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Author Topic: How does a 'field' become observer dependent?  (Read 200702 times)

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1325 on: 14/11/2014 14:46:37 »
Now, how big would it need to be :)

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1326 on: 14/11/2014 14:48:10 »
You're perfectly correct if you're asking yourself from 'where'. Inside?
Outside?

Would this field need a 'outside'?
Or is it created from a 'inside'?

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1327 on: 14/11/2014 14:57:53 »
Assuming there to be a 'outside' you will add to the complexity, especially with a inflationary, subsequently expanding universe, that also can be presumed 'infinite'. To go around this you will have to invent further dimensional tricks that allow it to become something similar to a möbius ring.

defining it from solely a inside it has no problems being 'infinite' that I can see? And I don't need dimensional tricks.

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Re: How does a 'field' become observer dependent?
« Reply #1328 on: 14/11/2014 15:01:18 »
With both a 'outside' (aka our Möbius ring) and a inside, it will not be enough with defining what our universe is. Actually it won't be enough even if you define that 'outside' containing it. Because, what contain the 'outside' you now have defined, ad infinitum.

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1329 on: 14/11/2014 15:06:45 »
It's all abstractions of course. We build the universe from abstractions. Using such the universe is as small, or 'infinite' you measure it to be. Then again, assuming LorentzFitzgerald contractions to be practically true, even when being infinitely close to 'c', would you expect a 'infinite universe' to have a end?

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1330 on: 14/11/2014 15:19:29 »
The question is, how much light do you need? Everything can be translated into 'energy', light too. A light quanta is measurable though, and we presume light, virtual or not, to be the force carriers that define us, as well as the universe we see. So the 'field of light' I'm wondering about, if so, is what you consist of. Without force carriers defining you you disappear.

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1331 on: 14/11/2014 15:23:36 »
Defining it as your clock (local arrow) is equivalent to 'c' it then seem to mean that stopping the clock should dissolve a universe. And the clock is you, as well as me.

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Re: How does a 'field' become observer dependent?
« Reply #1332 on: 14/11/2014 15:27:31 »
because, taking away a propagation will still leave us a clock. the assumption of propagation is a global description, a 'container model'. The clock is a local definition, a local constant equivalent to 'c' globally.
=

This one is tricky. 'c' is also a local definition, repeatably so. It's a statement about what 'speed' light will have in a two way (mirror) experiment, no matter to what you define your own 'inertially (uniformly) moving' speed (and the experiments naturally as long as you are at rest with it). It doesn't matter how fast you go relative something else, you will still get 'c'.

But you can only get 'c' through using what I call 'global' parameters', you must use the container (common universe) to get that measurement of a speed. So even though it's local defined by your clock and ruler, it's also 'global', from thinking of the universe as a 'container model' using it that way.

And yes, the 'clock' is a equivalence, splitting 'c' in 'even chunks', building on it. But it's your local clock and you 'vibrate' with it. It's also so that from an assumption of a 'clock' you can, as I do, assume that there still need to be a property of 'time', or 'arrow', existing when magnifying reality into the really small, which then should be quantum mechanically, also becoming a question of if there can be some smallest discrete 'length', 'time', 'mass', etc. Maybe that one isn't necessary? I think it is myself though, for the moment being that is :)

« Last Edit: 14/11/2014 18:04:44 by yor_on »

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1333 on: 14/11/2014 15:35:15 »
Can you see how I think writing " assuming LorentzFitzgerald contractions to be practically true, even when being infinitely close to 'c', would you expect a 'infinite universe' to have a end?"

You contract the universe, but it won't end, not even then. It can't, not as long as rest mass is involved.

Light is different, it's constantly at 'c', no acceleration. We define accelerations, from rest mass. We see a equivalence to it in red and blue shifts, but from our observer dependencies as rest mass.

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1334 on: 14/11/2014 15:42:03 »
Assume the universe to be infinite, accept that it consist of force carriers communicating it. How much 'energy' would such a universe consist of? Is that question meaningful?

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Re: How does a 'field' become observer dependent?
« Reply #1335 on: 14/11/2014 15:43:59 »
It's not a meaningful question, even though I could give that idea some symbol.

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Re: How does a 'field' become observer dependent?
« Reply #1336 on: 14/11/2014 15:53:27 »
Same with infinity, not a meaningful question. Doesn't mean it's impossible to build a logic from it, or from defining different infinities. But for me, being inside any of those infinite universes, it won't matter. And assuming there to be no 'outside' to compare this universe from, those definitions definitely loses any meaning.

Still, if you believe in a 'container model' they will mean something to you, I don't.

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Re: How does a 'field' become observer dependent?
« Reply #1337 on: 14/11/2014 17:10:56 »
To give the universe a shape you first need somewhere from you define it, as I see it. Assuming a 'inside' having a specific shape means you have to ask yourself, relative what? You are free to define it naturally, as soon as you created somewhere from you can compare. As for example using a container model. But you're not home free yet, you also need to define what this container is. In my case as I like to define it from local constants etc, turning the container model inside out sort of, the 'common container' we find should be a result of those, localities communicating, although I don't know how it can join? 'c' is communication, as well as a 'clock', as well as 'force carriers', but it doesn't state what makes it all possible.

