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Author Topic: Linear vector spaces and gravity  (Read 2048 times)

Offline jeffreyH

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Linear vector spaces and gravity
« on: 16/01/2016 18:26:53 »
Is there a vector space that can be used with linear combinations that is representative of a non-linear space such as that of the gravitational field? If this exists can it be formulated as an energy vector space?


 

Offline MichaelMD

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Re: Linear vector spaces and gravity
« Reply #1 on: 19/01/2016 13:59:50 »
Rather than trying to explain linear energy transmissions like light and electromagnetism with new theories about space, I believe it would be better to invoke a new theory of how individual energy units could act in a linear fashion.

In my model, elemental ether units which are of matching size (inasmuch as they are all "elemental,") acting via a vibrational resonating mechanism, and in intimate proximity to each other, is the best model to explain linear transmissions.

My ether model starts with original space (space prior to the first appearance of any forces) in which point-localities of space oscillated in perfect symmetry, Then oscillational fatigue caused adjacent "points" to combine in "Yin and Yang" fashion., Then, these point-pairs reversibly re-equilibrated with the initial oscillational setting, reverting to elemental singleton units, which broke the perfect symmetry of space, producing an ether matrix where matching-size elemental ether units resonate with each other vibrationally (as derived from the oscillational.)

Certainly present energy theory, where quantum units of varying sizes are acting via spin and across vectors of space, cannot rationally explain linear transmissions.

I believe quantum "waves" represent a "shoreline" effect, where the ether matrix, composed of elemental ether units, vibrating together with slightly larger "etheroidal" units, finally transitions fully to our quantum energy system, i.e., a "wavy line" zone where the vibrational etheric units all become spin/vector quantum units.

Occasionally, a vibrating etheroidal unit "escapes" from its vibrational mechanism and unexpectedly appears in an unusually designed quantum energy system. -This model would be able to account for the phenomenon called "Quasiparticles."
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #2 on: 19/01/2016 18:27:33 »
When I posed the questions I certainly didn't expect an answer. What you have described has nothing at all to do with what I am attempting so is irrelevant. I didn't understand most of it as it made no sense anyway.
 

Offline Thebox

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Re: Linear vector spaces and gravity
« Reply #3 on: 19/01/2016 19:34:19 »
When I posed the questions I certainly didn't expect an answer. What you have described has nothing at all to do with what I am attempting so is irrelevant. I didn't understand most of it as it made no sense anyway.
Not sure if this relevant to you Jeff, but when I read your question my first though was a bow and arrow.

 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #4 on: 19/01/2016 19:49:44 »
When I posed the questions I certainly didn't expect an answer. What you have described has nothing at all to do with what I am attempting so is irrelevant. I didn't understand most of it as it made no sense anyway.
Not sure if this relevant to you Jeff, but when I read your question my first though was a bow and arrow.

If you were to fire one arrow directly upwards and another at an angle then you would start to see the issues. If you are using exactly the same force to launch all arrows then you can propel an arrow the furthest distance by launching it at a 45 degree angle with respect to the ground. If you consider how fast and therefore how much energy an arrow would need to be able to sustain an orbit or escape the earth altogether then you can reach some interesting conclusions.
 

Offline Thebox

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Re: Linear vector spaces and gravity
« Reply #5 on: 19/01/2016 19:53:55 »
When I posed the questions I certainly didn't expect an answer. What you have described has nothing at all to do with what I am attempting so is irrelevant. I didn't understand most of it as it made no sense anyway.
Not sure if this relevant to you Jeff, but when I read your question my first though was a bow and arrow.

If you were to fire one arrow directly upwards and another at an angle then you would start to see the issues. If you are using exactly the same force to launch all arrows then you can propel an arrow the furthest distance by launching it at a 45 degree angle with respect to the ground. If you consider how fast and therefore how much energy an arrow would need to be able to sustain an orbit or escape the earth altogether then you can reach some interesting conclusions.

A bit like the Cannon ball idea and escape velocity.

I should hope the one at 45 degrees would travel further, the arrow fired vertically up will eventually slow , then turn, then travel back down vertically , excluding weather and air of course.

added - i drew it for you Jeff , It may help you



I added atmosphere to it



« Last Edit: 19/01/2016 20:16:43 by Thebox »
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #6 on: 19/01/2016 21:36:08 »
When I posed the questions I certainly didn't expect an answer. What you have described has nothing at all to do with what I am attempting so is irrelevant. I didn't understand most of it as it made no sense anyway.
Not sure if this relevant to you Jeff, but when I read your question my first though was a bow and arrow.

If you were to fire one arrow directly upwards and another at an angle then you would start to see the issues. If you are using exactly the same force to launch all arrows then you can propel an arrow the furthest distance by launching it at a 45 degree angle with respect to the ground. If you consider how fast and therefore how much energy an arrow would need to be able to sustain an orbit or escape the earth altogether then you can reach some interesting conclusions.

A bit like the Cannon ball idea and escape velocity.

