 104
 0
Main Question or Discussion Point
Can someone please have a look at my analysis here and tell me if there are any mistakes
the rest of the pistures are in my other posts
___________________________________________________________________________
Einstein’s explanation of time dilation is given here http://en.wikipedia.org/wiki/Time_dilation
Referring to fig 1
In Einstein’s explanation you have 2 identical clocks. Clock 1 is at rest wrt the observer. Clock 2 is in a space ship zooming along wrt the observer.
At the red end of the clock is a light source, at the blue end a detector.
If L is the length of the clock. The observer, observing the clock at rest wrt him would see that the light pulse would take L/c time to traverse the length of the clock.
Einstein postulated that as the observer would see the photons in clock 2 travel a longer path to move from the red end to the blue end when compared with length of the path the photons in the clock 1. Therefore clock 2 must tick slower (time passes slower) when compared to clock1.
Referring to fig 2
As in Einstein’s explanation again you have 2 identical clocks. Clock 1 and clock 2 are at rest wrt the observer. This is the trivial case. Again at the red end of each clock is a light source, at the blue end a detector.
If L is the length of the clock. The observer, observing both clocks would see that the light pulse would take L/c time to traverse the length of the clock.
Referring to fig 3
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a space ship that is moving wrt the observer.
As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.
If L is the length of the clock.
The observer, observing the clock 1 would see that the light pulse would take L/c time to traverse the length of the clock.
T = L/c
As can be seen the path travelled by the photons in the clock2 is shorter than the path of the photons in the clock1. This is also what the observer would see.
The time the photons in clock2 would take would be at most
T = L/(c + V)
As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c + V).
I note the limit of the function t = L/(c + V) as V tends to infinity so t (time) goes to 0
Referring to fig 4
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a space ship that is moving wrt the observer.
You should now note both clocks have been rotated through 180 degrees compared to fig 3. In Fig 3 the light source is at the right the detector at the left. In fig 4 the reverse.
As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.
If L is the length of the clock. The time taken in Clock 1 would be L/c time to traverse the length of the clock.
T = L/c
As can be seen the path travelled by the photons in the clock2 is longer than the path of the photons in the clock1. This is also what the observer would see.
The time the photons in clock2 would take would be at least
T = L/(c  V)
As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c  V).
I note the limit of the function t = L/(c  V) as V tends to infinity so t (time) goes to infinity
I have three questions
Is this analysis right?
How is it that the observer can get two different results simply by rotating the clock?
The theory is a moving clock ticks slower then a stationary clock. In fig 3 clock 2 is ticking faster than clock1 which refutes the accepted theory?
WHAT AN OBSERVER IN THE SPACE SHIP SEES
Referring to fig 5
I have a clock in the space ship.
The space ship is flying at constant velocity V.
At the red end of the clock is a light source, at the blue end a detector.
As the speed is constant the contraction is constant and the time rate is constant.
The observer, point A and point B form an isosceles triangle.
The length of the clock is L which equals the length from B to the observer and also A to the observer.
I NOTE THE DRAWING IS NOT TO SCALE.
At some initial time the ends of the clock will be located at points A and B. At some time in the future as the space ship is moving the ends of the clock will be located at points A hat and B hat.
At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.
By the time the pulse from the red end of the clock reaches the observer he will be located at point x.
When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.
By the time the pulse from the blue end reaches the observer the observer will be located at point z.
Analysis
Referring to fig 7
Using the cosine rule L3 can be calculated.
Times will be proportional to lengths as time is simply length/c
Angle g = 60 degrees
Angle e = 120 degrees
Cos 60 = 0.5
Cos 120 = 0.5
L3^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
t1 is the time it takes for the pulse of light to travel from B hat to point z
Referring to fig 8
Using the cosine rule L1 can be calculated.
Angle e = 60 degrees
Cos 60 = 0.5
L1^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
t2 is the time it takes for the pulse of light to travel from point A to point x
from fig 5
L2 = L – V(t3)
t3 is the time it takes for the pulse of light to travel from point A to point B hat.
