fdesilva
05-01-2010, 05:56 AM
Abstract: It is demonstrate by the use of a theoretical model that if the shape of an expanding universe was maintained and the ration of space/time is a constant equal to C, then from this fact alone is sufficient to derive all the laws of motion.
Consider the following hypothetical model of a universe that starts as a point (similar to the BigBang) and is expanding symmetrically, then it would result in an expanding 4 Dimensional sphere.
Consider this universe to be a 4 Dimensional sphere with time as the radial Dimension. Then all of space at “present” would be represented by the “3D Surface” of the sphere. Thus each point in space at “present” would be a boundary point. Now the expanding universe would correspond to an increase in the radial dimension or time. If the sphere is to remain a sphere then it would have to be accompanied by an equal increase of the surface or space.
Let us take a 2D cross section from this 4D sphere containing its centre. This would be a familiar circle with the radius representing time and the circumference 1D of space. Thus the size of this 1D of space at any time t would be given by
1D space = Constant*2*pi*t
Thus d(space)/d(time) = Constant*2*pi = CONSTANT. ---- 1
Could this CONSTANT = C = Speed of light?
Now take the square of both sides, then we obtain the metric for a photon travelling in empty space (special relativity).
d(space)^2 / d(time)^2 = C^2
(dx^2 + dy^2 + dz^2 ) = C^2dt^2
The above holds in our universe, in areas free of matter. Thus it could well be that C the speed of light, is the ratio in which space/time is created by the expansion of the universe.
Hubbels law
Consider again the case of the 2 dimensional universe. As stated previously d(space)\d(time) = C. Now consider 2 objects on the circumference. As this universe expands, these 2 objects will also move apart. However the angle or the arc will always remain constant.
So be the distance between them at some time To . Then
(So + ds)/(To + dt ) = So/To = angle between point at centre of circle = Constant
So + ds = So(To + dt ) /To
Dividing by dt and rearranging gives
ds/dt = So(To + dt ) /(To * dt) - So /dt
ds/dt = So / To --------2
That is Velocity = ds/dt = (1/To ) So
Which is Hubbles law where (1/To ) = H = Hubble’s Constant
Special Relativity
Consider again the case of the 2 dimensional universe.
Consider 2 reference frames O and O’ such that at some time T0 O’ is So distance from O. Thus from 2 above its velocity as seen by O is given by ds/dt = So / To = Vo which is a constant.
It is worth noting that this corresponds to newtons first law as Vo will remain constant with expansion.
Let the units of the Time dimension be such that the constant in 1 is 2pi. (It has the same unit’s as S that is space as such it correspond to a simple circle)
Thus we have 2piTo = Do the Circumference of the circle or total of all space at the time To .
D/T = 2pi will be maintained as this 2d universe expands.
Consider a beam of light leaving O at To and Travelling towards O’ at a velocity 2*pi or the speed of light in this universe.
In O’ this light beam would have to travel a distance So . Lets say it takes a time dT’.
The we get 2pi(dT’) = So = To*Vo ----- 3
Consider the passage of the light as seen by O. As in the time the light travels, O’ is moving away from O at a velocity Vo it will see as having to travel a greater distance. Thus we get
2pi(dT) = So + dS = To*Vo + (dT)Vo
Substituting from 3
2pi(dT) = 2pi(dT’) + (dT)Vo
Rearranging
2pi(dT’) = 2pi(dT) - (dT)Vo
Dividing both sides by 2pi(dT) we get
(dT’) / (dT) = 1 - Vo/2pi = gamma ------- 4
4 is the time dilation equation of special relativity. To see the correspondence of this coordinate and units with that normally used we can equate the constants gamma as it is a ratio and should be the same. Thus if O’ is moving away from O at a velocity Vx we get.
gamma = 1 - Vo/2pi = SQRT (1 - Vx^2/ C^2) ------5
Where C is the velocity of light in the common coordinate system .Squaring both sides and solving 5 for Vx
Vx = sqrt ( ( Vo/pi)( 1- Vo/4pi)) ---------6
It can be seen by substitution when Vo = 2pi the velocity of light in this coordinate system we get Vx = C the velocity of light, as is to be expected.
By substituting (dT’) = dS’/Vo and (dT) = dS/Vo
We get
(dS’)/(dS) = = 1 - Vo/2pi = gamma ------- 7
Which is length contraction of Special relativity.
The intent of the demonstration above is not to propose that the shape of the universe is spherical but rather to demonstrate that if the universe had a shape and maintained that shape with expansion, then the expansion alone can account for the laws of motion.
General Relativity
Speculating further, around matter this expansion is changed, that is d(space)/d(time) is not equal to C. It could well be that matter is none other than those areas in which this constant is made less than C. Thus matter would be seen to attract each other or gravity would be proportional to the extent that matter has altered this constant C around it. Thus the curvature of general relativity is a direct consequence of a reduction in the ratio of space/time in the expansion around matter.
To give an example, take the following solution by Sir Arthur Stanley Eddington to the General Relativity field equations.
ds^2 = ( 1 + Gm/2C^2r)^4 (dx^2 + dy^2 + dz^2 ) - C^2dt^2( 1 - Gm/2C^2r)^2 / ( 1 + Gm/2C^2r)^2
Setting ds^2 = 0 gives the metric for a beam of light, which is proposed to be also the ratio of space/time creation in the localised expansion around matter.
Thus
d(space)^2 / d(time)^2 = C^2( 1 - Gm/2C^2r)^2 / ( 1 + Gm/2C^2r)^6
The increase in entropy or disintegration of matter would be a direct result of the rate of expansion within matter tending towards the rate C which is the rate of empty space Time.
