unknown vector of a closed space

Jul 2019
74
1
Authour: Ross Blenkinsop
Title
Finding unknown velocity vector of a moving closed space.
Introduction
It is a long held theory that if one is in a closed space moving at a constant velocity, there is no experiment that can be carried out, within that closed space, without reference to phenomena external to that closed space, to find the unknown velocity of the closed space.
In this article I will propose an experiment to show how such an experiment could be conducted. I will also discuss some of the ramification of the experiment. In essence this problem involves finding an unknown velocity vector.
Attached are figures 1 to 9 , all within the one file.
Referring to figure 1.
This shows the general set up of the experiment. There is a closed space that is travelling at some unknown velocity depicted by the vector Vu (vector unknown) in the direction Vu at the magnitude of Vu.
The purpose of the experiment is to find the unknown vector Vu, from an experiment conducted within the closed space, without reference to any phenomena external to the closed space.
Within that space are 2 trains one red one black. Each train has a collimator attached to it. The red train has many identical collimators attached to it. The trains sit on tracks. It just so happens the tracks are parallel to Vu. The red train is moving to the left at a known constant velocity Vk (relative to the tracks). The black train is not moving relative to the tracks.
There is a strobe light in the ceiling depicted as a green square , lets call this point B. The strobe light is at rest wrt the black train. The collimator of the black train sits directly below the light at point B.
The black train has a collimator attached to it of length L. The red train has many collimators attached to it, all also of length L. There is an identical stop watch at the top and at the bottom of every collimator. Apart from the number of collimators the trains, collimators and stop watches are all identical in every way.
Each collimator has a detector at the top that can detect when a photon enters the collimator. Each collimator has a detector in its base that can detect when a photon arrives at the base of the collimator.
When a photon enters a collimator the top detector is triggered and it fires a light pulse to the top stop watch associated with that collimator, which starts the top stop watch running.
When a photon arrives at the base of a collimator the base detector is triggered and fires a light pulse to the bottom stop watch associated with that collimator, which starts the bottom stop watch running.
As shown in fig 1 the red train is moving at a constant velocity Vk towards the black train. There is a point on the track just before the black train. When the front of the red train reaches that point a pulse is sent to the strobe in the ceiling. When the pulse arrives at the strobe it is triggered. The point on the track makes an allowance for the time it takes for the pulse to transit to the strobe. The point on the track is positioned in such a way so that the strobe will trigger when the collimator on the black train is exactly adjacent to the front collimator of the red train, and both collimators will be precisely below point B. However, this is not precisely correct.
As the entire experiment is moving at a velocity Vu the strobe will move some distance in the direction Vu, in the time interval whilst the pulse is in transit from the point on the track to the strobe. So the strobe will actually trigger at some point to the right of point B. As long there is a portion of the red train below the strobe, allowing for the Vu movement, the experiment will work.
Allowing for the above the strobe goes off when the last collimator on the red train is directly below the strobe, see fig 2.
It just so happens that the speed of the red train Vk is equal to Vu. In the experiment the experimenters would try many values for Vk, I am simply showing the specific experiment where Vk is equal to Vu.
In trying many different values for Vk there are three possible permutations Vk < Vu, Vk > Vu and Vk = Vu.
Mechanics of collimators and stop watches
Referring to Fig 9. A photon from the strobe enters the aperture at the top of a given collimator, triggering the top detector, sending a light pulse to the top stop watch, thus starting the top stop watch.
The photon traverses the length of the collimator, eventually triggering the base detector, sending a light pulse to the bottom stop watch.
The time is then calculated by subtracting the top stop watch from the bottom stop watch.
Fig 2
This image is a short time after the strobe is triggered, let’s call this time T1.
The red square in the roof is the centre of the photon wave front, lets call this point A. This is the original position where the strobe was when it fired.
The photons spreading out from the centre of the wave front are shown as a red semi circle with point A at the centre of that semi circle.
As the closed space is moving at Vu in direction Vu the actual strobe is a distance to the right of point A, the red square. The actual strobe light is shown as the green square in the roof, just to the right of point A. I have called this point B.
If the strobe fired at time T0 then the new position of the strobe, from where it was at time T0, would be delta T, times the velocity of the closed space; magnitude of Vu. Or (T0 - T1) multiplied by the magnitude of Vu or Vu x (T1-T0).