The time difference the observer will see between the first pulse of light he sees and the second is:
Time_Diff1 = (L1 – (L2 + L3))/c
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L2 = L – V(t3)
Referring to fig 6
I have a clock in the space ship.
The space ship is flying at constant velocity V.
At the red end of the clock is a light source, at the blue end a detector.
As the speed is constant the contraction is constant and the time rate is constant.
The observer, point A and point B form an isosceles triangle.
The length of the clock is L which equals the length from B to the observer and also A to the observer.
I note the clock in fig 6 is rotated 180 degrees wrt the clock in fig 5 the red and blue ends have changed.
I NOTE THE DRAWING IS NOT TO SCALE.
At some initial time the ends of the clock will be located at points A and B. At some time in the future as the space ship is moving the ends of the clock will be located at points A hat and B hat.
At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.
By the time the pulse from the red end of the clock reaches the observer he will be located at point x.
When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.
By the time the pulse from the blue end reaches the observer the observer will be located at point z.
Analysis
Referring to fig 7
Using the cosine rule L1 can be calculated.
Times will be proportional to lengths as time is simply length/c
Angle g = 60 degrees
Angle e = 120 degrees
Cos 60 = 0.5
Cos 120 = 0.5
L1^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
t1 is the time it takes for the pulse of light to travel from point A to point x
Referring to fig 8
Using the cosine rule L3 can be calculated.
Angle e = 60 degrees
Cos 60 = 0.5
L3^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
t2 is the time it takes for the pulse of light to travel from point B hat to point z
from fig 6
L2 = L + V(t3)
t3 is the time it takes for the pulse of light to travel from point A to point B hat.
The time difference the observer will see between the first pulse of light he sees and the second is:
Time_Diff2 = (L1 – (L2 + L3))/c
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L2 = L + V(t3)
Observations
Further observation:
L3 in fig 6 = L1 in fig 5
L1 in fig 6 = L3 in fig 5
V(t3) in fig 5 = V(t3) in fig 6
The time difference measured in fig 5 must equal the time difference measured in fig 6.
Time_Diff1 = (L1 – (L2 + L3))/c
Time_Diff2 = (L1 – (L2 + L3))/c
Time_Diff1 = Time_Diff2
Time_Diff2 = (L1 – (L2 + L3))/c
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L2 = L + V(t3)
Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – (L + V(t3) + sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) ) )
Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L  V(t3)  sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
Time_Diff1 = (L1 – (L2 + L3))/c
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L3 = sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L2 = L – V(t3)
Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – (L – V(t3) + sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) ))
Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3)  sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
Time_Diff1 = Time_Diff2
sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3)  sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
=
sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L  V(t3)  sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
As can be seen Time_Diff1 does NOT equal Time_Diff2
As all frames of ref are completely contained within a larger frame I note for any observer, unless they are in the absolute frame of ref, Time_Diff1 will not equal Time_Diff2 as they will have some residual velocity due to the velocity of the larger containing frame of ref.
This analysis demonstrates that an experiment can be performed in a moving frame of reference to show the observer at rest with that frame of ref that they are in a moving frame of ref.
As all frames of ref are completely contained within a larger frame of ref. By using this apparatus it may be possible to determine if there is an absolute frame of ref. By using this apparatus and changing the direction and velocity of a frame of ref until Time_diff1 = Time_diff2 then the direction and velocity of that frame of ref would be the same as the absolute (or preferred) frame of ref.
the rest of the pistures are in my other posts
___________________________________________________________________________
Einstein’s explanation of time dilation is given here http://en.wikipedia.org/wiki/Time_dilation
Referring to fig 1
In Einstein’s explanation you have 2 identical clocks. Clock 1 is at rest wrt the observer. Clock 2 is in a space ship zooming along wrt the observer.
At the red end of the clock is a light source, at the blue end a detector.
If L is the length of the clock. The observer, observing the clock at rest wrt him would see that the light pulse would take L/c time to traverse the length of the clock.
