a gravitational acceleration then?  Something seeking a equilibrium? And is it mass that creates it? Think of it, 'gravity' has in reality one direction, inwards to a 'center', of sorts. Doesn't matter if you define gravitational 'fields bending' a SpaceTime. There's only one direction for gravity, one sign to it.
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Re: How does a 'field' become observer dependent?
« Reply #185 on: 27/12/2013 20:28:00 »
Actually, there's no 'up' or 'down' to a spacetime coordinate :) It's us defining it (directions), and we usually do it relative gravity.
=

If you like, I see this as allowing me to define any gravitational direction as 'down', or as I more prefer, towards a 'center'. Doesn't matter what frame of reference I use defining a 'gravity' from, for this.
« Last Edit: 27/12/2013 20:33:39 by yor_on »
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Re: How does a 'field' become observer dependent?
« Reply #186 on: 27/12/2013 20:55:28 »
I can transform away a gravity, coming into a 'free fall'. I can't transform away my own mass though, and that mass has its own gravitational direction. All 'directions' becoming one, if you think of it as me. Toward a 'center', and that 'center' is best represented by scaling.
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Offline CPT ArkAngel

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Re: How does a 'field' become observer dependent?
« Reply #187 on: 27/12/2013 21:33:42 »
Effectively, gravity reduced distances between two bodies. You can see it in terms of velocities too.

Follow Mach's Principle. Consider that there is no spacetime without matter. You must understand that Einstein introduced the concept of spacetime because he didn't have a valid description of matter, including the source of inertia and gravity. Spacetime replaces matter in GR, if you follow Mach. There is spacetime without matter in GR!

According to the Equivalence Principle, inertia and gravity are indistinguishable. Inertia pushes and spacetime pushes.
Maybe there are two forces or maybe two distinct interactions.

1- Two forces: Gravity attracts through the interaction of the graviton and inertial push is local but relative to the universe.

2- Two distinct interactions: Gravity is mediated somehow by the inertia of photons and inertia is mediated by photons locally.

If you throw out the entity of spacetime, photons must have a gravitational mass. If photons travel really at the speed of light, they can only have a transverse gravitational mass. It explains too many things, like why electrons and positrons have a gravitational mass, what is the Higgs boson... Unless the positron has a negative mass, which disagree with GR anyway.

The Casimir experiment gives a vacuum energy that is 10^120 order of magnitude higher than the observation of Dark energy. Maybe the Casimir effect is actually attraction between the plates...

« Last Edit: 27/12/2013 21:35:20 by CPT ArkAngel »

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Re: How does a 'field' become observer dependent?
« Reply #188 on: 28/12/2013 00:02:32 »
Well, I don't know what gravity is CPT :) It's mostly when I've made it into 'my own', things starts to make sense to me. I'm just defining it as something directed toward a 'center' for now. Using gravitons, combining that with such an idea doesn't work for me at this time? A little like I might agree on a Higgs field describing inertia, but not, representing 'mass'.
« Last Edit: 28/12/2013 00:07:43 by yor_on »
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Re: How does a 'field' become observer dependent?
« Reply #189 on: 28/12/2013 00:23:18 »
Think of SpaceTime as a place in where all gravity points to a 'center'. In terms of positions inside a 'containing universe' those 'centers' points everywhere, no defined preferred direction. it is related to mass, locally defined accelerations, 'energy'. From a point of scaling something down though it won't matter :) what I define as their directions 'globally', inside that container universe. From that point gravity have one same direction, toward a 'center', as I see it.
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Re: How does a 'field' become observer dependent?
« Reply #190 on: 28/12/2013 00:38:20 »
From a definition of using 'matter' as an anchor for gravity, I can ignore space for this idea. A pure vacuum can not become gravity's origin experimentally. If I first define a 'container universe', which somehow also seem to define limits, I can use a gravitational field, existing in a vacuum. But I can't define the vacuum as a origin of gravity although I can use 'energy', but still not experimentally prove a gravity to origin from a perfect vacuum. Even when assuming a vacuum to exist as a 'energy', defined by 'virtual particles', space should be neutral. I don't expect one patch of perfect vacuum to differ from another, 'energy wise'.