The red train is moving to the left (direction of Vk) at a velocity Vk . If the strobe triggered at timeT0 and Vk is equal to Vu. The position of the red train at time T1, compared to where it was at T0, would be would be the magnitude of vector Vu, minus the magnitude of vector Vk multiplied by delta T or (Vu - Vk) x (T1 – T0). The magnitude of Vk minus the magnitude of Vu is zero, as Vk is equal to and opposite to Vu.
The interpretation of this is that one of the collimators on the red train will at all times remain directly below point A, the point where the strobe triggered, the centre of the wavefront.
This collimator at rest wrt point A will interact with a photon moving vertically wrt point A. The photon will be perpendicular to the roof of the room. That photon will have no horizontal component. The red train will like wise have no component.
Each collimator on the red train has its own top stop watch and bottom stop watch.
For an explanation of “horizontal component” see fig 7.
Fig 2 to 5
The figures are a series of snapshots in time, early at figure 2 and time incrementally progressing figures 3 to 5. The trains move further apart. The photon wave front gets larger and larger. The figures are not to scale.
All the stop watches are a fixed distance D from their respective collimator.
According to current theory what should an observer at point B see?
An observer at point B would see the red train moving away from them at Vk. The conventional wisdom would dictate that the observer at point B, at rest wrt point B, should observe that the time difference on the red train stop watch should always be more than the time difference on the black train stop watch.
Red train summary analysis
Referring to fig 6 the photon moving vertically wrt point A, perpendicular to the roof, would be the first photon to reach a top detector on a collimator on the red train.
A vector equivalent of this photon would be a vector pointing vertically at the floor, perpendicular to the roof, and would have a magnitude of the speed of light C towards the floor.
As mentioned the length of the collimator is L. It is a trivial matter to calculate that the time taken for the photon to traverse the length of the red train collimator would be L/C.As the photon has no horizontal component and as the collimator has no horizontal component.
Whilst the photon is traversing the red train collimator there would be zero horizontal displacement of the collimator or the photon.
Finding : I submit for a collimator of length L, if the long axis of a collimator is lying on a vector that points to the centre of a wave front, if the collimator is at rest wrt that point. The time it takes for that photon to traverse the length of that collimator is the smallest possible time that action can be performed in any experiment.
Black train summary analysis
Referring to fig 7. The black train has a horizontal component which will be the magnitude of Vu. All the photons interacting with the black train will also have horizontal components.
For a photon to interact with the collimator on the black train it must also have a horizontal component equal to, or greater than, the magnitude of Vu.
It is trivial to demonstrate that whilst any photon is traversing the collimator on the black train the collimator will experience a horizontal displacement and the photon will also experience a corresponding horizontal displacement. The horizontal displacement will be proportional to Vu.
A photon interacting with the black train will naturally be moving at C. As the black train photon has a horizontal component, the magnitude of the photons vertical component will always be some value less than C, where C is the speed of light.
Referring to fig 7 the time taken for a photon to traverse the length of the collimator on the black train will be the length L divided by the magnitude of the vertical component of the photon (Vv), where the magnitude of the vertical component of the photon will always be some value less than C.
Therefore the black train photon will always traverse the length L, the length of the collimator on the black train, in a time greater than the equivalent time for the red train.
Collimator detectors delay red train
Referring to figure 8 and 9. These images are depictions of a collimator and the light on the collimator.
When a photon enters the aperture of the collimator the detector is triggered and a signal flows to the light on the collimator.
The arrangement of the top light is identical to the bottom light. As they are identical the delay in time for the signal to travel from aperture to the detector and detector to the light at the top will be the same as the bottom. As the delay is identical for both there will be no material effect on the transmission of the signal to the stop watch. The same is true for the black train.
I think it is reasonable to assume ignoring this delay on the red and black trains will have no material effect on the outcome of the experiment.
Red train detailed analysis
Each collimator on the red train has its own top stop watch and bottom stop watch. Each stop watch is a distance D from the collimator. D is equal to L, where L is the length of the collimators.
As Vk is equal to an opposite to Vu the red train will not undergo any horizontal displacement after the roof strobe has triggered.