What is interesting in the above model is that the arrow of time follows the expansion. That is when we step into the future, we also step into newly created space-time.
Consider the following hypothetical model of a universe that starts as a point (similar to the BigBang) and is expanding symmetrically, then it would result in an expanding 4 Dimensional sphere.
Consider this universe to be a 4 Dimensional sphere with time as the radial Dimension. Then all of space at “present” would be represented by the “3D Surface” of the sphere. Thus each point in space at “present” would be a boundary point. Now the expanding universe would correspond to an increase in the radial dimension or time. If the sphere is to remain a sphere then it would have to be accompanied by an equal increase of the surface or space.
Let us take a 2D cross section from this 4D sphere containing its centre. This would be a familiar circle with the radius representing time and the circumference 1D of space. Thus the size of this 1D of space at any time t would be given by
1D space = Constant*2*pi*t
Thus d(space)/d(time) = Constant*2*pi = CONSTANT. ---- 1
Could this CONSTANT = C = Speed of light?
Now take the square of both sides, then we obtain the metric for a photon travelling in empty space (special relativity).
d(space)^2 / d(time)^2 = C^2
(dx^2 + dy^2 + dz^2 ) = C^2dt^2
The above holds in our universe, in areas free of matter. Thus it could well be that C the speed of light, is the ratio in which space/time is created by the expansion of the universe.
Hubbels law
Consider again the case of the 2 dimensional universe. As stated previously d(space)\d(time) = C. Now consider 2 objects on the circumference. As this universe expands, these 2 objects will also move apart. However the angle or the arc will always remain constant.
So be the distance between them at some time To . Then
(So + ds)/(To + dt ) = So/To = angle between point at centre of circle = Constant
So + ds = So(To + dt ) /To
Dividing by dt and rearranging gives
ds/dt = So(To + dt ) /(To * dt) - So /dt
ds/dt = So / To --------2
That is Velocity = ds/dt = (1/To ) So
Which is Hubbles law where (1/To ) = H = Hubble’s Constant
Special Relativity
Consider again the case of the 2 dimensional universe.
Consider 2 reference frames O and O’ such that at some time T0 O’ is So distance from O. Thus from 2 above its velocity as seen by O is given by ds/dt = So / To = Vo which is a constant.
It is worth noting that this corresponds to newtons first law as Vo will remain constant with expansion.
Let the units of the Time dimension be such that the constant in 1 is 2pi. (It has the same unit’s as S that is space as such it correspond to a simple circle)
Thus we have 2piTo = Do the Circumference of the circle or total of all space at the time To .
D/T = 2pi will be maintained as this 2d universe expands.
Consider a beam of light leaving O at To and Travelling towards O’ at a velocity 2*pi or the speed of light in this universe.
In O’ this light beam would have to travel a distance So . Lets say it takes a time dT’.
The we get 2pi(dT’) = So = To*Vo ----- 3
Consider the passage of the light as seen by O. As in the time the light travels, O’ is moving away from O at a velocity Vo it will see as having to travel a greater distance. Thus we get
2pi(dT) = So + dS = To*Vo + (dT)Vo
Substituting from 3
2pi(dT) = 2pi(dT’) + (dT)Vo
Rearranging
2pi(dT’) = 2pi(dT) - (dT)Vo
Dividing both sides by 2pi(dT) we get
(dT’) / (dT) = 1 - Vo/2pi = gamma ------- 4
4 is the time dilation equation of special relativity. To see the correspondence of this coordinate and units with that normally used we can equate the constants gamma as it is a ratio and should be the same. Thus if O’ is moving away from O at a velocity Vx we get.
gamma = 1 - Vo/2pi = SQRT (1 - Vx^2/ C^2) ------5
Where C is the velocity of light in the common coordinate system .Squaring both sides and solving 5 for Vx
Vx = sqrt ( ( Vo/pi)( 1- Vo/4pi)) ---------6
It can be seen by substitution when Vo = 2pi the velocity of light in this coordinate system we get Vx = C the velocity of light, as is to be expected.
By substituting (dT’) = dS’/Vo and (dT) = dS/Vo
We get
(dS’)/(dS) = = 1 - Vo/2pi = gamma ------- 7
Which is length contraction of Special relativity.
The intent of the demonstration above is not to propose that the shape of the universe is spherical but rather to demonstrate that if the universe had a shape and maintained that shape with expansion, then the expansion alone can account for the laws of motion.
General Relativity
Speculating further, around matter this expansion is changed, that is d(space)/d(time) is not equal to C. It could well be that matter is none other than those areas in which this constant is made less than C. Thus matter would be seen to attract each other or gravity would be proportional to the extent that matter has altered this constant C around it. Thus the curvature of general relativity is a direct consequence of a reduction in the ratio of space/time in the expansion around matter.
To give an example, take the following solution by Sir Arthur Stanley Eddington to the General Relativity field equations.
ds^2 = ( 1 + Gm/2C^2r)^4 (dx^2 + dy^2 + dz^2 ) - C^2dt^2( 1 - Gm/2C^2r)^2 / ( 1 + Gm/2C^2r)^2
Setting ds^2 = 0 gives the metric for a beam of light, which is proposed to be also the ratio of space/time creation in the localised expansion around matter.
Thus
d(space)^2 / d(time)^2 = C^2( 1 - Gm/2C^2r)^2 / ( 1 + Gm/2C^2r)^6
The increase in entropy or disintegration of matter would be a direct result of the rate of expansion within matter tending towards the rate C which is the rate of empty space Time.
What is interesting in the above model is that the arrow of time follows the expansion. That is when we step into the future, we also step into newly created space-time.