# The Naked Scientists Forum

### Author Topic: How does a 'field' become observer dependent?  (Read 199155 times)

#### yor_on

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##### Re: How does a 'field' become observer dependent?
« Reply #175 on: 23/12/2013 19:35:09 »

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.

#### yor_on

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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.

#### yor_on

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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.

#### yor_on

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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.

#### yor_on

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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?

#### yor_on

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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.

#### 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.

#### yor_on

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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?

#### yor_on

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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?

#### yor_on

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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.

#### yor_on

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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 »

#### 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.

#### 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 »

#### yor_on

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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 »

#### 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.

#### yor_on

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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.

#### yor_on

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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).

#### yor_on

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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?

#### yor_on

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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?

#### yor_on

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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?

#### yor_on

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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.

#### yor_on

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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.

#### yor_on

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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.

#### yor_on

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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.

#### yor_on

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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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##### Re: How does a 'field' become observer dependent?
« Reply #199 on: 28/12/2013 11:42:39 »