Whilst the photon from the top light on the collimator is in transit to the top stop watch, the position of the collimator will not undergo any horizontal displacement and the position of the stop watch will not undergo any horizontal displacement, as the vector Vk is exactly cancelled by the vector Vu. The photon moving vertically wrt point also will not undergo any horizontal displacement.
If the photon moving vertically wrt point A was resolved to its vertical and horizontal vectors the vertical vector would have a magnitude of C and the horizontal vector would be 0.
Therefore the time it takes for the top photon from the light at the top of the collimator, to transit from the top light to the top watch will be
Ttop = D/C (where C is the speed of light)
In the time Ttop the photon will move a distance Ttop x C down the collimator. As D = L (the length of the collimator) It is a trivial matter to see that Ttop x C is equal to L. In the time the light travels to the stop watch the photon in the collimator traverse the entire length of the collimator.
As not point on the red train undergoes a horizontal displacement, and as a vector representation of the collimator photon has no horizontal component, only a vertical component of C. Once the photon arrives at the base of the collimator the distance from the collimator to the stop watches is still D.
No entity (the photons, stop watches or collimator) on the red train, undergoes horizontal displacement at any time.
So, whilst a photon is in flight to the top stop watch all entities will remain at rest wrt point A. Whilst a photon is in flight down the collimator all entities will remain at rest wrt point A. Whilst a photon is in flight to the bottom stop watch all entities will remain at rest wrt point A.
The time it takes for the photon from the bottom light to transit to the bottom stop watch is the same as for the top stop watch
Tbot = D/C
The time difference between the top stop watch and the bottom stop watch will be
T_Diff_Red = Tcol + Tbot
But Tcol = 0 so
T_Diff_Red = D/C
From the note below L = D so replacing gives
T_Diff_Red = L/C
Note the time difference is not a function of Vk or Vu.
So what has happened in this scenario is in the time photon has moved from the top of the collimator a distance L to the top stop watch, at velocity C, the photon in the collimator has traversed then entire length of the collimator (L) at velocity C. As the distance from the collimator to the stop watch is L, the length of the collimator is L and both photons are travelling at C. Not a single entity has a horizontal component as Vk is exactly cancelled by Vu.
 
Jul 2019
74
1
Authour: Ross Blenkinsop
Title
Finding unknown velocity vector of a moving closed space.
Introduction
It is a long held theory that if one is in a closed space moving at a constant velocity, there is no experiment that can be carried out, within that closed space, without reference to phenomena external to that closed space, to find the unknown velocity of the closed space.
In this article I will propose an experiment to show how such an experiment could be conducted. I will also discuss some of the ramification of the experiment. In essence this problem involves finding an unknown velocity vector.
Attached are figures 1 to 9 , all within the one file.
Referring to figure 1.
This shows the general set up of the experiment. There is a closed space that is travelling at some unknown velocity depicted by the vector Vu (vector unknown) in the direction Vu at the magnitude of Vu.
The purpose of the experiment is to find the unknown vector Vu, from an experiment conducted within the closed space, without reference to any phenomena external to the closed space.
Within that space are 2 trains one red one black. Each train has a collimator attached to it. The red train has many identical collimators attached to it. The trains sit on tracks. It just so happens the tracks are parallel to Vu. The red train is moving to the left at a known constant velocity Vk (relative to the tracks). The black train is not moving relative to the tracks.
There is a strobe light in the ceiling depicted as a green square , lets call this point B. The strobe light is at rest wrt the black train. The collimator of the black train sits directly below the light at point B.
The black train has a collimator attached to it of length L. The red train has many collimators attached to it, all also of length L. There is an identical stop watch at the top and at the bottom of every collimator. Apart from the number of collimators the trains, collimators and stop watches are all identical in every way.
Each collimator has a detector at the top that can detect when a photon enters the collimator. Each collimator has a detector in its base that can detect when a photon arrives at the base of the collimator.
When a photon enters a collimator the top detector is triggered and it fires a light pulse to the top stop watch associated with that collimator, which starts the top stop watch running.
When a photon arrives at the base of a collimator the base detector is triggered and fires a light pulse to the bottom stop watch associated with that collimator, which starts the bottom stop watch running.