Einstein postulated that as the observer would see the photons in clock 2 travel a longer path to move from the red end to the blue end when compared with length of the path the photons in the clock 1. Therefore clock 2 must tick slower (time passes slower) when compared to clock1.
Referring to fig 2
As in Einstein’s explanation again you have 2 identical clocks. Clock 1 and clock 2 are at rest wrt the observer. This is the trivial case. Again at the red end of each clock is a light source, at the blue end a detector.
If L is the length of the clock. The observer, observing both clocks would see that the light pulse would take L/c time to traverse the length of the clock.
Referring to fig 3
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a space ship that is moving wrt the observer.
As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.
If L is the length of the clock.
The observer, observing the clock 1 would see that the light pulse would take L/c time to traverse the length of the clock.
T = L/c
As can be seen the path travelled by the photons in the clock2 is shorter than the path of the photons in the clock1. This is also what the observer would see.
The time the photons in clock2 would take would be at most
T = L/(c + V)
As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c + V).
I note the limit of the function t = L/(c + V) as V tends to infinity so t (time) goes to 0
Referring to fig 4
As in Einstein’s explanation again you have 2 identical clocks clock 1 and clock 2. Again at the red end is a light source, at the blue end a detector.
Clock 1 is at rest wrt the observer, clock 2 is in a space ship that is moving wrt the observer.
You should now note both clocks have been rotated through 180 degrees compared to fig 3. In Fig 3 the light source is at the right the detector at the left. In fig 4 the reverse.
As clock 2 is moving wrt the observer. At an initial time the observer will see Clock 2 at points A and B. At some time in the future Clock 2 will be observed at position A hat and B hat.
If L is the length of the clock. The time taken in Clock 1 would be L/c time to traverse the length of the clock.
T = L/c
As can be seen the path travelled by the photons in the clock2 is longer than the path of the photons in the clock1. This is also what the observer would see.
The time the photons in clock2 would take would be at least
T = L/(c  V)
As clock2 is moving its length would be contracted and the time taken would actually be less than L/(c  V).
I note the limit of the function t = L/(c  V) as V tends to infinity so t (time) goes to infinity
I have three questions
Is this analysis right?
How is it that the observer can get two different results simply by rotating the clock?
The theory is a moving clock ticks slower then a stationary clock. In fig 3 clock 2 is ticking faster than clock1 which refutes the accepted theory?
WHAT AN OBSERVER IN THE SPACE SHIP SEES
Referring to fig 5
I have a clock in the space ship.
The space ship is flying at constant velocity V.
At the red end of the clock is a light source, at the blue end a detector.
As the speed is constant the contraction is constant and the time rate is constant.
The observer, point A and point B form an isosceles triangle.
The length of the clock is L which equals the length from B to the observer and also A to the observer.
I NOTE THE DRAWING IS NOT TO SCALE.
At some initial time the ends of the clock will be located at points A and B. At some time in the future as the space ship is moving the ends of the clock will be located at points A hat and B hat.
At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.
By the time the pulse from the red end of the clock reaches the observer he will be located at point x.
When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.
By the time the pulse from the blue end reaches the observer the observer will be located at point z.
Analysis
Referring to fig 7
Using the cosine rule L3 can be calculated.
Times will be proportional to lengths as time is simply length/c
Angle g = 60 degrees
Angle e = 120 degrees
Cos 60 = 0.5
Cos 120 = 0.5
L3^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
t1 is the time it takes for the pulse of light to travel from B hat to point z
Referring to fig 8
Using the cosine rule L1 can be calculated.
Angle e = 60 degrees
Cos 60 = 0.5
L1^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
t2 is the time it takes for the pulse of light to travel from point A to point x
from fig 5
L2 = L – V(t3)
t3 is the time it takes for the pulse of light to travel from point A to point B hat.
The time difference the observer will see between the first pulse of light he sees and the second is:
Time_Diff1 = (L1 – (L2 + L3))/c
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L3 =sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L2 = L – V(t3)
Referring to fig 6
I have a clock in the space ship.
The space ship is flying at constant velocity V.
At the red end of the clock is a light source, at the blue end a detector.