I should hope the one at 45 degrees would travel further, the arrow fired vertically up will eventually slow , then turn, then travel back down vertically , excluding weather and air of course.

added - i drew it for you Jeff , It may help you



I added atmosphere to it



I will get back to you on this. I don't have the time at the moment.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #7 on: 23/01/2016 20:17:01 »
Thebox's diagrams are suitable for the purpose and I will come back to them. For now what is needed is a linearly independent solution to the equation Ax = 0 for the energy space. That is that the only solution is the trivial solution where all scalars are zero. Once we have a span and basis for the set of vectors then maybe something interesting will reveal itself.

However there has been no definition yet of the format of the space equations.
 

Offline Thebox

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Re: Linear vector spaces and gravity
« Reply #8 on: 24/01/2016 06:59:00 »
Thebox's diagrams are suitable for the purpose and I will come back to them. For now what is needed is a linearly independent solution to the equation Ax = 0 for the energy space. That is that the only solution is the trivial solution where all scalars are zero. Once we have a span and basis for the set of vectors then maybe something interesting will reveal itself.

However there has been no definition yet of the format of the space equations.

What do you mean by a linearly equation? explaining what a straight line?

0rē =1  does this not rep a straight line?

What are you trying to rep with the equation ?
 

Offline Thebox

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Re: Linear vector spaces and gravity
« Reply #9 on: 25/01/2016 23:25:37 »
Jef this just came to me while I was sleeping, don't ask.


0+0=00=x

are you working on something like this?

https://www.sciencenews.org/article/quantum-histories-get-all-tangled
« Last Edit: 26/01/2016 01:26:29 by Thebox »
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #10 on: 29/01/2016 02:22:11 »
Jef this just came to me while I was sleeping, don't ask.


0+0=00=x

are you working on something like this?

https://www.sciencenews.org/article/quantum-histories-get-all-tangled

No. However, your diagrams above are interesting because they show straight line inertial paths. Whereas gravity will deflect this path. This may sound strange but is it possible to quantize the deflection? Certainly there is considered a smallest unit of length, the Planck length. Yet that is far too small to apply to particles. So is there a larger scale at which this could apply.

I will be posting matrix and vector maths here so if you get lost tell me.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #11 on: 29/01/2016 02:26:31 »
A final thought for anyone to ponder. Gravity can be thought of as adding or removing kinetic energy and imparting acceleration. However, in a perfectly circular orbit with constant velocity this is not the case. If there is no increase or decrease in velocity then the only reason this can be called acceleration is due to the change in path of an object. In which case the amount of deflection from a straight line is the only thing that can be measured to show any change. Hence the idea of quantization of deflection.
 

Offline alysdexia

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Re: Linear vector spaces and gravity
« Reply #12 on: 30/01/2016 10:38:28 »
The velocity components increase and decrease.  Gravity stratifies matter like any other centrifuge.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #13 on: 31/01/2016 17:35:23 »
Except the kinetic energy is a property of the orbiting object and not the central mass. The central mass may deflect the path but has no other effect on a perfectly circular orbit in an idealized system. This is theoretical and is in no way a true reflection of nature.
 

Offline Thebox

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Re: Linear vector spaces and gravity
« Reply #14 on: 02/02/2016 08:01:18 »
The velocity components increase and decrease.  Gravity stratifies matter like any other centrifuge.

mass is ghost particles
 
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Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #15 on: 02/02/2016 19:41:41 »
Consider, for a moment, a set of points at varying radial distances from a source. It may be a planet, a sun or other. At each of these points a particular escape velocity would be required to move away to infinity. As long as the path does not intersect the surface of the central object, that is. Say we start an object on a course that will collide with the central object. What happens to tidal forces if the initial velocity exceeds ascape velocity right down to the surface? If this were a compact, dense object, with a relativistic escape velocity at the surface, could the tidal forces be canceled if the velocity starts at surface escape velocity? Isn't that an interesting question?
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #16 on: 04/02/2016 00:04:43 »
If we consider the red vector in thebox's diagram this can describe the idealized path of an object. At the apex of this path the kinetic energy is zero and all the energy is then gravitational potential energy. Like the orbit described previously this path can be said to suffer a deflection. In this case a negative deflection. Unlike the perfect circular orbit the kinetic energy in this case does change. Both situations can be described as vector spaces as long as they can be scaled linearly.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #17 on: 18/02/2016 18:53:51 »
We can consider 3 situations in newton's version of gravity. 1. Free fall, 2 Orbit and 3 Escape. We can also arrange for each value to be orthogonal so we could have 1 on the x-axis, 2 on the y-axis and 3 on the z-axis. If we are considering kinetic energies we have 3 energy values E_1, E_2 and E_3. These can then be written as vectors. E_1 = [Ke_1 0 0], E_2 = [0 Ke_2 0] and E_3 = [0 0 Ke_3]. These are then the rows of 3 by 3 matrix. It can then be investigated as to whether or not a scalar modification of an identity matrix when applied to this matrix will give reasonable values. That is will they scale with equal radial distances from the source. These are simple newtonian equations. I will post these later.