As shown in fig 1 the red train is moving at a constant velocity Vk towards the black train. There is a point on the track just before the black train. When the front of the red train reaches that point a pulse is sent to the strobe in the ceiling. When the pulse arrives at the strobe it is triggered. The point on the track makes an allowance for the time it takes for the pulse to transit to the strobe. The point on the track is positioned in such a way so that the strobe will trigger when the collimator on the black train is exactly adjacent to the front collimator of the red train, and both collimators will be precisely below point B. However, this is not precisely correct.
As the entire experiment is moving at a velocity Vu the strobe will move some distance in the direction Vu, in the time interval whilst the pulse is in transit from the point on the track to the strobe. So the strobe will actually trigger at some point to the right of point B. As long there is a portion of the red train below the strobe, allowing for the Vu movement, the experiment will work.
Allowing for the above the strobe goes off when the last collimator on the red train is directly below the strobe, see fig 2.
It just so happens that the speed of the red train Vk is equal to Vu. In the experiment the experimenters would try many values for Vk, I am simply showing the specific experiment where Vk is equal to Vu.
In trying many different values for Vk there are three possible permutations Vk < Vu, Vk > Vu and Vk = Vu.
Mechanics of collimators and stop watches
Referring to Fig 9. A photon from the strobe enters the aperture at the top of a given collimator, triggering the top detector, sending a light pulse to the top stop watch, thus starting the top stop watch.
The photon traverses the length of the collimator, eventually triggering the base detector, sending a light pulse to the bottom stop watch.
The time is then calculated by subtracting the top stop watch from the bottom stop watch.
Fig 2
This image is a short time after the strobe is triggered, let’s call this time T1.
The red square in the roof is the centre of the photon wave front, lets call this point A. This is the original position where the strobe was when it fired.
The photons spreading out from the centre of the wave front are shown as a red semi circle with point A at the centre of that semi circle.
As the closed space is moving at Vu in direction Vu the actual strobe is a distance to the right of point A, the red square. The actual strobe light is shown as the green square in the roof, just to the right of point A. I have called this point B.
If the strobe fired at time T0 then the new position of the strobe, from where it was at time T0, would be delta T, times the velocity of the closed space; magnitude of Vu. Or (T0 - T1) multiplied by the magnitude of Vu or Vu x (T1-T0).
The red train is moving to the left (direction of Vk) at a velocity Vk . If the strobe triggered at timeT0 and Vk is equal to Vu. The position of the red train at time T1, compared to where it was at T0, would be would be the magnitude of vector Vu, minus the magnitude of vector Vk multiplied by delta T or (Vu - Vk) x (T1 – T0). The magnitude of Vk minus the magnitude of Vu is zero, as Vk is equal to and opposite to Vu.
The interpretation of this is that one of the collimators on the red train will at all times remain directly below point A, the point where the strobe triggered, the centre of the wavefront.
This collimator at rest wrt point A will interact with a photon moving vertically wrt point A. The photon will be perpendicular to the roof of the room. That photon will have no horizontal component. The red train will like wise have no component.
Each collimator on the red train has its own top stop watch and bottom stop watch.
For an explanation of “horizontal component” see fig 7.
Fig 2 to 5
The figures are a series of snapshots in time, early at figure 2 and time incrementally progressing figures 3 to 5. The trains move further apart. The photon wave front gets larger and larger. The figures are not to scale.
All the stop watches are a fixed distance D from their respective collimator.
According to current theory what should an observer at point B see?
An observer at point B would see the red train moving away from them at Vk. The conventional wisdom would dictate that the observer at point B, at rest wrt point B, should observe that the time difference on the red train stop watch should always be more than the time difference on the black train stop watch.
Red train summary analysis
Referring to fig 6 the photon moving vertically wrt point A, perpendicular to the roof, would be the first photon to reach a top detector on a collimator on the red train.
A vector equivalent of this photon would be a vector pointing vertically at the floor, perpendicular to the roof, and would have a magnitude of the speed of light C towards the floor.
As mentioned the length of the collimator is L. It is a trivial matter to calculate that the time taken for the photon to traverse the length of the red train collimator would be L/C.As the photon has no horizontal component and as the collimator has no horizontal component.
Whilst the photon is traversing the red train collimator there would be zero horizontal displacement of the collimator or the photon.
Finding : I submit for a collimator of length L, if the long axis of a collimator is lying on a vector that points to the centre of a wave front, if the collimator is at rest wrt that point. The time it takes for that photon to traverse the length of that collimator is the smallest possible time that action can be performed in any experiment.
