What Gravity Is

[Reality Check]

No the Gravitational force is merely the likely direction a particle will move in. This is defined by the relationships between G m and r.

Gravitational force acts between the masses not the grid particles. It is not a "likely direction". It is an acceleration of a mass.

 

[LordoftheLeftHand]

I don't have a physics degree or anything but I do have a question:

What advantages does this model provide? Does it allow for calculations that would not normally be possible? Does it make any predictions that differ from classical mechanics/relativity (or whatever the current ideas are)?

LLH

 

[martu]

An individual piece of matter can go anywhere.

Yes but on average it will tend towards the larger gaps in the Grid

You state that the grid had only one force: "just one the repelling of the particles of the Grid". What is causing the matter to go towrds the larger gaps in the grid

The expansion of the Grid. Take away the expansion of the Grid and nothing moves. The expansion of the Grid can be considered time – stop the expansion and you stop time and hence nothing can move.

Why are there larger gaps in some places and smaller in others?

The Grid is distorted by the presence of Matter.

Why is the grid expanding?

The Big Bang.

Why is this expansion not uniform?

The Expansion of the Grid is uniform. However put mass into the Grid and the Grid internally is distorted creating Gravity. No mass and yes the Grid would be uniform.

Don’t think of objects as such just think like this – Gravity is the most probable direction matter will move. This is defined by the amount of matter G, the direction defined by m and is proportional to r which is the distance covered by a mass in a slice of time.

 

[The Man]

It is proportional to the distortion in the Grid.



The expansion of the Grid is caused by the big bang – the edge of the universe is the edge of the Grid..

I am getting the impression that this is one of those “expansion” TOEs where everything is expanding at the same proportional rate (so we can not detect the expansion) but some how that expansion results in gravitational attraction. Is my surmise correct martu?

Because the maths works.

What math, did I miss a post where you actually calculated something with this?

 

[Lucky]

I do not have the capability of doing so. I know next to nothing about physics. However, you are right that my comments were not phrased as constructively or helpfully as they could have been. My point, which you helped clarify, is that it is a necessary first step in coming up with any useful theory that it be able to account for all observable phenomena. And learning all of the obsevable phenomena in the area of physics necessarily requires a great, great deal of hard work and (formal or informal) education.

I would suggest anyone with a pet theory spend a long time with the textbooks before shoving it into the spotlight for its debut.

I don't entirely agree – I think it depends on the spirit in which the new 'theory' is offered. In this case martu admitted his ideas could be wrong, and asked for input from more knowledgeable people. We need to be careful not to respond by immediately jumping on the person when we see this kind of OP – it's not the best way to encourage critical thinking.

Coming up – the really fatal objection to the model:
In order to explain any interaction between particles of matter and the 'gravitational particles' you need to postulate a gravitational force – the very thing you are trying to explain!

Back to the drawing board!

No the Gravitational force is merely the likely direction a particle will move in. This is defined by the relationships between G m and r.

You jumped the gun – I meant I would explain why this is so in my next post, when I have time.

In the meantime, you could read my post again – there's enough there (and in plenty of posts from other people) to show you beyond any doubt that your model is not tenable. Please try not to be so defensive. You made an interesting OP, which has led to an enjoyable discussion. Your ideas are wrong, but in a fruitful way – I, for one, appreciate the brain exercise it's given me to show why you are wrong.

All attempts to come up with simple physical pictures of gravity (or electromagnetism etc.) have failed – so far. The theories remain purely mathematical, and it does seem to be the case that the more we investigate the nature of the universe, the more we are forced to abandon the reliance on physical intuition and accept that the maths itself is the underlying reality. True, this may not always be the case (though I suspect it will be), but do consider what you are up against, and how unlikely it is that you are making a major contribution to theoretical physics.

If you continue defending your model without showing any sign of seriously considering our objections then the discussion is bound to degenerate. Some posters (unfortunately) will resort to 'hammering the opposition into the ground' mode – which would be a pity, because that's largely been avoided so far in this thread.

 

[Reality Check]

Yes but on average it will tend towards the larger gaps in the Grid

What about matter at rest? Your model predicts no gravitational pull. I am at rest and I feel gravity. Are you in a zero gravity environment?

The expansion of the Grid. Take away the expansion of the Grid and nothing moves. The expansion of the Grid can be considered time – stop the expansion and you stop time and hence nothing can move.

No: matter moves regardless of the grid.
Time has nothing to do with the expansion of the grid. If time does not exist independently of the grid then the grid cannot expand.