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Re: How does a 'field' become observer dependent?
« Reply #1338 on: 14/11/2014 20:22:46 »
It becomes a really weird universe, thinking this way. But if you accept 'c', and that one is tested, it already is as weird as it can be I think. Locality is an idea about where you find chain reactions, or 'waves', spreading from a center. And Einstein's relativity uses it too, when we look at how one define what is real for a observer aka 'observer dependencies'. But it's not the exact same, neither is the idea of a 'information universe' in where a entanglement is allowed because it breaks no laws of 'useful information' faster than 'c', although both is linked to the idea of what locality mean, as I read it.

the way I think of it (my homegrown version of locality) is as if the universe consist of one ground state, described by you being able to move anywhere inside it, to find your clock and ruler perfectly synchronized with wherever you are. In a wider context this relates to an idea of discreteness, although I'm not sure how that should express itself. You need some quantity that stays the same, wherever you go, and there constants are a nice choice.

In that manner those that think of LorentzFitzgerald contractions and time dilations as being 'illusionary' becomes correct, as long as we accept and hopefully can define, the limitations of that 'locality'. Superimposed you will fit perfectly everywhere, sort of, no place out of sync :) Einstein used being 'at rest'  describing it.

But the universe, looked at this way, (and I can't actually see any other way to look at it, without invalidating one of the foundations of physics, namely repeatable experiments) suddenly consist, in QM terms, of something 'discrete', what I then call 'locally equivalent points', or even weirder just one 'point' that then builds the rest. And the last one is very weird, but it comes from me wondering about infinity :) and 'insides', using imaginary 'outsides' to describe it. The universe is indeed spaced out :)
=

Or maybe not, just as time dilations and LorentzFitzgerald contractions to me seem complementary, and just as light has a particlewave duality, and linearity and non-linearity weave in and out of each other, maybe the idea of discreteness relative a flow also is complementary? Depending on how you measure? And there the scale you choose seems very important to me. I'm still not sure what really I think of dimensions, but I'm pretty sure the scale you use are more important than what one usually assume.
« Last Edit: 14/11/2014 20:49:26 by yor_on »

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1339 on: 14/11/2014 21:35:18 »
There's a difference between such a definition, versus one in where we instead assume each point different from the other. This one is simple, to me it explains why repeatable experiments must exist, why physics inside the measurable portion of a universe must be the same, and it demands constants to exist.

what it doesn't explain is how those points are joined. No matter if the universe can be seen as infinite, simultaneously having no defined size as measured from some imaginary 'outside'. Inside it, where we live we do find dimensions, we have length, width, height, and 'time'. We find a vacuum, mass (EM included) and 'energy'. We find a envelope consisting of the earliest light reaching us, 'c', also defining a arrow. We find conservation laws.

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Re: How does a 'field' become observer dependent?
« Reply #1340 on: 17/11/2014 16:46:58 »
So what is a property? When water becomes ice we call it a 'emergence'. Does it gain new properties? Was they there before?

The last question seems more important than what one might think. Was it there before? Would that be how you can identify a property? Think of 'c', as a clock. Split it, split it again, just keep splitting this 'clock speed' into smaller and smaller chunks. Does the property of 'c' disappear, the clock stop? Planck scale is defined as the place where light takes one Planck 'step' at one Planck 'time'.

does the clock disappear there?
What about the property of a clock?

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Re: How does a 'field' become observer dependent?
« Reply #1341 on: 17/11/2014 16:56:40 »
Properties are tricky, if you ever used a cup you know that it has a property, its form and function. The form and function is a result of our demands, and it becomes the property of a container. But physical properties then? Like spin? And 'c', and emergences, giving us new properties?

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Re: How does a 'field' become observer dependent?
« Reply #1342 on: 17/11/2014 17:01:19 »
It's about scales too. Using 'c' as a clock. If Planck scale is a real delimit of what's observable, and the clock stops, does the property disappear with it? Would you then call it a 'emergence' when it starts to 'tick' again?

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Re: How does a 'field' become observer dependent?
« Reply #1343 on: 20/11/2014 22:04:47 »
You know what, I don't really give a sh*
I want you to be brave, I want you to push the barriers.

I will not promise you that it won't cost you.

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1344 on: 20/11/2014 23:42:58 »
We seem to be losing out to life.
What will we blame it on? The last one of us, the one standing on the isotropic and homogeneous dung heap we will leave? the others? Those that's no longer here? I don't know, what I know is that bravery is ageless, and you have it.

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1345 on: 20/11/2014 23:46:24 »
Go for it.

Make something change, and give us a chance.

yor_on

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Re: How does a 'field' become observer dependent?
« Reply #1346 on: 20/11/2014 23:47:20 »
As they say in all good stories.

The end.

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Re: How does a 'field' become observer dependent?
« Reply #1346 on: 20/11/2014 23:47:20 »