Black train summary analysis
Referring to fig 7. The black train has a horizontal component which will be the magnitude of Vu. All the photons interacting with the black train will also have horizontal components.
For a photon to interact with the collimator on the black train it must also have a horizontal component equal to, or greater than, the magnitude of Vu.
It is trivial to demonstrate that whilst any photon is traversing the collimator on the black train the collimator will experience a horizontal displacement and the photon will also experience a corresponding horizontal displacement. The horizontal displacement will be proportional to Vu.
A photon interacting with the black train will naturally be moving at C. As the black train photon has a horizontal component, the magnitude of the photons vertical component will always be some value less than C, where C is the speed of light.
Referring to fig 7 the time taken for a photon to traverse the length of the collimator on the black train will be the length L divided by the magnitude of the vertical component of the photon (Vv), where the magnitude of the vertical component of the photon will always be some value less than C.
Therefore the black train photon will always traverse the length L, the length of the collimator on the black train, in a time greater than the equivalent time for the red train.
Collimator detectors delay red train
Referring to figure 8 and 9. These images are depictions of a collimator and the light on the collimator.
When a photon enters the aperture of the collimator the detector is triggered and a signal flows to the light on the collimator.
The arrangement of the top light is identical to the bottom light. As they are identical the delay in time for the signal to travel from aperture to the detector and detector to the light at the top will be the same as the bottom. As the delay is identical for both there will be no material effect on the transmission of the signal to the stop watch. The same is true for the black train.
I think it is reasonable to assume ignoring this delay on the red and black trains will have no material effect on the outcome of the experiment.
Red train detailed analysis
Each collimator on the red train has its own top stop watch and bottom stop watch. Each stop watch is a distance D from the collimator. D is equal to L, where L is the length of the collimators.
As Vk is equal to an opposite to Vu the red train will not undergo any horizontal displacement after the roof strobe has triggered.
Whilst the photon from the top light on the collimator is in transit to the top stop watch, the position of the collimator will not undergo any horizontal displacement and the position of the stop watch will not undergo any horizontal displacement, as the vector Vk is exactly cancelled by the vector Vu. The photon moving vertically wrt point also will not undergo any horizontal displacement.
If the photon moving vertically wrt point A was resolved to its vertical and horizontal vectors the vertical vector would have a magnitude of C and the horizontal vector would be 0.
Therefore the time it takes for the top photon from the light at the top of the collimator, to transit from the top light to the top watch will be
Ttop = D/C (where C is the speed of light)
In the time Ttop the photon will move a distance Ttop x C down the collimator. As D = L (the length of the collimator) It is a trivial matter to see that Ttop x C is equal to L. In the time the light travels to the stop watch the photon in the collimator traverse the entire length of the collimator.
As not point on the red train undergoes a horizontal displacement, and as a vector representation of the collimator photon has no horizontal component, only a vertical component of C. Once the photon arrives at the base of the collimator the distance from the collimator to the stop watches is still D.
No entity (the photons, stop watches or collimator) on the red train, undergoes horizontal displacement at any time.
So, whilst a photon is in flight to the top stop watch all entities will remain at rest wrt point A. Whilst a photon is in flight down the collimator all entities will remain at rest wrt point A. Whilst a photon is in flight to the bottom stop watch all entities will remain at rest wrt point A.
The time it takes for the photon from the bottom light to transit to the bottom stop watch is the same as for the top stop watch
Tbot = D/C
The time difference between the top stop watch and the bottom stop watch will be
T_Diff_Red = Tcol + Tbot
But Tcol = 0 so
T_Diff_Red = D/C
From the note below L = D so replacing gives
T_Diff_Red = L/C
Note the time difference is not a function of Vk or Vu.
So what has happened in this scenario is in the time photon has moved from the top of the collimator a distance L to the top stop watch, at velocity C, the photon in the collimator has traversed then entire length of the collimator (L) at velocity C. As the distance from the collimator to the stop watch is L, the length of the collimator is L and both photons are travelling at C. Not a single entity has a horizontal component as Vk is exactly cancelled by Vu.
Reading the red stop watches
The stop watches could be read by someone at rest wrt the red train, or at rest wrt point B.