So I'll stay with matter for this, for now. And doing so the only preferred direction for a gravitational field, for now, is its 'inward direction'. And that one is a local description of gravity.
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Re: How does a 'field' become observer dependent?
« Reply #191 on: 28/12/2013 01:03:16 »
Then again, I can't ignore motion, can I? And motion presumes a space to move in. Motion of matter gives relative mass, easily defined in accelerations/decelerations. But it must be existent in uniform motion too. Motion is one mass moving relative, either another mass, or as locally defined from infalling lights blueshift/redshift. And we can easily prove that there must be different 'relative motions', aka uniform motions (velocities/speeds).
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Re: How does a 'field' become observer dependent?
« Reply #192 on: 28/12/2013 01:10:06 »
But, assume yourself to get close to light speed, moving relativistically. As you accelerate you will find a new gravity, locally defined. Either that gravity's reach is infinite in which case we have to assume it to act on the space, and matter, existing. Or it is a illusion.

Now you stopped accelerating, defining your speed as relativistically close to lights speed in a vacuum. Do you still find a local gravity? You must still have a added relativistic mass, but the local definition of a gravity is gone, isn't it?
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Re: How does a 'field' become observer dependent?
« Reply #193 on: 28/12/2013 01:17:20 »
Depending on what anchors you use you can define earths, or any planets suns, speed differently. The best way, to me, seems to be using 'fixed stars' light for it, assuming 'candles' for it, measuring locally incoming blue shift. Alternatively one can use the CBR (Cosmic Background Radiation). But as you can't prove CBR to be 'still', and as you can't prove 'fixed stars' to be still either, although one can define them so due to their enormous distance from us, it's still relative speeds, isn't it?
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Re: How does a 'field' become observer dependent?
« Reply #194 on: 28/12/2013 01:23:12 »
A planets definition of a gravity will be the same, no matter how much you first accelerate it, to then return to a uniform motion. To me it becomes a question what motion is? Accelerations versus uniform motions. But I think I can define matters gravity as directed towards a 'center'. Can I do the same for a constant uniform acceleration?
==

differently expressed.
Can I differ inertia, from gravity?
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Re: How does a 'field' become observer dependent?
« Reply #195 on: 28/12/2013 11:20:21 »
Using a Higgs field we have one definition of inertia, as I see it. But a Higgs field do not describe a mass in uniform motion. A book 'at rest' with/on a table, on Earth, can not be defined as 'accelerating' by itself, can it? And so there can be no Higgs field to react on it.

The Higgs field is a definition relative accelerations/decelerations, to me.
And that I would call classical inertia.

But then we have the idea that inertia, and gravity, is inseparable.
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Re: How does a 'field' become observer dependent?
« Reply #196 on: 28/12/2013 11:31:06 »
You can define it as acting on all mass, by using the equivalence principle, possibly. But that one is limited to a very special type of motion, if we want to be strict. A uniform constant acceleration, as by some 'cosmic' elevator, equivalent to the experience you get on a (non spinning) Earth.

Gravity.

Don't like that. It's taking one approach, relativity, and then try to glue a Higgs field upon it, without explaining why it should be so. A Higgs field demands several archetypes, it demands real motion, and real accelerations, inside a containing universe defined by a existing geometry. It may fit relativity in that you still can allow an idea of observer dependencies for it, but it neither explains those observer dependencies, nor does it fit uniform motions.

Einsteins universe can be seen as using a universal container model, although observer dependent. In it a relativistic mass is defined by tensions, relations acting on something moving uniformly, creating a relativistic mass, and so a added energy, although not measurable locally.
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Re: How does a 'field' become observer dependent?
« Reply #197 on: 28/12/2013 11:38:41 »
That's the normal approach to relativity, defining it from a 'preexisting geometry'. From such a approach it becomes self evident why Einstein demands the moon to still exist, even when he's not looking. But you can labor with relativity, as I see it, from local definitions instead. And define a geometry from local properties, 'constants', principles. It should still leave the moon to exist when I'm not looking, to fit my ideas, but it's not a 'preexisting geometry' anymore.
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Re: How does a 'field' become observer dependent?
« Reply #198 on: 28/12/2013 11:40:08 »
Einstein defines the room as collaborating with time. That in its turn makes the universe 'plastic'. But it's still a container model.
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Re: How does a 'field' become observer dependent?
« Reply #199 on: 28/12/2013 11:42:39 »
Or it's not. Some archetypes are very hard to get rid of.
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