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Author Topic: How does a 'field' become observer dependent?  (Read 201151 times)

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #275 on: 04/01/2014 14:19:36 »
So, is the universe 'infinite'?
From what definition?

Go out to the left to come back at the right?
Or infinite as in there being no end at all, and when ignoring time for it meaning no 'repetitiveness', as the above example actually mean, ignoring a arrow?

Or 'infinite' as in connections, defining a universe, observer dependently?
« Last Edit: 04/01/2014 14:30:14 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #276 on: 04/01/2014 14:27:02 »
I better admit that I'm partial to my own definition.

If I define it from connections (locality) then there is no defined magnitude to it, except your local definition of clock and ruler. There is no end either, as long as connections exist, without those we won't exist either. so what we have is in one way a 'bubble', but not as definable from any thought up outside. It's existent to us inside it, but? There is no way to define what a 'outside' will mean that I can see from it. Unless you want to use 'constants, properties and principles' as what 'exist outside'?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #277 on: 04/01/2014 14:36:46 »
And a inflation does not use a arrow, initially. Not as I think of it. Let's call SpaceTime a 'rip' coming from a symmetry. That rip can only use a arrow if we also can define frames of reference interacting. We need oscillations for it :) sort of. Proofs of change.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #278 on: 04/01/2014 14:37:30 »
That would then be connections establishing themselves.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #279 on: 04/01/2014 14:40:39 »
And they use 'c', don't they?
But a inflationary space?

Well, that space should be connections too, shouldn't it? To exist I mean? Either that, or that 'space' is non existent, as defined classically, perfect vacuum.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #280 on: 04/01/2014 14:43:03 »
I'm not sure at all what a vacuum means. If I use a definition of light non-propagating, instead giving us a rhythm, and a pattern that we find to propagate, what would the vacuum we describe be?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #281 on: 04/01/2014 14:47:06 »
Think of it as layers upon layers, each layer subtly different. The pattern you find being those layers described from your 'position' inside a SpaceTime, giving you motion. Just as we find 'gravity' to act 'sideways', although I define it as a 'down welling' in each point.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #282 on: 04/01/2014 14:50:28 »
If it really is non existent, as in a classical definition, then there is no 'connections' to it either, and no definable 'speed'. If it uses connections though, I would expect those to be defined by 'c'. You want it to be 'energy? Don't you?

Well, does 'energy' obey 'c'?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #283 on: 04/01/2014 15:34:56 »
Sure, it obeys 'c', in any outcome you can describe. So, presuming 'energy' to be a added quantity (or 'property') to a vacuum, we still need to define that vacuum, as either obeying 'c', or not?
=

That is, if you want a inflation. Better point out that it does not matter what 'speed' you expect a inflation to have had. Any deviation from relativity is as bad to me. Doesn't matter how you define it as, tachyons or a 'outside', as long as you involve a definable 'speed' to this inflation form some 'origin'. Also, it is inconsistent to define it as there being no origin at the same time you expect something to have a 'origin'.

« Last Edit: 04/01/2014 15:39:52 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #284 on: 04/01/2014 15:41:18 »
Exchange 'energy' for constants, properties and principles. Then tell me, do they need to obey 'c'?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #285 on: 04/01/2014 15:42:34 »
Sure, inside they must, as in a transformation defined by them. But as a 'background', equivalent in all 'points'?
=

You need to remember my definition of a 'inside' for this, creating 'dimensions' through connections, obeying 'c'. You also need to remember that it will not make sense discussing a 'outside', as long as we're discussing connections (relations) defining a universe, observer dependently. It has no meaning from a thought up outside, it's not a 'tangible' thing.
« Last Edit: 04/01/2014 15:46:45 by yor_on »
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #286 on: 04/01/2014 15:49:58 »
That's why I think of it as a projection. You can see that projection different ways, as layers upon layers, or as something 'dimensionless' giving us 'infinities'. You can only define a distance relative a clock, and a ruler. You need both.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #287 on: 04/01/2014 15:58:10 »
So what would a inflation be from my point of view? I can easily accept it to have no origin, we actually need to stipulate that property of it. We also need to agree on that this is the only way it can keep 'c'. You need something, coming into a arrow of time 'everywhere', no defined origin. You also need it to establish connections at 'c'. Because that is how I define 'c', as equivalent to your very local arrow of time. And that is also transformations, and 'c'.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #288 on: 04/01/2014 16:00:14 »
So, what is this vacuum? 'Energies'? Or constants, properties and principles?
 