Because 2 stop watches are used the possibility exists that reading the stop watches could cause an error as the person reading the stop watches is subject to the vector Vu, if they are at rest wrt the point B. The top stop watch sit vertically above the bottom stop watch.
Despite who reads the stop watches, if both stop watches are compared from a point vertically mid way between each stop watch no error will occur and the true time will be recorded, which will agree with the above results.
 
Jul 2019
74
1
Black train stop watch analysis
Referring to fig 7. After the strobe triggers a photon will travel from the strobe to the aperture on the black train collimator.
Strobe photon - point A to collimator
As the black train is subject to the unknown vector Vu, in the time the photon transits from where the strobe triggered (point A) to the aperture on the black train collimator, the black train will undergo a horizontal displacement proportional to Vu.
The photon from the point A will trace a straight line, from the point A to the point where it eventually enters the aperture on the black train collimator. Lest call this line Q. The photon from point A will travel along Q at C. If a vector is drawn along Q , pointing in the direction of the black train, of a length C (the speed of light).That vector can be resolved into its vertical and horizontal components as drawn in fig 7.
As can be seen a photon on the line Q undergoes a horizontal and vertical displacement, both of which are less than C.
Collimator and top stop watch horizontal displacement
As the train and the stop watch are subject to the unknown vector Vu. As the photon transits from the top light on the collimator, to the top stop watch, the collimator and stop watch, and all points connected to the black train, will undergo a horizontal displacement in the direction of Vu of magnitude Vu.
Top light photon transit to top stop watch
As the top stop watch is the same height as the light on the top of the collimator. The photon travelling from the top light on the collimator to the top stop watch will not undergo a horizontal or vertical displacement. As photons are not ballistic in nature the vector Vu will have no influence on this photon.
Collimator horizontal displacement
As the train and collimator are subject to the unknown vector Vu. As the strobe photon transits down the collimator the collimator will undergo a horizontal displacement in the direction Vu and a magnitude Vu. As already mentioned the photon travelling down the collimator has a horizontal component in the direction of Vu and a magnitude of Vu.
The horizontal component of the photon travelling down the collimator is the same as the horizontal component of the collimator. If that was not the case the photon would collide with the sides of the collimator and not reach the base of the collimator. The collimator would not work.
Bottom light photon transit to bottom stop watch
As the bottom stop watch is the same height as the light on the top of the collimator. The photon travelling from the bottom light on the collimator, to the bottom stop watch will not undergo a horizontal or vertical displacement. As photons are not ballistic in nature the vector Vu will have no influence on this photon.
Collimator and bottom stop watch horizontal displacement
As the train and the stop watches are subject to the unknown vector Vu. As the photon transits from the bottom light on the collimator, to the bottom stop watch, the collimator and stop watch, and all points connected to the black train, will undergo a horizontal displacement in the direction of Vu of magnitude Vu.
Time of flight top light to top stop watch
The time it takes for the photon to transit from the top light to the top stop watch will be
Ttop = (D – (Ttop x Vu))/C
Note D = L so
Ttop = (L – (Ttop x Vu))/C
Ttop = (L – (Ttop x Vu))/C simplifies to
Ttop = L/(Vu + C)
Time of flight collimator
The distance the collimator photon will move down the collimator in time Ttop is
Tdist = Vv x Ttop
Or
Tdist = [L/(C + Vu)] x Vv
The distance reaming to be traverse by the photon in the collimator after time Ttop will be
Dremain = L – Tdist
The time to traverse the distance Dremain at Vv will be
Trem = Dremain /Vv
Or
Trem = [L – (Vv x [(L – (Ttop x Vu))/C]) ]/Vv
In the time Trem the distance between the collimator and the bottom stop watch will not change as the collimator experiences a horizontal displacement of Vu, as does the stop watch.
Time of fliht bottom stop watch
In the time that a photon from the bottom light transits to the bottom stop watch, the bottom stop watch will experience a horizontal displacement of in the direction Vu at Vu. The time taken for a photon to move from the bottom light to the bottom stop watch will be
Tbot = (D – (Tbot x Vu))/Vu
Tbot will be equal to Ttop.
The time difference on the black stop watches will be.