The Grid is distorted by the presence of Matter.
The Big Bang.
The Expansion of the Grid is uniform. However put mass into the Grid and the Grid internally is distorted creating Gravity. No mass and yes the Grid would be uniform.
Don’t think of objects as such just think like this – Gravity is the most probable direction matter will move. This is defined by the amount of matter G, the direction defined by m and is proportional to r which is the distance covered by a mass in a slice of time.

Please use the standard definitions for constants: G = the gravitational constant (not an amount of matter), m = mass, r = the distance of a mass from another mass.

What is 'm' and how does it define a direction?
Is the 'amount of matter' a mass and why use G to denote it?

Do you agree that 2 masses at rest will feel a gravitational force?
But in your model there is no "pushing" of the grid particles for masses at rest. How come your model predicts no gravitational force for 2 masses at rest?

 

[Reality Check]

Another problem: Consider a object moving through space at a certain velocity. The grid model predicts the object will "push" against the grid particles. Newton's third law of motion states that the grid particles will push against the object with an equal and opposite force. Thus the object will come to a stop. The grid particles act like friction. This does not depend on the presence of any other object. In fact since the force between the grid particles is spring-like ("think of an imaginary spring between each adjacent particle which can contract and expand caused by the particles repelling each other"), I would then expect the object to be pushed backward.

Oddly enough scientists do not have to account for the frictional force when planing missions to other planets.

 

[martu]

What about matter at rest? Your model predicts no gravitational pull. I am at rest and I feel gravity. Are you in a zero gravity environment?



No: matter moves regardless of the grid.
Time has nothing to do with the expansion of the grid. If time does not exist independently of the grid then the grid cannot expand.



Please use the standard definitions for constants: G = the gravitational constant (not an amount of matter), m = mass, r = the distance of a mass from another mass.

What is 'm' and how does it define a direction?
Is the 'amount of matter' a mass and why use G to denote it?

Do you agree that 2 masses at rest will feel a gravitational force?
But in your model there is no "pushing" of the grid particles for masses at rest. How come your model predicts no gravitational force for 2 masses at rest?

I shall try to explain what happens in 2D to see if that helps. This thread needs pictures.

In 2D you have F equals m over r where F can be considered to be the most likely direction a piece of matter will move in.

Consider a uniform grid which can be pictured thus:

[qimg]http://i299.photobucket.com/albums/mm309/themartu/GridNoGravity.jpg[/qimg]

Now picture an atom in the centre of this square – it will not move as atoms tend towards the longest line of the Grid, in this case all the lines are equal.

Now add Gravity. We have added two masses one above the particle and one below the particle. The Grid now looks like this:

[qimg]http://i299.photobucket.com/albums/mm309/themartu/GridGravity.jpg[/qimg]

The force or rather which direction the atom is likely to travel is defined by m over r where m is the longest line in the quadrilateral. D is the distance between the two masses that are influencing the Grid.

Now we double the distance between the two masses:

[qimg]http://i299.photobucket.com/albums/mm309/themartu/GridGravityMassesfurtherapart.jpg[/qimg]

The particle still experiences a force but it’s less as r has doubled while m has stayed the same. In 2D this relationship is linear.

First question - is this logially sound?

 

[martu]

the OP describes something that looks a lot like a viscous medium. I don't see other posts discussing that. However, contrary to the usual physics of viscous media, where smaller particles experience the viscosity more than larger particle, the OP posits a media where larger particles experience the viscosity more.

One thing that's not clear to me from the OP is the scale of disturbance in the media of a large object moving through it. In the 'iron sphere' example, does the media penetrate the iron sphere? So it is the iron atoms that cause the disturbance? If that's the case I don't see an explanation of how the disturbance can propagate beyond the atomic level.

Yes this is pretty close to what I picture, thank you.

In the iron sphere example the media does penetrate the iron yes. So yes the number of atoms and the amount of volume they take up is what disturbs the grid. The disturbance propagates because if you change the position of a Grid particle this effect influences the internal shape of the Grid as each particle repels the other - move one particle and it's getting closer to another particle which is repelling it. This other particle therefore moves.

 

[martu]

Because you could define something's location and velocity relative to the grid. Special relativity just doesn't work like that.

Only if the Grid is not expanding. If the grid is expanding how could define something’s location and velocity?

That would impose a minimum wavelength at any given point in space and mean all higher wavelengths would be multiples of this. To the best of our observations, this just isn't the case. Besides, the vector of propagation for photons is in a straight line - the transverse waves exist within electromagnetic field space.

Yes the minimum wavelength is the shortest distance that two particles can be apart.

 

[martu]

I am getting the impression that this is one of those “expansion” TOEs where everything is expanding at the same proportional rate (so we can not detect the expansion) but some how that expansion results in gravitational attraction. Is my surmise correct martu?