As the speed is constant the contraction is constant and the time rate is constant.
The observer, point A and point B form an isosceles triangle.
The length of the clock is L which equals the length from B to the observer and also A to the observer.
I note the clock in fig 6 is rotated 180 degrees wrt the clock in fig 5 the red and blue ends have changed.
I NOTE THE DRAWING IS NOT TO SCALE.
At some initial time the ends of the clock will be located at points A and B. At some time in the future as the space ship is moving the ends of the clock will be located at points A hat and B hat.
At t0 a pulse of light leaves the red end of the clock. The pulse simultaneously heads towards the blue end of the clock and towards the observer. Also at t0 the observer is located at point W.
By the time the pulse from the red end of the clock reaches the observer he will be located at point x.
When the pulse reaches the blue end of the clock the observer is located at point y. The pulse of light bounces of the blue end and now heads in the direction of the observer.
By the time the pulse from the blue end reaches the observer the observer will be located at point z.
Analysis
Referring to fig 7
Using the cosine rule L1 can be calculated.
Times will be proportional to lengths as time is simply length/c
Angle g = 60 degrees
Angle e = 120 degrees
Cos 60 = 0.5
Cos 120 = 0.5
L1^2 = L^2 + (Vt1)^2 – LV(t1)Cos(120)
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
t1 is the time it takes for the pulse of light to travel from point A to point x
Referring to fig 8
Using the cosine rule L3 can be calculated.
Angle e = 60 degrees
Cos 60 = 0.5
L3^2 = L^2 + (Vt2)^2 – LV(t2)Cos(60)
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
t2 is the time it takes for the pulse of light to travel from point B hat to point z
from fig 6
L2 = L + V(t3)
t3 is the time it takes for the pulse of light to travel from point A to point B hat.
The time difference the observer will see between the first pulse of light he sees and the second is:
Time_Diff2 = (L1 – (L2 + L3))/c
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L2 = L + V(t3)
Observations
Further observation:
L3 in fig 6 = L1 in fig 5
L1 in fig 6 = L3 in fig 5
V(t3) in fig 5 = V(t3) in fig 6
The time difference measured in fig 5 must equal the time difference measured in fig 6.
Time_Diff1 = (L1 – (L2 + L3))/c
Time_Diff2 = (L1 – (L2 + L3))/c
Time_Diff1 = Time_Diff2
Time_Diff2 = (L1 – (L2 + L3))/c
L1 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L3 = sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L2 = L + V(t3)
Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – (L + V(t3) + sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) ) )
Time_Diff2 = sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L  V(t3)  sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
Time_Diff1 = (L1 – (L2 + L3))/c
L1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) )
L3 = sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
L2 = L – V(t3)
Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – (L – V(t3) + sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) ))
Time_Diff1 = sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3)  sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
Time_Diff1 = Time_Diff2
sqrt( L^2 + (Vt2)^2 – LV(t2)Cos(60) ) – L + V(t3)  sqrt( L^2 + (Vt1)^2 – LV(t1)Cos(120) )
=
sqrt(L^2 + (Vt1)^2 – LV(t1)Cos(120) ) – L  V(t3)  sqrt(L^2 + (Vt2)^2 – LV(t2)Cos(60) )
As can be seen Time_Diff1 does NOT equal Time_Diff2
As all frames of ref are completely contained within a larger frame I note for any observer, unless they are in the absolute frame of ref, Time_Diff1 will not equal Time_Diff2 as they will have some residual velocity due to the velocity of the larger containing frame of ref.
This analysis demonstrates that an experiment can be performed in a moving frame of reference to show the observer at rest with that frame of ref that they are in a moving frame of ref.
As all frames of ref are completely contained within a larger frame of ref. By using this apparatus it may be possible to determine if there is an absolute frame of ref. By using this apparatus and changing the direction and velocity of a frame of ref until Time_diff1 = Time_diff2 then the direction and velocity of that frame of ref would be the same as the absolute (or preferred) frame of ref.
Attachments

20 KB Views: 268

17.8 KB Views: 244

18.6 KB Views: 227