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Re: How does a 'field' become observer dependent?
« Reply #289 on: 04/01/2014 16:04:00 »
And what is a symmetry break?
Temperatures?
 

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Re: How does a 'field' become observer dependent?
« Reply #290 on: 04/01/2014 16:08:51 »
So, that will in a way give us a speed, as you measure over frames of reference. The points need to connect, and they should do it at 'c', as I presume. Because 'c' is what rules locally, inside this universe. And a symmetry break will involve temperatures, and transformations, or as I think of it, 'frames of reference' interacting.
 

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Re: How does a 'field' become observer dependent?
« Reply #291 on: 04/01/2014 16:11:16 »
Even when thinking of it as there can be no single frame of reference possible inside this universe, I still can think of it as 'dual frames' connecting to dual frames at 'c', if you see my reasoning here.

but there is no 'time' to this universe, other than the one you get using 'c'.
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #292 on: 04/01/2014 16:18:11 »
The only presumption you need for this reasoning is 'c', hopefully :) Everything follows from that one. The arrow becomes a local 'c', frames of reference interacting is 'c'. Your definition of a age for this universe is 'c'. etc etc.
 

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Re: How does a 'field' become observer dependent?
« Reply #293 on: 04/01/2014 16:20:20 »
And the speed of communicating here is also 'c'.
Now prove me wrong :)
 

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Re: How does a 'field' become observer dependent?
« Reply #294 on: 04/01/2014 17:17:40 »
I guess it's slightly boring. But one of the biggest hurdles I found, is what I think of as the 'observer problem'. It goes like this. Assume yourself to 'observe' anything.

1. what you observe you will define as one frame of reference, yourself consisting of the other.
2. You will, in accordance to relativity, have to define what you see as dependent on your local clock and ruler.

Now imagine that what you observe is a interaction.

How many frames are we talking about?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #295 on: 04/01/2014 17:23:10 »
That depends on your 'system' doesn't it :)

You can define it as two, or you can assume that all interactions are a result of a frame of reference interacting with another, in which case you might want to consider it three, or even more. Three as we might assume that a decay is a result from frames of reference interacting, your observing becoming a third frame.

Or more. And that one has to do with how you think of SpaceTime.
Gravity? Is that as a 'net', updated at 'c'?

What did Mach mean, and why did Einstein find his ideas so interesting?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #296 on: 04/01/2014 17:27:26 »
It does matter. For me, as I'm looking for the 'holy grail', a proof for one singular frame of reference able to exist on its own it play a huge roll. If there is no such proof, then there are no 'singular frames' either. Everything becomes a question of connections defining those 'frames of reference'.
 

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Re: How does a 'field' become observer dependent?
« Reply #297 on: 04/01/2014 17:31:03 »
Why do I want to find a proof for one single frame of reference?
Well, what about a 'bit'

How would you define it quantum mechanically if there only are relations, defining those bits?
No 'bits' existing on its own?
 

Offline yor_on

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Re: How does a 'field' become observer dependent?
« Reply #298 on: 04/01/2014 17:32:42 »
That would mean that any assumption of something existing 'by itself' quantum mechanically is wrong.
 

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Re: How does a 'field' become observer dependent?
« Reply #299 on: 04/01/2014 17:40:38 »
No bits to it, still QM?
=

What about strings and loops then?
« Last Edit: 04/01/2014 17:42:35 by yor_on »
 

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Re: How does a 'field' become observer dependent?
« Reply #299 on: 04/01/2014 17:40:38 »

 

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