Yes the expansion defines time the rate of expansion is the rate of time. Stop the expansion and you stop time - nothing can move in the Grid.

What math, did I miss a post where you actually calculated something with this?

It was the equation that defines the dimensions. In 3D the Force is proportional to r squared which implies a cube.

 

[martu]

Look at an even simpler case: a 1D line of particles with spring-like forces acting between them. Add a mass at a point in the line. This moves some of the grid particles. The forces balance out - the distortion and the force between the grid particles. Thus there is no overall force. The mass at rest will stay at rest. Add another mass and that mass will also stay at rest. Therefore there is no force between the 2 masses.

This model is not a valid model of gravity in any dimension.

No 1D is merely a point therfore movement and force makes no sense.

I apreciate your input in this. Could you do me a favour? Could you consider what happens in this scenario in the 2D Grid? Have one mass then add another.

 

[martu]

What distorts the grid? You stated that there is only the force between the grid particles. There is no force between the mass and the gird according to you.

Matter. If you move the gravity particles you change the structure of the Grid.

Also m is never equal to r. 'm' is a mass. 'r' is a distance. A mass is never equal to a distance. How many pounds are there in a light year?
Or do you mean x and y (directions in 2D).

Yep that's nearly it - in 2D it's effectively the directions x and y which defines the movement, x equals m and y equals r.

 

[Hellbound]

It was the equation that defines the dimensions. In 3D the Force is proportional to r squared which implies a cube.

This is incorrect.

Any force that varies with radius (or r squared, or square root of r, or anything r-based) implies a circular/spherical shape. In other words, at any distance r, the force exerted will be identical, so the set of all distances equal to r forms a line around the source point. In other words, the force experienced is identical at all points that are equidistant from the source. In two dimensions, this describes a circle. In three dimensions, a sphere. In four dimensions, a hypersphere, and so on.

If I am correc,t then your grid model is wrong. The correct packing for circles in 2D is a triagnluar pattern: If you have particles A, B, anc C, which are all adjacant to each other in your maximized packing scheme, then the angle ABC is 60 degrees, the angle BCA is 60 degreees, and the angle CAB is 60 degrees, forming an equilateral triangle if you drew lines between each point.

This changes in three dimenssions, but I believe it would develop into a pyramidal structure...just as if you tried to stack ball bearings or marbles all close together...a cube is not the optimal solution (at least, not as you are considering).

See http://mathworld.wolfram.com/SpherePacking.html for infor on sphere packing. Your optimal solutions are close cubic packing and hexagonal packing. What you've been describing is cubic lattice.

 

[The Man]

Yes the expansion defines time the rate of expansion is the rate of time. Stop the expansion and you stop time - nothing can move in the Grid.

Unfortunately time does not pass at any universally constant rate. It is influenced by both gravity and relative motion (time dilation). With the expansion of the grid as the rate of time and the rate of time dependent on gravity and relative motion we should be able to see things expanding at different rates based on either of those dependences, which we do not see.

It was the equation that defines the dimensions. In 3D the Force is proportional to r squared which implies a cube.

No, R³ or R cubed is 3D, which does not necessarily imply a cube but requires units to be cubed to represent a 3D volume (like Meter³), R squared implies an area or 2D.

No 1D is merely a point therfore movement and force makes no sense.

No, a point is 0D as it has no dimensions, 1D would be a line as it has only one dimension, length. Is it as Reality Check suggested that you are deliberately making your dimensional consideration 1 off just to get your professed math to work?


0D = point = no units
1D = line = units
2D = plane = units squared
3D = volume = units cubed
4D = spacetime = units⁴

 

[X]

Because as you get closer to a mass the grid is distorted more - the more the grid is distorted the further you go in a slice if time. Going further in a slice of time is fundamentally what acceleration is.

As you get further away from a mass the grid is distorted less therefore you travel a shorter distance in a slice of time.

This is easier in 2D where the Grid can be considered this way. With no mass we have a square and therefore no direction (m is equal to r) Add two masses that influence each other. Now think of the grid as an isosceles trapezoid with the heavier mass below and the lighter mass above, the height is r and m is effectively our direction. You can consider the direction of movement as being represented by the line at the base of the trapezoid. So some matter directly between the centres of mass of two stationary objects will move towards the larger mass.

This is also why you feel gravity when you accelerate - You are ‘pushing’ past the particles, the faster you go the more particles you have to push past.

Okay, I think I understand what you're getting at now.

A mass and the gravity-particle-grid exert a force on each other (action - reaction).