T_diff_black = Tbot + Treml
Plugging in values
Time of flight to stop watch
Tbot = Tbot = L/(C + Vu)
If Vu = 200 000 000
C = 300 000 000
L = 10 m
Tbot = Ttop = 10/500 000 000 s
Vv
From fig 7
If L = 10
Vu = 200 000 000
As Vv = squareroot [(C x C) – (Vu x Vu)] the vertical displacement of the photon in the Black train (Vv) is
Vv = squareroot(300 000 0000 – 200 000 000) = 233 606 000 m/s
Time of flight collimator
The distance the photon will move down the collimator in time Ttop is
Tdist = Vv x Ttop
Or
Tdist = [L/(C + Vu)] x Vv
If L = 10
Vu = 200 000 000
Vv = 233 606 000
The result is
Tdist = (10/500 000 000) x 233 606 000 = 4.67212 m
The distance reaming to be traverse by the photon in the collimator will be if L = 10
Dremain = L – Tdist = 10 - 4.67212 = 5.32788m
When the top stop watch started the photon on the black train was = 4.67212 m down the black train collimator. The distance to the bottom of the collimator is 10 - 4.67212 m or 5.32788 m
The time it will take for the photon in the collimator to traverse the remaining portion of the collimator 5.32788 m is the distance to travel divided by the vertical displacement of the photon
Trem = Dremain /Vv
Or
Trem = [L – (Vv x [(L – (Ttop x Vu))/C]) ]/Vv
If L is 10 m
Vu is 200 000 000 m/s
Vv = 233 606 000
Trem = 5.32788 /223 606 000 s
Bottom stop watch
Once the photon reaches the bottom of the collimator a pulse is fired at the bottom stop watch. As the bottom stop watch is equidistant to the top stop watch the time of flight will be the same as the top stop watch.
Summary
From the above
T_diff_black = Tbot + Trem
Tbot = 10/500 000 000s
Trem = 5.32788 /223 606 000 s
so
T_diff_black = 5.32788 /223 606 000 s + 10/500 000 000s
And
T_diff_red was L/C or 10/300 000 000
Final analysis
The question is is T_diff_black > T_diff_red? or is 5.32788 /223 606 000 + 10/400 000 000 more than 10/300 000 000?
Converting to the same denominator
Trem = 5.32788 /223 606 000 = 7.1481266/300 000 000
Tbot = Ttop = 10/400 000 000 = 7.4/300 000 000
adding
T_diff_black = 7.1481266/300 000 000 + 7.5/300 000 000 = c 14.6/300 000 000
14.6/300 000 000 is circa 40% greater than 10/300 000 000 which means the time difference on the black train is substantially more than the time difference on the red train, which proves the theory.
Further observations
Wrong place
If the red train is moving at Vk , and Vk = Vu, but the red train is not at point A, and is instead at some other point. Then, the photons from the strobe will have to move horizontally to enter the aperture of the collimators on the red train.
However, as Vk = Vu the red train will not at any stage undergo a horizontal displacement, but the photon entering the aperture are. As a result the photon will collide with side of collimators, and the collimator will not work.
The collimators not working would possibly indicate that the red train is moving at Vu.
Optimisation
The experiment could be optimised by making the stop watches as close as possible to the collimators and making the collimators very long.
Conclusion
The red train is driven at all different velocities and the experiment is conducted over and over. As the speed of the red train approaches the value Vu the time difference in the red train will get less and less until it reaches a minimum value. This is the point where the red train’s velocity wrt the train tracks exactly equals Vu.
If the red train is going faster than Vu then the time difference will start to increase again. By finding the minimum value it is possible to find Vu without reference to external phenomena.
This means there is such a thing as absolute time and absolute position, absolute length, absolute direction, absolute measurement.
As vk approaches Vu the net horizontal displacement of the red train decreases gradually to zero. As a consequence the time difference on the red train becomes less and less until the minimum time is found. The same occurs but in reverse as Vk gradually , after finding the minimum, gradually increases as Vk becomes more than Vu.
The end
 

topsquark

Forum Staff
Apr 2008
3,024
638
On the dance floor, baby!
Please do us a favor. Do you have a quick synopsis of what it is you are trying to do here? Since you are writing a paper here you can consider this a request for an abstract.

-Dan
 
Jul 2019
74
1
couple of minor changes to the text and trying to make it clearer and easy to read, reattached it here in PDF
 

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