EDIT: These will all be proper values. No coordinate values will be used.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #18 on: 18/02/2016 22:45:42 »
I just want to post a conclusion without immediate justification. Gravity MUST increase the relativistic mass of any object under its influence. Justification to follow.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #19 on: 19/02/2016 09:24:09 »
We can write out the scalar matrix as such:

656c6ad911dd35bc917955b9cdba2978.gif

Only if all the lambda values, when multiplied by the energy matrix, scale each energy value proportionally will we have a vector space. I will show that this is only the case for escape and orbital energy as would be expected. I will also show why relativistic mass is an issue and the problems it brings with it.

EDIT: The above assumes we are considering lambda_1 = lambda_2 = lambda_3. Which will not be the case for all 3 energy values.
« Last Edit: 19/02/2016 11:10:23 by jeffreyH »
 

Offline Thebox

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Re: Linear vector spaces and gravity
« Reply #20 on: 19/02/2016 13:56:06 »
We can write out the scalar matrix as such:

656c6ad911dd35bc917955b9cdba2978.gif

Only if all the lambda values, when multiplied by the energy matrix, scale each energy value proportionally will we have a vector space. I will show that this is only the case for escape and orbital energy as would be expected. I will also show why relativistic mass is an issue and the problems it brings with it.

EDIT: The above assumes we are considering lambda_1 = lambda_2 = lambda_3. Which will not be the case for all 3 energy values.

Wow Jeffrey you are putting in a lot of effort, I have no idea what you are on about but it looks some real cool thought just by your diagram matrix, hope you get a scientist give this some attention, good luck.

P.s although I do not understand what you are on about, I think the center value of your matrix should be zero extending to zero. And also turned into a 3d matrix .

I don't know if this helps you but the visual universe is a bit like this

X=∞λ0λ1λ0λ1λ0

Y=∞ λ0λ1λ0λ1λ0

Z=∞λ0λ1λ0λ1λ0

t= ∞λ0λ1λ0λ1λ0

G=∞λ0λ1λ0λ1λ0

But of course the length of amounts of λ0 can be greater or less(λ0λ0λ0), and of course this is a uniform matrix where the real matrix would show a contorted shape and X≠Y etc .

« Last Edit: 19/02/2016 15:36:31 by Thebox »
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #21 on: 19/02/2016 19:57:42 »
Well, Thebox, I don't know what that was all about but look at the attached graph. There are 4 functions plotted. Three are the result of multiplying the initial calculated values of orbital, escape and free fall kinetic energies. Where the lambda multipliers range between 1 and 10 in steps of 1. As the energy increases then the radial distance from the source will decrease.

The equations used were

1 - Free fall Ke = mGMd/r^2

2 - Escape Ke = GMm/r

3 - Orbital Ke = GMm/(2r)

Immediately it can be seen that both escape and orbital are of the same basic form and will scale in direct proportion but that freefall Ke won't.

On the graph the blue, red and green plots represent lambda scaled versions of escape, orbital and free fall ke respectively. The pink line shows the actual values for free fall Ke at the same radial distances as the escape and orbital Kes.

Where the pink line crosses the orbital line is the light like orbit outside the event horizon. Where the pink line crosses the blue line we find the event horizon.

However, because these calculations are all Newtonian, no account has been taken of relativistic mass. This should increase exponentially, reaching an infinite amount at the event horizon. Which means the kinetic energy is also infinite. Meaning that the gravitational energy ALSO has to become infinite. Otherwise it will not intercept either the blue or red lines and the black hole will no longer trap light. This is where the mathematics breaks down. Or is it?

Now that gravitational waves have been detected I aim to show exactly why it doesn't break down.
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #22 on: 20/02/2016 14:09:29 »
If we organize our data so that the energy values for free fall, orbit and escape are E_1, E_2 and E_3 respectively then we can actually describe 3 linear vector spaces.

1) The free fall, orbital space.
5c14adf1d0e74dc390102b096f2344dd.gif

2) The free fall, escape space.
48c3d283ed77fc24cbf31a6d29f86ee0.gif

3) The orbital, escape space.
146698d461c41b036f5428f3c7f6d533.gif

These will all scale using the following scalar matrix.
0295eb97f9ca6c1e832c5ed3131ff981.gif

For all the spaces the radial distance of each energy value must be identical. Vector space 1 will scale to show the light-like surface of any viable black hole. Vector space 2 describes the surface of the event horizon of any viable black hole. Vector space 3 shows the energy difference between escape and orbit at any radial distance from the source.
« Last Edit: 20/02/2016 14:18:13 by jeffreyH »
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #23 on: 20/02/2016 16:01:45 »
As a starting point for a set of relativistic vector spaces this would be good.

https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechanics

In the first instance for special relativity but especially for the section "Lagrangian formulation in general relativity".

EDIT: In the above cases lambda is a function of the Lorentz transformation.
« Last Edit: 20/02/2016 16:05:57 by jeffreyH »
 

Offline jeffreyH

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Re: Linear vector spaces and gravity
« Reply #24 on: 21/02/2016 19:25:01 »
The next step in the formulation of the vector spaces starts with this.

https://en.wikipedia.org/wiki/Theoretical_motivation_for_general_relativity

Specifically the concept of a Lorentz scalar.
 

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Re: Linear vector spaces and gravity
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