The mass forces the particles to spread apart. This creates an attraction, which reduces with increasing distance as the grid tries to draw other particles into the less-dense area.


Unfortunately, aside from being cumbersome, this model has basic problems with reality.

One example:
Either the gravity-particle-grid exerts a force on matter traveling through it, or it does not.

If it does, then you can account for the acceleration of objects due to gravity. However, at the same time, matter traveling through directionally uniform regions of your grid (a.k.a. probe in deep space, or planet moving around a circular orbit) would be slowed by the force exerted by the particles as it moves them.
This is not observed.
(Well, there is the Pioneer anomaly, but aside from having explanations waiting in the wings, not enough is known about it to justify throwing away physics. Yet.)

If your gravity-particle-grid does not exert a force on objects, it allows for the observed constant tangential velocity of orbital motion and the behaviour of space probes. But it does not account for acceleration towards a mass. Nor does this model allow for the mass to distort the grid in the first place.

As a third alternative, you could specify that the motion is only caused by a density gradient, that is, from one density to another.
Even if you do this, however, you will still need to account for how mass can distort the grid in the first place. And it likely has other implications I haven't thought of.

 

[nathan]

It is proportional to the distortion in the Grid.

How is this distortion measured? Are you saying the force felt between two adjacent points A & B depends on more that just the distance between then? Are other points involved? Please be specific?

The expansion of the Grid is caused by the big bang – the edge of the universe is the edge of the Grid.

So, how does the expansion of the universe affect this? Were all the points close together initially, or were new vertices created during expansion.

Because the maths works.

What maths?

 

[Almo]

I've read the whole thread now, and I think that there's a few too many fundamental problems with the model. I think martu's time would be better spent getting a more thorough understanding of the current theory of gravity, rather than trying to patch the model or learn gravitational physics through questions about the model.

 

[nathan]

Yes this is pretty close to what I picture, thank you.

ok, you realize that viscosity in a static medium only provides a retarding force? It never provides acceleration.

In the iron sphere example the media does penetrate the iron yes. So yes the number of atoms and the amount of volume they take up is what disturbs the grid. The disturbance propagates because if you change the position of a Grid particle this effect influences the internal shape of the Grid as each particle repels the other - move one particle and it's getting closer to another particle which is repelling it. This other particle therefore moves.

Hm, your answer seems internally inconsistent. If the media interpenetrates the object, and the atoms of the object cause distortion of the grid, then the grid distortions are at the atomic level. But you also seem to claim the grid distortions propagate further than this.

Consider an atom at the surface of the object. There will be grid particles on both sides of it (I'm considering a 1 dimensional line perpendicular to the surface of the object). There will be other atoms of iron on one side of that line, and no atoms on the other side (in a vacuum). Now consider grid particles along that line. Will a grid particle on the outside of the object 'feel' any influence from iron atoms on the inside? They may feel some, but they must feel less than the iron atom right at the edge of the object. I think they must feel considerably less. If they felt a significant fraction of the edge iron atom's influence then it would be harder for the grid to interpenetrate the iron sphere in the first place. The more you make grid particles outside the sphere feel the influence of internal atoms, the more you make it harder for the grid particles to interpenetrate the iron sphere, and the harder you make it for the iron sphere to move through the grid.

Your theory clearly hints at a preferred coordinate system of free space. As others have pointed out, such an ethereal theory has been found wanting by experiment.

 

[Reality Check]

I shall try to explain what happens in 2D to see if that helps. This thread needs pictures.

In 2D you have F equals m over r where F can be considered to be the most likely direction a piece of matter will move in.

Consider a uniform grid which can be pictured thus:

http://i299.photobucket.com/albums/mm309/themartu/GridNoGravity.jpg

Now picture an atom in the centre of this square – it will not move as atoms tend towards the longest line of the Grid, in this case all the lines are equal.

Now add Gravity. We have added two masses one above the particle and one below the particle. The Grid now looks like this:

http://i299.photobucket.com/albums/mm309/themartu/GridGravity.jpg

The force or rather which direction the atom is likely to travel is defined by m over r where m is the longest line in the quadrilateral. D is the distance between the two masses that are influencing the Grid.

Now we double the distance between the two masses:

http://i299.photobucket.com/albums/mm309/themartu/GridGravityMassesfurtherapart.jpg

The particle still experiences a force but it’s less as r has doubled while m has stayed the same. In 2D this relationship is linear.

First question - is this logially sound?

F= m/r is not a spring. A spring is F = kr.
In 3D does your formula become F= m²/r², i.e. the Newtonian gravitational force between 2 masses of mass m with G = 1?

Thus this is not logically sound.